THE RISING SEA
Foundations of Algebraic Geometry

math216.wordpress.com
November 18, 2017 draftc 2010–2017 by Ravi Vakil.
⃝
Note to reader: the index and formatting have yet to be properly dealt with. Thereremain many issues still to be dealt with in the main part of the notes (including manyof your corrections and suggestions).

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

Contents
Preface
0.1.
0.2.
0.3.
0.4.

11
12
16
17
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For the reader
For the expert
Background and conventions⋆⋆ The goals of this bookPart I. Preliminaries

21

Chapter 1. Some category theory
1.1. Motivation
1.2. Categories and functors
1.3. Universal properties determine an object up to unique isomorphism
1.4. Limits and colimits
1.5. Adjoints
1.6. An introduction to abelian categories
1.7. ⋆ Spectral sequences

23
23
25
31
39
43
47
57

Chapter 2. Sheaves
2.1. Motivating example: The sheaf of differentiable functions
2.2. Definition of sheaf and presheaf
2.3. Morphisms of presheaves and sheaves
2.4. Properties determined at the level of stalks, and sheafification
2.5. Recovering sheaves from a “sheaf on a base”
2.6. Sheaves of abelian groups, and OX -modules, form abelian categories
2.7. The inverse image sheaf

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71
73
78
82
86
89
92

Part II. Schemes

97

Chapter 3. Toward affine schemes: the underlying set, and topological space 99
3.1. Toward schemes99
3.2. The underlying set of affine schemes
101
3.3. Visualizing schemes I: generic points
113
3.4. The underlying topological space of an affine scheme
115
3.5. A base of the Zariski topology on Spec A: Distinguished open sets 118
3.6. Topological (and Noetherian) properties
119
3.7. The function I(·), taking subsets of Spec A to ideals of A
127
Chapter 4. The structure sheaf, and the definition of schemes in general
4.1. The structure sheaf of an affine scheme
4.2. Visualizing schemes II: nilpotents

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133

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4.3.
4.4.
4.5.

Definition of schemes
Three examples
Projective schemes, and the Proj construction

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139
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Chapter 5. Some properties of schemes
5.1. Topological properties
5.2. Reducedness and integrality
5.3. Properties of schemes that can be checked “affine-locally”
5.4. Normality and factoriality
5.5. The crucial points of a scheme that control everything: Associatedpoints and primes

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153
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157
161

Part III. Morphisms

175

Chapter 6. Morphisms of schemes
6.1. Introduction
6.2. Morphisms of ringed spaces
6.3. From locally ringed spaces to morphisms of schemes
6.4. Maps of graded rings and maps of projective schemes
6.5. Rational maps from reduced schemes
6.6. ⋆ Representable functors and group schemes
6.7. ⋆⋆ The Grassmannian (initial construction)

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180
186
188
194
199

166

Chapter 7. Useful classes of morphisms of schemes
201
7.1. An example of a reasonable class of morphisms: Open embeddings 201
7.2. Algebraic interlude: Lying Over and Nakayama
203
7.3. A gazillion finiteness conditions on morphisms
207
7.4. Images of morphisms: Chevalley’s Theorem and elimination theory 216
Chapter 8. Closed embeddings and related notions
225
8.1. Closed embeddings and closed subschemes
225
8.2. More projective geometry
230
8.3. The (closed sub)scheme-theoretic image
236
8.4. Effective Cartier divisors, regular sequences and regular embeddings240
Chapter 9. Fibered products of schemes, and base change
9.1. They exist
9.2. Computing fibered products in practice
9.3. Interpretations: Pulling back families, and fibers of morphisms
9.4. Properties preserved by base change
9.5. ⋆ Properties not preserved by base change, and how to fix them
9.6. Products of projective schemes: The Segre embedding
9.7. Normalization

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256
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Chapter 10. Separated and proper morphisms, and (finally!) varieties
10.1. Separated morphisms (and quasiseparatedness done properly)
10.2. Rational maps to separated schemes
10.3. Proper morphisms

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Part IV. “Geometric” properties: Dimension and smoothness

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Chapter 11. Dimension
11.1. Dimension and codimension
11.2. Dimension, transcendence degree, and Noether normalization
11.3. Codimension one miracles: Krull’s and Hartogs’s Theorems
11.4. Dimensions of fibers of morphisms of varieties
11.5. ⋆⋆ Proof of Krull’s Principal Ideal and Height Theorems

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303
307
315
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Chapter 12. Regularity and smoothness
12.1. The Zariski tangent space
12.2. Regularity, and smoothness over a field
12.3. Examples
12.4. Bertini’s Theorem
12.5. Another (co)dimension one miracle: Discrete valuation rings
12.6. Smooth (and étale) morphisms (first definition)
12.7. ⋆ Valuative criteria for separatedness and properness
12.8. ⋆ More sophisticated facts about regular local rings
12.9. ⋆ Filtered rings and modules, and the Artin-Rees Lemma

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342
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349
354
358
362
364

Part V. Quasicoherent sheaves

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Chapter 13. Quasicoherent and coherent sheaves
13.1. Vector bundles and locally free sheaves
13.2. Quasicoherent sheaves
13.3. Characterizing quasicoherence using the distinguished affine base
13.4. Quasicoherent sheaves form an abelian category
13.5. Module-like constructions
13.6. Finite type and coherent sheaves
13.7. Pleasant properties of finite type and coherent sheaves
13.8. ⋆⋆ Coherent modules over non-Noetherian rings

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377
381
383
386
389
393

Chapter 14. Line bundles: Invertible sheaves and divisors
14.1. Some line bundles on projective space
14.2. Line bundles and Weil divisors
14.3. ⋆ Effective Cartier divisors “=” invertible ideal sheaves

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397
400
408

Chapter 15. Quasicoherent sheaves and projective A-schemes
15.1. The quasicoherent sheaf corresponding to a graded module
15.2. Invertible sheaves (line bundles) on projective A-schemes
15.3. Globally generated and base-point-free line bundles
15.4. Quasicoherent sheaves and graded modules

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411
412
413
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Chapter 16. Pushforwards and pullbacks of quasicoherent sheaves
16.1. Introduction
16.2. Pushforwards of quasicoherent sheaves
16.3. Pullbacks of quasicoherent sheaves
16.4. Line bundles and maps to projective schemes
16.5. The Curve-to-Projective Extension Theorem
16.6. Ample and very ample line bundles
16.7. ⋆ The Grassmannian as a moduli space

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Chapter 17. Relative versions of Spec and Proj, and projective morphisms
17.1. Relative Spec of a (quasicoherent) sheaf of algebras
17.2. Relative Proj of a sheaf of graded algebras
17.3. Projective morphisms
17.4. Applications to curves

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450
453
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Chapter 18. Čech cohomology of quasicoherent sheaves
18.1. (Desired) properties of cohomology
18.2. Definitions and proofs of key properties
18.3. Cohomology of line bundles on projective space
18.4. Riemann-Roch, degrees of coherent sheaves, and arithmetic genus
18.5. A first glimpse of Serre duality
18.6. Hilbert functions, Hilbert polynomials, and genus
18.7. ⋆ Serre’s cohomological characterization of ampleness18.8. Higher pushforward (or direct image) sheaves
18.9.
⋆ From projective to proper hypotheses: Chow’s Lemma andGrothendieck’s Coherence Theorem

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475
477
485
488
494
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Chapter 19. Application: Curves
19.1. A criterion for a morphism to be a closed embedding
19.2. A series of crucial tools
19.3. Curves of genus 019.4. Classical geometry arising from curves of positive genus
19.5. Hyperelliptic curves
19.6. Curves of genus 2
19.7. Curves of genus 3
19.8. Curves of genus 4 and 5
19.9. Curves of genus 1
19.10. Elliptic curves are group varieties
19.11. Counterexamples and pathologies using elliptic curves

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518
519
521
523
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Chapter 20. Application: A glimpse of intersection theory
20.1. Intersecting n line bundles with an n-dimensional variety
20.2. Intersection theory on a surface
20.3. The Grothendieck group of coherent sheaves, and an algebraic
version of homology
20.4. ⋆⋆ The Nakai-Moishezon and Kleiman criteria for ampleness

543
543
547

Chapter 21. Differentials
21.1. Motivation and game plan
21.2. Definitions and first properties
21.3. Smoothness of varieties revisited
21.4. Examples
21.5. Understanding smooth varieties using their cotangent bundles
21.6. Unramified morphisms
21.7. The Riemann-Hurwitz Formula

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578
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Chapter 22. ⋆ Blowing up
22.1. Motivating example: blowing up the origin in the plane22.2. Blowing up, by universal property

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599

501

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22.3.
22.4.

The blow-up exists, and is projective
Examples and computations

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608

Part VI. More

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Chapter 23. Derived functors
23.1. The Tor functors
23.2. Derived functors in general
23.3. Derived functors and spectral sequences
23.4. Derived functor cohomology of O-modules
23.5. Čech cohomology and derived functor cohomology agree

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625
629
634
637

Chapter 24. Flatness
24.1. Introduction
24.2. Easier facts
24.3. Flatness through Tor
24.4. Ideal-theoretic criteria for flatness
24.5. Topological aspects of flatness
24.6. Local criteria for flatness
24.7. Flatness implies constant Euler characteristic

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652
654
661
665
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Chapter 25. Smooth and étale morphisms, and flatness
25.1. Some motivation
25.2. Different characterizations of smooth and étale morphisms
25.3. Generic smoothness and the Kleiman-Bertini Theorem

673
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675
681

Chapter 26. Depth and Cohen-Macaulayness
26.1. Depth
26.2. Cohen-Macaulay rings and schemes
26.3. ⋆ Serre’s R1 + S2 criterion for normality

685
685
688
691

Chapter 27. Twenty-seven lines
27.1. Introduction
27.2. Preliminary facts
27.3. Every smooth cubic surface (over k) has 27 lines
27.4. Every smooth cubic surface (over k) is a blown up plane

697
697
699
700
704

Chapter 28. Cohomology and base change theorems
28.1. Statements and applications
28.2. ⋆ Proofs of cohomology and base change theorems
28.3. Applying cohomology and base change to moduli problems

707
707
713
720

Chapter 29. Power series and the Theorem on Formal Functions
29.1. Introduction
29.2. Algebraic preliminaries
29.3. Defining types of singularities
29.4. The Theorem on Formal Functions
29.5. Zariski’s Connectedness Lemma and Stein Factorization
29.6. Zariski’s Main Theorem
29.7. Castelnuovo’s Criterion for contracting (−1)-curves

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729
732
733
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29.8.

⋆⋆ Proof of the Theorem on Formal Functions 29.4.2

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Chapter 30. ⋆ Proof of Serre duality
30.1. Introduction
30.2. Ext groups and Ext sheaves for O-modules
30.3. Serre duality for projective k-schemes
30.4. The adjunction formula for the dualizing sheaf, and ωX = KX

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754
758
763

Bibliography

769

Index

775

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November 18, 2017 draft

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Je pourrais illustrer la ... approche, en gardant l’image de la noix qu’il s’agit d’ouvrir.
La première parabole qui m’est venue à l’esprit tantôt, c’est qu’on plonge la noix dans
un liquide émollient, de l’eau simplement pourquoi pas, de temps en temps on frotte pourqu’elle pénètre mieux, pour le reste on laisse faire le temps. La coque s’assouplit au fil dessemaines et des mois — quand le temps est mûr, une pression de la main suffit, la coques’ouvre comme celle d’un avocat mûr à point! . . .
L’image qui m’était venue il y a quelques semaines était différente encore, la choseinconnue qu’il s’agit de connaı̂tre m’apparaissait comme quelque étendue de terre ou demarnes compactes, réticente à se laisser pénétrer. ... La mer s’avance insensiblement etsans bruit, rien ne semble se casser rien ne bouge l’eau est si loin on l’entend à peine...
Pourtant elle finit par entourer la substance rétive...
I can illustrate the ... approach with the ... image of a nut to be opened. The firstanalogy that came to my mind is of immersing the nut in some softening liquid, and why
not simply water? From time to time you rub so the liquid penetrates better, and otherwise
you let time pass. The shell becomes more flexible through weeks and months — when thetime is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado! . . .
A different image came to me a few weeks ago. The unknown thing to be known
appeared to me as some stretch of earth or hard marl, resisting penetration ... the sea
advances insensibly in silence, nothing seems to happen, nothing moves, the water is so
far off you hardly hear it ... yet finally it surrounds the resistant substance.
— A. Grothendieck [Gr6, p. 552-3], translation by C. McLarty [Mc, p. 1]

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

Preface
This book is intended to give a serious and reasonably complete introduction
to algebraic geometry, not just for (future) experts in the field. The exposition
serves a narrow set of goals (see §0.4), and necessarily takes a particular point ofview on the subject.
It has now been four decades since David Mumford wrote that algebraic geometry “seems to have acquired the reputation of being esoteric, exclusive, andvery abstract, with adherents who are secretly plotting to take over all the rest ofmathematics! In one respect this last point is accurate ...” ([Mu4, preface] and[Mu7, p. 227]). The revolution has now fully come to pass, and has fundamentallychanged how we think about many fields of pure mathematics. A remarkable
number of celebrated advances rely in some way on the insights and ideas forcefully articulated by Alexander Grothendieck, Jean-Pierre Serre, and others.
For a number of reasons, algebraic geometry has earned a reputation of being
inaccessible. The power of the subject comes from rather abstract heavy machinery, and it is easy to lose sight of the intuitive nature of the objects and methods.
Many in nearby fields have only a vague sense of the fundamental ideas of thesubject. Algebraic geometry itself has fractured into many parts, and even within
algebraic geometry, new researchers are often unaware of the basic ideas in subfields removed from their own.
But there is another more optimistic perspective to be taken. The ideas that allow algebraic geometry to connect several parts of mathematics are fundamental,
and well-motivated. Many people in nearby fields would find it useful to develop
a working knowledge of the foundations of the subject, and not just at a superficial level. Within algebraic geometry itself, there is a canon (at least for those
approaching the subject from this particular direction), that everyone in the field
can and should be familiar with. The rough edges of scheme theory have been
sanded down over the past half century, although there remains an inescapableneed to understand the subject on its own terms.
0.0.1. The importance of exercises. This book has a lot of exercises. I have foundthat unless I have some problems I can think through, ideas don’t get fixed in mymind. Some exercises are trivial — some experts find this offensive, but I findthis desirable. A very few necessary ones may be hard, but the reader should havebeen given the background to deal with them — they are not just an excuse to push
hard material out of the text. The exercises are interspersed with the exposition,
not left to the end. Most have been extensively field-tested. The point of view here
is one I explored with Kedlaya and Poonen in [KPV], a book that was ostensibly
about problems, but secretly a case for how one should learn and do and think
about mathematics. Most people learn by doing, rather than just passively reading.
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The Rising Sea: Foundations of Algebraic Geometry

Judiciously chosen problems can be the best way of guiding the learner toward
enlightenment.
0.0.2. Structure. You will quickly notice that everything is numbered by chapter
and section, and everything is numbered the same way after that (for ease of reference), except exercises are indicated by letters (and are sprinkled throughout the
text, rather than at the end of sections). Individual paragraphs often get numbers
for ease of reference, or to indicate a new topic. Definitions are in bold, and aresometimes given in passing.
0.0.3. Acknowledgments.
This one is going to be really hard, so I’ll write this later. (Mike Stay is the author of Jokes 1.3.11 and 21.5.2.) The phrase “The Rising Sea” is due to Grothendieck[Gr6, p. 552-3], with this particular translation by McLarty [Mc, p. 1], and popularized as the title of Daniel Murfet’s excellent blog [Mur].

0.1 For the reader
This is your last chance. After this, there is no turning back. You take the blue pill,
the story ends, you wake up in your bed and believe whatever you want to believe. You
take the red pill, you stay in Wonderland and I show you how deep the rabbit-hole goes.
— Morpheus
The contents of this book are intended to be a collection of communal wisdom,
necessarily distilled through an imperfect filter. I wish to say a few words on how
you might use it, although it is not clear to me if you should or will follow thisadvice.
Before discussing details, I want to say clearly at the outset: the wonderfulmachine of modern algebraic geometry was created to understand basic and naive
questions about geometry (broadly construed). The purpose of this book is to
give you a thorough foundation in these powerful ideas. Do not be seduced by thelotus-eaters into infatuation with untethered abstraction. Hold tight to your geometricmotivation as you learn the formal structures which have proved to be so effectivein studying fundamental questions. When introduced to a new idea, always ask
why you should care. Do not expect an answer right away, but demand an answer
eventually. Try at least to apply any new abstraction to some concrete example
you can understand well.
Understanding algebraic geometry is often thought to be hard because it consists of large complicated pieces of machinery. In fact the opposite is true; to switch
metaphors, rather than being narrow and deep, algebraic geometry is shallow but
extremely broad. It is built out of a large number of very small parts, in keepingwith Grothendieck’s vision of mathematics. It is a challenge to hold the entire
organic structure, with its messy interconnections, in your head.
A reasonable place to start is with the idea of “affine complex varieties”: subsets of Cn cut out by some polynomial equations. Your geometric intuition can immediately come into play — you may already have some ideas or questions about
dimension, or smoothness, or solutions over subfields such as R or Q. Wiser heads

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would counsel spending time understanding complex varieties in some detail before learning about schemes. Instead, I encourage you to learn about schemes
immediately, learning about affine complex varieties as the central (but not exclusive) example. This is not ideal, but can save time, and is surprisingly workable.
An alternative is to learn about varieties elsewhere, and then come back later.
The intuition for schemes can be built on the intuition for affine complex varieties. Allen Knutson and Terry Tao have pointed out that this involves three different simultaneous generalizations, which can be interpreted as three large themesin mathematics. (i) We allow nilpotents in the ring of functions, which is basicallyanalysis (looking at near-solutions of equations instead of exact solutions). (ii) We
glue these affine schemes together, which is what we do in differential geometry(looking at manifolds instead of coordinate patches). (iii) Instead of working overC (or another algebraically closed field), we work more generally over a ring that
isn’t an algebraically closed field, or even a field at all, which is basically number
theory (solving equations over number fields, rings of integers, etc.).
Because our goal is to be comprehensive, and to understand everything one
should know after a first course, it will necessarily take longer to get to interesting
sample applications. You may be misled into thinking that one has to work thishard to get to these applications — it is not true! You should deliberately keep an
eye out for examples you would have cared about before. This will take some timeand patience.
As you learn algebraic geometry, you should pay attention to crucial steppingstones. Of course, the steps get bigger the farther you go.
Chapter 1. Category theory is only language, but it is language with an embedded logic. Category theory is much easier once you realize that it is designed
to formalize and abstract things you already know. The initial chapter on category theory prepares you to think cleanly. For example, when someone names
something a “cokernel” or a “product”, you should want to know why it deserves
that name, and what the name really should mean. The conceptual advantages ofthinking this way will gradually become apparent over time. Yoneda’s Lemma —and more generally, the idea of understanding an object through the maps to it —will play an important role.
Chapter 2. The theory of sheaves again abstracts something you already understand well (see the motivating example of §2.1), and what is difficult is understanding how one best packages and works with the information of a sheaf (stalks,
sheafification, sheaves on a base, etc.).
Chapters 1 and 2 are a risky gamble, and they attempt a delicate balance. Attempts
to explain algebraic geometry often leave such background to the reader, refer toother sources the reader won’t read, or punt it to a telegraphic appendix. Instead,this book attempts to explain everything necessary, but as little as possible, andtries to get across how you should think about (and work with) these fundamental
ideas, and why they are more grounded than you might fear.
Chapters 3–5. Armed with this background, you will be able to think cleanly
about various sorts of “spaces” studied in different parts of geometry (including differentiable real manifolds, topological spaces, and complex manifolds), as
ringed spaces that locally are of a certain form. A scheme is just another kind

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The Rising Sea: Foundations of Algebraic Geometryof “geometric space”, and we are then ready to transport lots of intuition from“classical geometry” to this new setting. (This also will set you up to later think
about other geometric kinds of spaces in algebraic geometry, such as complex analytic spaces, algebraic spaces, orbifolds, stacks, rigid analytic spaces, and formal
schemes.) The ways in which schemes differ from your geometric intuition can beinternalized, and your intuition can be expanded to accomodate them. There aremany properties you will realize you will want, as well as other properties thatwill later prove important. These all deserve names. Take your time becoming
familiar with them.
Chapters 6–10. Thinking categorically will lead you to ask about morphismsof schemes (and other spaces in geometry). One of Grothendieck’s fundamental
lessons is that the morphisms are central. Important geometric properties should
really be understood as properties of morphisms. There are many classes of morphisms with special names, and in each case you should think through why thatclass deserves a name.
Chapters 11–12. You will then be in a good position to think about fundamental geometric properties of schemes: dimension and smoothness. You may be surprised that these are subtle ideas, but you should keep in mind that they are subtle
everywhere in mathematics.
Chapters 13–21. Vector bundles are ubiquitous tools in geometry, and algebraic
geometry is no exception. They lead us to the more general notion of quasicoherent sheaves, much as free modules over a ring lead us to modules more generally.
We study their properties next, including cohomology. Chapter 19, applying theseideas to study curves, may help make clear how useful they are.
Chapters 23–30. With this in hand, you are ready to learn more advanced tools
widely used in the subject. Many examples of what you can do are given, andthe classical story of the 27 lines on a smooth cubic surface (Chapter 27) is a good
opportunity to see many ideas come together.
The rough logical dependencies among the chapters are shown in Figure 0.1.
(Caution: this should be taken with a grain of salt. For example, you can avoid
using much of Chapter 19 on curves in later chapters, but it is a crucial source of
examples, and a great way to consolidate your understanding. And Chapter 29 oncompletions uses Chapters 19, 20 and 22 only in the discussion of Castelnuovo’s
Criterion 29.7.1.)
In general, I like having as few hypotheses as possible. Certainly a hypothesisthat isn’t necessary to the proof is a red herring. But if a reasonable hypothesis canmake the proof cleaner and more memorable, I am willing to include it.
In particular, Noetherian hypotheses are handy when necessary, but are otherwise misleading. Even Noetherian-minded readers (normal human beings) are
better off having the right hypotheses, as they will make clearer why things aretrue.
We often state results particular to varieties, especially when there are techniques unique to this situation that one should know. But restricting to algebraically closed fields is useful surprisingly rarely. Geometers needn’t be afraidof arithmetic examples or of algebraic examples; a central insight of algebraic geometry is that the same formalism applies without change.

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November 18, 2017 draft

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29 ❚ ❚✡✡
28■■ ✝✝✽✝✽✽ ❥❥ 26■■✝ ❥❥✽❥
✡ ❚
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❥
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❚ ❚❥❥❥✝❥✝ ■■■ ✽✽✡✡
✡✡
❥❥ ✝❚✝ ❚ ■■ ✽✽
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-✡ ❥❥❥❥" .✝
❚ ❚■+ , " -✡
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0
30
25
27
F IGURE 0.1. Important logical dependences among chapters (ormore precisely, a directed graph showing which chapter should
be read before which other chapter)

Pathological examples are useful to know. On mountain highways, there aretall sticks on the sides of the road designed for bad weather. In winter, you cannot
see the road clearly, and the sticks serve as warning signs: if you cross this line,
you will die! Pathologies and (counter)examples serve a similar goal. They also
serve as a reality check, when confronting a new statement, theorem, or conjecture,
whose veracity you may doubt.
When working through a book in algebraic geometry, it is particularly helpful
to have other algebraic geometry books at hand, to see different approaches andto have alternate expositions when things become difficult. This book may serve
as a good secondary book. If it is your primary source, then two other excellent
books with what I consider a similar philosophy are [Liu] and [GW]. De Jong’s
encyclopedic online reference [Stacks] is peerless. There are many other outstanding sources out there, perhaps one for each approach to the subject; you shouldbrowse around and find one you find sympathetic.
If you are looking for a correct or complete history of the subject, you havecome to the wrong place. This book is not intended to be a complete guide to
the literature, and many important sources are ignored or left out, due to my own
ignorance and laziness.

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The Rising Sea: Foundations of Algebraic Geometry

Finally, if you attempt to read this without working through a significant number of exercises (see §0.0.1), I will come to your house and pummel you with
[Gr-EGA] until you beg for mercy. It is important to not just have a vague sense ofwhat is true, but to be able to actually get your hands dirty. To quote Mark Kisin:“You can wave your hands all you want, but it still won’t make you fly.” Note: The
hints may help you, but sometimes they may not.

0.2 For the expert
If you use this book for a course, you should of course adapt it to your own
point of view and your own interests. In particular, you should think about anapplication or theorem you want to reach at the end of the course (which may
well not be in this book), and then work toward it. You should feel no compulsion
to sprint to the end; I advise instead taking more time, and ending at the rightplace for your students. (Figure 0.1, showing large-scale dependencies among thechapters, may help you map out a course.) I have found that the theory of curves(Chapter 19) and the 27 lines on the cubic surface (Chapter 27) have served thispurpose well at the end of winter and spring quarters. This was true even if someof the needed background was not covered, and had to be taken by students assome sort of black box.
Faithfulness to the goals of §0.4 required a brutal triage, and I have made anumber of decisions you may wish to reverse. I will briefly describe some choices
made that may be controversial.
Decisions on how to describe things were made for the sake of the learners.
If there were two approaches, and one was “correct” from an advanced point ofview, and one was direct and natural from a naive point of view, I went with thelatter.
On the other hand, the theory of varieties (over an algebraically closed field,say) was not done first and separately. This choice brought me close to tears, but
in the end I am convinced that it can work well, if done in the right spirit.
Instead of spending the first part of the course on varieties, I spent the timein a different way. It is tempting to assume that students will either arrive withgreat comfort and experience with category theory and sheaf theory, or that theyshould pick up these ideas on their own time. I would love to live in that world.
I encourage you to not skimp on these foundational issues. I have found that
although these first lectures felt painfully slow to me, they were revelatory to anumber of the students, and those with more experience were not bored and did
not waste their time. This investment paid off in spades when I was able to relyon their ability to think cleanly and to use these tools in practice. Furthermore, ifthey left the course with nothing more than hands-on experience with these ideas,
the world was still better off for it.
For the most part, we will state results in the maximal generality that the proof
justifies, but we will not give a much harder proof if the generality of the stronger
result will not be used. There are a few cases where we work harder to provea somewhat more general result that many readers may not appreciate. For example, we prove a number of theorems for proper morphisms, not just projective

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17

morphisms. But in such cases, readers are invited or encouraged to ignore thesubtleties required for the greater generality.
I consider line bundles (and maps to projective space) more fundamental thandivisors. General Cartier divisors are not discussed (although effective Cartier divisors play an essential role).
Cohomology is done first using the Čech approach (as Serre first did), and derived functor cohomology is introduced only later. I am well aware that Grothendieckthinks of the fact that the agreement of Čech cohomology with derived functor cohomology “should be considered as an accidental phenomenon”, and that “it isimportant for technical reasons not to take as definition of cohomology the Čechcohomology”, [Gr4, p. 108]. But I am convinced that this is the right way for mostpeople to see this kind of cohomology for the first time. (It is certainly true thatmany topics in algebraic geometry are best understood in the language of derived
functors. But this is a view from the mountaintop, looking down, and not the best
way to explore the forests. In order to appreciate derived functors appropriately,
one must understand the homological algebra behind it, and not just take it as a
black box.)
We restrict to the Noetherian case only when it is necessary, or (rarely) when it
really saves effort. In this way, non-Noetherian people will clearly see where theyshould be careful, and Noetherian people will realize that non-Noetherian things
are not so terrible. Moreover, even if you are interested primarily in Noetherian
schemes, it helps to see “Noetherian” in the hypotheses of theorems only whennecessary, as it will help you remember how and when this property gets used.
There are some cases where Noetherian readers will suffer a little more thanthey would otherwise. As an inflammatory example, instead of using Noetherian
hypotheses, the notion of quasiseparatedness comes up early and often. The costis that one extra word has to be remembered, on top of an overwhelming numberof other words. But once that is done, it is not hard to remember that essentiallyevery scheme anyone cares about is quasiseparated. Furthermore, whenever the
hypotheses “quasicompact and quasiseparated” turn up, the reader will immediately guess a key idea of the proof. As another example, coherent sheaves andfinite type (quasicoherent) sheaves are the same in the Noetherian situation, but
are still worth distinguishing in statements of the theorems and exercises, for thesame reason: to be clearer on what is used in the proof.
Many important topics are not discussed. Valuative criteria are not proved(see §12.7), and their statement is relegated to an optional section. Completely
omitted: dévissage, formal schemes, and cohomology with supports. Sorry!

0.3 Background and conventions“Should you just be an algebraist or a geometer?” is like saying “Would you rather
be deaf or blind?”— M. Atiyah, [At2, p. 659]
All rings are assumed to be commutative unless explicitly stated otherwise.
All rings are assumed to contain a unit, denoted 1. Maps of rings must send 1 to
1. We don’t require that 0 ̸= 1; in other words, the “0-ring” (with one element)

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The Rising Sea: Foundations of Algebraic Geometry

is a ring. (There is a ring map from any ring to the 0-ring; the 0-ring only maps
to itself. The 0-ring is the final object in the category of rings.) The definition of“integral domain” includes 1 ̸= 0, so the 0-ring is not an integral domain. We
accept the Axiom of Choice. In particular, any proper ideal in a ring is containedin a maximal ideal. (The Axiom of Choice also arises in the argument that thecategory of A-modules has enough injectives, see Exercise 23.2.G.)The reader should be familiar with some basic notions in commutative ringtheory, in particular the notion of ideals (including prime and maximal ideals) andlocalization. Tensor products and exact sequences of A-modules will be important.
We will use the notation (A, m) or (A, m, k) for local rings (rings with a uniquemaximal ideal) — A is the ring, m its maximal ideal, and k = A/m its residue field.
We will use the structure theorem for finitely generated modules over a principal
ideal domain A: any such module can be written as the direct sum of principal
modules A/(a). Some experience with field theory will be helpful from time totime.
0.3.1. Caution about foundational issues. We will not concern ourselves with subtle
foundational issues (set-theoretic issues, universes, etc.). It is true that some people should be careful about these issues. But is that really how you want to live
your life? (If you are one of these rare people, a good start is [KS2, §1.1].)
0.3.2. Further background. It may be helpful to have books on other subjects at
hand that you can dip into for specific facts, rather than reading them in advance.
In commutative algebra, [E] is good for this. Other popular choices are [AtM] and[Mat2]. The book [Al] takes a point of view useful to algebraic geometry. For
homological algebra, [Weib] is simultaneously detailed and readable.
Background from other parts of mathematics (topology, geometry, complex
analysis, number theory, ...) will of course be helpful for intuition and grounding.
Some previous exposure to topology is certainly essential.
0.3.3. Nonmathematical conventions. “Unimportant” means “unimportant for thecurrent exposition”, not necessarily unimportant in the larger scheme of things.
Other words may be used idiosyncratically as well.
There are optional starred sections of topics worth knowing on a second or
third (but not first) reading. They are marked with a star: ⋆. Starred sections are
not necessarily harder, merely unimportant. You should not read double-starredsections (⋆⋆) unless you really really want to, but you should be aware of their
existence. (It may be strange to have parts of a book that should not be read!)Let’s now find out if you are taking my advice about double-starred sections.

0.4 ⋆⋆ The goals of this book
There are a number of possible introductions to the field of algebraic geometry: Riemann surfaces; complex geometry; the theory of varieties; a nonrigorous
examples-based introduction; algebraic geometry for number theorists; an abstract
functorial approach; and more. All have their place. Different approaches suit different students (and different advisors). This book takes only one route.

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Our intent is to cover a canon completely and rigorously, with enough examples and calculations to help develop intuition for the machinery. This is often
the content of a second course in algebraic geometry, and in an ideal world, people would learn this material over many years, after having background courses
in commutative algebra, algebraic topology, differential geometry, complex analysis, homological algebra, number theory, and French literature. We do not live inan ideal world. For this reason, the book is written as a first introduction, but achallenging one.
This book seeks to do a very few things, but to try to do them well. Our goals
and premises are as follows.
The core of the material should be digestible over a single year. After ayear of blood, sweat, and tears, readers should have a broad familiarity with thefoundations of the subject, and be ready to attend seminars, and learn more advanced material. They should not just have a vague intuitive understanding of
the ideas of the subject; they should know interesting examples, know why they
are interesting, and be able to work through their details. Readers in other fields
of mathematics should know enough to understand the algebro-geometric ideasthat arise in their area of interest.
This means that this book is not encyclopedic, and even beyond that, hard
choices have to be made. (In particular, analytic aspects are essentially ignored,
and are at best dealt with in passing without proof. This is a book about algebraic
algebraic geometry.)
This book is usable (and has been used) for a course, but the course should(as always) take on the personality of the instructor. With a good course, people
should be able to leave early and still get something useful from the experience.
With this book, it is possible to leave without regret after learning about category
theory, or about sheaves, or about geometric spaces, having become a better person.
The book is also usable (and has been used) for learning on your own. But
most mortals cannot learn algebraic geometry fully on their own; ideally you
should read in a group, and even if not, you should have someone you can ask
questions to (both stupid and smart questions).
There is certainly more than a year’s material here, but I have tried to makeclear which topics are essential, and which are not. Those teaching a class will
choose which “inessential” things are important for the point they wish to get
across, and use them.
There is a canon (at least for this particular approach to algebraic geometry). I
have been repeatedly surprised at how much people in different parts of algebraic
geometry agree on what every civilized algebraic geometer should know after afirst (serious) year. (There are of course different canons for different parts of thesubject, e.g., complex algebraic geometry, combinatorial algebraic geometry, computational algebraic geometry, etc.) There are extra bells and whistles that different
instructors might add on, to prepare students for their particular part of the fieldor their own point of view, but the core of the subject remains unified, despite thediversity and richness of the subject. There are some serious and painful compromises to be made to reconcile this goal with the previous one.

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The Rising Sea: Foundations of Algebraic Geometry

Algebraic geometry is for everyone (with the appropriate definition of “everyone”). Algebraic geometry courses tend to require a lot of background, whichmakes them inaccessible to all but those who know they will go deeply into thesubject. Algebraic geometry is too important for that; it is essential that many of
those in nearby fields develop some serious familiarity with the foundational ideas
and tools of the subject, and not just at a superficial level. (Similarly, algebraic geometers uninterested in any nearby field are necessarily arid, narrow thinkers. Do
not be such a person!)
For this reason, this book attempts to require as little background as possible.
The background required will, in a technical sense, be surprisingly minimal — ideally just some commutative ring theory and point-set topology, and some comfort
with things like prime ideals and localization. This is misleading of course — themore you know, the better. And the less background you have, the harder you will
have to work — this is not a light read. On a related note...
The book is intended to be as self-contained as possible. I have tried tofollow the motto: “if you use it, you must prove it”. I have noticed that most
students are human beings: if you tell them that some algebraic fact is in some latechapter of a book in commutative algebra, they will not immediately go and read
it. Surprisingly often, what we need can be developed quickly from scratch, andeven if people do not read it, they can see what is involved. The cost is that thebook is much denser, and that significant sophistication and maturity is demandedof the reader. The benefit is that more people can follow it; they are less likely toreach a point where they get thrown. On the other hand, people who already havesome familiarity with algebraic geometry, but want to understand the foundations
more completely, should not be bored, and can focus on more subtle issues.
As just one example, Krull’s Principal Ideal Theorem 11.3.3 is an important
tool. I have included an essentially standard proof (§11.5). I do not want people
to read it (unless they really really want to), and signal this by a double-star in thetitle: ⋆⋆. Instead, I want people to skim it and realize that they could read it, andthat it is not seriously hard.
This is an important goal because it is important not just to know what is true,
but to know why things are true, and what is hard, and what is not hard. Also,
this helps the previous goal, by reducing the number of prerequisites.
The book is intended to build intuition for the formidable machinery of algebraic geometry. The exercises are central for this (see §0.0.1). Informal language
can sometimes be helpful. Many examples are given. Learning how to thinkcleanly (and in particular categorically) is essential. The advantages of appropriate
generality should be made clear by example, and not by intimidation. The motivation is more local than global. For example, there is no introductory chapter
explaining why one might be interested in algebraic geometry, and instead there
is an introductory chapter explaining why you should want to think categorically(and how to actually do this).
Balancing the above goals is already impossible. We must thus give up any
hope of achieving any other desiderata. There are no other goals.

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Part I

Preliminaries

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CHAPTER 1

Some category theory
The introduction of the digit 0 or the group concept was general nonsense too, andmathematics was more or less stagnating for thousands of years because nobody wasaround to take such childish steps...
— A. Grothendieck, [BP, p. 4–5]
That which does not kill me, makes me stronger.
— F. Nietzsche

1.1 Motivation
Before we get to any interesting geometry, we need to develop a language
to discuss things cleanly and effectively. This is best done in the language ofcategories. There is not much to know about categories to get started; it is justa very useful language. Like all mathematical languages, category theory comeswith an embedded logic, which allows us to abstract intuitions in settings we know
well to far more general situations.
Our motivation is as follows. We will be creating some new mathematicalobjects (such as schemes, and certain kinds of sheaves), and we expect them toact like objects we have seen before. We could try to nail down precisely what
we mean by “act like”, and what minimal set of things we have to check in orderto verify that they act the way we expect. Fortunately, we don’t have to — otherpeople have done this before us, by defining key notions, such as abelian categories,
which behave like modules over a ring.
Our general approach will be as follows. I will try to tell you what you need toknow, and no more. (This I promise: if I use the word “topoi”, you can shoot me.) I
will begin by telling you things you already know, and describing what is essential
about the examples, in a way that we can abstract a more general definition. We
will then see this definition in less familiar settings, and get comfortable with using
it to solve problems and prove theorems.
For example, we will define the notion of product of schemes. We could justgive a definition of product, but then you should want to know why this precise
definition deserves the name of “product”. As a motivation, we revisit the notionof product in a situation we know well: (the category of) sets. One way to define
the product of sets U and V is as the set of ordered pairs {(u, v) : u ∈ U, v ∈ V}.
But someone from a different mathematical culture might reasonably define it asu
the set of symbols { v : u ∈ U, v ∈ V}. These notions are “obviously the same”.
Better: there is “an obvious bijection between the two”.
23

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The Rising Sea: Foundations of Algebraic Geometry

This can be made precise by giving a better definition of product, in termsof a universal property. Given two sets M and N, a product is a set P, along withmaps µ : P → M and ν : P → N, such that for any set P ′ with maps µ ′ : P ′ → M andν ′ : P ′ → N, these maps must factor uniquely through P:(1.1.0.1)

P ′✵ 00✵✵ 0000 ′✵✵ ∃! 00ν00
000
✵✵ 1
002
✵
%N
µ ′ ✵✵ P
ν
✵✵
✵✵ µ
3 "
M

(The symbol ∃ means “there exists”, and the symbol ! here means “unique”.) Thusa product is a diagram
P

ν

%N

µ

"
M
and not just a set P, although the maps µ and ν are often left implicit.
This definition agrees with the traditional definition, with one twist: there
isn’t just a single product; but any two products come with a unique isomorphism
between them. In other words, the product is unique up to unique isomorphism.
Here is why: if you have a product
P1

ν1

%N

µ1

"
M
and I have a product
P2

ν2

%N

µ2

"
M
then by the universal property of my product (letting (P2 , µ2 , ν2 ) play the role of(P, µ, ν), and (P1 , µ1 , ν1 ) play the role of (P ′ , µ ′ , ν ′ ) in (1.1.0.1)), there is a unique
map f : P1 → P2 making the appropriate diagram commute (i.e., µ1 = µ2 ◦ f andν1 = ν2 ◦ f). Similarly by the universal property of your product, there is a unique
map g : P2 → P1 making the appropriate diagram commute. Now consider theuniversal property of my product, this time letting (P2 , µ2 , ν2 ) play the role ofboth (P, µ, ν) and (P ′ , µ ′ , ν ′ ) in (1.1.0.1). There is a unique map h : P2 → P2 such

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November 18, 2017 draft

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that
P2✵ ❆00✵✵❆❆00000ν2✵✵ h❆❆ 0000
000
✵✵ ❆1
04
✵
%N
P
µ2 ✵
2 ν2
✵✵
✵✵ µ2
✵3 "
M
commutes. However, I can name two such maps: the identity map idP2 , and f ◦ g.
Thus f ◦ g = idP2 . Similarly, g ◦ f = idP1 . Thus the maps f and g arising fromthe universal property are bijections. In short, there is a unique bijection between
P1 and P2 preserving the “product structure” (the maps to M and N). This gives
us the right to name any such product M × N, since any two such products are
uniquely identified.
This definition has the advantage that it works in many circumstances, andonce we define categories, we will soon see that the above argument applies verbatim in any category to show that products, if they exist, are unique up to unique
isomorphism. Even if you haven’t seen the definition of category before, you canverify that this agrees with your notion of product in some category that you haveseen before (such as the category of vector spaces, where the maps are taken to
be linear maps; or the category of differentiable manifolds, where the maps aretaken to be submersions, i.e., differentiable maps whose differential is everywhere
surjective).
This is handy even in cases that you understand. For example, one way ofdefining the product of two manifolds M and N is to cut them both up into charts,
then take products of charts, then glue them together. But if I cut up the manifolds
in one way, and you cut them up in another, how do we know our resulting manifolds are the “same”? We could wave our hands, or make an annoying argument
about refining covers, but instead, we should just show that they are “categorical
products” and hence canonically the “same” (i.e., isomorphic). We will formalize
this argument in §1.3.
Another set of notions we will abstract are categories that “behave like modules”. We will want to define kernels and cokernels for new notions, and weshould make sure that these notions behave the way we expect them to. This
leads us to the definition of abelian categories, first defined by Grothendieck in his
Tôhoku paper [Gr1].
In this chapter, we will give an informal introduction to these and related notions, in the hope of giving just enough familiarity to comfortably use them inpractice.

1.2 Categories and functors
Before functoriality, people lived in caves. — B. Conrad
We begin with an informal definition of categories and functors.
1.2.1. Categories.

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The Rising Sea: Foundations of Algebraic Geometry

A category consists of a collection of objects, and for each pair of objects, a
set of morphisms (or arrows) between them. (For experts: technically, this is thedefinition of a locally small category. In the correct definition, the morphisms need
only form a class, not necessarily a set, but see Caution 0.3.1.) Morphisms are often
informally called maps. The collection of objects of a category C is often denoted
obj(C ), but we will usually denote the collection also by C . If A, B ∈ C , then theset of morphisms from A to B is denoted Mor(A, B). A morphism is often written
f : A → B, and A is said to be the source of f, and B the target of f. (Of course,
Mor(A, B) is taken to be disjoint from Mor(A ′ , B ′ ) unless A = A ′ and B = B ′ .)Morphisms compose as expected: there is a composition Mor(B, C)×Mor(A, B) →Mor(A, C), and if f ∈ Mor(A, B) and g ∈ Mor(B, C), then their composition is denoted g ◦ f. Composition is associative: if f ∈ Mor(A, B), g ∈ Mor(B, C), andh ∈ Mor(C, D), then h ◦ (g ◦ f) = (h ◦ g) ◦ f. For each object A ∈ C , there is always
an identity morphism idA : A → A, such that when you (left- or right-)compose amorphism with the identity, you get the same morphism. More precisely, for any
morphisms f : A → B and g : B → C, idB ◦f = f and g ◦ idB = g. (If you wish,
you may check that “identity morphisms are unique”: there is only one morphism
deserving the name idA .) This ends the definition of a category.
We have a notion of isomorphism between two objects of a category (a morphism f : A → B such that there exists some — necessarily unique — morphismg : B → A, where f ◦ g and g ◦ f are the identity on B and A respectively), and anotion of automorphism of an object (an isomorphism of the object with itself).

1.2.2. Example. The prototypical example to keep in mind is the category of sets,denoted Sets. The objects are sets, and the morphisms are maps of sets. (Because
Russell’s paradox shows that there is no set of all sets, we did not say earlier that
there is a set of all objects. But as stated in §0.3, we are deliberately omitting all
set-theoretic issues.)
1.2.3. Example. Another good example is the category Veck of vector spaces overa given field k. The objects are k-vector spaces, and the morphisms are linear
transformations. (What are the isomorphisms?)

1.2.A. U NIMPORTANT EXERCISE . A category in which each morphism is an isomorphism is called a groupoid. (This notion is not important in what we will
discuss. The point of this exercise is to give you some practice with categories, byrelating them to an object you know well.)(a) A perverse definition of a group is: a groupoid with one object. Make sense of
this.
(b) Describe a groupoid that is not a group.
1.2.B. E XERCISE . If A is an object in a category C , show that the invertible elements of Mor(A, A) form a group (called the automorphism group of A, denoted
Aut(A)). What are the automorphism groups of the objects in Examples 1.2.2
and 1.2.3? Show that two isomorphic objects have isomorphic automorphismgroups. (For readers with a topological background: if X is a topological space,then the fundamental groupoid is the category where the objects are points of X,
and the morphisms x → y are paths from x to y, up to homotopy. Then the automorphism group of x0 is the (pointed) fundamental group π1 (X, x0 ). In the case

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where X is connected, and π1 (X) is not abelian, this illustrates the fact that fora connected groupoid — whose definition you can guess — the automorphismgroups of the objects are all isomorphic, but not canonically isomorphic.)
1.2.4. Example: abelian groups.
phisms, form a category Ab.

The abelian groups, along with group homomor-

1.2.5. Important Example: Modules over a ring. If A is a ring, then the A-modules
form a category ModA . (This category has additional structure; it will be the prototypical example of an abelian category, see §1.6.) Taking A = k, we obtain Example 1.2.3; taking A = Z, we obtain Example 1.2.4.
1.2.6. Example: rings. There is a category Rings, where the objects are rings, andthe morphisms are maps of rings in the usual sense (maps of sets which respect
addition and multiplication, and which send 1 to 1 by our conventions, §0.3).
1.2.7. Example: topological spaces. The topological spaces, along with continuous
maps, form a category Top. The isomorphisms are homeomorphisms.
In all of the above examples, the objects of the categories were in obvious
ways sets with additional structure (a concrete category, although we won’t use
this terminology). This needn’t be the case, as the next example shows.
1.2.8. Example: partially ordered sets. A partially ordered set, (or poset), is a set S
along with a binary relation ≥ on S satisfying:(i) x ≥ x (reflexivity),(ii) x ≥ y and y ≥ z imply x ≥ z (transitivity), and(iii) if x ≥ y and y ≥ x then x = y (antisymmetry).
A partially ordered set (S, ≥) can be interpreted as a category whose objects are
the elements of S, and with a single morphism from x to y if and only if x ≥ y (andno morphism otherwise).
A trivial example is (S, ≥) where x ≥ y if and only if x = y. Another example
is(1.2.8.1)

•
•

% •"

"
•

% •"

Here there are three objects. The identity morphisms are omitted for convenience,and the two non-identity morphisms are depicted. A third example is%•(1.2.8.2)•

Here the “obvious” morphisms are again omitted: the identity morphisms, andthe morphism from the upper left to the lower right. Similarly,
%•
%•
%•
···

depicts a partially ordered set, where again, only the “generating morphisms” aredepicted.

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The Rising Sea: Foundations of Algebraic Geometry

1.2.9. Example: the category of subsets of a set, and the category of open subsets of a topological space. If X is a set, then the subsets form a partially ordered set, where theorder is given by inclusion. Informally, if U ⊂ V, then we have exactly one morphism U → V in the category (and otherwise none). Similarly, if X is a topologicalspace, then the open sets form a partially ordered set, where the order is given byinclusion.
1.2.10. Definition. A subcategory A of a category B has as its objects some of theobjects of B, and some of the morphisms, such that the morphisms of A include
the identity morphisms of the objects of A , and are closed under composition.
(For example, (1.2.8.1) is in an obvious way a subcategory of (1.2.8.2). Also, we
have an obvious “inclusion functor” i : A → B.)
1.2.11. Functors.
A covariant functor F from a category A to a category B, denoted F : A → B,
is the following data. It is a map of objects F : obj(A ) → obj(B), and for each A1 ,A2 ∈ A , and morphism m : A1 → A2 , a morphism F(m) : F(A1 ) → F(A2 ) in B. We
require that F preserves identity morphisms (for A ∈ A , F(idA ) = idF(A) ), and thatF preserves composition (F(m2 ◦ m1 ) = F(m2 ) ◦ F(m1 )). (You may wish to verify
that covariant functors send isomorphisms to isomorphisms.) A trivial example isthe identity functor id : A → A , whose definition you can guess. Here are someless trivial examples.
1.2.12. Example: a forgetful functor. Consider the functor from the category ofvector spaces (over a field k) Veck to Sets, that associates to each vector space its
underlying set. The functor sends a linear transformation to its underlying map ofsets. This is an example of a forgetful functor, where some additional structure isforgotten. Another example of a forgetful functor is ModA → Ab from A-modules
to abelian groups, remembering only the abelian group structure of the A-module.

1.2.13. Topological examples. Examples of covariant functors include the fundamental group functor π1 , which sends a topological space X with choice of a pointx0 ∈ X to a group π1 (X, x0 ) (what are the objects and morphisms of the source category?), and the ith homology functor Top → Ab, which sends a topological space
X to its ith homology group Hi (X, Z). The covariance corresponds to the fact thata (continuous) morphism of pointed topological spaces φ : X → Y with φ(x0 ) = y0
induces a map of fundamental groups π1 (X, x0 ) → π1 (Y, y0 ), and similarly forhomology groups.
1.2.14. Example. Suppose A is an object in a category C . Then there is a functor hA : C → Sets sending B ∈ C to Mor(A, B), and sending f : B1 → B2 toMor(A, B1 ) → Mor(A, B2 ) described by[g : A → B1 ] (→ [f ◦ g : A → B1 → B2 ].

This seemingly silly functor ends up surprisingly being an important concept.
1.2.15. Definitions. If F : A → B and G : B → C are covariant functors, then we
define a functor G ◦ F : A → C (the composition of G and F) in the obvious way.
Composition of functors is associative in an evident sense.

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A covariant functor F : A → B is faithful if for all A, A ′ ∈ A , the map
MorA (A, A ′ ) → MorB (F(A), F(A ′ )) is injective, and full if it is surjective. A functor that is full and faithful is fully faithful. A subcategory i : A → B is a fullsubcategory if i is full. (Inclusions are always faithful, so there is no need for thephrase “faithful subcategory”.) Thus a subcategory A ′ of A is full if and only if for
all A, B ∈ obj(A ′ ), MorA ′ (A, B) = MorA (A, B). For example, the forgetful functor Veck → Sets is faithful, but not full; and if A is a ring, the category of finitely
generated A-modules is a full subcategory of the category ModA of A-modules.

1.2.16. Definition. A contravariant functor is defined in the same way as a covariant functor, except the arrows switch directions: in the above language, F(A1 →
A2 ) is now an arrow from F(A2 ) to F(A1 ). (Thus F(m2 ◦ m1 ) = F(m1 ) ◦ F(m2 ), notF(m2 ) ◦ F(m1 ).)It is wise to state whether a functor is covariant or contravariant, unless thecontext makes it very clear. If it is not stated (and the context does not make it
clear), the functor is often assumed to be covariant.
(Sometimes people describe a contravariant functor C → D as a covariant
functor C opp → D, where C opp is the same category as C except that the arrows
go in the opposite direction. Here C opp is said to be the opposite category to C .)One can define fullness, etc. for contravariant functors, and you should do so.

1.2.17. Linear algebra example. If Veck is the category of k-vector spaces (introducedin Example 1.2.3), then taking duals gives a contravariant functor (·)∨ : Veck →Veck . Indeed, to each linear transformation f : V → W, we have a dual transformation f∨ : W ∨ → V ∨ , and (f ◦ g)∨ = g∨ ◦ f∨ .
1.2.18. Topological example (cf. Example 1.2.13) for those who have seen cohomology. The
ith cohomology functor Hi (·, Z) : Top → Ab is a contravariant functor.

1.2.19. Example. There is a contravariant functor Top → Rings taking a topologicalspace X to the ring of real-valued continuous functions on X. A morphism of
topological spaces X → Y (a continuous map) induces the pullback map fromfunctions on Y to functions on X.
1.2.20. Example (the functor of points, cf. Example 1.2.14). Suppose A is an objectof a category C . Then there is a contravariant functor hA : C → Sets sending
B ∈ C to Mor(B, A), and sending the morphism f : B1 → B2 to the morphism
Mor(B2 , A) → Mor(B1 , A) via[g : B2 → A] (→ [g ◦ f : B1 → B2 → A].

This example initially looks weird and different, but Examples 1.2.17 and 1.2.19
may be interpreted as special cases; do you see how? What is A in each case? Thisfunctor might reasonably be called the functor of maps (to A), but is actually known
as the functor of points. We will meet this functor again in §1.3.10 and (in the
category of schemes) in Definition 6.3.9.
1.2.21. ⋆ Natural transformations (and natural isomorphisms) of covariant functors, and equivalences of categories.
(This notion won’t come up in an essential way until at least Chapter 6, so you
shouldn’t read this section until then.) Suppose F and G are two covariant functors

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from A to B. A natural transformation of covariant functors F → G is the dataof a morphism mA : F(A) → G(A) for each A ∈ A such that for each f : A → A ′ inA , the diagram
F(A)

F(f)

mA ′

mA

"
G(A)

% F(A ′ )G(f)"% G(A ′ )commutes. A natural isomorphism of functors is a natural transformation suchthat each mA is an isomorphism. (We make analogous definitions when F and G
are both contravariant.)
The data of functors F : A → B and F ′ : B → A such that F ◦ F ′ is naturally
isomorphic to the identity functor idB on B and F ′ ◦ F is naturally isomorphic to
idA is said to be an equivalence of categories. “Equivalence of categories” is anequivalence relation on categories. The right notion of when two categories are“essentially the same” is not isomorphism (a functor giving bijections of objects andmorphisms) but equivalence. Exercises 1.2.C and 1.2.D might give you some vague
sense of this. Later exercises (for example, that “rings” and “affine schemes” areessentially the same, once arrows are reversed, Exercise 6.3.D) may help too.
Two examples might make this strange concept more comprehensible. The
double dual of a finite-dimensional vector space V is not V, but we learn early to
say that it is canonically isomorphic to V. We can make that precise as follows. Let
f.d.Veck be the category of finite-dimensional vector spaces over k. Note that thiscategory contains oodles of vector spaces of each dimension.
1.2.C. E XERCISE . Let (·)∨∨ : f.d.Veck → f.d.Veck be the double dual functor fromthe category of finite-dimensional vector spaces over k to itself. Show that (·)∨∨
is naturally isomorphic to the identity functor on f.d.Veck . (Without the finitedimensionality hypothesis, we only get a natural transformation of functors fromid to (·)∨∨ .)Let V be the category whose objects are the k-vector spaces kn for each n ≥ 0(there is one vector space for each n), and whose morphisms are linear transformations. The objects of V can be thought of as vector spaces with bases, and themorphisms as matrices. There is an obvious functor V → f.d.Veck , as each kn is a
finite-dimensional vector space.

1.2.D. E XERCISE . Show that V → f.d.Veck gives an equivalence of categories,
by describing an “inverse” functor. (Recall that we are being cavalier about settheoretic assumptions, see Caution 0.3.1, so feel free to simultaneously choose
bases for each vector space in f.d.Veck . To make this precise, you will need to useGödel-Bernays set theory or else replace f.d.Veck with a very similar small category,
but we won’t worry about this.)
1.2.22. ⋆⋆ Aside for experts. Your argument for Exercise 1.2.D will show that (modulo set-theoretic issues) this definition of equivalence of categories is the same as
another one commonly given: a covariant functor F : A → B is an equivalenceof categories if it is fully faithful and every object of B is isomorphic to an objectof the form F(A) for some A ∈ A (F is essentially surjective, a term we will not

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31need). Indeed, one can show that such a functor has a quasiinverse (another term
we will not use later), i.e., a functor G : B → A (necessarily also an equivalence∼ idA and F ◦ G =∼ idB ,
and unique up to unique isomorphism) for which G ◦ F =and conversely, any functor that has a quasiinverse is an equivalence.

1.3 Universal properties determine an object up to unique
isomorphism
Given some category that we come up with, we often will have ways of producing new objects from old. In good circumstances, such a definition can bemade using the notion of a universal property. Informally, we wish that there werean object with some property. We first show that if it exists, then it is essentiallyunique, or more precisely, is unique up to unique isomorphism. Then we go about
constructing an example of such an object to show existence.
Explicit constructions are sometimes easier to work with than universal properties, but with a little practice, universal properties are useful in proving things
quickly and slickly. Indeed, when learning the subject, people often find explicit
constructions more appealing, and use them more often in proofs, but as they become more experienced, they find universal property arguments more elegant andinsightful.
1.3.1. Products were defined by a universal property. We have seen one important example of a universal property argument already in §1.1: products. You
should go back and verify that our discussion there gives a notion of product inany category, and shows that products, if they exist, are unique up to unique isomorphism.
1.3.2. Initial, final, and zero objects. Here are some simple but useful concepts
that will give you practice with universal property arguments. An object of acategory C is an initial object if it has precisely one map to every object. It is a
final object if it has precisely one map from every object. It is a zero object if it isboth an initial object and a final object.
1.3.A. E XERCISE . Show that any two initial objects are uniquely isomorphic. Show
that any two final objects are uniquely isomorphic.
In other words, if an initial object exists, it is unique up to unique isomorphism,and similarly for final objects. This (partially) justifies the phrase “the initial object”
rather than “an initial object”, and similarly for “the final object” and “the zero
object”. (Convention: we often say “the”, not “a”, for anything defined up tounique isomorphism.)
1.3.B. E XERCISE . What are the initial and final objects in Sets, Rings, and Top (ifthey exist)? How about in the two examples of §1.2.9?
1.3.3. Localization of rings and modules. Another important example of a definition by universal property is the notion of localization of a ring. We first review aconstructive definition, and then reinterpret the notion in terms of universal property. A multiplicative subset S of a ring A is a subset closed under multiplication

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The Rising Sea: Foundations of Algebraic Geometrycontaining 1. We define a ring S−1 A. The elements of S−1 A are of the form a/s
where a ∈ A and s ∈ S, and where a1 /s1 = a2 /s2 if (and only if) for some s ∈ S,
s(s2 a1 − s1 a2 ) = 0. We define (a1 /s1 ) + (a2 /s2 ) = (s2 a1 + s1 a2 )/(s1 s2 ), and(a1 /s1 ) × (a2 /s2 ) = (a1 a2 )/(s1 s2 ). (If you wish, you may check that this equality of fractions really is an equivalence relation and the two binary operations on
fractions are well-defined on equivalence classes and make S−1 A into a ring.) We
have a canonical ring map(1.3.3.1)

A → S−1 A

given by a (→ a/1. Note that if 0 ∈ S, S−1 A is the 0-ring.
There are two particularly important flavors of multiplicative subsets. The
first is {1, f, f2 , . . . }, where f ∈ A. This localization is denoted Af . The second is
A − p, where p is a prime ideal. This localization S−1 A is denoted Ap . (Notational
warning: If p is a prime ideal, then Ap means you’re allowed to divide by elements
not in p. However, if f ∈ A, Af means you’re allowed to divide by f. This can be
confusing. For example, if (f) is a prime ideal, then Af ̸= A(f) .)Warning: sometimes localization is first introduced in the special case where A
is an integral domain and 0 ∈/ S. In that case, A "→ S−1 A, but this isn’t always true,as shown by the following exercise. (But we will see that noninjective localizations
needn’t be pathological, and we can sometimes understand them geometrically,see Exercise 3.2.L.)
1.3.C. E XERCISE . Show that A → S−1 A is injective if and only if S contains nozerodivisors. (A zerodivisor of a ring A is an element a such that there is a nonzero
element b with ab = 0. The other elements of A are called non-zerodivisors. For
example, an invertible element is never a zerodivisor. Counter-intuitively, 0 is a
zerodivisor in every ring but the 0-ring. More generally, if M is an A-module, thena ∈ A is a zerodivisor for M if there is a nonzero m ∈ M with am = 0. The otherelements of A are called non-zerodivisors for M.)
If A is an integral domain and S = A−{0}, then S−1 A is called the fraction fieldof A, which we denote K(A). The previous exercise shows that A is a subring of its
fraction field K(A). We now return to the case where A is a general (commutative)ring.
1.3.D. E XERCISE . Verify that A → S−1 A satisfies the following universal property:
S−1 A is initial among A-algebras B where every element of S is sent to an invertible element in B. (Recall: the data of “an A-algebra B” and “a ring map A → B”
are the same.) Translation: any map A → B where every element of S is sent to an
invertible element must factor uniquely through A → S−1 A. Another translation:a ring map out of S−1 A is the same thing as a ring map from A that sends every
element of S to an invertible element. Furthermore, an S−1 A-module is the samething as an A-module for which s × · : M → M is an A-module isomorphism forall s ∈ S.
In fact, it is cleaner to define A → S−1 A by the universal property, and toshow that it exists, and to use the universal property to check various properties
S−1 A has. Let’s get some practice with this by defining localizations of modules
by universal property. Suppose M is an A-module. We define the A-module map
φ : M → S−1 M as being initial among A-module maps M → N such that elements

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33

of S are invertible in N (s × · : N → N is an isomorphism for all s ∈ S). Moreprecisely, any such map α : M → N factors uniquely through φ:φ% S−1 M
M❋
❋❋
❋❋
❋
∃!
α ❋❋
❋$ "
N

(Translation: M → S−1 M is universal (initial) among A-module maps from M tomodules that are actually S−1 A-modules. Can you make this precise by defining
clearly the objects and morphisms in this category?)
Notice: (i) this determines φ : M → S−1 M up to unique isomorphism (youshould think through what this means); (ii) we are defining not only S−1 M, butalso the map φ at the same time; and (iii) essentially by definition the A-module
structure on S−1 M extends to an S−1 A-module structure.
1.3.E. E XERCISE . Show that φ : M → S−1 M exists, by constructing somethingsatisfying the universal property. Hint: define elements of S−1 M to be of theform m/s where m ∈ M and s ∈ S, and m1 /s1 = m2 /s2 if and only if for somes ∈ S, s(s2 m1 − s1 m2 ) = 0. Define the additive structure by (m1 /s1 ) + (m2 /s2 ) =(s2 m1 + s1 m2 )/(s1 s2 ), and the S−1 A-module structure (and hence the A-module
structure) is given by (a1 /s1 ) · (m2 /s2 ) = (a1 m2 )/(s1 s2 ).
1.3.F. E XERCISE .
(a) Show that localization commutes with finite products, or equivalently, withfinite direct sums. In other words, if M1 , . . . , Mn are A-modules, describe an isomorphism (of A-modules, and of S−1 A-modules) S−1 (M1 ×· · ·×Mn ) → S−1 M1 ×
· · · × S−1 Mn .
(b) Show that localization commutes with arbitrary direct sums.
(c) Show that “localization" does not" necessarily commute with infinite products”:
the obvious map S−1 ( i Mi ) → i S−1 Mi induced by the universal property of
localization is not always an isomorphism. (Hint: (1, 1/2, 1/3, 1/4, . . . ) ∈ Q × Q ×· · · .)
1.3.4. Remark.
Localization does not always commute with Hom, see Example 1.6.8. But Exercise 1.6.G will show that in good situations (if the first argumentof Hom is finitely presented), localization does commute with Hom.
1.3.5. Tensor products. Another important example of a universal property construction is the notion of a tensor product of A-modules⊗A :obj(ModA ) × obj(ModA )% obj(ModA )(M, N) ✤% M ⊗A N

The subscript A is often suppressed when it is clear from context. The tensor product is often defined as follows. Suppose you have two A-modules M and N. Thenelements of the tensor product M⊗A N are finite A-linear combinations of symbols
m ⊗ n (m ∈ M, n ∈ N), subject to relations (m1 + m2 ) ⊗ n = m1 ⊗ n + m2 ⊗ n,m ⊗ (n1 + n2 ) = m ⊗ n1 + m ⊗ n2 , a(m ⊗ n) = (am) ⊗ n = m ⊗ (an) (where a ∈ A,

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The Rising Sea: Foundations of Algebraic Geometry

m1 , m2 ∈ M, n1 , n2 ∈ N). More formally, M ⊗A N is the free A-module generated
by M × N, quotiented by the submodule generated by (m1 + m2 , n) − (m1 , n) −(m2 , n), (m, n1 +n2 )−(m, n1 )−(m, n2 ), a(m, n)−(am, n), and a(m, n)−(m, an)for a ∈ A, m, m1 , m2 ∈ M, n, n1 , n2 ∈ N. The image of (m, n) in this quotient ism ⊗ n.
If A is a field k, we recover the tensor product of vector spaces.
1.3.G. E XERCISE ( IF YOU HAVEN ’ T SEEN TENSOR PRODUCTS BEFORE ). Show that
∼ Z/(2). (This exercise is intended to give some hands-on pracZ/(10) ⊗Z Z/(12) =tice with tensor products.)
1.3.H. I MPORTANT E XERCISE : RIGHT- EXACTNESS OF (·) ⊗A N. Show that (·) ⊗A N
gives a covariant functor ModA → ModA . Show that (·) ⊗A N is a right-exact
functor, i.e., if
M ′ → M → M ′′ → 0
is an exact sequence of A-modules (which means f : M → M ′′ is surjective, andM ′ surjects onto the kernel of f; see §1.6), then the induced sequence
M ′ ⊗A N → M ⊗A N → M ′′ ⊗A N → 0

is also exact. This exercise is repeated in Exercise 1.6.F, but you may get a lot out of
doing it now. (You will be reminded of the definition of right-exactness in §1.6.5.)
In contrast, you can quickly check that tensor product is not left-exact: tensor
the exact sequence of Z-modules
0

%Z

×2

%Z

% Z/(2)%0with Z/(2).
The constructive definition ⊗ is a weird definition, and really the “wrong”
definition. To motivate a better one: notice that there is a natural A-bilinear map
M × N → M ⊗A N. (If M, N, P ∈ ModA , a map f : M × N → P is A-bilinear if
f(m1 + m2 , n) = f(m1 , n) + f(m2 , n), f(m, n1 + n2 ) = f(m, n1 ) + f(m, n2 ), andf(am, n) = f(m, an) = af(m, n).) Any A-bilinear map M × N → P factors throughthe tensor product uniquely: M × N → M ⊗A N → P. (Think this through!)
We can take this as the definition of the tensor product as follows. It is an
A-module T along with an A-bilinear map t : M × N → T , such that given any
A-bilinear map t ′ : M × N → T ′ , there is a unique A-linear map f : T → T ′ suchthat t ′ = f ◦ t.
t
%T
M×N
●●
●●
●●
●●
t′
●$
5 ∃!f
T′
1.3.I. E XERCISE . Show that (T, t : M×N → T ) is unique up to unique isomorphism.
Hint: first figure out what “unique up to unique isomorphism” means for suchpairs, using a category of pairs (T, t). Then follow the analogous argument for the
product.
In short: given M and N, there is an A-bilinear map t : M × N → M ⊗A N,
unique up to unique isomorphism, defined by the following universal property:

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for any A-bilinear map t ′ : M × N → T ′ there is a unique A-linear map f : M ⊗A
N → T ′ such that t ′ = f ◦ t.
As with all universal property arguments, this argument shows uniquenessassuming existence. To show existence, we need an explicit construction.
1.3.J. E XERCISE . Show that the construction of §1.3.5 satisfies the universal property of tensor product.
The three exercises below are useful facts about tensor products with which
you should be familiar.
1.3.K. I MPORTANT E XERCISE .
(a) If M is an A-module and A → B is a morphism of rings, give B ⊗A M the structure of a B-module (this is part of the exercise). Show that this describes a functor
ModA → ModB .
(b) If further A → C is another morphism of rings, show that B ⊗A C has a natural
structure of a ring. Hint: multiplication will be given by (b1 ⊗ c1 )(b2 ⊗ c2 ) =(b1 b2 ) ⊗ (c1 c2 ). (Exercise 1.3.U will interpret this construction as a fibered coproduct.)
1.3.L. I MPORTANT E XERCISE . If S is a multiplicative subset of A and M is an A∼ S−1 M (as S−1 A-modules
module, describe a natural isomorphism (S−1 A)⊗A M =and as A-modules).
1.3.M. E XERCISE (⊗ COMMUTES WITH ⊕). Show that tensor products commute
with arbitrary direct sums: if M and {Ni }i∈I are all A-modules, describe an isomorphism∼ %M ⊗ (⊕i∈I Ni )⊕i∈I (M ⊗ Ni ) .
1.3.6. Essential Example: Fibered products.
Suppose we have morphisms
α : X → Z and β : Y → Z (in any category). Then the fibered product (or fibred product) is an object X ×Z Y along with morphisms prX : X ×Z Y → X andprY : X ×Z Y → Y, where the two compositions α ◦ prX , β ◦ prY : X ×Z Y → Z agree,
such that given any object W with maps to X and Y (whose compositions to Z
agree), these maps factor through some unique W → X ×Z Y:
W✺ ❙❙❙✺✺ ❙❙❙❙✺✺ ∃! ❙❙❙❙❙
❙❙❙
✺✺
❙❙❙
✺✺ $
❙0
✺✺ X ×Z Y prY % Y✺✺✺✺ prX
β
✺6 "
"
α
%Z
X

(Warning: the definition of the fibered product depends on α and β, even though
they are omitted from the notation X ×Z Y.)
By the usual universal property argument, if it exists, it is unique up to unique
isomorphism. (You should think this through until it is clear to you.) Thus the useof the phrase “the fibered product” (rather than “a fibered product”) is reasonable,
and we should reasonably be allowed to give it the name X ×Z Y. We know whatmaps to it are: they are precisely maps to X and maps to Y that agree as maps to Z.

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The Rising Sea: Foundations of Algebraic Geometry

Depending on your religion, the diagram
X ×Z Y prY

%Y

prX

"
X

β
α

"
%Z

is called a fibered/pullback/Cartesian diagram/square (six possibilities — evenmore are possible if you prefer “fibred” to “fibered”).
The right way to interpret the notion of fibered product is first to think about
what it means in the category of sets.
1.3.N. E XERCISE ( FIBERED PRODUCTS OF SETS ). Show that in Sets,
X ×Z Y = {(x, y) ∈ X × Y : α(x) = β(y)}.
More precisely, show that the right side, equipped with its evident maps to X andY, satisfies the universal property of the fibered product. (This will help you build
intuition for fibered products.)
1.3.O. E XERCISE . If X is a topological space, show that fibered products always
exist in the category of open sets of X, by describing what a fibered product is.
(Hint: it has a one-word description.)
1.3.P. E XERCISE . If Z is the final object in a category C , and X, Y ∈ C , show that
“X ×Z Y = X × Y”: “the” fibered product over Z is uniquely isomorphic to “the”
product. Assume all relevant (fibered) products exist. (This is an exercise about
unwinding the definition.)
1.3.Q. U SEFUL E XERCISE : TOWERS OF C ARTESIAN DIAGRAMS ARE C ARTESIAN DI AGRAMS . If the two squares in the following commutative diagram are Cartesian
diagrams, show that the “outside rectangle” (involving U, V, Y, and Z) is also a
Cartesian diagram.
U

%V

"
W

"
%X

"
Y

"
%Z

1.3.R. E XERCISE . Given morphisms X1 → Y, X2 → Y, and Y → Z, show that there
is a natural morphism X1 ×Y X2 → X1 ×Z X2 , assuming that both fibered products
exist. (This is trivial once you figure out what it is saying. The point of this exercise
is to see why it is trivial.)
1.3.S. U SEFUL EXERCISE : T HE MAGIC DIAGRAM . Suppose we are given morphisms X1 , X2 → Y and Y → Z. Show that the following diagram is a Cartesian

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square.
X1 ×Y X2

% X1 × Z X2"Y"
% Y ×Z Y

Assume all relevant (fibered) products exist. This diagram is surprisingly useful— so useful that we will call it the magic diagram.
1.3.7. Coproducts. Define coproduct in a category by reversing all the arrows in
the definition of product. Define fibered coproduct in a category by reversing all
the arrows in the definition of fibered product.
1.3.T. E XERCISE . #Show that coproduct for Sets is disjoint union. This is why weuse the notation
for disjoint union.

1.3.U. E XERCISE . Suppose A → B and A → C are two ring morphisms, so inparticular B and C are A-modules. Recall (Exercise 1.3.K) that B ⊗A C has a ring
structure. Show that there is a natural morphism B → B ⊗A C given by b (→ b ⊗ 1.
(This is not necessarily an inclusion; see Exercise 1.3.G.) Similarly, there is a naturalmorphism C → B⊗A C. Show that this gives a fibered coproduct on rings, i.e., thatB ⊗7A C (

C7

B(

A

satisfies the universal property of fibered coproduct.
1.3.8. Monomorphisms and epimorphisms.
1.3.9. Definition. A morphism π : X → Y is a monomorphism if any two morphisms µ1 : Z → X and µ2 : Z → X such that π ◦ µ1 = π ◦ µ2 must satisfy µ1 = µ2 .
In other words, there is at most one way of filling in the dotted arrow so that the
diagram
Z❄
❄❄
❄❄
≤1
❄❄
❄8
"
%Y
X
π

commutes — for any object Z, the natural map Mor(Z, X) → Mor(Z, Y) is an injection. Intuitively, it is the categorical version of an injective map, and indeed
this notion generalizes the familiar notion of injective maps of sets. (The reason
we don’t use the word “injective” is that in some contexts, “injective” will havean intuitive meaning which may not agree with “monomorphism”. One example:
in the category of divisible groups, the map Q → Q/Z is a monomorphism but
not injective. This is also the case with “epimorphism” (to be defined shortly) vs.
“surjective”.)
1.3.V. E XERCISE . Show that the composition of two monomorphisms is a monomorphism.

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The Rising Sea: Foundations of Algebraic Geometry

1.3.W. E XERCISE . Prove that a morphism π : X → Y is a monomorphism if andonly if the fibered product X ×Y X exists, and the induced morphism X → X ×Y X
is an isomorphism. We may then take this as the definition of monomorphism.
(Monomorphisms aren’t central to future discussions, although they will come up
again. This exercise is just good practice.)
1.3.X. E ASY E XERCISE . We use the notation of Exercise 1.3.R. Show that if Y → Z
is a monomorphism, then the morphism X1 ×Y X2 → X1 ×Z X2 you described in
Exercise 1.3.R is an isomorphism. (Hint: for any object V, give a natural bijection
between maps from V to the first and maps from V to the second. It is also possible
to use the magic diagram, Exercise 1.3.S.)The notion of an epimorphism is “dual” to the definition of monomorphism,
where all the arrows are reversed. This concept will not be central for us, although
it turns up in the definition of an abelian category. Intuitively, it is the categorical version of a surjective map. (But be careful when working with categories ofobjects that are sets with additional structure, as epimorphisms need not be surjective. Example: in the category Rings, Z → Q is an epimorphism, but obviously notsurjective.)

1.3.10. Representable functors and Yoneda’s Lemma. Much of our discussion
about universal properties can be cleanly expressed in terms of representable functors, under the rubric of “Yoneda’s Lemma”. Yoneda’s lemma is an easy fact statedin a complicated way. Informally speaking, you can essentially recover an object
in a category by knowing the maps into it. For example, we have seen that thedata of maps to X × Y are naturally (canonically) the data of maps to X and to Y.
Indeed, we have now taken this as the definition of X × Y.
Recall Example 1.2.20. Suppose A is an object of category C . For any object
C ∈ C , we have a set of morphisms Mor(C, A). If we have a morphism f : B → C,
we get a map of sets
(1.3.10.1)

Mor(C, A) → Mor(B, A),

by composition: given a map from C to A, we get a map from B to A by precomposing with f : B → C. Hence this gives a contravariant functor hA : C → Sets.
Yoneda’s Lemma states that the functor hA determines A up to unique isomorphism. More precisely:
1.3.Y. I MPORTANT E XERCISE THAT YOU SHOULD DO ONCE IN YOUR LIFE (Y ONEDA’ S
L EMMA ).
(a) Suppose you have two objects A and A ′ in a category C , and morphisms(1.3.10.2)

iC : Mor(C, A) → Mor(C, A ′ )that commute with the maps (1.3.10.1). Show that the iC (as C ranges over the objects of C ) are induced from a unique morphism g : A → A ′ . More precisely, showthat there is a unique morphism g : A → A ′ such that for all C ∈ C , iC is u (→ g ◦ u.
(b) If furthermore the iC are all bijections, show that the resulting g is an isomorphism. (Hint for both: This is much easier than it looks. This statement is so
general that there are really only a couple of things that you could possibly try.
For example, if you’re hoping to find a morphism A → A ′ , where will you find

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it? Well, you are looking for an element Mor(A, A ′ ). So just plug in C = A to
(1.3.10.2), and see where the identity goes.)
There is an analogous statement with the arrows reversed, where instead ofmaps into A, you think of maps from A. The role of the contravariant functor hA
of Example 1.2.20 is played by the covariant functor hA of Example 1.2.14. Because
the proof is the same (with the arrows reversed), you needn’t think it through.
The phrase “Yoneda’s Lemma” properly refers to a more general statement.
Although it looks more complicated, it is no harder to prove.
1.3.Z. ⋆ E XERCISE .
(a) Suppose A and B are objects in a category C . Give a bijection between the natural transformations hA → hB of covariant functors C → Sets (see Example 1.2.14
for the definition) and the morphisms B → A.
(b) State and prove the corresponding fact for contravariant functors hA (see Example 1.2.20). Remark: A contravariant functor F from C to Sets is said to be
representable if there is a natural isomorphism
ξ: F∼% hA .

Thus the representing object A is determined up to unique isomorphism by thepair (F, ξ). There is a similar definition for covariant functors. (We will revisit
this in §6.6, and this problem will appear again as Exercise 6.6.C. The element
ξ−1 (idA ) ∈ F(A) is often called the “universal object”; do you see why?)(c) Yoneda’s Lemma. Suppose F is a covariant functor C → Sets, and A ∈ C .
Give a bijection between the natural transformations hA → F and F(A). (Thecorresponding fact for contravariant functors is essentially Exercise 9.1.C.)
In fancy terms, Yoneda’s lemma states the following. Given a category C , wecan produce a new category, called the functor category of C , where the objects arecontravariant functors C → Sets, and the morphisms are natural transformationsof such functors. We have a functor (which we can usefully call h) from C to its
functor category, which sends A to hA . Yoneda’s Lemma states that this is a fully
faithful functor, called the Yoneda embedding. (Fully faithful functors were definedin §1.2.15.)

1.3.11. Joke. The Yoda embedding, contravariant it is.

1.4 Limits and colimits
Limits and colimits are two important definitions determined by universal
properties. They generalize a number of familiar constructions. I will give the definition first, and then show you why it is familiar. For example, fractions will be
motivating examples of colimits (Exercise 1.4.C(a)), and the p-adic integers (Example 1.4.3) will be motivating examples of limits.
1.4.1. Limits. We say that a category is a small category if the objects and themorphisms are sets. (This is a technical condition intended only for experts.) Suppose I is any small category, and C is any category. Then a functor F : I → C(i.e., with an object Ai ∈ C for each element i ∈ I , and appropriate commuting

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The Rising Sea: Foundations of Algebraic Geometrymorphisms dictated by I ) is said to be a diagram indexed by I . We call I an
index category. Our index categories will usually be partially ordered sets (Example 1.2.8), in which in particular there is at most one morphism between any
two objects. (But other examples are sometimes useful.) For example, if ! is thecategory
%•
•
"
% •"
•
and A is a category, then a functor ! → A is precisely the data of a commuting
square in A .
Then the limit of the diagram is an object lim Ai of C along with morphisms←−Ifj : lim Ai → Aj for each j ∈ I , such that if m : j → k is a morphism in I , then←−I

(1.4.1.1)

lim Ai
←−
I
fj

"
Aj

❈❈ fk
❈❈
❈❈
❈
F(m) 9% Akcommutes, and this object and maps to each Ai are universal (final) with respect tothis property. More precisely, given any other object W along with maps gi : W →Ai commuting with the F(m) (if m : j → k is a morphism in I , then gk = F(m)◦gj ),then there is a unique map
g : W → lim Ai←−I

so that gi = fi ◦ g for all i. (In some cases, the limit is sometimes called the inverse
limit or projective limit. We won’t use this language.) By the usual universal
property argument, if the limit exists, it is unique up to unique isomorphism.
1.4.2. Examples: products. For example, if I is the partially ordered set•

we obtain the fibered product.
If I is

•

% •"

•

•

we obtain the product.
If I is a set (i.e., the only morphisms are"the identity maps), then the limit iscalled the product of the Ai , and is denoted i Ai . The special case where I has
two elements is the example of the previous paragraph.
1.4.A. E XERCISE ( REALITY CHECK ). Suppose that the partially ordered set I has
an initial object e. Show that the limit of any diagram indexed by I exists.
1.4.3. Example: the p-adic integers. For a prime number p, the p-adic integers(or more informally, p-adics), Zp , are often described informally (and somewhat

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unnaturally) as being of the form Zp = a0 + a1 p + a2 p2 + · · · (where 0 ≤ ai < p).
They are an example of a limit in the category of rings:
Zp ❱❘❉❉❘❱❘❱❘❱❘❱❘❱❱❱❱❉❉ ❘❘❘ ❱❱❱❱❉❉❘❘❘ ❱❱❱❱❱❉❉❱❘❘❘❘❘0 ❱❱❱❱❱❱❱❱
9:%% Z/p3% Z/p2···Z/p.
(Warning: Zp is sometimes is used to denote the integers modulo p, but Z/(p) or
Z/pZ is better to use for this, to avoid confusion. Worse: by §1.3.3, Zp also denotes
those rationals whose denominators are a power of p. Hopefully the meaningof Zp will be clear from the context.) The p-adic integers are an example of acompletion, the topic of Chapter 29.
Limits do not always exist for any index category I . However, you can often
easily check that limits exist if the objects of your category can be interpreted as
sets with additional structure, and arbitrary products exist (respecting the set-likestructure).
1.4.B. I MPORTANT E XERCISE . Show that in the category Sets,%(ai )i∈I ∈&i'Ai : F(m)(aj ) = ak for all m ∈ MorI (j, k) ∈ Mor(I ) ,along with the obvious projection maps to each Ai , is the limit lim Ai .
←−I

This clearly also works in the category ModA of A-modules (in particular Veck
and Ab), as well as Rings.
From this point of view, 2 + 3p + 2p2 + · · · ∈ Zp can be understood as thesequence (2, 2 + 3p, 2 + 3p + 2p2 , . . . ).
1.4.4. Colimits.
More immediately relevant for us will be the dual (arrowreversed version) of the notion of limit (or inverse limit). We just flip the arrows
fi in (1.4.1.1), and get the notion of a colimit, which is denoted limI Ai . (You−→should draw the corresponding diagram.) Again, if it exists, it is unique up tounique isomorphism. (In some cases, the colimit is sometimes called the direct
limit, inductive limit, or injective limit. We won’t use this language. I prefer using limit/colimit in analogy with kernel/cokernel and product/coproduct. Thisis more than analogy, as kernels and products may be interpreted as limits, andsimilarly with cokernels and coproducts. Also, I remember that kernels “map to”,
and cokernels are “mapped to”, which reminds me that a limit maps to all the objects in the big commutative diagram indexed by I ; and a colimit has a map fromall the objects.)
1.4.5. Joke. A comathematician is a device for turning cotheorems into ffee.
Even though we have just flipped the arrows, colimits behave quite differently
from limits.

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The Rising Sea: Foundations of Algebraic Geometry

1.4.6. Example. The set 5−∞ Z of rational numbers whose denominators are powersof 5 is a colimit lim 5−i Z. More precisely, 5−∞ Z is the colimit of the diagram−→Z% 5−1 Z% 5−2 Z% ···The colimit over an index set I is called the coproduct, denoted
the dual (arrow-reversed) notion to the product.

#

i Ai , and is

1.4.C. E XERCISE .
1(a) Interpret the statement “Q = lim nZ”.
−→(b) Interpret the union of some subsets of a given set as a colimit. (Dually, theintersection can be interpreted as a limit.) The objects of the category in question
are the subsets of the given set.
Colimits do not always exist, but there are two useful large classes of examples
for which they do.
1.4.7. Definition. A nonempty partially ordered set (S, ≥) is filtered (or is said tobe a filtered set) if for each x, y ∈ S, there is a z such that x ≥ z and y ≥ z. Moregenerally, a nonempty category I is filtered if:(i) for each x, y ∈ I , there is a z ∈ I and arrows x → z and y → z, and(ii) for every two arrows u : x → y and v : x → y, there is an arrow w : y → zsuch that w ◦ u = w ◦ v.

(Other terminologies are also commonly used, such as “directed partially orderedset” and “filtered index category”, respectively.)
1.4.D. E XERCISE . Suppose I is filtered. (We will
# almost exclusively use the casewhere I is a filtered set.) Recall the symbolfor disjoint union of sets. Show
that any diagram in Sets indexed by I has the following, with the obvious maps
to it, as a colimit:() *!" (ai , i) ∼ (aj , j) if and only if there are f : Ai → Ak and #(ai , i) ∈Ai
g : Aj → Ak in the diagram for which f(ai ) = g(aj ) in Aki∈I(You will see that the “filtered” hypothesis is there is to ensure that ∼ is an equivalence relation.)

For example, in Example 1.4.6, each element of the colimit is an element ofsomething upstairs, but you can’t say in advance what it is an element of. For
instance, 17/125 is an element of the 5−3 Z (or 5−4 Z, or later ones), but not 5−2 Z.
This idea applies to many categories whose objects can be interpreted as setswith additional structure (such as abelian groups, A-modules, groups, etc.). For
example, the colimit lim Mi in the category of A-modules ModA can be described−→as follows. The set underlying lim Mi is defined as in Exercise 1.4.D. To add the−→elements mi ∈ Mi and mj ∈ Mj , choose an ℓ ∈ I with arrows u : i → ℓ andv : j → ℓ, and then define the sum of mi and mj to be F(u)(mi ) + F(v)(mj ) ∈ Mℓ .
The element mi ∈ Mi is 0 if and only if there is some arrow u : i → k for whichF(u)(mi ) = 0, i.e., if it becomes 0 “later in the diagram”. Last, multiplication by anelement of A is defined in the obvious way. (You can now reinterpret Example 1.4.6
as a colimit of groups, not just of sets.)

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1.4.E. E XERCISE . Verify that the A-module described above is indeed the colimit.
(Make sure you verify that addition is well-defined, i.e., is independent of the
choice of representatives mi and mj , the choice of ℓ, and the choice of arrows uand v. Similarly, make sure that scalar multiplication is well-defined.)
1.4.F. U SEFUL E XERCISE ( LOCALIZATION AS A COLIMIT ). Generalize Exercise 1.4.C(a)to interpret localization of an integral domain as a colimit over a filtered set: suppose S is a multiplicative set of A, and interpret S−1 A = lim 1s A where the limit−→
is over s ∈ S, and in the category of A-modules. (Aside: Can you make someversion of this work even if A isn’t an integral domain, e.g., S−1 A = lim As ? This−→
will work in the category of A-algebras.)
A variant of this construction works without the filtered condition, if you haveanother means of “connecting elements in different objects of your diagram”. For
example:1.4.G. E XERCISE : COLIMITS OF A- MODULES WITHOUT THE FILTERED CONDITION .
Suppose you are given a diagram of A-modules indexed by I : F : I → ModA ,where we let Mi := F(i). Show that the colimit is ⊕i∈I Mi modulo the relations
mi −F(n)(mi ) for every n : i → j in I (i.e., for every arrow in the diagram). (Somewhat more precisely: “modulo” means “quotiented by the submodule generatedby”.)
1.4.8. Summary. One useful thing to informally keep in mind is the following. In
a category where the objects are “set-like”, an element of a limit can be thought ofas a family of elements of each object in the diagram, that are “compatible” (Exercise 1.4.B). And an element of a colimit can be thought of (“has a representativethat is”) an element of a single object in the diagram (Exercise 1.4.D). Even though
the definitions of limit and colimit are the same, just with arrows reversed, theseinterpretations are quite different.
1.4.9. Small remark. In fact, colimits exist in the category of sets for all reasonable(“small”) index categories (see for example [E, Thm. A6.1]), but that won’t matter
to us.

1.5 Adjoints
We next come to a very useful notion closely related to universal properties.
Just as a universal property “essentially” (up to unique isomorphism) determines
an object in a category (assuming such an object exists), “adjoints” essentially determine a functor (again, assuming it exists). Two covariant functors F : A → B
and G : B → A are adjoint if there is a natural bijection for all A ∈ A and B ∈ B(1.5.0.1)

τAB : MorB (F(A), B) → MorA (A, G(B)).

We say that (F, G) form an adjoint pair, and that F is left-adjoint to G (and G isright-adjoint to F). We say F is a left adjoint (and G is a right adjoint). By “natural”

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we mean the following. For all f : A → A ′ in A , we require(1.5.0.2)

MorB (F(A ′ ), B)

Ff∗% MorB (F(A), B)

τA ′ BτAB"
MorA (A ′ , G(B))∗

f"% MorA (A, G(B))

to commute, and for all g : B → B ′ in B we want a similar commutative diagram tocommute. (Here f∗ is the map induced by f : A → A ′ , and Ff∗ is the map induced
by Ff : F(A) → F(A ′ ).)1.5.A. E XERCISE . Write down what this diagram should be.

1.5.B. E XERCISE . Show that the map τAB (1.5.0.1) has the following properties.
For each A there is a map ηA : A → GF(A) so that for any g : F(A) → B, the corresponding τAB (g) : A → G(B) is given by the compositionAηA% GF(A)Gg% G(B).

Similarly, there is a map ϵB : FG(B) → B for each B so that for any f : A → G(B),the corresponding map τ−1
AB (f) : F(A) → B is given by the composition
F(A)

Ff

% FG(B)ϵB% B.

Here is a key example of an adjoint pair.
1.5.C. E XERCISE . Suppose M, N, and P are A-modules (where A is a ring). Describe a bijection HomA (M ⊗A N, P) ↔ HomA (M, HomA (N, P)). (Hint: try to use
the universal property of ⊗.)1.5.D. E XERCISE . Show that (·) ⊗A N and HomA (N, ·) are adjoint functors.

1.5.E. E XERCISE . Suppose B → A is a morphism of rings. If M is an A-module,
you can create a B-module MB by considering it as a B-module. This gives afunctor ·B : ModA → ModB . Show that this functor is right-adjoint to · ⊗B A. In
other words, describe a bijection∼ HomB (N, MB )HomA (N ⊗B A, M) =functorial in both arguments. (This adjoint pair is very important, and is the key
player in Chapter 16.)
1.5.1. ⋆ Fancier remarks we won’t use. You can check that the left adjoint determines the right adjoint up to natural isomorphism, and vice versa. The maps
ηA and ϵB of Exercise 1.5.B are called the unit and counit of the adjunction. This
leads to a different characterization of adjunction. Suppose functors F : A → B
and G : B → A are given, along with natural transformations η : idA → GF andϵ : FG → idB with the property that Gϵ ◦ ηG = idG (for each B ∈ B, the composition of ηG(B) : G(B) → GFG(B) and G(ϵB ) : GFG(B) → G(B) is the identity) andϵF ◦ Fη = idF . Then you can check that F is left-adjoint to G. These facts aren’t hard
to check, so if you want to use them, you should verify everything for yourself.

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1.5.2. Examples from other fields. For those familiar with representation theory:
Frobenius reciprocity may be understood in terms of adjoints. Suppose V is a
finite-dimensional representation of a finite group G, and W is a representation ofG
a subgroup H < G. Then induction and restriction are an adjoint pair (IndG
H , ResH )between the category of G-modules and the category of H-modules.
Topologists’ favorite adjoint pair may be the suspension functor and the loop
space functor.
1.5.3. Example: groupification of abelian semigroups. Here is another motivating example: getting an abelian group from an abelian semigroup. (An abelian
semigroup is just like an abelian group, except we don’t require an identity or an
inverse. Morphisms of abelian semigroups are maps of sets preserving the binary
operation. One example is the non-negative integers Z≥0 = {0, 1, 2, . . . } under addition. Another is the positive integers 1, 2, . . . under multiplication. You may
enjoy groupifying both.) From an abelian semigroup, you can create an abeliangroup. In our examples, from the nonnegative (Z≥0 , +), we create the integers Z,
and from the positive integers under multiplication (Z>0 , ×), we create the positive rationals Q>0 . Here is a formalization of that notion. A groupification of asemigroup S is a map of abelian semigroups π : S → G such that G is an abeliangroup, and any map of abelian semigroups from S to an abelian group G ′ factors
uniquely through G:
π %
S❅
G
❅❅
❅❅
❅❅ ∃!
1 "
G′
(Perhaps “abelian groupification” would be more precise than “groupification”.)
1.5.F. E XERCISE ( AN ABELIAN GROUP IS GROUPIFIED BY ITSELF ). Show that if anabelian semigroup is already a group then the identity morphism is the groupification. (More correct: the identity morphism is a groupification.) Note that you
don’t need to construct groupification (or even know that it exists in general) tosolve this exercise.
1.5.G. E XERCISE . Construct the “groupification functor” H from the categoryof nonempty abelian semigroups to the category of abelian groups. (One possible
construction: given an abelian semigroup S, the elements of its groupification H(S)are ordered pairs (a, b) ∈ S × S, which you may think of as a − b, with the equivalence that (a, b) ∼ (c, d) if a + d + e = b + c + e for some e ∈ S. Describe addition inthis group, and show that it satisfies the properties of an abelian group. Describe
the abelian semigroup map S → H(S).) Let F be the forgetful functor from thecategory of abelian groups Ab to the category of abelian semigroups. Show that H
is left-adjoint to F.
(Here is the general idea for experts: We have a full subcategory of a category.
We want to “project” from the category to the subcategory. We haveMorcategory (S, H) = Morsubcategory (G, H)
automatically; thus we are describing the left adjoint to the forgetful functor. How
the argument worked: we constructed something which was in the smaller category, which automatically satisfies the universal property.)

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1.5.H. E XERCISE ( CF. E XERCISE 1.5.E). The purpose of this exercise is to give
you more practice with “adjoints of forgetful functors”, the means by which we
get groups from semigroups, and sheaves from presheaves. Suppose A is a ring,and S is a multiplicative subset. Then S−1 A-modules are a fully faithful subcategory (§1.2.15) of the category of A-modules (via the obvious inclusion ModS−1 A "→ModA ). Then ModA → ModS−1 A can be interpreted as an adjoint to the forgetful
functor ModS−1 A → ModA . State and prove the correct statements.
(Here is the larger story. Every S−1 A-module is an A-module, and this is aninjective map, so we have a covariant forgetful functor F : ModS−1 A → ModA . In
fact this is a fully faithful functor: it is injective on objects, and the morphisms
between any two S−1 A-modules as A-modules are just the same when they are considered as S−1 A-modules. Then there is a functor G : ModA → ModS−1 A , whichmight reasonably be called “localization with respect to S”, which is left-adjoint
to the forgetful functor. Translation: If M is an A-module, and N is an S−1 Amodule, then Mor(GM, N) (morphisms as S−1 A-modules, which are the same asmorphisms as A-modules) are in natural bijection with Mor(M, FN) (morphisms
as A-modules).)
Here is a table of most of the adjoints that will come up for us.
situation
A-modules (Ex. 1.5.D)
ring maps
B → A (Ex. 1.5.E)(pre)sheaves on a
topological space
X (Ex. 2.4.L)(semi)groups (§1.5.3)
sheaves,
π : X → Y (Ex. 2.7.B)
sheaves of abeliangroups or O-modules,
open embeddings
π : U "→ Y (Ex. 2.7.G)
quasicoherent sheaves,
π : X → Y (Prop. 16.3.6)
ring maps
B → A (Ex. 30.3.A)
quasicoherent sheaves,affine π : X → Y(Ex. 30.3.B(b))

categoryA
ModA

category
B
ModA

ModB

ModA

presheaves
on X

sheaveson X

semigroups groups
sheavessheaves
on Y
on X

left adjoint
F: A → B(·) ⊗A N(·) ⊗B A(extensionof scalars)

right adjoint
G: B → A
HomA (N, ·)M (→ MB(restrictionof scalars)

sheafification

forgetfulgroupification
π−1forgetfulπ∗sheaveson U

sheaves
on Y

π!

π−1

QCohY

QCohX

π∗

π∗

ModA

ModB

QCohX

QCohY

M (→ MB(restrictionof scalars)

N (→HomB (A, N)

π∗

π!sh

Other examples will also come up, such as the adjoint pair (∼, Γ• ) between
graded modules over a graded ring, and quasicoherent sheaves on the corresponding projective scheme (§15.4).

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1.5.4. Useful comment for experts. One last comment only for people who have seenadjoints before: If (F, G) is an adjoint pair of functors, then F commutes with colimits, and G commutes with limits. Also, limits commute with limits and colimits
commute with colimits. We will prove these facts (and a little more) in §1.6.12.

1.6 An introduction to abelian categories
Ton papier sur l’Algèbre homologique a été lu soigneusement, et a converti tout lemonde (même Dieudonné, qui semble complètement fonctorisé!) à ton point de vue.
Your paper on homological algebra was read carefully and converted everyone (evenDieudonné, who seems to be completely functorised!) to your point of view.
— J.-P. Serre, letter to A. Grothendieck, Jul 13, 1955 [GrS, p. 17-18]
Since learning linear algebra, you have been familiar with the notions andbehaviors of kernels, cokernels, etc. Later in your life you saw them in the categoryof abelian groups, and later still in the category of A-modules.
We will soon define some new categories (certain sheaves) that will have familiarlooking behavior, reminiscent of that of modules over a ring. The notions of kernels, cokernels, images, and more will make sense, and they will behave “the way
we expect” from our experience with modules. This can be made precise throughthe notion of an abelian category. Abelian categories are the right general settingin which one can do “homological algebra”, in which notions of kernel, cokernel,
and so on are used, and one can work with complexes and exact sequences.
We will see enough to motivate the definitions that we will see in general:monomorphism (and subobject), epimorphism, kernel, cokernel, and image. But
in this book we will avoid having to show that they behave “the way we expect”
in a general abelian category because the examples we will see are directly interpretable in terms of modules over rings. In particular, it is not worth memorizing
the definition of abelian category.
Two central examples of an abelian category are the category Ab of abeliangroups, and the category ModA of A-modules. The first is a special case of thesecond (just take A = Z). As we give the definitions, you should verify that ModA
is an abelian category.
We first define the notion of additive category. We will use it only as a steppingstone to the notion of an abelian category. Two examples you can keep in mind
while reading the definition: the category of free A-modules (where A is a ring),
and real (or complex) Banach spaces.
1.6.1. Definition. A category C is said to be additive if it satisfies the following
properties.
Ad1. For each A, B ∈ C , Mor(A, B) is an abelian group, such that compositionof morphisms distributes over addition. (You should think about what
this means — it translates to two distinct statements.)
Ad2. C has a zero object, denoted 0. (This is an object that is simultaneously
an initial object and a final object, Definition 1.3.2.)
Ad3. It has products of two objects (a product A × B for any pair of objects),
and hence by induction, products of any finite number of objects.

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The Rising Sea: Foundations of Algebraic Geometry

In an additive category, the morphisms are often called homomorphisms, andMor is denoted by Hom. In fact, this notation Hom is a good indication that you’re
working in an additive category. A functor between additive categories preserving
the additive structure of Hom, is called an additive functor.
1.6.2. Remarks. It is a consequence of the definition of additive category that finite
direct products are also finite direct sums (coproducts) — the details don’t matter
to us. The symbol ⊕ is used for this notion. Also, it is quick to show that additive
functors send zero objects to zero objects (show that Z is a 0-object if and only if
idZ = 0Z ; additive functors preserve both id and 0), and preserve products.
One motivation for the name 0-object is that the 0-morphism in the abeliangroup Hom(A, B) is the composition A → 0 → B. (We also remark that the notionof 0-morphism thus makes sense in any category with a 0-object.)
The category of A-modules ModA is clearly an additive category, but it haseven more structure, which we now formalize as an example of an abelian category.
1.6.3. Definition. Let C be a category with a 0-object (and thus 0-morphisms). A
kernel of a morphism f : B → C is a map i : A → B such that f ◦ i = 0, and that isuniversal with respect to this property. Diagramatically:
Z ❄❖❖❖❄❄ ❖❖❄❄ ❖❖❖0∃!
❄❄ ❖❖❖❖❖❖❄8"i% B f 4%; CA
0(Note that the kernel is not just an object; it is a morphism of an object to B.) Hence
it is unique up to unique isomorphism by universal property nonsense. The kernel
is written ker f → B. A cokernel (denoted coker f) is defined dually by reversing
the arrows — do this yourself. The kernel of f : B → C is the limit (§1.4) of the
diagram(1.6.3.1)

0

B

f

"
%C

and similarly the cokernel is a colimit (see (2.6.0.1)).
If i : A → B is a monomorphism, then we say that A is a subobject of B, where
the map i is implicit. There is also the notion of quotient object, defined dually tosubobject.
An abelian category is an additive category satisfying three additional properties.
(1) Every map has a kernel and cokernel.
(2) Every monomorphism is the kernel of its cokernel.
(3) Every epimorphism is the cokernel of its kernel.
It is a nonobvious (and imprecisely stated) fact that every property you want
to be true about kernels, cokernels, etc. follows from these three. (Warning: inpart of the literature, additional hypotheses are imposed as part of the definition.)

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The image of a morphism f : A → B is defined as im(f) = ker(coker f) whenever it exists (e.g., in every abelian category). The morphism f : A → B factors
uniquely through im f → B whenever im f exists, and A → im f is an epimorphism and a cokernel of ker f → A in every abelian category. The reader may
want to verify this as a (hard!) exercise.
The cokernel of a monomorphism is called the quotient. The quotient of amonomorphism A → B is often denoted B/A (with the map from B implicit).
We will leave the foundations of abelian categories untouched. The key thing
to remember is that if you understand kernels, cokernels, images and so on in
the category of modules over a given ring, you can manipulate objects in any
abelian category. This is made precise by Freyd-Mitchell Embedding Theorem(Remark 1.6.4).
However, the abelian categories we will come across will obviously be related
to modules, and our intuition will clearly carry over, so we needn’t invoke a theorem whose proof we haven’t read. For example, we will show that sheaves ofabelian groups on a topological space X form an abelian category (§2.6), and theinterpretation in terms of “compatible germs” will connect notions of kernels, cokernels etc. of sheaves of abelian groups to the corresponding notions of abeliangroups.
1.6.4. Small remark on chasing diagrams. It is useful to prove facts (and solve exercises) about abelian categories by chasing elements. This can be justified by
the Freyd-Mitchell Embedding Theorem: If C is an abelian category such that
Hom(X, Y) is a set for all X, Y ∈ C , then there is a ring A and an exact, fully faithful
functor from C into ModA , which embeds C as a full subcategory. (Unfortunately,
the ring A need not be commutative.) A proof is sketched in [Weib, §1.6], andreferences to a complete proof are given there. A proof is also given in [KS1, §9.7].
The upshot is that to prove something about a diagram in some abelian category,
we may assume that it is a diagram of modules over some ring, and we may then“diagram-chase” elements. Moreover, any fact about kernels, cokernels, and so onthat holds in ModA holds in any abelian category.
If invoking a theorem whose proof you haven’t read bothers you, a short alternative is Mac Lane’s “elementary rules for chasing diagrams”, [Mac, Thm. 3,
p. 200]; [Mac, Lem. 4, p. 201] gives a proof of the Five Lemma (Exercise 1.7.6) as an
example.
But in any case, do what you need to do to put your mind at ease, so you canmove forward. Do as little as your conscience will allow.
1.6.5. Complexes, exactness, and homology.
(In this entire discussion, we assume we are working in an abelian category.)
We say a sequence(1.6.5.1)

···

%A

f

%B

g

%C

% ···

is a complex at B if g ◦ f = 0, and is exact at B if ker g = im f. (More specifically,g has a kernel that is an image of f. Exactness at B implies being a complex at B
— do you see why?) A sequence is a complex (resp. exact) if it is a complex (resp.
exact) at each (internal) term. A short exact sequence is an exact sequence withfive terms, the first and last of which are zeros — in other words, an exact sequence

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The Rising Sea: Foundations of Algebraic Geometryof the form
%A

0
For example, 0

%B

%C

% 0.

% 0 is exact if and only if A = 0;

%A

f

%A

0

%B

is exact if and only if f is a monomorphism (with a similar statement for A
%A

0

f

%B

f

%B

% 0 );%0is exact if and only if f is an isomorphism; and
%A

0

f

g

%B

%C

g
f %
%C
is exact if and only if f is a kernel of g (with a similar statement for A
B
To show some of these facts it may be helpful to prove that (1.6.5.1) is exact at B if
and only if the cokernel of f is a cokernel of the kernel of g.
If you would like practice in playing with these notions before thinking abouthomology, you can prove the Snake Lemma (stated in Example 1.7.5, with a stronger
version in Exercise 1.7.B), or the Five Lemma (stated in Example 1.7.6, with astronger version in Exercise 1.7.C). (I would do this in the category of A-modules,
but see [KS1, Lem. 12.1.1, Lem. 8.3.13] for proofs in general.)If (1.6.5.1) is a complex at B, then its homology at B (often denoted by H) is
ker g / im f. (More precisely, there is some monomorphism im f "→ ker g, and that
H is the cokernel of this monomorphism.) Therefore, (1.6.5.1) is exact at B if andonly if its homology at B is 0. We say that elements of ker g (assuming the objects of the category are sets with some additional structure) are the cycles, andelements of im f are the boundaries (so homology is “cycles mod boundaries”). If
the complex is indexed in decreasing order, the indices are often written as subscripts, and Hi is the homology at Ai+1 → Ai → Ai−1 . If the complex is indexed
in increasing order, the indices are often written as superscripts, and the homology
Hi at Ai−1 → Ai → Ai+1 is often called cohomology.
An exact sequence(1.6.5.2)

A• :

···

% Ai−1 f

i−1

% Ai

fi

i+1

% Ai+1 f

can be “factored” into short exact sequences% ker fi0

% ker fi+1

% Ai

% 0 ).

% ···%0

which is helpful in proving facts about long exact sequences by reducing them tofacts about short exact sequences.
More generally, if (1.6.5.2) is assumed only to be a complex, then it can be“factored” into short exact sequences.
(1.6.5.3)

0% ker fi

% Ai

% im fi%00% im fi−1% ker fi

% Hi (A• )%0

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1.6.A. E XERCISE . Describe exact sequences(1.6.5.4)

0% im fi% Ai+1% coker fi%00% Hi (A• )% coker fi−1

% im fi%0(These are somehow dual to (1.6.5.3). In fact in some mirror universe this might
have been given as the standard definition of homology.) Assume the category isthat of modules over a fixed ring for convenience, but be aware that the result istrue for any abelian category.
1.6.B. E XERCISE AND IMPORTANT DEFINITION . Suppose
0

d0

% A1

d1

n−1

% ··· d

% An

dn

%0

•
is a complex of finite-dimensional k-vector(often called
short). De, spaces, Ai for
i
•
i
•
i
i
fine h (A ) := dim H (A ).,Show that (−1) dim A = (−1) hi (A• ). In particular, if A• is exact, then (−1)i dim Ai = 0. (If you haven’t dealt much withcohomology, this will give you some practice.)

1.6.C. I MPORTANT E XERCISE . Suppose C is an abelian category. Define the category ComC of complexes as follows. The objects are infinite complexes
A• :

···

i−1

% Ai−1 f

% Ai

fi

i+1

% Ai+1 f% ···in C , and the morphisms A• → B• are commuting diagrams(1.6.5.5)

···
···

% Ai

fi

% Ai+1 f""i−1% Bi−1 g
% Bi

gi

"
i+1
% Bi+1 g% ···% Ai−1 f

i−1

i+1

% ···

Show that ComC is an abelian category. Feel free to deal with the special caseof modules over a fixed ring. (Remark for experts: Essentially the same argumentshows that the C I is an abelian category for any small category I and any abeliancategory C . This immediately implies that the category of presheaves on a topological space X with values in an abelian category C is automatically an abeliancategory, cf. §2.3.5.)
1.6.D. I MPORTANT E XERCISE . Show that (1.6.5.5) induces a map of homology
Hi (A• ) → Hi (B• ). Show furthermore that Hi is a covariant functor ComC → C .
(Again, feel free to deal with the special case ModA .)We will later define when two maps of complexes are homotopic (§23.1), andshow that homotopic maps induce the same map on cohomology (Exercise 23.1.A),
but we won’t need that any time soon.

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The Rising Sea: Foundations of Algebraic Geometry

1.6.6. Theorem (Long exact sequence). — A short exact sequence of complexes
0•

:

···

"
A•

:

···

"
B•

:

···

"
C•

:

···

"
0•

:

···

%0

%0

%0

"
"
i−1
% Ai−1 f
% Ai

fi

"
"
i−1
% Bi−1 g
% Bi

gi

"
"
i−1
% Ci−1 h
% Ci

hi

"
%0

% ···

"
i+1
% Ai+1 f
% ···
"
i+1
% Bi+1 g
% ···
"
i+1
% Ci+1 h
% ···

"
%0

"
%0

% ···

induces a long exact sequence in cohomology...

% Hi−1 (C• )%Hi (A• )% Hi (B• )% Hi (C• )%Hi+1 (A• )% ···(This requires a definition of the connecting homomorphism Hi−1 (C• ) →H (A• ), which is “natural” in an appropriate sense.) In the category of modules
over a ring, Theorem 1.6.6 will come out of our discussion of spectral sequences,
see Exercise 1.7.F, but this is a somewhat perverse way of proving it. For a proofin general, see [KS1, Theorem 12.3.3]. You may want to prove it yourself, by firstproving a weaker version of the Snake Lemma (Example 1.7.5), where in the hypotheses (1.7.5.1), the 0’s in the bottom left and top right are removed, and in the
conclusion (1.7.5.2), the first and last 0’s are removed.
i1.6.7. Exactness of functors. If F : A → B is a covariant additive functor from one
abelian category to another, we say that F is right-exact if the exactness ofA′%A% A ′′% 0,

in A implies that
F(A ′ )% F(A)% F(A ′′ )%0is also exact. Dually, we say that F is left-exact if the exactness of
0

% A′

%A

% A ′′

implies0% F(A ′ )% F(A)% F(A ′′ )is exact.

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A contravariant functor is left-exact if the exactness of%A% A ′′%0A′

0% F(A ′′ )% F(A)% F(A ′ )53

impliesis exact.

The reader should be able to deduce what it means for a contravariant functor to
be right-exact.
A covariant or contravariant functor is exact if it is both left-exact and rightexact.
1.6.E. E XERCISE . Suppose F is an exact functor. Show that applying F to an exact sequence preserves exactness. For example, if F is covariant, and A ′ → A → A ′′ is exact, then FA ′ → FA → FA ′′ is exact. (This will be generalized in Exercise 1.6.H(c).)

1.6.F. E XERCISE . Suppose A is a ring, S ⊂ A is a multiplicative subset, and M isan A-module.
(a) Show that localization of A-modules ModA → ModS−1 A is an exact covariant
functor.
(b) Show that (·) ⊗A M is a right-exact covariant functor ModA → ModA . (This is arepeat of Exercise 1.3.H.)(c) Show that Hom(M, ·) is a left-exact covariant functor ModA → ModA . If C is
any abelian category, and C ∈ C , show that Hom(C, ·) is a left-exact covariant
functor C → Ab.
(d) Show that Hom(·, M) is a left-exact contravariant functor ModA → ModA . If C
is any abelian category, and C ∈ C , show that Hom(·, C) is a left-exact contravariant functor C → Ab.
1.6.G. E XERCISE . Suppose M is a finitely presented A-module: M has a finite
number of generators, and with these generators it has a finite number of relations;or usefully equivalently, fits in an exact sequence(1.6.7.1)

A⊕q → A⊕p → M → 0

Use (1.6.7.1) and the left-exactness of Hom to describe an isomorphism∼ HomS−1 A (S−1 M, S−1 N).
S−1 HomA (M, N) =(You might be able to interpret this in light of a variant of Exercise 1.6.H below, forleft-exact contravariant functors rather than right-exact covariant functors.)
1.6.8. Example: Hom doesn’t always commute with localization.
Exercise 1.6.G, take A = N = Z, M = Q, and S = Z \ {0}.

In the language of

1.6.9. ⋆ Two useful facts in homological algebra.
We now come to two (sets of) facts I wish I had learned as a child, as theywould have saved me lots of grief. They encapsulate what is best and worst of
abstract nonsense. The statements are so general as to be nonintuitive. The proofs
are very short. They generalize some specific behavior it is easy to prove on an
ad hoc basis. Once they are second nature to you, many subtle facts will become
obvious to you as special cases. And you will see that they will get used (implicitly
or explicitly) repeatedly.

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The Rising Sea: Foundations of Algebraic Geometry

1.6.10. ⋆ Interaction of homology and (right/left-)exact functors.
You might wait to prove this until you learn about cohomology in Chapter 18,
when it will first be used in a serious way.
1.6.H. I MPORTANT E XERCISE ( THE FHHF T HEOREM ). This result can take you
far, and perhaps for that reason it has sometimes been called the Fernbahnhof(FernbaHnHoF) Theorem, notably in [Vak, Exer. 1.6.H]. Suppose F : A → B is acovariant functor of abelian categories, and C• is a complex in A .

% H• F ) If F is right-exact, describe a natu(a) (F right-exact yields FH•••
ral morphism FH → H F. (More precisely, for each i, the left side is F
applied to the cohomology at piece i of C• , while the right side is thecohomology at piece i of FC• .)(b) (F left-exact yields FH• (H• F ) If F is left-exact, describe a natural mor••phism H F → FH .
∼ %(c) (F exact yields FH• (H• F ) If F is exact, show that the morphisms of(a) and (b) are inverses and thus isomorphisms.
i

d % i+1% coker di% 0 to give an isomorphism
Hint for (a): use Ci
Ci ∼i
F coker d = coker Fd . Then use the first line of (1.6.5.4) to give a epimorphism% % im Fdi . Then use the second line of (1.6.5.4) to give the desired map
F im diFHi C•% Hi FC• . While you are at it, you may as well describe a map for the

fourth member of the quartet {coker, im, H, ker}: F ker di% ker Fdi .

1.6.11. If this makes your head spin, you may prefer to think of it in the following
specific case, where both A and B are the category of A-modules, and F is (·) ⊗ N
for some fixed N-module. Your argument in this case will translate without change
to yield a solution to Exercise 1.6.H(a) and (c) in general. If ⊗N is exact, then N iscalled a flat A-module. (The notion of flatness will turn out to be very important,and is discussed in detail in Chapter 24.)For example, localization is exact (Exercise 1.6.F(a)), so S−1 A is a flat A-algebra
for all multiplicative sets S. Thus taking cohomology of a complex of A-modules
commutes with localization — something you could verify directly.
1.6.12. Interaction of adjoints, (co)limits, and (left- and right-) exactness.
A surprising number of arguments boil down to the statement:
Limits commute with limits and right adjoints. In particular, in an abelian category,because kernels are limits, both right adjoints and limits are left-exact.
as well as its dual:
Colimits commute with colimits and left adjoints. In particular, because cokernels arecolimits, both left adjoints and colimits are right-exact.
These statements were promised in §1.5.4, and will be proved below. The latterhas a useful extension:
In ModA , colimits over filtered index categories are exact. “Filtered” was definedin §1.4.7.
1.6.13. ⋆⋆ Caution. It is not true that in abelian categories in general, colimits
over filtered index categories are exact. (Grothendieck realized the desirability of

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such colimits being exact, and formalized this as his “AB5” axiom, see for example [Stacks, tag 079A].) Here is a counterexample. Because the axioms of abeliancategories are self-dual, it suffices to give an example in which a filtered limit fails
to be exact, and we do this. Fix a prime p. In the category Ab of abelian groups,
for each positive integer n, we have an exact sequence Z → Z/(pn ) → 0. Taking
the limit over all n in the obvious way, we obtain Z → Zp → 0, which is certainly
not exact.)
See Unimportant Remark 1.6.16 for another dashed hope.
1.6.14. If you want to use these statements (for example, later in this book), youwill have to prove them. Let’s now make them precise.
1.6.I. E XERCISE ( KERNELS COMMUTE WITH LIMITS ). Suppose C is an abeliancategory, and a : I → C and b : I → C are two diagrams in C indexed by I .
For convenience, let Ai = a(i) and Bi = b(i) be the objects in those two diagrams.
Let hi : Ai → Bi be maps commuting with the maps in the diagram. (Translation:h is a natural transformation of functors a → b, see §1.2.21.) Then the ker hiform another diagram in C indexed by I . Describe a canonical isomorphism∼ ker(lim Ai → lim Bi ), assuming the limits exist.
lim ker hi =←−←−←−Implicit in the previous exercise is the idea that limits should somehow be
understood as functors.
1.6.J. E XERCISE . Make sense of the statement that “limits commute with limits” ina general category, and prove it. (Hint: recall that kernels are limits. The previous
exercise should be a corollary of this one.)
1.6.15. Proposition (right adjoints commute with limits). — Suppose (F : C →D, G : D → C ) is a pair of adjoint functors. If A = lim Ai is a limit in D of a diagram←−
indexed by I , then GA = lim GAi (with the corresponding maps GA → GAi ) is a limit←−
in C .

Proof. We must show that GA → GAi satisfies the universal property of limits.
Suppose we have maps W → GAi commuting with the maps of I . We wish toshow that there exists a unique W → GA extending the W → GAi . By adjointnessof F and G, we can restate this as: Suppose we have maps FW → Ai commuting
with the maps of I . We wish to show that there exists a unique FW → A extending
the FW → Ai . But this is precisely the universal property of the limit.
!
Of course, the dual statements to Exercise 1.6.J and Proposition 1.6.15 hold by
the dual arguments.
If F and G are additive functors between abelian categories, and (F, G) is anadjoint pair, then (as kernels are limits and cokernels are colimits) G is left-exact
and F is right-exact.

1.6.K. E XERCISE . Show that in ModA , colimits over filtered index categories areexact. (Your argument will apply without change to any abelian category whose
objects can be interpreted as “sets with additional structure”.) Right-exactness
follows from the above discussion, so the issue is left-exactness. (Possible hint:
After you show that localization is exact, Exercise 1.6.F(a), or stalkification is exact,

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Exercise 2.6.D, in a hands-on way, you will be easily able to prove this. Conversely,
if you do this exercise, those two will be easy.)
1.6.L. E XERCISE . Show that filtered colimits commute with homology in ModA .
Hint: use the FHHF Theorem (Exercise 1.6.H), and the previous Exercise.
In light of Exercise 1.6.L, you may want to think about how limits (and colimits) commute with homology in general, and which way maps go. The statementof the FHHF Theorem should suggest the answer. (Are limits analogous to leftexact functors, or right-exact functors?) We won’t directly use this insight.
Just as colimits are exact (not just right-exact) in especially good circumstances,
limits are exact (not just left-exact) too. The following will be used twice in Chapter 29.
1.6.M. E XERCISE . Suppose
..
.

..
.

..
.

0

"
% An+1"% Bn+1"% Cn+1

%0

0

"
% An

"
% Bn

"
% Cn

%0

"
..
.

"
..
.

"
..
.

"
% A0

"
% B0

"
% C0

"
0

"
0

"
0

0

%0

is an inverse system of exact sequences of modules over a ring, such that the maps
An+1 → An are surjective. (We say: “transition maps of the left term are surjective”.) Show that the limit(1.6.15.1)

0% lim An←−% lim Bn←−

% lim Cn←−%0is also exact. (You will need to define the maps in (1.6.15.1).)

1.6.16. Unimportant Remark. Based on these ideas, you may suspect that rightexact functors always commute with colimits. The fact that tensor product commutes with infinite direct sums (Exercise 1.3.M) may reinforce this idea. Unfortunately, it is not true — “double dual” ·∨∨ : Veck → Veck is covariant and rightexact (why is it right-exact?), but does not commute with infinite direct sums, as∨∨∨∨⊕∞) is not isomorphic to (⊕∞.
i=1 (ki=1 k)

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1.6.17. ⋆ Dreaming of derived functors. When you see a left-exact functor, you
should always dream that you are seeing the end of a long exact sequence. If
0 → M ′ → M → M ′′ → 0

is an exact sequence in abelian category A , and F : A → B is a left-exact functor,
then
0 → FM ′ → FM → FM ′′is exact, and you should always dream that it should continue in some naturalway. For example, the next term should depend only on M ′ , call it R1 FM ′ , and if it
is zero, then FM → FM ′′ is an epimorphism. This remark holds true for left-exact
and contravariant functors too. In good cases, such a continuation exists, and isincredibly useful. We will discuss this in Chapter 23.

1.7 ⋆ Spectral sequences
Je suis quelque peu affolé par ce déluge de cohomologie, mais j’ai courageusement tenu
le coup. Ta suite spectrale me paraı̂t raisonnable (je croyais, sur un cas particulier, l’avoir
mise en défaut, mais je m’étais trompé, et cela marche au contraire admirablement bien).
I am a bit panic-stricken by this flood of cohomology, but have borne up courageously.
Your spectral sequence seems reasonable to me (I thought I had shown that it was wrong
in a special case, but I was mistaken, on the contrary it works remarkably well).
— J.-P. Serre, letter to A. Grothendieck, March 14, 1956 [GrS, p. 38]
Spectral sequences are a powerful book-keeping tool for proving things involving complicated commutative diagrams. They were introduced by Leray inthe 1940’s at the same time as he introduced sheaves. They have a reputation forbeing abstruse and difficult. It has been suggested that the name ‘spectral’ wasgiven because, like spectres, spectral sequences are terrifying, evil, and dangerous. I have heard no one disagree with this interpretation, which is perhaps not
surprising since I just made it up.
Nonetheless, the goal of this section is to tell you enough that you can use
spectral sequences without hesitation or fear, and why you shouldn’t be frightenedwhen they come up in a seminar. What is perhaps different in this presentation isthat we will use spectral sequences to prove things that you may have already seen,
and that you can prove easily in other ways. This will allow you to get some handson experience for how to use them. We will also see them only in the special case of
double complexes (the version by far the most often used in algebraic geometry),and not in the general form usually presented (filtered complexes, exact couples,etc.). See [Weib, Ch. 5] for more detailed information if you wish.
You should not read this section when you are reading the rest of Chapter 1.
Instead, you should read it just before you need it for the first time. When you
finally do read this section, you must do the exercises up to Exercise 1.7.F.
For concreteness, we work in the category ModA of module over a ring A.
However, everything we say will apply in any abelian category. (And if it helps
you feel secure, work instead in the category Veck of vector spaces over a field k.)
1.7.1. Double complexes.

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The Rising Sea: Foundations of Algebraic Geometry

A double complex is a collection of A-modules Ep,q (p, q ∈ Z), and “rightp,qward” morphisms dp,q
→ Ep+1,q and “upward” morphisms dp,q
: Ep,q →→ : E↑
Ep,q+1 . In the superscript, the first entry denotes the column number (the “xcoordinate”), and the second entry denotes the row number (the “y-coordinate”).
(Warning: this is opposite to the convention for matrices.) The subscript is meant
to suggest the direction of the arrows. We will always write these as d→ and d↑
and ignore the superscripts. We require that d→ and d↑ satisfy (a) d2→ = 0, (b)d2↑ = 0, and one more condition: (c) either d→ d↑ = d↑ d→ (all the squares commute) or d→ d↑ + d↑ d→ = 0 (they all anticommute). Both come up in nature, andyou can switch from one to the other by replacing dp,qwith (−1)p dp,q↑
↑ . So I willassume that all the squares anticommute, but that you know how to turn the commuting case into this one. (You will see that there is no difference in the recipe,
basically because the image and kernel of a homomorphism f equal the image andkernel respectively of −f.)
Ep,q+1
7
dp,q
↑

Ep,q

dp,q+1→% Ep+1,q+1
7

anticommutes
dp,q→

dp+1,q↑% Ep+1,q

There are variations on this definition, where for example the vertical arrows
go downwards, or some subset of the Ep,q is required to be zero, but I will leavethese straightforward variations to you.
From the double complex we construct a corresponding (single) complex E•with Ek = ⊕i Ei,k−i , with d = d→ + d↑ . In other words, when there is a single
superscript k, we mean a sum of the kth antidiagonal of the double complex. The
single complex is sometimes called the total complex. Note that d2 = (d→ +d↑ )2 =d2→ + (d→ d↑ + d↑ d→ ) + d2↑ = 0, so E• is indeed a complex.
The cohomology of the single complex is sometimes called the hypercohomology of the double complex. We will instead use the phrase “cohomology ofthe double complex”.
Our initial goal will be to find the cohomology of the double complex. You
will see later that we secretly also have other goals.
A spectral sequence is a recipe for computing some information about thecohomology of the double complex. I won’t yet give the full recipe. Surprisingly,
this fragmentary bit of information is sufficent to prove lots of things.
1.7.2. Approximate Definition. A spectral sequence with rightward orientation
p,q
p,q
p,q
is a sequence of tables or pages → Ep,q=
0 , → E1 , → E2 , . . . (p, q ∈ Z), where → E0p,q
E , along with a differential
p,q: → Ep,q→ → Ep−r+1,q+r
→ dr
r
r

with → dp,q
◦ → dp+r−1,q−r= 0, and with an isomorphism of the cohomology ofrrp,qp+r−1,q−r(i.e., ker → dp,q
) with → Ep,qr / im → dr
→ dr at → Er
r+1 .
The orientation indicates that our 0th differential is the rightward one: d0 =d→ . The left subscript “→” is usually omitted.

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November 18, 2017 draft59

The order of the morphisms is best understood visually:(1.7.2.1)

• =✺
✺✺
✺✺
✺✺
✺✺ •
✺✺ <✴✴
✺✺ ✴✴
d3 ✺ ✴
✺✺ ✴
✺d✺2✴ •7✺✺✴✴
✺✴✺✴ d1
✺✴
•

d0

%•

(the morphisms each apply to different pages). Notice that the map always is“degree 1” in terms of the grading of the single complex E• . (You should figureout what this informal statement really means.)•,••,•
The actual definition describes what E•,•.
r and dr really are, in terms of E
We will describe d0 , d1 , and d2 below, and you should for now take on faith that
this sequence continues in some natural way.
Note that Ep,qis always a subquotient of the corresponding term on the ithrpage Ep,q
for
all
i
<
r. In particular, if Ep,q = 0, then Ep,q
= 0 for all r.
riSuppose now that E•,• is a first quadrant double complex, i.e., Ep,q = 0 for
p < 0 or q < 0 (so Ep,q
= 0 for all r unless p, q ∈ Z≥0 ). Then for any fixed p, q,r•,•
once r is sufficiently large, Ep,qr+1 is computed from (Er , dr ) using the complex
0 >✸
✸✸
✸✸
✸✸ p,q
✸✸dr
✸✸
✸✸
Ep,q
r >✸
✸✸
✸✸
✸✸ dp+r−1,q−r
✸✸ r
✸✸
✸✸
0
and thus we have canonical isomorphisms∼ Ep,q =∼ Ep,q =∼ ··· .
Ep,q
=
r
r+1
r+2
We denote this module Ep,q
∞ . The same idea works in other circumstances, forexample if the double complex is only nonzero in a finite number of rows — Ep,q =0 unless q0 < q < q1 . This will come up for example in the mapping cone andlong exact sequence discussion (Exercises 1.7.F and 1.7.E below).
We now describe the first few pages of the spectral sequence explicitly. As•,•
stated above, the differential d0 on E•,•is defined to be d→ . The rows are
0 = E

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The Rising Sea: Foundations of Algebraic Geometrycomplexes:

The 0th page E0 :

•

%•

%•

•

%•

%•

•

%•

%•

•7

•7

•7

•7

•7

•7

•

•

•

and so E1 is just the table of cohomologies of the rows. You should check thatp,qthere are now vertical maps dp,q
→ Ep,q+1of the row cohomology groups,
1 : E11
induced by d↑ , and that these make the columns into complexes. (This is essentially the fact that a map of complexes induces a map on homology.) We have“used up the horizontal morphisms”, but “the vertical differentials live on”.

The 1st page E1 :We take cohomology of d1 on E1 , giving us a new table, Ep,q
2 . It turns out that

there are natural morphisms from each entry to the entry two above and one to theleft, and that the composition of these two is 0. (It is a very worthwhile exercise
to work out how this natural morphism d2 should be defined. Your argument
may be reminiscent of the connecting homomorphism in the Snake Lemma 1.7.5
or in the long exact sequence in cohomology arising from a short exact sequenceof complexes, Exercise 1.6.C. This is no coincidence.)
•<✳
•<✳
•
✳✳
✳✳
✳✳
✳✳
✳✳
✳✳
• ✳✳✳ • ✳✳✳ •
✳✳
✳✳
✳✳
✳✳
✳
✳
•
•
•

The 2nd page E2 :This is the beginning of a pattern.
Then it is a theorem that there is a filtration of Hk (E• ) by Ep,q
∞ where p + q = k.
(We can’t yet state it as an official Theorem because we haven’t precisely defined
the pages and differentials in the spectral sequence.) More precisely, there is afiltration(1.7.2.2)

1,k−1

E0,k
∞

!E" ∞

2,k−2

%

?

! " E∞

%

k,0

···

! " E∞ %Hk (E• )where the quotients are displayed above each inclusion. (Here is a tip for remember which way the quotients are supposed to go. The differentials on later and later
pages point deeper and deeper into the filtration. Thus the entries in the directionof the later arrowheads are the subobjects, and the entries in the direction of the

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later “arrowtails” are quotients. This tip has the advantage of being independentof the details of the spectral sequence, e.g., the “quadrant” or the orientation.)
We say that the spectral sequence → E•,•
converges to H• (E• ). We often say
•
•,•
• •
that → E2 (or any other page) abuts to H (E ).
Although the filtration gives only partial information about H• (E• ), sometimes one can find H• (E• ) precisely. One example is if all Ei,k−i
are zero, or if∞
all but one of them are zero (e.g., if E•,•has
preciselyone
nonzerorow or colrumn, in which case one says that the spectral sequence collapses at the rth step,
although we will not use this term). Another example is in the category of vector
spaces over a field, in which case we can find the dimension of Hk (E• ). Also, inlucky circumstances, E2 (or some other small page) already equals E∞ .

1.7.A. E XERCISE : INFORMATION FROM THE SECOND PAGE . Show that H0 (E• ) =0,0
E0,0∞ = E2 and0% E0,1
2% H1 (E• )% E1,0d1,02

2% E0,22% H2 (E• )is exact.
1.7.3. The other orientation.
You may have observed that we could as well have done everything in the
opposite direction, i.e., reversing the roles of horizontal and vertical morphisms.
Then the sequences of arrows giving the spectral sequence would look like this(compare to (1.7.2.1)).
(1.7.3.1)•7•❖❏❏❖❖ % •
❏❏❖❖❖
❏❏❖❖❖
❏❏ ❖❖
❏❏ ❖❖❖
❏❏ 4
❏❏ •
❏❏
❏❏
❏❏
❏❏
❏#

•

This spectral sequence is denoted ↑ E•,•(“with the upward orientation”).
•
• •

Then
we would again get pieces of a filtration of H (E ) (where we have to be a bit
careful with the order with which ↑ Ep,q
∞ corresponds to the subquotients — it in
the opposite order to that of (1.7.2.2) for → Ep,q∞ ). Warning: in general there is nop,qisomorphism between → Ep,q
and
E
.
↑
∞
∞
In fact, this observation that we can start with either the horizontal or verticalmaps was our secret goal all along. Both algorithms compute information about
the same thing (H• (E• )), and usually we don’t care about the final answer — we
often care about the answer we get in one way, and we get at it by doing thespectral sequence in the other way.
1.7.4. Examples.
We are now ready to see how this is useful. The moral of these examples isthe following. In the past, you may have proved various facts involving various
sorts of diagrams, by chasing elements around. Now, you will just plug them into

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a spectral sequence, and let the spectral sequence machinery do your chasing foryou.
1.7.5. Example: Proving the Snake Lemma. Consider the diagram(1.7.5.1)

0

0

%D
7

%E
7

%F
7

α

β

γ

%A

%B

%C

%0

%0

where the rows are exact in the middle (at A, B, C, D, E, F) and the squares commute. (Normally the Snake Lemma is described with the vertical arrows pointing
downwards, but I want to fit this into my spectral sequence conventions.) We wish
to show that there is an exact sequence(1.7.5.2)

0 → ker α → ker β → ker γ → coker α → coker β → coker γ → 0.

We plug this into our spectral sequence machinery. We first compute the cohomology using the rightward orientation, i.e., using the order (1.7.2.1). Then because the rows are exact, Ep,q
= 0, so the spectral sequence has already converged:
1
Ep,q∞ = 0.
We next compute this “0” in another way, by computing the spectral sequence
using the upward orientation. Then ↑ E•,•1 (with its differentials) is:0% coker α% coker β% coker γ%00% ker α% ker β% ker γ% 0.

Then ↑ E•,•
2 is of the form:

0 ◆◆◆ 0 ❖❖❖◆◆◆❖◆◆◆ ❖❖❖❖❖
◆◆◆
❖❖❖
◆◆4
❖❖
0 ❖❖❖ ?? ◆◆ 4 ? ❖❖❖
?
0
◆
❖❖❖
❖❖❖
◆◆◆
❖
◆
❖❖❖
❖❖❖ ◆◆◆◆◆ ❖❖❖❖❖
◆◆4
❖❖4
❖❖4
0
?
? ❖❖❖ ?? ◆◆
0
◆◆◆
❖❖❖
◆
❖❖❖
◆
❖❖❖ ◆◆◆◆◆
❖❖4
◆◆4
0
0

We see that after ↑ E2 , all the terms will stabilize except for the double question
marks — all maps to and from the single question marks are to and from 0-entries.
And after ↑ E3 , even these two double-question-mark terms will stabilize. But inthe end our complex must be the 0 complex. This means that in ↑ E2 , all the entries
must be zero, except for the two double question marks, and these two must be
isomorphic. This means that 0 → ker α → ker β → ker γ and coker α → coker β →coker γ → 0 are both exact (that comes from the vanishing of the single question
marks), and∼ ker(coker α → coker β)coker(ker β → ker γ) =is an isomorphism (that comes from the equality of the double question marks).
Taken together, we have proved the exactness of (1.7.5.2), and hence the Snake

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Lemma! (Notice: in the end we didn’t really care about the double complex. We
just used it as a prop to prove the Snake Lemma.)
Spectral sequences make it easy to see how to generalize results further. For
example, if A → B is no longer assumed to be injective, how would the conclusionchange?

1.7.B. U NIMPORTANT E XERCISE ( GRAFTING EXACT SEQUENCES , A WEAKER VER SION OF THE S NAKE L EMMA ). Extend the Snake Lemma as follows. Suppose wehave a commuting diagram
% X′
7

% Y′
7

% Z′
7

a

b

c

%X

%Y

%Z

07
···

%W

% A′
7

% ···

% 0.

where the top and bottom rows are exact. Show that the top and bottom rows canbe “grafted together” to an exact sequence···%W% ker a% ker b% ker c% coker a% coker b% coker c% A′% ··· .

1.7.6. Example: the Five Lemma. Suppose(1.7.6.1)

%G
7

%H
7

%I
7

%J
7

α

β

γ

δ

ϵ

A

%B

%C

%D

%E

F7

where the rows are exact and the squares commute.
Suppose α, β, δ, ϵ are isomorphisms. We will show that γ is an isomorphism.
We first compute the cohomology of the total complex using the rightward
orientation (1.7.2.1). We choose this because we see that we will get lots of zeros.
Then → E•,•
1 looks like this:
?7

07

07

07

?7

?

0

0

0

?

Then → E2 looks similar, and the sequence will converge by E2 , as we will never get
any arrows between two nonzero entries in a table thereafter. We can’t conclude
that the cohomology of the total complex vanishes, but we can note that it vanishes in all but four degrees — and most important, it vanishes in the two degrees
corresponding to the entries C and H (the source and target of γ).
We next compute this using the upward orientation (1.7.3.1). Then ↑ E1 looks
like this:
%0
%?
%0
%0
0

0

%0

%?

%0

%0

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and the spectral sequence converges at this step. We wish to show that those twoquestion marks are zero. But they are precisely the cohomology groups of the total
complex that we just showed were zero — so we are done!
The best way to become comfortable with this sort of argument is to try it out
yourself several times, and realize that it really is easy. So you should do the following exercises! Many can readily be done directly, but you should deliberately
try to use this spectral sequence machinery in order to get practice and develop
confidence.
1.7.C. E XERCISE : A SUBTLER F IVE L EMMA . By looking at the spectral sequence
proof of the Five Lemma above, prove a subtler version of the Five Lemma, where
one of the isomorphisms can instead just be required to be an injection, and another can instead just be required to be a surjection. (I am deliberately not telling
you which ones, so you can see how the spectral sequence is telling you how to
improve the result.)
1.7.D. E XERCISE : ANOTHER SUBTLE VERSION OF THE FIVE LEMMA . If β and δ (in(1.7.6.1)) are injective, and α is surjective, show that γ is injective. Give the dual
statement (whose proof is of course essentially the same).
The next two exercises no longer involve first quadrant double complexes.
You will have to think a little to realize why there is no reason for confusion or
alarm.
1.7.E. E XERCISE ( THE MAPPING CONE ). Suppose µ : A• → B• is a morphism ofcomplexes. Suppose C• is the single complex associated to the double complex
A• → B• . (C• is called the mapping cone of µ.) Show that there is a long exact
sequence of complexes:· · · → Hi−1 (C• ) → Hi (A• ) → Hi (B• ) → Hi (C• ) → Hi+1 (A• ) → · · · .

(There is a slight notational ambiguity here; depending on how you index yourdouble complex, your long exact sequence might look slightly different.) In particular, we will use the fact that µ induces an isomorphism on cohomology if and only
if the mapping cone is exact. (We won’t use it until the proof of Theorem 18.2.4.)
1.7.F. E XERCISE . Use spectral sequences to show that a short exact sequence ofcomplexes gives a long exact sequence in cohomology (Exercise 1.6.C). (This is a
generalization of Exercise 1.7.E.)
The Grothendieck composition-of-functors spectral sequence (Theorem 23.3.5)
will be an important example of a spectral sequence that specializes in a numberof useful ways.
You are now ready to go out into the world and use spectral sequences to yourheart’s content!
1.7.7. ⋆⋆ Complete definition of the spectral sequence, and proof.
You should most definitely not read this section any time soon after reading
the introduction to spectral sequences above. Instead, flip quickly through it toconvince yourself that nothing fancy is involved.

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1.7.8. Remark: Spectral sequences are actually spectral functors. It is useful to noticethat the proof implies that spectral sequences are functorial in the 0th page: thespectral sequence formalism has good functorial properties in the double complex.
Unfortunately the terminology “spectral functor” that Grothendieck used in [Gr1,§2.4] did not catch on.
1.7.9. Goals. We consider the rightward orientation. The upward orientation is ofcourse a trivial variation of this. We wish to describe the pages and differentials of
the spectral sequence explicitly, and prove that they behave the way we said theydid. More precisely, we wish to:(a) describe Ep,q(and verify that Ep,qis indeed Ep,q ),
r
0
k •
p,k−p
(b) verify that H (E ) is filtered by E∞
as in (1.7.2.2),(c) describe dr and verify that d2r = 0, and(d) verify that Ep,qr+1 is given by cohomology using dr .
Before tackling these goals, you can impress your friends by giving this short
description of the pages and differentials of the spectral sequence. We say that
an element of E•,• is a (p, q)-strip if it is an element of ⊕l≥0 Ep−l,q+l (see Fig. 1.1).
Its nonzero entries lie on an “upper-leftwards” semi-infinite antidiagonal startingwith position (p, q). We say that the (p, q)-entry (the projection to Ep,q ) is theleading term of the (p, q)-strip. Let Sp,q ⊂ E•,• be the submodule of all the (p, q)strips. Clearly Sp,q ⊂ Ep+q , and Sk,0 = Ek .

..

.

0

0

0

0

0

∗p−2,q+2

0

0

0

0

0

∗p−1,q+1

0

0

0

0

0

∗p,q

0

0

0

0

0

0p+1,q−1

F IGURE 1.1. A (p, q)-strip (in Sp,q ⊂ Ep+q ). Clearly Sk,0 = Ek .
Note that the differential d = d↑ + d→ sends a (p, q)-strip x to a (p + 1, q)-strip
dx. If dx is furthermore a (p−r+1, q+r)-strip (r ∈ Z≥0 ), we say that x is an r-closed(p, q)-strip — “the differential knocks x at least r terms deeper into the filtration”.
We denote the set of r-closed (p, q)-strips Sp,q(so for example Sp,q
= Sp,q , andr
0

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Sk,0= Ek ). An element of Sp,qmay be depicted as:
r
0
..

.

%?
7
∗p−2,q+2%0
7∗p−1,q+1

%0
7
∗p,q

%0

1.7.10. Preliminary definition of Ep,qr . We are now ready to give a first definition ofp,qEp,q= Ep,qr , which by construction should be a subquotient of E
0 . We describe
p,q
p,q
p,q
it as such by describing two submodules Yr ⊂ Xr ⊂ E , and defining Ep,q
=
r
p,q
p,q
Xp,q
be those elements of Ep,q that are the leading terms of r-closed
r /Yr . Let Xr(p, q)-strips. Note that by definition, d sends (r−1)-closed (p+(r−1)−1, q−(r−1))strips to (p, q)-strips. Let Yrp,q be the leading ((p, q))-terms of the differential d of(r − 1)-closed (p + (r − 1) − 1, q − (r − 1))-strips (where the differential is considered
as a (p, q)-strip).
1.7.G. E XERCISE ( REALITY CHECK ). Verify that Ep,qis (canonically isomorphic to)0
Ep,q .
We next give the definition of the differential dr of such an element x ∈ Xp,qr .
We take any r-closed (p, q)-strip with leading term x. Its differential d is a (p − r +1, q + r)-strip, and we take its leading term. The choice of the r-closed (p, q)-strip
means that this is not a well-defined element of Ep,q . But it is well-defined modulo
the differentials of the (r − 1)-closed (p − 1, q + 1)-strips, and hence gives a map
Ep,q→ Ep−r+1,q+r.
rrThis definition is fairly short, but not much fun to work with, so we will forgetit, and instead dive into a snakes’ nest of subscripts and superscripts.
We begin with making some quick but important observations about (p, q)strips.
1.7.H. E XERCISE ( NOT HARD ). Verify the following.
(a) Sp,q = Sp−1,q+1 ⊕ Ep,q .
(b) (Any closed (p, q)-strip is r-closed for all r.) Any element x of Sp,q = Sp,q0
that is a cycle (i.e., dx = 0) is automatically in Sp,qfor all r. For example,rthis holds when x is a boundary (i.e., of the form dy).
(c) Show that for fixed p, q,
Sp,q⊃ Sp,q⊃ · · · ⊃ Sp,q
⊃ ···
r
0
1

stabilizes for r ≫ 0 (i.e., Sp,q
= Sp,qrr+1 = · · · ). Denote the stabilized modp,qp,qule S∞ . Show S∞ is the set of closed (p, q)-strips (those (p, q)-strips
annihilated by d, i.e., the cycles). In particular, Sk,0
∞ is the set of cycles in
Ek .

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1.7.11. Defining Ep,qr .
p−1,q+1
Define Xp,q
:=
Sp,q
and
r
r /Sr−1$%p+(r−1)−1,q−(r−1)
Yrp,q := dSr−1+ Sp−1,q+1/Sp−1,q+1.
r−1
r−1
Then Yrp,q ⊂ Xp,q
by Exercise 1.7.H(b). We definer(1.7.11.1)

Ep,q
=
r

Xp,q
Sp,q
r
r
p,q =
p+(r−1)−1,q−(r−1)
YrdS+ Sp−1,q+1
r−1

r−1

We have completed Goal 1.7.9(a).
You are welcome to verify that these definitions of Xp,qand Yrp,q and hence
r
p,q
p,q
Er agree with the earlier ones of §1.7.10 (and in particular Xr and Yrp,q are both
submodules of Ep,q ), but we won’t need this fact.
1.7.I. E XERCISE : Ep,k−p
GIVES SUBQUOTIENTS OF Hk (E• ). By Exercise 1.7.H(c),∞p+(r−1)−1,q−(r−1)
p,qEr stabilizes as r → ∞. For r ≫ 0, interpret Sp,qas ther /dSr−1
p,q
p+q
p+q−1
cycles in S∞ ⊂ E
modulo those boundary elements of dE
contained ink •Sp,q∞ . Finally, show that H (E ) is indeed filtered as described in (1.7.2.2).
We have completed Goal 1.7.9(b).

1.7.12. Definition of dr .
We shall see that the map dr : Ep,q
→ Ep−r+1,q+r
is just induced by our differrr
ential d. Notice that d sends r-closed (p, q)-strips Sp,qto (p − r + 1, q + r)-stripsr
Sp−r+1,q+r , by the definition “r-closed”. By Exercise 1.7.H(b), the image lies in
Sp−r+1,q+r.
r
1.7.J. E XERCISE . Verify that d sends
p+(r−1)−1,q−(r−1)

dSr−1(p−r+1)+(r−1)−1,(q+r)−(r−1)+Sp−1,q+1→ dSr−1r−1(p−r+1)−1,(q+r)+1

+Sr−1

.

(The first term on the left goes to 0 from d2 = 0, and the second term on the left
goes to the first term on the right.)
Thus we may define
dr : Ep,q
=
r

Sp,q
r
p+(r−1)−1,q−(r−1)

dSr−1+ Sp−1,q+1
r−1

Sp−r+1,q+rr
dSp−1,q+1+ Sp−r,q+r+1
r−1
r−1

→

= Ep−r+1,q+rr

and clearly d2r = 0 (as we may interpret it as taking an element of Sp,qand applyr
ing d twice).
We have accomplished Goal 1.7.9(c).

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1.7.13. Verifying that the cohomology of dr at Ep,qis Ep,qr
r+1 .
unpleasant job of verifying that the cohomology of
Sp+r−1,q−rr
dSp+2r−3,q−2r+1+Sp+r−2,q−r+1
r−1
r−1(1.7.13.1)

dr

%

dr

We are left with theSp,qr
dSp+r−2,q−r+1+Sp−1,q+1
r−1
r−1

%

Sp−r+1,q+rr
dSp−1,q+1+Sp−r,q+r+1
r−1
r−1

is naturally identified with
Sp,qr+1
dSp+r−1,q−r+ Sp−1,q+1r
r
and this will conclude our final Goal 1.7.9(d).
We begin by understanding the kernel of the right map of (1.7.13.1). Supposea ∈ Sp,q
is mapped to 0. This means that da = db + c, where b ∈ Sp−1,q+1
. If
r
r−1
p−r,q+r+1
u = a − b, then u ∈ Sp,q , while du = c ∈ Sp−r,q+r+1
⊂
S
,
from
which
r−1
p−1,q+1
u is (r + 1)-closed, i.e., u ∈ Sp,q
+ Sp,qr+1 . Thus a = b + u ∈ Sr−1
r+1 . Conversely,
p−1,q+1p,q
any a ∈ Sr−1
+ Sr+1 satisfies
p−1,q+1
da ∈ dSp−1,q+1+ dSp,q+ Sp−r,q+r+1
r−1
r+1 ⊂ dSr−1
r−1

p−r,q+r+1(using dSp,qand Exercise 1.7.H(b)) so any such a is indeed in ther+1 ⊂ S0kernel of
Sp−r+1,q+r
r
Sp,q
→
.
r
dSp−1,q+1+ Sp−r,q+r+1
r−1
r−1
Hence the kernel of the right map of (1.7.13.1) isker =Sp−1,q+1+ Sp,qr−1
r+1
dSp+r−2,q−r+1+ Sp−1,q+1
r−1
r−1

.

Next, the image of the left map of (1.7.13.1) is immediately
im =dSp+r−1,q−r+ dSp+r−2,q−r+1+ Sp−1,q+1
r
r−1
r−1
dSp+r−2,q−r+1+ Sp−1,q+1
r−1
r−1

=

dSp+r−1,q−r+ Sp−1,q+1r
r−1
dSp+r−2,q−r+1+ Sp−1,q+1
r−1
r−1(as Sp+r−1,q−r
contains Sp+r−2,q−r+1).
r
r−1
Thus the cohomology of (1.7.13.1) is
ker / im =Sp−1,q+1+ Sp,qr−1
r+1=
p−1,q+1

dSp+r−1,q−r+ Sr−1rSp,qr+1
p+r−1,q−rSp,q+ Sp−1,q+1)
r+1 ∩ (dSrr−1

where the equality on the right uses the fact that dSp+r−1,q−r⊂ Sp,qr
r+1 and an
isomorphism theorem. We thus must show
However,p+r−1,q−rSp,q+ Sp−1,q+1) = dSp+r−1,q−r+ Sp−1,q+1.
r
r
r+1 ∩ (dSrr−1

p−1,q+1
p+r−1,q−rSp,q+ Sp−1,q+1) = dSp+r−1,q−r+ Sp,qr
r+1 ∩ (dSrr−1
r+1 ∩ Sr−1

p+1,q−1and Sp,q
consists of (p − 1, q + 1)-strips whose differential vanishes up
r+1 ∩ Sr−1
p−1,q+1
to row p + r, from which Sp,q
= Sp−1,q+1
as desired.
r
r+1 ∩ Sr−1

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This completes the explanation of how spectral sequences work for a firstquadrant double complex. The argument applies without significant change tomore general situations, including filtered complexes.

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CHAPTER 2

Sheaves
It is perhaps surprising that geometric spaces are often best understood interms of (nice) functions on them. For example, a differentiable manifold that is
a subset of Rn can be studied in terms of its differentiable functions. Because“geometric spaces” can have few (everywhere-defined) functions, a more precise
version of this insight is that the structure of the space can be well understoodby considering all functions on all open subsets of the space. This informationis encoded in something called a sheaf. Sheaves were introduced by Leray in the
1940’s, and Serre introduced them to algebraic geometry. (The reason for the namewill be somewhat explained in Remark 2.4.4.) We will define sheaves and describe
useful facts about them. We will begin with a motivating example to convince youthat the notion is not so foreign.
One reason sheaves are slippery to work with is that they keep track of a hugeamount of information, and there are some subtle local-to-global issues. There arealso three different ways of getting a hold of them:• in terms of open sets (the definition §2.2) — intuitive but in some ways
the least helpful;• in terms of stalks (see §2.4.1); and• in terms of a base of a topology (§2.5).

(Some people strongly prefer the espace étalé interpretation, §2.2.11, as well.) Knowing which to use requires experience, so it is essential to do a number of exercises
on different aspects of sheaves in order to truly understand the concept.

2.1 Motivating example: The sheaf of differentiable functions
Consider differentiable functions on the topological space X = Rn (or moregenerally on a differentiable manifold X). The sheaf of differentiable functions on
X is the data of all differentiable functions on all open subsets on X. We will see
how to manage these data, and observe some of their properties. On each open set
U ⊂ X, we have a ring of differentiable functions. We denote this ring of functions
O(U).
Given a differentiable function on an open set, you can restrict it to a smaller
open set, obtaining a differentiable function there. In other words, if U ⊂ V is an
inclusion of open sets, we have a “restriction map” resV,U : O(V) → O(U).
Take a differentiable function on a big open set, and restrict it to a medium
open set, and then restrict that to a small open set. The result is the same as if you
restrict the differentiable function on the big open set directly to the small open set.
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In other words, if U "→ V "→ W, then the following diagram commutes:
resW,V% O(V)
O(W)
❍❍
✇
❍❍
✇
✇
❍❍
✇✇
resW,U ❍❍❍✇✇resV,U
✇
$
!✇
O(U)

Next take two differentiable functions f1 and f2 on a big open set U, and an
open cover of U by some collection of open subsets {Ui }. (We say {Ui } covers U, oris an open cover of U, if U = ∪Ui .) Suppose that f1 and f2 agree on each of these
Ui . Then they must have been the same function to begin with. In other words, if{Ui }i∈I is a cover of U, and f1 , f2 ∈ O(U), and resU,Ui f1 = resU,Ui f2 , then f1 = f2 .
Thus we can identify functions on an open set by looking at them on a covering bysmall open sets.
Finally, suppose you are given the same U and cover {Ui }, take a differentiable
function on each of the Ui — a function f1 on U1 , a function f2 on U2 , and soon — and assume they agree on the pairwise overlaps. Then they can be “glued
together” to make one differentiable function on all of U. In other words, givenfi ∈ O(Ui ) for all i, such that resUi ,Ui ∩Uj fi = resUj ,Ui ∩Uj fj for all i and j, thenthere is some f ∈ O(U) such that resU,Ui f = fi for all i.
The entire example above would have worked just as well with continuousfunctions, or smooth functions, or just plain functions. Thus all of these classesof “nice” functions share some common properties. We will soon formalize theseproperties in the notion of a sheaf.
2.1.1. The germ of a differentiable function. Before we do, we first give anotherdefinition, that of the germ of a differentiable function at a point p ∈ X. Intuitively,
it is a “shred” of a differentiable function at p. Germs are objects of the form{(f, open U) : p ∈ U, f ∈ O(U)}modulo the relation that (f, U) ∼ (g, V) if there is some open set W ⊂ U, V containing p where f|W = g|W (i.e., resU,W f = resV,W g). In other words, two functions
that are the same in an open neighborhood of p (but may differ elsewhere) havethe same germ. We call this set of germs the stalk at p, and denote it Op . Noticethat the stalk is a ring: you can add two germs, and get another germ: if you havea function f defined on U, and a function g defined on V, then f + g is defined on
U ∩ V. Moreover, f + g is well-defined: if f̃ has the same germ as f, meaning that
there is some open set W containing p on which they agree, and g̃ has the same
germ as g, meaning they agree on some open W ′ containing p, then f̃ + g̃ is thesame function as f + g on U ∩ V ∩ W ∩ W ′ .
Notice also that if p ∈ U, you get a map O(U) → Op . Experts may already seethat we are talking about germs as colimits.
We can see that Op is a local ring as follows. Consider those germs vanishingat p, which we denote mp ⊂ Op . They certainly form an ideal: mp is closed under
addition, and when you multiply something vanishing at p by any function, theresult also vanishes at p. We check that this ideal is maximal by showing that thequotient ring is a field:(2.1.1.1) 0% mp := ideal of germs vanishing at p

% Op f*→f(p)% R%0

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2.1.A. E XERCISE . Show that this is the only maximal ideal of Op . (Hint: show thatevery element of Op \ mp is invertible.)
Note that we can interpret the value of a function at a point, or the value ofa germ at a point, as an element of the local ring modulo the maximal ideal. (Wewill see that this doesn’t work for more general sheaves, but does work for things
behaving like sheaves of functions. This will be formalized in the notion of a locallyringed space, which we will see, briefly, in §6.3.)∼ R, i.e., it is a real vector
2.1.2. Aside. Notice that mp /m2p is a module over Op /mp =space. It turns out to be naturally (whatever that means) the cotangent space tothe smooth manifold at p. This insight will prove handy later, when we define
tangent and cotangent spaces of schemes.
2.1.B. ⋆ E XERCISE FOR THOSE WITH DIFFERENTIAL GEOMETRIC BACKGROUND .
Prove this. (Rhetorical question for experts: what goes wrong if the sheaf of continuous functions is substituted for the sheaf of differentiable functions?)

2.2 Definition of sheaf and presheaf
We now formalize these notions, by defining presheaves and sheaves. Presheaves
are simpler to define, and notions such as kernel and cokernel are straightforward.
Sheaves are more complicated to define, and some notions such as cokernel require more thought. But sheaves are more useful because they are in some vague
sense more geometric; you can get information about a sheaf locally.
2.2.1. Definition of sheaf and presheaf on a topological space X.
To be concrete, we will define sheaves of sets. However, in the definition the
category Sets can be replaced by any category, and other important examples areabelian groups Ab, k-vector spaces Veck , rings Rings, modules over a ring ModA ,and more. (You may have to think more when dealing with a category of objects
that aren’t “sets with additional structure”, but there aren’t any new complications.
In any case, this won’t be relevant for us, although people who want to do thisshould start by solving Exercise 2.2.C.) Sheaves (and presheaves) are often written
in calligraphic font. The fact that F is a sheaf on a topological space X is often
written asFX
2.2.2. Definition: Presheaf. A presheaf F on a topological space X is the following data.
• To each open set U ⊂ X, we have a set F (U) (e.g., the set of differentiable
functions in our motivating example). (Notational warning: Several notations arein use, for various good reasons: F (U) = Γ (U, F ) = H0 (U, F ). We will use themall.) The elements of F (U) are called sections of F over U. (§2.2.11 combined

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with Exercise 2.2.G gives a motivation for this terminology, although this isn’t soimportant for us.)
By convention, if the “U” is omitted, it is implicitly taken to be X: “sections of
F ” means “sections of F over X”. These are also called global sections.
• For each inclusion U "→ V of open sets, we have a restriction map resV,U : F (V) →F (U) (just as we did for differentiable functions).
The data is required to satisfy the following two conditions.
• The map resU,U is the identity: resU,U = idF (U) .
• If U "→ V "→ W are inclusions of open sets, then the restriction maps commute, i.e.,
resW,V% F (V)
F (W)
■■
✈
■■
✈
■■
✈✈
✈
resW,U ■■■✈✈resV,U
+
!✈✈
F (U)

commutes.
2.2.A. E XERCISE FOR CATEGORY- LOVERS : “A PRESHEAF IS THE SAME AS A CON Given any topological space X, we have a “categoryof open sets” (Example 1.2.9), where the objects are the open sets and the morphisms are inclusions. Verify that the data of a presheaf is precisely the data of acontravariant functor from the category of open sets of X to the category of sets.
(This interpretation is surprisingly useful.)
TRAVARIANT FUNCTOR ”.

2.2.3. Definition: Stalks and germs. We define the stalk of a presheaf at a pointin two equivalent ways. One will be hands-on, and the other will be as a colimit.
2.2.4. Define the stalk of a presheaf F at a point p to be the set of germs of F at p,denoted Fp , as in the example of §2.1.1. Germs correspond to sections over someopen set containing p, and two of these sections are considered the same if theyagree on some smaller open set. More precisely: the stalk is{(f, open U) : p ∈ U, f ∈ F (U)}modulo the relation that (f, U) ∼ (g, V) if there is some open set W ⊂ U, V wherep ∈ W and resU,W f = resV,W g. (To explain the agricultural terminology: theFrench name “germe” is meant to suggest a tiny shoot sprouting from a seed, cf.
“germinate”.)
2.2.5. A useful equivalent definition of a stalk is as a colimit of all F (U) over all
open sets U containing p:
Fp = lim F (U).
−→The index category is a filtered set (given any two such open sets, there is a thirdsuch set contained in both), so these two definitions are the same by Exercise 1.4.D.
Hence we can define stalks for sheaves of sets, groups, rings, and other things forwhich colimits exist for directed sets. It is very helpful to keep both definitions ofstalk in mind at the same time.
If p ∈ U, and f ∈ F (U), then the image of f in Fp is called the germ of f at p.
(Warning: unlike the example of §2.1.1, in general, the value of a section at a point
doesn’t make sense.)

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2.2.6. Definition: Sheaf. A presheaf is a sheaf if it satisfies two more axioms.
Notice that these axioms use the additional information of when some open sets
cover another.
Identity axiom. If {Ui }i∈I is an open cover of U, and f1 , f2 ∈ F (U), andresU,Ui f1 = resU,Ui f2 for all i, then f1 = f2 .
(A presheaf satisfying the identity axiom is called a separated presheaf, but
we will not use that notation in any essential way.)
Gluability axiom. If {Ui }i∈I is an open cover of U, then given fi ∈ F (Ui )for all i, such that resUi ,Ui ∩Uj fi = resUj ,Ui ∩Uj fj for all i, j, then there is somef ∈ F (U) such that resU,Ui f = fi for all i.
In mathematics, definitions often come paired: “at most one” and “at leastone”. In this case, identity means there is at most one way to glue, and gluability
means that there is at least one way to glue.
(For experts and scholars of the empty set only: an additional axiom sometimes included is that F (∅) is a one-element set, and in general, for a sheaf withvalues in a category, F (∅) is required to be the final object in the category. Thisactually follows from the above definitions, assuming that the empty product isappropriately defined as the final object.)Example. If U and V are disjoint, then F (U ∪ V) = F (U) × F (V). Here we use
the fact that F (∅) is the final object.
The stalk of a sheaf at a point is just its stalk as a presheaf — the same definition applies — and similarly for the germs of a section of a sheaf.
2.2.B. U NIMPORTANT EXERCISE : PRESHEAVES THAT ARE NOT SHEAVES . Show
that the following are presheaves on C (with the classical topology), but not sheaves:(a) bounded functions, (b) holomorphic functions admitting a holomorphic square
root.
Both of the presheaves in the previous Exercise satisfy the identity axiom. A“natural” example failing even the identity axiom is implicit in Remark 2.5.5.
We now make a couple of points intended only for category-lovers.
2.2.7. Interpretation in terms of the equalizer exact sequence. The two axioms for apresheaf to be a sheaf can be interpreted as “exactness” of the “equalizer exact%% " F (U ∩ U ). Identity is exact% F (U)% " F (Ui )sequence”: ·ij"ness at F (U), and gluability is exactness at
F (Ui ). I won’t make this precise,or even explain what the double right arrow means. (What is an exact sequence ofsets?!) But you may be able to figure it out from the context.
2.2.C. E XERCISE . The identity and gluability axioms may be interpreted as sayingthat F (∪i∈I Ui ) is a certain limit. What is that limit?
Here are a number of examples of sheaves.
2.2.D. E XERCISE .
(a) Verify that the examples of §2.1 are indeed sheaves (of differentiable functions,or continuous functions, or smooth functions, or functions on a manifold or Rn ).
(b) Show that real-valued continuous functions on (open sets of) a topologicalspace X form a sheaf.

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2.2.8. Important Example: Restriction of a sheaf. Suppose F is a sheaf on X, and U
is an open subset of X. Define the restriction of F to U, denoted F |U , to be thecollection F |U (V) = F (V) for all open subsets V ⊂ U. Clearly this is a sheaf on
U. (Unimportant but fun fact: §2.7 will tell us how to restrict sheaves to arbitrary
subsets.)
2.2.9. Important Example: the skyscraper sheaf. Suppose X is a topological space,with p ∈ X, and S is a set. Let ip : p → X be the inclusion. Then ip,∗ S defined by%S
if p ∈ U, andip,∗ S(U) ={e} if p ∈/Uforms a sheaf. Here {e} is any one-element set. (Check this if it isn’t clear to you— what are the restriction maps?) This is called a skyscraper sheaf, because theinformal picture of it looks like a skyscraper at p. (Mild caution: this informal picture suggests that the only nontrivial stalk of a skyscraper sheaf is at p, which isn’t
the case. Exercise 13.2.A(b) gives an example, although it certainly isn’t the simplest one.) There is an analogous definition for sheaves of abelian groups, except
ip,∗ (S)(U) = {0} if p ∈/ U; and for sheaves with values in a category more generally,
ip,∗ S(U) should be a final object.
(This notation is admittedly hideous, and the alternative (ip )∗ S is equally bad.
In §2.2.12 we explain this notation.)
2.2.10. Constant presheaves and constant sheaves. Let X be a topological space, and S
a set. Define Spre (U) = S for all open sets U. You will readily verify that Spre forms
a presheaf (with restriction maps the identity). This is called the constant presheaf
associated to S. This isn’t (in general) a sheaf. (It may be distracting to say why.
Lovers of the empty set will insist that the sheaf axioms force the sections over theempty set to be the final object in the category, i.e., a one-element set. But even if
we patch the definition by setting Spre (∅) = {e}, if S has more than one element,
and X is the two-point space with the discrete topology, i.e., where every subset
is open, you can check that Spre fails gluability.)
2.2.E. E XERCISE ( CONSTANT SHEAVES ). Now let F (U) be the maps to S thatare locally constant, i.e., for any point p in U, there is an open neighborhood of p
where the function is constant. Show that this is a sheaf. (A better description is
this: endow S with the discrete topology, and let F (U) be the continuous maps
U → S.) This is called the constant sheaf (associated to S); do not confuse it withthe constant presheaf. We denote this sheaf S.

2.2.F. E XERCISE (“ MORPHISMS GLUE ”). Suppose Y is a topological space. Show
that “continuous maps to Y” form a sheaf of sets on X. More precisely, to each
open set U of X, we associate the set of continuous maps of U to Y. Show that thisforms a sheaf. (Exercise 2.2.D(b), with Y = R, and Exercise 2.2.E, with Y = S withthe discrete topology, are both special cases.)

2.2.G. E XERCISE . This is a fancier version of the previous exercise.
(a) (sheaf of sections of a map) Suppose we are given a continuous map µ : Y → X.
Show that “sections of µ” form a sheaf. More precisely, to each open set U of X,
associate the set of continuous maps s : U → Y such that µ ◦ s = id|U . Show that

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this forms a sheaf. (For those who have heard of vector bundles, these are a good
example.) This is motivation for the phrase “section of a sheaf”.
(b) (This exercise is for those who know what a topological group is. If you don’tknow what a topological group is, you might be able to guess.) Suppose that Y isa topological group. Show that continuous maps to Y form a sheaf of groups.
2.2.11. ⋆ The space of sections (espace étalé) of a (pre)sheaf. Depending on your background, you may prefer the following perspective on sheaves. Suppose F is apresheaf (e.g., a sheaf) on a topological space X. Construct a topological space F
along with a continuous map π : F → X as follows: as a set, F is the disjoint unionof all the stalks of F . This naturally gives a map of sets π : F → X. Topologize F as
follows. Each s in F (U) determines a subset {(x, sx ) : x ∈ U} of F. The topology
on F is the weakest topology such that these subsets are open. (These subsets forma base of the topology. For each y ∈ F, there is an open neighborhood V of y andan open neighborhood U of π(y) such that π|V is a homeomorphism from V to U.
Do you see why these facts are true?) The topological space F could be thought ofas the space of sections of F (and in French is called the espace étalé of F ). We
will not discuss this construction at any length, but it can have some advantages:(a) It is always better to know as many ways as possible of thinking about a concept. (b) Pullback has a natural interpretation in this language (mentioned briefly
in Exercise 2.7.C). (c) Sheafification has a natural interpretation in this language(see Remark 2.4.8).
2.2.H. I MPORTANT E XERCISE /D EFINITION : THE PUSHFORWARD SHEAF OR DIRECT
IMAGE SHEAF. Suppose π : X → Y is a continuous map, and F is a presheaf on X.
Then define π∗ F by π∗ F (V) = F (π−1 (V)), where V is an open subset of Y. Show
that π∗ F is a presheaf on Y, and is a sheaf if F is. This is called the pushforward
or direct image of F . More precisely, π∗ F is called the pushforward of F by π.
2.2.12. As the notation suggests, the skyscraper sheaf (Example 2.2.9) can be interpreted as the pushforward of the constant sheaf S on a one-point space p, under
the inclusion morphism ip : {p} → X.
Once we realize that sheaves form a category, we will see that the pushforward
is a functor from sheaves on X to sheaves on Y (Exercise 2.3.B).
2.2.I. E XERCISE ( PUSHFORWARD INDUCES MAPS OF STALKS ). Suppose π : X → Y
is a continuous map, and F is a sheaf of sets (or rings or A-modules) on X. If
π(p) = q, describe the natural morphism of stalks (π∗ F )q → Fp . (You can use theexplicit definition of stalk using representatives, §2.2.4, or the universal property,§2.2.5. If you prefer one way, you should try the other.)
2.2.13. Important Example: Ringed spaces, and OX -modules. Suppose OX isa sheaf of rings on a topological space X (i.e., a sheaf on X with values in the
category of Rings). Then (X, OX ) is called a ringed space. The sheaf of rings isoften denoted by OX , pronounced “oh-X”. This sheaf is called the structure sheafof the ringed space. Sections of the structure sheaf OX over an open subset U
are called functions on U. (Caution: what we call “functions”, others sometimes call“regular functions”. Furthermore, we will later define “rational functions” on schemesin §5.5.6, which are not precisely functions in this sense; they are a particular type of“partially-defined function”.)

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The symbol OX will always refer to the structure sheaf of a ringed space X.
The restriction OX |U of OX to an open subset U ⊂ X is denoted OU . (We will latercall (U, OU ) → (X, OX ) an open embedding of ringed spaces, see Definition 6.2.1.)
The stalk of OX at a point p is written “OX,p ”, because this looks less hideous than
“OXp ”.
Just as we have modules over a ring, we have OX -modules over a sheaf ofrings OX . There is only one possible definition that could go with the name OX module — a sheaf of abelian groups F with the following additional structure.
For each U, F (U) is an OX (U)-module. Furthermore, this structure should behave
well with respect to restriction maps: if U ⊂ V, then

(2.2.13.1)

OX (V) × F (V)resV,U × resV,Uaction %F (V)resV,U""
action %OX (U) × F (U)F (U)commutes. (You should convince yourself that I haven’t forgotten anything.)
Recall that the notion of A-module generalizes the notion of abelian group,
because an abelian group is the same thing as a Z-module. Similarly, the notion of
OX -module generalizes the notion of sheaf of abelian groups, because the latter isthe same thing as a Z-module, where Z is the constant sheaf associated to Z. Hencewhen we are proving things about OX -modules, we are also proving things about
sheaves of abelian groups.
2.2.J. E XERCISE . If (X, OX ) is a ringed space, and F is an OX -module, describe
how for each p ∈ X, Fp is an OX,p -module.
2.2.14. For those who know about vector bundles. The motivating example of OX modules is the sheaf of sections of a vector bundle. If (X, OX ) is a differentiable
manifold (so OX is the sheaf of differentiable functions), and π : V → X is a vector
bundle over X, then the sheaf of differentiable sections σ : X → V is an OX -module.
Indeed, given a section s of π over an open subset U ⊂ X, and a function f on U,
we can multiply s by f to get a new section fs of π over U. Moreover, if U ′ is asmaller subset, then we could multiply f by s and then restrict to U ′ , or we could
restrict both f and s to U ′ and then multiply, and we would get the same answer.
That is precisely the commutativity of (2.2.13.1).

2.3 Morphisms of presheaves and sheaves
2.3.1. Definitions. Whenever one defines a new mathematical object, category theory teaches to try to understand maps between them. We now define morphismsof presheaves, and similarly for sheaves. In other words, we will describe thecategory of presheaves (of sets, abelian groups, etc.) and the category of sheaves.
A morphism of presheaves of sets (or indeed of presheaves with values inany category) on X, φ : F → G , is the data of maps φ(U) : F (U) → G (U) for all U

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behaving well with respect to restriction: if U "→ V thenF (V)resV,Uφ(V)% G (V)
resV,U""φ(U)% G (U)F (U)commutes. (Notice: the underlying space of both F and G is X.)
Morphisms of sheaves are defined identically: the morphisms from a sheaf F
to a sheaf G are precisely the morphisms from F to G as presheaves. (Translation:
The category of sheaves on X is a full subcategory of the category of presheaves on
X.) If (X, OX ) is a ringed space, then morphisms of OX -modules have the obvious
definition. (Can you write it down?)
An example of a morphism of sheaves is the map from the sheaf of differentiable functions on R to the sheaf of continuous functions. This is a “forgetfulmap”: we are forgetting that these functions are differentiable, and remembering
only that they are continuous.
2.3.2. Notation. We may as well set some notation: let SetsX , AbX , etc. denotethe category of sheaves of sets, abelian groups, etc. on a topological space X. Let
preModOX denote the category of OX -modules on a ringed space (X, OX ). Let SetsX ,etc. denote the category of presheaves of sets, etc. on X.
2.3.3. Aside for category-lovers. If you interpret a presheaf on X as a contravariant functor (from the category of open sets), a morphism of presheaves on X is anatural transformation of functors (§1.2.21).
2.3.A. E XERCISE : MORPHISMS OF ( PRE ) SHEAVES INDUCE MORPHISMS OF STALKS .
If φ : F → G is a morphism of presheaves on X, and p ∈ X, describe an induced
morphism of stalks φp : Fp → Gp . Translation: taking the stalk at p induces afunctor SetsX → Sets. (Your proof will extend in obvious ways. For example, if φ
is a morphism of OX -modules, then φp is a map of OX,p -modules.)

2.3.B. E XERCISE . Suppose π : X → Y is a continuous map of topological spaces(i.e., a morphism in the category of topological spaces). Show that pushforward
gives a functor π∗ : SetsX → SetsY . Here Sets can be replaced by other categories.
(Watch out for some possible confusion: a presheaf is a functor, and presheaves
form a category. It may be best to forget that presheaves are functors for now.)
2.3.C. I MPORTANT E XERCISE AND D EFINITION : “S HEAF Hom”. Suppose F and
G are two sheaves of sets on X. (In fact, it will suffice that F is a presheaf.) Let
Hom (F , G ) be the collection of data

Hom (F , G )(U) := Mor(F |U , G |U ).
(Recall the notation F |U , the restriction of the sheaf to the open set U, Example 2.2.8.) Show that this is a sheaf of sets on X. (To avoid a common confusion:
the right side does not say Mor(F (U), G (U)).) This sheaf is called “sheaf Hom ”.
(Strictly speaking, we should reserve Hom for when we are in an additive category, so this should possibly be called “sheaf Mor”. But the terminology “sheaf
Hom” is too established to uproot.) It will be clear from your construction that,

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like Hom, Hom is a contravariant functor in its first argument and a covariant functor in its second argument.
Warning: Hom does not commute with taking stalks. More precisely: it is
not true that Hom (F , G )p is isomorphic to Hom(Fp , Gp ). (Can you think of acounterexample? There is at least a map from one of these to the other — in which
direction?)
2.3.4. We will use many variants of the definition of Hom . For example, if F and
G are sheaves of abelian groups on X, then Hom AbX (F , G ) is defined by taking
Hom AbX (F , G )(U) to be the maps as sheaves of abelian groups F |U → G |U . (Notethat Hom AbX (F , G ) has the structure of a sheaf of abelian groups in a natural way.)
Similarly, if F and G are OX -modules, we define Hom ModOX (F , G ) in the analogous way (and it is an OX -module). Obnoxiously, the subscripts AbX and ModOX
are often dropped (here and in the literature), so be careful which category you areworking in! We call Hom ModOX (F , OX ) the dual of the OX -module F , and denote
it F ∨ .
2.3.D. U NIMPORTANT E XERCISE ( REALITY CHECK ).

∼ F , where {p}(a) If F is a sheaf of sets on X, then show that Hom ({p}, F ) =is the constant sheaf associated to the one element set {p}.
∼F(b) If F is a sheaf of abelian groups on X, then show that Hom AbX (Z, F ) =(an isomorphism of sheaves of abelian groups).
∼ F (an iso(c) If F is an OX -module, then show that Hom ModOX (OX , F ) =morphism of OX -modules).

A key idea in (b) and (c) is that 1 “generates” (in some sense) Z (in (b)) and OX (in(c)).
2.3.5. Presheaves of abelian groups (and even “presheaf OX -modules”) form an
abelian category.
We can make module-like constructions using presheaves of abelian groups
on a topological space X. (Throughout this section, all (pre)sheaves are of abeliangroups.) For example, we can clearly add maps of presheaves and get another mapof presheaves: if φ, ψ : F → G , then we define the map φ + ψ by (φ + ψ)(V) =φ(V) + ψ(V). (There is something small to check here: that the result is indeed amap of presheaves.) In this way, presheaves of abelian groups form an additive
category (Definition 1.6.1: the morphisms between any two presheaves of abeliangroups form an abelian group; there is a 0-object; and one can take finite products).
For exactly the same reasons, sheaves of abelian groups also form an additive
category.
If φ : F → G is a morphism of presheaves, define the presheaf kernel kerpre φby (kerpre φ)(U) := ker φ(U).

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2.3.E. E XERCISE . Show that kerpre φ is a presheaf. (Hint: if U "→ V, define therestriction map by chasing the following diagram:00% kerpre φ(V)"∃!

% kerpre φ(U)% F (V)
resV,U"% F (U)% G (V)
resV,U"% G (U)You should check that the restriction maps compose as desired.)Define the presheaf cokernel cokerpre φ similarly. It is a presheaf by essentiallythe same (dual) argument.
2.3.F. E XERCISE : THE COKERNEL DESERVES ITS NAME . Show that the presheaf cokernel satisfies the universal property of cokernels (Definition 1.6.3) in the categoryof presheaves.
Similarly, kerpre φ → F satisfies the universal property for kernels in the category of presheaves.
It is not too tedious to verify that presheaves of abelian groups form an abeliancategory, and the reader is free to do so. The key idea is that all abelian-categoricalnotions may be defined and verified “open set by open set”. We needn’t worry
about restriction maps — they “come along for the ride”. Hence we can define
terms such as subpresheaf, image presheaf (or presheaf image), and quotientpresheaf (or presheaf quotient), and they behave as you would expect. You construct kernels, quotients, cokernels, and images open set by open set. Homologicalalgebra (exact sequences and so forth) works, and also “works open set by openset”. In particular:
2.3.G. E ASY E XERCISE . Show (or observe) that for a topological space X with openpreset U, F (→ F (U) gives a functor from presheaves of abelian groups on X, AbX ,to abelian groups, Ab. Then show that this functor is exact.
2.3.H. E XERCISE . Show that a sequence of presheaves 0 → F1 → F2 → · · · →
Fn → 0 is exact if and only if 0 → F1 (U) → F2 (U) → · · · → Fn (U) → 0 is exact
for all U.

The above discussion essentially carries over without change to presheaves
with values in any abelian category. (Think this through if you wish.)
However, we are interested in more geometric objects, sheaves, where things
can be understood in terms of their local behavior, thanks to the identity and gluing axioms. We will soon see that sheaves of abelian groups also form an abeliancategory, but a complication will arise that will force the notion of sheafification on
us. Sheafification will be the answer to many of our prayers. We just haven’t yet
realized what we should be praying for.
To begin with, sheaves AbX form an additive category, as described in the firstparagraph of §2.3.5.
Kernels work just as with presheaves:

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2.3.I. I MPORTANT E XERCISE . Suppose φ : F → G is a morphism of sheaves. Show
that the presheaf kernel kerpre φ is in fact a sheaf. Show that it satisfies the universal property of kernels (Definition 1.6.3). (Hint: the second question follows
immediately from the fact that kerpre φ satisfies the universal property in the category of presheaves.)
Thus if φ is a morphism of sheaves, we define
ker φ := kerpre φ.
The problem arises with the cokernel.
2.3.J. I MPORTANT E XERCISE . Let X be C with the classical topology, let Z be theconstant sheaf on X associated to Z, OX the sheaf of holomorphic functions, andF the presheaf of functions admitting a holomorphic logarithm. Describe an exact
sequence of presheaves on X:
0

%Z

% OX

%F

%0

where Z → OX is the natural inclusion and OX → F is given by f (→ exp(2πif).
(Be sure to verify exactness.) Show that F is not a sheaf. (Hint: F does not satisfy
the gluability axiom. The problem is that there are functions that don’t have a logarithm but locally have a logarithm.) This will come up again in Example 2.4.10.
We will have to put our hopes for understanding cokernels of sheaves on holdfor a while. We will first learn to understand sheaves using stalks.

2.4 Properties determined at the level of stalks, and sheafification
2.4.1. Properties determined by stalks. We now come to the second way ofunderstanding sheaves mentioned at the start of the chapter. In this section, we
will see that lots of facts about sheaves can be checked “at the level of stalks”. This
isn’t true for presheaves, and reflects the local nature of sheaves. We will see that
sections and morphisms are determined “by their stalks”, and the property of amorphism being an isomorphism may be checked at stalks. (The last one is thetrickiest.)
2.4.A. I MPORTANT EASY EXERCISE (sections are determined by germs). Provethat a section of a sheaf of sets is determined by its germs, i.e., the natural map&(2.4.1.1)
F (U) →Fpp∈U

is injective. Hint 1: you won’t use the gluability axiom, so this is true for separated
presheaves. Hint 2: it is false for presheaves in general, see Exercise 2.4.F, so you
will use the identity axiom. (Your proof will also apply to sheaves of groups, rings,
etc. — to categories of “sets with additional structure”. The same is true of many
exercises in this section.)
2.4.2. Definition: support of a section. This motivates a concept we will find usefullater. Suppose F is a sheaf (or indeed separated presheaf) of abelian groups on X,

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and s is a global section of F . Then let the support of s, denoted Supp(s), be thepoints p of X where s has a nonzero germ:
Supp(s) := {p ∈ X : sp ̸= 0 in Fp }.

We think of this as the subset of X where “the section s lives” — the complement is
the locus where s is the 0-section. We could define this even if F is a presheaf, but
without the inclusion of Exercise 2.4.A, we could have the strange situation where
we have a nonzero section that “lives nowhere” (because it is 0 “near every point”,
i.e., is 0 in every stalk).
2.4.B. E XERCISE ( THE SUPPORT OF A SECTION IS CLOSED ). Show that Supp(s) isa closed subset of X.
Exercise 2.4.A suggests an important question: which elements of the rightside of (2.4.1.1) are in the image of the left side?
"
2.4.3. Important definition. We say that an element p∈U sp of the right side"p∈U Fp of (2.4.1.1) consists of compatible germs if for all p ∈ U, there is some
representative(Up open in U, s̃p ∈ F (Up ))for sp (where p ∈ Up ⊂ U) such that the germ of s̃p at all q ∈ Up is sq . Equivalently, there is an open cover {Ui } of U, and sections fi ∈ F (Ui ), such that if
p ∈ Ui , then sp is the germ of fi at p. Clearly any section s of F over U gives achoice of compatible germs for U.
2.4.C. I MPORTANT E XERCISE . Prove that any choice of compatible germs for asheaf of sets F over U is the image of a section of F over U. (Hint: you will use
gluability.)
We have thus completely described the image of (2.4.1.1), in a way that wewill find useful.
2.4.4. Remark. This perspective motivates the agricultural terminology “sheaf”: itis (the data of) a bunch of stalks, bundled together appropriately.
Now we throw morphisms into the mix. Recall Exercise 2.3.A: morphisms of(pre)sheaves induce morphisms of stalks.
2.4.D. E XERCISE (morphisms are determined by stalks). If φ1 and φ2 are morphisms from a presheaf of sets F to a sheaf of sets G that induce the same maps
on each stalk, show that φ1 = φ2 . Hint: consider the following diagram.
(2.4.4.1)

F (U)
"

"

p∈U Fp% G (U)"#%""p∈U Gp2.4.E. T RICKY E XERCISE (isomorphisms are determined by stalks). Show that amorphism of sheaves of sets is an isomorphism if and only if it induces an isomorphism of all stalks. Hint: Use (2.4.4.1). Once you have injectivity, show surjectivity,
perhaps using Exercise 2.4.C, or gluability in some other way; this is more subtle.

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Note: this question does not say that if two sheaves have isomorphic stalks, thenthey are isomorphic.
2.4.F. E XERCISE .
(a) Show that Exercise 2.4.A is false for general presheaves.
(b) Show that Exercise 2.4.D is false for general presheaves.
(c) Show that Exercise 2.4.E is false for general presheaves.
(General hint for finding counterexamples of this sort: consider a 2-point spacewith the discrete topology.)
2.4.5. Sheafification.
Every sheaf is a presheaf (and indeed by definition sheaves on X form a fullsubcategory of the category of presheaves on X). Just as groupification (§1.5.3)
gives an abelian group that best approximates an abelian semigroup, sheafification gives the sheaf that best approximates a presheaf, with an analogous universal property. (One possible example to keep in mind is the sheafification of thepresheaf of holomorphic functions admitting a square root on C with the classicaltopology. See also the exponential exact sequence, Exercise 2.4.10.)
2.4.6. Definition. If F is a presheaf on X, then a morphism of presheaves sh : F →F sh on X is a sheafification of F if F sh is a sheaf, and for any sheaf G , andany presheaf morphism g : F → G , there exists a unique morphism of sheaves
f : F sh → G making the diagramsh% F sh
F❉
❉❉
❉❉
❉
f
g ❉❉
❉9 "
G

commute.
We still have to show that it exists. The following two exercises require existence (which we will show shortly), but not the details of the construction.
2.4.G. E XERCISE . Show that sheafification is unique up to unique isomorphism,%F .
assuming it exists. Show that if F is a sheaf, then the sheafification is id : F(This should be second nature by now.)
2.4.H. E ASY E XERCISE ( SHEAFIFICATION IS A FUNCTOR ). Assume for now thatsheafification exists. Use the universal property to show that for any morphism ofpresheaves φ : F → G , we get a natural induced morphism of sheaves φsh : F sh →G sh . Show that sheafification is a functor from presheaves on X to sheaves on X.

2.4.7. Construction. We next show that any presheaf of sets (or groups, rings, etc.)
has a sheafification. Suppose F is a presheaf. Define F sh by defining F sh (U) as
the set of “compatible germs” of the presheaf F over U. Explicitly:
F sh (U) :={(fp ∈ Fp )p∈U : for all p ∈ U, there exists an open
neighborhood V of p, contained in U,
and s ∈ F (V) with sq = fq for all q ∈ V}.

Here sq means the germ of s at q — the image of s in the stalk Fq .

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2.4.I. E ASY E XERCISE .
forms a sheaf.

85

Show that F sh (using the tautological restriction maps)

2.4.J. E ASY E XERCISE . Describe a natural map of presheaves sh : F → F sh .

2.4.K. E XERCISE . Show that the map sh satisfies the universal property of sheafification (Definition 2.4.6). (This is easier than you might fear.)
2.4.L. U SEFUL EXERCISE , NOT JUST FOR CATEGORY- LOVERS . Show that the sheafification functor is left-adjoint to the forgetful functor from sheaves on X to presheaves
on X. This is not difficult — it is largely a restatement of the universal property.
But it lets you use results from §1.6.12, and can “explain” why you don’t need tosheafify when taking kernel (why the presheaf kernel is already the sheaf kernel),and why you need to sheafify when taking cokernel and (soon, in Exercise 2.6.J)⊗.
2.4.M. E XERCISE . Show F → F sh induces an isomorphism of stalks. (Possible
hint: Use the concrete description of the stalks. Another possibility once you read
Remark 2.7.3: judicious use of adjoints.)
As a reality check, you may want to verify that “the sheafification of a constantpresheaf is the corresponding constant sheaf” (see §2.2.10): if X is a topologicalspace and S is a set, then (Spre )sh may be naturally identified with S.
2.4.8. ⋆ Remark. The “space of sections” (or “espace étalé”) construction (§2.2.11)
yields a different-sounding description of sheafification which may be preferredby some readers. The main idea is identical: if F is a presheaf, let F be the space ofsections (or espace étalé) of F . You may wish to show that if F is a presheaf, thesheaf of sections of F → X (defined in Exercise 2.2.G(a)) is the sheafification of F .
Exercise 2.2.E may be interpreted as an example of this construction. The “spaceof sections” construction of the sheafification is essentially the same as Construction 2.4.7.
2.4.9. Subsheaves and quotient sheaves.
We now discuss subsheaves and quotient sheaves from the perspective of
stalks.
2.4.N. E XERCISE . Suppose φ : F → G is a morphism of sheaves of sets on a
topological space X. Show that the following are equivalent.

(a) φ is a monomorphism in the category of sheaves.
(b) φ is injective on the level of stalks: φp : Fp → Gp is injective for all p ∈ X.
(c) φ is injective on the level of open sets: φ(U) : F (U) → G (U) is injective
for all open U ⊂ X.

(Possible hints: for (b) implies (a), recall that morphisms are determined by stalks,
Exercise 2.4.D. For (a) implies (c), use the “indicator sheaf” with one section overevery open set contained in U, and no section over any other open set.) If these
conditions hold, we say that F is a subsheaf of G (where the “inclusion” φ issometimes left implicit).
(You may later wish to extend your solution to Exercise 2.4.N to show that forany morphism of presheaves, if all maps of sections are injective, then all stalk maps

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are injective. And furthermore, if φ : F → G is a morphism from a separated
presheaf to an arbitrary presheaf, then injectivity on the level of stalks implies
that φ is a monomorphism in the category of presheaves. This is useful in someapproaches to Exercise 2.6.C.)
2.4.O. E XERCISE . Continuing the notation of the previous exercise, show that thefollowing are equivalent.
(a) φ is an epimorphism in the category of sheaves.
(b) φ is surjective on the level of stalks: φp : Fp → Gp is surjective for allp ∈ X.

(Possible hint: use a skyscraper sheaf.)

If these conditions hold, we say that G is a quotient sheaf of F .
Thus monomorphisms and epimorphisms — subsheafiness and quotient sheafiness —can be checked at the level of stalks.
Both exercises generalize readily to sheaves with values in any reasonable category, where “injective” is replaced by “monomorphism” and “surjective” is replaced by “epimorphism”.
Notice that there was no part (c) to Exercise 2.4.O, and Example 2.4.10 shows
why. (But there is a version of (c) that implies (a) and (b): surjectivity on all open
sets in the base of a topology implies that the corresponding map of sheaves is an
epimorphism, Exercise 2.5.E.)
2.4.10. Example (cf. Exercise 2.3.J). Let X = C with the classical topology, and define
OX to be the sheaf of holomorphic functions, and OX∗ to be the sheaf of invertible(nowhere zero) holomorphic functions. This is a sheaf of abelian groups undermultiplication. We have maps of sheaves(2.4.10.1)

0% Z ×2πi % OX

exp

% O∗

X

%1

where Z is the constant sheaf associated to Z. (You can figure out what the sheaves
0 and 1 mean; they are isomorphic, and are written in this way for reasons that may
be clear.) We will soon interpret this as an exact sequence of sheaves of abeliangroups (the exponential exact sequence, see Exercise 2.6.E), although we don’t yet
have the language to do so.
% O ∗ describes O ∗ as
2.4.P. E NLIGHTENING E XERCISE . Show that exp : OX
X
Xa quotient sheaf of OX . Find an open set on which this map is not surjective.
This is a great example to get a sense of what “surjectivity” means for sheaves:
nowhere vanishing holomorphic functions have logarithms locally, but they need
not globally.

2.5 Recovering sheaves from a “sheaf on a base”
Sheaves are natural things to want to think about, but hard to get our hands on.
We like the identity and gluability axioms, but they make proving things trickier
than for presheaves. We have discussed how we can understand sheaves using

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stalks (using “compatible germs”). We now introduce a second way of getting ahold of sheaves, by introducing the notion of a sheaf on a base. Warning: this wayof understanding an entire sheaf from limited information is confusing. It may
help to keep sight of the central insight that this partial information is enoughto understand germs, and the notion of when they are compatible (with nearby
germs).
First, we define the notion of a base of a topology. Suppose we have a topological space X, i.e., we know which subsets Ui of X are open. Then a base of
a topology is a subcollection of the open sets {Bj } ⊂ {Ui }, such that each Ui is aunion of the Bj . Here is one example that you have seen early in your mathematical life. Suppose X = Rn . Then the way the classical topology is often first definedis by defining open balls Br (x) = {y ∈ Rn : |y − x| < r}, and declaring that any
union of open balls is open. So the balls form a base of the classical topology — wesay they generate the classical topology. As an application of how we use them, tocheck continuity of some map π : X → Rn , you need only think about the pullbackof balls on Rn — part of the traditional δ-ϵ definition of continuity.
Now suppose we have a sheaf F on a topological space X, and a base {Bi } ofopen sets on X. Then consider the information({F (Bi )}, {resBi ,Bj : F (Bi ) → F (Bj )}),

which is a subset of the information contained in the sheaf — we are only paying
attention to the information involving elements of the base, not all open sets.
We can recover the entire sheaf from this information. This is because we candetermine the stalks from this information, and we can determine when germs arecompatible.
2.5.A. I MPORTANT E XERCISE . Make this precise. How can you recover a sheaf F
from this partial information?
This suggests a notion, called a sheaf on a base. A sheaf of sets (or abeliangroups, rings, . . . ) on a base {Bi } is the following. For each Bi in the base, we havea set F(Bi ). If Bi ⊂ Bj , we have maps resBj ,Bi : F(Bj ) → F(Bi ), with resBi ,Bi =idF(Bi ) . (Things called “B” are always assumed to be in the base.) If Bi ⊂ Bj ⊂ Bk ,then resBk ,Bi = resBj ,Bi ◦ resBk ,Bj . So far we have defined a presheaf on a base{Bi }.
We also require the base identity axiom: If B = ∪Bi , then if f, g ∈ F(B) aresuch that resB,Bi f = resB,Bi g for all i, then f = g.
We require the base gluability axiom too: If B = ∪Bi , and we have fi ∈F(Bi ) such that fi agrees with fj on any basic open set contained in Bi ∩ Bj (i.e.,
resBi ,Bk fi = resBj ,Bk fj for all Bk ⊂ Bi ∩ Bj ) then there exists f ∈ F(B) such that
resB,Bi f = fi for all i.
2.5.1. Theorem. — Suppose {Bi } is a base on X, and F is a sheaf of sets on this base.
∼ F(Bi ) agreeing with
Then there is a sheaf F extending F (with isomorphisms F (Bi ) =the restriction maps). This sheaf F is unique up to unique isomorphism.
Proof. We will define F as the sheaf of compatible germs of F.
Define the stalk of a presheaf F on a base at p ∈ X by
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where the colimit is over all Bi (in the base) containing p.
We will say a family of germs in an open set U is compatible near p if there is asection s of F over some Bi containing p such that the germs over Bi are precisely
the germs of s. More formally, define
F (U) :={(fp ∈ Fp )p∈U : for all p ∈ U, there exists B with p ∈ B ⊂ U, s ∈ F(B),with sq = fq for all q ∈ B}

where each B is in our base.
This is a sheaf (for the same reasons that the sheaf of compatible germs was,cf. Exercise 2.4.I).
I next claim that if B is in our base, the natural map F(B) → F (B) is an isomorphism.

2.5.B. E XERCISE . Verify that F(B) → F (B) is an isomorphism, likely by showingthat it is injective and surjective (or else by describing the inverse map and verifying that it is indeed inverse). Possible hint: elements of F (B) are determined bystalks, as are elements of F(B).
It will be clear from your solution to Exercise 2.5.B that the restriction maps
for F are the same as the restriction maps of F (for elements of the base).
Finally, you should verify to your satisfaction that F is indeed unique up tounique isomorphism. (First be sure that you understand what this means!)!
Theorem 2.5.1 shows that sheaves on X can be recovered from their “restriction
to a base”. It is clear from the argument (and in particular the solution to the
Exercise 2.5.B) that if F is a sheaf and F is the corresponding sheaf on the base B,
then for any p ∈ X, Fp is naturally isomorphic to Fp .
Theorem 2.5.1 is a statement about objects in a category, so we should hope fora similar statement about morphisms.
2.5.C. I MPORTANT E XERCISE : M ORPHISMS OF SHEAVES CORRESPOND TO MOR PHISMS OF SHEAVES ON A BASE . Suppose {Bi } is a base for the topology of X. A
morphism F → G of sheaves on the base is a collection of maps F(Bk ) → G(Bk )such that the diagram% G(Bi )F(Bi )
resBi ,Bj"
F(Bj )

resBi ,Bj"
% G(Bj )commutes for all Bj "→ Bi .
(a) Verify that a morphism of sheaves is determined by the induced morphism ofsheaves on the base.
(b) Show that a morphism of sheaves on the base gives a morphism of the induced
sheaves. (Possible hint: compatible stalks.)
2.5.2. Remark. The above constructions and arguments describe an equivalence of
categories (§1.2.21) between sheaves on X and sheaves on a given base of X. Thereis no new content to this statement, but you may wish to think through what itmeans. What are the functors in each direction? Why aren’t their compositions
the identity?

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2.5.3. Remark. It will be useful to extend these notions to OX -modules (see for example Exercise 13.3.C). You will readily be able to verify that there is a correspondence (really, equivalence of categories) between OX -modules and “OX -modules
on a base”. Rather than working out the details, you should just informally think
through the main points: what is an “OX -module on a base”? Given an OX -module
on a base, why is the corresponding sheaf naturally an OX -module? Later, if you
are forced at gunpoint to fill in details, you will be able to.
2.5.D. I MPORTANT E XERCISE . Suppose X = ∪Ui is an open cover of X, and wehave sheaves Fi on Ui along with isomorphisms φij : Fi |Ui ∩Uj → Fj |Ui ∩Uj (withφii the identity) that agree on triple overlaps, i.e., φjk ◦ φij = φik on Ui ∩ Uj ∩ Uk
(this is called the cocycle condition, for reasons we ignore). Show that thesesheaves can be glued together into a sheaf F on X (unique up to unique isomor∼ F |U , and the isomorphisms over Ui ∩ Uj are the obvious
phism), such that Fi =iones. (Thus we can “glue sheaves together”, using limited patching information.)
Warning: we are not assuming this is a finite cover, so you cannot use induction.
For this reason this exercise can be perplexing. (You can use the ideas of this section to solve this problem, but you don’t necessarily need to. Hint: As the base,take those open sets contained in some Ui . Small observation: the hypothesis on
φii is extraneous, as it follows from the cocycle condition.)
2.5.4. Important remark. We will repeatedly see the theme of constructing some
object by gluing, in many different contexts. Keep an eye out for it! In each case,
we carefully consider what information we need in order to glue.
2.5.5. Remark for experts. Exercise 2.5.D almost says that the “set” of sheaves forms
a sheaf itself, but not quite. Making this precise leads one to the notion of a stack.
2.5.E. U NIMPORTANT EXERCISE . Suppose a morphism of sheaves F → G on abase {Bi } is surjective for all Bi (i.e., F(Bi ) → G(Bi ) is surjective for all i). Show
that the corresponding morphism of sheaves (not on the base) is surjective (or more
precisely: an epimorphism). The converse is not true, unlike the case for injectivity.
This gives a useful sufficient criterion for “surjectivity”: a morphism of sheaves is
an epimorphism (“surjective”) if it is surjective for sections on a base. You may
enjoy trying this out with Example 2.4.10 (dealing with holomorphic functions in
the classical topology on X = C), showing that the exponential map exp : OX → OX∗
is surjective, using the base of contractible open sets.

2.6 Sheaves of abelian groups, and OX -modules, form abeliancategories
We are now ready to see that sheaves of abelian groups, and their cousins, OX modules, form abelian categories. In other words, we may treat them similarly tovector spaces, and modules over a ring. In the process of doing this, we will seethat this is much stronger than an analogy; kernels, cokernels, exactness, and soforth can be understood at the level of stalks (which are just abelian groups), andthe compatibility of the germs will come for free.

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The category of sheaves of abelian groups on a topological space X is clearly an
additive category (Definition 1.6.1). In order to show that it is an abelian category,
we must begin by showing that any morphism φ : F → G has a kernel and acokernel. We have already seen that φ has a kernel (Exercise 2.3.I): the presheaf
kernel is a sheaf, and is a kernel in the category of sheaves.
2.6.A. E XERCISE . Show that the stalk of the kernel is the kernel of the stalks: forall p ∈ X, there is a natural isomorphism∼ ker(Fp → Gp ).
(ker(F → G ))p =We next address the issue of the cokernel. Now φ : F → G has a cokernel inthe category of presheaves; call it Hpre (where the subscript is meant to remind us
that this is a presheaf). Let sh : Hpre → H be its sheafification. Recall that thecokernel is defined using a universal property: it is the colimit of the diagram(2.6.0.1)

F

φ

%G

"
0
in the category of presheaves (cf. (1.6.3.1) and the comment thereafter).
2.6.1. Proposition. — The composition G → H is the cokernel of φ in the category ofsheaves.

Proof. We show that it satisfies the universal property. Given any sheaf E and acommutative diagram
F
"
0

φ

%G
"
%E

We construct
F

φ

%G

"
"
sh %
0 ❱❱❱❱❱ % Hpre
H
❱❱❱❱
❱❱❱❱
❱❱❱❱
❱❱❱❱
❱❱❱❱ ?
@E
We show that there is a unique morphism H → E making the diagram commute.
As Hpre is the cokernel in the category of presheaves, there is a unique morphismof presheaves Hpre → E making the diagram commute. But then by the universal
property of sheafification (Definition 2.4.6), there is a unique morphism of sheaves
H → E making the diagram commute.
!
2.6.B. E XERCISE . Show that the stalk of the cokernel is naturally isomorphic tothe cokernel of the stalk.

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We have now defined the notions of kernel and cokernel, and verified that they
may be checked at the level of stalks. We have also verified that the properties ofa morphism being a monomorphism or epimorphism are also determined at thelevel of stalks (Exercises 2.4.N and 2.4.O). Hence we have proved the following:
2.6.2. Theorem. — Sheaves of abelian groups on a topological space X form an abeliancategory.
That’s all there is to it — what needs to be proved has been shifted to the stalks,
where everything works because stalks are abelian groups!
And we see more: all structures coming from the abelian nature of this category may be checked at the level of stalks. For example:
2.6.C. E XERCISE . Suppose φ : F → G is a morphism of sheaves of abelian groups.
Show that the image sheaf im φ is the sheafification of the image presheaf. (Youmust use the definition of image in an abelian category. In fact, this gives theaccepted definition of image sheaf for a morphism of sheaves of sets.) Show thatthe stalk of the image is the image of the stalk.
As a consequence, exactness of a sequence of sheaves may be checked at thelevel of stalks. In particular:
2.6.D. I MPORTANT E XERCISE ( CF. E XERCISE 2.3.A). Show that taking the stalk ofa sheaf of abelian groups is an exact functor. More precisely, if X is a topologicalspace and p ∈ X is a point, show that taking the stalk at p defines an exact functor
AbX → Ab.
2.6.E. E XERCISE . Check that the exponential exact sequence (2.4.10.1) is exact.

2.6.F. E XERCISE : L EFT- EXACTNESS OF THE FUNCTOR OF “ SECTIONS OVER U”. Suppose U ⊂ X is an open set, and 0 → F → G → H is an exact sequence of sheavesof abelian groups. Show that
0 → F (U) → G (U) → H (U)is exact. (You should do this “by hand”, even if you realize there is a very fast
proof using the left-exactness of the “forgetful” right adjoint to the sheafification
functor.) Show that the section functor need not be exact: show that if 0 → F →G → H → 0 is an exact sequence of sheaves of abelian groups, then
0 → F (U) → G (U) → H (U) → 0

need not be exact. (Hint: the exponential exact sequence (2.4.10.1). But feel free tomake up a different example.)
2.6.G. E XERCISE : L EFT- EXACTNESS OF PUSHFORWARD . Suppose 0 → F → G →H is an exact sequence of sheaves of abelian groups on X. If π : X → Y is a continuous map, show that0 → π∗ F → π∗ G → π∗ H

is exact. (The previous exercise, dealing with the left-exactness of the global section functor can be interpreted as a special case of this, in the case where Y is apoint.)

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2.6.H. E XERCISE : L EFT- EXACTNESS OF Hom ( CF. E XERCISE 1.6.F( C ) AND ( D )). Suppose F is a sheaf of abelian groups on a topological space X. Show that Hom (F , ·)is a left-exact covariant functor AbX → AbX . Show that Hom (·, F ) is a left-exact
contravariant functor AbX → AbX .
2.6.3. OX -modules.

2.6.I. E XERCISE . Show that if (X, OX ) is a ringed space, then OX -modules form an
abelian category. (There is a fair bit to check, but there aren’t many new ideas.)
2.6.4. Many facts about sheaves of abelian groups carry over to OX -modules without change, because a sequence of OX -modules is exact if and only if the underlying sequence of sheaves of abelian groups is exact. You should be able to easily
check that all of the statements of the earlier exercises in §2.6 also hold for OX modules, when modified appropriately. For example (Exercise 2.6.H), Hom OX (·, ·)is a left-exact contravariant functor in its first argument and a left-exact covariant
functor in its second argument.
We end with a useful construction using some of the ideas in this section.
2.6.J. I MPORTANT E XERCISE : T ENSOR PRODUCTS OF OX - MODULES .
(a) Suppose OX is a sheaf of rings on X. Define (categorically) what we shouldmean by tensor product of two OX -modules. Give an explicit construction, andshow that it satisfies your categorical definition. Hint: take the “presheaf tensor
product” — which needs to be defined — and sheafify. Note: ⊗OX is often written⊗ when the subscript is clear from the context. (An example showing sheafification is necessary will arise in Example 14.1.1.)(b) Show that the tensor product of stalks is the stalk of the tensor product. (If you
can show this, you may be able to make sense of the phrase “colimits commute
with tensor products”.)
2.6.5. Conclusion. Just as presheaves are abelian categories because all abeliancategorical notions make sense open set by open set, sheaves are abelian categories
because all abelian-categorical notions make sense stalk by stalk.

2.7 The inverse image sheaf
We next describe a notion that is fundamental, but rather intricate. We will
not need it for some time, so this may be best left for a second reading. Suppose
we have a continuous map π : X → Y. If F is a sheaf on X, we have defined
the pushforward or direct image sheaf π∗ F , which is a sheaf on Y. There is also a
notion of inverse image sheaf. (We will not call it the pullback sheaf, reserving thatname for a later construction for quasicoherent sheaves, §16.3.) This is a covariant
functor π−1 from sheaves on Y to sheaves on X. If the sheaves on Y have some
additional structure (e.g., group or ring), then this structure is respected by π−1 .
2.7.1. Definition by adjoint: elegant but abstract. We define the inverse image π−1 asthe left adjoint to π∗ .
This isn’t really a definition; we need a construction to show that the adjoint
exists. Note that we then get canonical maps π−1 π∗ F → F (associated to the

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identity in MorY (π∗ F , π∗ F )) and G → π∗ π−1 G (associated to the identity in
MorX (π−1 G , π−1 G )).
%F
π−1 G
❦❦❦
③
❦
③
❦
❦
③③
❦❦❦
③③ ❦❦❦❦❦
③
❦
③ ❦
③
❦❦❦
% π∗ F
X
G
❦❦❦
②②
❦
❦
②
❦❦
②
π
②② ❦❦❦❦" ②②❦②❦❦❦❦Y❦

2.7.2. Construction: concrete but ugly. Define the temporary notation
π−1
pre G (U) =lim G (V).
−→V⊃π(U)(Recall the explicit description of colimit: sections of π−1
pre over U are sections on
open sets containing π(U), with an equivalence relation. Note that π(U) won’t be
an open set in general.)
2.7.A. E XERCISE . Show that this defines a presheaf on X. Show that it needn’t
form a sheaf. (Hint: map 2 points to 1 point.)sh−1
Now define the inverse image of G by π−1 G := (π−1pre G ) . Note that π
is a functor from sheaves on Y to sheaves on X. The next exercise shows that
π−1 is indeed left-adjoint to π∗ . But you may wish to try the later exercises first,
and come back to Exercise 2.7.B at another time. (For the later exercises, try togive two proofs, one using the universal property, and the other using the explicit
description.)

2.7.B. I MPORTANT T RICKY E XERCISE . If π : X → Y is a continuous map, and F isa sheaf on X and G is a sheaf on Y, describe a bijection
MorX (π−1 G , F ) ↔ MorY (G , π∗ F ).

Observe that your bijection is “natural” in the sense of the definition of adjoints(i.e., functorial in both F and G ). Thus Construction 2.7.2 satisfies the universal
property of Definition 2.7.1. Possible hint: Show that both sides agree with thefollowing third construction, which we denote MorYX (G , F ). A collection of maps
φVU : G (V) → F (U) (as U runs through all open sets of X, and V runs through allopen sets of Y containing π(U)) is said to be compatible if for all open U ′ ⊂ U ⊂ X
and all open V ′ ⊂ V ⊂ Y with π(U) ⊂ V, π(U ′ ) ⊂ V ′ , the diagram(2.7.2.1)

G (V)resV,V ′φVU% F (U)
resU,U ′""φV ′ U ′% F (U ′ )G (V ′ )commutes. Define MorYX (G , F ) to be the set of all compatible collections φ ={φVU }.

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2.7.3. Remark (“stalk and skyscraper are an adjoint pair”). As a special case, if X is apoint p ∈ Y, we see that π−1 G is the stalk Gp of G , and maps from the stalk Gp toa set S are the same as maps of sheaves on Y from G to the skyscraper sheaf with
set S supported at p. You may prefer to prove this special case by hand directly
before solving Exercise 2.7.B, as it is enlightening. (It can also be useful — can you
use it to solve Exercises 2.4.M and 2.4.O?)
2.7.C. E XERCISE . Show that the stalks of π−1 G are the same as the stalks of G .
∼ %More precisely, if π(p) = q, describe a natural isomorphism Gq(π−1 G )p .
(Possible hint: use the concrete description of the stalk, as a colimit. Recall thatstalks are preserved by sheafification, Exercise 2.4.M. Alternatively, use adjointness.) This, along with the notion of compatible germs, may give you a simple
way of thinking about (and perhaps visualizing) inverse image sheaves. (Those
preferring the “espace étalé” or “space of sections” perspective, §2.2.11, can checkthat the pullback of the “space of sections” is the “space of sections” of the pullback.)
2.7.D. E XERCISE ( EASY BUT USEFUL ). If U is an open subset of Y, i : U → Y is theinclusion, and G is a sheaf on Y, show that i−1 G is naturally isomorphic to G |U
(the restriction of G to U, §2.2.8).
2.7.E. E XERCISE . Show that π−1 is an exact functor from sheaves of abelian groups
on Y to sheaves of abelian groups on X (cf. Exercise 2.6.D). (Hint: exactness can bechecked on stalks, and by Exercise 2.7.C, the stalks are the same.) Essentially thesame argument will show that π−1 is an exact functor from OY -modules (on Y) to(π−1 OY )-modules (on X), but don’t bother writing that down. (Remark for experts:
π−1 is a left adjoint, hence right-exact by abstract nonsense, as discussed in §1.6.12.
Left-exactness holds because colimits of abelian groups over filtered index sets areexact, Exercise 1.6.K.)
2.7.F. E XERCISE .
(a) Suppose Z ⊂ Y is a closed subset, and i : Z "→ Y is the inclusion. If F is a sheafof sets on Z, then show that the stalk (i∗ F )q is a one element-set if q ∈/ Z, and Fq
if q ∈ Z.
(b) Definition (cf. Definition 2.4.2, the support of a section): Define the support of asheaf G of sets, denoted Supp G , as the locus where the stalks are not the oneelement set:
Supp G := {p ∈ X : |Gp | ̸= 1}.
(More generally, if the sheaf has value in some category, the support consists ofpoints where the stalk is not the final object. For a sheaf G of abelian groups,
the support consists of points with nonzero stalks — Supp G = {p ∈ X : Gp ̸=0} — or equivalently is the union of supports of sections over all open sets, see
Definition 2.4.2.) Suppose Supp G ⊂ Z where Z is closed. Show that the natural
map G → i∗ i−1 G is an isomorphism. Thus a sheaf supported on a closed subset
can be considered a sheaf on that closed subset. (“Support of a sheaf” is a useful
notion, and will arise again in §13.7.D.)

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The final exercise of this section may be best left for later, when you will realize
why you care about it. (We will start to use it in Chapter 23 — more precisely, in
Exercise 23.4.F.)
2.7.G. E XERCISE ( EXTENSION BY ZERO π! : AN OCCASIONAL left adjoint TO π−1 ).
In addition to always being a left adjoint, π−1 can sometimes be a right adjoint.
Suppose i : U "→ Y is an inclusion of an open set into Y. Define the extension of iby zero i! : ModOU → ModOY as follows. Suppose F is an OU -module. For openpreW ⊂ Y, define (i! F )(W) = F (W) if W ⊂ U, and 0 otherwise (with the obvious
prerestriction maps). This is clearly a presheaf OY -module. Define i! as (i! )sh . Notethat i! F is an OY -module, and that this defines a functor. (The symbol “!” is readas “shriek”. I have no idea why, but I suspect it is because people often shriek
when they see it. Thus “i! ” is read as “i-lower-shriek”.)pre(a) Show that i! F need not be a sheaf. (We won’t need this, but it may give someinsight into why this is called “extension by zero”. Possible source for an example:
continuous functions on R.)(b) For q ∈ Y, show that (i! F )q = Fq if q ∈ U, and 0 otherwise.
(c) Show that i! is an exact functor.
(d) If G is an OY -module, describe an inclusion i! i−1 G "→ G . (Interesting remark
we won’t need: Let Z be the complement of U, and j : Z → Y the natural inclusion.
Then there is a short exact sequence
0 → i! i−1 G → G → j∗ j−1 G → 0.

This is best checked by describing the maps, then checking exactness at stalks.)(e) Show that (i! , i−1 ) is an adjoint pair, so there is a natural bijection HomOY (i! F , G ) ↔
HomOU (F , G |U ) for any OU -module F and OY -module G . (In particular, the sections of G over U can be identified with HomOY (i! OU , G ).)Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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Part II

Schemes

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L’idée même de schéma est d’une simplicité enfantine — si simple, si humble, quepersonne avant moi n’avait songé à se pencher si bas. Si “bébête” même, pour tout dire,que pendant des années encore et en dépit de l’évidence, pour beaucoup de mes savantscollègues, ça faisait vraiment “pas sérieux”!
The very idea of scheme is of infantile simplicity — so simple, so humble, that no one
before me thought of stooping so low. So childish, in short, that for years, despite all theevidence, for many of my erudite colleagues, it was really “not serious”!
— A. Grothendieck [Gr6, p. P32], translated by C. McLarty [Mc, p. 313]

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CHAPTER 3

Toward affine schemes: the underlying set, andtopological space
There is no serious historical question of how Grothendieck found his definition ofschemes. It was in the air. Serre has well said that no one invented schemes... . The
question is, what made Grothendieck believe he should use this definition to simplify an 80
page paper by Serre into some 1000 pages of Élements de Géométrie Algébrique?
— C. McLarty [Mc, p. 313]

3.1 Toward schemes
We are now ready to consider the notion of a scheme, which is the type of geometricspace central to algebraic geometry. We should first think through what we meanby “geometric space”. You have likely seen the notion of a manifold, and we wish
to abstract this notion so that it can be generalized to other settings, notably so that
we can deal with nonsmooth and arithmetic objects.
The key insight behind this generalization is the following: we can understanda geometric space (such as a manifold) well by understanding the functions on
this space. More precisely, we will understand it through the sheaf of functions
on the space. If we are interested in differentiable manifolds, we will consider
differentiable functions; if we are interested in smooth manifolds, we will consider
smooth functions; and so on.
Thus we will define a scheme to be the following data
• The set: the points of the scheme• The topology: the open sets of the scheme• The structure sheaf: the sheaf of “algebraic functions” (a sheaf of rings) on
the scheme.
Recall that a topological space with a sheaf of rings is called a ringed space (§2.2.13).
We will try to draw pictures throughout. Pictures can help develop geometricintuition, which can guide the algebraic development (and, eventually, vice versa).
Some people find pictures very helpful, while others are repulsed or confused.
We will try to make all three notions as intuitive as possible. For the set, inthe key example of complex (affine) varieties (roughly, things cut out in Cn bypolynomials), we will see that the points are the “traditional points” (n-tuples ofcomplex numbers), plus some extra points that will be handy to have around. For
the topology, we will require that “the subset where an algebraic function vanishesmust be closed”, and require nothing else. For the sheaf of algebraic functions (thestructure sheaf), we will expect that in the complex plane, (3x2 +y2 )/(2x+4xy+1)
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should be an algebraic function on the open set consisting of points where thedenominator doesn’t vanish, and this will largely motivate our definition.
3.1.1. Example: Differentiable manifolds.
As motivation, we return to our
example of differentiable manifolds, reinterpreting them in this light. We will bequite informal in this discussion. Suppose X is a differentiable manifold. It is a
topological space, and has a sheaf of differentiable functions OX (see §2.1). This gives
X the structure of a ringed space. We have observed that evaluation at a pointp ∈ X gives a surjective map from the stalk to R
OX,p

% % R,

so the kernel, the (germs of) functions vanishing at p, is a maximal ideal mX,p (see§2.1.1).
We could define a differentiable real manifold as a topological space X with asheaf of rings. We would require that there is a cover of X by open sets such thaton each open set the ringed space is isomorphic to a ball around the origin in Rn(with the sheaf of differentiable functions on that ball). With this definition, theball is the basic patch, and a general manifold is obtained by gluing these patches
together. (Admittedly, a great deal of geometry comes from how one chooses topatch the balls together!) In the algebraic setting, the basic patch is the notion of anaffine scheme, which we will discuss soon. (In the definition of manifold, there is anadditional requirement that the topological space be Hausdorff, to avoid pathologies. Schemes are often required to be “separated” to avoid essentially the same
pathologies. Separatedness will be discussed in Chapter 10.)
Functions are determined by their values at points. This is an obvious statement,
but won’t be true for schemes in general. We will see an example in Exercise 3.2.A(a),and discuss this behavior further in §3.2.11.
Morphisms of manifolds. How can we describe differentiable maps of manifolds
π : X → Y? They are certainly continuous maps — but which ones? We can pullback functions along continuous maps. Differentiable functions pull back to differentiable functions. More formally, we have a map π−1 OY → OX . (The inverse
image sheaf π−1 was defined in §2.7.) Inverse image is left-adjoint to pushforward,
so we also get a map π♯ : OY → π∗ OX .
Certainly given a differentiable map of differentiable manifolds, differentiable
functions pull back to differentiable functions. It is less obvious that this is a sufficient condition for a continuous function to be differentiable.
3.1.A. I MPORTANT E XERCISE FOR THOSE WITH A LITTLE EXPERIENCE WITH MAN IFOLDS . Suppose that π : X → Y is a continuous map of differentiable manifolds(as topological spaces — not a priori differentiable). Show that π is differentiable
if differentiable functions pull back to differentiable functions, i.e., if pullback by
π gives a map OY → π∗ OX . (Hint: check this on small patches. Once you figureout what you are trying to show, you will realize that the result is immediate.)
3.1.B. E XERCISE .
Show that a morphism of differentiable manifolds π : X →Y with π(p) = q induces a morphism of stalks π♯ : OY,q → OX,p . Show thatπ♯ (mY,q ) ⊂ mX,p . In other words, if you pull back a function that vanishes at

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q, you get a function that vanishes at p — not a huge surprise. (In §6.3, we formalize this by saying that maps of differentiable manifolds are maps of locally ringed
spaces.)3.1.2. Aside. Here is a little more for experts: Notice that π induces a map on
tangent spaces (see Aside 2.1.2)(mX,p /m2X,p )∨ → (mY,q /m2Y,q )∨ .

This is the tangent map you would geometrically expect. Again, it is interestingthat the cotangent map mY,q /m2Y,q → mX,p /m2X,p is algebraically more natural than
the tangent map (there are no “duals”).
Experts are now free to try to interpret other differential-geometric information using only the map of topological spaces and map of sheaves. For example:
how can one check if π is a submersion of manifolds? How can one check if f is
an immersion? (We will see that the algebro-geometric version of these notions
are smooth morphism and unramified morphism; see Chapter 25, although they will
be defined earlier, in Definition 12.6.2 and §21.6, respectively.)
3.1.3. Side Remark. Manifolds are covered by disks that are all isomorphic. This
isn’t true for schemes (even for “smooth complex varieties”). There are examplesof two “smooth complex curves” (the algebraic version of Riemann surfaces) X
and Y so that no nonempty open subset of X is isomorphic to a nonempty open
subset of Y (see Exercise 6.5.K). And there is a Riemann surface X such that no two
open subsets of X are isomorphic (see Exercise 19.7.D). Informally, this is because
in the Zariski topology on schemes, all nonempty open sets are “huge” and havemore “structure”.
3.1.4. Other examples. If you are interested in differential geometry, you will beinterested in differentiable manifolds, on which the functions under consideration
are differentiable functions. Similarly, if you are interested in topology, you will beinterested in topological spaces, on which you will consider the continuous function. If you are interested in complex geometry, you will be interested in complex
manifolds (or possibly “complex analytic varieties”), on which the functions areholomorphic functions. In each of these cases of interesting “geometric spaces”,
the topological space and sheaf of functions is clear. The notion of scheme fitsnaturally into this family.

3.2 The underlying set of affine schemes
For any ring A, we are going to define something called Spec A, the spectrumof A. In this section, we will define it as a set, but we will soon endow it with a
topology, and later we will define a sheaf of rings on it (the structure sheaf). Suchan object is called an affine scheme. Later Spec A will denote the set along withthe topology, and a sheaf of functions. But for now, as there is no possibility ofconfusion, Spec A will just be the set.

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3.2.1. The set Spec A is the set of prime ideals of A. The prime ideal p of A when
considered as an element of Spec A will be denoted [p], to avoid confusion. Elements a ∈ A will be called functions on Spec A, and their value at the point [p]will be a (mod p). This is weird: a function can take values in different rings at different points — the function 5 on Spec Z takes the value 1 (mod 2) at [(2)] and 2 (mod 3)
at [(3)].
“An element a of the ring lying in a prime ideal p” translates to “a function a thatis 0 at the point [p]” or “a function a vanishing at the point [p]”, and we will use
these phrases interchangeably. Notice that if you add or multiply two functions,
you add or multiply their values at all points; this is a translation of the fact that
A → A/p is a ring morphism. These translations are important — make sure you
are very comfortable with them! They should become second nature.
If A is generated over a base field (or base ring) by elements x1 , . . . , xr , theelements x1 , . . . , xr are often called coordinates, because we will later be able toreinterpret them as restrictions of “coordinates on r-space”, via the idea of §3.2.9,
made precise in Exercise 6.2.D.
3.2.2. Glimpses of the future. In §4.1.3: we will interpret functions on Spec A as
global sections of the “structure sheaf”, i.e., as a function on a ringed space, in thesense of §2.2.13. We repeat a caution from §2.2.13: what we will call “functions”,
others may call “regular functions”. And we will later define “rational functions”(§5.5.6), which are not precisely functions in this sense; they are a particular typeof “partially-defined function”.
The notion of “value of a function” will be later interpreted as a value of afunction on a particular locally ringed space, see Definition 4.3.6.
3.2.3. We now give some examples.
Example 1 (the complex affine line): A1C := Spec C[x]. Let’s find the prime
ideals of C[x]. As C[x] is an integral domain, 0 is prime. Also, (x − a) is prime, for
any a ∈ C: it is even a maximal ideal, as the quotient by this ideal is a field:0% (x − a)% C[x]

f*→f(a)%C%0(This exact sequence may remind you of (2.1.1.1) in our motivating example ofmanifolds.)
We now show that there are no other prime ideals. We use the fact that C[x]has a division algorithm, and is a unique factorization domain. Suppose p is aprime ideal. If p ̸= (0), then suppose f(x) ∈ p is a nonzero element of smallest
degree. It is not constant, as prime ideals can’t contain 1. If f(x) is not linear,then factor f(x) = g(x)h(x), where g(x) and h(x) have positive degree. (Here weuse that C is algebraically closed.) Then g(x) ∈ p or h(x) ∈ p, contradicting theminimality of the degree of f. Hence there is a linear element x − a of p. Then I
claim that p = (x − a). Suppose f(x) ∈ p. Then the division algorithm would givef(x) = g(x)(x − a) + m where m ∈ C. Then m = f(x) − g(x)(x − a) ∈ p. If m ̸= 0,
then 1 ∈ p, giving a contradiction.
Thus we can and should (and must!) make a picture of A1C = Spec C[x] (seeFigure 3.1). This is just the first illustration of a point of view of Sophie Germain[Ge]: “L’algèbre n’est qu’une géométrie écrite; la géométrie n’est qu’une algèbre
figurée.” (Algebra is but written geometry; geometry is but drawn algebra.)

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There is one “traditional” point for each complex number, plus one extra (“bonus”)
point [(0)]. We can mostly picture A1C as C: the point [(x − a)] we will reasonably
associate to a ∈ C. Where should we picture the point [(0)]? The best way of thinking about it is somewhat zen. It is somewhere on the complex line, but nowhere inparticular. Because (0) is contained in all of these prime ideals, we will somehow
associate it with this line passing through all the other points. This new point [(0)]is called the “generic point” of the line. (We will formally define “generic point”
in §3.6.) It is “generically on the line” but you can’t pin it down any further thanthat. It is not at any particular place on the line. (This is misleading too — we will
see in Easy Exercise 3.6.N that it is “near” every point. So it is near everything, but
located nowhere precisely.) We will place it far to the right for lack of anywhere
better to put it. You will notice that we sketch A1C as one-(real-)dimensional (eventhough we picture it as an enhanced version of C); this is to later remind ourselvesthat this will be a one-dimensional space, where dimensions are defined in an algebraic (or complex-geometric) sense. (Dimension will be defined in Chapter 11.)

F IGURE 3.1. A picture of A1C = Spec C[x]To give you some feeling for this space, we make some statements that are
currently undefined, but suggestive. The functions on A1C are the polynomials. So
f(x) = x2 − 3x + 1 is a function. What is its value at [(x − 1)], which we think of as
the point 1 ∈ C? Answer: f(1)! Or equivalently, we can evaluate f(x) modulo x − 1— this is the same thing by the division algorithm. (What is its value at [(0)]? It isf(x) (mod 0), which is just f(x).)
Here is a more complicated example: g(x) = (x − 3)3 /(x − 2) is a “rational
function”. It is defined everywhere but x = 2. (When we know what the structure
sheaf is, we will be able to say that it is an element of the structure sheaf on the
open set A1C \ {2}.) We want to say that g(x) has a triple zero at 3, and a single pole
at 2, and we will be able to after §12.5.

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Example 2 (the affine line over k = k): A1k := Spec k[x] where k is an algebraically closed field. This is called the affine line over k. All of our discussion in
the previous example carries over without change. We will use the same picture,
which is after all intended to just be a metaphor.
Example 3: Spec Z. An amazing fact is that from our perspective, this will
look a lot like the affine line A1k . The integers, like k[x], form a unique factorization
domain, with a division algorithm. The prime ideals are: (0), and (p) where p
is prime. Thus everything from Example 1 carries over without change, even the
picture. Our picture of Spec Z is shown in Figure 3.2.

(2)

(3)

(5)

···

(0)

F IGURE 3.2. A “picture” of Spec Z, which looks suspiciously like Figure 3.1
Let’s blithely carry over our discussion of functions to this space. 100 is afunction on Spec Z. Its value at (3) is “1 (mod 3)”. Its value at (2) is “0 (mod 2)”,and in fact it has a double zero. 27/4 is a “rational function” on Spec Z, definedaway from (2). We want to say that it has a double pole at (2), and a triple zero at(3). Its value at (5) is
27 × 4−1 ≡ 2 × (−1) ≡ 3 (mod 5).

(We will gradually make this discussion precise over time.)
Example 4: silly but important examples, and the German word for bacon.
The set Spec k where k is any field is boring: one point. Spec 0, where 0 is thezero-ring, is the empty set, as 0 has no prime ideals.
3.2.A. A SMALL EXERCISE ABOUT SMALL SCHEMES .
(a) Describe the set Spec k[ϵ]/(ϵ2 ). The ring k[ϵ]/(ϵ2 ) is called the ring of dual
numbers, and will turn out to be quite useful. You should think of ϵ as a very small
number, so small that its square is 0 (although it itself is not 0). It is a nonzero function whose value at all points is zero, thus giving our first example of functions
not being determined by their values at points. We will discuss this phenomenon
further in §3.2.11.
(b) Describe the set Spec k[x](x) (see §1.3.3 for a discussion of localization). We
will see this scheme again repeatedly, starting with §3.2.8 and Exercise 3.4.K. Youmight later think of it as a shred of a particularly nice “smooth curve”.
In Example 2, we restricted to the case of algebraically closed fields for a reason: things are more subtle if the field is not algebraically closed.
Example 5 (the affine line over R): A1R = Spec R[x]. Using the fact that R[x] is a
Euclidean domain domain, similar arguments to those of Examples 1–3 show thatthe prime ideals are (0), (x − a) where a ∈ R, and (x2 + ax + b) where x2 + ax + b
is an irreducible quadratic. The latter two are maximal ideals, i.e., their quotients∼ R, R[x]/(x2 + 1) =∼ C.
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3.2.B. U NIMPORTANT E XERCISE . Show that for the last type of prime, of the form
(x2 + ax + b), the quotient is always isomorphic to C.
So we have the points that we would normally expect to see on the real line,
corresponding to real numbers; the generic point 0; and new points which we mayinterpret as conjugate pairs of complex numbers (the roots of the quadratic). This
last type of point should be seen as more akin to the real numbers than to thegeneric point. You can picture A1R as the complex plane, folded along the real axis.
But the key point is that Galois-conjugate points (such as i and −i) are considered
glued.
Let’s explore functions on this space. Consider the function f(x) = x3 − 1. Itsvalue at the point [(x − 2)] is f(x) = 7, or perhaps better, 7 (mod x − 2). How aboutat (x2 + 1)? We getx3 − 1 ≡ −x − 1 (mod x2 + 1),
which may be profitably interpreted as −i − 1.
One moral of this example is that we can work over a non-algebraically closed
field if we wish. It is more complicated, but we can recover much of the information we care about.
3.2.C. I MPORTANT E XERCISE . Describe the set A1Q . (This is harder to picture in a
way analogous to A1R . But the rough cartoon of points on a line, as in Figure 3.1,
remains a reasonable sketch.)
Example 6 (the affine line over Fp ): A1Fp = Spec Fp [x]. As in the previous
examples, Fp [x] is a Euclidean domain, so the prime ideals are of the form (0) or(f(x)) where f(x) ∈ Fp [x] is an irreducible polynomial, which can be of any degree.
Irreducible polynomials correspond to sets of Galois conjugates in Fp .
Note that Spec Fp [x] has p points corresponding to the elements of Fp , butalso many more (infinitely more, see Exercise 3.2.D). This makes this space muchricher than simply p points. For example, a polynomial f(x) is not determined by
its values at the p elements of Fp , but it is determined by its values at the points ofSpec Fp [x]. (As we have mentioned before, this is not true for all schemes.)
You should think about this, even if you are a geometric person — this intuition will later turn up in geometric situations. Even if you think you are interested
only in working over an algebraically closed field (such as C), you will have nonalgebraically closed fields (such as C(x)) forced upon you.
3.2.D. E XERCISE . If k is a field, show that Spec k[x] has infinitely many points.
(Hint: Euclid’s proof of the infinitude of primes of Z.)
Example 7 (the complex affine plane): A2C = Spec C[x, y]. (As with Examples 1
and 2, our discussion will apply with C replaced by any algebraically closed field.)
Sadly, C[x, y] is not a principal ideal domain: (x, y) is not a principal ideal. We
can quickly name some prime ideals. One is (0), which has the same flavor as the(0) ideals in the previous examples. (x − 2, y − 3) is prime, and indeed maximal,∼ C, where this isomorphism is via f(x, y) (→ f(2, 3).
because C[x, y]/(x − 2, y − 3) =More generally, (x − a, y − b) is prime for any (a, b) ∈ C2 . Also, if f(x, y) is anirreducible polynomial (e.g., y − x2 or y2 − x3 ) then (f(x, y)) is prime.

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3.2.E. E XERCISE . Show that we have identified all the prime ideals of C[x, y]. Hint:
Suppose p is a prime ideal that is not principal. Show you can find f(x, y), g(x, y) ∈p with no common factor. By considering the Euclidean algorithm in the Euclidean
domain C(x)[y], show that you can find a nonzero h(x) ∈ (f(x, y), g(x, y)) ⊂ p.
Using primality, show that one of the linear factors of h(x), say (x − a), is in p.
Similarly show there is some (y − b) ∈ p.

F IGURE 3.3. Picturing A2C = Spec C[x, y]
We now attempt to draw a picture of A2C (see Figure 3.3). The maximal prime
ideals of C[x, y] correspond to the traditional points in C2 : [(x − a, y − b)] corresponds to (a, b) ∈ C2 . We now have to visualize the “bonus points”. [(0)] somehow lives behind all of the traditional points; it is somewhere on the plane, but
nowhere in particular. So for example, it does not lie on the parabola y = x2 . The
point [(y − x2 )] lies on the parabola y = x2 , but nowhere in particular on it. (Figure 3.3 is a bit misleading. For example, the point [(0)] isn’t in the fourth quadrant;
it is somehow near every other point, which is why it is depicted as a somewhat
diffuse large dot.) You can see from this picture that we already are implicitly
thinking about “dimension”. The prime ideals (x − a, y − b) are somehow of dimension 0, the prime ideals (f(x, y)) are of dimension 1, and (0) is of dimension
2. (All of our dimensions here are complex or algebraic dimensions. The complex
plane C2 has real dimension 4, but complex dimension 2. Complex dimensions
are in general half of real dimensions.) We won’t define dimension precisely until
Chapter 11, but you should feel free to keep it in mind before then.
Note too that maximal ideals correspond to the “smallest” points. Smaller
ideals correspond to “bigger” points. “One prime ideal contains another” meansthat the points “have the opposite containment.” All of this will be made precise
once we have a topology. This order-reversal is a little confusing, and will remain
so even once we have made the notions precise.
We now come to the obvious generalization of Example 7.
Example 8 (complex affine n-space — important!): Let An
C := Spec C[x1 , . . . , xn ].
(Important definition: More generally, An
isdefinedtobeSpecA[x1 , . . . , xn ], whereA
A is an arbitrary ring. When the base ring is clear from context, the subscript A is

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often omitted. For pedants: the notation An
A implicitly includes the data of the n“coordinate functions” x1 , . . . , xn .) For concreteness, let’s consider n = 3. We now
have an interesting question in what at first appears to be pure algebra: What are
the prime ideals of C[x, y, z]?
Analogously to before, (x − a, y − b, z − c) is a prime ideal. This is a maximal
ideal, because its residue ring is a field (C); we think of these as “0-dimensional
points”. We will often write (a, b, c) for [(x − a, y − b, z − c)] because of our geometric interpretation of these ideals. There are no more maximal ideals, by Hilbert’s
Weak Nullstellensatz.
3.2.4. Hilbert’s Weak Nullstellensatz. — If k is an algebraically closed field, then the
maximal ideals of k[x1 , . . . , xn ] are precisely those ideals of the form (x1 − a1 , . . . , xn −an ), where ai ∈ k.
We may as well state a slightly stronger version now.

3.2.5. Hilbert’s Nullstellensatz. — If k is any field, every maximal ideal of k[x1 , . . . , xn ]has residue field a finite extension of k. Translation: any field extension of k that is finitely
generated as a ring is necessarily also finitely generated as a module (i.e., is a finite extension of fields).
This statement is also often called Zariski’s Lemma.
3.2.F. E XERCISE . Show that the Nullstellensatz 3.2.5 implies the Weak Nullstellensatz 3.2.4.
We will prove the Nullstellensatz in §7.4.3, and again in Exercise 11.2.B.
The following fact is a useful accompaniment to the Nullstellensatz.
3.2.G. E XERCISE ( NOT REQUIRING THE N ULLSTELLENSATZ ). Any integral domain A which is a finite k-algebra (i.e., a k-algebra that is a finite-dimensional vector space over k) must be a field. Hint: for any nonzero x ∈ A, show ×x : A → A
is an isomorphism. (Thus, in combination the Nullstellensatz 3.2.5, we see thatprime ideals of k[x1 , . . . , xn ] with finite residue ring are the same as maximal ideals ofk[x1 , . . . , xn ]. This is worth remembering.)
There are other prime ideals of C[x, y, z] too. We have (0), which corresponds
to a “3-dimensional point”. We have (f(x, y, z)), where f is irreducible. To this
we associate the “hypersurface” f = 0, so this is “2-dimensional” in nature. But
we have not found them all! One clue: we have prime ideals of “dimension” 0,2, and 3 — we are missing “dimension 1”. Here is one such prime ideal: (x, y).
We picture this as the locus where x = y = 0, which is the z-axis. This is a prime∼ C[z] is an integral domain
ideal, as the corresponding quotient C[x, y, z]/(x, y) =(and should be interpreted as the functions on the z-axis). There are lots of “1dimensional prime ideals”, and it is not possible to classify them in a reasonable
way. It will turn out that they correspond to things that we think of as irreducible
curves. Thus remarkably the answer to the purely algebraic question (“what are
the prime ideals of C[x, y, z]”) is fundamentally geometric!
The fact that the points of A1Q corresponding to maximal ideals of the ring Q[x](what we will soon call “closed points”, see Definition 3.6.8) can be interpreted as
points of Q where Galois-conjugates are glued together (Exercise 3.2.C) extends to

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√ √
√
√
√
√
2
An
Q . For example, in AQ , ( 2, 2) is glued to (− 2, − 2) but not to ( 2, − 2).
The following exercise will give you some idea of how this works.
√ √3.2.H. E√XERCISE.
Describethemaximal
ideal
of
Q[x,
y]
correspondingto(√√ 2, √2)and (− √2, −√ 2). Describe the maximal ideal of Q[x, y] corresponding to ( 2, − 2)and (− 2, 2). What are the residue fields in each case?
The description of “closed points” of A2Q (those points corresponding to maxi2

mal ideals of the ring Q[x, y]) as Galois-orbits of points in Q can even be extended
to other “nonclosed” points, as follows.
3.2.I. U NIMPORTANT AND TRICKY BUT FUN EXERCISE . Consider the map of sets
φ : C2 → A2Q defined as follows. (z1 , z2 ) is sent to the prime ideal of Q[x, y] consisting of polynomials vanishing at (z1 , z2 ).
(a) What is the image of (π, π2 )?
⋆ (b) Show that φ is surjective. (Warning: You will need some ideas we haven’t
discussed in order to solve this. Once we define the Zariski topology on A2Q , youwill be able to check that φ is continuous, where we give C2 the classical topology.
This example generalizes. For example, you may later be able to generalize this toarbitrary dimension.)
3.2.6. Quotients and localizations. Two natural ways of getting new rings fromold — quotients and localizations — have interpretations in terms of spectra.
3.2.7. Quotients: Spec A/I as a subset of Spec A. It is an important fact that theprime ideals of A/I are in bijection with the prime ideals of A containing I.
3.2.J. E SSENTIAL ALGEBRA EXERCISE ( MANDATORY IF YOU HAVEN ’ T SEEN IT BE FORE ). Suppose A is a ring, and I an ideal of A. Let φ : A → A/I. Show that φ−1
gives an inclusion-preserving bijection between prime ideals of A/I and prime
ideals of A containing I. Thus we can picture Spec A/I as a subset of Spec A.
As an important motivational special case, you now have a picture of affine
complex varieties. Suppose A is a finitely generated C-algebra, generated by x1 , . . . ,xn , with relations f1 (x1 , . . . , xn ) = · · · = fr (x1 , . . . , xn ) = 0. Then this descriptionin terms of generators and relations naturally gives us an interpretation of Spec A
as a subset of An
C , which we think of as “traditional points” (n-tuples of complex
numbers) along with some “bonus” points we haven’t yet fully described. To see
which of the traditional points are in Spec A, we simply solve the equations f1 =· · · = fr = 0. For example, Spec C[x, y, z]/(x2 +y2 −z2 ) may be pictured as shown in
Figure 3.4. (Admittedly this is just a “sketch of the R-points”, but we will still findit helpful later.) This entire picture carries over (along with the Nullstellensatz)with C replaced by any algebraically closed field. Indeed, the picture of Figure 3.4
can be said to depict k[x, y, z]/(x2 + y2 − z2 ) for most algebraically closed fields k(although it is misleading in characteristic 2, because of the coincidence x2 + y2 −z2 = (x + y + z)2 ).
3.2.8. Localizations: Spec S−1 A as a subset of Spec A. The following exercise shows
how prime ideals behave under localization.

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F IGURE 3.4. A “picture” of Spec C[x, y, z]/(x2 + y2 − z2 )3.2.K. E SSENTIAL ALGEBRA EXERCISE ( MANDATORY IF YOU HAVEN ’ T SEEN IT BE FORE ). Suppose S is a multiplicative subset of A. Describe an order-preserving
bijection of the prime ideals of S−1 A with the prime ideals of A that don’t meet themultiplicative set S.
Recall from §1.3.3 that there are two important flavors of localization. The
first is Af = {1, f, f2 , . . . }−1 A where f ∈ A. A motivating example is A = C[x, y],
f = y − x2 . The second is Ap = (A − p)−1 A, where p is a prime ideal. A motivating
example is A = C[x, y], S = A − (x, y).
If S = {1, f, f2 , . . . }, the prime ideals of S−1 A are just those prime ideals not containing f — the points where “f doesn’t vanish”. (In §3.5, we will call this a distinguished open set, once we know what open sets are.) So to picture Spec C[x, y]y−x2 ,we picture the affine plane, and throw out those points on the parabola y − x2 —the points (a, a2 ) for a ∈ C (by which we mean [(x − a, y − a2 )]), as well as the“new kind of point” [(y − x2 )].
It can be initially confusing to think about localization in the case where zerodivisors are inverted, because localization A → S−1 A is not injective (Exercise 1.3.C).
Geometric intuition can help. Consider the case A = C[x, y]/(xy) and f = x. Whatis the localization Af ? The space Spec C[x, y]/(xy) “is” the union of the two axesin the plane. Localizing means throwing out the locus where x vanishes. So weare left with the x-axis, minus the origin, so we expect Spec C[x]x . So there should
be some natural isomorphism(C[x, y]/(xy))x∼% C[x]x .

3.2.L. E XERCISE . Show that these two rings are isomorphic. (You will see that yon the left goes to 0 on the right.)
If S = A−p, the prime ideals of S−1 A are just the prime ideals of A contained inp. In our example A = C[x, y], p = (x, y), we keep all those points corresponding to“things through the origin”, i.e., the 0-dimensional point (x, y), the 2-dimensional
point (0), and those 1-dimensional points (f(x, y)) where f(0, 0) = 0, i.e., those“irreducible curves through the origin”. You can think of this being a shred of theplane near the origin; anything not actually “visible” at the origin is discarded (seeFigure 3.5).

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Spec C[x, y](x,y)

F IGURE 3.5. Picturing Spec C[x, y](x,y) as a “shred of A2C ”. Only
those points near the origin remain.

Another example is when A = k[x], and p = (x) (or more generally when p isany maximal ideal). Then Ap has only two prime ideals (Exercise 3.2.A(b)). You
should see this as the germ of a “smooth curve”, where one point is the “classicalpoint”, and the other is the “generic point of the curve”. This is an example of adiscrete valuation ring, and indeed all discrete valuation rings should be visualized in such a way. We will discuss discrete valuation rings in §12.5. By then we
will have justified the use of the words “smooth” and “curve”. (Reality check: try
to picture Spec of Z localized at (2) and at (0). How do the two pictures differ?)
3.2.9. Important fact: Maps of rings induce maps of spectra (as sets). We now
make an observation that will later grow up to be the notion of morphisms ofschemes.
3.2.M. I MPORTANT EASY EXERCISE . If φ : B → A is a map of rings, and p is aprime ideal of A, show that φ−1 (p) is a prime ideal of B.

Hence a map of rings φ : B → A induces a map of sets Spec A → Spec B “inthe opposite direction”. This gives a contravariant functor from the category ofrings to the category of sets: the composition of two maps of rings induces thecomposition of the corresponding maps of spectra.

3.2.N. E ASY E XERCISE ( REALITY CHECK ). Let B be a ring.
(a) Suppose I ⊂ B is an ideal. Show that the map Spec B/I → Spec B is the inclusion of §3.2.7.
(b) Suppose S ⊂ B is a multiplicative set. Show that the map Spec S−1 B → Spec B
is the inclusion of §3.2.8.

3.2.10. An explicit example. In the case of “affine complex varieties” (or indeed
affine varieties over any algebraically closed field), the translation between maps
given by explicit formulas and maps of rings is quite direct. For example, consider
a map from the parabola in C2 (with coordinates a and b) given by b = a2 , to the“curve” in C3 (with coordinates x, y, and z) cut out by the equations y = x2 andz = y2 . Suppose the map sends the point (a, b) ∈ C2 to the point (a, b, b2 ) ∈ C3 .

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In our new language, we have a map
Spec C[a, b]/(b − a2 )% Spec C[x, y, z]/(y − x2 , z − y2 )given byC[a, b]/(b − a2 ) (C[x, y, z]/(y − x2 , z − y2 )(a, b, b2 ) (✤ (x, y, z),i.e., x (→ a, y (→ b, and z (→ b2 . If the idea is not yet clear, the following two
exercises are very much worth doing — they can be very confusing the first time
you see them, and very enlightening (and finally, trivial) when you finally figure
them out.

y-linex-lineF IGURE 3.6. The map C → C given by x (→ y = x2

3.2.O. I MPORTANT E XERCISE ( SPECIAL CASE ).
Consider the map of complex
manifolds sending C → C via x (→ y = x2 . We interpret the “source” C as the“x-line”, and the “target” C the “y-line”. You can picture it as the projection of
the parabola y = x2 in the xy-plane to the y-axis (see Figure 3.6). Interpret thecorresponding map of rings as given by C[y] → C[x] by y (→√x2 . Verify that thepreimage (the fiber) above the point a ∈ C is the point(s) ± a ∈ C, using thedefinition given above. (A more sophisticated version of this example appears in
Example 9.3.3. Warning: the roles of x and y are swapped there, in order to picture
double covers in a certain way.)
3.2.P. I MPORTANT E XERCISE ( GENERALIZING E XAMPLE 3.2.10). Suppose k is afield, and f1 , . . . , fn ∈ k[x1 , . . . , xm ] are given. Let φ : k[y1 , . . . , yn ] → k[x1 , . . . , xm ]be the ring morphism defined by yi (→ fi .
(a) Show that φ induces a map of sets Spec k[x1 , . . . , xm ]/I → Spec k[y1 , . . . , yn ]/J
for any ideals I ⊂ k[x1 , . . . , xm ] and J ⊂ k[y1 , . . . , yn ] such that φ(J) ⊂ I. (You may
wish to consider the case I = 0 and J = 0 first. In fact, part (a) has nothing to dowith k-algebras; you may wish to prove the statement when the rings k[x1 , . . . , xm ]and k[y1 , . . . , yn ] are replaced by general rings A and B.)

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The Rising Sea: Foundations of Algebraic Geometry(b) Show that the map of part (a) sends the point (a1 , . . . , am ) ∈ km (or more
precisely, [(x1 − a1 , . . . , xm − am )] ∈ Spec k[x1 , . . . , xm ]) to(f1 (a1 , . . . , am ), . . . , fn (a1 , . . . , am )) ∈ kn .
3.2.Q. E XERCISE : P ICTURING An
Consider the map of sets π : AnZ.
Z → Spec Z,
given by the ring map Z → Z[x1 , . . . , xn ]. If p is prime, describe a bijection between
the fiber π−1 ([(p)]) and AFnp . (You won’t need to describe either set! Which is good
because you can’t.) This exercise may give you a sense of how to picture mapsn(see Figure 3.7), and in particular why you can think of An
Z as an “A -bundle”
nover Spec Z. (Can you interpret the fiber over [(0)] as Ak for some field k?)n
F IGURE 3.7. A picture of An
Z → Spec Z as a “family of A ’s”, ornan “A -bundle over Spec Z”. What is the field k? How should
you “geometrically” think of the three points indicated?

3.2.11. Functions are not determined by their values at points: the fault of nilpotents. We conclude this section by describing some strange behavior. We are developing machinery that will let us bring our geometric intuition to algebra. Thereis one serious serious point where your intuition will be false, so you should know
now, and adjust your intuition appropriately. As noted by Mumford ([Mu2, p. 12]),“it is this aspect of schemes which was most scandalous when Grothendieck defined them.”
Suppose we have a function (ring element) vanishing at all points. Then it is
not necessarily the zero function! The translation of this question is: is the intersection of all prime ideals necessarily just 0? The answer is no, as is shown by theexample of the ring of dual numbers k[ϵ]/(ϵ2 ): ϵ ̸= 0, but ϵ2 = 0. (We saw thisring in Exercise 3.2.A(a).) Any function whose power is zero certainly lies in theintersection of all prime ideals.

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113

3.2.R. E XERCISE . Ring elements that have a power that is 0 are called nilpotents.
(a) Show that if I is an ideal of nilpotents, then the inclusion Spec B/I → Spec B
of Exercise 3.2.J is a bijection. Thus nilpotents don’t affect the underlying set. (Wewill soon see in §3.4.5 that they won’t affect the topology either — the difference
will be in the structure sheaf.)(b) Show that the nilpotents of a ring B form an ideal. This ideal is called thenilradical, and is denoted N = N(B).
Thus the nilradical is contained in the intersection of all the prime ideals. The
converse is also true:
3.2.12. Theorem. — The nilradical N(A) is the intersection of all the prime ideals of A.
Geometrically: a function on Spec A vanishes at every point if and only if it is nilpotent.
3.2.S. E XERCISE . If you don’t know this theorem, then look it up, or better yet,prove it yourself. (Hint: Use the fact that any proper ideal of A is contained ina maximal ideal, which requires Zorn’s Lemma. Possible further hint: Supposex∈/ N(A). We wish to show that there is a prime ideal not containing x. Show that
Ax is not the 0-ring, by showing that 1 ̸= 0.)
3.2.13. In particular, although it is upsetting that functions are not determined bytheir values at points, we have precisely specified what the failure of this intuition
is: two functions have the same values at points if and only if they differ by anilpotent. You should think of this geometrically: a function vanishes at every
point of the spectrum of a ring if and only if it has a power that is zero. And if
there are no nonzero nilpotents — if N = (0) — then functions are determined bytheir values at points. If a ring has no nonzero nilpotents, we say that it is reduced.
3.2.T. F UN UNIMPORTANT EXERCISE : DERIVATIVES WITHOUT DELTAS AND EPSILONS( OR AT LEAST WITHOUT DELTAS ). Suppose we have a polynomial f(x) ∈ k[x]. Instead, we work in k[x, ϵ]/(ϵ2 ). What then is f(x + ϵ)? (Do a couple of examples,
then prove the pattern you observe.) This is a hint that nilpotents will be importantin defining differential information (Chapter 21).

3.3 Visualizing schemes I: generic points
A heavy warning used to be given that pictures are not rigorous; this has never had
its bluff called and has permanently frightened its victims into playing for safety. Some
pictures, of course, are not rigorous, but I should say most are (and I use them wheneverpossible myself).
— J. E. Littlewood, [Lit, p. 54]
For years, you have been able to picture x2 + y2 = 1 in the plane, and you
now have an idea of how to picture Spec Z. If we are claiming to understand rings
as geometric objects (through the Spec functor), then we should wish to develop
geometric insight into them. To develop geometric intuition about schemes, it is
helpful to have pictures in your mind, extending your intuition about geometricspaces you are already familiar with. As we go along, we will empirically develop

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some idea of what schemes should look like. This section summarizes what wehave gleaned so far.
Some mathematicians prefer to think completely algebraically, and never thinkin terms of pictures. Others will be disturbed by the fact that this is an art, not a science. And finally, this hand-waving will necessarily never be used in the rigorous
development of the theory. For these reasons, you may wish to skip these sections.
However, having the right picture in your mind can greatly help understanding
what facts should be true, and how to prove them.
Our starting point is the example of “affine complex varieties” (things cut out
by equations involving a finite number variables over C), and more generally similar examples over arbitrary algebraically closed fields. We begin with notions thatare intuitive (“traditional” points behaving the way you expect them to), and thenadd in the two features which are new and disturbing, generic points and nonreduced behavior. You can then extend this notion to seemingly different spaces,such as Spec Z.
Hilbert’s Weak Nullstellensatz 3.2.4 shows that the “traditional points” arepresent as points of the scheme, and this carries over to any algebraically closed
field. If the field is not algebraically closed, the traditional points are glued together into clumps by Galois conjugation, as in Examples 5 (the real affine line)
and 6 (the affine line over Fp ) in §3.2. This is a geometric interpretation of Hilbert’s
Nullstellensatz 3.2.5.
But we have some additional points to add to the picture. You should remember that they “correspond” to “irreducible” “closed” (algebraic) subsets. As
motivation, consider the case of the complex affine plane (Example 7): we had
one for each irreducible polynomial, plus one corresponding to the entire plane.
We will make “closed” precise when we define the Zariski topology (in the next
section). You may already have an idea of what “irreducible” should mean; wemake that precise at the start of §3.6. By “correspond” we mean that each closed
irreducible subset has a corresponding point sitting on it, called its generic point(defined in §3.6). It is a new point, distinct from all the other points in the subset.
(The correspondence is described in Exercise 3.7.E for Spec A, and in Exercise 5.1.B
for schemes in general.) We don’t know precisely where to draw the generic point,
so we may stick it arbitrarily anywhere, but you should think of it as being “almost
everywhere”, and in particular, near every other point in the subset.
In §3.2.7, we saw how the points of Spec A/I should be interpreted as a subsetof Spec A. So for example, when you see Spec C[x, y]/(x + y), you should picture
this not just as a line, but as a line in the xy-plane; the choice of generators x and yof the algebra C[x, y] implies an inclusion into affine space.
In §3.2.8, we saw how the points of Spec S−1 A should be interpreted as subsetsof Spec A. The two most important cases were discussed. The points of Spec Af
correspond to the points of Spec A where f doesn’t vanish; we will later (§3.5)
interpret this as a distinguished open set.
If p is a prime ideal, then Spec Ap should be seen as a “shred of the space
Spec A near the subset corresponding to p”. The simplest nontrivial case of thisis p = (x) ⊂ Spec k[x] = A (see Exercise 3.2.A, which we discuss again in Exercise 3.4.K).

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115“If any one of them can explain it,” said Alice, (she had grown so large in the last few
minutes that she wasn’t a bit afraid of interrupting him), “I’ll give him sixpence. I don’t
believe there’s an atom of meaning in it.” ...
“If there’s no meaning in it,” said the King, “that saves a world of trouble, you know,
as we needn’t try to find any.”— Lewis Carroll [Carr, Ch. XII]

3.4 The underlying topological space of an affine scheme
We next introduce the Zariski topology on the spectrum of a ring. When you
first hear the definition, it seems odd, but with a little experience it becomes reasonable. As motivation, consider A2C = Spec C[x, y], the complex plane (with a few
extra points). In algebraic geometry, we will only be allowed to consider algebraic
functions, i.e., polynomials in x and y. The locus where a polynomial vanishes
should reasonably be a closed set, and the Zariski topology is defined by sayingthat the only sets we should consider closed should be these sets, and other sets
forced to be closed by these. In other words, it is the coarsest topology where thesesets are closed.
In particular, although topologies are often described using open subsets, itwill be more convenient for us to define this topology in terms of closed subsets.
If S is a subset of a ring A, define the Vanishing set of S by
V(S) := {[p] ∈ Spec A : S ⊂ p}.

It is the set of points on which all elements of S are zero. (It should now be second
nature to equate “vanishing at a point” with “contained in a prime”.) We declarethat these — and no others — are the closed subsets.
For example, consider V(xy, yz) ⊂ A3C = Spec C[x, y, z]. Which points are contained in this locus? We think of this as solving xy = yz = 0. Of the “traditional”
points (interpreted as ordered triples of complex numbers, thanks to the Hilbert’s
Nullstellensatz 3.2.4), we have the points where y = 0 or x = z = 0: the xz-plane
and the y-axis respectively. Of the “new” points, we have the generic point of thexz-plane (also known as the point [(y)]), and the generic point of the y-axis (alsoknown as the point [(x, z)]). You might imagine that we also have a number of“one-dimensional” points contained in the xz-plane.
3.4.A. E ASY E XERCISE . Check that the x-axis is contained in V(xy, yz). (The x-axis
is defined by y = z = 0, and the y-axis and z-axis are defined analogously.)
Let’s return to the general situation. The following exercise lets us restrict
attention to vanishing sets of ideals.
3.4.B. E ASY E XERCISE . Show that if (S) is the ideal generated by S, then V(S) =V((S)).
We define the Zariski topology by declaring that V(S) is closed for all S. Let’scheck that this is a topology:
3.4.C. E XERCISE .
(a) Show that ∅ and Spec A are both open subsets of Spec A.

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The Rising Sea: Foundations of Algebraic Geometry(b),
If Ii is a collection of ideals (as i runs over some index set), show that ∩i V(Ii ) =V( i Ii ). Hence the union of any collection of open sets is open.
(c) Show that V(I1 ) ∪ V(I2 ) = V(I1 I2 ). (The product of two ideals I1 and I2 of A
are finite A-linear,n combinations of products of elements of I1 and I2 , i.e., elementsof the form j=1 i1,j i2,j , where ik,j ∈ Ik . Equivalently, it is the ideal generated byproducts of elements of I1 and I2 . You should quickly check that this is an ideal,
and that products are associative, i.e., (I1 I2 )I3 = I1 (I2 I3 ).) Hence the intersectionof any finite number of open sets is open.
3.4.1. Properties of the “vanishing set” function V(·). The function V(·) is obviously inclusion-reversing: If S1 ⊂ S2 , then V(S2 ) ⊂ V(S1 ). Warning: We could
have equality in the second inclusion without equality in the first, as the next exercise shows.
3.4.D. E XERCISE /D EFINITION . If I ⊂ A is an ideal, then define its radical by√

I := {r ∈ A : rn ∈ I for some n ∈ Z>0 }.

&√For example, the nilradical N(§3.2.R)
is
(0).
Show
that
I is an ideal (cf. Exer√cise 3.2.R(b)). Show that&V( I) = V(I). We say an ideal is radical if it equals its√√
own radical. Show thatI = I, and that prime ideals are radical.

Here are two useful consequences. As (I ∩ J)2 ⊂ IJ ⊂ I ∩ J (products of ideals
were defined in Exercise 3.4.C), we have that V(IJ) = V(I ∩ J) (= V(I) ∪ V(J)by Exercise &3.4.C(c)). Also, combining this with Exercise 3.4.B, we see V(S) =V((S)) = V( (S)).

3.4.E. E XERCISE ( RADICALS COMMUTE& WITH FINITE√INTERSECTIONS ). If I1 , . . . ,n
In are ideals of a ring A, show that ∩nI=∩
i=1 Ii . We will use this property
i=1 i
repeatedly without referring back to this exercise.
√
3.4.F. E XERCISE FOR LATER USE . Show that I is the intersection of all the prime
ideals containing I. (Hint: Use Theorem 3.2.12 on an appropriate ring.)

3.4.2. Examples. Let’s see how this meshes with our examples from the previous
section.
Recall that A1C , as a set, was just the “traditional” points (corresponding tomaximal ideals, in bijection with a ∈ C), and one “new” point [(0)]. The Zariski
topology on A1C is not that exciting: the open sets are the empty set, and A1C minus a
finite number of maximal ideals. (It “almost” has the cofinite topology. Notice thatthe open sets are determined by their intersections with the “traditional points”.
The “new” point [(0)] comes along for the ride, which is a good sign that it isharmless. Ignoring the “new” point, observe that the topology on A1C is a coarser
topology than the classical topology on C.)
3.4.G. E XERCISE . Describe the topological space A1k (cf. Exercise 3.2.D). (Noticethat the strange new point [(0)] is “near every other point” — every neighborhoodof every point contains [(0)]. This is typical of these new points, see Easy Exercise 3.6.N.)

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The case Spec Z is similar. The topology is “almost” the cofinite topology inthe same way. The open sets are the empty set, and Spec Z minus a finite numberof “ordinary” ((p) where p is prime) primes.
3.4.3. Closed subsets of A2C . The case A2C is more interesting. You should think
through where the “one-dimensional prime ideals” fit into the picture. In Exercise 3.2.E, we identified all the prime ideals of C[x, y] (i.e., the points of A2C ) as the
maximal ideals [(x − a, y − b)] (where a, b ∈ C — “zero-dimensional points”), the“one-dimensional points” [(f(x, y))] (where f(x, y) is irreducible), and the “twodimensional point” [(0)].
Then the closed subsets are of the following form:(a) the entire space (the closure of the “two-dimensional point” [(0)]), and(b) a finite number (possibly none) of “curves” (each the closure of a “onedimensional point” — the “one-dimensional point” along with the “zero-dimensional
points” “lying on it”) and a finite number (possibly none) of “zero-dimensional”
points (what we will soon call “closed points”, see Definition 3.6.8).
We will soon know enough to verify this using general theory, but you canprove it yourself now, using ideas in Exercise 3.2.E. (The key idea: if f(x, y) andg(x, y) are irreducible polynomials that are not multiples of each other, why dotheir zero sets intersect in a finite number of points?)
3.4.4. Important fact: Maps of rings induce continuous maps of topologicalspaces. We saw in §3.2.9 that a map of rings φ : B → A induces a map of sets
π : Spec A → Spec B.
3.4.H. I MPORTANT EASY EXERCISE .
By showing that closed sets pull back toclosed sets, show that π is a continuous map. Interpret Spec as a contravariant
functor Rings → Top.

Not all continuous maps arise in this way. Consider for example the continuous map on A1C that is the identity except 0 and 1 (i.e., [(x)] and [(x − 1)]) areswapped; no polynomial can manage this marvellous feat.
In §3.2.9, we saw that Spec B/I and Spec S−1 B are naturally subsets of Spec B.
It is natural to ask if the Zariski topology behaves well with respect to these inclusions, and indeed it does.
3.4.I. I MPORTANT E XERCISE ( CF. E XERCISE 3.2.N). Suppose that I, S ⊂ B are an
ideal and multiplicative subset respectively.
(a) Show that Spec B/I is naturally a closed subset of Spec B. If S = {1, f, f2 , . . . }(f ∈ B), show that Spec S−1 B is naturally an open subset of Spec B. Show that forarbitrary S, Spec S−1 B need not be open or closed. (Hint: Spec Q ⊂ Spec Z, or
possibly Figure 3.5.)(b) Show that the Zariski topology on Spec B/I (resp. Spec S−1 B) is the subspace
topology induced by inclusion in Spec B. (Hint: compare closed subsets.)
3.4.5. In particular, if I ⊂ N is an ideal of nilpotents, the bijection Spec B/I →Spec B (Exercise 3.2.R) is a homeomorphism. Thus nilpotents don’t affect the topological space. (The difference will be in the structure sheaf.)

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3.4.J. U SEFUL EXERCISE FOR LATER√ . Suppose I ⊂ B is an ideal. Show that f vanishes on V(I) if and only if f ∈ I (i.e., fn ∈ I for some n ≥ 1). (Hint: Exercise 3.4.F.
If you are stuck, you will get another hint when you see Exercise 3.5.E.)
3.4.K. E ASY E XERCISE ( CF. E XERCISE 3.2.A). Describe the topological space Spec k[x](x) .

3.5 A base of the Zariski topology on Spec A: Distinguished opensets
If f ∈ A, define the distinguished open set
D(f) :=={[p] ∈ Spec A : f ∈/ p}{[p] ∈ Spec A : f([p]) ̸= 0}.

It is the locus where f doesn’t vanish. (I often privately write this as D(f ̸= 0)
to remind myself of this. I also privately call this the “Doesn’t-vanish set” of f
in analogy with V(f) being the Vanishing set of f.) We have already seen this set
when discussing Spec Af as a subset of Spec A. For example, we have observedthat the Zariski-topology on the distinguished open set D(f) ⊂ Spec A coincides
with the Zariski topology on Spec Af (Exercise 3.4.I).
The reason these sets are important is that they form a particularly nice basefor the (Zariski) topology:
3.5.A. E ASY E XERCISE . Show that the distinguished open sets form a base for the(Zariski) topology. (Hint: Given a subset S ⊂ A, show that the complement of
V(S) is ∪f∈S D(f).)
Here are some important but not difficult exercises to give you a feel for thisconcept.

3.5.B. E XERCISE . Suppose fi ∈ A as i runs over some index set J. Show that∪i∈J D(fi ) = Spec A if and only if ({fi }i∈J ) = A, or equivalently
and very usefully,,
if there are ai (i ∈ J), all but finitely many 0, such that i∈J ai fi = 1. (One of thedirections will use the fact that any proper ideal of A is contained in some maximal
ideal.)
3.5.C. E XERCISE . Show that if Spec A is an infinite union of distinguished open
sets ∪j∈J D(fj ), then in fact it is a union of a finite number of these, i.e., there is a
finite subset J ′ so that Spec A = ∪j∈J ′ D(fj ). (Hint: Exercise 3.5.B.)
3.5.D. E ASY E XERCISE . Show that D(f) ∩ D(g) = D(fg).
3.5.E. I MPORTANT E XERCISE ( CF. E XERCISE 3.4.J). Show that D(f) ⊂ D(g) if andonly if fn ∈ (g) for some n ≥ 1, if and only if g is an invertible element of Af .

We will use Exercise 3.5.E often. You can solve it thinking purely algebraically,
but the following geometric interpretation may be helpful. (You should try to
draw your own picture to go with this discussion.) Inside Spec A, we have theclosed subset V(g) = Spec A/(g), where g vanishes, and its complement D(g),where g doesn’t vanish. Then f is a function on this closed subset V(g) (or more

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precisely, on Spec A/(g)), and by assumption it vanishes at all points of the closed
subset. Now any function vanishing at every point of the spectrum of a ring mustbe nilpotent (Theorem 3.2.12). In other words, there is some n such that fn = 0 in
A/(g), i.e., fn ≡ 0 (mod g) in A, i.e., fn ∈ (g).
3.5.F. E ASY E XERCISE . Show that D(f) = ∅ if and only if f ∈ N.

3.6 Topological (and Noetherian) properties
Many topological notions are useful when applied to the topological space
Spec A, and later, to schemes.
3.6.1. Possible topological attributes of Spec A: connectedness, irreducibility,
quasicompactness.
3.6.2. Connectedness.
A topological space X is connected if it cannot be written as the disjoint unionof two nonempty open sets. Exercise 3.6.A below gives an example of a nonconnected Spec A, and the subsequent remark explains that all examples are of thisform.
3.6.A. E #XERCISE .
#If A#= A1 × A2 × · · · × An , describe a homeomorphism
Spec A1 Spec A2 · · · Spec An → Spec A for which each
"nSpec Ai is#mappednonto a distinguished open subset D(fi ) of Spec A. Thus Spec i=1 Ai = i=1 Spec Ai
as topological spaces. (Hint: reduce to n = 2 for convenience. Let f1 = (1, 0) and
f2 = (0, 1).)
3.6.3. Remark. An extension of Exercise 3.6.A (that you can prove if you wish)
is that Spec A is not connected if and only if A is isomorphic to the product ofnonzero rings A1 and A2 . The key idea is to show that both conditions are equivalent to there existing nonzero a1 , a2 ∈ A for which a21 = a1 , a22 = a2 , a1 + a2 = 1,
and hence a1 a2 = 0. An element a ∈ A satisfying a2 = a is called an idempotent.
This will appear as Exercise 9.5.J.
3.6.4. Irreducibility.
A topological space is said to be irreducible if it is nonempty, and it is not theunion of two proper closed subsets. In other words, a nonempty topological space
X is irreducible if whenever X = Y ∪ Z with Y and Z closed in X, we have Y = X or
Z = X. This is a less useful notion in classical geometry — C2 is reducible (i.e., notirreducible), but we will see that A2C is irreducible (Exercise 3.6.C).
3.6.B. E ASY E XERCISE .
(a) Show that in an irreducible topological space, any nonempty open set is dense.
(For this reason, you will see that unlike in the classical topology, in the Zariski
topology, nonempty open sets are all “huge”.)(b) If X is a topological space, and Z (with the subspace topology) is an irreducible
subset, then the closure Z in X is irreducible as well.

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3.6.C. E ASY E XERCISE . If A is an integral domain, show that Spec A is irreducible.
(Hint: pay attention to the generic point [(0)].) We will generalize this in Exercise 3.7.F.
3.6.D. E XERCISE . Show that an irreducible topological space is connected.
3.6.E. E XERCISE . Give (with proof!) an example of a ring A where Spec A isconnected but reducible. (Possible hint: a picture may help. The symbol “×” has
two “pieces” yet is connected.)
3.6.F. T RICKY E XERCISE .
(a) Suppose I = (wz−xy, wy−x2 , xz−y2 ) ⊂ k[w, x, y, z]. Show that Spec k[w, x, y, z]/Iis irreducible, by showing that k[w, x, y, z]/I is an integral domain. (This is hard,so here is one of several possible hints: Show that k[w, x, y, z]/I is isomorphic tothe subring of k[a, b] generated by monomials of degree divisible by 3. There areother approaches as well, some of which we will see later. This is an example ofa hard question: how do you tell if an ideal is prime?) We will later see this as
the cone over the twisted cubic curve (the twisted cubic curve is defined in Exercise 8.2.A, and is a special case of a Veronese embedding, §8.2.6).
⋆ (b) Note that the generators of the ideal of part (a) may be rewritten as the equations ensuring that
"
#
w x y
rank
≤ 1,
x y z
i.e., as the determinants of the 2 × 2 submatrices. Generalize part (a) to the ideal ofrank one 2 × n matrices. This notion will correspond to the cone (§8.2.12) over the
degree n rational normal curve (Exercise 8.2.J).
3.6.5. Quasicompactness.
A topological space X is quasicompact if given any cover X = ∪i∈I Ui by open
sets, there is a finite subset S of the index set I such that X = ∪i∈S Ui . Informally:
every open cover has a finite subcover. We will like this condition, because we areafraid of infinity. Depending on your definition of “compactness”, this is the definition of compactness, minus possibly a Hausdorff condition. However, this isn’t
really the algebro-geometric analog of “compact” (we certainly wouldn’t want A1C
to be compact) — the right analog is “properness” (§10.3).
3.6.G. E XERCISE .
(a) Show that Spec A is quasicompact. (Hint: Exercise 3.5.C.)⋆ (b) (less important) Show that in general Spec A can have nonquasicompact open
sets. Possible hint: let A = k[x1 , x2 , x3 , . . . ] and m = (x1 , x2 , . . . ) ⊂ A, and consider
the complement of V(m). This example will be useful to construct other “counterexamples” later, e.g., Exercises 7.1.C and 5.1.J. In Exercise 3.6.T, we will see thatsuch weird behavior doesn’t happen for “suitably nice” (Noetherian) rings.
3.6.H. E XERCISE .
(a) If X is a topological space that is a finite union of quasicompact spaces, showthat X is quasicompact.
(b) Show that every closed subset of a quasicompact topological space is quasicompact.

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#n
∼
3.6.6."⋆⋆ Fun but irrelevant remark. Exercise 3.6.A shows that i=1 Spec Ai =nSpec i=1 Ai , but this never holds if “n is infinite” and all Ai are nonzero, asSpec of any ring is quasicompact "(Exercise 3.6.G(a)). This leads to aninteresting#∞∞
phenomenon. We show that Spec i=1 Ai is “strictly bigger” than i=1 Spec Ai
where
inclusion of sets#∞ each Ai is isomorphic"∞ to the field k. First, we have an"
Spec
A
"→
Spec
A
,
as
there
is
a
maximalidealof
Ai correspondi
i
i=1
i=1
ing to each i (precisely,
thoseelements
0
in
the
ith
component.)
But
there are other"maximal ideals of Ai . Hint: describe a proper
ideal
not
contained
in any of these"maximal ideals. (One idea: consider elements ai that are “eventually zero”, i.e.,
ai = 0 for i ≫ 0.) This leads to the notion of ultrafilters, which are very useful, butirrelevant to our current discussion.
3.6.7. Possible topological properties of points of Spec A.
3.6.8. Definition. A point of a topological space p ∈ X is said to be a closed point
if {p} is a closed subset. In the classical topology on Cn , all points are closed. In
Spec Z and Spec k[t], all the points are closed except for [(0)].
3.6.I. E XERCISE . Show that the closed points of Spec A correspond to the maximal
ideals. (In particular, non-empty affine schemes have closed points, as nonzerorings have maximal ideals, §0.3.)
3.6.9. Connection to the classical theory of varieties. Hilbert’s Nullstellensatz lets us
interpret the closed points of An
C as the n-tuples of complex numbers. More generally, the closed points of Spec k[x1 , . . . , xn ]/(f1 , . . . , fr ) are naturally interpretedn
as those points in k satisfying the equations f1 = · · · = fr = 0 (see Exercises 3.2.J
and 3.2.N(a) for example). Hence from now on we will say “closed point” insteadof “traditional point” and “non-closed point” instead of “bonus” point when discussing subsets of An.
k
3.6.J. E XERCISE .
(a) Suppose that k is a field, and A is a finitely generated k-algebra. Show thatclosed points of Spec A are dense, by showing that if f ∈ A, and D(f) is a nonempty(distinguished) open subset of Spec A, then D(f) contains a closed point of Spec A.
Hint: note that Af is also a finitely generated k-algebra. Use the Nullstellensatz 3.2.5
to recognize closed points of Spec of a finitely generated k-algebra B as those forwhich the residue field is a finite extension of k. Apply this to both B = A and
B = Af .
(b) Show that if A is a k-algebra that is not finitely generated the closed points
need not be dense. (Hint: Exercise 3.4.K.)
3.6.K. E XERCISE . Suppose k is an algebraically closed field, and A = k[x1 , . . . , xn ]/I
is a finitely generated k-algebra with N(A) = {0} (so the discussion of §3.2.13 apn
plies). Consider the set X = Spec A as a subset of An
k . The space Ak containsnthe “classical” points k . Show that functions on X are determined by their values on the closed points (by the weak Nullstellensatz 3.2.4, the “classical” points
kn ∩ Spec A of Spec A). Hint: if f and g are different functions on X, then f − g isnowhere zero on an open subset of X. Use Exercise 3.6.J(a).

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Once we know what a variety is (Definition 10.1.7), this will immediately imply that a function on a variety over an algebraically closed field is determined by its values
on the “classical points”. (Before the advent of scheme theory, functions on varieties— over algebraically closed fields — were thought of as functions on “classical”
points, and Exercise 3.6.K basically shows that there is no harm in thinking of “traditional” varieties as a particular flavor of schemes.)
3.6.10. Specialization and generization. Given two points x, y of a topological space
X, we say that x is a specialization of y, and y is a generization of x, if x ∈ {y}.
This (and Exercise 3.6.L) now makes precise our hand-waving about “one point
containing another”. It is of course nonsense for a point to contain another. But it
is not nonsense to say that the closure of a point contains another. For example, in
A2C = Spec C[x, y], [(y − x2 )] is a generization of [(x − 2, y − 4)] = (2, 4) ∈ C2 , and(2, 4) is a specialization of [(y − x2 )] (see Figure 3.8).

F IGURE 3.8. (2, 4) = [(x − 2y − 4)] is a specialization of [(y − x2 )].
[(y − x2 )] is a generization of (2, 4).
3.6.L. E XERCISE . If X = Spec A, show that [q] is a specialization of [p] if and only
if p ⊂ q. Hence show that V(p) = {[p]}.
3.6.11. Definition. We say that a point p ∈ X is a generic point for a closed subset
K if {p} = K.
This important notion predates Grothendieck. The early twentieth-century
Italian algebraic geometers had a notion of “generic points” of a variety, by which
they meant points with no special properties, so that anything proved of “a genericpoint” was true of “almost all” the points on that variety. The modern “genericpoint” has the same intuitive meaning. If something is “generically” or “mostly”
true for the points of an irreducible subset, in the sense of being true for a dense
open subset (for “almost all points”), then it is true for the generic point, and viceversa. (This is a statement of principle, not of fact. An interesting case is “reducedness”, for which this principle does not hold in general, but does hold for “reasonable” schemes such as varieties, see Remark 5.2.2.) For example, a function has
value zero at the generic point of an integral scheme if and only if it has the value

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zero at all points. (See Exercise 5.5.C and the paragraph following it, although you
will be able to prove this yourself well before then.) You should keep an eye out
for other examples of this.
The phrase general point does not have the same meaning. The phrase “thegeneral point of K satisfies such-and-such a property” means “every point of somedense open subset of X satisfies such-and-such a property”. Be careful not to
confuse “general” and “generic”. But be warned that accepted terminology does
not always follow this convention; witness “generic freeness” (see Lemma 7.4.4),“generic flatness” (§24.5.9), and “generic smoothness” (§25.3, for example).
3.6.M. E XERCISE . Verify that [(y − x2 )] ∈ A2 is a generic point for V(y − x2 ).

As more motivation for the terminology “generic”: we think of [(y − x2 )] as
being some nonspecific point on the parabola (with the closed points (a, a2 ) ∈C2 , i.e., (x − a, y − a2 ) for a ∈ C, being “specific points”); it is “generic” in theconventional sense of the word. We might “specialize it” to a specific point of
the parabola; hence for example (2, 4) is a specialization of [(y − x2 )]. (Again, see
Figure 3.8.) To make this somewhat more precise:
3.6.N. E ASY E XERCISE . Suppose p is a generic point for the closed subset K. Show
that it is “near every point q of K” (every neighborhood of q contains p), and “notnear any point r not in K” (there is a neighborhood of r not containing p). (Thisidea was mentioned in §3.2.3 Example 1 and in Exercise 3.4.G.)

We will soon see (Exercise 3.7.E) that there is a natural bijection between pointsof Spec A and irreducible closed subsets of Spec A, sending each point to its closure, and each irreducible closed subset to its (unique) generic point. You canprove this now, but we will wait until we have developed some convenient terminology.
3.6.12. Irreducible and connected components, and Noetherian conditions.
An irreducible component of a topological space is a maximal irreducible
subset (an irreducible subset not contained in any larger irreducible subset). irreducible components are closed (as the closure of irreducible subsets are irreducible,
Exercise 3.6.B(b)), and it can be helpful to think of irreducible components of a
topological space X as maximal among the irreducible closed subsets of X. We
think of these as the “pieces of X” (see Figure 3.9).
Similarly, a subset Y of a topological space X is a connected component if it
is a maximal connected subset (a connected subset not contained in any larger
connected subset).
3.6.O. E XERCISE ( EVERY TOPOLOGICAL SPACE IS THE UNION OF IRREDUCIBLE
COMPONENTS ).
Show that every point x of a topological space X is containedin an irreducible component of X. Hint: Zorn’s Lemma. More precisely, consider
the partially ordered set S of irreducible closed subsets of X containing x. Show
that there exists a maximal totally ordered subset {Zα } of S . Show that ∪Zα isirreducible.
3.6.13. Remark. Every point is contained in a connected component, and connected components are always closed. You can prove this now, but we deliberately
postpone this until we need it, in an optional starred section (Exercise 9.5.H). On

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F IGURE 3.9. This closed subset of A2C has six irreducible components
the other hand, connected components need not be open, see [Stacks, tag 004T].
An example"∞ of an affine scheme with connected components that are not open isSpec( 1 F2 ).
3.6.14. In the examples we have considered, the spaces have naturally broken up
into a finite number of irreducible components. For example, the locus xy = 0
in A2C we think of as having two “pieces” — the two axes. The reason for this isthat their underlying topological spaces (as we shall soon establish) are Noetherian. A topological space X is called Noetherian if it satisfies the descending chain
condition for closed subsets: any sequence Z1 ⊃ Z2 ⊃ · · · ⊃ Zn ⊃ · · · of closed
subsets eventually stabilizes: there is an r such that Zr = Zr+1 = · · · . Here is a
first example (which you should work out explicitly, not using Noetherian rings).
3.6.P. E XERCISE . Show that A2C is a Noetherian topological space: any decreasing
sequence of closed subsets of A2C = Spec C[x, y] must eventually stabilize. Notethat it can take arbitrarily long to stabilize. (The closed subsets of A2C were described in §3.4.3.) Show that C2 with the classical topology is not a Noetherian
topological space.
3.6.15. Proposition. — Suppose X is a Noetherian topological space. Then every
nonempty closed subset Z can be expressed uniquely as a finite union Z = Z1 ∪ · · · ∪ Zn
of irreducible closed subsets, none contained in any other.
Translation: any closed subset Z has a finite number of “pieces”.
Proof. The following technique is called Noetherian induction, for reasons thatwill be clear. We will use it again, many times.
Consider the collection of closed subsets of X that cannot be expressed as afinite union of irreducible closed subsets. We will show that it is empty. Otherwise,
let Y1 be one such. If Y1 properly contains another such, then choose one, and callit Y2 . If Y2 properly contains another such, then choose one, and call it Y3 , and so

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on. By the descending chain condition, this must eventually stop, and we musthave some Yr that cannot be written as a finite union of irreducible closed subsets,
but every closed subset properly contained in it can be so written. But then Yr is
not itself irreducible, so we can write Yr = Y ′ ∪Y ′′ where Y ′ and Y ′′ are both proper
closed subsets. Both of these by hypothesis can be written as the union of a finite
number of irreducible subsets, and hence so can Yr , yielding a contradiction. Thus
each closed subset can be written as a finite union of irreducible closed subsets.
We can assume that none of these irreducible closed subsets contain any others, bydiscarding some of them.
We now show uniqueness. Suppose
Z = Z1 ∪ Z2 ∪ · · · ∪ Zr = Z1′ ∪ Z2′ ∪ · · · ∪ Zs′
are two such representations. Then Z1′ ⊂ Z1 ∪ Z2 ∪ · · · ∪ Zr , so Z1′ = (Z1 ∩ Z1′ ) ∪· · · ∪ (Zr ∩ Z1′ ). Now Z1′ is irreducible, so one of these is Z1′ itself, say (without
loss of generality) Z1 ∩ Z1′ . Thus Z1′ ⊂ Z1 . Similarly, Z1 ⊂ Za′ for some a; but
because Z1′ ⊂ Z1 ⊂ Za′ , and Z1′ is contained in no other Zi′ , we must have a = 1,
and Z1′ = Z1 . Thus each element of the list of Z’s is in the list of Z ′ ’s, and viceversa, so they must be the same list.
!
3.6.Q. E XERCISE . Show that every connected component of a topological space
X is the union of irreducible components of X. Show that any subset of X that issimultaneously open and closed must be the union of some of the connected components of X. If X is a Noetherian topological space, show that the union of any
subset of the connected components of X is always open and closed in X. (In particular, connected components of Noetherian topological spaces are always open,
which is not true for more general topological spaces, see Remark 3.6.13.)
3.6.16. Noetherian rings. It turns out that all of the spectra we have considered(except in starred Exercise 3.6.G(b)) are Noetherian topological spaces, but that
isn’t true of the spectra of all rings. The key characteristic all of our examples have
had in common is that the rings were Noetherian. A ring is Noetherian if every
ascending sequence I1 ⊂ I2 ⊂ · · · of ideals eventually stabilizes: there is an r suchthat Ir = Ir+1 = · · · . (This is called the ascending chain condition on ideals.)
Here are some quick facts about Noetherian rings. You should be able to provethem all.
• Fields are Noetherian. Z is Noetherian.
• (Noetherianness is preserved by quotients.) If A is Noetherian, and I is any
ideal of A, then A/I is Noetherian.
• (Noetherianness is preserved by localization.) If A is Noetherian, and S is any
multiplicative set, then S−1 A is Noetherian.
3.6.R. I MPORTANT E XERCISE . Show that a ring A is Noetherian if and only ifevery ideal of A is finitely generated.
The next fact is nontrivial.
3.6.17. The Hilbert Basis Theorem. — If A is Noetherian, then so is A[x].

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Hilbert proved this in the epochal paper [Hil] where he also proved the Hilbert
Syzygy Theorem (§15.3.2), and defined Hilbert functions and showed that they are
eventually polynomial (§18.6).
3.6.18. By the results described above, any polynomial ring over any field, or overthe integers, is Noetherian — and also any quotient or localization thereof. Hencefor example any finitely generated algebra over k or Z, or any localization thereof,is Noetherian. Most “nice” rings are Noetherian, but not all rings are Noetherian: k[x1 , x2 , . . . ] is not, because (x1 ) ⊂ (x1 , x2 ) ⊂ (x1 , x2 , x3 ) ⊂ · · · is a strictly
ascending chain of ideals (cf. Exercise 3.6.G(b)).
3.6.19. Proof of the Hilbert Basis Theorem 3.6.17. We show that any ideal I ⊂ A[x]is finitely generated. We inductively produce a set of generators f1 , . . . as follows.
For n > 0, if I ̸= (f1 , . . . , fn−1 ), let fn be any nonzero element of I − (f1 , . . . , fn−1 )of lowest degree. Thus f1 is any element of I of lowest degree, assuming I ̸= (0).
If this procedure terminates, we are done. Otherwise, let an ∈ A be the initialcoefficient of fn for n > 0. Then as A is Noetherian, (a1 , a2 , . . . ) = (a1 , . . . , aN ),N
for some N. Say aN+1 = i=1 bi ai . Then
fN+1 −

N.

bi fi xdeg fN+1 −deg fi

i=1

is an element of I that is nonzero (as fN+1 ∈/ (f1 , . . . , fN )), and of lower degree
than fN+1 , yielding a contradiction.
!
We now connect Noetherian rings and Noetherian topological spaces.
3.6.S. E XERCISE . If A is Noetherian, show that Spec A is a Noetherian topologicalspace. Describe a ring A such that Spec A is not a Noetherian topological space.
(Aside: if Spec A is a Noetherian topological space, A need not be Noetherian.
One example is A = k[x1 , x2 , x3 , . . . ]/(x1 , x22 , x33 , . . . ). Then Spec A has one point,
so is Noetherian. But A is not Noetherian as (x1 ) " (x1 , x2 ) " (x1 , x2 , x3 ) " · · · inA.)
3.6.T. E XERCISE ( PROMISED IN E XERCISE 3.6.G( B )). Show that every open subset
of a Noetherian topological space is quasicompact. Hence if A is Noetherian, every
open subset of Spec A is quasicompact.
3.6.20. For future use: Noetherian conditions for modules. An important related
notion is that of a Noetherian module. Although we won’t use this notion for sometime (§9.7.3), we will develop their most important properties, while Noetherianideas are still fresh in your mind. If A is any ring, not necessarily Noetherian,
we say an A-module is Noetherian if it satisfies the ascending chain condition
for submodules. Thus for example a ring A is Noetherian if and only if it is a
Noetherian A-module.
3.6.U. E XERCISE ( CF. I MPORTANT E XERCISE 3.6.R). Show that if M is a Noetherian
A-module, then any submodule of M is a finitely generated A-module.
3.6.V. E XERCISE . If 0 → M ′ → M → M ′′ → 0 is exact, show that M ′ and M ′′ are
Noetherian if and only if M is Noetherian. (Hint: Given an ascending chain in M,

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we get two simultaneous ascending chains in M ′ and M ′′ . Possible further hint:
prove that if
M′

%M

φ

% M ′′

is exact, and N ⊂ N ′ ⊂ M, and N ∩ M ′ = N ′ ∩ M ′ and φ(N) = φ(N ′ ), thenN = N ′ .)3.6.W. E XERCISE . Show that if A is a Noetherian ring, then A⊕n is a Noetherian
A-module.
3.6.X. E XERCISE . Show that if A is a Noetherian ring and M is a finitely generated
A-module, then M is a Noetherian module. Hence by Exercise 3.6.U, any submodule of a finitely generated module over a Noetherian ring is finitely generated.
3.6.21. Why you should not worry about Noetherian hypotheses. Should you
work hard to eliminate Noetherian hypotheses? Should you worry about Noetherian hypotheses? Should you stay up at night thinking about non-Noetherian rings?
For the most part, the answer to these questions is “no”. Most people will neverneed to worry about non-Noetherian rings, but there are reasons to be open tothem. First, they can actually come up. For example, fibered products of Noetherian schemes over Noetherian schemes (and even fibered products of Noetherian
points over Noetherian points!) can be non-Noetherian (Warning 9.1.4), and thenormalization of Noetherian rings can be non-Noetherian (Warning 9.7.4). You
can either work hard to show that the rings or schemes you care about don’t havethis pathology, or you can just relax and not worry about it. Second, there is often
no harm in working with schemes in general. Knowing when Noetherian conditions are needed will help you remember why results are true, because you will
have some sense of where Noetherian conditions enter into arguments. Finally,
for some people, non-Noetherian rings naturally come up. For example, adeles
are not Noetherian. And many valuation rings that naturally arise in arithmetic
and tropical geometry are not Noetherian.

3.7 The function I(·), taking subsets of Spec A to ideals of A
We now introduce a notion that is in some sense “inverse” to the vanishing set
function V(·). Given a subset' S ⊂ Spec A, I(S) is the set of functions vanishing on
S. In other words, I(S) = [p]∈S p ⊂ A (at least when S is nonempty).
We make three quick observations. (Do you see why they are true?)• I(S) is clearly an ideal of A.
• I(·) is inclusion-reversing: if S1 ⊂ S2 , then I(S2 ) ⊂ I(S1 ).
• I(S) = I(S).
3.7.A. E XERCISE . Let A = k[x, y]. If S = {[(y)], [(x, y − 1)]} (see Figure 3.10), thenI(S) consists of those polynomials vanishing on the y-axis, and at the point (1, 0).
Give generators for this ideal.

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F IGURE 3.10. The set S of Exercise 3.7.A, pictured as a subset of A2
3.7.B. E XERCISE . Suppose S ⊂ A3C is the union of the three axes. Give generators
for the ideal I(S). Be sure to prove it! We will see in Exercise 12.1.F that this ideal
is not generated by less than three elements.
3.7.C. E XERCISE . Show that V(I(S)) = S. Hence V(I(S)) = S for a closed set S.
&
Note that I(S) is always a radical ideal — if f ∈ I(S), then fn vanishes on S
for some n > 0, so then f vanishes on S, so f ∈ I(S).
√
3.7.D. E ASY E XERCISE . Prove that if J ⊂ A is an ideal, then I(V(J)) = J. (Hugehint: Exercise 3.4.J.)
Exercises 3.7.C and 3.7.D show that V and I are “almost” inverse. More precisely:
3.7.1. Theorem. — V(·) and I(·) give an inclusion-reversing bijection between closed
subsets of Spec A and radical ideals of A (where a closed subset gives a radical ideal byI(·), and a radical ideal gives a closed subset by V(·)).
Theorem 3.7.1 is sometimes called Hilbert’s Nullstellensatz, but we reservethat name for Theorem 3.2.5.
3.7.E. I MPORTANT E XERCISE ( CF. E XERCISE 3.7.F). Show that V(·) and I(·) givea bijection between irreducible closed subsets of Spec A and prime ideals of A. From
this conclude that in Spec A there is a bijection between points of Spec A and irreducible closed subsets of Spec A (where a point determines an irreducible closed
subset by taking the closure). Hence each irreducible closed subset of Spec A has precisely one generic point — any irreducible closed subset Z can be written uniquely
as {z}.
3.7.F. E XERCISE / DEFINITION . A prime ideal of a ring A is a minimal prime ideal(or more simply, minimal prime) if it is minimal with respect to inclusion. (Forexample, the only minimal prime of k[x, y] is (0).) If A is any ring, show that theirreducible components of Spec A are in bijection with the minimal prime ideals of
A. In particular, Spec A is irreducible if and only if A has only one minimal prime
ideal; this generalizes Exercise 3.6.C.
Proposition 3.6.15, Exercise 3.6.S, and Exercise 3.7.F imply that every Noetherian ring has a finite number of minimal prime ideals: an algebraic fact is now
revealed to be really a “geometric” fact!
3.7.G. E XERCISE . What are the minimal prime ideals of k[x, y]/(xy) (where k is afield)?

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CHAPTER 4

The structure sheaf, and the definition of schemes ingeneral

4.1 The structure sheaf of an affine scheme
The final ingredient in the definition of an affine scheme is the structure sheaf
OSpec A , which we think of as the “sheaf of algebraic functions”. You should keep
in your mind the example of “algebraic functions” on Cn , which you understandwell. For example, in A2 , we expect that on the open set D(xy) (away from thetwo axes), (3x4 + y + 4)/x7 y3 should be an algebraic function.
These functions will have values at points, but won’t be determined by theirvalues at points. But like all sections of sheaves, they will be determined by theirgerms (see §4.3.5).
It suffices to describe the structure sheaf as a sheaf (of rings) on the base of
distinguished open sets (Theorem 2.5.1 and Exercise 3.5.A).
4.1.1. Definition. Define OSpec A (D(f)) to be the localization of A at the multiplicative set of all functions that do not vanish outside of V(f) (i.e., those g ∈ A suchthat V(g) ⊂ V(f), or equivalently D(f) ⊂ D(g), cf. Exercise 3.5.E). This depends
only on D(f), and not on f itself. (Scholars of the empty set might notice that by
Exercise 3.5.F, we have that OSpec A (∅) = {0}.)4.1.A. G REAT E XERCISE . Show that the natural map Af → OSpec A (D(f)) is an
isomorphism. (Possible hint: Exercise 3.5.E.)
If D(f ′ ) ⊂ D(f), define the restriction map

resD(f),D(f ′ ) : OSpec A (D(f)) → OSpec A (D(f ′ ))in the obvious way: the latter ring is a further localization of the former ring.
The restriction maps obviously commute: this is a “presheaf on the distinguishedbase”.
4.1.2. Theorem. — The data just described give a sheaf on the distinguished base, andhence determine a sheaf on the topological space Spec A.
4.1.3. This sheaf is called the structure sheaf, and will be denoted OSpec A , or sometimes O if the subscript is clear from the context. Such a topological space, withsheaf, will be called an affine scheme (Definition 4.3.1). The notation Spec A will
hereafter denote the data of a topological space with a structure sheaf. An important lesson of Theorem 4.1.2 is not just that OSpec A is a sheaf, but also that the
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The Rising Sea: Foundations of Algebraic Geometry

distinguished base provides a good way of working with OSpec A . Notice also that
we have justified interpreting elements of A as functions on Spec A.
Proof. We must show the base identity and base gluability axioms hold (§2.5). We
show that they both hold for the open set that is the entire space Spec A, and leave
to you the trick which extends them to arbitrary distinguished open sets (Exercises 4.1.B and 4.1.C). Suppose Spec A = ∪i∈I D(fi ), or equivalently (Exercise 3.5.B)
the ideal generated by the fi is the entire ring A.
(Aside: experts familiar with the equalizer exact sequence of §2.2.7 will realizethat we are showing exactness of&&(4.1.3.1)
0→A→
Afi →
Afi fji∈Ii̸=j∈I

where {fi }i∈I is a set of functions with (fi )i∈I = A. Signs are involved in the righthand map: the map Afi → Afi fj is the “obvious one” if i < j, and negative of the“obvious one” if i > j. Base identity
corresponds to injectivity at A, and gluability"
corresponds to exactness at i Afi .)We check identity on the base. Suppose that Spec A = ∪i∈I D(fi ) where iruns over some index set I. Then there is some finite subset of I, which we name{1, . . . , n}, such that Spec A = ∪ni=1 D(fi ), i.e., (f1 , . . . , fn ) = A (quasicompactness
of Spec A, Exercise 3.5.C). Suppose we are given s ∈ A such that resSpec A,D(fi ) s =0 in Afi for all i. We wish to show that s = 0. The fact that resSpec A,D(fi ) s = 0 in
Afi implies that there is some m such that for each i ∈ {1, . . . , n}, fm
i s = 0. Nowm(fm= A (for example, from Spec A = ∪D(fi ) = ∪D(fm
1 , . . . , fn ),i )), so there arenri ∈ A with i=1 ri fm
i = 1 in A, from which
$.
%
.
s=
ri fm
s=
ri (fmii s) = 0.
Thus we have checked the “base identity” axiom for Spec A.

4.1.B. E XERCISE . Make tiny changes to the above argument to show base identity
for any distinguished open D(f). (Hint: judiciously replace A by Af in the above
argument.)
We next show base gluability. (Serre has described this as a “partition of unity”
argument, and if you look at it in the right way, his insight is very enlightening.)
Suppose again ∪i∈I D(fi ) = Spec A, where I is an index set (possibly horribly infinite). Suppose we are given elements in each Afi that agree on the overlaps Afi fj .
Note that intersections of distinguished open sets are also distinguished open sets.
Assume first that I is finite, say I = {1, . . . , n}. We have elements ai /flii ∈ Afi
agreeing on overlaps Afi fj (see Figure 4.1(a)). Letting gi = flii , using D(fi ) =D(gi ), we can simplify notation by considering our elements as of the form ai /gi ∈Agi (Figure 4.1(b)).
The fact that ai /gi and aj /gj “agree on the overlap” (i.e., in Agi gj ) means thatfor some mij ,(gi gj )mij (gj ai − gi aj ) = 0
in A. By taking m = max mij (here we use the finiteness of I), we can simplify
notation:(gi gj )m (gj ai − gi aj ) = 0

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November 18, 2017 draft131

F IGURE 4.1. Base gluability of the structure sheaf
m+1
for all i, j (Figure 4.1(c)). Let bi = ai gm(so D(hi ) = D(gi )).
i for all i, and hi = gi
Then we can simplify notation even more (Figure 4.1(d)): on each D(hi ), we havea function bi /hi , and the overlap condition is(4.1.3.2)

hj bi = hi bj .
,n
Now ∪i D(hi ) = Spec A, implying that 1 = i=1 ri hi for some ri ∈ A. Define.
(4.1.3.3)r=ri bi .

This will be the element of A that restricts to each bj /hj . Indeed, from the overlap
condition (4.1.3.2),
.
.
rhj =
ri bi hj =ri hi bj = bj .
ii

We next deal with the case where I is infinite. Choose a finite subset {1, . . . , n} ⊂I with (f1 , . . . , fn ) = A (or equivalently, use quasicompactness of Spec A to choose
a finite subcover by D(fi )). Construct r as above, using (4.1.3.3). We will show thatfor any z ∈ I − {1, . . . , n}, r restricts to the desired element az of Afz . Repeat theentire process above with {1, . . . , n, z} in place of {1, . . . , n}, to obtain r ′ ∈ A which
restricts to ai for i ∈ {1, . . . , n, z}. Then by the base identity axiom, r ′ = r. (Notethat we use base identity to prove base gluability. This is an example of how theidentity axiom is somehow “prior” to the gluability axiom.) Hence r restricts to
az /flzz as desired.
4.1.C. E XERCISE . Alter this argument appropriately to show base gluability forany distinguished open D(f).
We have now completed the proof of Theorem 4.1.2.

!

The following generalization of Theorem 4.1.2 will be essential in the definition of a quasicoherent sheaf in Chapter 13.
4.1.D. I MPORTANT E XERCISE /D EFINITION . Suppose M is an A-module. Show
( on the distinguished base. Dethat the following construction describes a sheaf M(
fine M(D(f)) to be the localization of M at the multiplicative set of all functions

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The Rising Sea: Foundations of Algebraic Geometrythat do not vanish outside of V(f). Define restriction maps resD(f),D(g) in the analogous way to OSpec A . Show that this defines a sheaf on the distinguished base,and hence a sheaf on Spec A. Then show that this is an OSpec A -module.
4.1.4. Remark. In the course of answering the previous exercise, you will showthat if (fi )i∈I = A,
&
&
0→M→
Mfi →
Mfi fji∈Ii̸=j∈I(cf. (4.1.3.1)) is exact. In particular, M can be identified with a specific submoduleof Mf1 × · · · × Mfr . Even though M → Mfi may not be an inclusion for any fi ,M → Mf1 × · · · × Mfr is an inclusion. This will be useful later: we will want toshow that if M has some nice property, then Mf does too, which will be easy. We
will also want to show that if (f1 , . . . , fn ) = A, and the Mfi have this property,
then M does too. This idea will be made precise in the Affine Communication
Lemma 5.3.2.
4.1.5. ⋆ Remark. Definition 4.1.1 and Theorem 4.1.2 suggests a potentially slick
way of describing sections of OSpec A over any open subset: perhaps OSpec A (U) is
the localization of A at the multiplicative set of all functions that do not vanish at
any point of U. This is not true. A counterexample (that you will later be able to
make precise): let Spec A be two copies of A2k glued together at their origins (see(24.4.8.1) for explicit equations) and let U be the complement of the origin(s). Thenthe function which is 1 on the first copy of A2k \ {(0, 0)} and 0 on the second copyof A2k \ {(0, 0)} is not of this form. (Follow-up question: why would this discussion
not work for two copies of A1k glued at a point?)
4.1.6. List of five recurring (counter)examples. The example of two planes meeting at a point will appear many times as an example or a counterexample. This is
as good a time as any to collect a list of those examples that will repeatedly comeup (often in counterexamples). Many of these you will not have the language tounderstand yet, but you can come back to this list later.
(The cone over) the quadric surface, Spec A and Proj A where A = k[w, x, y, z]/(wz −xy): §4.4.12, Exercise 5.4.D, Exercise 5.4.I, Exercise 5.4.L, Exercise 6.5.H, §8.2.2,§8.2.9, §9.6.2, Exercise 11.2.D, §12.1.D, Exercise 12.1.E, §12.2.14, Exercise 12.3.D,§12.5.13, Exercise 14.2.O, Exercise 14.2.R, Exercise 14.2.U, Exercise 19.8.B, Exercise 19.10.J, §19.10.7, Exercise 22.4.N, §22.4.11 .

Two planes meeting at a point: §4.1.5, Exercise 18.5.B, Exercise 18.6.U, Exercise 24.4.I,§24.4.10, Exercise 24.6.E, Exercise 26.1.G, §26.1.6, §26.2.2, §26.2.12, §26.3.2, §26.3.4.
Affine space minus the origin: An −0; its inclusion in An ; and An with doubled origin;each with n = 1, 2, ∞: Exercise 3.6.G, §3.6.18, §4.4.1, §4.4.3, §4.4.5, Exercise 4.4.B,
Exercise 4.4.C, Exercise 5.1.J, §7.3.9, §10.1.1, Exercise 10.1.D, §10.1.8, Exercise 10.2.C,
Exercise 10.2.D, Exercise 12.7.B, Exercise 12.7.C, Exercise 18.3.A.
#
Infinite disjoint unions of schemes (especially Spec k[x]/(xn )): Exercise 4.3.E, §4.4.4,
Exercise 5.2.E, Example 4 of §8.3.2, Caution/example 8.3.11, Exercise 13.3.I, §29.2.7.

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133

Spec Q → Spec Q: comment after Corollary 7.2.2, §7.3.10, §7.3.14, Warning 9.1.4,
Exercise 9.2.E, Exercise 9.4.C.

4.2 Visualizing schemes II: nilpotents
The price of metaphor is eternal vigilance.
— A. Rosenbluth and N. Wiener (attribution by [Leo, p. 4])
In §3.3, we discussed how to visualize the underlying set of schemes, addingin generic points to our previous intuition of “classical” (or closed) points. Ourlater discussion of the Zariski topology fit well with that picture. In our definitionof the “affine scheme” (Spec A, OSpec A ), we have the additional information of
nilpotents, which are invisible on the level of points (§3.2.11), so now we figureout to picture them. We will then readily be able to glue them together to picture
schemes in general, once we have made the appropriate definitions. As we arebuilding intuition, we cannot be rigorous or precise.
As motivation, note that we have incidence-reversing bijections
maximal ideals of A (% closed points of Spec A(Exercise 3.6.I)

prime ideals of A (% irreducible closed subsets of Spec A(Exercise 3.7.E)

radical ideals of A (% closed subsets of Spec A(Theorem 3.7.1)

If we take the things on the right as “pictures”, our goal is to figure out how to
picture ideals that are not radical:
ideals of A (% ???

(We will later fill this in rigorously in a different way with the notion of a closed
subscheme, the scheme-theoretic version of closed subsets, §8.1. But our goal now
is to create a picture.)
As motivation, when we see the expression, Spec C[x]/(x(x − 1)(x − 2)), we
immediately interpret it as a closed subset of A1C , namely {0, 1, 2}. In particular,that the map C[x] → C[x]/(x(x − 1)(x − 2)) can be interpreted (via the Chinese
Remainder Theorem) as: take a function on A1 , and restrict it to the three points 0,
1, and 2.
This will guide us in how to visualize a nonradical ideal. The simplest example to consider is Spec C[x]/(x2 ) (Exercise 3.2.A(a)). As a subset of A1 , it is justthe origin 0 = [(x)], which we are used to thinking of as Spec C[x]/(x) (i.e., corresponding to the ideal (x), not (x2 )). We want to enrich this picture in some way.
We should picture C[x]/(x2 ) in terms of the information the quotient remembers.
The image of a polynomial f(x) is the information of its value at 0, and its derivative (cf. Exercise 3.2.T). We thus picture this as being the point, plus a little bit more
— a little bit of infinitesimal “fuzz” on the point (see Figure 4.2). The sequence of

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restrictions C[x] → C[x]/(x2 ) → C[x]/(x) should be interpreted as nested pictures.
C[x]% % C[x]/(x2 )% % C[x]/(x)f(x) ✤% f(0),

Similarly, C[x]/(x3 ) remembers even more information — the second derivative as
well. Thus we picture this as the point 0 with even more fuzz.

Spec C[x]/(x)Spec C[x]/(x2 )Spec C[x]/(x3 )Spec C[x] = A1C

F IGURE 4.2. Picturing quotients of C[x]More subtleties arise in two dimensions (see Figure 4.3). Consider
Spec C[x, y]/(x, y)2 ,which is sandwiched between two rings we know well:C[x, y]f(x, y) ✤% % C[x, y]/(x, y)2% % C[x, y]/(x, y)% f(0).

Again, taking the quotient by (x, y)2 remembers the first derivative, “in all directions”. We picture this as fuzz around the point, in the shape of a circle (nodirection is privileged). Similarly, (x, y)3 remembers the second derivative “in all
directions” — bigger circular fuzz.
Consider instead the ideal (x2 , y). What it remembers is the derivative onlyin the x direction — given a polynomial, we remember its value at 0, and thecoefficient of x. We remember this by picturing the fuzz only in the x direction.
This gives us some handle on picturing more things of this sort, but now it
becomes more an art than a science. For example, Spec C[x, y]/(x2 , y2 ) we might
picture as a fuzzy square around the origin. (Could you believe that this square iscircumscribed by the circular fuzz Spec C[x, y]/(x, y)3 , and inscribed by the circular fuzz Spec C[x, y]/(x, y)2 ?) One feature of this example is that given two ideals I
and J of a ring A (such as C[x, y]), your fuzzy picture of Spec A/(I, J) should be the“intersection” of your picture of Spec A/I and Spec A/J in Spec A. (You will makethis precise in Exercise 8.1.J(a).) For example, Spec C[x, y]/(x2 , y2 ) should be the

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November 18, 2017 draftSpec C[x, y]/(x, y)Spec C[x, y]/(x2 , y)Spec C[x, y]/(x2 , y2 )135Spec C[x, y]/(x, y)2Spec C[x, y]/(y2 )F IGURE 4.3. Picturing quotients of C[x, y]intersection of two thickened lines. (How would you picture Spec C[x, y]/(x5 , y3 )?
Spec C[x, y, z]/(x3 , y4 , z5 , (x + y + z)2 )? Spec C[x, y]/((x, y)5 , y3 )?)One final example that will motivate us in §5.5 is Spec C[x, y]/(y2 , xy). Knowing what a polynomial in C[x, y] is modulo (y2 , xy) is the same as knowing its
value on the x-axis, as well as first-order differential information around the origin. This is worth thinking through carefully: do you see how this information is
captured (however imperfectly) in Figure 4.4?

F IGURE 4.4. A picture of the scheme Spec k[x, y]/(y2 , xy). The
fuzz at the origin indicates where “there is nonreducedness”.
Our pictures capture useful information that you already have some intuitionfor. For example, consider the intersection of the parabola y = x2 and the x-axis(in the xy-plane), see Figure 4.5. You already have a sense that the intersection has
multiplicity two. In terms of this visualization, we interpret this as intersecting (inSpec C[x, y]):Spec C[x, y]/(y − x2 ) ∩ Spec C[x, y]/(y) ==Spec C[x, y]/(y − x2 , y)Spec C[x, y]/(y, x2 )which we interpret as the fact that the parabola and line not just meet with multiplicity two, but that the “multiplicity 2” part is in the direction of the x-axis. You
will make this example precise in Exercise 8.1.J(b).
4.2.1. We will later make the location of the fuzz somewhat more precise when we
discuss associated points (§5.5). We will see that in reasonable circumstances, thefuzz is concentrated on closed subsets (Remark 13.7.2).
On a bien souvent répété que la Géométrie est l’art de bien raisonner sur des figures
mal faites; encore ces figures, pour ne pas nous tromper, doivent-elles satisfaire à certaines
conditions; les proportions peuvent être grossièrement altérées, mais les positions relatives
des diverses parties ne doivent pas être bouleversées.

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intersect=

F IGURE 4.5. The “scheme-theoretic” intersection of the parabolay = x2 and the x-axis is a nonreduced scheme (with fuzz in thex-direction)
It is often said that geometry is the art of reasoning well from badly made figures;
however, these figures, if they are not to deceive us, must satisfy certain conditions; theproportions may be grossly altered, but the relative positions of the different parts mustnot be upset.
— H. Poincaré, [Po1, p. 2] (see J. Stillwell’s translation [Po2, p. ix])

4.3 Definition of schemes
4.3.1. Definitions. We can now define scheme in general. First, define an isomorphism of ringed spaces (X, OX ) and (Y, OY ) as (i) a homeomorphism π : X → Y,
and (ii) an isomorphism of sheaves OX and OY , considered to be on the same
space via π. (Part (ii), more precisely, is an isomorphism OY → π∗ OX of sheaves
on Y, or equivalently by adjointness, π−1 OY → OX of sheaves on X.) In otherwords, we have a “correspondence” of sets, topologies, and structure sheaves. An
affine scheme is a ringed space that is isomorphic to (Spec A, OSpec A ) for some A.
A scheme (X, OX ) is a ringed space such that any point of X has an open neighborhood U such that (U, OX |U ) is an affine scheme. The topology on a scheme iscalled the Zariski topology. The scheme can be denoted (X, OX ), although it isoften denoted X, with the structure sheaf implicit.
An isomorphism of two schemes (X, OX ) and (Y, OY ) is an isomorphism as
ringed spaces. Recall the definition of Γ (·, ·) in §2.2.2. If U ⊂ X is an open subset,
then the elements of Γ (U, OX ) are said to be the functions on U; this generalizes inan obvious way the definition of functions on an affine scheme, §3.2.1.
4.3.2. Remark. From the definition of the structure sheaf on an affine scheme,
several things are clear. First of all, if we are told that (X, OX ) is an affine scheme,
we may recover its ring (i.e., find the ring A such that Spec A = X) by taking thering of global sections, as X = D(1), so:Γ (X, OX ) ==Γ (D(1), OSpec A )as D(1) = Spec A

A.

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137(You can verify that we get more, and can “recognize X as the scheme Spec A”: weget an isomorphism π : (Spec Γ (X, OX ), OSpec Γ (X,OX ) ) → (X, OX ). For example, if m
is a maximal ideal of Γ (X, OX ), {π([m])} = V(m).) The following exercise will give
you a chance to make these ideas rigorous — they are subtler than they appear.
4.3.A. E NLIGHTENING E XERCISE ( WHICH CAN BE STRANGELY CONFUSING ). Describe a bijection between the isomorphisms Spec A → Spec A ′ and the ring isomorphisms A ′ → A. Hint: the hardest part is to show that if an isomorphism
π : Spec A → Spec A ′ induces an isomorphism π♯ : A ′ → A, which in turn induces
an isomorphism ρ : Spec A → Spec A ′ , then π = ρ. First show this on the levelof points; this is (surprisingly) the trickiest part. Then show π = ρ as maps of
topological spaces. Finally, to show π = ρ on the level of structure sheaves, use
the distinguished base. Feel free to use insights from later in this section, but be
careful to avoid circular arguments. Even struggling with this exercise and failing(until reading later sections) will be helpful.
∼ Af . Thus under the naturalMore generally, given f ∈ A, Γ (D(f), OSpec A ) =inclusion of sets Spec Af "→ Spec A, the Zariski topology on Spec A restricts togive the Zariski topology on Spec Af (Exercise 3.4.I), and the structure sheaf ofSpec A restricts to the structure sheaf of Spec Af , as the next exercise shows.
4.3.B. I MPORTANT BUT EASY EXERCISE . Suppose f ∈ A. Show that under theidentification of D(f) in Spec A with Spec Af (§3.5), there is a natural isomorphism∼ (Spec Af , OSpec A ). Hint: notice that distinof ringed spaces (D(f), OSpec A |D(f) ) =fguished open sets of Spec Af are already distinguished open sets in Spec A.
4.3.C. E ASY E XERCISE . If X is a scheme, and U is any open subset, prove that
(U, OX |U ) is also a scheme.
4.3.3. Definitions. We say (U, OX |U ) is an open subscheme of X. If U is also anaffine scheme, we often say U is an affine open subset, or an affine open subscheme, or sometimes informally just an affine open. For example, D(f) is anaffine open subscheme of Spec A. (Unimportant remark: it is not true that every
affine open subscheme of Spec A is of the form D(f); see §19.11.10 for an example.)
4.3.D. E ASY E XERCISE . Show that if X is a scheme, then the affine open sets forma base for the Zariski topology.
4.3.E. E ASY E XERCISE . The disjoint union of schemes is defined as you would
expect: it is the disjoint union of sets, with the expected topology (thus it is the disjoint union of topological spaces), with the expected sheaf. Once we know whatmorphisms are, it will be immediate (Exercise 9.1.A) that (just as for sets and topological spaces) disjoint union is the coproduct in the category of schemes.
(a) Show that the disjoint union of a finite number of affine schemes is also an affine
scheme. (Hint: Exercise 3.6.A.)(b) (a first example of a non-affine scheme) Show that an infinite disjoint union of(nonempty) affine schemes is not an affine scheme. (Hint: affine schemes are quasicompact, Exercise 3.6.G(a). This is basically answered in Remark 3.6.6.)

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4.3.4. Remark: A first glimpse of closed subschemes. Open subsets of a scheme comewith a natural scheme structure (Definition 4.3.3). For comparison, closed subsets
can have many “natural” scheme structures. We will discuss this later (in §8.1), but
for now, it suffices for you to know that a closed subscheme of X is, informally, aparticular kind of scheme structure on a closed subset of X. As an example: if I ⊂ A
is an ideal, then Spec A/I endows the closed subset V(I) ⊂ Spec A with a scheme
structure; but note that there can be different ideals with the same vanishing set(for example (x) and (x2 ) in k[x]).
4.3.5. Stalks of the structure sheaf: germs, values at a point, and the residue fieldof a point. Like every sheaf, the structure sheaf has stalks, and we shouldn’t be
surprised if they are interesting from an algebraic point of view. In fact, we haveseen them before.
4.3.F. I MPORTANT EASY EXERCISE . Show that the stalk of OSpec A at the point [p] isthe local ring Ap .
( (defined
Essentially the same argument will show that the stalk of the sheaf M
in Exercise 4.1.D) at [p] is Mp . Here is an interesting consequence, or if you prefer, ageometric interpretation of an algebraic"fact. A section is determined by its germs(Exercise 2.4.A), meaning that M →p Mp is an inclusion. So for example an
A-module is zero if and only if all its localizations at prime ideals are zero.

4.3.6. Definition. We say a ringed space is a locally ringed space if its stalks
are local rings. Thus Exercise 4.3.F shows that schemes are locally ringed spaces.
Manifolds are another example of locally ringed spaces, see §2.1.1. In both cases,
taking quotient by the maximal ideal may be interpreted as evaluating at the point.
The maximal ideal of the local ring OX,p is denoted mX,p or mp , and the residue
field OX,p /mp is denoted κ(p). Functions on an open subset U of a locally ringed
space have values at each point of U. The value at p of such a function lies inκ(p). As usual, we say that a function vanishes at a point p if its value at p is
0. (This generalizes our notion of the value of a function on Spec A, defined in§3.2.1.) Notice that we can’t even make sense of the phrase of “function vanishing”
on ringed spaces in general.

4.3.G. U SEFUL E XERCISE .
(a) If f is a function on a locally ringed space X, show that the subset of X where fvanishes is closed. (Hint: show that if f is a function on a ringed space X, show thatthe subset of X where the germ of f is invertible is open.)(b) Show that if f is a function on a locally ringed space that vanishes nowhere,
then f is invertible.
Consider a point [p] of an affine scheme Spec A. (Of course, any point of ascheme can be interpreted in this way, as each point has an affine open neighborhood.) The residue field at [p] is Ap /pAp , which is isomorphic to K(A/p), thefraction field of the quotient. It is useful to note that localization at p and taking

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quotient by p “commute”, i.e., the following diagram commutes.
(4.3.6.1)

Ap ❖❖
❖❖❖
❖❖quotient
❖❖❖
⑥
⑥
❖❖❖
⑥
⑥
4
⑥
A❆
Ap /pAp = K(A/p)
❆❆
♦B
❆❆
♦♦♦
♦
♦
❆
♦♦♦localize, i.e., K(·)
quotient ❆❆
1
♦♦♦
A/p
⑥A

localize ⑥⑥

For example, consider the scheme A2k = Spec k[x, y], where k is a field of characteristic not 2. Then (x2 + y2 )/x(y2 − x5 ) is a function away from the y-axis
and the curve y2 − x5 . Its value at (2, 4) (by which we mean [(x − 2, y − 4)]) is(22 + 42 )/(2(42 − 25 )), as
x2 + y2
22 + 42≡25x(y − x )2(42 − 25 )in the residue field — check this if it seems mysterious. And its value at [(y)], thex2
4generic point of the x-axis, is −x
6 = −1/x , which we see by setting y to 0. This
is indeed an element of the fraction field of k[x, y]/(y), i.e., k(x). (If you think
you care only about algebraically closed fields, let this example be a first warning:
Ap /pAp won’t be algebraically closed in general, even if A is a finitely generated
C-algebra!)
If anything makes you nervous, you should make up an example to make you
feel better. Here is one: 27/4 is a function on Spec Z − {[(2)], [(7)]} or indeed on aneven bigger open set. What is its value at [(5)]? Answer: 2/(−1) ≡ −2 (mod 5).
What is its value at the generic point [(0)]? Answer: 27/4. Where does it vanish?
At [(3)].
4.3.7. Stray definition: the fiber of an OX -module at a point. If F is an OX -module ona scheme X (or more generally, a locally ringed space), define the fiber (or fibre) of
F at a point p ∈ X by
F |p := Fp ⊗OX,p κ(p).
For example, OX |p is κ(p). (This notion will start to come into play in §13.7.)

4.4 Three examples
We now give three extended examples. Our short-term goal is to see that we
can really work with the structure sheaf, and can compute the ring of sections ofinteresting open sets that aren’t just distinguished open sets of affine schemes. Our
long-term goal is to meet interesting examples that will come up repeatedly in the
future.
4.4.1. First example: The plane minus the origin. This example will show youthat the distinguished base is something that you can work with. Let A = k[x, y],so Spec A = A2k . Let’s work out the space of functions on the open set U = A2 −{(0, 0)} = A2 − {[(x, y)]}.

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It is not immediately obvious whether this is a distinguished open set. (In fact
it is not — you may be able to figure out why within a few paragraphs, if you
can’t right now. It is not enough to show that (x, y) is not a principal ideal.) But
in any case, we can describe it as the union of two things which are distinguishedopen sets: U = D(x) ∪ D(y). We will find the functions on U by gluing together
functions on D(x) and D(y).
The functions on D(x) are, by Definition 4.1.1, Ax = k[x, y, 1/x]. The functionson D(y) are Ay = k[x, y, 1/y]. Note that A injects into its localizations (if 0 is
not inverted), as it is an integral domain (Exercise 1.3.C), so A injects into both
Ax and Ay , and both inject into Axy (and indeed k(x, y) = K(A)). So we arelooking for functions on D(x) and D(y) that agree on D(x) ∩ D(y) = D(xy), i.e.,
we are interpreting Ax ∩ Ay in Axy (or in k(x, y)). Clearly those rational functions
with only powers of x in the denominator, and also with only powers of y in the
denominator, are the polynomials. Translation: Ax ∩ Ay = A. Thus we conclude:(4.4.1.1)

Γ (U, OA2 ) ≡ k[x, y].

In other words, we get no extra functions by removing the origin. Notice how easythat was to calculate!
4.4.2. Aside. Notice that any function on A2 − {(0, 0)} extends over all of A2 . This isan analog of Hartogs’s Lemma in complex geometry: you can extend a holomorphic
function defined on the complement of a set of codimension at least two on a complex manifold over the missing set. This will work more generally in the algebraic
setting: you can extend over points in codimension at least 2 not only if they are“smooth”, but also if they are mildly singular — what we will call normal. We will
make this precise in §11.3.11. This fact will be very useful for us.
4.4.3. We now show an interesting fact: (U, OA2 |U ) is a scheme, but it is not anaffine scheme. (This is confusing, so you will have to pay attention.) Here’s
why: otherwise, if (U, OA2 |U ) = (Spec A, OSpec A ), then we can recover A by taking
global sections:A = Γ (U, OA2 |U ),which we have already identified in (4.4.1.1) as k[x, y]. So if U is affine, then
∼ A2 . But this bijection between prime ideals in a ring and points of the specU=ktrum is more constructive than that: given the prime ideal I, you can recover the point
as the generic point of the closed subset cut out by I, i.e., V(I), and given the point p, you
can recover the ideal as those functions vanishing at p, i.e., I(p). In particular, the prime
ideal (x, y) of A should cut out a point of Spec A. But on U, V(x) ∩ V(y) = ∅. Conclusion: U is not an affine scheme. (If you are ever looking for a counterexample
to something, and you are expecting one involving a non-affine scheme, keep thisexample in mind!)
4.4.4. Gluing two copies of A1 together in two different ways. We have now
seen two examples of non-affine schemes: an infinite disjoint union of nonempty
schemes (Exercise 4.3.E) and A2 − {(0, 0)}. I want to give you two more examples.
They are important because they are the first examples of fundamental behavior,
the first pathological, and the second central.
First, I need to tell you how to glue two schemes together. Before that, you
should review how to glue topological spaces together along isomorphic open

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sets. Given two topological spaces X and Y, and open subsets U ⊂ X and V ⊂ Y∼ %along with a homeomorphism U (V , we can create a new topological space∼ %W, that we think of as gluing
X and Y together along U (V . It is the quotient#of the disjoint union X Y by the equivalence relation U ∼ V, where the quotientis given the quotient topology. Then X and Y are naturally (identified with) open
subsets of W, and indeed cover W. Can you restate this cleanly with an arbitrary(not necessarily finite) number of topological spaces?
Now that we have discussed gluing topological spaces, let’s glue schemes together. (This applies without change more generally to ringed spaces.) Suppose
you have two schemes (X, OX ) and (Y, OY ), and open subsets U ⊂ X and V ⊂ Y,∼
% V , and an isomorphism of structure
along with a homeomorphism π : U∼ %∼ (V, OY |V )).
sheaves OVπ∗ OU (i.e., an isomorphism of schemes (U, OX |U ) =Then we can glue these together to get a single scheme. Reason: let W be X and∼ V. Then Exercise 2.5.D shows that
Y glued together using the isomorphism U =the structure sheaves can be glued together to get a sheaf of rings. Note that thisis indeed a scheme: any point has an open neighborhood that is an affine scheme.
(Do you see why?)

4.4.A. E SSENTIAL E XERCISE ( CF. E XERCISE 2.5.D). Show that you can glue anarbitrary collection of schemes together. Suppose we are given:• schemes Xi (as i runs over some index set I, not necessarily finite),• open subschemes Xij ⊂ Xi with Xii = Xi ,• isomorphisms fij : Xij → Xji with fii the identity

such that• (the cocycle condition) the isomorphisms “agree on triple intersections”,
i.e., fik |Xij ∩Xik = fjk |Xji ∩Xjk ◦ fij |Xij ∩Xik (so implicitly, to make sense of
the right side, fij (Xik ∩ Xij ) ⊂ Xjk ).
(The cocycle condition ensures that fij and fji are inverses. In fact, the hypothesisthat fii is the identity also follows from the cocycle condition.) Show that there is aunique scheme X (up to unique isomorphism) along with open subsets isomorphic
to the Xi respecting this gluing data in the obvious sense. (Hint: what is X as a set?
What is the topology on this set? In terms of your description of the open sets of
X, what are the sections of this sheaf over each open set?)
I will now give you two non-affine schemes. Both are handy to know. In bothcases, I will glue together two copies of the affine line A1k . Let X = Spec k[t], andY = Spec k[u]. Let U = D(t) = Spec k[t, 1/t] ⊂ X and V = D(u) = Spec k[u, 1/u] ⊂Y. We will get both examples by gluing X and Y together along U and V. The
difference will be in how we glue.
4.4.5. Second example: the affine line with the doubled origin. Consider the∼ V via the isomorphism k[t, 1/t] =∼ k[u, 1/u] given by t ↔ u (cf.
isomorphism U =Exercise 4.3.A). The resulting scheme is called the affine line with doubled origin.
Figure 4.6 is a picture of it.
As the picture suggests, intuitively this is an analog of a failure of Hausdorffness.
Now A1 itself is not Hausdorff, so we can’t say that it is a failure of Hausdorffness.

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F IGURE 4.6. The affine line with doubled origin
We see this as weird and bad, so we will want to make a definition that will prevent this from happening. This will be the notion of separatedness (to be discussed
in Chapter 10).(In fact, in some sense, separatedness is the right definition of Hausdorfness!) This will answer other of our prayers as well. For example, on a separated scheme, the “affine base of the Zariski topology” is nice — the intersectionof two affine open sets will be affine (Proposition 10.1.8).
4.4.B. E XERCISE . Show that the affine line with doubled origin is not affine. Hint:
calculate the ring of global sections, and look back at the argument for A2 − {(0, 0)}.
4.4.C. E ASY E XERCISE . Do the same construction with A1 replaced by A2 . You will
have defined the affine plane with doubled origin. Describe two affine open subsets of this scheme whose intersection is not an affine open subset. (An “infinitedimensional” version comes up in Exercise 5.1.J.)∼ V via
4.4.6. Third example: the projective line. Consider the isomorphism U =∼the isomorphism k[t, 1/t] = k[u, 1/u] given by t ↔ 1/u. Figure 4.7 is a suggestive
picture of this gluing. The resulting scheme is called the projective line over thefield k, and is denoted P1k .

F IGURE 4.7. Gluing two affine lines together to get P1
Notice how the points glue. Let me assume that k is algebraically closed forconvenience. (You can think about how this changes otherwise.) On the first affine
line, we have the closed (“traditional”) points [(t − a)], which we think of as “aon the t-line”, and we have the generic point [(0)]. On the second affine line, wehave closed points that are “b on the u-line”, and the generic point. Then a on
the t-line is glued to 1/a on the u-line (if a ̸= 0 of course), and the generic point
is glued to the generic point (the ideal (0) of k[t] becomes the ideal (0) of k[t, 1/t]
upon localization, and the ideal (0) of k[u] becomes the ideal (0) of k[u, 1/u]. And(0) in k[t, 1/t] is (0) in k[u, 1/u] under the isomorphism t ↔ 1/u).

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4.4.7. If k is algebraically closed, we can interpret the closed points of P1k in thefollowing way, which may make this sound closer to the way you have seen projective space defined earlier. The points are of the form [a, b], where a and b are
not both zero, and [a, b] is identified with [ac, bc] where c ∈ k× . Then if b ̸= 0, thisis identified with a/b on the t-line, and if a ̸= 0, this is identified with b/a on theu-line.
4.4.8. Proposition. — P1k is not affine.
Proof. We do this by calculating the ring of global sections. The global sections
correspond to sections over X and sections over Y that agree on the overlap. A
section on X is a polynomial f(t). A section on Y is a polynomial g(u). If we restrictf(t) to the overlap, we get something we can still call f(t); and similarly for g(u).
Now we want them to be equal: f(t) = g(1/t). But the only polynomials in t thatare at the same time polynomials in 1/t are the constants k. Thus Γ (P1 , OP1 ) = k.
If P1 were affine, then it would be Spec Γ (P1 , OP1 ) = Spec k, i.e., one point. But it
isn’t — it has lots of points.
!
We have proved an analog of an important theorem: the only holomorphic
functions on CP1 are the constants! (See §10.3.7 for a serious yet easy generalization.)
4.4.9. Important Example: Projective space. We now make a preliminary definition of projective n-space over a field k, denoted Pn
k , by gluing together n + 1
open sets each isomorphic to An
k . Judicious choice of notation for these open sets
will make our life easier. Our motivation is as follows. In the construction of P1
above, we thought of points of projective space as [x0 , x1 ], where (x0 , x1 ) are only
determined up to scalars, i.e., (x0 , x1 ) is considered the same as (λx0 , λx1 ). Thenthe first patch can be interpreted by taking the locus where x0 ̸= 0, and then we
consider the points [1, t], and we think of t as x1 /x0 ; even though x0 and x1 are notwell-defined, x1 /x0 is. The second corresponds to where x1 ̸= 0, and we consider
the points [u, 1], and we think of u as x0 /x1 . It will be useful to instead use thenotation x1/0 for t and x0/1 for u.
For Pn , we glue together n + 1 open sets, one for each of i = 0, . . . , n. The ith
open set Ui will have coordinates x0/i , . . . , x(i−1)/i , x(i+1)/i , . . . , xn/i . It will be
convenient to write this as(4.4.9.1)

Spec k[x0/i , x1/i , . . . , xn/i ]/(xi/i − 1)(so we have introduced a “dummy variable” xi/i which we immediately set to1). We glue the distinguished open set D(xj/i ) of Ui to the distinguished open set
D(xi/j ) of Uj , by identifying these two schemes by describing the identification ofrings∼
Spec k[x0/i , x1/i , . . . , xn/i , 1/xj/i ]/(xi/i − 1) =Spec k[x0/j , x1/j , . . . , xn/j , 1/xi/j ]/(xj/j − 1)
via xk/i = xk/j /xi/j and xk/j = xk/i /xj/i (which implies xi/j xj/i = 1). We need tocheck that this gluing information agrees over triple overlaps.
4.4.D. E XERCISE . Check this, as painlessly as possible. (Possible hint: the tripleintersection is affine; describe the corresponding ring.)

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4.4.10. Definition: Pn
Note that our definition does not use the fact that k is aA.
field. Hence we may as well define Pn
A for any ring A. This will be useful later.
n∼
4.4.E. E XERCISE . Show that the only functions on Pn
k are constants (Γ (Pk , O) = k),nand hence that Pk is not affine if n > 0. Hint: you might fear that you will needsome delicate interplay among all of your affine open sets, but you will only need
two of your open sets to see this. There is even some geometric intuition behind
this: the complement of the union of two open sets has codimension 2. But “Algebraic Hartogs’s Lemma” (discussed informally in §4.4.2, and to be stated rigorously in Theorem 11.3.11) says that any function defined on this union extends
to be a function on all of projective space. Because we are expecting to see only
constants as functions on all of projective space, we should already see this for thisunion of our two affine open sets.

4.4.F. E XERCISE ( GENERALIZING §4.4.7). Show that if k is algebraically closed,
the closed points of Pn
k may be interpreted in the traditional way: the points areof the form [a0 , . . . , an ], where the ai are not all zero, and [a0 , . . . , an ] is identified
with [λa0 , . . . , λan ] where λ ∈ k× .

We will later give other definitions of projective space (Definition 4.5.8, §16.4.2).
Our first definition here will often be handy for computing things. But there issomething unnatural about it — projective space is highly symmetric, and that
isn’t clear from our current definition. Furthermore, as noted by Herman Weyl
[Wey, p. 90], “The introduction of numbers as coordinates is an act of violence.”

4.4.11. Fun aside: The Chinese Remainder Theorem is a geometric fact. The
Chinese Remainder Theorem is embedded in what we have done. We will see thisin a single example, but you should then figure out the general statement. The
Chinese Remainder Theorem says that knowing an integer modulo 60 is the sameas knowing an integer modulo 3, 4, and 5. Here is how to see this in the languageof schemes. What is Spec Z/(60)? What are the prime ideals of this ring? Answer:
those prime ideals containing (60), i.e., those primes dividing 60, i.e., (2), (3), and(5). Figure 4.8 is a sketch of Spec Z/(60). They are all closed points, as these are all
maximal ideals, so the topology is the discrete topology. What are the stalks? You
can check that they are Z/4, Z/3, and Z/5. The nilpotents “at (2)” are indicated by
the “fuzz” on that point. (We discussed visualizing nilpotents with “infinitesimalfuzz” in §4.2.) So what are global sections on this scheme? They are sections on
this open set (2), this other open set (3), and this third open set (5). In other words,
we have a natural isomorphism of rings
Z/60

[(2)]

% Z/4 × Z/3 × Z/5.

[(3)]

[(5)]

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4.4.12. ⋆ Example. Here is an example of a function on an open subset of a schemewith some surprising behavior. On X = Spec k[w, x, y, z]/(wz − xy), consider the
open subset D(y) ∪ D(w). Clearly the function z/y on D(y) agrees with x/w on
D(w) on the overlap D(y) ∩ D(w). Hence they glue together to give a section. You
may have seen this before when thinking about analytic continuation in complex
geometry — we have a “holomorphic” function which has the description z/y onan open set, and this description breaks down elsewhere, but you can still “analytically continue” it by giving the function a different definition on different parts ofthe space.
Follow-up for curious experts. This function has no “single description” as awell-defined expression in terms of w, x, y, z! There is a lot of interesting geometry here, and this scheme will be a constant source of (counter)examples for us(look in the index under “cone over smooth quadric surface”). Here is a glimpse,
in terms of words we have not yet defined. The space Spec k[w, x, y, z] is A4 , andis, not surprisingly, 4-dimensional. We are working with the subset X, which is a
hypersurface, and is 3-dimensional. It is a cone over a smooth quadric surface inP3 (flip to Figure 8.2). The open subset D(y) ⊂ X is X minus some hypersurface,
so we are throwing away a codimension 1 locus. D(w) involves throwing away
another codimension 1 locus. You might think that the intersection of these two
discarded loci is then codimension 2, and that failure of extending this weird function to a global polynomial comes because of a failure of our Hartogs’s Lemmatype theorem. But that’s not true — V(y) ∩ V(w) is in fact codimension 1. Hereis what is actually going on. The space V(y) is obtained by throwing away the(cone over the) union of two lines ℓ and m1 , one in each “ruling” of the surface,and V(w) also involves throwing away the (cone over the) union of two lines ℓand m2 . The intersection is the (cone over the) line ℓ, which is a codimension 1
set. Remarkably, despite being “pure codimension 1” the cone over ℓ is not cut out
even set-theoretically by a single equation (see Exercise 14.2.R). This means thatany expression f(w, x, y, z)/g(w, x, y, z) for our function cannot correctly describe
our function on D(y)∪D(w) — at some point of D(y)∪D(w) it must be 0/0. Here’s
why. Our function can’t be defined on V(y) ∩ V(w), so g must vanish here. But g
can’t vanish just on the cone over ℓ — it must vanish elsewhere too.

4.5 Projective schemes, and the Proj construction
Projective schemes are important for a number of reasons. Here are a few.
Schemes that were of “classical interest” in geometry — and those that you would
have cared about before knowing about schemes — are all projective or an open
subset thereof (basically, quasiprojective, see §4.5.9). Moreover, most “schemes ofinterest” tend to be projective or quasiprojective. In fact, it is very hard to evengive an example of a scheme satisfying basic properties — for example, finite type
and “Hausdorff” (“separated”) over a field — that is provably not quasiprojective.
For complex geometers: it is hard to find a compact complex variety that is provably not projective (see Remark 10.3.6), and it is quite hard to come up with acomplex variety that is provably not an open subset of a projective variety. So projective schemes are really ubiquitous. Also, the notion of “projective k-scheme” is

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a good approximation of the algebro-geometric version of compactness (“properness”, see §10.3).
Finally, although projective schemes may be obtained by gluing together affine
schemes, and we know that keeping track of gluing can be annoying, there is a
simple means of dealing with them without worrying about gluing. Just as there
is a rough dictionary between rings and affine schemes, we will have a (slightly
looser) analogous dictionary between graded rings and projective schemes. Just
as one can work with affine schemes by instead working with rings, one can workwith projective schemes by instead working with graded rings.
4.5.1. Motivation from classical geometry.
For geometric intuition, we recall how one thinks of projective space “classically” (in the classical topology, over the real numbers). Pn can be interpreted as
the lines through the origin in Rn+1 . Thus subsets of Pn correspond to unions oflines through the origin of Rn+1 , and closed subsets correspond to such unions
which are closed. (The same is not true with “closed” replaced by “open”!)
One often pictures Pn as being the “points at infinite distance” in Rn+1 , where
the points infinitely far in one direction are associated with the points infinitely far
in the opposite direction. We can make this more precise using the decomposition)
Pn+1 = Rn+1
Pn
by which we mean that there is an open subset in Pn+1 identified with Rn+1 (thepoints with last “projective coordinate” nonzero), and the complementary closed
subset identified with Pn (the points with last “projective coordinate” zero). (Thephrase “projective coordinate” will be formally defined in §4.5.8, but we will use
it even before then, in Exercises 4.5.A and 4.5.B.)
Then for example any equation cutting out some set V of points in Pn will also
cut out some set of points in Rn+1 that will be a closed union of lines. We call thisthe affine cone of V. These equations will cut out some union of P1 ’s in Pn+1 , andwe call this the projective cone of V. The projective cone is the disjoint union of theaffine cone and V. For example, the affine cone over x2 + y2 = z2 in P2 is justthe “classical” picture of a cone in R3 , see Figure 4.9. We will make this analogy
precise in our algebraic setting in §8.2.12.
x2 + y2 = z2 in P2
projective cone in P3
affine cone: x2 + y2 = z2 in R3

F IGURE 4.9. The affine and projective cone of x2 + y2 = z2 inclassical geometry

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4.5.2. Projective schemes, a first description.
We now describe a construction of projective schemes, which will help motivate the Proj construction. We begin by giving an algebraic interpretation of thecone just described. We switch coordinates from x, y, z to x0 , x1 , x2 in order to use
the notation of §4.4.9.
4.5.A. E XERCISE ( WORTH DOING BEFORE READING THE REST OF THIS SECTION ).
Consider P2k , with projective coordinates x0 , x1 , and x2 . (The terminology “projective coordinate” will not be formally defined until §4.5.8, but you should be able tosolve this problem anyway.) Think through how to define a scheme that should beinterpreted as x20 + x21 − x22 = 0 “in P2k ”. Hint: in the affine open subset corresponding to x2 ̸= 0, it should (in the language of 4.4.9) be cut out by x20/2 + x21/2 − 1 = 0,i.e., it should “be” the scheme Spec k[x0/2 , x1/2 ]/(x20/2 + x21/2 − 1). You can similarly guess what it should be on the other two standard open sets, and show thatthe three schemes glue together.
4.5.3. Remark: Degree d hypersurfaces in Pn . We informally observe that degreed homogeneousin n + 1 variables over a field form a vector space) polynomials*of dimension n+d
.
(This
is
essentially the content of Exercise 8.2.K and Exerd
cise 14.1.C.) It is almost true that two polynomials cut out the same subset of Pn
k if
one is a nonzero multiple of the other. Unfortunately, the examples of x2 y = 0 and
xy2 = 0 show that this isn’t quite right. You will later be able to check that two
polynomials cut out the same closed subscheme (whatever that means) if and only if
one is a nonzero multiple of the other. (This is some evidence that the notion of a“closed subscheme” is better than that of a “closed subset”.) The zero polynomial
doesn’t really cut out a hypersurface in any reasonable sense of the word. Thus
we informally imagine that “degree d hypersurfaces in Pn are parametrized byn+dP( d )−1 ”. This intuition will come up repeatedly (in special cases), and we will
give it some precise meaning in §28.3.5. (We will properly define hypersurfaces in§8.2.2, once we have the language of closed subschemes. At that time we will also
define line, hyperplane, quadric hypersurfaces, conic curves, and other wondrous
notions.)
4.5.B. E XERCISE . More generally, consider Pn
A , with projective coordinates x0 ,. . . , xn . Given a collection of homogeneous polynomials fi ∈ A[x0 , . . . , xn ], makesense of the scheme “cut out in Pn
A by the fi .” (This will later be made precise as
an example of a “vanishing scheme”, see Exercise 4.5.P.) Hint: you will be able topiggyback on Exercise 4.4.D to make this quite straightforward.
This can be taken as the definition of a projective A-scheme, but we will wait
until §4.5.9 to state it a little better.
4.5.4. Preliminaries on graded rings.
The Proj construction produces a scheme out of a graded ring. We now givesome background on graded rings.
4.5.5. Z-graded rings. A Z-graded ring is a ring S• = ⊕n∈Z Sn (the subscript iscalled the grading), where multiplication respects the grading, i.e., sends Sm × Sn
to Sm+n . Clearly S0 is a subring, each Sn is an S0 -module, and S• is a S0 -algebra.
Suppose for the remainder of §4.5.5 that S• is a Z-graded ring. Those elements of

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some Sn are called homogeneous elements of S• ; nonzero homogeneous elements
have an obvious degree. An ideal I of S• is a homogeneous ideal (or a graded
ideal, although we won’t use this terminology) if it is generated by homogeneous
elements.
4.5.C. E XERCISE .
(a) Show that an ideal I is homogeneous if and only if it contains the degree n pieceof each of its elements for each n. (Hence I can be decomposed into homogeneouspieces, I = ⊕In , and S• /I has a natural Z-grading. This is the reason for the namehomogeneous ideal.)(b) Show that the set of homogeneous ideals of a given Z-graded ring S• is closed
under sum, product, intersection, and radical.
(c) Show that a homogeneous ideal I ⊂ S• is prime if I ̸= S• , and if for any homogeneous a, b ∈ S• , if ab ∈ I, then a ∈ I or b ∈ I.
If T is a multiplicative subset of S• containing only homogeneous elements,
then T −1 S• has a natural structure as a Z-graded ring.
(Everything in §4.5.5 can be generalized: Z can be replaced by an arbitrary
abelian group.)
4.5.6. Z≥0 -graded rings, graded ring over A, and finitely generated graded rings.
Z≥0 -graded ring is a Z-graded ring with no elements of negative degree.

A

For the remainder of the book, graded ring will refer to a Z≥0 -graded ring.
Warning: this convention is nonstandard (for good reason).
From now on, unless otherwise stated, S• is assumed to be a graded ring. Fixa ring A, which we call the base ring. If S0 = A, we say that S• is a graded ringover A. A key example is A[x0 , . . . , xn ], or more generally A[x0 , . . . , xn ]/I where
I is a homogeneous ideal with I ∩ S0 = 0 (cf. Exercise 4.5.B). Here we take theconventional grading on A[x0 , . . . , xn ], where each xi has weight 1.
The subset S+ := ⊕i>0 Si ⊂ S• is an ideal, called the irrelevant ideal. The reason for the name “irrelevant” will be clearer in a few paragraphs. If the irrelevant
ideal S+ is a finitely generated ideal, we say that S• is a finitely generated gradedring over A. If S• is generated by S1 as an A-algebra, we say that S• is generated indegree 1. (We will later find it useful to interpret “S• is generated in degree 1” as“the natural map Sym• S1 → S• is a surjection”. The symmetric algebra construction
will be briefly discussed in §13.5.3.)
4.5.D. E XERCISE .
(a) Show that a graded ring S• over A is a finitely generated graded ring (overA) if and only if S• is a finitely generated graded A-algebra, i.e., generated over
A = S0 by a finite number of homogeneous elements of positive degree. (Hint
for the forward implication: show that the generators of S+ as an ideal are also
generators of S• as an algebra.)(b) Show that a graded ring S• over A is Noetherian if and only if A = S0 is
Noetherian and S• is a finitely generated graded ring.
4.5.7. The Proj construction.
We now define a scheme Proj S• , where S• is a (Z≥0 -)graded ring. Here are two
examples, to provide a light at the end of the tunnel. If S• = A[x0 , . . . , xn ], we will

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recover Pn
A ; and if S• = A[x0 , . . . , xn ]/(f(x0 , . . . , xn )) where f is homogeneous, wewill construct something “cut out in Pn
A by the equation f = 0” (cf. Exercise 4.5.B).
As we did with Spec of a ring, we will build Proj S• first as a set, then as a
topological space, and finally as a ringed space. In our preliminary definition ofPn
A , we glued together n + 1 well-chosen affine pieces, but we don’t want to makeany choices, so we do this by simultaneously considering “all possible” affine open
sets. Our affine building blocks will be as follows. For each homogeneous f ∈ S+ ,note that the localization (S• )f is naturally a Z-graded ring, where deg(1/f) =− deg f. Consider(4.5.7.1)

Spec((S• )f )0 .

where ((S• )f )0 means the 0-graded piece of the graded ring (S• )f . (These will be
our affine building blocks, as f varies over the homogeneous elements of S+ .) The
notation ((S• )f )0 is admittedly horrible — the first and third subscripts refer to thegrading, and the second refers to localization. As motivation for considering thisconstruction: applying this to S• = k[x0 , . . . , xn ], with f = xi , we obtain the ringappearing in (4.4.9.1):
k[x0/i , x1/i , . . . , xn/i ]/(xi/i − 1).
(Before we begin the construction: another possible way of defining Proj S• is
by gluing together affines of this form, by jumping straight to Exercises 4.5.K and
4.5.L. If you prefer that, by all means do so.)The points of Proj S• are the set of homogeneous prime ideals of S• not containing the irrelevant ideal S+ (the “relevant prime ideals”).
4.5.E. I MPORTANT AND TRICKY EXERCISE . Suppose f ∈ S+ is homogeneous.
(a) Give a bijection between the prime ideals of ((S• )f )0 and the homogeneous
prime ideals of (S• )f . Hint: Avoid notational confusion by proving instead that if
A is a Z-graded ring with a homogeneous invertible element f in positive degree,then there is a bijection between prime ideals of A0 and homogeneous prime idealsof A. Using the ring map A0 → A, from each homogeneous prime ideal of A wefind a prime ideal of A0 . The reverse direction is the harder one. Given a prime
ideal P0 ⊂ A0 , define P ⊂ A (a priori only a subset) as ⊕Qi , where Qi ⊂ Ai , anda ∈ Qi if and only if adeg f /fi ∈ P0 . Note that Q0 = P0 . Show that a ∈ Qi if andonly if a2 ∈ Q2i ; show that if a1 , a2 ∈ Qi then a21 + 2a1 a2 + a22 ∈ Q2i and hence
a1 + a2 ∈ Qi ; then show that P is a homogeneous ideal of A; then show that P isprime.
(b) Interpret the set of prime ideals of ((S• )f )0 as a subset of Proj S• .
The correspondence of the points of Proj S• with homogeneous prime ideals
helps us picture Proj S• . For example, if S• = k[x, y, z] with the usual grading, thenwe picture the homogeneous prime ideal (z2 − x2 − y2 ) first as a subset of Spec S• ;it is a cone (see Figure 4.9). As in §4.5.1, we picture P2k as the “plane at infinity”.
Thus we picture this equation as cutting out a conic “at infinity” (in Proj S• ). We
will make this intuition somewhat more precise in §8.2.12.
Motivated by the affine case, if T is a set of homogeneous elements of S• of positive degree, define the (projective) vanishing set of T , V(T ) ⊂ Proj S• , to be thosehomogeneous prime ideals containing T but not S+ . Define V(f) if f is a homogeneous element of positive degree, and V(I) if I is a homogeneous ideal containedin S+ , in the obvious way. Let D(f) := Proj S• \ V(f) (the projective distinguished

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open set) be the complement of V(f). Once we define a scheme structure on Proj S• ,we will (without comment) use D(f) to refer to the open subscheme, not just the
open subset. (These definitions can certainly be extended to remove the positive
degree hypotheses. For example, the definition of V(T ) makes sense for any subset T of S• , and the definition of D(f) makes sense even if f has degree 0. In what
follows, we deliberately stick to these narrower definitions. For example, we will
want the D(f) to form an affine cover, and if f has degree 0, then D(f) needn’t beaffine.)4.5.F. E XERCISE . Show that D(f) “is” (or more precisely, “corresponds to”) the
subset Spec((S• )f )0 you described in Exercise 4.5.E(b). For example, in §4.4.9, theD(xi ) are the standard open sets covering projective space.
As in the affine case, the V(I)’s satisfy the axioms of the closed set of a topology, and we call this the Zariski topology on Proj S• . (Other definitions given in
the literature may look superficially different, but can be easily shown to be thesame.) Many statements about the Zariski topology on Spec of a ring carry overto this situation with little extra work. Clearly D(f) ∩ D(g) = D(fg), by the same
immediate argument as in the affine case (Exercise 3.5.D).
4.5.G. E ASY E XERCISE . Verify that the projective distinguished open sets D(f)(as f runs through the homogeneous elements of S+ ) form a base of the Zariski
topology.
4.5.H. E XERCISE . Fix a graded ring S• .
(a) Suppose I is any homogeneous ideal of S• contained in S+ , and f is a homogeneous element of positive degree. Show that f vanishes on V(I) (i.e., V(I) ⊂ V(f))
if and only if fn ∈ I for some n. (Hint: Mimic the affine case; see Exercise 3.4.J.) In
particular, as in the affine case (Exercise 3.5.E), if D(f) ⊂ D(g), then fn ∈ (g) forsome n, and vice versa. (Here g is also homogeneous of positive degree.)(b) If Z ⊂ Proj S• , define I(Z) ⊂ S+ . Show that it is a homogeneous ideal of S• . For
any two subsets, show that I(Z1 ∪ Z2 ) = I(Z1 ) ∩ I(Z2 ).
(c) For any subset Z ⊂ Proj S• , show that V(I(Z)) = Z.
4.5.I. E XERCISE ( CF. E XERCISE 3.5.B). Fix a graded ring S• , and a homogeneous
ideal I. Show that the following are equivalent.
(a) V(I) = ∅.
(b) For any fi (as i runs through some index set) generating I, ∪D(fi ) =Proj√ S• .
(c) I ⊃ S+ .

This is more motivation for the ideal S+ being “irrelevant”: any ideal whose radical contains it is “geometrically irrelevant”.
We now construct Proj S• as a scheme.
4.5.J. E XERCISE . Suppose some homogeneous f ∈ S+ is given. Via the inclusion
D(f) = Spec((S• )f )0 "→ Proj S•
of Exercise 4.5.F, show that the Zariski topology on Proj S• restricts to the Zariski
topology on Spec((S• )f )0 .

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Now that we have defined Proj S• as a topological space, we are ready to define the structure sheaf. On D(f), we wish it to be the structure sheaf of Spec((S• )f )0 .
We will glue these sheaves together using Exercise 2.5.D on gluing sheaves.
4.5.K. E XERCISE . If f, g ∈ S+ are homogeneous and nonzero, describe an isomorphism between Spec((S• )fg )0 and the distinguished open subset D(gdeg f /fdeg g )of Spec((S• )f )0 .
Similarly, Spec((S• )fg )0 is identified with a distinguished open subset of Spec((S• )g )0 .
We then glue the various Spec((S• )f )0 (as f varies) altogether, using these pairwise
gluings.
4.5.L. E XERCISE . By checking that these gluings behave well on triple overlaps(see Exercise 2.5.D), finish the definition of the scheme Proj S• .
4.5.M. E XERCISE . (Some will find this essential, others will prefer to ignore it.)(Re)interpret the structure sheaf of Proj S• in terms of compatible germs.
4.5.8. Definition. We (re)define projective space (over a ring A) by Pn
A := Proj A[x0 , . . . , xn ].
This definition involves no messy gluing, or special choice of patches. Note thatthe variables x0 , . . . , xn , which we call the projective coordinates on Pn
A , are partof the definition. (They may have other names than x’s, depending on the context.)
4.5.N. E XERCISE .
Check that this agrees with our earlier construction of Pn
A(§4.4.9). (How do you know that the D(xi ) cover Proj A[x0 , . . . , xn ]?)Notice that with our old definition of projective space, it would have been anontrivial exercise to show that D(x2 + y2 − z2 ) ⊂ P2k (the complement of a plane
conic) is affine; with our new perspective, it is immediate — it is Spec(k[x, y, z](x2 +y2 −z2 ) )0 .
4.5.O. E XERCISE .
Suppose that k is an algebraically closed field. We know
from Exercise 4.4.F that the closed points of Pn
k , as defined in §4.4.9, are in bijection with the points of “classical” projective space. By Exercise 4.5.N, the schemePnk as defined in §4.4.9 is isomorphic to Proj k[x0 , . . . , xn ]. Therefore, each point[a0 , . . . , an ] of classical projective space corresponds to a homogeneous prime
ideal of k[x0 , . . . , xn ]. Which homogeneous prime ideal is it?
We now figure out the “right definition” of the vanishing scheme, in analogy
with the vanishing set V(·) defined at the start of §3.4. You will be defining a closed
subscheme (mentioned in Remark 4.3.4, and to be properly defined in §8.1).
4.5.P. E XERCISE . If S• is generated in degree 1, and f ∈ S+ is homogeneous,
explain how to define V(f) “in” Proj S• , the vanishing scheme of f. (Warning: fin general isn’t a function on Proj S• . We will later interpret it as something close:a section of a line bundle, see for example §14.1.2.) Hence define V(I) for any
homogeneous ideal I of S+ . (Another solution in more general circumstances will
be given in Exercise 13.1.I.)
4.5.9. Projective and quasiprojective schemes.
We call a scheme of the form (i.e., isomorphic to) Proj S• , where S• is a finitely
generated graded ring over A, a projective scheme over A, or a projective Ascheme. A quasiprojective A-scheme is a quasicompact open subscheme of a

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projective A-scheme. The “A” is omitted if it is clear from the context; often A isa field.
4.5.10. Unimportant remarks. (i) Note that Proj S• makes sense even when S• is not
finitely generated. This can be useful. For example, you will later be able to do
Exercise 6.4.D without worrying about Exercise 6.4.H.
(ii) The quasicompact requirement in the definition of quasiprojectivity is ofcourse redundant in the Noetherian case (cf. Exercise 3.6.T), which is all that matters to most.
∼ Spec A. Thus “Spec A is a
4.5.11. Silly example. Note that P0A = Proj A[T ] =projective A-scheme”.
4.5.12. Example: projectivization of a vector space PV. We can make this definition of projective space even more choice-free as follows. Let V be an (n + 1)dimensional vector space over k. (Here k can be replaced by any ring A as usual.)
DefineSym• V ∨ = k ⊕ V ∨ ⊕ Sym2 V ∨ ⊕ · · · .
(The reason for the dual is explained by the next exercise. For a reminder of the definition of Sym, flip to §13.5.3.) If for example V is the dual of the vector space withbasis associated to x0 , . . . , xn , we would have Sym• V ∨ = k[x0 , . . . , xn ]. Then we
can define PV := Proj(Sym• V ∨ ). In this language, we have an interpretation forx0 , . . . , xn : they are linear functionals on the underlying vector space V. (Warning:some authors use the definition PV = Proj(Sym• V), so be cautious.)
4.5.Q. U NIMPORTANT EXERCISE . Suppose k is algebraically closed. Describe anatural bijection between one-dimensional subspaces of V and the closed pointsof PV. Thus this construction canonically (in a basis-free manner) describes the
one-dimensional subspaces of the vector space V.
Unimportant remark: you may be surprised at the appearance of the dual in
the definition of PV. This is partially explained by the previous exercise. Most
normal (traditional) people define the projectivization of a vector space V to be
the space of one-dimensional subspaces of V. Grothendieck considered the projectivization to be the space of one-dimensional quotients. One motivation for this isthat it gets rid of the annoying dual in the definition above. There are better reasons, that we won’t go into here. In a nutshell, quotients tend to be better-behaved
than subobjects for coherent sheaves, which generalize the notion of vector bundle. (Coherent sheaves are discussed in Chapter 13.)On another note related to Exercise 4.5.Q: you can also describe a natural bijection between points of V and the closed points of Spec(Sym• V ∨ ). This construction respects the affine/projective cone picture of §8.2.12.
4.5.13. The Grassmannian. At this point, we could describe the fundamental geometric object known as the Grassmannian, and give the “wrong” (but correct) definition of it. We will instead wait until §6.7 to give the wrong definition, when we
will know enough to sense that something is amiss. The right definition will begiven in §16.7.

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CHAPTER 5

Some properties of schemes

5.1 Topological properties
We will now define some useful properties of schemes. As you see each example, you should try these out in specific examples of your choice, such as particular
schemes of the form Spec C[x1 . . . , xn ]/(f1 , . . . , fr ).
The definitions of connected, connected component, (ir)reducible, irreducible component, quasicompact, generization, specialization, generic point, Noetherian topologicalspace, and closed point, were given in §3.6. You should have pictures in your mindof each of these notions.
Exercise 3.6.C shows that An is irreducible (it was easy). This argument “behaves well under gluing”, yielding:
5.1.A. E ASY E XERCISE . Show that Pn
k is irreducible.
5.1.B. E XERCISE . Exercise 3.7.E showed that there is a bijection between irreducible closed subsets and points for affine schemes (the map sending a point p tothe closed subset {p} is a bijection). Show that this is true of schemes in general.
5.1.C. E ASY E XERCISE . Prove that if X is a scheme that has a finite cover X =∪ni=1 Spec Ai where Ai is Noetherian, then X is a Noetherian topological space(§3.6.14). (We will soon call a scheme with such a cover a Noetherian scheme, §5.3.4.)
Hint: show that a topological space that is a finite union of Noetherian subspaces
is itself Noetherian.
n
Thus Pnk and PZ are Noetherian topological spaces: we built them by gluing
together a finite number of spectra of Noetherian rings.

5.1.D. E ASY E XERCISE . Show that a scheme X is quasicompact if and only if it canbe written as a finite union of affine open subschemes. (Hence Pn
A is quasicompact
for any ring A.)
5.1.E. I MPORTANT E XERCISE : Q UASICOMPACT SCHEMES HAVE CLOSED POINTS .
Show that if X is a quasicompact scheme, then every point has a closed point in its
closure. Show that every nonempty closed subset of X contains a closed point of
X. In particular, every nonempty quasicompact scheme has a closed point. (Warning: there exist nonempty schemes with no closed points — see for example [Liu,Exer. 3.27], or [Schw], or Hochster’s thesis [Ho1, Ho2] — so your argument had
better use the quasicompactness hypothesis!)153

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This exercise will often be used in the following way. If there is some property P of points of a scheme that is “open” (if a point p has P, then there is someopen neighborhood U of p such that all the points in U have P), then to check if
all points of a quasicompact scheme have P, it suffices to check only the closed
points. (A first example of this philosophy is Exercise 5.2.D.) This provides a connection between schemes and the classical theory of varieties — the points of traditional varieties are the closed points of the corresponding schemes (essentiallyby the Nullstellensatz, see §3.6.9 and Exercise 5.3.F). In many good situations, theclosed points are dense (such as for varieties, see §3.6.9 and Exercise 5.3.F again),but this is not true in some fundamental cases (see Exercise 3.6.J(b)).
5.1.1. Quasiseparated schemes. Quasiseparatedness is a weird notion that comesin handy for certain people. (Warning: we will later realize that this is really aproperty of morphisms, not of schemes §7.3.1.) Most people, however, can ignore
this notion, as the schemes they will encounter in real life will all have this property. A topological space is quasiseparated if the intersection of any two quasicompact open subsets is quasicompact. (The motivation for the “separatedness”
in the name is that it is a weakened version of separated, for which the intersectionof any two affine open sets is affine, see Proposition 10.1.8.)
5.1.F. S HORT EXERCISE . Show that a scheme is quasiseparated if and only if theintersection of any two affine open subsets is a finite union of affine open subsets.
5.1.G. E XERCISE .
Exercise 5.1.F.)Show that affine schemes are quasiseparated. (Possible hint:

We will see that quasiseparatedness will be a useful hypothesis in theorems (inconjunction with quasicompactness), and that various interesting kinds of schemes(affine, locally Noetherian, separated, see Exercises 5.1.G, 5.3.A, and 10.1.H respectively) are quasiseparated, and this will allow us to state theorems more succinctly(e.g., “if X is quasicompact and quasiseparated” rather than “if X is quasicompact,
and either this or that or the other thing hold”).
“Quasicompact and quasiseparated” means something concrete:
5.1.H. E XERCISE . Show that a scheme X is quasicompact and quasiseparated if
and only if X can be covered by a finite number of affine open subsets, any two of
which have intersection also covered by a finite number of affine open subsets.
So when you see “quasicompact and quasiseparated” as hypotheses in a theorem, you should take this as a clue that you will use this interpretation, and that
finiteness will be used in an essential way.
5.1.I. E ASY E XERCISE . Show that all projective A-schemes are quasicompact andquasiseparated. (Hint: use the fact that the graded ring in the definition is finitely
generated — those finite number of generators will lead you to a covering set.)
5.1.J. E XERCISE ( A NONQUASISEPARATED SCHEME ). Let X = Spec k[x1 , x2 , . . . ],and let U be X − [m] where m is the maximal ideal (x1 , x2 , . . . ). Take two copiesof X, glued along U (“affine ∞-space with a doubled origin”, see Example 4.4.5
and Exercise 4.4.C for “finite-dimensional” versions). Show that the result is not

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quasiseparated. Hint: This “open embedding” U ⊂ X came up earlier in Exercise 3.6.G(b) as an example of a nonquasicompact open subset of an affine scheme.
5.1.2. Dimension. One very important topological notion is dimension. (It isamazing that this is a topological idea.) But despite being intuitively fundamental,
it is more difficult, so we postpone it until Chapter 11.

5.2 Reducedness and integrality
Recall that one of the alarming things about schemes is that functions are not determined by their values at points, and that was because of the presence of nilpotents(§3.2.11).
5.2.1. Definition. Recall that a ring is said to be reduced if it has no nonzero nilpotents (§3.2.13). We say that a scheme X is reduced if OX (U) is reduced for every
open set U of X.
5.2.A. E XERCISE ( REDUCEDNESS IS A stalk-local PROPERTY, I . E ., CAN BE CHECKED
AT STALKS ).
Show that a scheme is reduced if and only if none of the stalks

have nonzero nilpotents. Hence show that if f and g are two functions (global
sections of OX ) on
" a reduced scheme that agree at all points, then f = g. (Twohints: OX (U) "→ p∈U OX,p from Exercise 2.4.A, and the nilradical is intersectionof all prime ideals from Theorem 3.2.12.)
5.2.B. E XERCISE . If A is a reduced ring, show that Spec A is reduced. Show thatnAnk and Pk are reduced.

The scheme Spec k[x, y]/(y2 , xy) is nonreduced. When we sketched it in Figure 4.4, we indicated that the fuzz represented nonreducedness at the origin. The
following exercise is a first stab at making this precise.
)*
5.2.C. E XERCISE . Show that k[x, y]/(y2 , xy) x has no nonzero nilpotent elements.
(Possible hint: show that it is isomorphic to another ring, by considering the geometric picture. Exercise 3.2.L may give another hint.) Show that the only point ofSpec k[x, y]/(y2 , xy) with a nonreduced stalk is the origin.
5.2.D. U NIMPORTANT E XERCISE . If X is a quasicompact scheme, show that it suffices to check reducedness at closed points. Hint: Do not try to show that reducedness is an open condition (see Remark 5.2.2). Instead show that any nonreduced
point has a nonreduced closed point in its closure, using Exercise 5.1.E. (This result
is interesting, but we won’t use it.)
5.2.2. ⋆ Remark. Reducedness is not in general an open condition. You may beable to identify the underlying topological space of
X = Spec C[x, y1 , y2 , . . . ]/(y21 , y22 , y23 , . . . , (x − 1)y1 , (x − 2)y2 , (x − 3)y3 , · · · )with that of Spec C[x], and then to show that the nonreduced points of X are precisely the closed points corresponding to the positive integers. The complement of
this set is not Zariski open. (This ring will come up again in §5.5, in the paragraph

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after the statement of property (B).) However, we will see in Remark 5.5.5 (andagain in Remark 13.7.2) that if X is a locally Noetherian scheme, then the reduced
locus is indeed open.
5.2.3. ⋆ Warning for experts. If a scheme X is reduced, then from the definition of
reducedness, its ring of global sections is reduced. However, the converse is nottrue. You already know enough to verify that if X is the scheme cut out by x2 = 0∼ k, and that X is nonreduced. (This example will come up
in P2k , then Γ (X, OX ) =again in §18.1.6.)
5.2.E. E XERCISE . Suppose X is quasicompact, and f is a function that vanishes
at all points of X. Show that there is some n such that fn = 0. Show that thismay fail if X is not quasicompact. (This exercise is less important, but shows why
we like quasicompactness, and gives a standard pathology when quasicompactness doesn’t hold.) Hint: take an infinite disjoint union of Spec An with An :=k[ϵ]/(ϵn ). This scheme arises again in §8.3.2 (see Figure 8.4 for a picture) and in
Caution/Example 8.3.11.
5.2.4. Definition. A scheme X is integral if it is nonempty, and OX (U) is an integral
domain for every nonempty open subset U of X.
5.2.F. I MPORTANT E XERCISE . Show that a scheme X is integral if and only if it
is irreducible and reduced. (Thus we picture integral schemes as: “one piece, nofuzz”. Possible hint: Exercise 4.3.G.)
5.2.G. E XERCISE . Show that an affine scheme Spec A is integral if and only if A is
an integral domain.
5.2.H. E XERCISE . Suppose X is an integral scheme. Then X (being irreducible)
has a generic point η. Suppose Spec A is any nonempty affine open subset of
X. Show that the stalk at η, OX,η is naturally identified with K(A), the fraction
field of A. This is called the function field K(X) of X. It can be computed on any
nonempty open set of X, as any such open set contains the generic point. The
reason for the name: we will soon think of this as the field of rational functions on
X (Definition 5.5.6 and Exercise 5.5.Q).
5.2.I. E XERCISE . Suppose X is an integral scheme. Show that the restriction maps
resU,V : OX (U) → OX (V) are inclusions so long as V ̸= ∅. Suppose Spec A is any
nonempty affine open subset of X (so A is an integral domain). Show that thenatural map OX (U) → OX,η = K(A) (where U is any nonempty open subset) is an
inclusion.
Thus irreducible varieties (an important example of integral schemes definedlater) have the convenient property that sections over different open sets can beconsidered subsets of the same ring. In particular, restriction maps (except to theempty set) are always inclusions, and gluing is easy: functions fi on a cover Ui
of U (as i runs over an index set) glue if and only if they are the same element of
K(X). This is one reason why (irreducible) varieties are usually introduced before
schemes.

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5.2.5. Caution. Integrality is not#stalk-local (the disjoint union of two integral
schemes is not integral, as Spec A Spec B = Spec(A × B) by Exercise 3.6.A), but
it almost is, see Exercise 5.3.C.

5.3 Properties of schemes that can be checked “affine-locally”
This section is intended to address something tricky in the definition of schemes.
We have defined a scheme as a topological space with a sheaf of rings, that can becovered by affine schemes. Hence we have all of the affine open sets in the cover,
but we don’t know how to communicate between any two of them. Somewhat
more explicitly, if I have an affine cover, and you have an affine cover, and wewant to compare them, and I calculate something on my cover, there should besome way of us getting together, and figuring out how to translate my calculation over to your cover. The Affine Communication Lemma 5.3.2 will provide aconvenient machine for doing this.
Thanks to this lemma, we can define a host of important properties of schemes.
All of these are “affine-local”, in that they can be checked on any affine cover, i.e.,a covering by open affine sets. (We state this more formally after the statement of
the Affine Communication Lemma.) We like such properties because we can checkthem using any affine cover we like. If the scheme in question is quasicompact,
then we need only check a finite number of affine open sets.
5.3.1. Proposition. — Suppose Spec A and Spec B are affine open subschemes of ascheme X. Then Spec A ∩ Spec B is the union of open sets that are simultaneously distinguished open subschemes of Spec A and Spec B.

Spec ASpec AfSpec Bg

Spec B[p]

F IGURE 5.1. A trick to show that the intersection of two affine
open sets may be covered by open sets that are simultaneously
distinguished in both affine open sets
Proof. (See Figure 5.1.) Given any point p ∈ Spec A ∩ Spec B, we produce an open
neighborhood of p in Spec A ∩ Spec B that is simultaneously distinguished in bothSpec A and Spec B. Let Spec Af be a distinguished open subset of Spec A containedin Spec A ∩ Spec B and containing p. Let Spec Bg be a distinguished open subsetof Spec B contained in Spec Af and containing p. Then g ∈ Γ (Spec B, OX ) restricts
to an element g ′ ∈ Γ (Spec Af , OX ) = Af . The points of Spec Af where g vanishes

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are precisely the points of Spec Af where g ′ vanishes, so
Spec Bg= Spec Af \ {[p] : g ′ ∈ p}
= Spec(Af )g ′ .

If g ′ = g ′′ /fn (g ′′ ∈ A) then Spec(Af )g ′ = Spec Afg ′′ , and we are done.

!

The following easy result will be crucial for us.

5.3.2. Affine Communication Lemma. — Let P be some property enjoyed by someaffine open subsets of a scheme X, such that(i) if an affine open subset Spec A "→ X has property P then for any f ∈ A,
Spec Af "→ X does too.
(ii) if (f1 , . . . , fn ) = A, and Spec Afi "→ X has P for all i, then so does Spec A "→X.
Suppose that X = ∪i∈I Spec Ai where Spec Ai has property P. Then every affine open
subset of X has P too.
We say such a property is affine-local: a property is affine-local if we can checkit on any affine cover. (For example, reducedness is an affine-local property. Do
you understand why?) Note that if U is an open subscheme of X, then U inherits
any affine-local property of X. Note also that any property that is stalk-local (ascheme has property P if and only if all its stalks have property Q) is necessarily
affine-local (a scheme has property P if and only if all of its affine open sets haveproperty R, where an affine scheme has property R if and only if all its stalks haveproperty Q). But it is sometimes not so obvious what the right definition of Q is;see for example the discussion of normality in the next section.
Proof. Let Spec A be an affine subscheme of X. Cover Spec A with a finite number of distinguished open sets Spec Agj , each of which is distinguished in someSpec Ai . This is possible by Proposition 5.3.1 and the quasicompactness of Spec A(Exercise 3.6.G(a)). By (i), each Spec Agj has P. By (ii), Spec A has P.
!
By choosing property P appropriately, we define some important propertiesof schemes.
5.3.3. Proposition. — Suppose A is a ring, and (f1 , . . . , fn ) = A.
(a) If A is a Noetherian ring, then so is Afi . If each Afi is Noetherian, then so is A.
(b) Suppose B is a ring, and A is a B-algebra. (Hence Ag is a B-algebra for allg ∈ A.) If A is a finitely generated B-algebra, then so is Afi . If each Afi is a
finitely generated B-algebra, then so is A.
We will prove these shortly (§5.3.9). But let’s first motivate you to read theproof by giving some interesting definitions and results assuming Proposition 5.3.3
is true.
5.3.4. Important Definition. Suppose X is a scheme. If X can be covered by affine
open sets Spec A where A is Noetherian, we say that X is a locally Noetherian
scheme. If in addition X is quasicompact, or equivalently can be covered by finitely
many such affine open sets, we say that X is a Noetherian scheme. (We will see anumber of definitions of the form “if X has this property, we say that it is locally Q;
if further X is quasicompact, we say that it is Q.”) By Exercise 5.1.C, the underlying

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topological space of a Noetherian scheme is Noetherian. Hence by Exercise 3.6.T,
all open subsets of a Noetherian scheme are quasicompact.
5.3.A. E XERCISE . Show that locally Noetherian schemes are quasiseparated.
5.3.B. E XERCISE . Show that a Noetherian scheme has a finite number of irreducible components. (Hint: Proposition 3.6.15.) Show that a Noetherian scheme
has a finite number of connected components, each a finite union of irreducible
components.
5.3.C. E XERCISE . Show that a Noetherian scheme X is integral if and only if
X is nonempty and connected and all stalks OX,p are integral domains. Thus in“good situations”, integrality is the union of local (stalks are integral domains) andglobal (connected) conditions. Hint: Recall that integral = irreducible + reduced(Exercise 5.2.F). If a scheme’s stalks are integral domains, then it is reduced (reducedness is a stalk-local condition, Exercise 5.2.A). If a scheme X has underlying
topological space that is Noetherian, then X has finitely many irreducible components (by the previous exercise); if two of them meet at a point p, then OX,p is
not an integral domain. (You can readily extend this from Noetherian schemes tolocally Noetherian schemes, by showing that a connected scheme is irreducible if
and only if it is nonempty and has a cover by open irreducible subsets. But some
Noetherian hypotheses are necessary, see [Stacks, tag 0568].
5.3.5. Unimportant caution. The ring of sections of a Noetherian scheme need not
be Noetherian, see Exercise 19.11.H.
5.3.6. Schemes over a given field k, or more generally over a given ring A (Aschemes). You may be particularly interested in working over a particular field,
such as C or Q, or over a ring such as Z. Motivated by this, we define the notionof A-scheme, or scheme over A, where A is a ring, as a scheme where all therings of sections of the structure sheaf (over all open sets) are A-algebras, andall restriction maps are maps of A-algebras. (Like some earlier notions such as
quasiseparatedness, this will later in Exercise 6.3.G be properly understood as a“relative notion”; it is the data of a morphism X → Spec A.) Suppose now X is an
A-scheme. If X can be covered by affine open sets Spec Bi where each Bi is a finitely
generated A-algebra, we say that X is locally of finite type over A, or that it is alocally finite type A-scheme. (This is admittedly cumbersome terminology; it will
make more sense later, once we know about morphisms in §7.3.12.) If furthermore
X is quasicompact, X is (of) finite type over A, or a finite type A-scheme. Notethat a scheme locally of finite type over k or Z (or indeed any Noetherian ring) islocally Noetherian, and similarly a scheme of finite type over any Noetherian ringis Noetherian. As our key “geometric” examples: (i) Spec C[x1 , . . . , xn ]/I is a finite
type C-scheme; and (ii) Pn
C is a finite type C-scheme. (The field C may be replacedby an arbitrary ring A.)
5.3.D. E XERCISE .
(a) (quasiprojective implies finite type) If X is a quasiprojective A-scheme (Definition 4.5.9), show that X is of finite type over A. If A is furthermore assumed to
be Noetherian, show that X is a Noetherian scheme, and hence has a finite numberof irreducible components.

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The Rising Sea: Foundations of Algebraic Geometry(b) Suppose U is an open subscheme of a projective A-scheme. Show that U is
locally of finite type over A. If A is Noetherian, show that U is quasicompact, andhence quasiprojective over A, and hence by (a) of finite type over A. Show thisneed not be true if A is not Noetherian. Better: give an example of an open subscheme of a projective A-scheme that is not quasicompact, necessarily for somenon-Noetherian A. (Hint: Silly example 4.5.11.)
5.3.7. Varieties. We now make a connection to the classical language of varieties.
An affine scheme that is reduced and of finite type over k is said to be an affine
variety (over k), or an affine k-variety. A reduced (quasi)projective k-scheme isa (quasi)projective variety (over k), or a (quasi)projective k-variety. (Warning:in the literature, it is sometimes also assumed in the definition of variety that thescheme is irreducible, or that k is algebraically closed.)
5.3.E. E XERCISE .
(a) Show that Spec k[x1 , . . . , xn ]/I is an affine k-variety if and only if I ⊂ k[x1 , . . . , xn ]is a radical ideal.
(b) Suppose I ⊂ k[x0 , . . . , xn ] is a radical graded ideal. Show that Proj k[x0 , . . . , xn ]/I
is a projective k-variety. (Caution: The example of I = (x20 , x0 x1 , . . . , x0 xn ) showsthat Proj k[x0 , . . . , xn ]/I can be a projective k-variety without I being radical.)
We will not define varieties in general until §10.1.7; we will need the notion of
separatedness first, to exclude abominations like the line with the doubled origin(Example 4.4.5). But many of the statements we will make in this section about
affine k-varieties will automatically apply more generally to k-varieties.
5.3.F. E XERCISE . Show that a point of a locally finite type k-scheme is a closed
point if and only if the residue field of the stalk of the structure sheaf at that pointis a finite extension of k. Show that the closed points are dense on such a scheme(even though it needn’t be quasicompact, cf. Exercise 5.1.E). Hint: §3.6.9. (Warning: closed points need not be dense even on quite reasonable schemes, see Exercise 3.6.J(b).)
5.3.8. Definition. The degree of a closed point p of a locally finite type k-scheme(e.g., a variety over k) is the degree of the field extension κ(p)/k. For example,
in A1k = Spec k[t], the point [(p(t))] (p(t) ∈ k[t] irreducible) is deg p(t). If k is
algebraically closed, the degree of every closed point is 1.
5.3.G. ⋆⋆ E XERCISE (analytification OF COMPLEX VARIETIES ). (Warning: Any
discussion of analytification will be only for readers who are familiar with the notion of complex analytic varieties, or willing to develop it on their own in parallelwith our development of schemes.) Suppose X is a reduced, finite type C-scheme.
Define the corresponding complex analytic prevariety Xan . (The definition of ananalytic prevariety is the same as the definition of a variety without the Hausdorff condition.) Caution: your definition should not depend on a choice of anaffine cover of X. (Hint: First explain how to analytify reduced finite type affine
C-schemes. Then glue.) Give a bijection between the closed points of X and thepoints of Xan , using the weak Nullstellensatz 3.2.4. (In fact one may construct a
continuous map of sets Xan → X generalizing Exercise 3.2.I.) In Exercise 6.3.N, we
will see that analytification can be made into a functor. As mentioned there, two

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nonisomorphic complex varieties can have isomorphic analytifications, but not ifthey are compact.
5.3.9. Proof of Proposition 5.3.3. We divide each part into (i) and (ii) following thestatement of the Affine Communication Lemma 5.3.2.
Proof of Proposition 5.3.3(a). (i) If I1 " I2 " I3 " · · · is a strictly increasing chainof ideals of Af , then we can verify that J1 " J2 " J3 " · · · is a strictly increasing
chain of ideals of A, where
Jj = {r ∈ A : r ∈ Ij }where r ∈ Ij means “the image of r in Af lies in Ij ”. (We think of this as Ij ∩ A,
except in general A needn’t inject into Afi .) Clearly Jj is an ideal of A. If x/fn ∈Ij+1 \ Ij where x ∈ A, then x ∈ Jj+1 , and x ∈/ Jj (or else x(1/f)n ∈ Ij as well).
(ii) Suppose I1 " I2 " I3 " · · · is a strictly increasing chain of ideals of A.
Then for each 1 ≤ i ≤ n,Ii,1 ⊂ Ii,2 ⊂ Ii,3 ⊂ · · ·is an increasing chain of ideals in Afi , where Ii,j = Ij ⊗A Afi . It remains to showthat for each j, Ii,j " Ii,j+1 for some i; the result will then follow.
"
5.3.H. E XERCISE . Finish the proof of Proposition 5.3.3(a). (Hint: A "→Afi by(4.1.3.1).)

Proof of Proposition 5.3.3(b). Part (i) is clear: if A is generated over B by r1 , . . . ,rn , then Af is generated over B by r1 , . . . , rn , 1/f.
,(ii) Here is the idea. As the fi generate A, we can write 1 =ci fi for ci ∈ A.
k
We have generators of Afi : rij /fi j , where rij ∈ A. I claim that {fi }i ∪ {ci } ∪ {rij }ij
generate A as a B-algebra. Here is why. Suppose you have any r ∈ A. Then in
Afi , we can write r as some polynomial in the rij ’s and fi , divided by some huge
power of fi . So “in each Afi , we have described r in the desired way”, except forthis annoying denominator. Now use a “partition of unity” type argument as in
the proof of Theorem 4.1.2 to combine all of these into a single expression, killingthe denominator. Show that the resulting expression you build still agrees withr in each of the Afi . Thus it is indeed r (by the identity axiom for the structure
sheaf).
5.3.I. E XERCISE . Make this argument precise.
This concludes the proof of Proposition 5.3.3.

!

5.4 Normality and factoriality
5.4.1. Normality.
We can now define a property of schemes that says that they are “not too
far from smooth”, called normality, which will come in very handy. We will see
later that “locally Noetherian normal schemes satisfy Hartogs’s Lemma” (Algebraic Hartogs’s Lemma 11.3.11 for Noetherian normal schemes): functions definedaway from a set of codimension 2 extend over that set. (We saw a first glimpse of

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this in §4.4.2.) As a consequence, rational functions that have no poles (certain setsof codimension one where the function isn’t defined) are defined everywhere. We
need definitions of dimension and poles to make this precise. See §12.8.7 and§26.3.5 for the fact that “smoothness” (really, “regularity”) implies normality.
Recall that an integral domain A is integrally closed if the only zeros in K(A)to any monic polynomial in A[x] must lie in A itself. The basic example is Z (seeExercise 5.4.F for a reason). We say a scheme X is normal if all of its stalks OX,p arenormal, i.e., are integral domains, and integrally closed in their fraction fields. (Soby definition, normality is a stalk-local property.) As reducedness is a stalk-local
property (Exercise 5.2.A), normal schemes are reduced.
5.4.A. E XERCISE . Show that integrally closed domains behave well under localization: if A is an integrally closed domain, and S is a multiplicative subset notcontaining 0, show that S−1 A is an integrally closed domain. (Hint: assume thatxn + an−1 xn−1 + · · · + a0 = 0 where ai ∈ S−1 A has a root in the fraction field.
Turn this into another equation in A[x] that also has a root in the fraction field.)
It is no fun checking normality at every single point of a scheme. Thanks
to this exercise, we know that if A is an integrally closed domain, then Spec A
is normal. Also, for quasicompact schemes, normality can be checked at closed
points, thanks to this exercise, and the fact that for such schemes, any point is a
generization of a closed point (see Exercise 5.1.E).
It is not true that normal schemes are integral. #For example, the disjoint∼ Spec(k × k) =∼union of two normal schemes is normal. Thus Spec k Spec k =Spec k[x]/(x(x − 1)) is normal, but its ring of global sections is not an integral domain.
5.4.B. U NIMPORTANT EXERCISE . Show that a Noetherian scheme is normal if andonly if it is the finite disjoint union of integral Noetherian normal schemes. (Hint:
Exercise 5.3.C.)
We are close to proving a useful result in commutative algebra, so we may as
well go all the way.
5.4.2. Proposition. — If A is an integral domain, then the following are equivalent.
(i) A is integrally closed.
(ii) Ap is integrally closed for all prime ideals p ⊂ A.
(iii) Am is integrally closed for all maximal ideals m ⊂ A.
Proof. Exercise 5.4.A shows that integral closedness is preserved by localization, so(i) implies (ii). Clearly (ii) implies (iii).
It remains to show that (iii) implies (i). This argument involves a pretty construction that we will use again. Suppose A is not integrally closed. We show that
there is some m such that Am is also not integrally closed. Suppose(5.4.2.1)

xn + an−1 xn−1 + · · · + a0 = 0(with ai ∈ A) has a solution s in K(A) \ A. Let I be the ideal of denominators of s:
I := {r ∈ A : rs ∈ A}.

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163(Note that I is clearly an ideal of A.) Now I ̸= A, as 1 ∈/ I. Thus there is somemaximal ideal m containing I. Then s ∈/ Am , so equation (5.4.2.1) in Am [x] showsthat Am is not integrally closed as well, as desired.
!
5.4.C. U NIMPORTANT E XERCISE .
If A is an integral domain, show that A =∩Am , where the intersection runs over all maximal ideals of A. (We won’t use thisexercise, but it gives good practice with the ideal of denominators.)
5.4.D. U NIMPORTANT E XERCISE RELATING TO THE IDEAL OF DENOMINATORS .
One might naively hope from experience with unique factorization domains thatthe ideal of denominators is principal. This is not true. As a counterexample,
consider our new friend A = k[w, x, y, z]/(wz − xy) (which we first met in Example 4.4.12, and which we will later recognize as the cone over the quadric surface),
and w/y = x/z ∈ K(A). Show that the ideal of denominators of this element ofK(A) is (y, z).
We will see that the I in the above exercise is not principal (Exercise 12.1.D —you may be able to show it directly, using the fact that I is a homogeneous ideal of agraded ring). But we will also see that in good situations (Noetherian, normal), theideal of denominators is “pure codimension 1” — this is the content of Algebraic
Hartogs’s Lemma 11.3.11. In its proof, §11.3.12, we give a geometric interpretationof the ideal of denominators.
5.4.3. Factoriality.
We define a notion which implies normality.
5.4.4. Definition. If all the stalks of a scheme X are unique factorization domains,
we say that X is factorial. (Unimportant remark: This is sometimes called locallyfactorial, which may falsely suggest that this notion is affine-local, which it isn’t,
see Exercise 5.4.N. Another unimportant remark: the locus of points on an affine
variety over an algebraically closed field that are factorial is an open subset, [BGS,
p. 1].)
5.4.E. E XERCISE . Show that any nonzero localization of a unique factorization
domain is a unique factorization domain.
5.4.5. Thus if A is a unique factorization domain, then Spec A is factorial. The
converse need not hold, see Exercise 5.4.N. In fact, we will see that elliptic curves
are factorial, yet no affine open set is the Spec of a unique factorization domain,§19.11.1. Hence one can show factoriality by finding an appropriate affine cover,
but there need not be such a cover of a factorial scheme.
5.4.6. Remark: How to check if a ring is a unique factorization domain. There are very
few means of checking that a Noetherian integral domain is a unique factorization
domain. Some useful ones are: (0) elementary means: rings with a Euclidean
algorithm such as Z, k[t], and Z[i]; polynomial rings over a unique factorization
domain, by Gauss’s Lemma (see e.g., [Lan, IV.2.3]). (1) Exercise 5.4.E, that the
localization of a unique factorization domain is also a unique factorization domain.
(2) height 1 prime ideals are principal (Proposition 11.3.5). (3) normal and Cl = 0(Exercise 14.2.T). (4) Nagata’s Lemma (Exercise 14.2.U). (Caution: even if A is a

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unique factorization domain, A[[x]] need not be; see [Mat2, p. 165]. The first such
example, due to P. Salmon, was A = k(u)[[x, y, z]]/(z2 + x3 + uz6 ), see [Sa, Dan].)
5.4.7. Factoriality implies normality. One of the reasons we like factoriality is that it
implies normality.
5.4.F. I MPORTANT E XERCISE . Show that unique factorization domains are integrally closed. Hence factorial schemes are normal, and if A is a unique factorization domain, then Spec A is normal. (However, rings can be integrally closed
without being unique factorization domains, as we will see in Exercise 5.4.L. Another example is given without proof in Exercise 5.4.N; in that example, Spec of
the ring is factorial. A variation on Exercise 5.4.L will show that schemes can benormal without being factorial, see Exercise 12.1.E.)
5.4.8. Examples.
n5.4.G. E ASY E XERCISE . Show that the following schemes are normal: An
k , Pk ,Spec Z. (As usual, k is a field. Although it is true that if A is integrally closed thenA[x] is as well — see [Bo, Ch. 5, §1, no. 3, Cor. 2] or [E, Ex. 4.18] — this is not aneasy fact, so do not use it here.)

5.4.H. H ANDY E XERCISE ( YIELDING MANY ENLIGHTENING EXAMPLES LATER ). Suppose A is a unique factorization domain with 2 invertible, and z2 − f is irreducible
in A[z].
(a) Show that if f ∈ A has no repeated prime factors, then Spec A[z]/(z2 − f) is
normal. Hint: B := A[z]/(z2 − f) is an integral domain, as (z2 − f) is prime
in A[z]. Suppose we have monic F(T ) ∈ B[T ] so that F(T ) = 0 has a root α inK(B) \ K(A). Then by replacing F(T ) by F(T )F(T ), we can assume F(T ) ∈ A[T ].
Also, α = g + hz where g, h ∈ K(A). Now α is the root of Q(T ) = 0 for monicQ(T ) = T 2 − 2gT + (g2 − h2 f) ∈ K(A)[T ], so we can factor F(T ) = P(T )Q(T ) inK(A)[T ]. By Gauss’s lemma, 2g, g2 − h2 f ∈ A. Say g = r/2, h = s/t (s and t haveno common factors, r, s, t ∈ A). Then g2 − h2 f = (r2 t2 − 4s2 f)/4t2 . Then t isinvertible.
(b) Show that if f ∈ A has repeated prime factors, then Spec A[z]/(z2 − f) is notnormal.
5.4.I. E XERCISE . Show that the following schemes are normal:(a) Spec Z[x]/(x2 −n) where n is a square-free integer congruent to 3 modulo
4. Caution: the hypotheses of Exercise 5.4.H do not apply, so you will
have to do this directly. (Your argument may also show
√ the result when 3
is replaced by 2. A similar argument shows that Z[(1+ n)/2] is integrally
closed if n ≡ 1 (mod 4) is square-free.)(b) Spec k[x1 , . . . , xn ]/(x21 + x22 + · · · + x2m ) where char k ̸= 2, n ≥ m ≥ 3.
(c) Spec k[w, x, y, z]/(wz − xy) where char k ̸= 2. This is our cone over aquadric surface example from Example 4.4.12 and Exercise 5.4.D. Hint:
Exercise 5.4.J may help. (The result also holds for char k = 2, but don’t
worry about this.)
5.4.J. E XERCISE ( DIAGONALIZING QUADRATIC FORMS ). Suppose k is an algebraically closed field of characteristic not 2. (The hypothesis that k is algebraically

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closed is not necessary, so feel free to deal with this more general case.)(a) Show that any quadratic form in n variables can be “diagonalized” by changing coordinates to be a sum of at most n squares (e.g., uw − v2 = ((u + w)/2)2 +(i(u − w)/2)2 + (iv)2 ), where the linear forms appearing in the squares are linearly independent. (Hint: use induction on the number of variables, by “completing thesquare” at each step.)(b) Show that the number of squares appearing depends only on the quadratic.
For example, x2 + y2 + z2 cannot be written as a sum of two squares. (Possible
approach: given a basis x1 , . . . , xn of the linear forms, write the quadratic form as

)

x1

···

⎛

⎞
x1
* ⎜
⎟
xn M ⎝ ... ⎠xn

where M is a symmetric matrix. Determine how M transforms under a change ofbasis, and show that the rank of M is independent of the choice of basis.)
The rank of the quadratic form is the number of (“linearly independent”)
squares needed. If the number of squares equals the number of variables, thequadratic form is said to be full rank or (of) maximal rank.
5.4.K. E ASY E XERCISE ( RINGS CAN BE INTEGRALLY CLOSED BUT NOT√ UNIQUE
FACTORIZATION DOMAINS , ARITHMETIC VERSION ).
Show that Z[ −5] is integrally closed√but not a √unique factorization domain. (Hints: Exercise 5.4.I(a) and2 × 3 = (1 + −5)(1 − −5).)
5.4.L. E ASY E XERCISE ( RINGS CAN BE INTEGRALLY CLOSED BUT NOT UNIQUE
FACTORIZATION DOMAINS , GEOMETRIC VERSION ).
Suppose char k ̸= 2. Let
A = k[w, x, y, z]/(wz − xy), so Spec A is the cone over the smooth quadric surface(cf. Exercises 4.4.12 and 5.4.D).
(a) Show that A is integrally closed. (Hint: Exercises 5.4.I(c) and 5.4.J.)(b) Show that A is not a unique factorization domain. (Clearly wz = xy. But why
are w, x, y, and z irreducible? Hint: A is a graded integral domain. Show that if a
homogeneous element factors, the factors must be homogeneous.)
The previous two exercises look similar, but there is a difference. The coneover the quadric surface is normal (by√ Exercise 5.4.L) but not factorial; see Exercise 12.1.E. On the other hand, Spec Z[ −5] is factorial — all of its stalks are unique√factorization domains. (You will later be able to show this by showing that Z[ −5]is a Dedekind domain, §12.5.15, whose stalks are necessarily unique factorization
domains by Theorem 12.5.8(f).)
5.4.M. E XERCISE . Suppose A is a k-algebra, and l/k is a finite extension of fields.
(Most likely your proof will not use finiteness; this hypothesis is included to avoid
distraction by infinite-dimensional vector spaces.) Show that if A ⊗k l is a normal
integral domain, then A is a normal integral domain as well. (Although we won’t
need this, a version of the converse is true if l/k is separable, [Gr-EGA, IV2 .6.14.2].)
Hint: fix a k-basis for l, b1 = 1, . . . , bd . Explain why 1 ⊗ b1 , . . . , 1 ⊗ bd forms a free

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A-basis for A ⊗k l. Explain why we have injectionsA% K(A)"A ⊗k l"% K(A) ⊗k l.

Show that K(A) ⊗k l = K(A ⊗k l). (Idea: A ⊗k l ⊂ K(A) ⊗k l ⊂ K(A ⊗k l). Why
is K(A) ⊗k l a field?) Show that (A ⊗k l) ∩ K(A) = A. Now assume P(T ) ∈ A[T ] ismonic and has a root α ∈ K(A), and proceed from there.
5.4.N. E XERCISE (UFD- NESS IS NOT AFFINE - LOCAL ). Let A = (Q[x, y]x2 +y2 )0
denote the homogeneous degree 0 part of the ring Q[x, y]x2 +y2 . In other words, itconsists of quotients f(x, y)/(x2 + y2 )n , where f has pure degree 2n. Show that the22
distinguished open sets D( x2x+y2 ) and D( x2y+y2 ) cover Spec A. (Hint: the sum of
those two fractions is 1.) Show that A x2 and A y2 are unique factorization
x2 +y2x2 +y2domains. (Hint: show that both rings are isomorphic to Q[t]t2 +1 ; this is a localization of the unique factorization domain Q[t].) Finally, show that A is not a uniquefactorization domain. Possible hint:
"
#2 "
#"
#
xy
x2
y2
=
.
x2 + y2x2 + y2x2 + y2(This is generalized in Exercise 14.2.L . It is also related to Exercise 14.2.Q.)√
Number theorists may prefer the example of Exercise 5.4.K: Z[ −5] is not aunique factorization domain, but it turns out that you can cover it with two affine
open subsets D(2) and D(3), each corresponding
to√unique factorization domains.
√(For number theorists: to show that Z[ −5]2 andZ[−5]3 are unique factorization√
domains, first show that the class group of Z[ −5] is Z/2 using the geometryof√numbers,
as
in
[Ar4,
Ch.
11,
Thm.
7.9].
Then
show
that
the
ideals
(1
+
−5,
2)
and
√
√
(1 + −5, 3) are not principal, using the usual norm in C.) The ring Z[ −5] is an
example of a Dedekind domain, as we will discuss in §12.5.15.
5.4.9. Remark. For an example of k-algebra A that is not a unique factorization
domain, but becomes one after a certain field extension, see Exercise 14.2.N.

5.5 The crucial points of a scheme that control everything:
Associated points and primes
The associated points of a scheme are the few crucial points of the schemethat capture essential information about its (sheaf of) functions. There are several
quite different ways of describing them, most of which are quite algebraic. We willtake a nonstandard approach, beginning with geometric motivation. Because they
involve both nilpotents and generic points — two concepts not part of your priorgeometric intuition — it can take some time to make them “geometric” in yourhead. We will first meet them in a motivating example in two ways. We will thendiscuss their key properties. Finally, we give proper (algebraic) definitions and

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proofs. As is almost always the case in mathematics, it is much more important toremember the properties than it is to remember their proofs.
There are other approaches to associated points. Most notably, the algebraicallymost central view is via a vitally important algebraic construction, primary decomposition, mentioned only briefly in Aside 5.5.13.
5.5.1. Associated points as “fuzz attractors”. Recall Figure 4.4, our “fuzzy” picture of
the nonreduced scheme Spec k[x, y]/(y2 , xy). When this picture was introduced,
we mentioned that the “fuzz” at the origin indicated that the nonreduced behavior was concentrated there. This was justified in Exercise 5.2.C: the origin is the
only point where the stalk of the structure sheaf is nonreduced. Thus the different
levels of reducedness are concentrated along two irreducible closed subsets — theorigin, and the entire x-axis. Since irreducible closed subsets are in bijection withpoints, we may as well say that the two key points with respect to “levels of nonreducedness” were the generic point [(y)], and the origin [(x, y)]. These will be theassociated points of this scheme.
5.5.2. Better: associated points as generic points of irreducible components of the supportof sections.
We now give a seemingly unrelated exercise about the same scheme. Recall
that the support of a function on a scheme (Definition 2.4.2) is a closed subset.
5.5.A. E XERCISE . Suppose f is a function on Spec k[x, y]/(y2 , xy) (i.e., f ∈ k[x, y]/(y2 , xy)).
Show that Supp f is either the empty set, or the origin, or the entire space.
The fact that the same closed subsets arise in two different ways is no coincidence — their generic points are the associated points of Spec k[x, y]/(y2 , xy).
We discuss associated points first in the affine case Spec A. We assume that
A is Noetherian, and we take this as a standing assumption when discussing associated points. More generally, we will discuss associated points of M where M
is a finitely generated A module (and A is Noetherian). When speaking of rings
rather than schemes, we speak of associated prime ideals rather than associated points.
Associated primes and associated points can be defined more generally, and we
discuss one easy case (the integral case) in Exercise 5.5.Q.
We now state three essential properties, to be justified later. The first is themost important.
(A) The associated primes/points of M are precisely the generic points of irreducible
components of the support of some element of M (on Spec A).
For example, by Exercise 5.5.A, Spec k[x, y]/(y2 , xy) has two associated points.
As another example:
5.5.B. E XERCISE ( ASSUMING (A)). Suppose A is an integral domain. Show that thegeneric point is the only associated point of Spec A.
(Important note: Exercises 5.5.B–5.5.H require you to work directly from someaxioms, not from our later definitions. If this troubles you, feel free to workthrough the definitions, and use the later exercises rather than the geometric axioms (A)–(C) to solve these problems. But you may be surprised at how short thearguments actually are, assuming the geometric axioms.)

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We could take (A) as the definition, although in our rigorous development
below, we will take a different (but logically equivalent) starting point. (Unimportant aside: if A is a ring that is not necessarily Noetherian, then (A) is the definitionof a weakly associated prime, see [Stacks, tag 0547].)
The next property makes (A) more striking.
(B) M has a finite number of associated primes/points.
For example, there are only a finite number of irreducible closed subsets ofSpec A, such that the only possible supports of functions of Spec A are unions ofthese. You may find this unexpected, although the examples above may have
prepared you for it. You should interpret this as another example of Noetherianness forcing some sort of finiteness. (For example, we will see that this generalizes“finiteness of irreducible components”, cf. Proposition 3.6.15. And to get an idea ofwhat can go wrong without Noetherian hypotheses, you can ponder the scheme of
Remark 5.2.2.) This gives some meaning to the statement that their generic points
are the few crucial points of the scheme.
We will see (in Exercise 5.5.O) that we can completely describe which subsetsof Spec A are the support of an element of M: precisely those subsets which are
the closure of a subset of the associated points.
5.5.3. We immediately see from (A) that if M = A, the generic points of the irreducible
components of Spec A are associated points of M = A, by considering the function
1. The other associated points of Spec A are called embedded points. Thus in thecase of Spec k[x, y]/(y2 , xy) (Figure 4.4), the origin is the only embedded point (byExercise 5.5.A).
5.5.C. E XERCISE ( ASSUMING (A)). Show that if A is reduced, Spec A has no embedded points. Hints: (i) first deal with the case where A is integral, i.e., where
Spec A is irreducible. (ii) Then deal with the general case. If f is a nonzero function
on a reduced affine scheme, show that Supp f = D(f): the support is the closureof the locus where f doesn’t vanish. Show that D(f) is the union of the irreducible
components meeting D(f), using (i).
Furthermore, the natural map(5.5.3.1)

M→

&

Mp

associated p

is an injection. (This is an important property. Once again, the associated points are“where all the action happens”.) We show this by showing that the kernel is zero.
Suppose a function f has a germ of zero at each associated point, so its support
contains no associated points. It is supported on a closed subset, which by (A)must be the union of closures of associated points. Thus it must be supported
nowhere, and thus be the zero function.
5.5.4. Unimportant Side Remark. An open subscheme U of a scheme X is said to be
scheme-theoretically dense if any function on any open set V is 0 if it restricts to
0 on U ∩ V. (We won’t use this phrase after this remark.) If X is locally Noetherian, then you can use the injection (5.5.3.1) (with M = A) to show that an open
subscheme U ⊂ X is scheme-theoretically dense if and only if it contains all the
associated points of X.

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5.5.D. E XERCISE ( ASSUMING (A)). Suppose m ∈ M. Show that Supp m is theclosure of those associated points of M where m has nonzero germ. (Hint: Supp m
is a closed set containing the points described, and thus their closure. Why does it
contain no other points?)
5.5.E. E XERCISE ( ASSUMING (A) AND (B)). Show that the locus on Spec A of points[p] where OSpec A,[p] = Ap is nonreduced is the closure of those associated points ofSpec A whose stalks are nonreduced. (Hint: why do points in the closure of theseassociated points all have nonreduced stalks? Why can’t any other point have anonreduced stalk?)
5.5.5. Remark. Exercise 5.5.E partially explains the link between associated points
and fuzzy pictures. (Primary decomposition, see Aside 5.5.13, gives a more explicit connection, but we won’t discuss it properly.) Note for future reference thatonce we establish these properties (later in this section), we will have shown that
if Y is a locally Noetherian scheme, the “reduced locus” of Y is an open subset of
Y. This fulfills a promise made in Remark 5.2.2.
(C) An element f of A is a zerodivisor of M (i.e., there exists m ̸= 0 with fm = 0) if
and only if it vanishes at some associated point of M (i.e., is contained in some associatedprime of M).
One direction is clear from the previous properties. (Do you see which?)
The next property allows us to globalize the construction of associated points
to arbitrary (locally Noetherian) schemes.
5.5.F. I MPORTANT E XERCISE ( ASSUMING (A)). Show that the definition in (A) of associated primes/points behaves well with respect to localizing: if S is a multiplicative
subset of A, then the associated primes/points of S−1 M are precisely those associated primes/points of M that lie in Spec S−1 A, i.e., associated primes of M that
do not meet S.
Thus the associated primes/points can be “determined locally”. For example,
associated points of A can be checked by looking at stalks of the structure sheaf(the notion is “stalk-local”). As another example, the associated prime ideals of M
may be determined by working on a distinguished open cover of Spec A. Thanks
to Important Exercise 5.5.F, we can (and do) define the associated points of a locally Noetherian scheme X to be those points p ∈ X such that, on any affine open
set Spec A containing p, p corresponds to an associated prime of A. This notion
is independent of choice of affine open neighborhood Spec A: if p has two affine
open neighborhoods Spec A and Spec B (say corresponding to prime ideals p ⊂ A
and q ⊂ B respectively), then p corresponds to an associated prime of A if andonly if it corresponds to an associated prime of Ap = OX,p = Bq if and only if it
corresponds to an associated prime of B, by Important Exercise 5.5.F.
(Here we are “globalizing” only the special case M = A. Once we define
quasicoherent sheaves, we will be able to globalize the case of a general M, see§13.6.5.)
By combining the above properties, we immediately have a number of facts,
including the following. (i) A Noetherian scheme has finitely many associatedpoints. (ii) Each of the irreducible components of the support of any function
on a locally Noetherian scheme is the union of the closures of some subset of the

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associated points. (iii) The generic points of the irreducible components of a locally
Noetherian scheme are associated points. (The remaining associated points arestill called embedded points.) (iv) A reduced locally Noetherian scheme has noembedded points. (v) The nonreduced locus of a locally Noetherian scheme (thelocus of points p ∈ X where OX,p is nonreduced) is the closure of those associatedpoints that have nonreduced stalk.
Furthermore, recall that one nice property of integral schemes X (such as irreducible affine varieties) not shared by all schemes is that for any nonempty open
subset U ⊂ X, the natural map Γ (U, OX ) → K(X) is an inclusion (Exercise 5.2.I).
Thus all sections over any nonempty open set, and elements of all stalks, can bethought of as lying in a single field K(X), which is the stalk at the generic point.
Associated points allow us to generalize this idea.
5.5.G. E XERCISE ( CF. (5.5.3.1)).
Assuming the above properties of associatedpoints, show that if X is a locally Noetherian scheme, then for any open subset
U ⊂ X, the natural map&(5.5.5.1)
Γ (U, OX ) →OX,passociated p in U

is an injection.

We can use these properties to refine our ability to visualize schemes in a waythat captures precise mathematical information. As a first check, you should be
able to understand Figure 5.2. As a second, you should be able to do the following
exercise.

F IGURE 5.2. This scheme has 6 associated points, of which 3 areembedded points. A function is a zerodivisor if it vanishes at anyof these six points.

5.5.H. E XERCISE ( PRACTICE WITH FUZZY PICTURES ). Assume the properties (A)–(C) of associated points. Suppose X = Spec C[x, y]/I, and that the associated pointsof X are [(y − x2 )], [(x − 1, y − 1)], and [(x − 2, y − 2)].
(a) Sketch X as a subset of A2C = Spec C[x, y], including fuzz.
(b) Do you have enough information to know if X is reduced?
(c) Do you have enough information to know if x + y − 2 is a zerodivisor? How

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about x + y − 3? How about y − x2 ? (Exercise 5.5.R will verify that such an X
actually exists.)
The following exercise shows that “hypersurfaces have no embedded points”.
(Of course, thanks to Exercise 5.5.C, this is interesting only when the hypersurface
is nonreduced.)
5.5.I. E XERCISE .
Assume the properties (A)–(C) of associated points. If f ∈k[x1 , . . . , xn ] is nonzero, show that A := k[x1 , . . . , xn ]/(f) has no embedded points.
Hint: suppose g ∈ A is a zerodivisor, and choose a lift g ∈ k[x1 , . . . , xn ] of g. Show
that g has a common factor with f. (We will use this exercise in §18.6.3. All you
should use is that k[x1 , . . . , xn ] is a Noetherian unique factorization domain. We
will generalize this in §26.2.7.)
5.5.6. Definitions: Rational functions. A rational function on a locally Noetherian
scheme is an element of the image of Γ (U, OU ) in (5.5.5.1) for some U containing
all the associated points. Equivalently, the set of rational functions is the colimit
of OX (U) over all open sets containing the associated points. Or if you prefer, arational function is a function defined on an open set containing all associatedpoints, i.e., an ordered pair (U, f), where U is an open set containing all associatedpoints, and f ∈ Γ (U, OX ). Two such data (U, f) and (U ′ , f ′ ) define the same rational
function if and only if the restrictions of f and f ′ to U ∩ U ′ are the same. If X is
reduced, this is the same as requiring that they are defined on an open set of eachof the irreducible components.
x−2
For example, on Spec k[x, y]/(y2 , xy) (Figure 4.4), (x−1)(x−3)
is a rational funcx−2tion, but x(x−1) is not.
A rational function has a maximal domain of definition, because any two actual functions on an open set (i.e., sections of the structure sheaf over that openset) that agree as “rational functions” (i.e., on small enough open sets containing
associated points) must be the same function, by the injectivity of (5.5.5.1). We saythat a rational function f is regular at a point p if p is contained in this maximal domain of definition (or equivalently, if there is some open set containing p where fx−2
is defined). For example, on Spec k[x, y]/(y2 , xy), the rational function (x−1)(x−3)
has domain of definition consisting of everything but 1 and 3 (i.e., [(x − 1)] and[(x − 3)]), and is regular away from those two points. A rational function is regular if it is regular at all points. (Unfortunately, “regular” is an overused word inmathematics, and in algebraic geometry in particular.)
5.5.7. The rational functions form a ring, called the total fraction ring or total
quotient ring of X. If X = Spec A is affine, then this ring is called the total fraction(or quotient) ring of A. If X is integral, the total fraction ring is the function field
K(X) — the stalk at the generic point — so this extends our earlier Definition 5.2.Hof K(·).
5.5.8. Definition and proofs.
We finally define associated points, and show that they have the desired properties (A)–(C) (and their consequences) for locally Noetherian schemes. Because
the definition is a useful property to remember (on the same level as (A)–(C)), we

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dignify it with a letter. We make the definition in more generality than we will use.
Suppose M is an A-module, and A is an arbitrary ring.
(D) A prime p ⊂ A is said to be associated to M if p is the annihilator of an element
m of M (p = {a ∈ A : am = 0}).
5.5.9. Equivalently, p is associated to M if and only if M has a submodule isomorphic to A/p. The set of prime ideals associated to M is denoted Ass M (or AssA M).
Awkwardly, if I is an ideal of A, the associated prime ideals of the module A/I aresaid to be the associated prime ideals of I. This is not my fault.
5.5.10. Theorem (properties of associated prime ideals). — Suppose A is a Noetherian ring, and M ̸= 0 is finitely generated.
(a) The set Ass M is finite (property(B)) and nonempty.
"(b) The natural map M → p∈Ass M Mp is an injection (cf. (5.5.3.1)).
(c) The set of zerodivisors of M is ∪p∈Ass M p (property (C)).
(d) (association commutes with localization, cf. Important Exercise 5.5.F) If S is amultiplicative set, then
AssS−1 A S−1 M = AssA M ∩ Spec S−1 A(= {p ∈ AssA M : p ∩ S = ∅}).
We prove Theorem 5.5.10 in a series of exercises.

5.5.J. I MPORTANT E XERCISE . Suppose M ̸= 0 is an A-module. Show that if I ⊂ A
is maximal among all proper ideals that are annihilators of elements of M, then I
is prime, and hence I ∈ Ass M. Thus if A is Noetherian, then Ass M is nonempty(part of Theorem 5.5.10(a)). (This is a good excuse to state a general philosophy:“Quite generally, proper ideals maximal with respect to some property have an
uncanny tendency to be prime,” [E, p. 70].)
5.5.K. E XERCISE . Suppose that M is a module over a Noetherian ring A. Show
that m = 0 if and only if m is 0 in Mp for each of the maximal associated prime
ideals p of M. (Hint: use the previous exercise.)This immediately implies Theorem 5.5.10(b). It also implies Theorem 5.5.10(c):Any nonzero element of ∪p∈Ass M p is clearly a zerodivisor. Conversely, if a annihilates a nonzero element of M, then a is contained in a maximal annihilator ideal.
5.5.L. E XERCISE . If 0 → M ′ → M → M ′′ → 0 is a short exact sequence ofA-modules, show that
Ass M ′ ⊂ Ass M ⊂ Ass M ′ ∪ Ass M ′′ .

∼(Possible hint for the second containment: if m ∈ M has annihilator p, then Am =A/p.)
5.5.M. E XERCISE .
(a) If M is a finitely generated module over Noetherian A, show that M has afiltration
0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M∼
where Mi+1 /Mi = A/pi for some prime ideal pi . (If the pi are all maximal, thefiltration is called a composition series, see Definition 18.4.7.)

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173(b) Show that the associated prime ideals are among the pi , and thus prove Theorem 5.5.10(a).
(c) Show that for each i, Supp A/pi is contained in Supp M, or equivalently, thatevery pi contains an associated prime. Hint: if pi does not contain an associatedprime, then localize at pi to “make M disappear”. (Caution: non-associated prime
ideals may be among the pi : take M = A = Z, and witness 0 ⊂ 2Z ⊂ Z.)
5.5.N. E XERCISE . Prove Theorem 5.5.10(d) as follows.
(a) Show that
AssA M ∩ Spec S−1 A ⊂ AssS−1 A S−1 M.
(Hint: suppose p ∈ AssA M ∩ Spec S−1 A, with p = Ann m for m ∈ M.)(b) Suppose q ∈ AssS−1 A S−1 M, which corresponds to p ∈ Spec A (i.e., q =p(S−1 A)). Then q = AnnS−1 A m (m ∈ S−1 M), which yields a nonzero elementof
HomS−1 A (S−1 A/q, S−1 M).
Argue that this group is isomorphic to S−1 HomA (A/p, M) (see Exercise 1.6.G),and hence HomA (A/p, M) ̸= 0.
This concludes the proof of Theorem 5.5.10. The remaining important loose
end is to understand associated points in terms of support.
5.5.O. E XERCISE . Show that those subsets of Spec A which are the support of anelement of M are precisely those subsets which are the closure of a subset of theassociated points. Hint: show that for any associated point p, there is a section
supported precisely on p. Remark: This can be used to solve Exercise 5.5.P, butsome people prefer to do Exercise 5.5.P first, and obtain this as a consequence.
5.5.P. I MPORTANT E XERCISE . Suppose A is a Noetherian ring, and M is a finitely
generated A-module. Show that associated points/primes of M satisfy property(A) as follows.
(a) Show that every associated point is the generic point of an irreducible
component of Supp m for some m ∈ M. Hint: if p ∈ A is associated, thenp = Ann m for some m ∈ M; this is useful in Exercise 5.5.O as well.
(b) If m ∈ M, show that the support of m is the closure of those associatedpoints at which m has nonzero germ (cf. Exercise 5.5.D, which relied on(A) and (B)). Hint: if p is in the closure of such an associated point, showthat m has nonzero germ at p. If p is not in the closure of such an associated point, show that m is 0 in Mp by localizing at p, and using Theorem 5.5.10(b) in the localized ring Ap (using Theorem 5.5.10(d)).
5.5.11. Loose ends.
We can easily extend the theory of associated points of schemes to a (veryspecial) setting without Noetherian hypotheses: integral domains, and integral
schemes.
5.5.Q. E XERCISE ( EASY VARIATION : ASSOCIATED POINTS OF INTEGRAL SCHEMES ).
Define the notion of associated points for integral domains and integral schemes.
More precisely, take (A) as the definition, and establish (B) and (C). (Hint: theunique associated prime of an integral domain is (0), and the unique associatedpoint of an integral scheme is its generic point.) In particular, rational functions

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on an integral scheme X are precisely elements of the function field K(X) (Definition 5.2.H).
Now that we have defined associated points, we can verify that there is an
example of the form described in Exercise 5.5.H
5.5.R. E XERCISE . Let I = (y − x2 )3 ∩ (x − 1, y − 1)15 ∩ (x − 2, y − 2). Show that
X = Spec C[x, y]/I satisfies the hypotheses of Exercise 5.5.H. (Rhetorical question:
Is there a “smaller” example? Is there a “smallest”?)
5.5.12. A non-Noetherian remark. By combining §5.5.3 with (C), we see that if A isa Noetherian ring, then any element of any minimal prime p is a zerodivisor. Thisis true without Noetherian hypotheses: suppose s ∈ p. Then by minimality of p,pAp is the unique prime ideal in Ap , so the element s/1 of Ap is nilpotent (because
it is contained in all prime ideals of Ap , Theorem 3.2.12). Thus for some t ∈ A \ p,tsn = 0, so s is a zerodivisor. We will use this in Exercise 11.1.G.
5.5.13. Aside: Primary ideals. The notion of primary ideals and primary decomposition is important, although we won’t use it. (An ideal I ⊂ A in a ring is primary
if I ̸= A and if xy ∈ I implies either x ∈ I or yn ∈ I for some n > 0.) The associatedprime ideals of an ideal turn out to be precisely the radicals of ideals in any primary decomposition. Primary decomposition was first introduced by the world
chess champion Lasker in 1905, and later axiomatized by Noether in the 1920’s.
See [E, §3.3], for example, for more on this topic.

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Part III

Morphisms

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CHAPTER 6Morphisms of schemes

6.1 Introduction
We now describe the morphisms between schemes. We will define some easyto-state properties of morphisms, but leave more subtle properties for later.
Recall that a scheme is (i) a set, (ii) with a topology, (iii) and a (structure) sheafof rings, and that it is sometimes helpful to think of the definition as having threesteps. In the same way, the notion of morphism of schemes X → Y may be defined(i) as a map of sets, (ii) that is continuous, and (iii) with some further information involving the sheaves of functions. In the case of affine schemes, we havealready seen the map as sets (§3.2.9) and later saw that this map is continuous(Exercise 3.4.H).
Here are two motivations for how morphisms should behave. The first is algebraic, and the second is geometric.
6.1.1. Algebraic motivation. We will want morphisms of affine schemes Spec A →Spec B to be precisely the ring maps B → A. We have already seen that ring maps
B → A induce maps of topological spaces in the opposite direction (Exercise 3.4.H);
the main new ingredient will be to see how to add the structure sheaf of functions
into the mix. Then a morphism of schemes should be something that “on the levelof affine open sets, looks like this”.
6.1.2. Geometric motivation. Motivated by the theory of differentiable manifolds(§3.1.1), which like schemes are ringed spaces, we want morphisms of schemesat the very least to be morphisms of ringed spaces; we now motivate what theseare. (We will formalize this in the next section.) Notice that if π : X → Y is a
map of differentiable manifolds, then a differentiable function on Y pulls back to
a differentiable function on X. More precisely, given an open subset U ⊂ Y, there
is a natural map Γ (U, OY ) → Γ (π−1 (U), OX ). This behaves well with respect torestriction (restricting a function to a smaller open set and pulling back yields
the same result as pulling back and then restricting), so in fact we have a mapof sheaves on Y: OY → π∗ OX . Similarly a morphism of schemes π : X → Y should
induce a map OY → π∗ OX . But in fact in the category of differentiable manifolds
a continuous map π : X → Y is a map of differentiable manifolds precisely whendifferentiable functions on Y pull back to differentiable functions on X (i.e., thepullback map from differentiable functions on Y to functions on X in fact lies in thesubset of differentiable functions, i.e., the continuous map π induces a pullback ofdifferential functions, which can be interpreted as a map OY → π∗ OX ), so this map
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The Rising Sea: Foundations of Algebraic Geometryof sheaves characterizes morphisms in the differentiable category. So we could use
this as the definition of morphism in the differentiable category (see Exercise 3.1.A).
But how do we apply this to the category of schemes? In the category of differentiable manifolds, a continuous map π : X → Y induces a pullback of (the sheaf of)functions, and we can ask when this induces a pullback of differentiable functions.
However, functions are odder on schemes, and we can’t recover the pullback map
just from the map of topological spaces. The right patch is to hardwire this into
the definition of morphism, i.e., to have a continuous map π : X → Y, along with apullback map π♯ : OY → π∗ OX . This leads to the definition of the category of ringed
spaces.
One might hope to define morphisms of schemes as morphisms of ringed
spaces. This isn’t quite right, as then Motivation 6.1.1 isn’t satisfied: as desired,to each morphism A → B there is a morphism Spec B → Spec A, but there can beadditional morphisms of ringed spaces Spec B → Spec A not arising in this way(see Exercise 6.2.E). A revised definition as morphisms of ringed spaces that locally look of this form will work, but this is awkward to work with, and we take adifferent approach. However, we will check that our eventual definition actuallyis equivalent to this (Exercise 6.3.C).
We begin by formally defining morphisms of ringed spaces.

6.2 Morphisms of ringed spaces
6.2.1. Definition. A morphism of ringed spaces π : X → Y is a continuous mapof topological spaces (which we unfortunately also call π) along with a map OY →π∗ OX , which we think of as a “pullback map”. By adjointness (§2.7.1), this is thesame as a map π−1 OY → OX . (It can be convenient to package this information as
in the diagram (2.7.2.1).) There is an obvious notion of composition of morphisms,
so ringed spaces form a category. Hence we have notion of automorphisms andisomorphisms. You can easily verify that an isomorphism of ringed spaces means
the same thing as it did before (Definition 4.3.1).
If U ⊂ Y is an open subset, then there is a natural morphism of ringed spaces(U, OY |U ) → (Y, OY ) (which implicitly appeared earlier in Exercise 2.7.G). Moreprecisely, if U → Y is an isomorphism of U with an open subset V of Y, and we are∼ (V, OY |V ) (via the isomorphism U =∼ V), then thegiven an isomorphism (U, OU ) =resulting map of ringed spaces is called an open embedding (or open immersion)of ringed spaces, and the morphism U → Y is often written U "→ Y.

6.2.A. E XERCISE ( MORPHISMS OF RINGED SPACES GLUE ). Suppose (X, OX ) and(Y, OY ) are ringed spaces, X = ∪i Ui is an open cover of X, and we have morphismsof ringed spaces πi : Ui → Y that “agree on the overlaps”, i.e., πi |Ui ∩Uj = πj |Ui ∩Uj .
Show that there is a unique morphism of ringed spaces π : X → Y such that π|Ui =πi . (Exercise 2.2.F essentially showed this for topological spaces.)

6.2.B. E ASY IMPORTANT EXERCISE : O - MODULES PUSH FORWARD . Given a morphism of ringed spaces π : X → Y, show that sheaf pushforward induces a functor
ModOX → ModOY .

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6.2.C. E ASY IMPORTANT EXERCISE . Given a morphism of ringed spaces π : X → Y
with π(p) = q, show that there is a map of stalks (OY )q → (OX )p .

6.2.D. K EY E XERCISE . Suppose π♯ : B → A is a morphism of rings. Define amorphism of ringed spaces π : Spec A → Spec B as follows. The map of topological spaces was given in Exercise 3.4.H. To describe a morphism of sheaves
OSpec B → π∗ OSpec A on Spec B, it suffices to describe a morphism of sheaves on the
distinguished base of Spec B. On D(g) ⊂ Spec B, we define
OSpec B (D(g)) → OSpec A (π−1 D(g)) = OSpec A (D(π♯ g))by Bg → Aπ♯ g . Verify that this makes sense (e.g., is independent of g), and that
this describes a morphism of sheaves on the distinguished base. (This is the third
in a series of exercises. We saw that a morphism of rings induces a map of setsin §3.2.9, a map of topological spaces in Exercise 3.4.H, and now a map of ringed
spaces here.)
The map of ringed spaces of Key Exercise 6.2.D is really not complicated. Hereis an example. Consider the ring map C[y] → C[x] given by y (→ x2 (see Figure 3.6).
We are mapping the affine line with coordinate x to the affine line with coordinate
y. The map is (on closed points) a (→ a2 . For example, where does [(x − 3)] go to?
Answer: [(y − 9)], i.e., 3 (→ 9. What is the preimage of [(y − 4)]? Answer: thoseprime ideals in C[x] containing [(x2 − 4)], i.e., [(x − 2)] and [(x + 2)], so the preimageof 4 is indeed ±2. This is just about the map of sets, which is old news (§3.2.9), so
let’s now think about functions pulling back. What is the pullback of the function
3/(y − 4) on D([(y − 4)]) = A1 − {4}? Of course it is 3/(x2 − 4) on A1 − {−2, 2}.
The construction of Key Exercise 6.2.D will soon be an example of morphismof schemes! In fact we could make that definition right now. Before we do, we
point out (via the next exercise) that not every morphism of ringed spaces between
affine schemes is of the form of Key Exercise 6.2.D. (In the language of §6.3, thismorphism of ringed spaces is not a morphism of locally ringed spaces.)
6.2.E. U NIMPORTANT E XERCISE . Recall (Exercise 3.4.K) that Spec k[y](y) has twopoints, [(0)] and [(y)], where the second point is closed, and the first is not. Describe a map of ringed spaces Spec k(x) → Spec k[y](y) sending the unique pointof Spec k(x) to the closed point [(y)], where the pullback map on global sections
sends k to k by the identity, and sends y to x. Show that this map of ringed spaces
is not of the form described in Key Exercise 6.2.D.
6.2.2. Tentative Definition we won’t use (cf. Motivation 6.1.1 in §6.1). A morphism of schemes π : (X, OX ) → (Y, OY ) is a morphism of ringed spaces that “locally looks like” the maps of affine schemes described in Key Exercise 6.2.D. Precisely, for each choice of affine open sets Spec A ⊂ X, Spec B ⊂ Y, such that
π(Spec A) ⊂ Spec B, the induced map of ringed spaces should be of the formshown in Key Exercise 6.2.D.
We would like this definition to be checkable on an affine cover, and we mighthope to use the Affine Communication Lemma to develop the theory in this way.
This works, but it will be more convenient to use a clever trick: in the next section,
we will use the notion of locally ringed spaces, and then once we have used it, we
will discard it like yesterday’s garbage.

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6.3 From locally ringed spaces to morphisms of schemes
In order to prove that morphisms behave in a way we hope, we will use thenotion of a locally ringed space. It will not be used later, although it is useful elsewhere in geometry. The notion of locally ringed spaces (and maps between them)is inspired by what we know about manifolds (see Exercise 3.1.B). If π : X → Y
is a morphism of manifolds, with π(p) = q, and f is a function on Y vanishingat q, then the pulled back function π♯ (f) on X should vanish on p. Put differently:
germs of functions (at q ∈ Y) vanishing at q should pull back to germs of functions(at p ∈ X) vanishing at p.
6.3.1. Definition. Recall (Definition 4.3.6) that a locally ringed space is a ringed
space (X, OX ) such that the stalks OX,p are all local rings. A morphism of locallyringed spaces π : X → Y is a morphism of ringed spaces such that the induced mapof stalks π♯ : OY,q → OX,p (Exercise 6.2.C) sends the maximal ideal of the former
into the maximal ideal of the latter (a “morphism of local rings”). This means
something rather concrete and intuitive: “if p (→ q, and g is a function vanishingat q, then it will pull back to a function vanishing at p.” (Side Remark: you would
also want: “if p (→ q, and g is a function not vanishing at q, then it will pull backto a function not vanishing at p.” This follows from our definition — can you seewhy?) Note that locally ringed spaces form a category.
To summarize: we use the notion of locally ringed space only to define morphisms of schemes, and to show that morphisms have reasonable properties. The
main things you need to remember about locally ringed spaces are (i) that the functions have values at points, and (ii) that given a map of locally ringed spaces, thepullback of where a function vanishes is precisely where the pulled back function
vanishes.
6.3.A. E XERCISE . Show that morphisms of locally ringed spaces glue (cf. Exercise 6.2.A). (Hint: your solution to Exercise 6.2.A may work without change.)
6.3.B. E ASY IMPORTANT EXERCISE .
(a) Show that Spec A is a locally ringed space. (Hint: Exercise 4.3.F.)(b) Show that the morphism of ringed spaces π : Spec A → Spec B defined by aring morphism π♯ : B → A (Exercise 6.2.D) is a morphism of locally ringed spaces.
6.3.2. Key Proposition. — If π : Spec A → Spec B is a morphism of locally ringed
spaces then it is the morphism of locally ringed spaces induced by the map π♯ : B =Γ (Spec B, OSpec B ) → Γ (Spec A, OSpec A ) = A as in Exercise 6.3.B(b).

(Aside: Exercise 4.3.A is a special case of Key Proposition 6.3.2. You shouldlook back at your solution to Exercise 4.3.A, and see where you implicitly usedideas about locally ringed spaces.)
Proof. Suppose π : Spec A → Spec B is a morphism of locally ringed spaces. We
wish to show that it is determined by its map on global sections π♯ : B → A. We
first need to check that the map of points is determined by global sections. Now
a point p of Spec A can be identified with the prime ideal of global functions vanishing on it. The image point π(p) in Spec B can be interpreted as the unique point
q of Spec B, where the functions vanishing at q are precisely those that pull back

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to functions vanishing at p. (Here we use the fact that π is a map of locally ringed
spaces.) This is precisely the way in which the map of sets Spec A → Spec B induced by a ring map B → A was defined (§3.2.9).
Note in particular that if b ∈ B, π−1 (D(b)) = D(π♯ b), again using the hypothesis that π is a morphism of locally ringed spaces.
It remains to show that π♯ : OSpec B → π∗ OSpec A is the morphism of sheaves
given by Exercise 6.2.D (cf. Exercise 6.3.B(b)). It suffices to check this on the distinguished base (Exercise 2.5.C(a)). We now want to check that for any map of
locally ringed spaces inducing the map of sheaves OSpec B → π∗ OSpec A , the map ofsections on any distinguished open set D(b) ⊂ Spec B is determined by the mapof global sections B → A.
Consider the commutative diagramBΓ (Spec B, OSpec B )π♯Spec B

A

resSpec A,D(π♯ b)

resSpec B,D(b)Bb% Γ (Spec A, OSpec A )"Γ (D(b), OSpec B )π♯D(b)"% Γ (D(π♯ b), OSpec A )Aπ♯ b = A ⊗B Bb .

The vertical arrows (restrictions to distinguished open sets) are localizations byb, so the lower horizontal map π♯D(b) is determined by the upper map (it is just
localization by b).
!
We are ready for our definition.
6.3.3. Definition. If X and Y are schemes, then a morphism π : X → Y as locallyringed spaces is called a morphism of schemes. We have thus defined the categoryof schemes, which we denote Sch. (We then have notions of isomorphism — justthe same as before, §4.3.6 — and automorphism. The target Y of π is sometimes
called the base scheme or the base, when we are interpreting π as a “family ofschemes parametrized by Y” — this may become clearer once we have defined thefibers of morphisms in §9.3.2.)
The definition in terms of locally ringed spaces easily implies Tentative Definition 6.2.2:
6.3.C. I MPORTANT E XERCISE . Show that a morphism of schemes π : X → Y isa morphism of ringed spaces that looks locally like morphisms of affine schemes.
Precisely, if Spec A is an affine open subset of X and Spec B is an affine open subset
of Y, and π(Spec A) ⊂ Spec B, then the induced morphism of ringed spaces is amorphism of affine schemes. (In case it helps, note: if W ⊂ X and Y ⊂ Z are both
open embeddings of ringed spaces, then any morphism of ringed spaces X → Y
induces a morphism of ringed spaces W → Z, by composition W → X → Y → Z.)
Show that it suffices to check on a set (Spec Ai , Spec Bi ) where the Spec Ai form
an open cover of X and the Spec Bi form an open cover of Y.
In practice, we will use the affine cover interpretation, and forget completely
about locally ringed spaces. In particular, put imprecisely, the category of affine
schemes is the category of rings with the arrows reversed. More precisely:

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6.3.D. E XERCISE . Show that the category of rings and the opposite category ofaffine schemes are equivalent (see §1.2.21 to read about equivalence of categories).
In particular: there can be many different maps from one point to another! For
example, here are two different maps from the point Spec C to the point Spec C: theidentity (corresponding to the identity C → C), and complex conjugation. (Thereare even more such maps!)
It is clear (from the corresponding facts about locally ringed spaces) that morphisms glue (Exercise 6.3.A), and the composition of two morphisms is a morphism. Isomorphisms in this category are precisely what we defined them to beearlier (§4.3.6).
6.3.E. E NLIGHTENING E XERCISE . (This exercise can give you some practice with
understanding morphisms of schemes by cutting up into affine open sets.) Make
sense of the following sentence: “An+1\ {⃗0} → Pn
k given byk(x0 , x1 , . . . , xn ) (→ [x0 , x1 , . . . , xn ]is a morphism of schemes.” Caution: you can’t just say where points go; you haveto say where functions go. So you may have to divide these up into affines, anddescribe the maps, and check that they glue. (Can you generalize to the case where
k is replaced by a general ring B? See Exercise 6.3.M for an answer.)
6.3.4. Morphisms to affine schemes.
The following result shows that it is easy to describe morphisms to an affine
scheme without working hard to cover the source with affine open sets.
6.3.F. E SSENTIAL EXERCISE . Show that morphisms X → Spec A are in naturalbijection with ring morphisms A → Γ (X, OX ). Hint: Show that this is true when X
is affine. Use the fact that morphisms glue, Exercise 6.3.A. (This is even true in the
category of locally ringed spaces. You are free to prove it in this generality, but it
is easier in the category of schemes.)
In particular, there is a canonical morphism from a scheme to Spec of its ring of
global sections. (Warning: Even if X is a finite type k-scheme, the ring of global sections might be nasty! In particular, it might not be finitely generated, see 19.11.13.)
The canonical morphism X → Spec Γ (X, OX ) is an isomorphism if and only if X isaffine (i.e., isomorphic to Spec A for some ring A), and in this case it is the isomorphism hinted at in Remark 4.3.2.
6.3.G. E ASY E XERCISE . Show that this definition of A-scheme given in §6.3.7
agrees with the earlier definition of §5.3.6.
6.3.H. E ASY E XERCISE . If S• is a finitely generated graded A-algebra, describe anatural “structure morphism” Proj S• → Spec A.

6.3.I. E ASY E XERCISE . Show that Spec Z is the final object in the category ofschemes. In other words, if X is any scheme, there exists a unique morphism
to Spec Z. (Hence the category of schemes is isomorphic to the category of Zschemes.) If k is a field, show that Spec k is the final object in the category of
k-schemes.

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6.3.5. Remark. From Essential Exercise 6.3.F, it is one small step to show that someproducts of schemes exist: if A and B are rings, then Spec A × Spec B = Spec(A ⊗Z
B); and if A and B are C-algebras, then Spec A ×Spec C Spec B = Spec(A ⊗C B). But
we are in no hurry, so we wait until Exercise 9.1.B to discuss this properly.
6.3.6. ⋆⋆ Side fact for experts: Γ and Spec are adjoints. We have a contravariant
functor Spec from rings to locally ringed spaces, and a contravariant functor Γfrom locally ringed spaces to rings. In fact (Γ, Spec) is an adjoint pair! (Caution:
we have only discussed adjoints for covariant functors; if you care, you will haveto figure out how to define adjoints for contravariant functors.) Thus we could
have defined Spec by requiring it to be right-adjoint to Γ . (Fun but irrelevant sidequestion: if you used ringed spaces rather than locally ringed spaces, Γ again has
a right adjoint. What is it?)
Our ability to easily describe morphisms to affine schemes will allow us torevisit our earlier discussions of schemes over a given field or ring (§5.3.6).
6.3.7. The category of complex schemes (or more generally the category of kschemes where k is a field, or more generally the category of A-schemes where
A is a ring, or more generally the category of S-schemes where S is a scheme).
The category of S-schemes SchS (where S is a scheme) is defined as follows. The
objects (S-schemes) are morphisms of the form
X"
S
(The morphism to S is called the structure morphism. A motivation for this terminology is the fact that if S = Spec A, the structure morphism gives the functions
on each open set of X the structure of an A-algebra, cf. §5.3.6.) The morphisms in
the category of S-schemes are defined to be commutative diagrams
%Y

X
"
S

=

"
%S

which is more conveniently written as a commutative diagram
X❅
❅❅
❅❅
❅❅
8

S.

%Y
⑧
⑧⑧
⑧⑧
⑧
5⑧

When there is no confusion (if the base scheme is clear), simply the top row ofthe diagram is given. In the case where S = Spec A, where A is a ring, we get
the notion of an A-scheme, which is the same as the same definition as in §5.3.6(Exercise 6.3.G), but in a more satisfactory form. For example, complex geometers
may consider the category of C-schemes.
6.3.8. Morphisms from (some) affine schemes.

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Morphisms from affine schemes are not quite as simple as morphisms to affine
schemes, but some cases are worth pointing out.
6.3.J. E XERCISE .
(a) Suppose p is a point of a scheme X. Describe a canonical (choice-free) morphism
Spec OX,p → X. (Hint: do this for affine X first. But then for general X be sure toshow that your morphism is independent of choice.)(b) Define a canonical morphism Spec κ(p) → X. (This is often written p → X; one
gives p the obvious interpretation as a scheme, Spec κ(p).)

6.3.K. E XERCISE ( MORPHISMS FROM Spec OF A LOCAL RING TO X). Suppose
X is a scheme, and (A, m) is a local ring. Suppose we have a scheme morphism
π : Spec A → X sending [m] to p. Show that any open set containing p contains theimage of π. Show that there is a bijection between Mor(Spec A, X) and{p ∈ X, local homomorphisms OX,p → A}.

(Possible hint: Exercise 6.3.J(a).)

These exercises lead us to the notion of field-valued points, and more generally
ring-valued points, and more generally still, scheme-valued points.
6.3.9. Definition: The functor of points, and scheme-valued points (and ringvalued points, and field-valued points) of a scheme. If Z is a scheme, thenZ-valued points of a scheme X, denoted X(Z), are defined to be maps Z → X. If A
is a ring, then A-valued points of a scheme X, denoted X(A), are defined to be the(Spec A)-valued points of the scheme. (The most common case of this is when A
is a field.)
If you are working over a base scheme B — for example, complex algebraic
geometers will consider only schemes and morphisms over B = Spec C — then in
the above definition, there is an implicit structure map Z → B (or Spec A → B in
the case of X(A)). For example, for a complex geometer, if X is a scheme over C,
the C(t)-valued points of X correspond to commutative diagrams of the form
%X
Spec C(t)
❑❑❑
③
③
❑❑❑
③③
❑
③③π
③
ξ ❑❑❑
#
C③③
Spec C
where π : X → Spec C is the structure map for X, and ξ corresponds to the obvious
inclusion of rings C → C(t). (Warning: a k-valued point of a k-scheme X is sometimes called a “rational point” of X, which is dangerous, as for most of the world,“rational” refers to Q. We will use the safer phrase “k-valued point” of X.)
The terminology “Z-valued point” (and A-valued point) is unfortunate, because we earlier defined the notion of points of a scheme, and Z-valued points(and A-valued points) are not (necessarily) points! But these usages are wellestablished in the literature. (Look in the index under “point” to see even moreinconsistent use of adjectives that modify this word.)
6.3.L. E XERCISE .
(a) (easy) Show that a morphism of schemes X → Y induces a map of Z-valued
points X(Z) → Y(Z).

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185(b) Note that morphisms of schemes X → Y are not determined by their “underlying” map of points. (What is an example?) Show that they are determined by theirinduced maps of Z-valued points, as Z varies over all schemes. (Hint: pick Z = X.
In the course of doing this exercise, you will largely prove Yoneda’s Lemma in theguise of Exercise 9.1.C.)
6.3.10. Furthermore, we will see that “products of Z-valued points” behave as you
might hope (§9.1.3). A related reason this language is suggestive: the notation X(Z)suggests the interpretation of X as a (contravariant) functor hX from schemes tosets — the functor of (scheme-valued) points of the scheme X (cf. Example 1.2.20).
Here is a low-brow reason A-valued points are a useful notion: the A-valued
points of an affine scheme Spec Z[x1 , . . . , xn ]/(f1 , . . . , fr ) (where fi ∈ Z[x1 , . . . , xn ] arerelations) are precisely the solutions to the equations
f1 (x1 , . . . , xn ) = · · · = fr (x1 , . . . , xn ) = 0
in the ring A. For example, the rational solutions to x2 + y2 = 16 are precisely the
Q-valued points of Spec Z[x, y]/(x2 + y2 − 16). The integral solutions are precisely
the Z-valued points. So A-valued points of an affine scheme (finite type over Z)
can be interpreted simply. In the special case where A is local, A-valued points ofa general scheme have a good interpretation too:On the other hand, Z-valued points of projective space can be subtle. There
are some maps we can write down easily, as shown by applying the next exercise
in the case X = Spec A, where A is a B-algebra.
6.3.M. E ASY ( BUT SURPRISINGLY ENLIGHTENING ) E XERCISE ( CF. E XERCISE 6.3.E).
(a) Suppose B is a ring. If X is a B-scheme, and f0 , . . . , fn are n + 1 functions
on X with no common zeros, then show that [f0 , . . . , fn ] gives a morphism of Bschemes X → Pn
B.
(b) Suppose g is a nowhere vanishing function on X, and fi are as in part (a). Show
that the morphisms [f0 , . . . , fn ] and [gf0 , . . . , gfn ] to Pn
B are the same.
6.3.11. Example (Exercise 6.3.E revisited). Consider the n + 1 functions x0 , . . . , xn
on An+1 (otherwise known as n + 1 sections of the trivial bundle). They have nocommon zeros on An+1 − 0. Hence they determine a morphism An+1 − 0 → Pn .

6.3.12. You might hope that Exercise 6.3.M(a) gives all morphisms to projective
space (over B). But this isn’t the case. Indeed, even the identity morphism X =P1k → P1k isn’t of this form, as the source P1 has no nonconstant global functions
with which to build this map. (There are similar examples with an affine source.)
However, there is a correct generalization (characterizing all maps from schemes
to projective schemes) in Theorem 16.4.1. This result roughly states that this works,
so long as the fi are not quite functions, but sections of a line bundle. Our desire
to understand maps to projective schemes in a clean way will be one important
motivation for understanding line bundles.
We will see more ways to describe maps to projective space in the next section. A different description directly generalizing Exercise 6.3.M(a) will be given
in Exercise 15.3.F, which will turn out (in Theorem 16.4.1) to be a “universal” description.

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Incidentally, before Grothendieck, it was considered a real problem to figureout the right way to interpret points of projective space with “coordinates” in aring. These difficulties were due to a lack of functorial reasoning. And the clues
to the right answer already existed (the same problems arise for maps from a manifold to RPn ) — if you ask such a geometric question (for projective space is geometric), the answer is necessarily geometric, not purely algebraic!
6.3.13. Visualizing schemes III: picturing maps of schemes when nilpotents are present.
You now know how to visualize the points of schemes (§3.3), and nilpotents (§4.2
and §5.5). The following imprecise exercise will give you some sense of how to visualize maps of schemes when nilpotents are involved. Suppose a ∈ C. Consider
the map of rings C[x] → C[ϵ]/ϵ2 given by x (→ aϵ. Recall that Spec C[ϵ]/(ϵ2 ) may
be pictured as a point with a tangent vector (§4.2). How would you picture thismap if a ̸= 0? How does your picture change if a = 0? (The tangent vector shouldbe “crushed” in this case.)
Exercise 12.1.I will extend this considerably. You may enjoy reading its statement now.
6.3.N. ⋆⋆ E XERCISE FOR THOSE WITH APPROPRIATE BACKGROUND : THE ANALYTI FICATION FUNCTOR . Recall the analytification construction of Exercise 5.3.G. For
each morphism of reduced finite type C-schemes π : X → Y (over C), define a morphism of complex analytic prevarieties πan : Xan → Yan (the analytification ofπ). Show that analytification gives a functor from the category of reduced finite
type C-schemes to the category of complex analytic prevarieties. (Remark: Twononisomorphic varieties can have isomorphic analytifications. For example, Serredescribed two different algebraic structures on the complex manifold C∗ × C∗ , see[Ha2, p. 232] and [MO68421]; one is “the obvious one”, and the other is a P1 bundle over an elliptic curve, with a section removed. For an example of a smoothcomplex surface with infinitely many algebraic structures, see §19.11.3. On theother hand, a compact complex variety can have only one algebraic structure (see[Se3, §19]).)

6.4 Maps of graded rings and maps of projective schemes
As maps of rings correspond to maps of affine schemes in the opposite direction, maps of graded rings (over a base ring A) sometimes give maps of projective
schemes in the opposite direction. This is an imperfect generalization: not every
map of graded rings gives a map of projective schemes (§6.4.2); not every mapof projective schemes comes from a map of graded rings (§19.11.9); and different
maps of graded rings can yield the same map of schemes (Exercise 6.4.C).
You may find it helpful to think through Examples 6.4.1 and 6.4.2 while working through the following exercise.
% R• is a morphism of (Z≥0 6.4.A. E SSENTIAL EXERCISE . Suppose that φ : S•
)graded rings. (By map of graded rings, we mean a map of rings that preserves
the grading as a map of “graded semigroups” (or “graded monoids”). In otherwords, there is a d > 0 such that Sn maps to Rdn for all n.) Show that this induces

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a morphism of schemes (Proj R• ) \ V(φ(S+ )) → Proj S• . (Hint: Suppose f is ahomogeneous element of S+ . Define a map D(φ(f)) → D(f). Show that they glue
together (as f runs over all homogeneous elements of S+ ). Show that this defines
a map from all of Proj R• \ V(φ(S+ )).) In particular, if(6.4.0.1)

V(φ(S+ )) = ∅,then we have a morphism Proj R• → Proj S• . From your solution, it will be clearthat if φ is furthermore a morphism of A-algebras, then the induced morphism
Proj R• \ V(φ(S• )) → Proj S• is a morphism of A-schemes.

6.4.1. Example. Let’s see Exercise 6.4.A in action. We will scheme-theoreticallyinterpret the map of complex projective manifolds CP1 to CP2 given by
CP1

% CP2

[s, t] ✤% [s20 , s9 t11 , t20 ]Notice first that this is well-defined: [λs, λt] is sent to the same point of CP2 as[s, t]. The reason for it to be well-defined is that the three polynomials s20 , s9 t11 ,and t20 are all homogeneous of degree 20.
Algebraically, this corresponds to a map of graded rings in the opposite direction
C[x, y, z] → C[s, t]given by x (→ s20 , y (→ s9 t11 , z (→ t20 . You should interpret this in light of yoursolution to Exercise 6.4.A, and compare this to the affine example of §3.2.10.

6.4.2. Example. Notice that there is no map of complex manifolds CP2 → CP1
given by [x, y, z] (→ [x, y], because the map is not defined when x = y = 0. This
corresponds to the fact that the map of graded rings C[s, t] → C[x, y, z] given bys (→ x and t (→ y, doesn’t satisfy hypothesis (6.4.0.1).
&
6.4.B. E XERCISE . Show that if φ : S• → R• satisfies (φ(S+ )) = R+ , then hypothesis (6.4.0.1) is satisfied. (Hint: Exercise 4.5.I.) This algebraic formulation of themore geometric hypothesis can sometimes be easier to verify.
6.4.C. U NIMPORTANT EXERCISE .
This exercise shows that different maps ofgraded rings can give the same map of schemes. Let R• = k[x, y, z]/(xz, yz, z2 )and S• = k[a, b, c]/(ac, bc, c2 ), where every variable has degree 1. Show that
∼ Proj S• =∼ P1 . Show that the maps S• → R• given by (a, b, c) (→ (x, y, z)
Proj R• =kand (a, b, c) (→ (x, y, 0) give the same (iso)morphism Proj R• → Proj S• . (The realreason is that all of these constructions are insensitive to what happens in a finite
number of degrees. This will be made precise in a number of ways later, mostimmediately in Exercise 6.4.F.)
6.4.3. Unimportant remark. Exercise 16.4.H shows that not every morphism ofschemes Proj R• → Proj S• comes from a map of graded rings S• → R• , even inquite reasonable circumstances.
6.4.4. Veronese subrings.

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Here is a useful construction. Suppose S• is a finitely generated graded ring.
Define the nth Veronese subring of S• by Sn• = ⊕∞j=0 Snj . (The “old degree” n is“new degree” 1.)
6.4.D. E XERCISE . Show that the map of graded rings Sn• "→ S• induces an isomorphism Proj S• → Proj Sn• . (Hint: if f ∈ S+ is homogeneous of degree divisible by n,identify D(f) on Proj S• with D(f) on Proj Sn• . Why do such distinguished open
sets cover Proj S• ?)6.4.E. E XERCISE . If S• is generated in degree 1, show that Sn• is also generated indegree 1. (You may want to consider the case of the polynomial ring first.)
6.4.F. E XERCISE . Show that if R• and S• are the same finitely generated graded∼rings except in a finite number of nonzero degrees (make this precise!), then Proj R• =Proj S• .
6.4.G. E XERCISE . Suppose S• is generated over S0 by f1 , . . . , fn . Find a d such that
Sd• is finitely generated in “new” degree 1 (= “old” degree d). (This is surprisingly
tricky, so here is a hint. Suppose there are generators x1 , . . . , xn of degrees d1 , . . . ,an1
dn respectively.
Show that any monomial xa
1 · · · xn of degree at least nd1 . . . dn"
has ai ≥ ( j dj )/di for some i. Show that the (nd1 . . . dn )th Veronese subring is
generated by elements in “new” degree 1.)

Exercise 6.4.G, in combination with Exercise 6.4.D, shows that there is little
harm in assuming that finitely generated graded rings are generated in degree 1,
as after a regrading (or more precisely, keeping only terms of degree a multiple ofd, then dividing the degree by d), this is indeed the case. This is handy, as it means
that, using Exercise 6.4.D, we can assume that any finitely generated graded ringis generated in degree 1. Exercise 8.2.G will later imply as a consequence that we
can embed every Proj in some projective space.
6.4.H. L ESS IMPORTANT EXERCISE . Suppose S• is a finitely generated ring. Show
that Sn• is a finitely generated graded ring. (Possible approach: use the previous
exercise, or something similar, to show there is some N such that SnN• is generated
in degree 1, so the graded ring SnN• is finitely generated. Then show that for each0 < j < N, SnN•+nj is a finitely generated module over SnN• .)6.5 Rational maps from reduced schemes
Informally speaking, a “rational map” is “a morphism defined almost everywhere”, much as a rational function (Definition 5.5.6) is a name for a function
defined almost everywhere. We will later see that in good situations, just as withrational functions, where a rational map is defined, it is uniquely defined (theReduced-to-Separated Theorem 10.2.2), and has a largest “domain of definition” (§10.2.3).
For this section only, we assume X to be reduced. A key example will be irreducible
varieties (§6.5.6), and the language of rational maps is most often used in this case.
6.5.1. Definition. A rational map π from X to Y, denoted π : X ""# Y, is a morphism
on a dense open set, with the equivalence relation (α : U → Y) ∼ (β : V → Y) if

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there is a dense open set Z ⊂ U∩V such that α|Z = β|Z . (In §10.2.3, we will improve
this to: if α|U∩V = β|U∩V in good circumstances — when Y is separated.) People
often use the word “map” for “morphism”, which is quite reasonable, except that arational map need not be a map. So to avoid confusion, when one means “rationalmap”, one should never just say “map”.
We will also discuss rational maps of S-schemes for a scheme S. The definitionis the same, except now X and Y are S-schemes, and α : U → Y is a morphism ofS-schemes.

6.5.2. ⋆ Rational maps more generally. Just as with rational functions, Definition 6.5.1
can be extended to where X is not reduced, as is (using the same name, “rationalmap”), or in a version that imposes some control over what happens over thenonreduced locus (pseudo-morphisms, [Stacks, tag 01RX]). We will see in §10.2 thatrational maps from reduced schemes to separated schemes behave particularly
well, which is why they are usually considered in this context. The reason forthe definition of pseudo-morphisms is to extend these results to when X is nonreduced. We will not use the notion of pseudo-morphism.
6.5.3. An obvious example of a rational map is a morphism. Another important
n−1
example is the projection Pn
given by [x0 , · · · , xn ] (→ [x0 , · · · , xn−1 ].
A ""# PA
(How precisely is this a rational map in the sense of Definition 6.5.1? What is its
domain of definition?)
A rational map π : X ""# Y is dominant (or in some sources, dominating) if
for some (and hence every) representative U → Y, the image is dense in Y. A
morphism is a dominant morphism (or dominating morphism) if it is dominant asa rational map.
6.5.A. E XERCISE . Show that a rational map π : X ""# Y of irreducible schemes isdominant if and only if π sends the generic point of X to the generic point of Y.
A little thought will convince you that you can compose (in a well-definedway) a dominant map π : X ""# Y from an irreducible scheme X to an irreducible
scheme Y with a rational map ρ : Y ""# Z. Furthermore, the composition ρ ◦ πwill be dominant if ρ is dominant. Integral schemes and dominant rational maps
between them form a category which is geometrically interesting.
6.5.B. E ASY E XERCISE . Show that dominant rational maps of integral schemes
give morphisms of function fields in the opposite direction.
In “suitably classical situations” (integral finite type k-schemes — and in particular, irreducible varieties, to be defined in §10.1.7) — this is reversible: dominantrational maps correspond to inclusions of function fields in the opposite direction.
We make this precise in §6.5.6 below. But it is not true that morphisms of function
fields always give dominant rational maps, or even rational maps. For example,
Spec k[x] and Spec k(x) have the same function field k(x), but there is no corresponding rational map Spec k[x] ""# Spec k(x) of k-schemes. Reason: such a rational map would correspond to a morphism from an open subset U of Spec k[x],say Spec k[x, 1/f(x)], to Spec k(x). But there is no map of rings k(x) → k[x, 1/f(x)](sending k identically to k and x to x) for any one f(x).

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The Rising Sea: Foundations of Algebraic Geometry(If you want more evidence that the topologically-defined notion of dominance is simultaneously algebraic, you can show that if φ : A → B is a ring morphism, then the corresponding morphism Spec B → Spec A is dominant if andonly if φ has kernel contained in the nilradical of A.)

6.5.4. Definition. A rational map π : X ""# Y is said to be birational if it is dominant,and there is another rational map (a “rational inverse”) ψ that is also dominant,such that π ◦ ψ is (in the same equivalence class as) the identity on Y, and ψ ◦π is (in the same equivalence class as) the identity on X. This is the notion of
isomorphism in the category of integral schemes and dominant rational maps. (Wenote in passing that in the differentiable category, this is a useless notion — any
connected differentiable manifolds of dimension d have isomorphic dense open
sets. To show this, use the fact that every smooth manifold admits a triangulation,see [Wh, p. 124-135].) A morphism is birational if it is birational as a rational map.
(Note that the “inverse” to a birational morphism may be only a rational map, nota morphism.)
We say X and Y are birational (to each other) if there exists a birational map
X ""# Y. If X and Y are irreducible, then birational maps induce isomorphismsof function fields. The fact that maps of function fields correspond to rational
maps in the opposite direction for integral finite type k-schemes, to be proved in
Proposition 6.5.7, shows that a map between integral finite type k-schemes thatinduces an isomorphism of function fields is birational. An integral finite typek-scheme is said to be rational if it is birational to An
k for some n.
6.5.5. Proposition. — Suppose X and Y are reduced schemes. Then X and Y are birational if and only if there is a dense open subscheme U of X and a dense open subscheme V∼ V.
of Y such that U =Proposition 6.5.5 tells you how to think of birational maps. Just as a rational
map is a “mostly defined function”, two birational reduced schemes are “mostly
isomorphic”. For example, a reduced finite type k-scheme (such as a reduced
affine variety over k) is rational if it has a dense open subscheme isomorphic toan open subscheme of Ank.
Proof. The “if” direction is trivial, so we prove the “only if” direction.
Step 1. Because X and Y are birational, we can find some dense open subschemes X1 ⊂ X and Y1 ⊂ Y, along with F : X1 → Y and G : Y1 → X whose composition in either order is the identity morphism on some dense open subscheme
where it makes sense. Replace X1 and Y1 by those dense open subschemes.
We have thus found dense open subschemes X1 ⊂ X and Y1 ⊂ Y, along withmorphisms F : X1 → Y and G : Y1 → X, whose composition in either order is theidentity on the open subset where it is defined. (More precisely, if X2 = F−1 (Y1 ),and Y2 = G−1 (X1 ), then G ◦ F|X2 = idX2 , and F ◦ G|Y2 = idY2 .)Step 2. For n > 1, inductively define Xn+1 = F−1 (Yn ) and Yn+1 = G−1 (Xn ).
Informally, Xn is the (dense) open subset of points of X that can be mapped ntimes by F and G alternately, and analogously for Yn . Define X∞ = ∩n≥1 Xn , andY∞ = ∩n≥1 Yn . Then X∞ = X2 , as G ◦ F is the identity on X2 (so any point of
X2 can be acted on by F and G alternately any number of times), and similarly
Y∞ = Y2 . Thus F and G define maps between X2 and Y2 , and these are inverse
maps by assumption.
!

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6.5.6. Rational maps of irreducible varieties.
6.5.7. Proposition. — Suppose X is an integral k-scheme and Y is an integral finite typek-scheme, and we are given an extension of function fields φ♯ : K(Y) "→ K(X) preserving
k. Then there exists a dominant rational map of k-schemes φ : X ""# Y inducing φ♯ .
Proof. By replacing Y with an open subset, we may assume that Y is affine, saySpec B, where B is generated over k by finitely many elements y1 , . . . , yn . Since
we only need to define φ on an open subset of X, we may similarly assume that
X = Spec A is affine. Then φ♯"gives an inclusion φ♯ : B "→ K(A). Write φ♯ (yi ) as
fi /gi (fi , gi ∈ A), and let g :=gi . Then φ♯ further induces an inclusion B "→ Ag .
Therefore φ : Spec Ag → Spec B induces φ♯ . The morphism φ is dominant because
the inverse image of the zero ideal under the inclusion B "→ Ag is the zero ideal,
so φ takes the generic point of X to the generic point of Y.
!
6.5.C. E XERCISE . Let K be a finitely generated field extension of k. (Recall that afield extension K over k is finitely generated if there is a finite “generating set” x1 ,. . . , xn in K such that every element of K can be written as a rational function in x1 ,. . . , xn with coefficients in k.) Show that there exists an irreducible affine k-variety
with function field K. (Hint: Consider the map k[t1 , . . . , tn ] → K given by ti (→ xi ,and show that the kernel is a prime ideal p, and that k[t1 , . . . , tn ]/p has fraction
field K. Interpreted geometrically: consider the map Spec K → Spec k[t1 , . . . , tn ]given by the ring map ti (→ xi , and take the closure of the one-point image.)
6.5.D. E XERCISE . Describe equivalences of categories among the following.

(a) the category with objects “integral k-varieties”, and morphisms “dominant rational maps defined over k”;(b) the category with objects “integral affine k-varieties”, and morphisms “dominant rational maps defined over k”; and(c) the opposite (“arrows-reversed”) category with objects “finitely generated field extensions of k”, and morphisms “inclusions extending theidentity on k”.
In particular, an integral affine k-variety X is rational if its function field K(X)
∼ k(x1 , . . . , xn ) for some n.
is a purely transcendental extension of k, i.e., K(X) =(This needs to be said more precisely: the map k "→ K(X) induced by X → Spec kshould agree with the “obvious” map k "→ k(x1 , . . . , xn ) under this isomorphism.)
6.5.8. More examples of rational maps.
A recurring theme in these examples is that domains of definition of rational
maps to projective schemes extend over regular codimension one points. We will
make this precise in the Curve-to-Projective Extension Theorem 16.5.1, when we
discuss curves.
The first example is the classical formula for Pythagorean triples. Suppose you
are looking for rational points on the circle C given by x2 +y2 = 1 (Figure 6.1). One
rational point is p = (1, 0). If q is another rational point, then pq is a line of rational(non-infinite) slope. This gives a rational map from the conic C (now interpreted
as Spec Q[x, y]/(x2 + y2 − 1)) to A1Q , given by (x, y) (→ y/(x − 1). (Something subtle
just happened: we were talking about Q-points on a circle, and ended up with a

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y
q

p

x

slope mC

F IGURE 6.1. Finding primitive Pythagorean triples using geometry

rational map of schemes.) Conversely, given a line of slope m through p, where mis rational, we can recover q by solving the equations y = m(x − 1), x2 + y2 = 1.
We substitute the first equation into the second, to get a quadratic equation in x.
We know that we will have a solution x = 1 (because the line meets the circle at(x, y) = (1, 0)), so we expect to be able to factor this out, and find the other factor.
This indeed works:
x2 + (m(x − 1))2

=

1

2

(m + 1)x + (−2m )x + (m − 1) =

0

2

0

2

=⇒
=⇒

2

2

2

(x − 1)((m + 1)x − (m − 1)) =The other solution is x = (m2 − 1)/(m2 + 1), which gives y = −2m/(m2 + 1). Thus
we get a birational map between the conic C and A1 with coordinate m, given byf : (x, y) (→ y/(x − 1) (which is defined for x ̸= 1), and with inverse rational map
given by m (→ ((m2 − 1)/(m2 + 1), −2m/(m2 + 1)) (which is defined away fromm2 + 1 = 0).
We can extend this to a rational map C ""# P1Q via the “inclusion” A1Q → P1Q(which we later call an open embedding). Then f is given by (x, y) (→ [y, x − 1]. We
then have an interesting question: what is the domain of definition of f? It appears
to be defined everywhere except for where y = x − 1 = 0, i.e., everywhere but p.
But in fact it can be extended over p! Note that (x, y) (→ [x + 1, −y] (where (x, y) ̸=(−1, 0)) agrees with f on their common domains of definition, as [x + 1, −y] =[y, x − 1]. Hence this rational map can be extended farther than we at first thought.
This will be a special case of the Curve-to-Projective Extension Theorem 16.5.1.
6.5.E. E XERCISE .
triples.

Use the above to find a “formula” yielding all Pythagorean

6.5.F. E XERCISE . Show that the conic x2 + y2 = z2 in P2k is isomorphic to P1k for
any field k of characteristic not 2. (Aside: What happens in characteristic 2?)

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6.5.9. In fact, any conic in P2k with a k-valued point (i.e., a point with residue fieldk) of rank 3 (after base change to k, so “rank” makes sense, see Exercise 5.4.J) is isomorphic to P1k . (The hypothesis of having a k-valued point is certainly necessary:x2 + y2 + z2 = 0 over k = R is a conic that is not isomorphic to P1k .
6.5.G. E XERCISE . Find all rational solutions to y2 = x3 +x2 , by finding a birational
map to A1Q , mimicking what worked with the conic. Hint: what point should you
project from? (In Exercise 19.10.F, we will see that these points form a group, andthat this is a degenerate elliptic curve.)
You will obtain a rational map to P1Q that is not defined over the node x =y = 0, and cannot be extended over this codimension 1 set. This is an example ofthe limits of our future result, the Curve-to-Projective Extension Theorem 16.5.1,
showing how to extend rational maps to projective space over codimension 1 sets:
the codimension 1 sets have to be regular.
6.5.H. E XERCISE . Use a similar idea to find a birational map from the quadricsurface Q = {x2 + y2 = w2 + z2 } ⊂ P3Q to P2Q . Use this to find all rational points
on Q. (This illustrates a good way of solving Diophantine equations. You will finda dense open subset of Q that is isomorphic to a dense open subset of P2 , where
you can easily find all the rational points. There will be a closed subset of Q where
the rational map is not defined, or not an isomorphism, but you can deal with thissubset in an ad hoc fashion.)
6.5.I. E XERCISE ( THE C REMONA TRANSFORMATION , A USEFUL CLASSICAL CON 2
STRUCTION ). Consider the rational map P2
k ""# Pk , given by [x, y, z] → [1/x, 1/y, 1/z].
What is the the domain of definition? (It is bigger than the locus where xyz ̸= 0!)
You will observe that you can extend it over “codimension 1 sets” (ignoring thefact that we don’t yet know what codimension means). This again foreshadows
the Curve-to-Projective Extension Theorem 16.5.1.
6.5.10. ⋆ Complex curves that are not rational (fun but inessential).
We now describe two examples of curves C that do not admit a nonconstantrational map from P1C . (Admittedly, we do not yet know what “curve” means, but
no matter.) Both proofs are by Fermat’s method of infinite descent. These results
can be interpreted (as you will later be able to check using Theorem 17.4.3) as thefact that these curves have no “nontrivial” C(t)-valued points, where by this, wemean that any C(t)-valued point is secretly a C-valued point. You may noticethat if you consider the same examples with C(t) replaced by Q (and where C
is a curve over Q rather than C), you get two fundamental questions in number
theory and geometry. The analog of Exercise 6.5.L is the question of rational points
on elliptic curves, and you may realize that the analog of Exercise 6.5.J is evenmore famous. Also, the arithmetic analog of Exercise 6.5.L(a) is the “Four Squares
Theorem” (there are not four integer squares in arithmetic progression), first statedby Fermat. These examples will give you a glimpse of how and why facts overnumber fields are often parallelled by facts over function fields of curves. This
parallelism is a recurring deep theme in the subject.
6.5.J. E XERCISE . If n > 2, show that P1C has no dominant rational maps to the“Fermat curve” xn + yn = zn in P2C . Hint: reduce this to showing that there is

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no “nonconstant” solution (f(t), g(t), h(t)) to f(t)n + g(t)n = h(t)n , where f(t),g(t), and h(t) are rational functions in t (that is, elements of C(t)). By clearing
denominators, reduce this to showing that there is no nonconstant solution wheref(t), g(t), and h(t) are relatively prime polynomials. For this, assume there is asolution, and consider one of the lowest positive degree. Then"n use the fact that
C[t] is a unique factorization domain, and h(t)n − g(t)n = i=1 (h(t) − ζi g(t)),
where ζ is a primitive nth root of unity. Argue that each h(t) − ζi g(t) is an nth
power. Then use)*(h(t) − g(t)) + α (h(t) − ζg(t)) = β h(t) − ζ2 g(t)for suitably chosen α and β to get a solution of smaller degree. (How does thisargument fail for n = 2?)

6.5.K. E XERCISE . Give two smooth complex curves X and Y so that no nonempty
open subset of X is isomorphic to a nonempty open subset of Y. (Try not to be bothered by the fact that we have not yet defined “smoothness”.) Hint: Exercise 6.5.J.
6.5.L. E XERCISE .
Suppose a, b, and c are distinct complex numbers. By thefollowing steps, show that if x(t) and y(t) are two rational functions of t such that
(6.5.10.1)

y(t)2 = (x(t) − a)(x(t) − b)(x(t) − c),then x(t) and y(t) are constants (x(t), y(t) ∈ C). (Here C may be replaced by any
field K of characteristic not 2; slight extra care is needed if K is not algebraicallyclosed.)(a) Suppose P, Q ∈ C[t] are relatively prime polynomials such that four linear
combinations of them are perfect squares, no two of which are constant multiples of each other. Show that P and Q are constant (i.e., P, Q ∈ C). Hint: By
renaming P and Q, show that you may assume that the perfect squares are P, Q,
P − Q, P − λQ (for some λ ∈ C). Define u and√ v to be√ square roots of P and
Q respectively. Show that u − v, u + v, u − λv, u + λv are perfect squares,
and that u and v are relatively prime. If P and Q are not both constant, note that0 < max(deg u, deg v) < max(deg P, deg Q). Assume from the start that P and Q
were chosen as a counterexample with minimal max(deg P, deg Q) to obtain a contradiction. (Aside: It is possible to have three distinct linear combinations that are
perfect squares. Such examples essentially correspond to primitive Pythagorean
triples in C(t) — can you see how?)(b) Suppose (x, y) = (p/q, r/s) is a solution to (6.5.10.1), where p, q, r, s ∈ C[t],and p/q and r/s are in lowest terms. Clear denominators to show that r2 q3 =s2 (p − aq)(p − bq)(p − cq). Show that s2 |q3 and q3 |s2 , and hence that s2 = δq3 forsome δ ∈ C. From r2 = δ(p − aq)(p − bq)(p − cq), show that (p − aq), (p − bq),(p − cq) are perfect squares. Show that q is also a perfect square, and then apply
part (a).
A much better geometric approach to Exercises 6.5.J and 6.5.L is given in Exercise 21.7.H.

6.6 ⋆ Representable functors and group schemes

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6.6.1. Maps to A1 correspond to functions. If X is a scheme, there is a bijection
between the maps X → A1 and global sections of the structure sheaf: by Exercise 6.3.F, maps π : X → A1Z correspond to maps to ring maps π♯ : Z[t] → Γ (X, OX ),and π♯ (t) is a function on X; this is reversible.
This map is very natural in an informal sense: you can even picture this map
to A1 as being given by the function. (By analogy, a function on a manifold is a
map to R.) But it is natural in a more precise sense: this bijection is functorial in X.
We will ponder this example at length, and see that it leads us to two importantsophisticated notions: representable functors and group schemes.
6.6.A. E ASY E XERCISE . Suppose X is a C-scheme. Verify that there is a naturalbijection between maps X → A1C in the category of C-schemes and functions on X.
(Here the base ring C can be replaced by any ring A.)
This interpretation can be extended to rational maps, as follows.

6.6.B. U NIMPORTANT E XERCISE . Interpret rational functions on an integral scheme(Exercise 5.5.Q, see also Definition 5.5.6) as rational maps to A1Z .
6.6.2. Representable functors. We restate the bijection of §6.6.1 as follows. We
have two different contravariant functors from Sch to Sets: maps to A1 (i.e., H : X (→Mor(X, A1Z )), and functions on X (F : X (→ Γ (X, OX )). The “naturality” of the bijection — the functoriality in X — is precisely the statement that the bijection gives anatural isomorphism of functors (§1.2.21): given any π : X → X ′ , the diagram
H(X ′ )"
F(X ′ )

H(π)

F(π)

% H(X)"% F(X)(where the vertical maps are the bijections given in §6.6.1) commutes.
More generally, if Y is an element of a category C (we care about the special case C = Sch), recall the contravariant functor hY : C → Sets defined by
hY (X) = Mor(X, Y) (Example 1.2.20). We say a contravariant functor from C to
Sets is represented by Y if it is naturally isomorphic to the functor hY . We say it isrepresentable if it is represented by some Y.
The bijection of §6.6.1 may now be restated as: the global section functor is represented by A1 .
6.6.C. I MPORTANT E ASY E XERCISE ( REPRESENTING OBJECTS ARE UNIQUE UP TOUNIQUE ISOMORPHISM ). Show that if a contravariant functor F is represented by
Y and by Z, then we have a unique isomorphism Y → Z induced by the natural
isomorphism of functors hY → hZ . Hint: this is a version of the universal property arguments of §1.3: once again, we are recognizing an object (up to unique
isomorphism) by maps to that object. This exercise is essentially Exercise 1.3.Z(b).
(This extends readily to Yoneda’s Lemma in this setting, Exercise 9.1.C. You arewelcome to try that now.)
You have implicitly seen this notion before: you can interpret the existence of
products and fibered products in a category as examples of representable functors.

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The Rising Sea: Foundations of Algebraic Geometry∼ hY × hZ induces(You may wish to work out how a natural isomorphism hY×Z =the projection maps Y × Z → Y and Y × Z → Z.)

6.6.D. E XERCISE ( WARM - UP ). Suppose F is the contravariant functor Sch → Sets
defined by F(X) = {Grothendieck, A.} for all schemes X. Show that F is representable. (What is it representable by?)

6.6.E. E XERCISE . In this exercise, Z may be replaced by any ring.
(a) (Affine n-space represents the functor of n functions.) Show that the contravariant
functor from (Z-)schemes to Sets
X (→ {(f1 , . . . , fn ) : fi ∈ Γ (X, OX )}

1
1 ∼
2
2
is represented by An
Z . Show that AZ ×Z AZ = AZ , i.e., that A satisfies the universal
11property of $A × A . (You will undoubtedly be able to immediately show that" mi ∼m
AZ = AZ i .)(b) (The functor of invertible functions is representable.) Show that the contravariant
functor from (Z-)schemes to Sets taking X to invertible functions on X is representable by Spec Z[t, t−1 ].

6.6.3. Definition. The scheme defined in Exercise 6.6.E(b) is called the multiplicative group Gm . “Gm over a field k” (“the multiplicative group over k”) means
Spec k[t, t−1 ], with the same group operations. Better: it represents the group of
invertible functions in the category of k-schemes. We can similarly define Gm overan arbitrary ring or even arbitrary scheme.
6.6.F. L ESS IMPORTANT EXERCISE . Fix a ring A. Consider the functor H from thecategory of locally ringed spaces to Sets given by H(X) = {A → Γ (X, OX )}. Show
that this functor is representable (by Spec A). This gives another (admittedly odd)
motivation for the definition of Spec A, closely related to that of §6.3.6.
6.6.4. ⋆⋆ Group schemes (or more generally, group objects in a category).
(The rest of §6.6 should be read only for entertainment.) We return again to
Example 6.6.1. Functions on X are better than a set: they form a group. (Indeed
they even form a ring, but we will worry about this later.) Given a morphism
X → Y, pullback of functions Γ (Y, OY ) → Γ (X, OX ) is a group homomorphism.
So we should expect A1 to have some group-like structure. This leads us to thenotion of group scheme, or more generally a group object in a category, which we
now define.
Suppose C is a category with a final object Z and with products. (We knowthat Sch has a final object Z = Spec Z, by Exercise 6.3.I. We will later see that it has
products, §9.1. But in Exercise 6.6.K we will give an alternative characterization ofgroup objects that applies in any category, so we won’t worry about this.)
A group object in C is an element X along with three morphisms:• multiplication: m : X × X → X• inverse: i : X → X• identity element: e : Z → X (not the identity map)

These morphisms are required to satisfy several conditions.

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197(i) associativity axiom:
X×X×X(m,id)% X×X(id,m)

"
X×X

m
m

"
%X

commutes. (Here id means the equality X → X.)(ii) identity axiom:
X∼% Z × X e×id % X × X m

%X

X

∼

% X × Z id×e % X × X m%Xandare both the identity map X → X. (This corresponds to the group axiom: “multiplication by the identity element is the identity map”.)
i,id %m% X and X id,i % X × X m % X are(iii) inverse axiom: X
X×X%Z e %X.
both the map that is the composition X
As motivation, you can check that a group object in the category of sets is infact the same thing as a group. (This is symptomatic of how you take some notion
and make it categorical. You write down its axioms in a categorical way, and if
all goes well, if you specialize to the category of sets, you get your original notion.
You can apply this to the notion of “rings” in an exercise below.)
A group scheme is defined to be a group object in the category of schemes. A
group scheme over a ring A (or a scheme S) is defined to be a group object in the
category of A-schemes (or S-schemes).
6.6.G. E XERCISE . Give A1Z the structure of a group scheme, by describing thethree structural morphisms, and showing that they satisfy the axioms. (Hint: themorphisms should not be surprising. For example, inverse is given by t (→ −t.
Note that we know that the product A1Z × A1Z exists, by Exercise 6.6.E(a).)

6.6.H. E XERCISE . Show that if G is a group object in a category C , then for any X ∈C , Mor(X, G) has the structure of a group, and the group structure is preserved bypullback (i.e., Mor(·, G) is a contravariant functor to Groups).

6.6.I. E XERCISE . Show that the group structure described by the previous exercise
translates the group scheme structure on A1Z to the group structure on Γ (X, OX ),via the bijection of §6.6.1.
6.6.J. E XERCISE . Define the notion of abelian group scheme, and ring scheme.
(You will undoubtedly at the same time figure out how to define the notion ofabelian group object and ring object in any category C. You may discover a more
efficient approach to such questions after reading §6.6.5.)
The language of scheme-valued points (Definition 6.3.9) has the following advantage: notice that the points of a group scheme need not themselves form a group(consider A1Z ). But Exercise 6.6.H shows that the Z-valued points of a group scheme(where Z is any given scheme) indeed form a group.

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6.6.5. Group schemes, more functorially. There was something unsatisfactory about
our discussion of the “group-respecting” nature of the bijection in §6.6.1: we observed that the right side (functions on X) formed a group, then we developed
the axioms of a group scheme, then we cleverly figured out the maps that madeA1Z into a group scheme, then we showed that this induced a group structure on
the left side of the bijection (Mor(X, A1 )) that precisely corresponded to the groupstructure on the right side (functions on X).
The picture is more cleanly explained as follows.
6.6.K. E XERCISE . Suppose we have a contravariant functor F from Sch (or indeed
any category) to Groups. Suppose further that F composed with the forgetful functor Groups → Sets is represented by an object Y. Show that the group operations
on F(X) (as X varies through Sch) uniquely determine m : Y × Y → Y, i : Y → Y,
e : Z → Y satisfying the axioms defining a group scheme, such that the group operation on Mor(X, Y) is the same as that on F(X).

In particular, the definition of a group object in a category was forced upon
us by the definition of group. More generally, you should expect that any class ofobjects that can be interpreted as sets with additional structure should fit into thispicture.
You should apply this exercise to A1Z , and see how the explicit formulas you
found in Exercise 6.6.G are forced on you.
6.6.L. E XERCISE .
cise 6.6.E.

Work out the maps m, i, and e in the group schemes of Exer-

6.6.M. E XERCISE . Explain why the product of group objects in a category can benaturally interpreted as a group object in that category.
6.6.N. E XERCISE .
(a) Define morphism of group schemes.
(b) Recall that if A is a ring, then GLn (A) (the general linear group over A) is thegroup of invertible n × n matrices with entries in the ring A. Figure out the rightdefinition of the group scheme GLn (over a ring A), and describe the determinant
map det : GLn → Gm .
(c) Make sense of the statement: “(·n ) : Gm → Gm given by t (→ tn is a morphismof group schemes.”

The language of Exercise 6.6.N(a) suggests that group schemes form a category; feel free to prove this if you want. In fact, the category of group schemes hasa zero object. What is it?
6.6.O. E XERCISE ( KERNELS OF MAPS OF GROUP SCHEMES ). Suppose F : G1 → G2
is a morphism of group schemes. Consider the contravariant functor Sch → Groups
given by X (→ ker(Mor(X, G1 ) → Mor(X, G2 )). If this is representable, by a groupscheme G0 , say, show that G0 → G1 is the kernel of F in the category of groupschemes.
6.6.P. E XERCISE . Show that the kernel of (·n ) (Exercise 6.6.N) is representable. If
n > 0, show that over a field k of characteristic p dividing n, this group scheme is

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nonreduced. (This group scheme, denoted µn , is important, although we will notuse it.)6.6.Q. E XERCISE . Show that the kernel of det : GLn → Gm is representable. Thisis the group scheme SLn . (You can do this over Z, or over a field k, or even overan arbitrary ring A; the algebra is the same.)
6.6.R. E XERCISE . Show (as easily as possible) that A1k is a ring k-scheme. (Here k
can be replaced by any ring.)
6.6.S. E XERCISE .
(a) Define the notion of a (left) group scheme action (of a group scheme on ascheme).
(b) Suppose A is a ring. Show that specifying an integer-valued grading on A
is equivalent to specifying an action of Gm on Spec A. (This interpretation of a
grading is surprisingly enlightening. Caution: there are two possible choices of
the integer-valued grading, and there are reasons for both. Both are used in theliterature.)
6.6.6. Aside: Hopf algebras. Here is a notion that we won’t use, but it is easy
enough to define now. Suppose G = Spec A is an affine group scheme, i.e., a groupscheme that is an affine scheme. The categorical definition of group scheme can berestated in terms of the ring A. (This requires thinking through Remark 6.3.5; see
Exercise 9.1.B.) Then these axioms define a Hopf algebra. For example, we have a“comultiplication map” A → A ⊗ A.

6.6.T. E XERCISE . As A1Z is a group scheme, Z[t] has a Hopf algebra structure.
Describe the comultiplication map Z[t] → Z[t] ⊗Z Z[t].

6.7 ⋆⋆ The Grassmannian (initial construction)
The Grassmannian is a useful geometric construction that is “the geometricobject underlying linear algebra”. In (classical) geometry over a field K = R orC, just as projective space parametrizes one-dimensional subspaces of a givenn-dimensional vector space, the Grassmannian parametrizes k-dimensional subspaces of n-dimensional space. The Grassmannian G(k, n) is a manifold of dimension k(n − k) (over the field). The manifold structure is given as follows. Given abasis (v1 , . . . , vn ) of n-space, “most” k-planes can be described as the span of thek vectors
1
2
n
n
n
.
.
.
(6.7.0.1)v1 +a1i vi , v2 +a2i vi , . . . , vk +aki vi .
i=k+1

i=k+1

i=k+1

(Can you describe which k-planes are not of this form? Hint: row reduced echelon
form. Aside: the stratification of G(k, n) by normal form is the decomposition of
the Grassmannian into Schubert cells. You may be able to show using the normal
form that each Schubert cell is isomorphic to an affine space.) Any k-plane ofthis form can be described in such a way uniquely. We use this to identify those kplanes of this form with the manifold Kk(n−k) (with coordinates aji ). This is a large

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affine patch on the Grassmannian (called the “open Schubert cell” with respect tothis basis). As the vi vary, these patches cover the Grassmannian (why?), and the
manifold structures agree (a harder fact).
We now define the Grassmannian in algebraic geometry, over a ring A. Suppose v = (v1 , . . . , vn ) is a basis for A⊕n . More precisely: vi ∈ A⊕n , and the map
A⊕n → A⊕n given by (a1 , . . . , an ) (→ a1 v1 + · · · + an vn is an isomorphism.
6.7.A. E XERCISE . Show that any two bases are related by an invertible n × nmatrix over A — a matrix with entries in A whose determinant is an invertible
element of A.

∼ Ak(n−k) , with coordinates
For each such basis v, we consider the scheme Uv =A
aji (k + 1 ≤ i ≤ n, 1 ≤ j ≤ k), which we imagine as corresponding to the k-plane
spanned by the vectors (6.7.0.1).
6.7.B. E XERCISE . Given two bases v and w, explain how to glue Uv to Uw along
appropriate open sets. You may find it convenient to work with coordinates aji
where i runs from 1 to n, not just k + 1 to n, but imposing aji = δji (i.e., 1 wheni = j and 0 otherwise) when i ≤ k. This convention is analogous to coordinatesxi/j on the patches of projective space (§4.4.9). Hint: the relevant open subset of
Uv will be where a certain determinant doesn’t vanish.
6.7.C. E XERCISE /D EFINITION . By checking triple intersections, verify that thesepatches (over all possible bases) glue together to a single scheme (Exercise 4.4.A).
This is the Grassmannian G(k, n) over the ring A. Because it can be interpreted asn−1
a space of linear “Pk−1
A ’s” in PA , it is often also written G(k − 1, n − 1). (You will
see that this is wise notation in Exercise 11.2.K, for example.)
Although this definition is pleasantly explicit (it is immediate that the Grassmannian is covered by Ak(n−k) ’s), and perhaps more “natural” than our original
definition of projective space in §4.4.9 (we aren’t making a choice of basis; we use
all bases), there are several things unsatisfactory about this definition of the Grassmannian. In fact the Grassmannian is projective; this isn’t obvious with this defin
nition. Furthermore, the Grassmannian comes with a natural morphism to P( k )−1(which is an example of a closed embedding, to be defined later), called the Plücker
embedding. Finally, there is an action of GLn on the space of k-planes in n-space, so
we should be able to see this in our algebraic incarnation. We will address theseissues in §16.7, by giving a better description, as a moduli space.
6.7.1. (Partial) flag varieties. Just as the Grassmannian “parametrizes” k-planes in
n-space, the flag variety parametrizes “flags”: nested sequences of subspaces of
n-spaceF0 ⊂ F1 ⊂ · · · ⊂ Fn
where dim Fi = i. Generalizing both of these is the notion of a partial flag variety
associated to some data 0 ≤ a1 < · · · < aℓ ≤ n, which parametrizes nested
sequences of subspaces of n-spaceFa1 ⊂ · · · ⊂ Faℓwhere dim Fai = ai . You should be able to generalize all of the discussion in §6.7
to this setting.

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CHAPTER 7

Useful classes of morphisms of schemes
We now define an excessive number of types of morphisms. Some (often
finiteness properties) are useful because every “reasonable” morphism has suchproperties, and they will be used in proofs in obvious ways. Others correspond togeometric behavior, and you should have a picture of what each means.
7.0.1. Reasonably behaved classes of morphisms. One of Grothendieck’s lessons isthat things that we often think of as properties of objects are better understood as
properties of morphisms. One way of turning properties of objects into propertiesof morphisms is as follows. If P is a property of schemes, we often (but not always)
say that a morphism π : X → Y has P if for every affine open subset U ⊂ Y, π−1 (U)has P. We will see this for P = quasicompact, quasiseparated, affine, and more.
(As you might hope, in good circumstances, P will satisfy the hypotheses of theAffine Communication Lemma 5.3.2, so we don’t have to check every affine open
subset.) Informally, you can think of such a morphism as one where all the fibers
have P. (You can quickly define the fiber of a morphism as a topological space,but once we define fibered product, we will define the scheme-theoretic fiber, andthen this discussion will make sense.) But it means more than that: it means that“being P” is really not just fiber-by-fiber, but behaves well as the fiber varies. (Forcomparison, a smooth morphism of manifolds means more than that the fibers aresmooth.)7.1 An example of a reasonable class of morphisms: Openembeddings
7.1.1. What to expect of any “reasonable” type of morphism. You will noticethat essentially all classes of morphisms have three properties.
(i) They are local on the target. In other words, (a) if π : X → Y is in the class,then for any open subset V of Y, the restricted morphism π−1 (V) → V isin the class; and (b) for a morphism π : X → Y, if there is an open cover{Vi } of Y for which each restricted morphism π−1 (Vi ) → Vi is in the class,then π is in the class. In particular, as schemes are built out of affine
schemes, it should be possible to check on an affine cover, as described in§7.0.1.
(ii) They are closed under composition: if π : X → Y and ρ : Y → Z are bothin this class, then so is ρ ◦ π.
(iii) They are closed under “base change” or “pullback” or “fibered product”.
We will discuss fibered products of schemes in Chapter 9.1.
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When anyone tells you a new class of morphism, you should immediately ask
yourself (or them) whether these three properties hold. And it is essentially truethat a class of morphism is “reasonable” if and only if it satisfies these three properties. Here is a first example.
An open embedding (or open immersion) of schemes is defined to be an open
embedding as ringed spaces (§6.2.1). In other words, a morphism π : (X, OX ) →(Y, OY ) of schemes is an open embedding if π factors as(X, OX )ρ

∼% (U, OY |U ) ! " τ% (Y, OY )where ρ is an isomorphism, and τ : U "→ Y is an inclusion of an open set. It is
immediate that isomorphisms are open embeddings. The symbol "→ is often used
to indicate that a morphism is an open embedding (or more generally, a locallyclosed embedding, see §8.1.2).
If X is actually a subset of Y (and π is the inclusion, i.e., ρ is the identity),
then we say (X, OX ) is an open subscheme of (Y, OY ). This is a bit confusing, andnot too important: at the level of sets, open subschemes are subsets, while open
embeddings are bijections onto subsets.
“Open subschemes” are scheme-theoretic analogs of open subsets. “Closed
subschemes” are scheme-theoretic analogs of closed subsets, but they have a surprisingly different flavor, as we will see in §8.1.
7.1.A. E XERCISE ( PROPERTIES ( I ) AND ( II )). Verify that the class of open embeddings satisfies properties (i) and (ii) of §7.1.1.
7.1.B. I MPORTANT BUT EASY EXERCISE ( PROPERTY ( III )). Verify that the classof open embeddings satisfies property (iii) of §7.1.1. More specifically: suppose
i : U → Z is an open embedding, and ρ : Y → Z is any morphism. Show thatU ×Z Y exists and U ×Z Y → Y is an open embedding. (Hint: I’ll even tell you
what U ×Z Y is: (ρ−1 (U), OY |ρ−1 (U) ).) In particular, if U "→ Z and V "→ Z are open
∼ U ∩ V.
embeddings, U ×Z V =7.1.C. E ASY E XERCISE . Suppose π : X → Y is an open embedding. Show that if
Y is locally Noetherian, then X is too. Show that if Y is Noetherian, then X is too.
However, show that if Y is quasicompact, X need not be. (Hint: let Y be affine but
not Noetherian, see Exercise 3.6.G(b).)
7.1.2. Definition. In analogy with “local on the target” (§7.1.1), we define what
it means for a property P of morphisms to be local on the source: to check if amorphism π : X → Y has P, it suffices to check on any open cover {Ui } of X. We thendefine affine-local on the source (and affine-local on the target) in the obvious
way: it suffices to check on any affine cover of the source (resp. target). The useof “affine-local” rather than “local” is to emphasize that the criterion on affine
schemes is simple to describe. (Clearly the Affine Communication Lemma 5.3.2
will be very handy.)
7.1.D. E XERCISE ( PRACTICE WITH THE CONCEPT ). Show that the notion of “openembedding” is not local on the source.

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7.2 Algebraic interlude: Lying Over and Nakayama
Algebra is the offer made by the devil to the mathematician. The devil says: I will give
you this powerful machine, it will answer any question you like. All you need to do is giveme your soul: give up geometry and you will have this marvelous machine.
— M. Atiyah, [At2, p. 659]; but see the Atiyah quote at the start of §0.3

To set up our discussion in the next section on integral morphisms, we develop some algebraic preliminaries. A clever trick we use can also be used toshow Nakayama’s Lemma, so we discuss this as well.
Suppose φ : B → A is a ring morphism. We say a ∈ A is integral over B if asatisfies some monic polynomial
an + ?an−1 + · · · + ? = 0
where the coefficients lie in φ(B). A ring homomorphism φ : B → A is integral if every element of A is integral over φ(B). An integral ring morphism φ is an integral
extension if φ is an inclusion of rings. You should think of integral homomorphisms
and integral extensions as ring-theoretic generalizations of the notion of algebraic
extensions of fields.
7.2.A. E XERCISE . Show that if φ : B → A is a ring morphism, (b1 , . . . , bn ) = 1
in B, and Bbi → Aφ(bi ) is integral for all i, then φ is integral. Hint: replace B by
φ(B) to reduce to the case where B is a subring of A. Suppose a ∈ A. Show that
there is some t and m such that bti am ∈ B + Ba + Ba2 + · · · + Bam−1 for somet and m independent of i. Use a “partition of unity” argument as in the proof of
Theorem 4.1.2 to show that am ∈ B + Ba + Ba2 + · · · + Bam−1 .
7.2.B. E XERCISE .
(a) Show that the property of a homomorphism φ : B → A being integral is always
preserved by localization and quotient of B, and quotient of A, but not localization of A. More precisely: suppose φ is integral. Show that the induced maps
T −1 B → φ(T )−1 A, B/J → A/φ(J)A, and B → A/I are integral (where T is a multiplicative subset of B, J is an ideal of B, and I is an ideal of A), but B → S−1 A neednot be integral (where S is a multiplicative subset of A). (Hint for the latter: showthat k[t] → k[t] is an integral homomorphism, but k[t] → k[t](t) is not.)(b) Show that the property of φ being an integral extension is preserved by localization of B, but not localization or quotient of A. (Hint for the latter: k[t] → k[t] is an
integral extension, but k[t] → k[t]/(t) is not.)(c) In fact the property of φ being an integral extension (as opposed to integral homomorphism) is not preserved by taking quotients of B either. (Let B = k[x, y]/(y2 )and A = k[x, y, z]/(z2 , xz − y). Then B injects into A, but B/(x) doesn’t inject intoA/(x).) But it is in some cases. Suppose φ : B → A is an integral extension, andJ ⊂ B is the restriction of an ideal I ⊂ A. (Side Remark: you can show that thisholds if J is prime.) Show that the induced map B/J → A/JA is an integral extension. (Hint: show that the composition B/J → A/JA → A/I is an injection.)
The following lemma uses a useful but sneaky trick.

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7.2.1. Lemma. — Suppose φ : B → A is a ring homomorphism. Then a ∈ A is
integral over B if and only if it is contained in a subalgebra of A that is a finitely generated
B-module.
Proof. If a satisfies a monic polynomial equation of degree n, then the B-submoduleof A generated by 1, a, . . . , an−1 is closed under multiplication, and hence a subalgebra of A.
Assume conversely that a is contained in a subalgebra A ′ of A that is a finitely
generated B-module. ,Choose a finite generating set m1 , . . . , mn of A ′ (as a Bmodule). Then ami = bij mj , for some bij ∈ B. Thus
⎛
⎞ ⎛ ⎞
m1
0
⎜ .. ⎟ ⎜ .. ⎟(7.2.1.1)(aIdn×n − [bij ]ij ) ⎝ . ⎠ = ⎝ . ⎠ ,mn0where Idn is the n × n identity matrix in A. We can’t invert the matrix (aIdn×n −[bij ]ij ), but we almost can. Recall that an n × n matrix M has an adjugate matrixadj(M) such that adj(M)M = det(M)Idn . (The (i, j)th entry of adj(M) is the determinant of the matrix obtained from M by deleting the ith column and jth row,
times (−1)i+j . You have likely seen this in the form of a formula for M−1 when
there is an inverse; see for example [DF, p. 440].) The coefficients of adj(M) arepolynomials in the coefficients of M. Multiplying (7.2.1.1) by adj(aIdn×n − [bij ]ij ),we get⎛⎞ ⎛ ⎞m1
0⎜ .. ⎟ ⎜ .. ⎟det(aIdn×n − [bij ]ij ) ⎝ . ⎠ = ⎝ . ⎠ .
mn0

So det(aI − [bij ]) annihilates the generating elements mi , and hence every elementof A ′ , i.e., det(aI − [bij ]) = 0. But expanding the determinant yields an integral
equation for a with coefficients in B.
!
7.2.2. Corollary (finite implies integral). — If A is a finite B-algebra (a finitely
generated B-module), then φ is an integral homomorphism.
The converse is false: integral does not imply finite, as Q "→ Q is an integral
homomorphism, but Q is not a finite Q-module. (A field extension is integral if it
is algebraic.)

7.2.C. E XERCISE . Show that if C → B and B → A are both integral homomorphisms, then so is their composition.
7.2.D. E XERCISE . Suppose φ : B → A is a ring morphism. Show that the elementsof A integral over B form a subalgebra of A.

7.2.3. Remark: Transcendence theory. These ideas lead to the main facts abouttranscendence theory we will need for a discussion of dimension of varieties, see
Exercise/Definition 11.2.A.
7.2.4. The Lying Over and Going-Up Theorems. The Lying Over Theorem is a
useful property of integral extensions.

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7.2.5. The Lying Over Theorem. — Suppose φ : B → A is an integral extension.
Then for any prime ideal q ⊂ B, there is a prime ideal p ⊂ A such that p ∩ B = q.
To be clear on how weak the hypotheses are: B need not be Noetherian, andA need not be finitely generated over B.

7.2.6. Geometric translation: Spec A → Spec B is surjective. (A map of schemes issurjective if the underlying map of sets is surjective.)
Although this is a theorem in algebra, the name can be interpreted geometrically: the theorem asserts that the corresponding morphism of schemes is surjective, and that “above” every prime q “downstairs”, there is a prime p “upstairs”,
see Figure 7.1. (For this reason, it is often said that p “lies over” q if p ∩ B = q.) The
following exercise sets up the proof.

F IGURE 7.1. A picture of the Lying Over Theorem 7.2.5: if φ : B →A is an integral extension, then Spec A → Spec B is surjective

7.2.E. ⋆ E XERCISE . Show that the special case where A is a field translates to: if
B ⊂ A is a subring with A integral over B, then B is a field. Prove this. (Hint: you
must show that all nonzero elements in B have inverses in B. Here is the start: Ifb ∈ B, then 1/b ∈ A, and this satisfies some integral equation over B.)⋆ Proof of the Lying Over Theorem 7.2.5. We first make a reduction: by localizing at
q (preserving integrality by Exercise 7.2.B(b)), we can assume that (B, q) is a local
ring. Then let p be any maximal ideal of A. Consider the following diagram.
A7% % A/p
7!$
B!$% % B/(p ∩ B)

field

The right vertical arrow is an integral extension by Exercise 7.2.B(c). By Exercise 7.2.E, B/(p ∩ B) is a field too, so p ∩ B is a maximal ideal, hence it is q.
!
7.2.F. I MPORTANT E XERCISE ( THE Going-Up Theorem).
(a) Suppose φ : B → A is an integral homomorphism (not necessarily an integral

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extension). Show that if q1 ⊂ q2 ⊂ · · · ⊂ qn is a chain of prime ideals of B, andp1 ⊂ · · · ⊂ pm is a chain of prime ideals of A such that pi “lies over” qi (and1 ≤ m < n), then the second chain can be extended to p1 ⊂ · · · ⊂ pn so that thisremains true. (Hint: reduce to the case m = 1, n = 2; reduce to the case whereq1 = (0) and p1 = (0); use the Lying Over Theorem 7.2.5.)(b) Draw a picture of this theorem (akin to Figure 7.1).
There are analogous “Going-Down” results (requiring quite different hypotheses); see for example Theorem 11.2.12 and Exercise 24.5.E.
7.2.7. Nakayama’s Lemma.
The trick in the proof of Lemma 7.2.1 can be used to quickly prove Nakayama’s
Lemma, which we will use repeatedly in the future. This name is used for several
different but related results, which we discuss here. (A geometric interpretation
will be given in Exercise 13.7.E.) We may as well prove it while the trick is fresh in
our minds.
7.2.8. Nakayama’s Lemma version 1. — Suppose A is a ring, I is an ideal of A, andM is a finitely generated A-module, such that M = IM. Then there exists an a ∈ A witha ≡ 1 (mod I) with aM = 0. (Equivalently, there is some i ∈ I for which multiplication
by i induces the identity on M: im = m for all m ∈ M.)
Proof.
Say M is generated by m1 , . . . , mn . Then as M = IM, we have mi =,j aij mj for some aij ∈ I. Thus
(7.2.8.1)

⎛

⎞
m1
⎜
⎟
(Idn − Z) ⎝ ... ⎠ = 0
mn

where Z = (aij ). Multiplying both sides of (7.2.8.1) on the left by adj(Idn − Z), we
obtain⎛⎞m1⎜⎟
det(Idn − Z) ⎝ ... ⎠ = 0.
mn

But when you expand out det(Idn − Z), as Z has entries in I, you get somethingthat is 1 (mod I).
!
Here is why you care. Suppose I is contained in all maximal ideals of A. (Theintersection of all the maximal ideals is called the Jacobson radical, but we won’t
use this phrase. For comparison, recall that the nilradical was the intersection of
the prime ideals of A.) Then any a ≡ 1 (mod I) is invertible. (We are not using
Nakayama yet!) Reason: otherwise (a) ̸= A, so the ideal (a) is contained in somemaximal ideal m — but a ≡ 1 (mod m), contradiction. As a is invertible, we havethe following.
7.2.9. Nakayama’s Lemma version 2. — Suppose A is a ring, I is an ideal of A
contained in all maximal ideals, and M is a finitely generated A-module. (The most interesting case is when A is a local ring, and I is the maximal ideal.) Suppose M = IM. Then
M = 0.

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7.2.G. E XERCISE (N AKAYAMA’ S L EMMA VERSION 3). Suppose A is a ring, and I
is an ideal of A contained in all maximal ideals. Suppose M is a finitely generated
A-module, and N ⊂ M is a submodule. If N/IN → M/IM is surjective, thenM = N.

7.2.H. I MPORTANT E XERCISE (N AKAYAMA’ S L EMMA VERSION 4: GENERATORS OF
M/mM LIFT TO GENERATORS OF M). Suppose (A, m) is a local ring. Suppose M isa finitely generated A-module, and f1 , . . . , fn ∈ M, with (the images of) f1 , . . . , fn
generating M/mM. Then f1 , . . . , fn generate M. (In particular, taking M = m, if
we have generators of m/m2 , they also generate m.)

7.2.I. I MPORTANT E XERCISE GENERALIZING L EMMA 7.2.1. Recall that a B-module
N is said to be faithful if the only element of B acting on N by the identity is 1 (orequivalently, if the only element of B acting as the 0-map on N is 0). Suppose S isa subring of a ring A, and r ∈ A. Suppose there is a faithful S[r]-module M that is
finitely generated as an S-module. Show that r is integral over S. (Hint: change a
few words in the proof of version 1 of Nakayama, Lemma 7.2.8.)
3 is the integral closure
7.2.J. E XERCISE . Suppose A is an integral domain, and A
of A in K(A), i.e., those elements of K(A) integral over A, which form a subalgebra
3 is integrally closed in K(A)3 = K(A).
by Exercise 7.2.D. Show that A

7.3 A gazillion finiteness conditions on morphisms
By the end of this section, you will have seen the following types of morphisms: quasicompact, quasiseparated, affine, finite, integral, closed, (locally) offinite type, quasifinite — and possibly, (locally) of finite presentation.
7.3.1. Quasicompact and quasiseparated morphisms.
A morphism π : X → Y of schemes is quasicompact if for every open affine
subset U of Y, π−1 (U) is quasicompact. (Equivalently, the preimage of any quasicompact open subset is quasicompact. This is the right definition in other parts ofgeometry.)
We will like this notion because (i) finite sets have advantages over infinite
sets (e.g., a finite set of integers has a maximum; also, things can be proved inductively), and (ii) most reasonable schemes will be quasicompact.
Along with quasicompactness comes the weird notion of quasiseparatedness.
A morphism π : X → Y is quasiseparated if for every affine open subset U of Y,
π−1 (U) is a quasiseparated scheme (§5.1.1). This will be a useful hypothesis intheorems, usually in conjunction with quasicompactness. (For this reason, “quasicompact and quasiseparated” are often abbreviated as qcqs, as we do in the nameof the Qcqs Lemma 7.3.5.) Various interesting kinds of morphisms (locally Noetherian source, affine, separated, see Exercises 7.3.B(b), 7.3.D, and 10.1.H resp.)
are quasiseparated, and having the word “quasiseparated” will allow us to state
theorems more succinctly.
7.3.A. E ASY E XERCISE . Show that the composition of two quasicompact morphisms is quasicompact. (It is also true — but not easy — that the composition of

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two quasiseparated morphisms is quasiseparated. This is not impossible to show
directly, but will in any case follow easily once we understand it in a more sophisticated way, see Proposition 10.1.13(b).)
7.3.B. E ASY E XERCISE .
(a) Show that any morphism from a Noetherian scheme is quasicompact.
(b) Show that any morphism from a quasiseparated scheme is quasiseparated.
Thus by Exercise 5.3.A, any morphism from a locally Noetherian scheme is quasiseparated. Thus readers working only with locally Noetherian schemes may
take quasiseparatedness as a standing hypothesis.
7.3.2. Caution. The two parts of the Exercise 7.3.B may lead you to suspect thatany morphism π : X → Y with quasicompact source and target is necessarily quasicompact. This is false, and you may verify that the following is a counterexample. Let Z be the nonquasiseparated scheme constructed in Exercise 5.1.J, and let
X = Spec k[x1 , x2 , . . . ] as in Exercise 5.1.J. The obvious open embedding π : X → Z(identifying X with one of the two pieces glued together to get Z) is not quasicompact. (But once you see the Cancellation Theorem 10.1.19, you will quickly see thatany morphism from a quasicompact source to a quasiseparated target is necessarily
quasicompact.)
7.3.C. E XERCISE . (Obvious hint for both parts: the Affine Communication Lemma 5.3.2.)(a) (quasicompactness is affine-local on the target) Show that a morphism π : X → Y is
quasicompact if there is a cover of Y by affine open sets Ui such that π−1 (Ui ) is
quasicompact.
(b) (quasiseparatedness is affine-local on the target) Show that a morphism π : X → Y
is quasiseparated if there is a cover of Y by affine open sets Ui such that π−1 (Ui )is quasiseparated.
Following Grothendieck’s philosophy of thinking that the important notions
are properties of morphisms, not of objects (§7.0.1), we can restate the definitionof quasicompact (resp. quasiseparated) scheme as a scheme that is quasicompact(resp. quasiseparated) over the final object Spec Z in the category of schemes (Exercise 6.3.I).
7.3.3. Affine morphisms.
A morphism π : X → Y is affine if for every affine open set U of Y, π−1 (U)(interpreted as an open subscheme of X) is an affine scheme. Trivially, the composition of two affine morphisms is affine.
7.3.D. FAST E XERCISE . Show that affine morphisms are quasicompact and quasiseparated. (Hint for the second: Exercise 5.1.G.)
7.3.4. Proposition (the property of “affineness” of a morphism is affine-local on
the target). — A morphism π : X → Y is affine if there is a cover of Y by affine open sets
U such that π−1 (U) is affine.
This proof is the hardest part of this section. For part of the proof (which will
start in §7.3.7), it will be handy to have a lemma.

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7.3.5. Qcqs Lemma. — If X is a quasicompact quasiseparated scheme and s ∈ Γ (X, OX ),then the natural map Γ (X, OX )s → Γ (Xs , OX ) is an isomorphism.

Here Xs means the locus on X where s doesn’t vanish. (By Exercise 4.3.G(a),Xs is open.) We avoid the notation D(s) to avoid any suggestion that X is affine.
(If X is affine, then Xs is D(s), and we already know the theorem is true — do you
see why?)
7.3.E. E XERCISE ( REALITY CHECK ). What is the natural map Γ (X, OX )s → Γ (Xs , OX )of the Qcqs Lemma 7.3.5? (Hint: the universal property of localization, Exercise 1.3.D.)
To repeat the earlier reassuring comment on the “quasicompact quasiseparated” hypothesis: this just means that X can be covered by a finite number ofaffine open subsets, any two of which have intersection also covered by a finite
number of affine open subsets (Exercise 5.1.H). The hypothesis applies in lots ofinteresting situations, such as if X is affine (Exercise 5.1.G) or Noetherian (Exercise 5.3.A). And conversely, whenever you see quasicompact quasiseparated hypotheses (e.g., Exercises 13.3.E, 13.3.H), they are most likely there because of thislemma. To remind ourselves of this fact, we call it the Qcqs Lemma.
7.3.6. Proof of the Qcqs Lemma 7.3.5. Cover X with finitely many affine open sets
Ui = Spec Ai . Let Uij = Ui ∩ Uj . Then&&
0 → Γ (X, OX ) →Ai →Γ (Uij , OX )
i

{i,j}

is exact. (See the discussion after (4.1.3.1) for the signs arising in the last map.)
By the quasiseparated hypotheses, we can cover each Uij with a finite number ofaffine open sets Uijk = Spec Aijk , so we have that&&
0 → Γ (X, OX ) →
Ai →
Aijk
i

{i,j},kis exact. Localizing at s (an exact functor, Exercise 1.6.F(a)) gives
⎛
⎞
4
5
&
&
0 → Γ (X, OX )s →Ai→⎝
Aijk ⎠is{i,j},ks

As localization commutes with finite products (Exercise 1.3.F(a)),&&(7.3.6.1)
0 → Γ (X, OX )s →(Ai )si →(Aijk )sijk
i{i,j},kis exact, where the global function s induces functions si ∈ Ai and sijk ∈ Aijk .
But similarly, the scheme Xs can be covered by affine opens Spec(Ai )si , andSpec(Ai )si ∩Spec(Aj )sj are covered by a finite number of affine opens Spec(Aijk )sijk ,so we have&&
(7.3.6.2)
0 → Γ (Xs , OX ) →(Ai )si →(Aijk )sijk .
i"{i,j},k"Notice that the maps i (Ai )si → {i,j},k (Aijk )sijk in (7.3.6.1) and (7.3.6.2) are
the same, and we have described the kernel of the map in two ways, so Γ (X, OX )s →Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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Γ (Xs , OX ) is indeed an isomorphism. (Notice how the quasicompact and quasiseparated hypotheses were used in an easy way: to obtain finite products, which
would commute with localization.)!
7.3.7. Proof of Proposition 7.3.4.
As usual, we use the Affine Communication
Lemma 5.3.2. (We apply it to the condition “π is affine over”.) We check our two
criteria. First, suppose π : X → Y is affine over Spec B, i.e., π−1 (Spec B) = Spec A.
Then for any s ∈ B, π−1 (Spec Bs ) = Spec Aπ♯ s .
Second, suppose we are given π : X → Spec B and (s1 , . . . , sn ) = B with Xπ♯ siaffine (Spec Ai , say). (As in the statement of the Qcqs Lemma 7.3.5, Xπ♯ si is thesubset of X where π♯ si doesn’t vanish.) We wish to show that X is affine too. Let
A = Γ (X, OX ). Then X → Spec B factors through the tautological map α : X →Spec A (arising from the (iso)morphism A → Γ (X, OX ), Exercise 6.3.F).
α% Spec A∪i Xπ♯ si = X
◗◗◗
♥
◗◗◗
♥♥♥
◗◗
♥♥β♥
♥
π ◗◗◗◗♥◗2
D♥♥∪i D(si ) = Spec B∼ Spec Aβ♯ s (the preimage of a distinguished open
Then β−1 (D(si )) = D(β♯ si ) =i
set is a distinguished open set), and π−1 (D(si )) = Spec Ai . Now X is quasicompact and quasiseparated by the affine-locality of these notions (Exercise 7.3.C), so
the hypotheses of the Qcqs Lemma 7.3.5 are satisfied. Hence we have an induced∼ Γ (Xβ♯ s , OX ) = Ai . Thus α induces
isomorphism of Aβ♯ si = Γ (X, OX )β♯ si =ian isomorphism Spec Ai → Spec Aβ♯ si (an isomorphism of rings induces an isomorphism of affine schemes, Exercise 4.3.A). Thus α is an isomorphism over each
∼ Spec A,
Spec Aβ♯ si , which cover Spec A, and thus α is an isomorphism. Hence X =so is affine as desired.
!
The affine-locality of affine morphisms (Proposition 7.3.4) has some nonobvious consequences, as shown in the next exercise.
7.3.F. U SEFUL E XERCISE . Suppose Z is a closed subset of an affine scheme Spec A
locally cut out by one equation. (In other words, Spec A can be covered by smaller
open sets, and on each such set Z is cut out by one equation.) Show that thecomplement Y of Z is affine. (This is clear if Z is globally cut out by one equationf, even set-theoretically; then Y = Spec Af . However, Z is not always of this form,see §19.11.10.)
7.3.8. Finite and integral morphisms.
Before defining finite and integral morphisms, we give an example to keep inmind. If L/K is a field extension, then Spec L → Spec K (i) is always affine; (ii) is
integral if L/K is algebraic; and (iii) is finite if L/K is finite.
A morphism π : X → Y is finite if for every affine open set Spec B of Y, π−1 (Spec B)
is the spectrum of a B-algebra that is a finitely generated B-module. By definition,
finite morphisms are affine. Warning about terminology (finite vs. finitely generated): Recall that if we have a ring morphism B → A such that A is a finitely
generated B-module then we say that A is a finite B-algebra. This is stronger thanbeing a finitely generated B-algebra.

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7.3.G. E XERCISE ( THE PROPERTY OF FINITENESS IS AFFINE - LOCAL ON THE TAR GET ). Show that a morphism π : X → Y is finite if there is a cover of Y by affine
open sets Spec A such that π−1 (Spec A) is the spectrum of a finite A-algebra.

The following four examples will give you some feeling for finite morphisms.
In each example, you will notice two things. In each case, the maps are always
finite-to-one (as maps of sets). We will verify this in general in Exercise 7.3.K. You
will also notice that the morphisms are closed as maps of topological spaces, i.e.,
the images of closed sets are closed. We will show that finite morphisms are always
closed in Exercise 7.3.M (and give a second proof in §8.2.5). Intuitively, you should
think of finite as being closed plus finite fibers, although this isn’t quite true. We
will make this precise in Theorem 29.6.2.
Example 1: Branched covers. Consider the morphism Spec k[t] → Spec k[u]given by u (→ p(t), where p(t) ∈ k[t] is a degree n polynomial (see Figure 7.2).
This is finite: k[t] is generated as a k[u]-module by 1, t, t2 , . . . , tn−1 .

F IGURE 7.2. The “branched cover” A1k → A1k of the “u-line” by
the “t-line” given by u (→ p(t) is finite

Example 2: Closed embeddings (to be defined soon, in §8.1.1). If I is an ideal ofa ring A, consider the morphism Spec A/I → Spec A given by the obvious map
A → A/I (see Figure 7.3 for an example, with A = k[t], I = (t)). This is a finite
morphism (A/I is generated as a A-module by the element 1 ∈ A/I).

0

F IGURE 7.3. The “closed embedding” Spec k → Spec k[t] given
by t (→ 0 is finite

Example 3: Normalization (to be defined in §9.7). Consider the morphism Spec k[t] →Spec k[x, y]/(y2 − x2 − x3 ) corresponding to k[x, y]/(y2 − x2 − x3 ) → k[t] given by

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The Rising Sea: Foundations of Algebraic Geometryx (→ t2 − 1, y (→ t3 − t (check that this is a well-defined ring map!), see Figure 7.4.
This is a finite morphism, as k[t] is generated as a (k[x, y]/(y2 − x2 − x3 ))-moduleby 1 and t. (The figure suggests that this is an isomorphism away from the “node”
of the target. You can verify this, by checking that it induces an isomorphism between D(t2 − 1) in the source and D(x) in the target. We will meet this example
again!)F IGURE 7.4. The “normalization” Spec k[t] → Spec k[x, y]/(y2 −x2 − x3 ) given by (x, y) (→ (t2 − 1, t3 − t) is finite
7.3.H. I MPORTANT E XERCISE (E XAMPLE 4, FINITE MORPHISMS TO Spec k). Show
that if X → Spec k is a finite morphism, then X is a finite union of points withthe discrete topology, each point with residue field a finite extension of k, see
Figure 7.5. (An example is Spec(F8 ×F4 [x, y]/(x2 , y4 )×F4 [t]/(t9 )×F2 ) → Spec F2 .)Do not just quote some fancy theorem! Possible approach: By Exercise 3.2.G, any
integral domain which is a finite k-algebra must be a field. If X = Spec A, showthat every prime p of A is maximal. Show that the irreducible components ofSpec A are closed points. Show Spec A is discrete and hence finite. Show that theresidue fields K(A/p) of A are finite extensions of k. (See Exercise 7.4.D for an
extension to quasifinite morphisms.)

F IGURE 7.5. A picture of a finite morphism to Spec k. Bigger
fields are depicted as bigger points.

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7.3.I. E ASY E XERCISE ( CF. E XERCISE 7.2.C). Show that the composition of two
finite morphisms is also finite.
7.3.J. E XERCISE (“ FINITE MORPHISMS TO Spec A ARE PROJECTIVE ”). If R is an
A-algebra, define a graded ring S• by S0 = A, and Sn = R for n > 0. (What is themultiplicative structure? Hint: you know how to multiply elements of R together,and how to multiply elements of A with elements of R.) Describe an isomorphism∼ Spec R. Show that if R is a finite A-algebra (finitely generated as an AProj S• =module) then S• is a finitely generated graded ring over A, and hence that Spec R
is a projective A-scheme (§4.5.9).
7.3.K. I MPORTANT E XERCISE . Show that finite morphisms have finite fibers. (Thisis a useful exercise, because you will have to figure out how to get at points in afiber of a morphism: given π : X → Y, and q ∈ Y, what are the points of π−1 (q)?
This will be easier to do once we discuss fibers in greater detail, see Remark 9.3.4,
but it will be enlightening to do it now.) Hint: if X = Spec A and Y = Spec B
are both affine, and q = [q], then we can throw out everything in B outside q
by modding out by q; show that the preimage is Spec(A/π♯ qA). Then you have
reduced to the case where Y is the Spec of an integral domain B, and [q] = [(0)]is the generic point. We can throw out the rest of the points of B by localizing at(0). Show that the preimage is Spec of A localized at π♯ (B \ {0}). Show that thecondition of finiteness is preserved by the constructions you have done, and thus
reduce the problem to Exercise 7.3.H.
There is more to finiteness than finite fibers, as is shown by the following two
examples.
7.3.9. Example. The open embedding A2k − {(0, 0)} → A2k has finite fibers, but is notaffine (as A2k − {(0, 0)} isn’t affine, §4.4.1) and hence not finite.
7.3.L. E ASY E XERCISE . Show that the open embedding A1C − {0} → A1C has finite
fibers and is affine, but is not finite.

7.3.10. Definition. A morphism π : X → Y of schemes is integral if π is affine,and for every affine open subset Spec B ⊂ Y, with π−1 (Spec B) = Spec A, the induced map B → A is an integral ring morphism. This is an affine-local condition
by Exercises 7.2.A and 7.2.B, and the Affine Communication Lemma 5.3.2. It isclosed under composition by Exercise 7.2.C. Integral morphisms are mostly useful
because finite morphisms are integral by Corollary 7.2.2. Note that the converse
implication doesn’t hold (witness Spec Q → Spec Q, as discussed after the statement of Corollary 7.2.2).

7.3.M. E XERCISE . Prove that integral morphisms are closed, i.e., that the imageof closed subsets are closed. (Hence finite morphisms are closed. A second proof
will be given in §8.2.5.) Hint: Reduce to the affine case. If π♯ : B → A is a ring map,
inducing the integral morphism π : Spec A → Spec B, then suppose I ⊂ A cuts out
a closed set of Spec A, and J = (π♯ )−1 (I), then note that B/J ⊂ A/I, and apply the
Lying Over Theorem 7.2.5 here.
7.3.N. U NIMPORTANT EXERCISE . Suppose B → A is integral. Show that for any
ring homomorphism B → C, the induced map C → A ⊗B C is integral. (Hint: We

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The Rising Sea: Foundations of Algebraic Geometry,nwish to show that any i=1 ai ⊗ ci ∈ A ⊗B C is integral over C. Use the fact thateach of the finitely many ai are integral over B, and then Exercise 7.2.D.) Once weknow what “base change” is, this will imply that the property of integrality of amorphism is preserved by base change, Exercise 9.4.B(e).
7.3.11. Fibers of integral morphisms. Unlike finite morphisms (Exercise 7.3.K), integral morphisms don’t always have finite fibers. (Can you think of an example?)
However, once we make sense of fibers as topological spaces (or even schemes)in §9.3.2, you can check (Exercise 11.1.D) that the fibers have the property that nopoint is in the closure of any other point.
7.3.12. Morphisms (locally) of finite type.
A morphism π : X → Y is locally of finite type if for every affine open set
Spec B of Y, and every affine open subset Spec A of π−1 (Spec B), the induced
morphism B → A expresses A as a finitely generated B-algebra. By the affinelocality of finite-typeness of B-schemes (Proposition 5.3.3(b)), this is equivalent to:π−1 (Spec B) can be covered by affine open subsets Spec Ai so that each Ai is a
finitely generated B-algebra.
A morphism π is of finite type if it is locally of finite type and quasicompact.
Translation: for every affine open set Spec B of Y, π−1 (Spec B) can be covered witha finite number of open sets Spec Ai so that the induced morphism B → Ai expresses
Ai as a finitely generated B-algebra.
7.3.13. Linguistic side remark. It is a common practice to name properties as follows:P equals locally P plus quasicompact. Two exceptions are “ringed space” (§6.3)
and “finite presentation” (§7.3.17).
7.3.O. E XERCISE ( THE NOTIONS “ LOCALLY FINITE TYPE ” AND “ FINITE TYPE ” ARE
Show that a morphism π : X → Y is locally offinite type if there is a cover of Y by affine open sets Spec Bi such that π−1 (Spec Bi )is locally finite type over Bi .

AFFINE - LOCAL ON THE TARGET ).

n
Example: the “structure morphism” Pn
A → Spec A is of finite type, as PA is
covered by n + 1 open sets of the form Spec A[x1 , . . . , xn ].
Our earlier definition of schemes of “finite type over k” (or “finite type kschemes”) from §5.3.6 is now a special case of this more general notion: the phrase“a scheme X is of finite type over k” means that we are given a morphism X →Spec k (the “structure morphism”) that is of finite type.
Here are some properties enjoyed by morphisms of finite type.

7.3.P. E XERCISE ( FINITE = INTEGRAL + FINITE TYPE ).
(a) (easier) Show that finite morphisms are of finite type.
(b) Show that a morphism is finite if and only if it is integral and of finite type.
7.3.Q. E XERCISES ( NOT HARD , BUT IMPORTANT ).
(a) Show that every open embedding is locally of finite type, and hence that every
quasicompact open embedding is of finite type. Show that every open embedding
into a locally Noetherian scheme is of finite type.
(b) Show that the composition of two morphisms locally of finite type is locally of

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finite type. (Hence as the composition of two quasicompact morphisms is quasicompact, Easy Exercise 7.3.A, the composition of two morphisms of finite type isof finite type.)(c) Suppose π : X → Y is locally of finite type, and Y is locally Noetherian. Show
that X is also locally Noetherian. If π : X → Y is a morphism of finite type, and Y is
Noetherian, show that X is Noetherian.

7.3.14. Definition. A morphism π is quasifinite if it is of finite type, and for
all q ∈ Y, π−1 (q) is a finite set. The main point of this definition is the “finite
fiber” part; the “finite type” hypothesis will ensure that this notion is “preserved
by fibered product,” Exercise 9.4.C.
Combining Exercise 7.3.K with Exercise 7.3.P(a), we see that finite morphisms
are quasifinite. There are quasifinite morphisms which are not finite, such as
A2k − {(0, 0)} → A2k (Example 7.3.9). However, we will soon see that quasifinite
morphisms to Spec k are finite (Exercise 7.4.D). A key example of a morphismwith finite fibers that is not quasifinite is Spec C(t) → Spec C. Another is Spec Q →Spec Q. (For interesting behavior caused by the fact that Spec Q → Spec Q is not of
finite type, see Warning 9.1.4.)
7.3.15. How to picture quasifinite morphisms. If π : X → Y is a finite morphism of
locally Noetherian schemes, then for any quasicompact open subset U ⊂ X, theinduced morphism π|U : U → Y is quasifinite. In fact every reasonable quasifinite
morphism arises in this way. (This simple-sounding statement is in fact a deep andimportant result — a form of Zariski’s Main Theorem, see Exercise 29.6.E.) Thus
the right way to visualize quasifiniteness is as a finite map with some (closed locusof) points removed.
7.3.16. Frobenius.
7.3.R. E XERCISE . Suppose p is prime and r ∈ Z+ . Let q = pr , and k = Fq . Definennφ : k[x1 , . . . , xn ] → k[x1 , . . . , xn ] by φ(xi ) = xp
i for each i, and let F : Ak → Ak be
the map of schemes corresponding to φ.
(a) Show that Fr is the identity on the level of sets, but is not the identity morphism.
(b) Show that F is a bijection, but is not an isomorphism of schemes.
n(c) If K = Fp , show that the morphism F : An
K → AK of K-schemes correspondingp
to xi (→ xi is a bijection, but no power of F is the identity on the level of sets.
7.3.S. E XERCISE . Suppose X is a scheme over Fp . Explain how to define (without
choice) an endomorphism F : X → X such that for each affine open subset Spec A ⊂X, F corresponds to the map A → A given by f (→ fp for all f ∈ A. (The morphism
F is called the absolute Frobenius morphism.)

7.3.17. ⋆ Morphisms (locally) of finite presentation.
There is a variant often useful to non-Noetherian people. A ring A is a finitely
presented B-algebra (or B → A is finitely presented) if∼ B[x1 , . . . , xn ]/(r1 (x1 , . . . , xn ), . . . , rj (x1 , . . . , xn ))A=(“A has a finite number of generators and a finite number of relations over B”).
If B is Noetherian, then finitely presented is the same as finite type, as the “finite
number of relations” comes for free, so most of you will not care. A morphism

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π : X → Y is locally of finite presentation (or locally finitely presented) if foreach affine open set Spec B of Y, π−1 (Spec B) = ∪i Spec Ai with B → Ai finitely
presented. A morphism is of finite presentation (or finitely presented) if it islocally of finite presentation and quasicompact and quasiseparated. If Y is locally
Noetherian, then locally of finite presentation is the same as locally of finite type,and finite presentation is the same as finite type. So if you are a Noetherian person,
you don’t need to worry about this notion.
This definition is a violation of the general principle that erasing “locally” isthe same as adding “quasicompact and” (Remark 7.3.13). But it is well motivated:finite presentation means “finite in all possible ways” (the algebra corresponding
to each affine open set has a finite number of generators, and a finite number ofrelations, and a finite number of such affine open sets cover, and their intersections
are also covered by a finite number affine open sets) — it is all you would hopefor in a scheme without it actually being Noetherian. Exercise 9.3.H makes this
precise, and explains how this notion often arises in practice.
7.3.T. E XERCISE . Show that the notion of “locally of finite presentation” is affinelocal on the target.
7.3.U. ⋆⋆ H ARD E XERCISE ( BUT REASSURING TO KNOW ). Show that if π : X → Y is
locally of finite presentation, then for any affine open subscheme Spec B of Y andany affine open subscheme Spec A of X with π(Spec A) ⊂ Spec B, A is a finitely
presented B-algebra. In particular, the notion of “locally of finite presentation” isaffine-local on the source.
7.3.V. E XERCISE . Show that open embeddings are locally finitely presented.
7.3.W. E XERCISE . Show that the composition of two locally finitely presented
morphisms is locally finitely presented. (Then once we show that the composition of two quasiseparated morphisms is quasiseparated in Proposition 10.1.13(b),we will know that the composition of two finitely presented morphisms is finitely
presented — recall that the composition of two quasicompact morphisms is quasicompact, by Easy Exercise 7.3.A.)
7.3.18. ⋆⋆ Remark. A morphism π : X → Y is locally of finite presentation if andonly if for every projective system of Y-schemes {Sλ }λ∈I with each Sλ an affine
scheme, the natural map
lim HomY (Sλ , X) → HomY (lim Sλ , X)
−→
←−
λ

λ

is a bijection (see [Gr-EGA, IV3 .8.14.2]). This characterization of locally finitely
presented morphisms as “limit-preserving” can be useful.

7.4 Images of morphisms: Chevalley’s Theorem and elimination
theoryIn this section, we will answer a question that you may have wondered about
long before hearing the phrase “algebraic geometry”. If you have a number ofpolynomial equations in a number of variables with indeterminate coefficients,

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you would reasonably ask what conditions there are on the coefficients for a (common) solution to exist. Given the algebraic nature of the problem, you might hopethat the answer should be purely algebraic in nature — it shouldn’t be “random”,
or involve bizarre functions like exponentials or cosines. You should expect the answer to be given by “algebraic conditions”. This is indeed the case, and it can be
profitably interpreted as a question about images of maps of varieties or schemes,
in which guise it is answered by Chevalley’s Theorem 7.4.2 (see 7.4.5 for a more
precise proof). Chevalley’s Theorem will give an immediate proof of the Nullstellensatz 3.2.5 (§7.4.3).
In special cases, the image is nicer still. For example, we have seen that finite morphisms are closed (the image of closed subsets under finite morphisms
are closed, Exercise 7.3.M). We will prove a classical result, the Fundamental Theorem of Elimination Theory 7.4.7, which essentially generalizes this (as explainedin §8.2.5) to maps from projective space. We will use it repeatedly. In a different direction, in the distant future we will see that in certain good circumstances (“flat”
plus a bit more, see Exercise 24.5.G), morphisms are open (the image of open subsets is open); one example (which you can try to show directly) is An
B → Spec B.

7.4.1. Chevalley’s Theorem.
If π : X → Y is a morphism of schemes, the notion of the image of π as sets isclear: we just take the points in Y that are the image of points in X. We know thatthe image can be open (open embeddings), and we have seen examples where it isclosed, and more generally, locally closed. But it can be weirder still: consider themorphism A2k → A2k given by (x, y) (→ (x, xy). The image is the plane, with they-axis removed, but the origin put back in (see Figure 7.6). This isn’t so horrible.
We make a definition to capture this phenomenon. A constructible subset of a
Noetherian topological space is a subset which belongs to the smallest family ofsubsets such that (i) every open set is in the family, (ii) a finite intersection of family
members is in the family, and (iii) the complement of a family member is also in
the family. For example the image of (x, y) (→ (x, xy) is constructible.

F IGURE 7.6. The image of (x, y) (→ (x, xy) = (u, v).

7.4.A. E XERCISE : CONSTRUCTIBLE SUBSETS ARE FINITE DISJOINT UNIONS OF LO CALLY CLOSED SUBSETS . Recall that a subset of a topological space X is locally

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The Rising Sea: Foundations of Algebraic Geometryclosed if it is the intersection of an open subset and a closed subset. (Equivalently,
it is an open subset of a closed subset, or a closed subset of an open subset. We
will later have trouble extending this to open and closed and locally closed subschemes, see Exercise 8.1.M.) Show that a subset of a Noetherian topological space
X is constructible if and only if it is the finite disjoint union of locally closed subsets.
As a consequence, if X → Y is a continuous map of Noetherian topological spaces,
then the preimage of a constructible set is a constructible set. (Important remark:the only reason for the hypothesis of the topological space in question being Noetherian is because this is the only setting in which we have defined constructible
sets. An extension of the notion of constructibility to more general topologicalspaces is mentioned in Exercise 9.3.I.)
7.4.B. E XERCISE ( USED IN P ROOF 7.4.3 OF THE N ULLSTELLENSATZ 3.2.5). Show
that the generic point of A1k does not form a constructible subset of A1k (where k isa field).
7.4.C. E XERCISE ( USED IN E XERCISE 24.5.G).
(a) Show that a constructible subset of a Noetherian scheme is closed if and only if
it is “stable under specialization”. More precisely, if Z is a constructible subset of a
Noetherian scheme X, then Z is closed if and only if for every pair of points y1 andy2 with y1 ∈ y2 , if#y2 ∈ Z, then y1 ∈ Z. Hint for the “if” implication: show that Zncan be written as i=1 U#i ∩ Zi where Ui ⊂ X is open and Zi ⊂ X is closed. Shownthat Z can be written as i=1 Ui ∩ Zi (with possibly different n, Ui , Zi ) where
each Zi is irreducible and meets Ui . Now use “stability under specialization” andthe generic point of Zi to show that Zi ⊂ Z for all i, so Z = ∪Zi .
(b) Show that a constructible subset of a Noetherian scheme is open if and only if
it is “stable under generization”. (Hint: this follows in one line from (a).)
The image of a morphism of schemes can be stranger than a constructible set.
Indeed if S is any subset of a scheme Y, it can be the image of a morphism: let X
be the disjoint union of spectra of the residue fields of all the points of S, and let
π : X → Y be the natural map. This is quite pathological, but in any reasonablesituation, the image is essentially no worse than arose in the previous example of(x, y) (→ (x, xy). This is made precise by Chevalley’s Theorem.
7.4.2. Chevalley’s Theorem. — If π : X → Y is a finite type morphism of Noetherian
schemes, the image of any constructible set is constructible. In particular, the image of π
is constructible.

(For the minority who might care: see §9.3.7 for an extension to locally finitely
presented morphisms.) We discuss the proof after giving some important consequences that may seem surprising, in that they are algebraic corollaries of aseemingly quite geometric and topological theorem. The first is a proof of the
Nullstellensatz.
7.4.3. Proof of the Nullstellensatz 3.2.5. We wish to show that if K is a field extensionof k that is finitely generated as a k-algebra, say by x1 , . . . , xn , then it is a finite
extension of fields. It suffices to show that each xi is algebraic over k. But if xi is notalgebraic over k, then we have an inclusion of rings k[xi ] → K, corresponding toa dominant morphism π : Spec K → A1k of finite type k-schemes. Of course Spec K

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is a single point, so the image of π is one point. By Chevalley’s Theorem 7.4.2 andExercise 7.4.B, the image of π is not the generic point of A1k , so im(π) is a closed
point of A1k , and thus π is not dominant.
!
A similar idea can be used in the following exercise.
7.4.D. E XERCISE ( QUASIFINITE MORPHISMS TO A FIELD ARE FINITE ). Suppose
π : X → Spec k is a quasifinite morphism. Show that π is finite. (Hint: deal firstwith the affine case, X = Spec A, where A is finitely generated over k. Suppose
A contains an element x that is not algebraic over k, i.e., we have an inclusionk[x] "→ A. Exercise 7.3.H may help.)
7.4.E. E XERCISE (“ FOR MAPS OF VARIETIES , SURJECTIVITY CAN BE CHECKED ON
CLOSED POINTS ”). Assume Chevalley’s Theorem 7.4.2. Show that a morphismof affine k-varieties π : X → Y is surjective if and only if it is surjective on closed
points (i.e., if every closed point of Y is the image of a closed point of X). (Once wedefine varieties in general, in Definition 10.1.7, you will see that your argument
works without change with the adjective “affine” removed.)
In order to prove Chevalley’s Theorem 7.4.2 (in Exercise 7.4.O), we introduce
a useful idea of Grothendieck’s. For the purposes of this discussion only, we say a
B-algebra A satisfies (†) if for each finitely generated A-module M, there exists a nonzerof ∈ B such that Mf is a free Bf -module.
7.4.4. Grothendieck’s Generic Freeness Lemma. — Suppose B is a Noetherian
integral domain. Then every finitely generated B-algebra satisfies (†).
Proof. We prove the Generic Freeness Lemma 7.4.4 in a series of exercises. We
assume that B is a Noetherian integral domain until Lemma 7.4.4 is proved, at theend of Exercise 7.4.K.
7.4.F. E XERCISE . Show that B itself satisfies (†).
7.4.G. E XERCISE . Reduce the proof of Lemma 7.4.4 to the following statement: if
A is a finitely-generated B-algebra satisfying (†), then A[T ] does too. (Hint: induct
on the number of generators of A as a B-algebra.)
We now prove this statement. Suppose A satisfies (†), and let M be a finitely
generated A[T ]-module, generated by the finite set S. Let M0 = 0, and let M1 bethe sub-A-module of M generated by S. For n > 0, inductively define
Mn+1 = Mn + TMn ,a sub-A-module of M. Note that M is the increasing union of the A-modules Mn .
7.4.H. E XERCISE . Show that multiplication by T induces a surjection
ψn : Mn /Mn−1 → Mn+1 /Mn .

7.4.I. E XERCISE . Show that for n ≫ 0, ψn is an isomorphism. Hint: use theascending chain condition on M1 .

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7.4.J. E XERCISE . Show that there is a nonzero f ∈ B such that (Mi+1 /Mi )f is free
as a Bf -module, for all i. Hint: as i varies, Mi+1 /Mi passes through only finitely
many isomorphism classes.
The following result concludes the proof of the Generic Freeness Lemma 7.4.4.
7.4.K. E XERCISE ( NOT REQUIRING N OETHERIAN HYPOTHESES ). Suppose M isa B-module that is an increasing union of submodules Mi , with M0 = 0, andthat every Mi+1 /Mi is free. Show that M is free. Hint: first construct compatible
isomorphisms φn : ⊕n−1
i=0 Mi+1 /Mi → Mn by induction on n. Then show thatthe colimit φ := lim φn : ⊕∞i=0 Mi+1 /Mi → M is an isomorphism. Side Remark:−→More generally, your argument will show that if the Mi+1 /Mi are all projective
modules (to be defined in §23.2.1), then M is (non-naturally) isomorphic to theirdirect sum.
!
We now set up the proof of Chevalley’s Theorem 7.4.2.
7.4.L. E XERCISE . Suppose π : X → Y is a finite type morphism of Noetherian
schemes, and Y is irreducible. Show that there is a dense open subset U of Y
such that the image of π either contains U or else does not meet U. (Hint: suppose π : Spec A → Spec B is such a morphism. Then by the Generic FreenessLemma 7.4.4, there is a nonzero f ∈ B such that Af is a free Bf -module. It musthave zero rank or positive rank. In the first case, show that the image of π does notmeet D(f) ⊂ Spec B. In the second case, show that the image of π contains D(f).)

There are more direct ways of showing the content of the above hint. For
example, another proof in the case of varieties will turn up in the proof of Theorem 11.4.1. We only use the Generic Freeness Lemma because we will use it again
in the future (§24.5.9).
7.4.M. E XERCISE . Show that to prove Chevalley’s Theorem, it suffices to provethat if π : X → Y is a finite type morphism of Noetherian schemes, the image of π
is constructible.

7.4.N. E XERCISE . Reduce further to the case where Y is affine, say Y = Spec B.
Reduce further to the case where X is affine.
We now give the rest of the proof by waving our hands, and leave it to you
to make it precise. The idea is to use Noetherian induction, and to reduce theproblem to Exercise 7.4.L.
We can deal with each of the components of Y separately, so we may assume
that Y is irreducible. We can then take B to be an integral domain. By Exercise 7.4.L,
there is a dense open subset U of Y where either the image of π includes it, or is
disjoint from it. If U = Y, we are done. Otherwise, it suffices to deal with thecomplement of U. Renaming this complement Y, we return to the start of theparagraph.
7.4.O. E XERCISE . Complete the proof of Chevalley’s Theorem 7.4.2, by making
the above argument precise.

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7.4.5. ⋆ Elimination of quantifiers. A basic sort of question that arises in any
number of contexts is when a system of equations has a solution. Suppose forexample you have some polynomials in variables x1 , . . . , xn over an algebraicallyclosed field k, some of which you set to be zero, and some of which you set to
be nonzero. (This question is of fundamental interest even before you know any
scheme theory!) Then there is an algebraic condition on the coefficients which willntell you if there is a solution. Define the Zariski topology on k in the obvious way:closed subsets are cut out by equations. (A mild generalization of this appears in
Exercise 11.2.I.)
7.4.P. E XERCISE ( ELIMINATION OF QUANTIFIERS , OVER AN ALGEBRAICALLY CLOSEDFIELD ). Fix an algebraically closed field k. Suppose
f1 , . . . , fp , g1 , . . . , gq ∈ k[W1 , . . . , Wm , X1 , . . . Xn ]are given. Show that there is a (Zariski-)constructible subset Y of k
(7.4.5.1)
andmsuch that

f1 (w1 , . . . , wm , X1 , . . . , Xn ) = · · · = fp (w1 , . . . , wm , X1 , . . . , Xn ) = 0(7.4.5.2) g1 (w1 , . . . , wm , X1 , . . . , Xn ) ̸= 0···n

gq (w1 , . . . , wm , X1 , . . . , Xn ) ̸= 0

has a solution (X1 , . . . , Xn ) = (x1 , . . . , xn ) ∈ k if and only if (w1 , . . . , wm ) ∈ Y.
Hints: if Z is a finite type scheme over k, and the closed points are denoted Zcl (“cl”
is for either “closed” or “classical”), then under the inclusion of topological spaces
Zcl "→ Z, the Zariski topology on Z induces the Zariski topology on Zcl . Note thatp
we can identify (Ap)cl with k by the Nullstellensatz (Exercise 5.3.F). If X is theklocally closed subset of Am+n cut out by the equalities and inequalities (7.4.5.1)
and (7.4.5.2), we have the diagram
Xcl !
πcl

"

"
m "
k !

% X ! " loc. cl. % Am+n
✇
✇✇
✇✇
π
✇
✇
" !✇✇
% Am

where Y = im πcl . By Chevalley’s Theorem 7.4.2, im π is constructible, and hencemm
so is (im π) ∩ k . It remains to show that (im π) ∩ k = Y (= im πcl ). You might
use the Nullstellensatz.
This is called “elimination of quantifiers” because it gets rid of the quantifier“there exists a solution”. The analogous statement for real numbers, where inequalities are also allowed, is a special case of Tarski’s celebrated theorem of eliminationof quantifiers for real closed fields (see, for example, [Ta]).
7.4.6. The Fundamental Theorem of Elimination Theory.
In the case of projective space (and later, projective morphisms), one can dobetter than Chevalley (at least in describing images of closed subsets).
7.4.7. Theorem (Fundamental Theorem of Elimination Theory). — The morphism
π : PnA → Spec A is closed (sends closed sets to closed sets).
Note that no Noetherian hypotheses are needed.

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A great deal of classical algebra and geometry is contained in this theorem as
special cases. Here are some examples.
First, let A = k[a, b, c, . . . , i], and consider the closed subset of P2A (taken withcoordinates x, y, z) corresponding to ax + by + cz = 0, dx + ey + fz = 0, gx + hy +iz = 0. Then we are looking for the locus in Spec A where these equations have anontrivial solution. This indeed corresponds to a Zariski-closed set — where
⎛
⎞
a b c
det ⎝d e f ⎠ = 0.
g h iThus the idea of the determinant is embedded in elimination theory.
As a second example, let A = k[a0 , a1 , . . . , am , b0 , b1 , . . . , bn ]. Now consider
the closed subset of P1A (taken with coordinates x and y) corresponding to a0 xm +a1 xm−1 y + · · · + am ym = 0 and b0 xn + b1 xn−1 y + · · · + bn yn = 0. Then there is apolynomial in the coefficients a0 , . . . , bn (an element of A) which vanishes if andonly if these two polynomials have a common nonzero root — this polynomial iscalled the resultant.
More generally, given a number of homogeneous equations in n + 1 variables
with indeterminate coefficients, Theorem 7.4.7 implies that one can write down
equations in the coefficients that precisely determine when the equations have anontrivial solution.
7.4.8. Proof of the Fundamental Theorem of Elimination Theory 7.4.7. Suppose Z "→ Pn
A
is a closed subset. We wish to show that π(Z) is closed.
By the definition of the Zariski topology on Proj A[x0 , . . . , xn ] (§4.5.7), Z is cutout (set-theoretically) by some homogeneous elements f1 , f2 , · · · ∈ A[x0 , . . . , xn ].
We wish to show that the points p ∈ Spec A that are in π(Z) form a closed subset.
Equivalently, we want to show that those p for which f1 , f2 , . . . have a common
zero in Proj κ(p)[x0 , . . . , xn ] form a closed subset of Spec A.
To motivate our argument, we consider a related question. Suppose that S• :=k[x0 , . . . , xn ], and that g1 , g2 , · · · ∈ k[x0 , . . . , xn ] are homogeneous polynomials.
How can we tell if g1 , g2 , . . . have a common zero in Proj S• = Pnk?
They would have a common zero if and only if in
An+1= Spec S• ,kthey cut out (set-theoretically) more than the origin, i.e., if set-theoreticallyV(g1 , g2 , . . . ) ̸⊂ V(x0 , . . . , xn ).

By the inclusion-reversing bijection between closed subsets and radical ideals (Theorem 3.7.1), this is true if and only if&&(g1 , g2 , . . . ) ̸⊃ (x0 , . . . , xn ).
But (x0 , . . . , xn ) is radical, so this is true if and only if&(g1 , g2 , . . . ) ̸⊃ (x0 , . . . , xn ).

This is&true if and only if not all of the generators x0 , . . . , xn of the ideal (x0 , . . . , xn )are in (g1 , g2 , . . . ), which is true if only if there is some i such that no power ofxi is in (g1 , g2 , . . . ). This is true if and only if(x0 , . . . , xn )N ̸⊂ (g1 , g2 , . . . ) for all N.

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November 18, 2017 draft223(Do you see why this implication goes both ways?) This is equivalent toSN ̸⊂ (g1 , g2 , . . . ) for all N,

which may be rewritten as(7.4.8.1)

SN ̸⊂ g1 SN−deg g1 ⊕ g2 SN−deg g2 ⊕ · · ·for all N. In other words, this is equivalent to the statement that the k-linear map
SN−deg g1 ⊕ SN−deg g2 ⊕ · · · → SN

is not surjective. This map is given by a matrix with dim SN rows. (It may have an
infinite number of columns, but this will not bother us.) To check that this linear
map is not surjective, we need only check that all the “maximal” (dim SN ×dim SN )determinants are zero. (Of course, we need to check this for all N.) Thus thecondition that g1 , g2 , . . . have no common zeros in Pn
k is the same as checking
some (admittedly infinite) number of equations. In other words, it is a Zariskiclosed condition on the coefficients of the polynomials g1 , g2 , . . . .
7.4.Q. E XERCISE . Complete the proof of the Fundamental Theorem of Elimination
Theory 7.4.7. (Hint: follow precisely the same argument, with k replaced by A, andthe gi replaced by the fi . How and why does this prove the theorem?)!
Notice that projectivity was crucial to the proof: we used graded rings in an
essential way. Notice also that the proof is essentially just linear algebra.

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CHAPTER 8

Closed embeddings and related notions

8.1 Closed embeddings and closed subschemes
The scheme-theoretic analog of closed subsets has a surprisingly different flavor from the analog of open sets (open embeddings). However, just as open embeddings (the scheme-theoretic version of open set) are locally modeled on open
sets U ⊂ Y, the analog of closed subsets also has a local model. This was foreshadowed by our understanding of closed subsets of Spec B as roughly corresponding
to ideals. If I ⊂ B is an ideal, then Spec B/I "→ Spec B is a morphism of schemes,
and we have checked that on the level of topological spaces, this describes Spec B/I
as a closed subset of Spec B, with the subspace topology (Exercise 3.4.I). This morphism is our “local model” of a closed embedding.
8.1.1. Definition. A morphism π : X → Y is a closed embedding (or closed immersion) if it is an affine morphism, and for every affine open subset Spec B ⊂ Y,
∼ Spec A, the map B → A is surjective (i.e., of the form B → B/I,
with π−1 (Spec B) =our desired local model). (“Embedding” is preferable to “immersion”, because thedifferential geometric notion of immersion is closer to what algebraic geometerscall unramified, which we will define in §21.6.) The symbol "→ often is used to indicate that a morphism is a closed embedding (or more generally, a locally closed
embedding, §8.1.2).
If X is a subset of Y (and π on the level of sets is the inclusion), we say that
X is a closed subscheme of Y. The difference between a closed embedding and aclosed subscheme is confusing and unimportant; the same issue for open embeddings/subschemes was discussed in §7.1.1. In case it helps: a closed embedding is
the same thing as an isomorphism with a closed subscheme.
8.1.A. E XERCISE . Show that a closed embedding identifies the topological spaceof X with a closed subset of the topological space of Y. (Caution: The closed embeddings Spec k[x]/(x) "→ Spec k[x] and Spec k[x]/(x2 ) "→ Spec k[x] show that theclosed subset does not determine the closed subscheme. The “infinitesimal” information, or “fuzz”, is lost.)
8.1.B. E ASY E XERCISE . Show that closed embeddings are finite morphisms, henceof finite type.
8.1.C. E ASY E XERCISE . Show that the composition of two closed embeddings is a
closed embedding.
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8.1.D. E XERCISE . Show that the property of being a closed embedding is affinelocal on the target.
8.1.E. E XERCISE . Suppose B → A is a surjection of rings. Show that the induced
morphism Spec A → Spec B is a closed embedding. (Our definition would be aterrible one if this were not true!)
A closed embedding π : X "→ Y determines an ideal sheaf on Y, as the kernel
IX/Y of the map of OY -modules
OY → π∗ OX .

An ideal sheaf on Y is what it sounds like: it is a sheaf of ideals. It is a subOY -module I of OY . On each open subset, it gives an ideal I (U) of the ring
OY (U). We thus have an exact sequence (of OY -modules) 0 → IX/Y → OY →π∗ OX → 0. (On Spec B, the epimorphism OY → π∗ OX is the surjection B → A of
Definition 8.1.1.)
Thus for each affine open subset Spec B "→ Y, we have an ideal I(B) ⊂ B, andwe can recover X from this information: the I(B) (as Spec B "→ Y varies over theaffine open subsets) defines an O-module on the base, hence an OY -module on
Y, and the cokernel of I "→ OY is π∗ OX . It will be useful to understand when
the information of the I(B) (for all affine opens Spec B "→ Y) actually determines
a closed subscheme. Our life is complicated by the fact that the answer is “notalways”, as shown by the following example.
8.1.F. U NIMPORTANT EXERCISE . Let X = Spec k[x](x) , the germ of the affine line at
the origin, which has two points, the closed point and the generic point η. Define
I (X) = {0} ⊂ OX (X) = k[x](x) , and I (η) = k(x) = OX (η). Show that this sheafof ideals does not correspond to a closed subscheme. (Possible approach: do thenext exercise first.)
The next exercise gives a necessary condition.
8.1.G. E XERCISE . Suppose IX/Y is a sheaf of ideals corresponding to a closed
embedding X "→ Y. Suppose Spec B "→ Y is an affine open subscheme, and f ∈ B.
Show that the natural map I(B)f → I(Bf ) is an isomorphism. (First state what the“natural map” is!)
It is an important and useful fact that this is also sufficient:

8.1.H. E SSENTIAL ( HARD ) EXERCISE : A USEFUL CRITERION FOR WHEN IDEALS IN
Suppose Y is a scheme, and foreach affine open subset Spec B of Y, I(B) ⊂ B is an ideal. Suppose further that foreach affine open subset Spec B "→ Y and each f ∈ B, restriction of functions from∼ I(B)f . Show that these data arise from a
B → Bf induces an isomorphism I(Bf ) =(unique) closed subscheme X "→ Y by the above construction. In other words, theclosed embeddings Spec B/I "→ Spec B glue together in a well-defined manner toobtain a closed embedding X "→ Y.
This is a hard exercise, so as a hint, here are three different ways of proceeding; some combination of them may work for you. Approach 1. For each affine
open Spec B, we have a closed subscheme Spec B/I "→ Spec B. (i) For any two
affine open subschemes Spec A and Spec B, show that the two closed subschemes
AFFINE OPEN SETS DEFINE A CLOSED SUBSCHEME .

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Spec A/I(A) "→ Spec A and Spec B/I(B) "→ Spec B restrict to the same closed subscheme of their intersection. (Hint: cover their intersection with open sets simultaneously distinguished in both affine open sets, Proposition 5.3.1.) Thus for example we can glue these two closed subschemes together to get a closed subschemeof Spec A ∪ Spec B. (ii) Use Exercise 4.4.A on gluing schemes (or the ideas therein)
to glue together the closed embeddings in all affine open subschemes simultaneously. You will only need to worry about triple intersections. Approach 2. (i) Use
the data of the ideals I(B) to define a sheaf of ideals I "→ O. (ii) For each affine
open subscheme Spec B, show that I (Spec B) is indeed I(B), and (O/I )(Spec B)
is indeed B/I(B), so the data of I recovers the closed subscheme on each Spec B
as desired. Approach 3. (i) Describe X first as a subset of Y. (ii) Check that X is
closed. (iii) Define the sheaf of functions OX on this subset, perhaps using compatible germs. (iv) Check that this resulting ringed space is indeed locally the closed
subscheme given by Spec B/I "→ Spec B.
We will see (in §13.5.4) that closed subschemes correspond to quasicoherent
sheaves of ideals; the mathematical content of this statement will turn out to beprecisely Exercise 8.1.H.
8.1.I. I MPORTANT E XERCISE /D EFINITION : THE VANISHING S CHEME .
(a) Suppose Y is a scheme, and s ∈ Γ (Y, OY ). Define the closed subscheme cut out
by s. We call this the vanishing scheme V(s) of s, as it is the scheme-theoretic
version of our earlier (set-theoretical) version of V(s) (§3.4). (Hint: on affine open
Spec B, we just take Spec B/(sB ), where sB is the restriction of s to Spec B. Use
Exercise 8.1.H to show that this yields a well-defined closed subscheme.)(b) If u is an invertible function, show that V(s) = V(su).
(c) If S is a set of functions, define V(S).
8.1.J. I MPORTANT E XERCISE .
(a) In analogy with closed subsets, define the notion of a finite union of closed
subschemes of X, and an arbitrary (not necessarily finite) intersection of closed
subschemes of X. (Exercise 8.1.H may help.) Hint: If X is affine, then you might
expect that the union of closed subschemes corresponding to I1 and I2 would be
the closed subscheme corresponding to either I1 ∩ I2 or I1 I2 — but which one? We
would want the union of a closed subscheme with itself to be itself, so the rightchoice is I1 ∩ I2 .
(b) Describe the scheme-theoretic intersection of V(y − x2 ) and V(y) in A2 . See
Figure 4.5 for a picture. (For example, explain informally how this corresponds
to two curves meeting at a single point with multiplicity 2 — notice how the 2 isvisible in your answer. Alternatively, what is the nonreducedness telling you —both its “size” and its “direction”?) Describe their scheme-theoretic union.
(c) Show that the underlying set of a finite union of closed subschemes is the finite
union of the underlying sets, and similarly for arbitrary intersections.
(d) Describe the scheme-theoretic intersection of V(y2 − x2 ) and V(y) in A2 . Drawa picture. (Did you expect the intersection to have multiplicity one or multiplicity
two?) Hence show that if X, Y, and Z are closed subschemes of W, then (X ∩ Z) ∪(Y ∩ Z) ̸= (X ∪ Y) ∩ Z in general. In particular, not all properties of intersection andunion carry over from sets to schemes.

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8.1.K. ⋆ H ARD E XERCISE ( NOT USED LATER ). In the literature, the usual definition of a closed embedding is a morphism π : X → Y such that π induces ahomeomorphism of the underlying topological space of X onto a closed subset of
the topological space of Y, and the induced map π♯ : OY → π∗ OX of sheaves on Y
is surjective. (By “surjective” we mean that the ring homomorphism on stalks issurjective.) Show that this definition agrees with the one given above. (To showthat our definition involving surjectivity on the level of affine open sets implies
this definition, you can use the fact that surjectivity of a morphism of sheaves canbe checked on a suitably chosen base, Exercise 2.5.E.)
We have now defined the analog of open subsets and closed subsets in theland of schemes. Their definition is slightly less “symmetric” than in the classicaltopological setting: the “complement” of a closed subscheme is a unique open
subscheme, but there are many “complementary” closed subschemes to a given
open subscheme in general. (We will soon define one that is “best”, that has areduced structure, §8.3.9.)
8.1.2. Locally closed embeddings and locally closed subschemes.
Now that we have defined analogs of open and closed subsets, it is natural todefine the analog of locally closed subsets. Recall that locally closed subsets areintersections of open subsets and closed subsets. Hence they are closed subsets ofopen subsets, or equivalently open subsets of closed subsets. The analog of theseequivalences will be a little problematic in the land of schemes.
We say a morphism π : X → Y is a locally closed embedding (or locally closed
immersion) if π can be factored into
X

ρ

%Z

τ

%Y

where ρ is a closed embedding and τ is an open embedding. For example, themorphism Spec k[t, t−1 ] → Spec k[x, y] given by (x, y) (→ (t, 0) is a locally closed
embedding (Figure 8.1). The symbol "→ is often used to indicate that a morphism
is a locally closed embedding.

F IGURE 8.1. The locally closed embedding Spec k[t, t−1 ] →Spec k[x, y] (t (→ (t, 0) = (x, y), i.e., (x, y) → (t, 0))

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November 18, 2017 draft229(Warning: The term immersion is often used instead of locally closed embedding
or locally closed immersion, but this is unwise terminology, for reasons that alreadyarose for closed embeddings in Definition 8.1.1: The differential geometric notionof immersion is closer to what algebraic geometers call unramified, which we willdefine in §21.6. Also, the naked term embedding should be avoided, because it isneedlessly imprecise.)
If X is a subset of Y (and π on the level of sets is the inclusion), we say X is alocally closed subscheme of Y.
8.1.L. E ASY EXERCISE . Show that locally closed embeddings are locally of finite
type.
At this point, you could define the intersection of two locally closed embeddings in a scheme X (which will also be a locally closed embedding in X). But
it would be awkward, as you would have to show that your construction is independent of the factorizations of each locally closed embedding into a closed
embedding and an open embedding. Instead, we wait until Exercise 9.2.C, whenrecognizing the intersection as a fibered product will make this easier.
Clearly an open subscheme U of a closed subscheme V of X can be interpreted
as a closed subscheme of an open subscheme: as the topology on V is induced
from the topology on X, the underlying set of U is the intersection of some open
subset U ′ on X with V. We can take V ′ = V ∩ U ′ , and then V ′ → U ′ is a closed
embedding, and U ′ → X is an open embedding.
It is not clear that a closed subscheme V ′ of an open subscheme U ′ can be
expressed as an open subscheme of a closed subscheme V. In the category of
topological spaces, we would take V as the closure of V ′ , so we are now motivatedto define the analogous construction, which will give us an excuse to introduce
several related ideas, in §8.3. We will then resolve this issue in good cases (e.g., if
X is Noetherian) in Exercise 8.3.C.
We formalize our discussion in an exercise.
8.1.M. E XERCISE . Suppose V → X is a morphism. Consider three conditions:(i) V is the intersection of an open subscheme of X and a closed subschemeof X (you will have to define the meaning of “intersection” here, see Exercise 7.1.B, or else see the hint below).
(ii) V is an open subscheme of a closed subscheme of X, i.e., it factors into an
open embedding followed by a closed embedding.
(iii) V is a closed subscheme of an open subscheme of X, i.e., V is a locallyclosed embedding.

Show that (i) and (ii) are equivalent, and both imply (iii). (Remark: (iii) does notalways imply (i) and (ii), see the pathological example [Stacks, tag 01QW].) Hint:
It may be helpful to think of the problem as follows. You might hope to think of a
locally closed embedding as a fibered diagram
V" ! #"

open emb.

closed emb.

" "
U!

% K"
#

closed emb.
open emb.

"% X.

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Interpret (i) as the existence of the diagram. Interpret (ii) as this diagram minus
the lower left corner. Interpret (iii) as the diagram minus the upper right corner.
8.1.N. E XERCISE . Show that the composition of two locally closed embeddings isa locally closed embedding. (Hint: you might use (ii) implies (iii) in the previous
exercise.)
8.1.3. Unimportant remark. It may feel odd that in the definition of a locally closed
embedding, we had to make a choice (as a composition of a closed embedding
followed by an open embedding, rather than vice versa), but this type of issue
comes up earlier: a subquotient of a group can be defined as the quotient of a subgroup, or a subgroup of a quotient. Which is the right definition? Or are they thesame? (Hint: compositions of two subquotients should certainly be a subquotient,
cf. Exercise 8.1.N.)

8.2 More projective geometry
We now interpret closed embeddings in terms of graded rings. Don’t worry;most of the annoying foundational discussion of graded rings is complete, and we
now just take advantage of our earlier work.
8.2.1. Example: Closed embeddings in projective space Pn
A . Recall the definitionof projective space Pn
given
in§4.4.10(andtheterminology
defined there). Any
Ahomogeneous polynomial f in x0 , . . . , xn defines a closed subscheme. (Thus even if
f doesn’t make sense as a function, its vanishing scheme still makes sense.) On
the open set Ui , the closed subscheme is V(f(x0/i , . . . , xn/i )), which we privately
deg fthink of as V(f(x0 , . . . , xn )/xi). On the overlap

Ui ∩ Uj = Spec A[x0/i , . . . , xn/i , x−1
j/i ]/(xi/i − 1),these functions on Ui and Uj don’t exactly agree, but they agree up to a nonvanishing scalar, and hence cut out the same closed subscheme of Ui ∩ Uj (Exercise 8.1.I(b)):
deg ff(x0/i , . . . , xn/i ) = xj/i f(x0/j , . . . , xn/j ).
Similarly, a collection of homogeneous polynomials in A[x0 , . . . , xn ] cuts out aclosed subscheme of Pn
A . (Exercise 8.2.C will show that all closed subschemes of
Pn
are
of
this
form.)
A
8.2.2. Definition. As usual, let k be a field. A closed subscheme of Pn
k cut out
by a single (nonzero, homogeneous) equation is called a hypersurface (in Pnk ). Ofcourse, a hypersurface is not cut out by a single global function on Pn
:
there
are
k
no nonconstant global functions (Exercise 4.4.E). The degree of a hypersurface isthe degree of the polynomial. (Implicit in this is that this notion can be determinedfrom the subscheme itself. You may have the tools to prove this now, but we won’t
formally prove it until Exercise 18.6.H.) A hypersurface of degree 1 (resp. degree
2, 3, . . . ) is called a hyperplane (resp. quadric, cubic, quartic, quintic, sextic,
septic, octic, . . . hypersurface). If n = 2, hypersurfaces are called curves; a degree
1 hypersurface is called a line, and a degree 2 hypersurface is called a conic curve,

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or a conic for short. If n = 3, a hypersurface is called a surface. (In Chapter 11, we
will justify the terms curve and surface.)
8.2.A. E XERCISE .
(a) Show that wz = xy, x2 = wy, y2 = xz describes an irreducible subscheme in
P3k . In fact it is a curve, a notion we will define once we know what dimension
is. This curve is called the twisted cubic. (The twisted cubic is a good nontrivial
example of many things, so you should make friends with it as soon as possible.
It implicitly appeared earlier in Exercise 3.6.F.)(b) Show that the twisted cubic is isomorphic to P1k .
We now extend this discussion to projective schemes in general.
% % R• is a surjection of graded rings. Show
8.2.B. E XERCISE . Suppose that S•that the domain of the induced morphism (Exercise 6.4.A) is Proj R• , and that theinduced morphism Proj R• → Proj S• is a closed embedding.
8.2.C. E XERCISE ( CONVERSE TO E XERCISE 8.2.B). Suppose that X "→ Proj S• isa closed embedding in a projective A-scheme (where S• is a finitely generated
graded A-algebra). Show that X is projective by describing it as Proj(S• /I), where
I is a homogeneous ideal, of “projective functions” vanishing on X. Many find thiseasier if S• is generated in degree 1, and this case is the most important, so you
may wish to deal only with this case. (Another approach to this problem is given
in Exercise 15.4.H.)
8.2.D. E XERCISE .
Show that an injective linear map of k-vector spaces V "→W induces a closed embedding PV "→ PW. (This is another justification for thedefinition of PV in Example 4.5.12 in terms of the dual of V.)
8.2.3. Definition. The closed subscheme defined in Exercise 8.2.D is called a linear space. Once we know about dimension, we will call this closed subscheme alinear space of dimension dim V − 1 = dim PV. More explicitly, a linear space ofdimension n in PN is any closed subscheme cut out by N − n k-linearly independent homogeneous linear polynomials in x0 , . . . , xN . A linear space of dimension
1 (resp. 2, n, dim PW − 1) is called a line (resp. plane, n-plane, hyperplane). (Ifthe linear map in the previous exercise is not injective, then the hypothesis (6.4.0.1)
of Exercise 6.4.A fails.)
8.2.E. E XERCISE ( A SPECIAL CASE OF B ÉZOUT ’ S T HEOREM ). Suppose X ⊂ Pn
k isa degree d hypersurface cut out by f = 0, and ℓ is a line not contained in X. A
very special case of Bézout’s Theorem (Exercise 18.6.K) implies that X and ℓ meetwith multiplicity d, “counted correctly”. Make sense of this, by restricting thehomogeneous degree d polynomial f to the line ℓ, and using the fact that a degreed polynomial in k[x] has d roots, counted properly. (If it makes you feel better,assume k = k.)
8.2.F. E XERCISE . Show that the map of graded rings k[w, x, y, z] → k[s, t] givenby (w, x, y, z) (→ (s3 , s2 t, st2 , t3 ) induces a closed embedding P1k "→ P3k , whichyields an isomorphism of P1k with the twisted cubic (defined in Exercise 8.2.A —in fact, this will solve Exercise 8.2.A(b)). Doing this in a hands-on way will set

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you up well for the general Veronese construction of §8.2.6; see Exercise 8.2.J for ageneralization.
8.2.4. A particularly nice case: when S• is generated in degree 1.
Suppose S• is a finitely generated graded ring generated in degree 1. Then
S1 is a finitely generated S0 -module, and the irrelevant ideal S+ is generated indegree 1 (cf. Exercise 4.5.D(a)).
8.2.G. E XERCISE . Show that if S• is generated (as an A-algebra) in degree 1 by n+1
elements x0 , . . . , xn , then Proj S• may be described as a closed subscheme of Pn
A asfollows. Consider A⊕(n+1) as a free module with generators t0 , . . . , tn associatedto x0 , . . . , xn . The surjection of)*Sym• A⊕(n+1) = A[t0 , t1 , . . . , tn ]% % S•ti ✤% xi

implies S• = A[t0 , t1 , . . . tn ]/I, where I is a homogeneous ideal. (In particular,
Proj S• can always be interpreted as a closed subscheme of some Pn
A if S• is finitely
generated in degree 1. Then using Exercises 6.4.D and 6.4.G, you can remove thehypothesis of generation in degree 1.)
This is analogous to the fact that if R is a finitely generated A-algebra, thenchoosing n generators of R as an algebra is the same as describing Spec R as aclosed subscheme of An
A . In the affine case this is “choosing coordinates”; in the
projective case this is “choosing projective coordinates”.
Recall (Exercise 4.4.F) that if k is algebraically closed, then we can interpret theclosed points of Pn as the lines through the origin in (n + 1)-space. The following
exercise states this more generally.
8.2.H. E XERCISE . Suppose S• is a finitely generated graded ring over an algebraically closed field k, generated in degree 1 by x0 , . . . , xn , inducing closed embeddings Proj S• "→ Pn and Spec S• "→ An+1 . Give a bijection between the closed
points of Proj S• and the “lines through the origin” in Spec S• ⊂ An+1 .
8.2.5. A second proof that finite morphisms are closed. This interpretation of Proj S• asa closed subscheme of projective space (when it is generated in degree 1) yields thefollowing second proof of the fact (shown in Exercise 7.3.M) that finite morphisms
are closed. Suppose π : X → Y is a finite morphism. The question is local on thetarget, so it suffices to consider the affine case Y = Spec B. It suffices to show that
π(X) is closed. Then by Exercise 7.3.J, X is a projective B-scheme, and hence by theFundamental Theorem of Elimination Theory 7.4.7, its image is closed.
8.2.6. Important classical construction: The Veronese embedding.
Suppose S• = k[x, y], so Proj S• = P1k . Then S2• = k[x2 , xy, y2 ] ⊂ k[x, y] (see§6.4.4 on the Veronese subring). We identify this subring as follows.
∼ k[u, v, w]/(uw − v2 ),8.2.I. E XERCISE . Let u = x2 , v = xy, w = y2 . Show that S2• =2
2
by mapping u, v, w to x , xy, y , respectively.

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We have a graded ring generated by three elements in degree 1. Thus we thinkof it as sitting “in” P2 , via the construction of §8.2.G. This can be interpreted as “P1
as a conic in P2 ”.
8.2.7. Thus if k is algebraically closed of characteristic not 2, using the fact that
we can diagonalize quadratics (Exercise 5.4.J), the conics in P2 , up to change of
coordinates, come in only a few flavors: sums of 3 squares (e.g., our conic of theprevious exercise), sums of 2 squares (e.g., y2 −x2 = 0, the union of 2 lines), a single
square (e.g., x2 = 0, which looks set-theoretically like a line, and is nonreduced),and 0 (perhaps not a conic at all). Thus we have proved: any plane conic (over analgebraically closed field of characteristic not 2) that can be written as the sum of
three nonzero squares is isomorphic to P1 . (See Exercise 6.5.F for a closely related
fact.)
We now soup up this example.
8.2.J. E XERCISE . We continue to take S• = k[x, y]. Show that Proj Sd• is given by
the equations that"#y0 y1 · · · yd−1
y1 y2 · · ·ydis rank 1 (i.e., that all the 2 × 2 minors vanish). This is called the degree d rational
normal curve “in” Pd . You did the twisted cubic case d = 3 in Exercises 8.2.A and8.2.F.
8.2.8. Definition. More generally, if S• = k[x0 , . . . , xn ], then Proj Sd• ⊂ PN−1
(where N is the dimension of the vector space of homogeneous degree d polynomials in x0 , . . . , xn ) is called the d-uple embedding or d-uple Veronese embedding. (Combining Exercise 6.4.E with Exercise 8.2.G shows that Proj S• → PN−1 isa closed embedding.))*
8.2.K. C OMBINATORIAL EXERCISE ( CF. R EMARK 4.5.3). Show that N = n+dd .

8.2.L. U NIMPORTANT EXERCISE . Find six linearly independent quadratic equations vanishing on the Veronese surface Proj S2• where S• = k[x0 , x1 , x2 ], whichsits naturally in P5 . (You needn’t show that these equations generate all the equations cutting out the Veronese surface, although this is in fact true.) Possible hint:
use the identity⎛⎞
x0 x0 x0 x1 x0 x2
det ⎝ x1 x0 x1 x1 x1 x2 ⎠ = 0.
x2 x0 x2 x1 x2 x2

8.2.9. Rulings on the quadric surface. We return to rulings on the quadric surface,which first appeared in the optional (starred) section §4.4.12.

8.2.M. U SEFUL GEOMETRIC EXERCISE : THE RULINGS ON THE QUADRIC SURFACE
wz = xy. This exercise is about the lines on the quadric surface X given by
wz − xy = 0 in P3k (where the projective coordinates on P3k are ordered w, x, y, z).
This construction arises all over the place in nature.
(a) Suppose a0 and b0 are given elements of k, not both zero. Make sense of thestatement: as [c, d] varies in P1 , [a0 c, b0 c, a0 d, b0 d] is a line in the quadric surface.
(This describes “a family of lines parametrized by P1 ”, although we can’t yet make

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this precise.) Find another family of lines. These are the two rulings of the smoothquadric surface.
(b) Show that through every k-valued point of the quadric surface X, there passes
one line from each ruling.
(c) Show there are no other lines. (There are many ways of proceeding. At riskof predisposing you to one approach, here is a germ of an idea. Suppose L is aline on the quadric surface, and [1, x, y, z] and [1, x ′ , y ′ , z ′ ] are distinct points on
it. Because they are both on the quadric, z = xy and z ′ = x ′ y ′ . Because all of L
is on the quadric, (1 + t)(z + tz ′ ) − (x + tx ′ )(y + ty ′ ) = 0 for all t. After somealgebraic manipulation, this translates into (x − x ′ )(y − y ′ ) = 0. How can this bemade watertight? Another possible approach uses Bézout’s Theorem, in the formof Exercise 8.2.E.)

F IGURE 8.2. One of the two rulings on the quadric surface V(wz−xy) ⊂ P3 . One ruling contains the line V(w, x), and the othercontains the line V(w, y).
Hence by Exercise 5.4.J, if we are working over an algebraically closed fieldof characteristic not 2, we have shown that all rank 4 quadric surfaces have tworulings of lines. (In Example 9.6.2, we will recognize this quadric as P1 × P1 .)8.2.10. Side Remark. The existence of these two rulings is the first chapter of anumber of important and beautiful stories. The second chapter is often the following. If k is an algebraically closed field, then a “maximal rank” (Exercise 5.4.J)
quadric hypersurface X of dimension m contains no linear spaces of dimension
greater than m/2. (We will see in Exercise 12.3.D that the maximal rank quadrichypersurfaces)a+2*are the “smooth” quadrics.) If m = 2a + 1, then X contains an irreducible 2 -dimensional family of a-planes. If m = 2a, then X contains two)*irreducible a+1
2 -dimensional families of a-planes, and furthermore two a-planes

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Λ and Λ ′ are in the same family if and only if dim(Λ ∩ Λ ′ ) ≡ a (mod 2). Thesefamilies of linear spaces are also called rulings. (For more information, see [GH1,§6.1, p. 735, Prop.].) You already know enough to think through the examples ofm = 0, 1, and 2. The case m = 3 is discussed in Exercise 16.7.K.
8.2.11. Weighted projective space. If we put a nonstandard weighting on thevariables of k[x1 , . . . , xn ] — say we give xi degree di — then Proj k[x1 , . . . , xn ] iscalled weighted projective space P(d1 , d2 , . . . , dn ).
∼8.2.N. E XERCISE . Show that P(m, n) is isomorphic to P1 . Show that P(1, 1, 2) =Proj k[u, v, w, z]/(uw − v2 ). Hint: do this by looking at the even-graded parts ofk[x0 , x1 , x2 ], cf. Exercise 6.4.D. (This is a projective cone over a conic curve. Over a
field of characteristic not 2, it is isomorphic to the traditional cone x2 + y2 = z2 in
P3 , see Figure 8.3.)

F IGURE 8.3. The cone x2 + y2 = z2 .
8.2.12. Affine and projective cones.
If S• is a finitely generated graded ring, then the affine cone of Proj S• isSpec S• . Caution: this terminology is not ideal, as this construction depends on S• ,not just on Proj S• . As motivation, consider the graded ring S• = C[x, y, z]/(z2 −x2 − y2 ). Figure 8.3 is a sketch of Spec S• . (Here we draw the “real picture” ofz2 = x2 + y2 in R3 .) It is a cone in the traditional sense; the origin (0, 0, 0) is the“cone point”.
This gives a useful way of picturing Proj (even over arbitrary rings, not justC). Intuitively, you could imagine that if you discarded the origin, you would
get something that would project onto Proj S• . The following exercise makes thatprecise.
8.2.O. E XERCISE . If Proj S• is a projective scheme over a field k, describe a naturalmorphism Spec S• \ V(S+ ) → Proj S• . (Can you see why V(S+ ) is a single point,and should reasonably be called the origin?)
This readily generalizes to the following exercise, which again motivates theterminology “irrelevant”.

8.2.P. E ASY E XERCISE . If S• is a finitely generated graded ring over a base ring A,
describe a natural morphism Spec S• \ V(S+ ) → Proj S• .

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In fact, it can be made precise that Proj S• is the quotient (by the multiplicative
group of scalars) of the affine cone minus the origin.
8.2.13. Definition. The projective cone of Proj S• is Proj S• [T ], where T is a new variable of degree 1. For example, the cone corresponding to the conic Proj k[x, y, z]/(z2 −x2 − y2 ) is Proj k[x, y, z, T ]/(z2 − x2 − y2 ). The projective cone is sometimes called
the projective completion of Spec S• . (Note: this depends on S• , not just on
Proj S• .)8.2.Q. L ESS IMPORTANT EXERCISE ( CF. §4.5.1). Show that the “projective cone”
Proj S• [T ] of Proj S• has a closed subscheme isomorphic to Proj S• (informally, corresponding to T = 0), whose complement (the distinguished open set D(T )) is
isomorphic to the affine cone Spec S• .
This construction can be usefully pictured as the affine cone union some points“at infinity”, and the points at infinity form the Proj. (The reader may wish to startwith Figure 8.3, and try to visualize the conic curve “at infinity”, and then compare
this visualization to Figure 4.9.)
We have thus completely described the algebraic analog of the classical pictureof §4.5.1.

8.3 The (closed sub)scheme-theoretic image
We now define a series of notions that are all of the form “the smallest closed
subscheme such that something or other is true”. One example will be the notionof scheme-theoretic closure of a locally closed embedding, which will allow us
to interpret locally closed embeddings in three equivalent ways (open subschemeintersect closed subscheme; open subscheme of closed subscheme; and closed subscheme of open subscheme — cf. Exercise 8.1.M).
8.3.1. Scheme-theoretic image.
We start with the notion of scheme-theoretic image. Set-theoretic images arebadly behaved in general (§7.4.1), and even with reasonable hypotheses such as
those in Chevalley’s Theorem 7.4.2, things can be confusing. For example, there is
no reasonable way to impose a scheme structure on the image of A2k → A2k givenby (x, y) (→ (x, xy). It will be useful (e.g., Exercise 8.3.C) to define a notion of
a closed subscheme of the target that “best approximates” the image. This will
incorporate the notion that the image of something with nonreduced structure(“fuzz”) can also have nonreduced structure. As usual, we will need to impose
reasonable hypotheses to make this notion behave well (see Theorem 8.3.4 andCorollary 8.3.5).
8.3.2. Definition. Suppose i : Z "→ Y is a closed subscheme, giving an exact sequence 0 → IZ/Y → OY → i∗ OZ → 0. We say that the image of π : X → Y liesin Z if the composition IZ/Y → OY → π∗ OX is zero. Informally, locally, functions
vanishing on Z pull back to the zero function on X. If the image of π lies in somesubschemes Zj (as j runs over some index set), it clearly lies in their intersection(cf. Exercise 8.1.J(a) on intersections of closed subschemes). We then define thescheme-theoretic image of π, a closed subscheme of Y, as the “smallest closed

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subscheme containing the image”, i.e., the intersection of all closed subschemes
containing the image. In particular (and in our first examples), if Y is affine, thescheme-theoretic image is cut out by functions on Y that are 0 when pulled back to
X.
Example 1. Consider π : Spec k[ϵ]/(ϵ2 ) → Spec k[x] = A1k given by x (→ ϵ.
Then the scheme-theoretic image of π is given by Spec k[x]/(x2 ) (the polynomials
pulling back to 0 are precisely multiples of x2 ). Thus the image of the fuzzy point
still has some fuzz.
Example 2. Consider π : Spec k[ϵ]/(ϵ2 ) → Spec k[x] = A1k given by x (→ 0. Thenthe scheme-theoretic image is given by Spec k[x]/(x): the image is reduced. In thispicture, the fuzz is “collapsed” by π.
Example 3. Consider π : Spec k[t, t−1 ] = A1 − {0} → A1 = Spec k[u] given byu (→ t. Any function g(u) which pulls back to 0 as a function of t must be thezero-function. Thus the scheme-theoretic image is everything. The set-theoretic
image, on the other hand, is the distinguished open set A1 − {0}. Thus in not-toopathological cases, the underlying set of the scheme-theoretic image is not the settheoretic image. But the situation isn’t terrible: the underlying set of the schemetheoretic image must be closed, and indeed it is the closure of the set-theoretic
image. We might imagine that in reasonable cases this will be true, and in evennicer cases, the underlying set of the scheme-theoretic image will be set-theoretic
image. We will later see that this is indeed the case (§8.3.6).
But sadly pathologies can sometimes happen in, well, pathological situations.
#
Example 4 (see Figure 8.4). Let X =Spec k[ϵn ]/((ϵn )n ) (a scheme whichappeared in the hint to Exercise 5.2.E) and Y = Spec k[x], and define X → Y by
x → ϵn on the nth component of X. If a function g(x) on Y pulls back to 0 on
X, then its Taylor expansion is 0 to order n (by examining the pullback to thenth component of X) for all n, so g(x) must be 0. (This argument will be vastly
generalized in Exercise 12.9.A(b).) Thus the scheme-theoretic image is V(0) on Y,
i.e., Y itself, while the set-theoretic image is easily seen to be just the origin (theclosed point 0). (This morphism implicitly arises in Caution/Example 8.3.11.)

F IGURE 8.4. Yuck (Example 4 of Definition 8.3.2)

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8.3.3. Criteria for computing scheme-theoretic images affine-locally. Example 4 clearly
is weird though, and we can show that in “reasonable circumstances” such pathology doesn’t occur.
In the special case where the target Y is affine (which covers Examples 1 through3), say Y = Spec B, almost by definition the scheme-theoretic image of π : X → Y is
cut out by the ideal I ⊂ B of functions on Y which pull back to zero on X.
It would be great to use this to compute the scheme-theoretic image affine
by affine (affine-locally). On the affine open set Spec B ⊂ Y, define the ideal
I(B) ⊂ B of functions which pull back to 0 on X. Formally, I(B) := ker(B →Γ (Spec B, π∗ (OX )). Then if for each such B, and each g ∈ B, I(B) ⊗B Bg → I(Bg )is an isomorphism, then we will have defined the scheme-theoretic image as aclosed subscheme (see Exercise 8.1.H). Clearly each function on Spec B that vanishes when pulled back to π−1 (Spec B) also vanishes when restricted to D(g) andthen pulled back to π−1 (D(g)). So the question is: given a function r/gn on D(g)that pulls back to zero on π−1 (D(g)), is it true that for some m, rgm = 0 whenpulled back to π−1 (Spec B)? Here are three cases where the answer is “yes”. (Iwould like to add a picture here, but I can’t think of one that would enlighten
more people than it would confuse. So you should try to draw one that suits you.)For each affine in the source, there is some m which works. There is one that works
for all affines in a cover (i) if m = 1 always works, or (ii) if there are only a finite
number of affines in the cover.
(i) The answer is yes if π−1 (Spec B) is reduced: we simply take m = 1 (as rvanishes on Spec Bg and g vanishes on V(g), so rg vanishes on Spec B = Spec Bg ∪V(g).)(ii) The answer is also yes if π−1 (Spec B) is affine, say Spec A: if r ′ = π♯ r and′g = π♯ g in A, then if r ′ = 0 on D(g ′ ), then there is an m such that r ′ (g ′ )m = 0 (asthe statement r ′ = 0 in D(g ′ ) means precisely this fact — the functions on D(g ′ )are Ag ′ ).
(ii)’ More generally, the answer is yes if π−1 (Spec B) is quasicompact: cover−1
π (Spec B) with finitely many affine open sets. For each one there will be somemi so that rgmi = 0 when pulled back to this open set. Then let m = max(mi ).
(We see again that quasicompactness is our friend!)
In conclusion, we have proved the following (subtle) theorem.
8.3.4. Theorem. — Suppose π : X → Y is a morphism of schemes. If X is reduced or π
is quasicompact, then the scheme-theoretic image of π may be computed affine-locally: on
Spec A ⊂ Y, it is cut out by the functions (elements of A) that pull back to the function 0(on π−1 (Spec A)).
8.3.5. Corollary. — Under the hypotheses of Theorem 8.3.4, the closure of the settheoretic image of π is the underlying set of the scheme-theoretic image.
(Example 4 above shows that we cannot excise these hypotheses.)
8.3.6. In particular, if the set-theoretic image is closed (e.g., if π is finite or projective), the set-theoretic image is the underlying set of the scheme-theoretic image,as promised in Example 3 above.

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8.3.7. Proof of Corollary 8.3.5. The set-theoretic image is in the underlying setof the scheme-theoretic image. (Check this!) The underlying set of the schemetheoretic image is closed, so the closure of the set-theoretic image is contained in
the underlying set of the scheme-theoretic image. On the other hand, if U is thecomplement of the closure of the set-theoretic image, π−1 (U) = ∅. Under thesehypotheses, the scheme theoretic image can be computed locally, so the schemetheoretic image is the empty set on U.
!
We conclude with a few stray remarks.
8.3.A. E ASY E XERCISE . If X is reduced, show that the scheme-theoretic image of
π : X → Y is also reduced.

More generally, you might expect there to be no unnecessary nonreduced
structure on the image not forced by nonreduced structure on the source. We
make this precise in the locally Noetherian case, when we can talk about associated points.
8.3.B. ⋆ U NIMPORTANT EXERCISE . If π : X → Y is a quasicompact morphism of
locally Noetherian schemes, show that the associated points of the image subscheme
are a subset of the image of the associated points of X. (The example of#Spec
C[t]/(t − a) → Spec C[t] shows what can go wrong if you give up quaa∈Csicompactness — note that reducedness of the source doesn’t help.) Hint: reduceto the case where X and Y are affine. (Can you develop your geometric intuition
so that this becomes plausible to you?)
8.3.8. Scheme-theoretic closure of a locally closed subscheme.
We define the scheme-theoretic closure of a locally closed embedding π : X →Y as the scheme-theoretic image of π. (A shorter phrase for this is schematic closure,although this more elegant nomenclature has not caught on.)
8.3.C. E XERCISE . If a locally closed embedding V → X is quasicompact (e.g., if
V is Noetherian, Exercise 7.3.B(a)), or if V is reduced, show that (iii) implies (i)and (ii) in Exercise 8.1.M. Thus in this fortunate situation, a locally closed embedding can be thought of in three different ways, whichever is convenient. (Hint:
Corollary 8.3.5.)
8.3.D. U NIMPORTANT EXERCISE , USEFUL FOR INTUITION . If π : X → Y is a locallyclosed embedding into a locally Noetherian scheme (so X is also locally Noetherian), then the associated points of the scheme-theoretic closure are (naturally inbijection with) the associated points of X. (Hint: Exercise 8.3.B.) Informally, we get
no nonreduced structure on the scheme-theoretic closure not “forced by” that on
X.
8.3.9. The (reduced) subscheme structure on a closed subset.
Suppose Xset is a closed subset of a scheme Y. Then we can define a canonicalscheme structure X on Xset that is reduced. We could describe it as being cut out
by those functions whose values are zero at all the points of Xset . On the affine
open set Spec B of Y, if the set Xset corresponds to the radical ideal I = I(Xset )(recall the I(·) function from §3.7), the scheme X corresponds to Spec B/I. You canquickly check that this behaves well with respect to any distinguished inclusion

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Spec Bf "→ Spec B. We could also consider this construction as an example of ascheme-theoretic image in the following crazy way: let W be the scheme that is adisjoint union of all the points of Xset , where the point corresponding to p in Xset
is Spec of the residue field of OY,p . Let ρ : W → Y be the “canonical” map sending“p to p”, and giving an isomorphism on residue fields. Then the scheme structure
on X is the scheme-theoretic image of ρ. A third definition: it is the smallest closed
subscheme whose underlying set contains Xset .
8.3.E. E XERCISE . Show that all three definitions are the same.
This construction is called the (induced) reduced subscheme structure on theclosed subset Xset . (Vague exercise: Make a definition of the reduced subscheme
structure precise and rigorous to your satisfaction.)
8.3.F. E XERCISE . Show that the underlying set of the induced reduced subscheme
X → Y is indeed the closed subset Xset . Show that X is reduced.

8.3.10. Reduced version of a scheme.
In the main interesting case where Xset is all of Y, we obtain a reduced closed
subscheme Y red → Y, called the reduction of Y. On the affine open subset Spec B "→Y, Y red "→ Y corresponds to the nilradical N(B) of B. The reduction of a scheme is
the “reduced version” of the scheme, and informally corresponds to “shearing off
the fuzz”.
An alternative equivalent definition: on the affine open subset Spec B "→ Y,
the reduction of Y corresponds to the ideal N(B) ⊂ B of nilpotents. As for any
f ∈ B, N(B)f = N(Bf ), by Exercise 8.1.H this defines a closed subscheme.
8.3.G. E XERCISE . Show that this alternative definition is indeed equivalent to theactual one.
8.3.11. ⋆ Caution/Example. It is not true that for every open subset# U ⊂ Y, Γ (U,nOY red )is Γ (U, OY ) modulo its nilpotents. For example, on Y =Spec k[x]/(x ), thefunction x is not nilpotent, but is 0 on Y red , as it is “locally nilpotent”. This may
remind you of Example 4 after Definition 8.3.2.
8.3.12. Scheme-theoretic support of a quasicoherent sheaf. Similar ideas areused in the definition of the scheme-theoretic support of a quasicoherent sheaf,
see Exercise 18.9.B.

8.4 Effective Cartier divisors, regular sequences and regular
embeddings
We now introduce regular embeddings, an important class of locally closed
embeddings. Locally closed embeddings of regular schemes in regular schemes
are one important example of regular embeddings (Exercise 12.2.L(b)). Effective
Cartier divisors, basically the codimension 1 case, will turn out to be repeatedly
useful as well — to see how useful, notice how often it appears in the index. We
begin with this case.

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8.4.1. Locally principal closed subschemes, and effective Cartier divisors. (Themotivating example to keep in mind of a locally principal closed subscheme is theclosed subscheme cut out by x2 = 0 in P2 , defined in Example 8.2.1.) A closed subscheme is locally principal if on each open set in a small enough open cover it iscut out by a single equation (i.e., by a principal ideal, hence the terminology). Morespecifically, a locally principal closed subscheme is a closed embedding π : X → Y
for which there is an open cover {Ui : i ∈ I} of Y for which for each i, the restricted
morphism π−1 (Ui ) → Ui is (isomorphic as a Ui -scheme to) the closed subscheme
V(si ) of Ui for some si ∈ Γ (Ui , OX ). In particular, if π : X → Y is a locally principal
closed subscheme, then the open cover of Y can be chosen to be affine opens: Simply cover each open set Ui by open affines, and restrict each si to each affine open
subset.
For example, hypersurfaces in Pnk (Definition 8.2.2) are locally principal: eachhomogeneous polynomial in n + 1 variables defines a locally principal closed
subscheme of Pn
A . (Warning: unlike “local principality”, “principality” is not anaffine-local condition, see §19.11.10! Also, the example of a projective hypersurface, §8.2.1, shows that a locally principal closed subscheme need not be cut out by
a globally-defined function.)If the ideal sheaf is locally generated near every point by a function that is
not a zerodivisor, we call the closed subscheme an effective Cartier divisor. Moreprecisely: if π : X → Y is a closed embedding, and there is a cover Y by affine
open subsets Spec Ai ⊂ Y, and there exist non-zerodivisors ti ∈ Ai with V(ti ) =X|Spec Ai (scheme-theoretically — i.e., the ideal sheaf of X over Spec Ai is generated
by ti ), then we say that X is an effective Cartier divisor on Y. (We will not explain
the origin of the phrase, as it is not relevant for this point of view.)
8.4.A. E XERCISE . Suppose t ∈ A is a non-zerodivisor. Show that t is a nonzerodivisor in Ap for each prime p.
8.4.2. Caution. If D is an effective Cartier divisor on an affine scheme Spec A,
it is not necessarily true that D = V(t) for some t ∈ A (see Exercise 14.2.M —§19.11.10 gives a different flavor of example). In other words, the condition of aclosed subscheme being an effective Cartier divisor can be verified on an affine
cover, but cannot be checked on an arbitrary affine cover — it is not an affine-local
condition in this obvious a way.
8.4.B. E XERCISE . Suppose X is a locally Noetherian scheme, and t ∈ Γ (X, OX ) isa function on it. Show that t (or more precisely the closed subscheme V(t)) is aneffective Cartier divisor if and only if it doesn’t vanish on any associated point of
X.
8.4.C. U NIMPORTANT E XERCISE . Suppose V(t) = V(t ′ ) "→ Spec A is an effective
Cartier divisor, with t and t ′ non-zerodivisors in A. Show that t is an invertible
function times t ′ .
The idea of an effective Cartier divisor leads us to the notion of regular sequences. (We will close the loop in Exercise 8.4.H, where we will interpret effective
Cartier divisors on any reasonable scheme as regular embeddings of codimension
1.)

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8.4.3. Regular sequences.
The notion of “regular sequence”, like so much else, is due to Serre, [Se2]. The
definition of regular sequence is the algebraic version of the following geometricidea: locally, we take an effective Cartier divisor (a non-zerodivisor); then an effective Cartier divisor on that; then an effective Cartier divisor on that; and so on,a finite number of times. A little care is necessary; for example, we might want
this to be independent of the order of the equations imposed, and this is true only
when we say this in the right way.
We make the definition of regular sequence for a ring A, and more generally
for an A-module M.
8.4.4. Definition. If M is an A-module, a sequence x1 , . . . , xr ∈ A is called an Mregular sequence (or a regular sequence for M) if the following two conditions
are satisfied.
(i) For each i, xi is not a zerodivisor for M/(x1 , . . . , xi−1 )M. (The case i =1 should be interpreted as: “x1 is not a zerodivisor of M.” Recall thatzerodivisors were defined in Exercise 1.3.C.)(ii) We have a proper inclusion (x1 , . . . , xr )M " M.
In the case most relevant to us, when M = A, this should be seen as a reasonable approximation of a “complete intersection”, and indeed we will use this as
the definition (§8.4.7). An A-regular sequence is just called a regular sequence.
8.4.D. E XERCISE . If M is an A-module, show that an M-regular sequence continues to satisfy condition (i) of the Definition 8.4.4 of regular sequence upon any
localization.
8.4.E. E XERCISE . If x, y is an M-regular sequence, show that xN , y is an Mregular sequence. Hint: the difficult part is showing that y is not a zerodivisor ofM/(xN M). Show this by induction on N. If y is a zerodivisor of M/(xN M), thenym ≡ 0 (mod xN ) for some m ∈ M \ xN M. Hence ym = xN k for some k ∈ M.
Use the fact that x, y is a regular sequence to show that m is a multiple of x. (Yourargument will easily extend to show more generally that if x1 , . . . , xn is a regularan1sequence, and a1 , . . . , an ∈ Z+ , then xa
1 , . . . , xn is a regular sequence.)
8.4.5. Interesting example. We now give an example ([E, Example 17.3]) showingthat the order of a regular sequence matters. Suppose A = k[x, y, z]/(x−1)z, so X =Spec A is the union of the z = 0 plane and the x = 1 plane — X is reduced and has
two components (see Figure 8.5). You can readily verify that x is a non-zerodivisorof A (x = 0 misses one component of X, and doesn’t vanish entirely on the other),and that the corresponding effective Cartier divisor X ′ = Spec k[x, y, z]/(x, z) (onX) is integral. Then (x − 1)y gives an effective Cartier divisor on X ′ (it doesn’t
vanish entirely on X ′ ), so x, (x−1)y is a regular sequence for A. However, (x−1)y is
not a non-zerodivisor of A, as it does vanish entirely on one of the two components.
Thus (x − 1)y, x is not a regular sequence. The reason that reordering the regularsequence x, (x − 1)y ruins regularity is clear: there is a locus on which (x − 1)y isn’t
effective Cartier, but it disappears if we enforce x = 0 first. The problem is one of“nonlocality” — “near” x = y = z = 0 there is no problem. This may motivate
the fact that in the (Noetherian) local situation, this problem disappears. We now
make this precise.

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F IGURE 8.5. Order matters in a regular sequence (in the “nonlocal” situation), for silly reasons
8.4.6. Theorem. — Suppose (A, m) is a Noetherian local ring, and M is a finitely
generated A-module. Then any M-regular sequence (x1 , . . . , xr ) in m remains a regularsequence upon any reordering.
(In [Di], Dieudonné gives a half-page example showing that Noetherian hypotheses are necessary in Theorem 8.4.6.)
Before proving Theorem 8.4.6 (in Exercise 8.4.F), we prove the first nontrivial
case, when r = 2. This discussion is secretly a baby case of the Koszul complex.
Suppose x, y ∈ m, and x, y is an M-regular sequence. Translation: x ∈ m is a
non-zerodivisor on M, and y ∈ m is a non-zerodivisor on M/xM.
Consider the double complex(8.4.6.1)

M
7

×(−x)

%M
7

×y

M

×y
×x

%M

where the bottom left is considered to be in position (0, 0). (The only reason forthe minus sign in the top row is solely our arbitrary preference for anti-commuting
rather than commuting squares in §1.7.1, but it really doesn’t matter.)
We compute the cohomology of the total complex using a (simple) spectral
sequence, beginning with the rightward orientation. (The gratuitous use of spectral sequences here, as in many of our other applications, is overkill; we do thispartially in order to get practice with the machine.) On the first page, we have
(0 : (x))
7

M/xM
7

×y

(0 : (x))

×y

M/xM

The entries (0 : (x)) in the first column are 0, as x is a non-zerodivisor on M. Taking
homology in the vertical direction to obtain the second page, we find(8.4.6.2)

0M/(x, y)M00

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using the fact that y is a non-zerodivisor on M/xM. The sequence clearly converges here. Thus the original double complex (8.4.6.1) only has nonzero cohomology in degree 2, where it is M/(x, y)M.
Now we run the spectral sequence on (8.4.6.1) using the upward orientation.
The first page of the sequence is:
M/yM

×(−x)

(0 : (y))×x% M/yM% (0 : (y))

The sequence must converge to (8.4.6.2) after the next step. From the top row, we
see that multiplication by x must be injective on M/yM, so x is a non-zerodivisor
on M/yM. From the bottom row, multiplication by x gives an isomorphism of(0 : (y)) with itself. As x ∈ m, by version 2 of Nakayama’s Lemma (Lemma 7.2.9),
this implies that (0 : (y)) = 0, so y is a non-zerodivisor on M. Thus we have shownthat y, x is a regular sequence on M — the r = 2 case of Theorem 8.4.6.
8.4.F. E ASY E XERCISE . Prove Theorem 8.4.6. (Hint: show it first in the case ofa reordering where only two adjacent xi are swapped, using the r = 2 case justdiscussed.) Where are the Noetherian hypotheses used?
8.4.7. Regular embeddings.
Suppose π : X "→ Y is a locally closed embedding. We say that π is a regular
embedding (of codimension r) at a point p ∈ X if in the local ring OY,p , the idealof X is generated by a regular sequence (of length r). We say that π is a regular
embedding (of codimension r) if it is a regular embedding (of codimension r) at
all p ∈ X. (Another reasonable name for a regular embedding might be “local
complete intersection”. Unfortunately, “local complete intersection morphism”,
or “lci morphism”, is already used for a related notion, see [Stacks, tag 068E].)
Our terminology uses the word “codimension”, which we have not defined asa word on its own. The reason for using this word will become clearer once you
meet Krull’s Principal Ideal Theorem 11.3.3 and Krull’s Height Theorem 11.3.7.
8.4.G. E XERCISE ( THE CONDITION OF A LOCALLY CLOSED EMBEDDING BEING A
REGULAR EMBEDDING IS OPEN ). Show that if a locally closed embedding π : X "→Y of locally Noetherian schemes is a regular embedding at p, then it is a regular
embedding in some open neighborhood of p in X. Hint: reduce to the case where
π is a closed embedding, and then where Y (hence X) is affine — say Y = Spec B,
X = Spec B/I, and p = [p] — and there are f1 , . . . , fr such that in OY,p , the imagesof the fi are a regular sequence generating Ip . We wish to show that (f1 , . . . , fr ) = I“in an open neighborhood of p”. Prove the following fact in algebra: if I and J are
ideals of a Noetherian ring A, and p ⊂ A is a prime ideal such that Ip = Jp , showthat there exists a ∈ A \ p such that Ia = Ja in Aa . To do this, show that it suffices
to consider the special case I ⊂ J, by considering I ∩ J and J instead of I and J. To
show this special case, let K = J/I, a finitely generated module, and show that if
Kp = 0 then Ka = 0 for some a ∈ A \ p.

Hence if X is quasicompact, then to check that a closed embedding π is a regular embedding it suffices to check at closed points of X.

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Exercise 12.1.F(b) will show that not all closed embeddings are regular embeddings.
8.4.H. E XERCISE . Show that a closed embedding X "→ Y of locally Noetherian
schemes is a regular embedding of codimension 1 if and only if X is an effective
Cartier divisor on Y. Unimportant remark: the Noetherian hypotheses can be replaced by requiring OY to be coherent, and essentially the same argument applies.
It is interesting to note that “effective Cartier divisor” implies “regular embeddingof codimension 1” always, but that the converse argument requires Noetherian(like) assumptions. (See [MO129242] for a counterexample to the converse.)
8.4.8. Definition. A codimension r complete intersection in a scheme Y is a
closed subscheme X that can be written as the scheme-theoretic intersection of reffective Cartier divisors D1 , . . . , Dr , such that at every point p ∈ X, the equations corresponding to D1 , . . . , Dr form a regular sequence. The phrase complete
intersection means “codimension r complete intersection for some r”.

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CHAPTER 9

Fibered products of schemes, and base change

9.1 They exist
Before we get to products, we note that coproducts exist in the category ofschemes: just as with the category of sets (Exercise 1.3.T), coproduct is disjoint
union. The next exercise makes this precise (and directly extends to coproducts ofan infinite number of schemes).
#
9.1.A. E ASY E XERCISE . Suppose X and Y are schemes. Let X Y be the scheme
whose underlying topological space is the disjoint union of the topological spacesof X and Y, and with structure sheaf# on (the part corresponding to) X given by OX ,and similarly for Y. Show thatX
Y is the coproduct of X and Y (justifying the#use of the coproduct symbol ).
We will now construct the fibered product in the category of schemes.

9.1.1. Theorem: Fibered products exist. — Suppose α : X → Z and β : Y → Z aremorphisms of schemes. Then the fibered product
X ×Z Y

α′

β′

%Y
β

"
X

α

"
%Z

exists in the category of schemes.
Note: if A is a ring, people often sloppily write ×A for ×Spec A . If B is an Aalgebra, and X is an A-scheme, people often write XB or X×A B for X×Spec A Spec B.
9.1.2. Warning: products of schemes aren’t products of sets. Before showing existence,here is a warning: the product of schemes isn’t a product of sets (and more generally for fibered products). We have made a big deal about schemes being sets,endowed with a topology, upon which we have a structure sheaf. So you might
think that we will construct the product in this order. But we won’t, because products behave oddly on the level of sets. You may have checked (Exercise 6.6.E(a))
that the product of two affine lines over your favorite algebraically closed field k∼ A2 . But the underlying set of the latter is not theis the affine plane: A1k ×k A1k =kunderlying set of the former —- we get additional points, corresponding to curves
in A2 that are not lines parallel to the axes!
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9.1.3. On the other hand, W-valued points (where W is a scheme, Definition 6.3.9)
do behave well under (fibered) products (as mentioned in §6.3.10). This is just theuniversal property definition of fibered product: an W-valued point of a scheme
X is defined as an element of Hom(W, X), and the fibered product is defined (via
Yoneda, see §1.3.1) by(9.1.3.1)

Hom(W, X ×Z Y) = Hom(W, X) ×Hom(W,Z) Hom(W, Y).

This is one justification for making the definition of scheme-valued point. For
this reason, those classical people preferring to think only about varieties overan algebraically closed field k (or more generally, finite type schemes over k), andpreferring to understand them through their closed points — or equivalently, the kvalued points, by the Nullstellensatz (Exercise 5.3.F) — needn’t worry: the closed
points of the product of two finite type k-schemes over k are (naturally identifiedwith) the product of the closed points of the factors. This will follow from the factthat the product is also finite type over k, which we verify in Exercise 9.2.D. Thisis one of the reasons that varieties over algebraically closed fields can be easier
to work with. But over a nonalgebraically closed field, things become even moreinteresting; Example 9.2.3 is a first glimpse.
(Fancy remark: You may feel that (i) “products of topological spaces are products on the underlying sets” is natural, while (ii) “products of schemes are notnecessarily products on the underlying sets” is weird. But really (i) is the lucky
consequence of the fact that the underlying set of a topological space can be interpreted as set of p-valued points, where p is a point, so it is best seen as a consequence of paragraph 9.1.3, which is the “more correct” — i.e., more general —fact.)
9.1.4. Warning on Noetherianness. The fibered product of Noetherian schemes neednot be Noetherian. You will later be able to verify that Exercise 9.2.E gives an example, i.e., that A := Q ⊗Q Q is not Noetherian, as follows. By Exercise 11.1.G(a),dim A = 0. A Noetherian dimension 0 scheme has a finite number of points (Exercise 11.1.C). But by Exercise 9.2.E, Spec A has an infinite number of points.
On the other hand, the fibered product of finite type k-schemes over finite
type k-schemes is a finite type k-scheme (Exercise 9.2.D), so this pathology does
not arise for varieties.
9.1.5. Philosophy behind the proof of Theorem 9.1.1. The proof of Theorem 9.1.1
can be confusing. The following comments may help a little.
We already basically know existence of fibered products in two cases: the casewhere X, Y, and Z are affine (stated explicitly below), and the case where β : Y → Z
is an open embedding (Exercise 7.1.B).

9.1.B. E XERCISE ( PROMISED IN R EMARK 6.3.5). Use Exercise 6.3.F (HomSch (W, Spec A) =HomRings (A, Γ (W, OW ))) to show that given ring maps C → A and C → B,
∼ Spec A ×Spec C Spec B.
Spec(A ⊗C B) =(Interpret tensor product as the “fibered coproduct” in the category of rings.) Hencethe fibered product of affine schemes exists (in the category of schemes). (This generalizes the fact that the product of affine lines exist, Exercise 6.6.E(a).)

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The main theme of the proof of Theorem 9.1.1 is that because schemes arebuilt by gluing affine schemes along open subsets, these two special cases will be
all that we need. The argument will repeatedly use the same ideas — roughly,that schemes glue (Exercise 4.4.A), and that morphisms of schemes glue (Exercise 6.3.A). This is a sign that something more structural is going on; §9.1.6 describes this for experts.
Proof of Theorem 9.1.1. The key idea is this: we cut everything up into affine open
sets, do fibered products there, and show that everything glues nicely. The conceptually difficult part of the proof comes from the gluing, and the realization that wehave to check almost nothing. We divide the proof up into a number of bite-sizedpieces.
Step 1: fibered products of affine with “almost-affine” over affine. We begin by combining the affine case with the open embedding case as follows. Suppose X and Z" ι % ′
% Z where ι is an open embedare affine, and β : Y → Z factors as Y !
Yding and Y ′ is affine. Then X ×Z Y exists. This is because if the two small squares
of

W" #
open

% Y"
#

open

"
W′

"
% Y′

"
X

"
%Z

are fibered diagrams, then the “outside rectangle” is also a fibered diagram. (Thiswas Exercise 1.3.Q, although you should be able to see this on the spot.) It will be
important to remember (from Important Exercise 7.1.B) that “open embeddings”
are “preserved by fibered product”: the fact that Y → Y ′ is an open embedding
implies that W → W ′ is an open embedding.
Key Step 2: fibered product of affine with arbitrary over affine exists. We now cometo the key part of the argument: if X and Z are affine, and Y is arbitrary. This isconfusing when you first see it, so we first deal with a special case, when Y is theunion of two affine open sets Y1 ∪ Y2 . Let Y12 = Y1 ∩ Y2 .
Now for i = 1 and 2, X ×Z Yi exists by the affine case, Exercise 9.1.B. Callthis Wi . Also, X ×Z Y12 exists by Step 1 (call it W12 ), and comes with canonicalopen embeddings into W1 and W2 (by construction of fibered products with open
embeddings, see the last sentence of Step 1). Thus we can glue W1 to W2 along
W12 ; call this resulting scheme W.
We check that the result is the fibered product by verifying that it satisfies theuniversal property. Suppose we have maps α ′′ : V → X, β ′′ : V → Y that compose(with α and β respectively) to the same map V → Z. We need to construct a unique

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map γ : V → W, so that α ′ ◦ γ = β ′′ and β ′ ◦ γ = α ′′ .
(9.1.5.1)

V✵❆000✵✵❆❆ 000 β ′′00✵∃!γ?
✵✵ ❆❆❆ 00000
004
✵✵ 1
%Y
α ′′ ✵✵ W α ′✵✵β✵✵ β ′E""
X α %ZFor i = 1, 2, define Vi := (β ′′ )−1 (Yi ). Define V12 := (β ′′ )−1 (Y12 ) = V1 ∩ V2 . Then
there is a unique map Vi → Wi such that the composed maps Vi → X and Vi → Yi
are as desired (by the universal property of the fibered product X ×Z Yi = Wi ),hence a unique map γi : Vi → W. Similarly, there is a unique map γ12 : V12 → Wsuch that the composed maps V12 → X and V12 → Y are as desired. But the
restriction of γi to V12 is one such map, so it must be γ12 . Thus the maps γ1 andγ2 agree on V12 , and glue together to a unique map γ : V → W. We have shown
existence and uniqueness of the desired γ.
We have thus shown that if Y is the union of two affine open sets, and X andZ are affine, then X ×Z Y exists.
We now tackle the general case. (You may prefer to first think through thecase where “two” is replaced by “three”.) We now cover Y with open sets Yi , asi runs over some index set (not necessarily finite!). As before, we define Wi and
Wij . We can glue these together to produce a scheme W along with open sets
we identify with Wi (Exercise 4.4.A — you should check the triple intersection“cocycle” condition).
As in the two-affine case, we show that W is the fibered product by showing that it satisfies the universal property. Suppose we have maps α ′′ : V → X,
β ′′ : V → Y that compose to the same map V → Z. We construct a unique map
γ : V → W, so that α ′ ◦ γ = β ′′ and β ′ ◦ γ = α ′′ . Define Vi = (β ′′ )−1 (Yi ) and
Vij := (β ′′ )−1 (Yij ) = Vi ∩ Vj . Then there is a unique map Vi → Wi such thatthe composed maps Vi → X and Vi → Yi are as desired, hence a unique map
γi : Vi → W. Similarly, there is a unique map γij : Vij → W such that the composed maps Vij → X and Vij → Y are as desired. But the restriction of γi to Vij
is one such map, so it must be γij . Thus the maps γi and γj agree on Vij . Thus
the γi glue together to a unique map γ : V → W. We have shown existence anduniqueness of the desired γ, completing this step.
Step 3: Z affine, X and Y arbitrary. We next show that if Z is affine, and X andY are arbitrary schemes, then X ×Z Y exists. We just follow Step 2, with the rolesof X and Y reversed, using the fact that by the previous step, we can assume thatthe fibered product of an affine scheme with an arbitrary scheme over an affine
scheme exists.
Step 4: Z “almost-affine”, X and Y arbitrary. This is akin to Step 1. Let Z "→ Z ′be an open embedding into an affine scheme. Then X ×Z Y satisfies the universal
property of X ×Z ′ Y.
Step 5: the general case. We employ the same trick yet again. Suppose α : X → Z,
β : Y → Z are two morphisms of schemes. Cover Z with affine open subschemes
Zi , and let Xi = α−1 (Zi ) and Yi = β−1 (Zi ). Define Zij := Zi ∩ Zj , Xij := α−1 (Zij ),and Yij := β−1 (Zij ). Then Wi := Xi ×Zi Yi exists for all i (Step 3), and Wij :=Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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Xij ×Zij Yij exists for all i, j (Step 4), and for each i and j, Wij comes with a canonically open immersion into both Wi and Wj (see the last sentence in Step 1). As Wi
satisfies the universal property of X ×Z Yi (do you see why?), we may canonically
identify Wi (which we know to exist by Step 3) with X×Z Yi . Similarly, we identify
Wij with X ×Z Yij .
We then proceed exactly as in Step 2: the Wi ’s can be glued together along the
Wij (the cocycle condition can be readily checked to be satisfied), and W can bechecked to satisfy the universal property of X×Z Y (again, exactly as in Step 2). !
9.1.6. ⋆⋆ Describing the existence of fibered products using the high-falutin’
language of representable functors. The proof above can be described morecleanly in the language of representable functors (§6.6). This will be enlightening
only after you have absorbed the above argument and meditated on it for a longtime. It may be most useful to shed light on representable functors, rather than on
the existence of the fibered product.
Until the end of §9.1 only, by functor, we mean contravariant functor from the category Sch of schemes to the category of Sets. For each scheme X, we have a functor hX ,taking a scheme Y to the set Mor(Y, X) (§1.2.20). Recall (§1.3.10, §6.6) that a functor
is representable if it is naturally isomorphic to some hX . If a functor is representable,
then the representing scheme is unique up to unique isomorphism (Exercise 6.6.C).
This can be usefully extended as follows:
9.1.C. E XERCISE (Yoneda’s Lemma). If X and Y are schemes, describe a bijection
between morphisms of schemes X → Y and natural transformations of functors
hX → hY . Hence show that the category of schemes is a fully faithful subcategory(§1.2.15) of the “functor category” (all contravariant functors Sch → Sets). Hint:
this has nothing to do with schemes; your argument will work in any category.
This is the contravariant version of Exercise 1.3.Z(c).
One of Grothendieck’s insights is that we should try to treat such functors as“geometric spaces”, without worrying about representability. Many notions carryover to this more general setting without change, and some notions are easier. For
example, fibered products of functors always exist: h ×h ′′ h ′ may be defined by(h ×h ′′ h ′ )(W) = h(W) ×h ′′ (W) h ′ (W),where the fibered product on the right is a fibered product of sets, which always
exists. (This isn’t quite enough to define a functor; we have only described where
objects go. You should work out where morphisms go too.) We didn’t use anything about schemes; this works with Sch replaced by any category.
Then “X ×Z Y exists” translates to “hX ×hZ hY is representable”.
9.1.7. Representable functors are Zariski sheaves. Because “morphisms to schemesglue” (Exercise 6.3.A), we have a necessary condition for a functor to be representable. We know that if {Ui } is an open cover of Y, a morphism Y → X is determined by its restrictions Ui → X, and given morphisms Ui → X that agree on theoverlap Ui ∩ Uj → X, we can glue them together to get a morphism Y → X. In thelanguage of equalizer exact sequences (§2.2.7),·% hX (Y)% " hX (Ui )%% " h (U ∩ U )Xij

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The Rising Sea: Foundations of Algebraic Geometryis exact. Thus morphisms to X (i.e., the functor hX ) form a sheaf on every scheme
Y. If this holds, we say that the functor is a Zariski sheaf. (You can impress yourfriends by telling them that this is a sheaf on the big Zariski site.) We can repeat this
discussion with Sch replaced by the category SchS of schemes over a given basescheme S. We have proved (or observed) that in order for a functor to be representable,
it is necessary for it to be a Zariski sheaf.
The fibered product passes this test:
9.1.D. E XERCISE . If X, Y → Z are schemes, show that hX ×hZ hY is a Zariski sheaf.
(Do not use the fact that hX ×hZ hY is representable! The point of this section is torecover representability from a more sophisticated perspective.)
We can make some other definitions that extend notions from schemes tofunctors. We say that a map (i.e., natural transformation) of functors h ′ → h expresses h ′ as an open subfunctor of h if for all representable functors hX and maps
hX → h, the fibered product hX ×h h ′ is representable, by U say, and hU → hX
corresponds to an open embedding of schemes U → X. The following Cartesiansquare may help.
hU
"
h′

open

% hX
"
%h

9.1.E. E XERCISE . Show that a map of representable functors hW → hZ is an open
subfunctor if and only if W → Z is an open embedding, so this indeed extends thenotion of open embedding to (contravariant) functors (Sch → Sets).

9.1.F. E XERCISE ( THE GEOMETRIC NATURE OF THE NOTION OF “ OPEN SUBFUNC -TOR ”).

(a) Show that an open subfunctor of an open subfunctor is also an open subfunctor.
(b) Suppose h ′ → h and h ′′ → h are two open subfunctors of h. Define the intersection of these two open subfunctors, which should also be an open subfunctorof h.
(c) Suppose U and V are two open subschemes of a scheme X, so hU → hX andhV → hX are open subfunctors. Show that the intersection of these two open
subfunctors is, as you would expect, hU∩V .

9.1.G. E XERCISE . Suppose α : X → Z and β : Y → Z are morphisms of schemes,and U ⊂ X, V ⊂ Y, W ⊂ Z are open embeddings, where U and V map to W.
Interpret hU ×hW hV as an open subfunctor of hX ×hZ hY . (Hint: given a map
hT → hX×Z Y , what open subset of T should correspond to U ×W V?)

A collection hi of open subfunctors of h is said to cover h if for every map
hX → h from a representable subfunctor, the corresponding open subsets Ui "→ X
cover X.
Given that functors do not have an obvious underlying set (let alone a topology), it is rather amazing that we are talking about when one is an “open subset”
of another, or when some functors “cover” another!

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9.1.H. E XERCISE . Suppose {Zi }i is an affine cover of Z, {Xij }j is an affine cover ofthe preimage of Zi in X, and {Yik }k is an affine cover of the preimage of Zi in Y.
Show that {hXij ×hZi hYik }ijk is an open cover of the functor hX ×hZ hY . (Hint:
consider a map hT → hX ×hZ hY , and extend your solution to Exercise 9.1.G.)

We now come to a key point: a Zariski sheaf that is “locally representable”
must be representable:
9.1.I. K EY E XERCISE . If a functor h is a Zariski sheaf that has an open cover byrepresentable functors (“is covered by schemes”), then h is representable. (Hint:
use Exercise 4.4.A to glue together the schemes representing the open subfunctors.)
This immediately leads to the existence of fibered products as follows. Exercise 9.1.D shows that hX ×hZ hY is a Zariski sheaf. But hXij ×hZi hYik is representable for each i, j, k (fibered products of affines over an affine exist, Exercise 9.1.B), and these functors are an open cover of hX ×hZ hY by Exercise 9.1.H, so
by Key Exercise 9.1.I we are done.

9.2 Computing fibered products in practice
Before giving some examples, we first see how to compute fibered products
in practice. There are four types of morphisms (1)–(4) that it is particularly easy
to take fibered products with, and all morphisms can be built from these atomic
components. More precisely, (1) will imply that we can compute fibered products
locally on the source and target. Thus to understand fibered products in general,
it suffices to understand them on the level of affine sets, i.e., to be able to compute
A ⊗B C given ring maps B → A and B → C. Any map B → A (and similarly
B → C) may be expressed as B → B[t1 , . . . ]/I, so if we know how to base change by“adding variables” (2) and “taking quotients” (3), we can “compute” any fiberedproduct (at least in theory). The fourth type of morphism (4), corresponding tolocalization, is useful to understand explicitly as well.
(1) Base change by open embeddings.
We have already done this (Exercise 7.1.B), and we used it in the proof thatfibered products of schemes exist.
(2) Adding an extra variable.
∼ A[t], so the following is a
9.2.A. E ASY ALGEBRA EXERCISE . Show that A ⊗B B[t] =fibered diagram. (Your argument might naturally extend to allow the addition of
infinitely many variables, but we won’t need this generality.) Hint: show that A[t]satisfies an appropriate universal property.
Spec A[t]% Spec B[t]"Spec A"% Spec B

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9.2.1. Definition (affine space over an arbitrary scheme). If X is any scheme, we definennAnX as X ×Z Spec Z[x1 , . . . , xn ]. Clearly ASpec A is canonically the same as AA as
defined in Example 8 of §3.2.3.

(3) Base change by closed embeddings.

9.2.B. E XERCISE . Suppose φ : B → A is a ring morphism, and I ⊂ B is an ideal.
Let Ie := ⟨φ(i)⟩i∈I ⊂ A be the extension of I to A. Describe a natural isomorphism∼ A ⊗B (B/I). (Hint: consider I → B → B/I → 0, and use the right-exactness
A/Ie =
of ⊗B A, Exercise 1.3.H.)

9.2.2. As an immediate consequence: the fibered product with a closed subscheme
is a closed subscheme of the fibered product in the obvious way. We say that“closed embeddings are preserved by base change”.

9.2.C. E XERCISE .
(a) Interpret the intersection of two closed embeddings into X (cf. Exercise 8.1.J) as
their fibered product over X.
(b) Show that “locally closed embeddings” are preserved by base change.
(c) Define the intersection of n locally closed embeddings Xi "→ Z (1 ≤ i ≤ n) by
the fibered product of the Xi over Z (mapping to Z). Show that the intersection of(a finite number of) locally closed embeddings is also a locally closed embedding.
As an application of Exercise 9.2.B, we can compute tensor products of finitely
generated k-algebras over k. For example, we have
∼ k[x1 , x2 , y1 , y2 ]/(x2 − x2 , y3 + y3 ).
k[x1 , x2 ]/(x2 − x2 ) ⊗k k[y1 , y2 ]/(y3 + y3 ) =
1

1

2

1

1

2

9.2.D. E XERCISE . Suppose X and Y are locally of finite type A-schemes. Show that
X ×A Y is also locally of finite type over A. Prove the same thing with “locally”
removed from both the hypothesis and conclusion.
9.2.3. Example. We can use these ideas to compute C ⊗R C:∼ C ⊗R (R[x]/(x2 + 1))
C ⊗R C =
∼
=
∼
=
∼
=
∼
=

(C ⊗R R[x])/(x2 + 1)

by §9.2(3)C[x]/(x + 1)
by §9.2(2)C[x]/ ((x − i)(x + i))
2

C[x]/(x − i) × C[x]/(x + i)by the Chinese Remainder Theorem
C×C∼ Spec C # Spec C. This example is the first example ofThus Spec C ×R Spec C =many different behaviors. Notice for example that two points somehow correspond to the Galois group of C over R; for one of them, x (the “i” in one of thecopies of C) equals i (the “i” in the other copy of C), and in the other, x = −i.
∼=

9.2.4. ⋆ Remark. Here is a clue that there is something deep going on behind Example 9.2.3. If L/K is a (finite) Galois extension with Galois group G, then L ⊗K L
is isomorphic to LG (the product of |G| copies of L). This turns out to be a restatement of the classical form of linear independence of characters! In the language of

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schemes, Spec L ×K Spec L is a union of a number of copies of Spec L that naturally
form a torsor over the Galois group G; but we will not define torsor here.
9.2.E. ⋆ H ARD BUT FASCINATING EXERCISE FOR THOSE FAMILIAR WITH Gal(Q/Q).
Show that the points of Spec Q ⊗Q Q are in natural bijection with Gal(Q/Q), andthe Zariski topology on the former agrees with the profinite topology on the latter.
(Some hints: first do the case of finite Galois extensions. Relate the topology on
Spec of a direct limit of rings to the inverse limit of Specs. Can you see which point
corresponds to the identity of the Galois group?)
At this point, we can compute any A⊗B C (where A and C are B-algebras): any
map of rings φ : B → A can be interpreted by adding variables (perhaps infinitely
many) to B, and then imposing relations. But in practice (4) is useful, as we will
see in examples.
(4) Base change of affine schemes by localization.
9.2.F. E XERCISE . Suppose φ : B → A is a ring morphism, and S ⊂ B is a multiplicative subset of B, which implies that φ(S) is a multiplicative subset of A. Describe∼ A ⊗B (S−1 B).
a natural isomorphism φ(S)−1 A =Informal translation: “the fibered product with a localization is the localization of the fibered product in the obvious way”. We say that “localizations arepreserved by base change”. This is handy if the localization is of the form B "→ Bf(corresponding to taking distinguished open sets) or (if B is an integral domain)
B "→ K(B) (corresponding to taking the generic point), and various things in between.
9.2.5. Examples. These four facts let you calculate lots of things in practice, andwe will use them freely.
9.2.G. E XERCISE : THE THREE IMPORTANT TYPES OF MONOMORPHISMS OF SCHEMES .
Show that the following are monomorphisms (Definition 1.3.9): open embeddings,
closed embeddings, and localization of affine schemes. As monomorphisms areclosed under composition, Exercise 1.3.V, compositions of the above are also monomorphisms — for example, locally closed embeddings, or maps from “Spec of stalks
at points of X” to X. (Caution: if p is a point of a scheme X, the natural morphism
Spec OX,p → X, cf. Exercise 6.3.K, is a monomorphism but is not in general anopen embedding.)
n ∼ n
∼ n
9.2.H. E XERCISE . Prove that An
A = AZ ×Spec Z Spec A. Prove that PA = PZ ×Spec Z
Spec A. Thus affine space and projective space are pulled back from their “universal manifestation” over the final object Spec Z.

9.2.6. Extending the base field. One special case of base change is called extending
the base field: if X is a k-scheme, and ℓ is a field extension (often ℓ is the algebraic
closure of k), then X ×Spec k Spec ℓ (sometimes informally written X ×k ℓ or Xℓ ) isan ℓ-scheme. Often properties of X can be checked by verifying them instead on
Xℓ . This is the subject of descent — certain properties “descend” from Xℓ to X. We
have already seen that the property of being the Spec of a normal integral domain
descends in this way (Exercise 5.4.M). Exercises 9.2.I and 9.2.J give other examples
of properties which descend: the property of two morphisms being equal, and the

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property of a(n affine) morphism being a closed embedding, both descend in thisway. Those interested in schemes over non-algebraically closed fields will use thisrepeatedly, to reduce results to the algebraically closed case.
9.2.I. E XERCISE . Suppose π : X → Y and ρ : X → Y are morphisms of k-schemes,ℓ/k is a field extension, and πℓ : X ×Spec k Spec ℓ → Y ×Spec k Spec ℓ and ρℓ : X ×Spec k
Spec ℓ → Y ×Spec k Spec ℓ are the induced maps of ℓ-schemes. (Be sure you understand what this means!) Show that if πℓ = ρℓ then π = ρ. (Hint: show that π and ρ
are the same on the level of sets. To do this, you may use that X ×Spec k Spec ℓ → X
is surjective, which we will soon prove in Exercise 9.4.D. Then reduce to the casewhere X and Y are affine.)
9.2.J. E ASY E XERCISE . Suppose π : X → Y is an affine morphism over k, and ℓ/k isa field extension. Show that π is a closed embedding if and only if π ×k ℓ : X ×k ℓ →Y ×k ℓ is. (The affine hypothesis is not necessary for this result, but it makes theproof easier, and this is the situation in which we will most need it.)9.2.K. U NIMPORTANT BUT FUN EXERCISE . Show that Spec Q(t) ⊗Q C has closed
points in natural correspondence with the transcendental complex numbers. (Ifthe description Spec Q(t)⊗Q[t] C[t] is more striking, you can use that instead.) This
scheme doesn’t come up in nature, but it is certainly neat! A related idea comesup in Remark 11.2.16.

9.3 Interpretations: Pulling back families, and fibers ofmorphisms
9.3.1. Pulling back families.
We can informally interpret fibered product in the following geometric way.
Suppose Y → Z is a morphism. We interpret this as a “family of schemes parametrized
by a base scheme (or just plain base) Z.” Then if we have another morphismψ : X → Z, we interpret the induced map X ×Z Y → X as the “pulled back family”(see Figure 9.1).
%Y
X ×Z Ypulled back familyfamily""ψ%ZX
We sometimes say that X ×Z Y is the scheme-theoretic pullback of Y, schemetheoretic inverse image, or inverse image scheme of Y. (Our forthcoming discussion of fibers may give some motivation for this.) For this reason, fibered productis often called base change or change of base or pullback. In addition to the various names for a Cartesian diagram given in §1.3.6, in algebraic geometry it is often
called a base change diagram or a pullback diagram, and X ×Z Y → X is called
the pullback of Y → Z by ψ, and X ×Z Y is called the pullback of Y by ψ. One
often uses the phrase “over X” or “above X” when discussing X ×Z Y, especially if
X is a locally closed subscheme of Z. (Random side remark: scheme-theoretic pullback always makes sense, while the notion of scheme-theoretic image is somehow
problematic, as discussed in §8.3.1.)

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F IGURE 9.1. A picture of a pulled back family
Before making any definitions, we give a motivating informal example. Consider the “family of curves” y2 = x3 + tx in the xy-plane parametrized by t. Translation: consider Spec k[x, y, t]/(y2 − x3 − tx) → Spec k[t]. If we pull back to a family parametrized by the uv-plane via uv = t (i.e., Spec k[u, v] → Spec k[t] given byt (→ uv), we get y2 = x3 + uvx, i.e., Spec k[x, y, u, v]/(y2 − x3 − uvx) → Spec k[u, v].
If instead we set t to 3 (i.e., pull back by Spec k[t]/(t − 3) → Spec k[t], we get the
curve y2 = x3 + 3x (i.e., Spec k[x, y]/(y2 − x3 − 3x) → Spec k), which we interpretas the fiber of the original family above t = 3. We will soon be able to interpretthese constructions in terms of fibered products.
9.3.2. Fibers of morphisms.
(If you did Exercise 7.3.K, that finite morphisms have finite fibers, you will notfind the following discussion surprising.) A special case of pullback is the notionof a fiber of a morphism. We motivate this with the notion of fiber in the categoryof topological spaces.
9.3.A. E XERCISE . Show that if Y → Z is a continuous map of topological spaces,
and X is a point p of Z, then the fiber of Y over p (the set-theoretic fiber, with theinduced topology) is naturally identified with X ×Z Y.

More generally, for any π : X → Z, the fiber of X ×Z Y → X over a point p of X
is naturally identified with the fiber of Y → Z over π(p).
Motivated by topology, we return to the category of schemes. Suppose p → Z
is the inclusion of a point (not necessarily closed). More precisely, if p is a pointwith residue field K, consider the map Spec K → Z sending Spec K to p, with thenatural isomorphism of residue fields. Then if g : Y → Z is any morphism, the basechange with p → Z is called the (scheme-theoretic) fiber (or fibre) of g above p orthe (scheme-theoretic) preimage of p, and is denoted g−1 (p). If Z is irreducible,

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the fiber above the generic point of Z is called the generic fiber (of g). In an affine
open subscheme Spec A containing p, p corresponds to some prime ideal p, andthe morphism Spec K → Z corresponds to the ring map A → Ap /pAp . This is thecomposition of localization and closed embedding, and thus can be computed by
the tricks above. (Note that p → Z is a monomorphism, by Exercise 9.2.G.)

9.3.B. E XERCISE . Show that the underlying topological space of the (schemetheoretic) fiber of X → Y above a point p is naturally identified with the topologicalfiber of X → Y above p.
9.3.C. E XERCISE ( ANALOG OF E XERCISE 9.3.A). Suppose that π : Y → Z andτ : X → Z are morphisms, and p ∈ X is a point. Show that the fiber of X ×Z Y → X
over p is (isomorphic to) the base change to p of the fiber of π : Y → Z over τ(p).

F IGURE 9.2. The map C → C given by y (→ y2

9.3.3. Example (enlightening in several ways). Consider the projection of the parabolay2 = x to the x-axis over Q, corresponding to the map of rings Q[x] → Q[y], withx (→ y2 . If Q alarms you, replace it with your favorite field and see what happens.
(You should look at Figure 9.2, which is a flipped version of the parabola of Figure 3.6, and figure out how to edit it to reflect what we glean here.) Writing Q[y]as Q[x, y]/(y2 − x) helps us interpret the morphism conveniently.
(i) Then the preimage of 1 is two points:∼ Spec Q[x, y]/(y2 − x, x − 1)
Spec Q[x, y]/(y2 − x) ⊗Q[x] Q[x]/(x − 1) =∼=∼=Spec Q[y]/(y2 − 1))
Spec Q[y]/(y − 1)
Spec Q[y]/(y + 1).

(ii) The preimage of 0 is one nonreduced point:∼ Spec Q[y]/(y2 ).
Spec Q[x, y]/(y2 − x, x) =(iii) The preimage of −1 is one reduced point, but of “size 2 over the basefield”.
∼ Spec Q[y]/(y2 + 1) =∼ Spec Q[i] = Spec Q(i).
Spec Q[x, y]/(y2 − x, x + 1) =(iv) The preimage of the generic point is again one reduced point, but of “size2 over the residue field”, as we verify now.
∼ Spec Q[y] ⊗Q[y2 ] Q(y2 )Spec Q[x, y]/(y2 − x) ⊗Q[x] Q(x) =Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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i.e., (informally) the Spec of the ring of polynomials in y divided by polynomials
in y2 . A little thought shows you that in this ring you may invert any polynomial
in y, as if f(y) is any polynomial in y, then
1
f(−y)
=
,
f(y)
f(y)f(−y)
and the latter denominator is a polynomial in y2 . Thus
∼ Q(y)
Q[x, y]/(y2 − x) ⊗Q[x] Q(x) =which is a degree 2 field extension of Q(x) (note that Q(x) = Q(y2 )).
(You might want to work through the preimages of other points of A1Q , suchas [(x3 − 4)] and [(x2 − 2)].)
Notice the following interesting fact: in each of the four cases, the number ofpreimages can be interpreted as 2, where you count to two in several ways: you
can count points (as in the case of the preimage of 1); you can get nonreduced
behavior (as in the case of the preimage of 0); or you can have a field extension ofdegree 2 (as in the case of the preimage of −1 or the generic point). In each case, thefiber is an affine scheme whose dimension as a vector space over the residue fieldof the point is 2. Number theoretic readers may have seen this behavior before.
We will discuss this example again in §17.4.9. This is going to be symptomatic ofa very important kind of morphism (a finite flat morphism, see Remark 24.4.8 and§24.4.12).
Try to draw a picture of this morphism if you can, so you can develop a pictorial shorthand for what is going on. A good first approximation is the parabola ofFigure 9.2, but you will want to somehow depict the peculiarities of (iii) and (iv).
9.3.4. Remark: Finite morphisms have finite fibers. If you haven’t done Exercise 7.3.K,
that finite morphisms have finite fibers, now would be a good time to do it, as you
will find it more straightforward given what you know now.
9.3.D. E XERCISE ( IMPORTANT FOR THOSE WITH MORE ARITHMETIC BACKGROUND ).
What is the scheme-theoretic fiber of Spec Z[i] → Spec Z over the prime (p)? Youranswer will depend on p, and there are four cases, corresponding to the four casesof Example 9.3.3. (Can you draw a picture?)
9.3.E. E XERCISE . (This exercise will give you practice in computing a fiberedproduct over something that is not a field.) Consider the morphism of schemes
X = Spec k[t] → Y = Spec k[u] corresponding to k[u] → k[t], u (→ t2 , wherechar k ̸= 2. Show that X ×Y X has two irreducible components. (What happens if
char k = 2? See Exercise 9.5.A for a clue.)
9.3.5. A first view of a blow-up.
9.3.F. I MPORTANT C ONCRETE E XERCISE . (The discussion here immediately gen2
2
1eralizes to An
A .) Define a closed subscheme Bl(0,0) Ak of Ak ×k Pk as follows (see2
Figure 9.3). If the coordinates on Ak are x, y, and the projective coordinates on P1k
are u, v, this subscheme is cut out in A2k ×k P1k by the single equation xv = yu. (Youmay wish to interpret Bl(0,0) A2k as follows. The P1k parametrizes lines through theorigin. The blow-up corresponds to ordered pairs of (point p, line ℓ) such that(0, 0) and p both lie on ℓ.) Describe the fiber of the morphism Bl(0,0) A2k → P1k over

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The Rising Sea: Foundations of Algebraic Geometryeach closed point of P1k . Show that the morphism Bl(0,0) A2k → A2k is an isomorphism away from (0, 0) ∈ A2k . Show that the fiber over (0, 0) is an effective Cartierdivisor (§8.4.1, a closed subscheme that is locally cut out by a single equation,
which is not a zerodivisor). It is called the exceptional divisor. We will discuss
blow-ups in Chapter 22. This particular example will come up in the motivating
example of §22.1, and in Exercise 20.2.D.

F IGURE 9.3. A first example of a blow-up
We haven’t yet discussed regularity, but here is a hand-waving argument suggesting that the Bl(0,0) A2k is “smooth”: the preimage above either standard open
set Ui ⊂ P1 is isomorphic to A2 . Thus “the blow-up is a surgery that takes thesmooth surface A2k , cuts out a point, and glues back in a P1 , in such a way that theoutcome is another smooth surface.”
9.3.6. General fibers, generic fibers, generically finite morphisms.
The phrases “generic fiber” and “general fiber” parallel the phrases “genericpoint” and “general point” (Definition 3.6.11). Suppose π : X → Y is a morphism ofschemes. When one says the general fiber (or a general fiber) of π a has a certain
property, this means that there exists a dense open subset U ⊂ Y such that thefibers above any point in U have that property.
When one says the generic fiber of π : X → Y, this implicitly means that Y
is irreducible, and the phrase refers to the fiber over the generic point. Generalfiber and generic fiber are not the same thing! Clearly if something holds for thegeneral fiber, then it holds for the generic fiber, but the converse is not always true.
However, in good circumstances, it can be — properties of the generic fiber extend
to an honest open neighborhood. For example, if Y is irreducible and Noetherian,and π is finite type, then if the generic fiber of π is empty (resp. nonempty), thenthe general fiber is empty (resp. nonempty), by Chevalley’s Theorem 7.4.2 (or more
simply, by Exercise 7.4.L).
If π : X → Y is finite type, we say π is generically finite if π is finite after basechange to the generic point of each irreducible component (or equivalently, by
Exercise 7.4.D, if the preimage of the generic point of each irreducible componentof Y is a finite set). (The notion of generic finiteness can be defined in more generalcircumstances, see [Stacks, tag 073A].)

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9.3.G. E XERCISE (“ GENERICALLY FINITE ” USUALLY MEANS “ GENERALLY FINITE ”).
Suppose π : X → Y is an affine, finite type, generically finite morphism of locally
Noetherian schemes, and Y is reduced. Show that there is an open neighborhoodof each generic point of Y over which π is actually finite. (The hypotheses can be
weakened considerably, see [Stacks, tag 02NW].) Hint: reduce to the case where Y
is Spec B, where B is an integral domain. Then X is affine, say X = Spec A. Write
A = B[x1 , . . . , xn ]/I. Now A ⊗B K(B) is a finite K(B)-module (finite-dimensional
vector space) by hypothesis, so there are monic polynomials fi (t) ∈ K(B)[t] suchthat fi (xi ) = 0 in A ⊗B K(B). Let b be the product of the (finite number of) denominators appearing in the coefficients in the fi (x). By replacing B by Bb , argue thatyou can assume that fi (t) ∈ B[t]. Then fi (xi ) = 0 in A⊗B K(B), meaning that fi (xi )is annihilated by some nonzero element of B. By replacing B by its localization at
the product of these n nonzero elements (“shrinking Spec B further”), argue thatfi (xi ) = 0 in A. Then conclude.
9.3.7. ⋆⋆ Finitely presented families (morphisms) are locally pullbacks of particularly nice families. If you are macho and are embarrassed by Noetherian rings,
the following exercise can be used to extend results from the Noetherian case to
locally finitely presented situations. Exercise 9.3.I, an extension of Chevalley’s Theorem 7.4.2, is a good example.
9.3.H. E XERCISE . Suppose π : X → Spec B is a finitely presented morphism. Show
that there exists a base change diagram of the form
% X′

X

π′

π

"
Spec Bρ"% Spec Z[x1 , . . . , xN ]where N is some integer, and π ′ is finitely presented (= finite type as the target
is Noetherian, see §7.3.17). Thus each finitely presented morphism is locally (onthe base) a pullback of a finite type morphism to a Noetherian scheme. Henceany result proved for Noetherian schemes and stable under base change is automatically proved for locally finitely presented morphisms to arbitrary schemes.
Hint: think about the case where X is affine first. If X = Spec A, then A =B[y1 , . . . , yn ]/(f1 , . . . , fr ). Choose one variable xi for each coefficient of fi ∈ B[y1 , . . . , yn ].
What is X ′ in this case? Then consider the case where X is the union of two affine
open sets, that intersect in an affine open set. Then consider more general cases
until you solve the full problem. You will need to use every part of the definitionof finite presentation. (Exercise 28.2.L extends this result.)
9.3.I. E XERCISE (C HEVALLEY ’ S T HEOREM FOR LOCALLY FINITELY PRESENTED MOR PHISMS ).
(a) Suppose that A is a finitely presented B-algebra (B not necessarily Noetherian),so A = B[x1 , . . . , xn ]/(f1 , . . . , fr ). Show that the image of Spec A → Spec B is a
finite union of locally closed subsets of Spec B. Hint: Exercise 9.3.H (the simpler
affine case).
(b) Show that if π : X → Y is a quasicompact locally finitely presented morphism,
and Y is quasicompact, then π(X) is a finite union of locally closed subsets. (For

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hardened experts only: [Gr-EGA, 0III .9.1] gives a definition of local constructibility, and of constructibility in more generality. The general form of Chevalley’s
Constructibility Theorem [Gr-EGA, IV1 .1.8.4] is that the image of a locally constructible subset, under a finitely presented map, is also locally constructible.)

9.4 Properties preserved by base change
All reasonable properties of morphisms are preserved under base change. (Infact, one might say that a property of morphisms cannot be reasonable if it is notpreserved by base change, cf. §7.1.1.) We discuss this, and in §9.5 we will explain
how to fix those that don’t fit this pattern.
We have already shown that the notion of “open embedding” is preserved bybase change (Exercise 7.1.B). We did this by explicitly describing what the fiberedproduct of an open embedding is: if Y "→ Z is an open embedding, and ψ : X → Z
is any morphism, then we checked that the open subscheme ψ−1 (Y) of X satisfies
the universal property of fibered products.
We have also shown that the notion of “closed embedding” is preserved bybase change (§9.2 (3)). In other words, given a Cartesian diagram
W

% Y"
#

cl. emb.

"
X

"
%Z

where Y "→ Z is a closed embedding, W → X is as well.

9.4.A. E ASY E XERCISE . Show that locally principal closed subschemes (Definition 8.4.1) pull back to locally principal closed subschemes.
Similarly, other important properties are preserved by base change.
9.4.B. E XERCISE . Show that the following properties of morphisms are preserved
by base change.
(a) quasicompact(b) quasiseparated(c) affine morphism(d) finite(e) integral(f) locally of finite type(g) finite type⋆⋆ (h) locally of finite presentation⋆⋆ (i) finite presentation
9.4.C. ⋆ E XERCISE . Show that the notion of “quasifinite morphism” (finite type +finite fibers, Definition 7.3.14) is preserved by base change. (Warning: the notionof “finite fibers” is not preserved by base change. Spec Q → Spec Q has finite
fibers, but Spec Q ⊗Q Q → Spec Q has one point for each element of Gal(Q/Q),
see Exercise 9.2.E.) Hint: reduce to the case Spec A → Spec B. Reduce to the case

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φ : Spec A → Spec k. By Exercise 7.4.D, such φ are actually finite, and finiteness ispreserved by base change.

9.4.D. E XERCISE . Show that surjectivity is preserved by base change. (Surjectivity
has its usual meaning: surjective as a map of sets.) You may end up showing thatfor any fields k1 and k2 containing k3 , k1 ⊗k3 k2 is nonzero, and using the Axiomof Choice to find a maximal ideal in k1 ⊗k3 k2 .
9.4.1. On the other hand, injectivity is not preserved by base change — witness thebijection Spec C → Spec R, which loses injectivity upon base change by Spec C →Spec R (see Example 9.2.3). This can be rectified (see §9.5.24).

9.4.E. E XERCISE ( CF. E XERCISE 9.2.D). Suppose X and Y are integral finite type kschemes. Show that X×k Y is an integral finite type k-scheme. (Once we define “variety”, this will become the important fact that the product of irreducible varieties
over an algebraically closed field is an irreducible variety, Exercise 10.1.E. The factthat the base field k is algebraically closed is important, see §9.5. See Exercise 9.5.M
for an improvement.) Hint: reduce to the case where X and Y are both affine, say
X = Spec A and Y = Spec B with A and B integral domains.
You might), ′ flip ahead*,
to Easy Exercise 9.5.L to see how to do this. Suppose ( ai ⊗ bi )aj ⊗ bj′ = 0
in A ⊗k B with ai , aj′ ∈ A, bi , bj′ ∈ B, where both {bi } and {bj′ } are linearly independent over k, and a1 and a1′ are nonzero. Show that D(a1 a1′ ) ⊂ Spec A is′nonempty. By the Weak Nullstellensatz 3.2.4, there is a maximal), m ⊂ A in
* D(a1 a1 ),′′with A/m = k. By reducing modulo m, deduce ( ai ⊗ bi )a j ⊗ bj = 0 in B,
where the overline indicates residue modulo m. Show that this contradicts the factthat B is an integral domain.
9.4.F. E XERCISE . If P is a property of morphisms preserved by base change andcomposition, and X → Y and X ′ → Y ′ are two morphisms of S-schemes withproperty P, show that X ×S X ′ → Y ×S Y ′ has property P as well.

9.5 ⋆ Properties not preserved by base change, and how to fix
them
There are some notions that you should reasonably expect to be preserved bypullback based on your geometric intuition. Given a family in the topological category, fibers pull back in reasonable ways. So for example, any pullback of a family
in which all the fibers are irreducible will also have this property; ditto for connected. Unfortunately, both of these fail in algebraic geometry, as Example 9.2.3
shows:#% Spec C
Spec C Spec C"
Spec C"% Spec R

The family on the right (the vertical map) has irreducible and connected fibers, andthe one on the left doesn’t. The same example shows that the notion of “integral

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fibers” also doesn’t behave well under pullback. And we used it in 9.4.1 to showthat injectivity isn’t preserved by Base Change.
9.5.A. E XERCISE . Suppose k is a field of characteristic p, so k(u)/k(up ) is an
inseparable extension. By considering k(u) ⊗k(up ) k(u), show that the notion of“reduced fibers” does not necessarily behave well under pullback. (We will soonsee that this happens only in characteristic p, in the presence of inseparability.)
We rectify this problem as follows.
9.5.1. A geometric point of a scheme X is defined to be a morphism Spec k → X
where k is an algebraically closed field. Awkwardly, this is now the third kindof “point” of a scheme! There are just plain points, which are elements of theunderlying set; there are Z-valued points (Z a scheme), which are maps Z → X,
§6.3.9; and there are geometric points. Geometric points are clearly a flavor of ascheme-valued point, but they are also an enriched version of a (plain) point: they
are the data of a point with an inclusion of the residue field of the point in an
algebraically closed field.
A geometric fiber of a morphism X → Y is defined to be the fiber over a geometric point of Y, i.e., the fibered product with the geometric point Spec k → Y. A
morphism has connected (resp. irreducible, integral, reduced) geometric fibers
if all its geometric fibers are connected (resp. irreducible, integral, reduced). One
usually says that the morphism has geometrically connected (resp. geometricallyirreducible, geometrically integral, geometrically reduced) fibers. A k-scheme X
is geometrically connected (resp. geometrically irreducible, geometrically integral, geometrically reduced) if the structure morphism X → Spec k has geometrically connected (resp. irreducible, integral, reduced) fibers. We will soon see thatto check any of these conditions, we need only base change to k.
(Warning: in some sources, in the definition of “geometric point”, “algebraicallyclosed” is replaced by “separably closed”.)
9.5.B. E XERCISE . Show that the notion of “connected (resp. irreducible, integral,
reduced) geometric fibers” behaves well under base change.
9.5.C. E XERCISE FOR THE ARITHMETICALLY- MINDED . Show that for the morphism Spec C → Spec R, all geometric fibers consist of two reduced points. (Cf.
Example 9.2.3.) Thus Spec C is a geometrically reduced but not geometrically irreducible R-scheme.
9.5.D. E ASY EXERCISE . Give examples of k-schemes that:(a) are reduced but not geometrically reduced;(b) are connected but not geometrically connected;(c) are integral but not geometrically integral.
9.5.E. E XERCISE ( TO CONVINCE GEOMETERS WHY GEOMETRIC FIBERS ARE MEAN INGFUL ). Recall Example 9.3.3, the projection of the parabola y2 = x to the x-axis,
corresponding to the map of rings Q[x] → Q[y], with x (→ y2 . Show that the geometric fibers of this map are always two points, except for those geometric fibers“over 0 = [(x)]”. (Note that Spec C → Spec Q[x] and Spec Q → Spec Q[x], bothcorresponding to ring maps with x (→ 0, are both geometric points “above 0”.)

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Checking whether a k-scheme is geometrically connected etc. seems annoying:
you need to check every single algebraically closed field containing k. However, ineach of these four cases, the failure of nice behavior of geometric fibers can alreadybe detected after a finite extension of fields. For example, Spec Q(i) → Spec Q is
not geometrically connected, and in fact you only need to base change by Spec Q(i)to see this. We make this precise as follows.
Suppose X is a k-scheme. If K/k is a field extension, define XK = X ×k Spec K.
Consider the following twelve statements.
• XK is reduced:(Ra ) for all fields K,(Rb ) for all algebraically closed fields K (X is geometrically reduced),(Rc ) for K = k,(Rd ) for K = kp (where kp is the perfect closure of k)• XK is irreducible:(Ia ) for all fields K,(Ib ) for all algebraically closed fields K (X is geometrically irreducible),(Ic ) for K = k,(Id ) for K = ks (where ks is the separable closure of k).
• XK is connected:(Ca ) for all fields K,(Cb ) for all algebraically closed fields K (X is geometrically connected),(Cc ) for K = k,(Cd ) for K = ks .
Trivially (Ra ) implies (Rb ) implies (Rc ), and (Ra ) implies (Rd ), and similarly with
“reduced” replaced by “irreducible” and “connected”. Note also that if char k = 0,then k = kp = ks .
9.5.F. E XERCISE .
(a) Suppose that E/F is a field extension, and A is an F-algebra. Show that A is asubalgebra of A ⊗F E. (Hint: think of these as vector spaces over F.)(b) Show that: (Rb ) implies (Ra ) and (Rc ) implies (Rd ).
(c) Show that: (Ib ) implies (Ia ) and (Ic ) implies (Id ).
(d) Show that: (Cb ) implies (Ca ) and (Cc ) implies (Cd ).
Possible hint: You may use the fact that if Y is a nonempty F-scheme, then Y ×FSpec E is nonempty, cf. Exercise 9.4.D.
Thus for example a k-scheme is geometrically integral if and only if it remains
integral under any field extension.
9.5.2. Hard fact. In fact, (Rd ) implies (Ra ), and thus (Ra ) through (Rd ) are all equivalent, and similarly for the other two rows. The explanation is below. On a firstreading, you may want to read only Corollary 9.5.11 on connectedness, Proposition 9.5.14 on irreducibility, Proposition 9.5.20 on reducedness, and Theorem 9.5.23
on varieties, and then to use them to solve Exercise 9.5.N. You can later come backand read the proofs, which include some useful tricks turning questions about
general schemes over a field to questions about finite type schemes.

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The following exercise may help even the geometrically-minded reader appreciate the utility of these notions. (There is nothing important about the dimension
2 and the degree 4 in this exercise!)
9.5.G. ⋆ E XERCISE . Recall from Remark 4.5.3 that the quartic curves in P2k areparametrized by a P14
k . (This will be made much more precise in §28.3.5.) Show
that the points of P14
k corresponding to geometrically irreducible curves form an
open subset. Explain the necessity of the modifier “geometrically” (even if k is
algebraically closed).
9.5.3. ⋆⋆ The rest of §9.5 is double-starred.
9.5.4. Proposition. — Suppose A and B are finite type k-algebras. Then Spec A ×k
Spec B → Spec B is an open map.

This is the one fact we will not prove here. We could (it isn’t too hard), butinstead we leave it until Remark 24.5.3.
9.5.5. Preliminary discussion.
9.5.6. Lemma. — Suppose X is a k-scheme. Then X → Spec k is universally open, i.e.,
remains open after any base change.

Proof. If S is an arbitrary k-scheme, we wish to show that XS → S is open. It suffices
to consider the case X = Spec A and S = Spec B. To show that φ : Spec A ⊗k B →Spec B is open, it suffices to show that the image of a distinguished open set D(f)(f ∈ A ⊗k B) is open.
We come to a trick we, will use repeatedly, which we will call the tensorfiniteness trick. Write f =ai ⊗ bi , where the sum is finite. It suffices to replace
A by the subring generated by the ai . (Reason: if this ring is A ′ , then factor φthrough Spec(A ′ ⊗k B).) Thus we may assume A is finitely generated over k. Thenuse Proposition 9.5.4.
!
9.5.7. Lemma. — Suppose the field extension E/F is purely inseparable polynomial
over F with only one root, perhaps with multiplicity). Suppose X is any F-scheme. Thenφ : XE → X is a homeomorphism.

Proof. The morphism φ is a bijection, so we may identify the points of X and XE .
(Reason: for any point p ∈ X, the scheme-theoretic fiber φ−1 (p) is a single point,by the definition of pure inseparability.) The morphism φ is continuous (so opens
in X are open in XE ), and by Lemma 9.5.6, φ is open (so opens in X are open in
XE ).
!
9.5.8. Connectedness.
Recall that a connected component of a topological space is a maximal connected subset §3.6.12.
9.5.H. E XERCISE ( PROMISED IN R EMARK 3.6.13). Show that every point is contained in a connected component, and that connected components are closed. (Hint:
see the hint for Exercise 3.6.O.)

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9.5.I. T OPOLOGICAL EXERCISE . Suppose φ : X → Y is open, and has nonempty
connected fibers. Then φ induces a bijection of connected components.
9.5.9. Lemma. — Suppose X is geometrically connected over k. Then for any scheme
Y/k, X ×k Y → Y induces a bijection of connected components.
!

Proof. Combine Lemma 9.5.6 and Exercise 9.5.I.

9.5.J. E XERCISE ( PROMISED IN R EMARK 3.6.3). Show that a scheme X is disconnected if and only if there exists a function e ∈ Γ (X, OX ) that is an idempotent(e2 = e) distinct from 0 and 1. (Hint: if X is the disjoint union of two open sets X0
and X1 , let e be the function that is 0 on X0 and 1 on X1 . Conversely, given suchan idempotent, define X0 = V(e) and X1 = V(1 − e).)9.5.10. Proposition. — Suppose k is separably closed, and A is a k-algebra with Spec A
connected. Then Spec A is geometrically connected over k.
Proof. We wish to show that Spec A ⊗k K is connected for any field extension K/k.
It suffices to assume that K is algebraically closed (as Spec A ⊗k K → Spec A ⊗k K
is surjective). By choosing an embedding k "→ K and considering the diagram
Spec A ⊗k K

% Spec A ⊗k k"
Spec K"% Spec k

homeo.

by Lem. 9.5.7% Spec A"% Spec k

it suffices to assume k is algebraically closed.
If Spec A⊗k K is disconnected, then A⊗k K contains an idempotent e ̸= 0, 1 (byExercise 9.5.J). By the tensor-finiteness trick, we may assume that A is a finitely generated algebra over k, and K is a finitely generated field extension. Write K = K(B)for some integral domain B of finite type over k. Then by the tensor-finitenesstrick, by considering the finite number of denominators appearing in a representative of e as a sum of decomposable tensors, e ∈ A ⊗k B[1/b] for some nonzerob#∈ B, so Spec A ⊗k B[1/b] is disconnected, say with open subsets U and V with
U V = Spec A ⊗k B[1/b].
Now φ : Spec A ⊗k B[1/b] → Spec B[1/b] is an open map (Proposition 9.5.4),
so φ(U) and φ(V) are nonempty open sets. As Spec B[1/b] is connected, the intersection φ(U) ∩ φ(V) is a nonempty open set, which has a closed point p (with∼ Spec A, and we have covered Spec A
residue field k, as k = k). But then φ−1 (p) =with two disjoint open sets, yielding a contradiction.
!
9.5.11. Corollary. — If k is separably closed, and Y is a connected k-scheme, then Y isgeometrically connected.
Proof. We wish to show that for any field extension K/k, YK is connected. By Proposition 9.5.10, Spec K is geometrically connected over k. Then apply Lemma 9.5.9with X = Spec K.
!
9.5.12. Irreducibility.

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9.5.13. Proposition. — Suppose k is separably closed, A is a k-algebra with Spec A
irreducible, and K/k is a field extension. Then Spec A ⊗k K is irreducible.
Proof. We follow the philosophy of the proof of Proposition 9.5.10. As in the firstparagraph of that proof, it suffices to assume that K and k are algebraically closed.
If A⊗k K is not irreducible, then we can find x and y with V(x), V(y) ̸= Spec A⊗k K
and V(x) ∪ V(y) = Spec A ⊗k K. As in the second paragraph of the proof of
Proposition 9.5.10, we may assume that A is a finitely generated algebra over k,
and K = K(B) for an integral domain B of finite type over k, and x, y ∈ A ⊗k B[1/b]for some nonzero b ∈ B. Then D(x) and D(y) are nonempty open subsets ofSpec A ⊗k B[1/b], whose image in Spec B[1/b] are nonempty opens, and thus their∼ Spec A,intersection is nonempty and contains a closed point p. But then φ−1 (p) =and we have covered Spec A with two proper closed sets (the restrictions of V(x)and V(y)), yielding a contradiction.
!
9.5.K. E XERCISE . Suppose k is separably closed, and A and B are k-algebras, both
irreducible (with irreducible Spec, i.e., with one minimal prime). Show that A⊗k B
is irreducible too. (Hint: reduce to the case where A and B are finite type over k.
Extend the proof of the previous proposition.)
9.5.L. E ASY EXERCISE . Show that a scheme X is irreducible if and only if there
exists an open cover X = ∪Ui with Ui irreducible for all i, and Ui ∩ Uj ̸= ∅ for alli, j.
9.5.14. Proposition. — Suppose K/k is a field extension of a separably closed field and
Xk is irreducible. Then XK is irreducible.
Proof. Take an open cover Xk = ∪Ui by pairwise intersecting irreducible affine
open subsets. The base change of each Ui to K is irreducible by Proposition 9.5.13,
and they pairwise intersect. The result then follows from Exercise 9.5.L.
!
9.5.M. E XERCISE . Suppose B is a geometrically integral k-algebra, and A is an
integral k-algebra. Show that A ⊗k B is integral. (Once we define “variety”, thiswill imply that the product of a geometrically integral variety with an integral
variety is an integral variety.) Hint: revisit the proof of Exercise 9.4.E.
9.5.15. Reducedness.
We recall the following fact from field theory, which is a refined version of
the basics of transcendence theory developed in Exercise 11.2.A. Because this is astarred section, we content ourselves with a reference rather than a proof.
9.5.16. Algebraic Fact: finitely generated extensions of perfect fields are separably generated, see [E, Cor. 16.17(b)] or [vdW, §19.7]. — Suppose E/F is a finitely
generated extension of a perfect field. Then it can be factored into a finite separable partand a purely transcendent part: E/F(t1 , . . . , tn )/F.
9.5.17. Proposition. — Suppose B is a geometrically reduced k-algebra, and A is areduced k-algebra. Then A ⊗k B is reduced.
(Compare this to Exercise 9.5.M.)

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Proof. Reduce to the case where A is finitely generated over k using the, tensorfiniteness trick. (Suppose we have x ∈ A ⊗k B with xn = 0. Then x =ai ⊗ bi .
Let A ′ be the finitely generated subring of A generated by the ai . Then A ′ ⊗k B′
is"a subring of A ⊗k B. Replace A by A .) Then A is a subring of the productKi of the function fields of"its irreducible components. (One can see this asfollows: the kernel of A →A/pi is the intersection of the primes pi , whichis the ideal of nilpotents N by Theorem 3.2.12; but n = 0 as A is reduced. And
A/pi "→ K(A/pi ) = Ki clearly. Alternatively, this is clear from our discussion of
associated points: Theorem 5.5.10(b), see also Exercise 5.5.G). So it suffices to prove
it for A a product of fields. Then it suffices to prove it when A is a field. But thenwe are done, by the definition of geometric reducedness.
!
9.5.18. Proposition. — Suppose A is a reduced k-algebra. Then:(a) A ⊗k k(t) is reduced.
(b) If E/k is a finite separable extension, then A ⊗k E is reduced.
Proof. (a) Clearly A ⊗ k[t] is reduced, and localization preserves reducedness (asreducedness is stalk-local, Exercise 5.2.A).
(b) Working inductively, we can assume E is generated by a single element,with minimal polynomial p(t). By the tensor-finiteness trick, we can assume A
is finitely generated over k. Then by the same trick as in the proof of Proposition 9.5.17, we can replace A by the product of its function fields of its components,
and then we can assume A is a field. But then A[t]/p(t) is reduced by the definitionof separability of p.
!
9.5.19. Lemma. — Suppose E/k is a field extension of a perfect field, and A is a reduced
k-algebra. Then A ⊗k E is reduced.
Proof. By the tensor-finiteness trick, we may assume E is finitely generated overk. By Algebraic Fact 9.5.16, we can factor E/k into extensions of the forms of
Proposition 9.5.18 (a) and (b). We then apply Proposition 9.5.18.
!
9.5.20. Proposition. — Suppose E/k is an extension of a perfect field, and X is a reduced
k-scheme. Then XE is reduced.
Proof. Reduce to the case where X is affine. Use Lemma 9.5.19.

!

9.5.21. Corollary. — Suppose k is perfect, and A and B are reduced k-algebras. ThenA ⊗k B is reduced.
Proof. By Lemma 9.5.19, A is a geometrically reduced k-algebra. Then apply
Lemma 9.5.17.
!
9.5.N. E XERCISE ( COMPLETING H ARD FACT 9.5.2). Show that (Rd ) implies (Ra ),(Id ) implies (Ia ), and (Cd ) implies (Ca ).
9.5.O. E XERCISE . Suppose that A and B are two integral domains that are kalgebras. Show that A ⊗k B is an integral domain.

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9.5.22. Varieties.
9.5.23. Theorem. —(a) If k is perfect, the product of affine k-varieties (over Spec k) is an affine k-variety.
(b) If k is separably closed, the product of irreducible affine k-varieties is an irreducible
affine k-variety.
(c) If k is separably closed, the product of connected affine k-varieties is a connected affine
k-variety.
Once we define varieties in general, in Definition 10.1.7, you will be able toremove the adjective “affine” throughout this statement of Theorem 9.5.23. (Wealso remark that (b) was proven in Exercise 9.4.E under the stronger hypothesisthat k is algebraically closed.)
Proof. (a) The finite type and separated statements are straightforward, as bothproperties are preserved by base change and composition. For reducedness, reduce to the affine case, then use Corollary 9.5.21.
(b) It only remains to show irreducibility. Reduce to the affine case using Exercise 9.5.L (as in the proof of Proposition 9.5.14). Then use Exercise 9.5.K.
(c) This follows from Corollary 9.5.11.
!
9.5.24. Universally injective (radicial) morphisms. As remarked in §9.4.1, injectivity is not preserved by base change. A better notion is that of universally
injective morphisms: morphisms that are injections of sets after any base change.
In keeping with the traditional agricultural terminology (sheaves, germs, ..., cf.
Remark 2.4.4), these morphisms were named radicial after one of the lesser root
vegetables. As a first example: all locally closed embeddings are universally injective (as they are injective, and remain locally closed embeddings upon any basechange). If you wish, you can show more generally that all monomorphisms areuniversally injective. (Hint: show that monomorphisms are injective, and that theclass of monomorphisms is preserved by base change.)
Universal injectivity is the algebro-geometric generalization of the notion ofpurely inseparable extensions of fields. A field extension K/L is purely inseparable if L is separably closed in K, or equivalently, if for every α ∈ K \ L, the minimal
polynomial of α is not a separable polynomial. This is obviously more interestingin positive characteristic.
9.5.P. E XERCISE . If K/L is an extension of fields. Show that Spec K → Spec L isuniversally injective if and only if K/L is purely inseparable.

9.5.Q. E XERCISE . Suppose π : X → Y is a morphism of schemes. Show that thefollowing are equivalent.
(i) The morphism π is universally injective.
(ii) For every field K, the induced map Hom(Spec K, X) → Hom(Spec K, Y) is
injective.
(iii) The morphism π is injective, and for each p ∈ X, the field extension
κ(p)/κ(π(p)) is purely inseparable.

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Possible hint: Connect i) and ii), and connect ii) and iii). Side Remark: in ii), you
can show that it suffices to take K to be algebraically closed, so “universally injective” is the same as “injective on geometric points”.
9.5.R. E ASY E XERCISE . Show that the class of universally injective morphisms
is stable under composition, base change, and products. Show that this notion is
local on the target. Thus the class of universally injective morphisms is reasonable
in the sense of §7.1.1.

Exercise 10.1.P shows that universal injectivity is really a property of the diagonal morphism, and explores the consequences of this.

9.6 Products of projective schemes: The Segre embedding
We next describe products of projective A-schemes over A. (The case of greatn
est initial interest is if A = k.) To do this, we need only describe Pm
A ×A PA , becausemany projective A-scheme has a closed embedding in some PA , and closed embedn
dings behave well under base change, so if X "→ Pm
A and Y "→ PA are closedmnembeddings, then X ×A Y "→ PA ×A PA is also a closed embedding, cut out bynthe equations of X and Y (§9.2(3)). We will describe Pm
A ×A PA , and see that it too
is a projective A-scheme. (Hence if X and Y are projective A-schemes, then theirproduct X ×A Y over A is also a projective A-scheme.)
Before we do this, we will get some motivation from classical projective spaces(nonzero vectors modulo nonzero scalars, Exercise 4.4.F) in a special case. Our
map will send [x0 , x1 , x2 ] × [y0 , y1 ] to a point in P5 , whose coordinates we think ofas being entries in the “multiplication table”[ x0 y0 , x1 y0 , x2 y0 ,x0 y1 , x1 y1 , x2 y1 ].
This is indeed a well-defined map of sets. Notice that the resulting matrix is rank
one, and from the matrix, we can read off [x0 , x1 , x2 ] and [y0 , y1 ] up to scalars. For
example, to read off the point [x0 , x1 , x2 ] ∈ P2 , we take the first row, unless it is
all zero, in which case we take the second row. (They can’t both be all zero.) In
conclusion: in classical projective geometry, given a point of Pm and Pn , we haveproduced a point in Pmn+m+n , and from this point in Pmn+m+n , we can recover
the points of Pm and Pn .
Suitably motivated, we return to algebraic geometry. We define a map
bynmn+m+nPm
A × A PA → PA([x0 , . . . , xm ], [y0 , . . . , yn ]) (→ [z00 , z01 , . . . , zij , . . . , zmn ]= [x0 y0 , x0 y1 , . . . , xi yj , . . . xm yn ].

More explicitly, we consider the map from the affine open set Ui × Vj (where Ui =D(xi ) and Vj = D(yj ) to the affine open set Wij = D(zij ) by
(x0/i , . . . , xm/i , y0/j , . . . , yn/j ) (→ (x0/i y0/j , . . . , xi/i yj/j , . . . , xm/i yn/j )or, in terms of algebras, zab/ij (→ xa/i yb/j .

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9.6.A. E XERCISE . Check that these maps glue to give a well-defined morphism
mn+m+nnPm.
A ×A PA → PA

9.6.1. We next show that this morphism is a closed embedding. We can check thison an open cover of the target (the notion of being a closed embedding is affinelocal, Exercise 8.1.D). Let’s check this on the open set where zij ̸= 0. The preimagenof this open set in Pm
A × PA is the locus where xi ̸= 0 and yj ̸= 0, i.e., Ui × Vj . As
described above, the map of rings is given by zab/ij (→ xa/i yb/j ; this is clearly asurjection, as zaj/ij (→ xa/i and zib/ij (→ yb/j . (A generalization of this ad hoc
description will be given in Exercise 16.4.D.)
This map is called the Segre morphism or Segre embedding. If A is a field,
the image is called the Segre variety.
9.6.B. E XERCISE . Show that the Segre scheme (the image of the Segre embedding)is cut out (scheme-theoretically) by the equations corresponding to⎛⎞
a00 · · · a0n⎜.. ⎟ = 1,..
rank ⎝ ...
.
. ⎠
am0

···

amn

i.e., that all 2 × 2 minors vanish. Hint: suppose you have a polynomial in the aijthat becomes zero upon the substitution aij = xi yj . Give a recipe for subtracting
polynomials of the form “monomial times 2 × 2 minor” so that the end result is 0.
(The analogous question for the Veronese embedding in special cases is the content
of Exercises 8.2.J and 8.2.L.)
9.6.2. Important Example. Let’s consider the first nontrivial example, when m =n = 1. We get P1k ×k P1k "→ P3k (where k is a field). We get a single equation"#
a00 a01rank= 1,
a10 a11i.e., a00 a11 − a01 a10 = 0. We again meet our old friend, the quadric surface(§8.2.9)! Hence: the smooth quadric surface wz − xy = 0 (Figure 8.2) is isomorphic
to P1k ×k P1k . Recall from Exercise 8.2.M that the quadric surface has two families(rulings) of lines. You may wish to check that one family of lines corresponds tothe image of {x} ×k P1k as x varies, and the other corresponds to the image P1k ×k {y}as y varies.
If k is an algebraically closed field of characteristic not 2, then by diagonalizability of quadratics (Exercise 5.4.J), all rank 4 (“full rank”) quadratics in 4 variables are isomorphic, so all rank 4 quadric surfaces over an algebraically closed
field of characteristic not 2 are isomorphic to P1k ×k P1k .
Note that this is not true over a field that is not algebraically closed. For example, over R, w2 + x2 + y2 + z2 = 0 (in P3R ) is not isomorphic to P1R ×R P1R . Reason:
the former has no real points, while the latter has lots of real points.
You may wish to do the next two exercises in either order. The second can beused to show the first, but the first may give you insight into the second.
9.6.C. E XERCISE : A COORDINATE - FREE DESCRIPTION OF THE S EGRE EMBEDDING .
Show that the Segre embedding can be interpreted as PV × PW → P(V ⊗ W) via

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the surjective map of graded rings
Sym• (V ∨ ⊗ W ∨ )“in the opposite direction”.

% % ⊕∞

i=0

$

% $
%
Symi V ∨ ⊗ Symi W ∨9.6.D. E XERCISE : A COORDINATE - FREE DESCRIPTION OF PRODUCTS OF PROJEC TIVE A- SCHEMES IN GENERAL .
Suppose that S• and T• are finitely generated
graded rings over A. Describe an isomorphism∼ Proj ⊕∞(Proj S• ) ×A (Proj T• ) =n=0 (Sn ⊗A Tn )(where hopefully the definition of multiplication in the graded ring ⊕∞n=0 (Sn ⊗ATn ) is clear).

9.7 Normalization
We discuss normalization now only because the central construction gives
practice with the central idea behind the construction in §9.1 of the fibered product(see Exercises 9.7.B and 9.7.I).
Normalization is a means of turning a reduced scheme into a normal scheme.
(Unimportant remark: reducedness is not a necessary hypothesis, and is included
only to avoid distraction. If you care, you can follow through the construction,and realize that the normalization of a scheme factors through the reduction, andis the normalization of the reduction.) A normalization of a reduced scheme X is a
3 → X from a normal scheme, where ν induces a bijection of irremorphism ν : X
3 and X, and ν gives a birational morphism on each of the
ducible components of X
irreducible components. It will satisfy a universal property, and hence it will be
unique up to unique isomorphism. Figure 7.4 is an example of a normalization.
We begin with the case where X is irreducible, and hence (by Exercise 5.2.F)
integral. (We will then deal with a more general case, and also discuss normalization in a function field extension.) In this case of irreducible X, the normalization
3 → X is a dominant morphism (not just dominant rational map!) from an irreν: X
ducible normal scheme to X, such that any other such morphism factors through
ν:
∃!
3
%X
normalY❂
normal
❂❂
✁
✁
❂❂
✁
✁
❂
✁✁
dominant ❂❂F G✁✁ ν dominantX
Thus if the normalization exists, then it is unique up to unique isomorphism.
We now have to show that it exists, and we do this in a way that will look familiar.
We deal first with the case where X is affine, say X = Spec A, where A is an integral
3 be the integral closure of A in its fraction field K(A).
domain. Then let A
3 →9.7.A. E XERCISE . (Recall that A is an integral domain.) Show that ν : Spec A
Spec A satisfies the universal property of normalization. (En route, you mightshow that the global sections of an irreducible normal scheme are also “normal”,
i.e., integrally closed.)

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9.7.B. I MPORTANT ( BUT SURPRISINGLY EASY ) E XERCISE . Show that normalizations of integral schemes exist in general. (Hint: Ideas from the existence of fiberedproducts, §9.1, may help.)
9.7.C. E ASY E XERCISE .
Show that normalizations are integral and surjective.
(Hint for surjectivity: the Lying Over Theorem, see §7.2.6.)
We will soon see that normalization of integral finite type k-schemes is always
a birational morphism, in Exercise 9.7.N.

9.7.D. E XERCISE . Explain (by defining a universal property) how to extend thenotion of normalization to the case where X is a reduced scheme, with possibly
more than one component, but under the hypothesis that every affine open subsetof X has finitely many irreducible components. Note that this includes all locally
Noetherian schemes. (If you wish, you can show that the normalization exists inthis case. See [Stacks, tag 035Q] for more.)
Here are some examples.
9.7.E. E XERCISE ( NORMALIZATION OF THE NODAL CUBIC ). Suppose that char k ̸=
2. Show that Spec k[t] → Spec k[x, y]/(y2 − x2 (x + 1)) given by (x, y) (→ (t2 −1, t(t2 − 1)) (see Figure 7.4) is a normalization. The target curve is called the nodal
cubic curve. (Hint: show that k[t] and k[x, y]/(y2 − x2 (x + 1)) have the same
fraction field. Show that k[t] is integrally closed. Show that k[t] is contained in the
integral closure of k[x, y]/(y2 − x2 (x + 1)).)
You will see from the previous exercise that once we guess what the normalization is, it isn’t hard to verify that it is indeed the normalization. Perhaps a few
words are in order as to where the polynomials t2 − 1 and t(t2 − 1) arose in theprevious exercise. The key idea is to guess t = y/x. (Then t2 = x + 1 and y = xt
quickly.) This idea comes from three possible places. We begin by sketching the
curve, and noticing the “node” at the origin. (“Node” will be formally defined in§29.3.1.) (a) The function y/x is well-defined away from the node, and at the node,
the two branches have “values” y/x = 1 and y/x = −1. (b) We can also note that if
t = y/x, then t2 is a polynomial, so we will need to adjoin t in order to obtain thenormalization. (c) The curve is cubic, so we expect a general line to meet the cubicin three points, counted with multiplicity. (We will make this precise when we discuss Bézout’s Theorem, Exercise 18.6.K, but in this case we have already gotten ahint of this in Exercise 6.5.G.) There is a P1 parametrizing lines through the origin(with coordinate equal to the slope of the line, y/x), and most such lines meet the
curve with multiplicity two at the origin, and hence meet the curve at precisely
one other point of the curve. So this “coordinatizes” most of the curve, and we try
adding in this coordinate.
9.7.F. E XERCISE . Find the normalization of the cusp y2 = x3 (see Figure 9.4).
(“Cusp” will be formally defined in Definition 29.3.3.)
9.7.G. E XERCISE . Suppose char k ̸= 2. Find the normalization of the tacnodey2 = x4 , and draw a picture analogous to Figure 9.4. (“Tacnode” will be formallydefined in Definition 29.3.3.)

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F IGURE 9.4. Normalization of a cusp
(Although we haven’t defined “singularity”, “smooth”, “curve”, or “dimension”, you should still read this.) Notice that in the previous examples, normalization “resolves” the singularities (“non-smooth” points) of the curve. In general, it
will do so in dimension one (in reasonable Noetherian circumstances, as normal local Noetherian integral domains of dimension one are all discrete valuation rings,§12.5), but won’t do so in higher dimension (the cone z2 = x2 + y2 over a field kof characteristic not 2 is normal, Exercise 5.4.I(b)).
3 → X.
9.7.H. E XERCISE . Suppose X = Spec Z[15i]. Describe the normalization X(Hint: Z[i] is a unique factorization domain, §5.4.6(0), and hence is integrally
closed by Exercise 5.4.F.) Over what points of X is the normalization not an isomorphism?
Another exercise in a similar vein is the normalization of the “pinched plane”,
Exercise 12.5.I.
9.7.I. E XERCISE ( NORMALIZATION IN A FUNCTION FIELD EXTENSION , AN IMPOR TANT GENERALIZATION ). Suppose X is an integral scheme. The normalization
3 → X of X in a given algebraic field extension L of the function field K(X) of
ν: X
3 with function field L,
X is a dominant morphism from a normal integral scheme X
such that ν induces the inclusion K(X) "→ L, and that is universal with respect tothis property.
Spec L❑Spec K(Y)
❑❑❑❑
❑❑❑❑
❑❑❑❑
❑❑❑❑ ∼❑"
3
Spec K(X)"Spec K(X)%Y

normal∃!

"3%X

normal" 5%XShow that the normalization in a finite extension of fields exists.
The following two examples, one arithmetic and one geometric, show that thisis an interesting construction.

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9.7.J. E XERCISE . Suppose X = Spec Z (with function field Q). Find its integral closure in the field extension Q(i). (There is no “geometric” way to do this; it is purely
an algebraic problem, although the answer should be understood geometrically.)
9.7.1. Remark: Rings of integers in number fields. A finite extension K of Q is called anumber field, and the integral closure of Z in K the ring of integers in K, denoted
OK . (This notation is awkward given our other use of the symbol O.)
Spec K% Spec OK"
Spec Q"% Spec Z

By the previous exercises, OK is a normal integral domain, and we will see soon
(Theorem 9.7.3(a)) that it is Noetherian, and later (Exercise 11.1.F) that it has “dimension 1”. This is an example of a Dedekind domain, see §12.5.15. We will thinkof it as a “smooth” curve as soon as we define what “smooth” (really, regular) and“curve” mean.
√
9.7.K. E XERCISE . Find the ring of integers in Q( n), where n is square-free andn ≡ 3 (mod 4). (Hint: Exercise 5.4.I(a), where you will also be able to figure out
the answer for square-free n in general.)
9.7.L. E XERCISE . Suppose char k ̸= 2 for convenience (although it isn’t necessary).
(a) Suppose X = Spec k[x] (with function field k(x)). Find its normalization in thefield extension k(y), where y2 = x2 + x. (Again we get a Dedekind domain.) Hint:
this can be done without too much pain. Show that Spec k[x, y]/(x2 + x − y2 )is normal, possibly by identifying it as an open subset of P1k , or possibly using
Exercise 5.4.H.
(b) Suppose X = P1 , with distinguished open Spec k[x]. Find its normalizationin the field extension k(x, y), where y2 = x2 + x. (Part (a) involves computing
the normalization over one affine open set; now figure out what happens over the“other” affine open set, and how to glue.)
9.7.2. Fancy fact: Finiteness of integral closure.
The following fact is useful.
9.7.3. Theorem (finiteness of integral closure). — Suppose A is a Noetherian integral
domain, K = K(A), L/K is a finite extension of fields, and B is the integral closure of A in
L (“the integral closure of A in the field extension L/K”, i.e., those elements of L integral
over A).
(a) If A is integrally closed and L/K is separable, then B is a finitely generated A-module.
(b) (E. Noether) If A is a finitely generated k-algebra, then B is a finitely generated Amodule.
Eisenbud gives a proof in a page and a half: (a) is [E, Prop. 13.14] and (b) is [E,Cor. 13.13]. A sketch is given in §9.7.5.
9.7.4. Warning. Part (b) does not hold for Noetherian A in general. In fact, theintegral closure of a Noetherian ring need not be Noetherian (see [E, p. 299] for

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some discussion). This is alarming. The existence of such an example is a sign that
Theorem 9.7.3 is not easy.
9.7.M. E XERCISE .
3→(a) Show that if X is an integral finite type k-scheme, then its normalization ν : X
X is a finite morphism.
(b) Suppose X is a locally Noetherian integral scheme. Show that if X is normal,then the normalization in a finite separable field extension is a finite morphism.
Show that if X is an integral finite type k-scheme, then the normalization in a finite
extension of fields is a finite morphism. In particular, once we define “variety”(Definition 10.1.7), you will see that this implies that the normalization of a variety(including in a finite field extension) is a variety.
9.7.N. E XERCISE . Show that if X is an integral finite type k-scheme, then the normalization morphism is birational. (Hint: Proposition 6.5.7; or solve Exercise 9.7.O
first.)
9.7.O. E XERCISE . Suppose that X is an integral finite type k-scheme. Show that thenormalization map of X is an isomorphism on an open dense subset of X. Hint:
Proposition 6.5.5.
9.7.P. E XERCISE .
Suppose ρ : Z → X is a finite birational morphism from anirreducible variety to an irreducible normal variety. Show that ρ extends to an
isomorphism.
9.7.5. ⋆⋆ Sketch of proof of finiteness of integral closure, Theorem 9.7.3. Here is a
sketch to show the structure of the argument. It uses commutative algebra ideas
from Chapter 11, so you should only glance at this to see that nothing fancy isgoing on. Part (a): reduce to the case where L/K is Galois, with group {σ1 , . . . , σn }.
Choose b1 , . . . , bn ∈ B forming a K-vector space basis of L. Let M be the matrix(familiar from Galois theory) with ijth entry σi bj , and let d = det M. Show thatthe entries of M lie in B, and that d2 ∈ K (as d2 is Galois-fixed). Show that d ̸= 0
using linear independence of characters. Then complete the proof by showingthat B ⊂ d−2 (Ab1 + · · · + Abn ) (submodules of finitely generated modules over
Noetherian rings,are also Noetherian, Exercise 3.6.X) as follows. Suppose b ∈ B,
and write b =ci bi (ci ∈ K). If c is the column vector with entries ci , showthat the ith entry of the column vector Mc is σi b ∈ B. Multiplying Mc on theleft by adj M (see the trick of the proof of Lemma 7.2.1), show that dci ∈ B. Thus
d2 ci ∈ B ∩ K = A (as A is integrally closed), as desired.
For (b), use the Noether Normalization Lemma 11.2.4 to reduce to the caseA = k[x1 , . . . , xn ]. Reduce to the case where L is normally closed over K. Let L ′ be
the subextension of L/K so that L/L ′ is Galois and L ′ /K is purely inseparable. Usepart (a) to reduce to the case L = L ′ . If L ′ ̸= K, then for some q, L ′ is generated
over K by the qth root of a finite set of rational functions. Reduce to the case L ′ =1/q
1/qk ′ (x1 , . . . , xn ) where k ′ /k is a finite purely inseparable extension. In this case,
1/q
1/q
show that B = k ′ [x1 , . . . , xn ], which is indeed finite over k[x1 , . . . , xn ].
!

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CHAPTER 10

Separated and proper morphisms, and (finally!)
varieties

10.1 Separated morphisms (and quasiseparatedness done
properly)
Separatedness is a fundamental notion. It is the analog of the Hausdorff condition
for manifolds (see Exercise 10.1.A), and as with Hausdorffness, this geometricallyintuitive notion ends up being just the right hypothesis to make theorems work.
Although the definition initially looks odd, in retrospect it is just perfect.
10.1.1. Motivation. Let’s review why we like Hausdorffness. Recall that a topological space is Hausdorff if for every two points x and y, there are disjoint open
neighborhoods of x and y. The real line is Hausdorff, but the “real line with doubled origin” (of which Figure 4.6 may be taken as a sketch) is not. Many proofs
and results about manifolds use Hausdorffness in an essential way. For example,
the classification of compact one-dimensional manifolds is very simple, but if theHausdorff condition were removed, we would have a very wild set.
So once armed with this definition, we can cheerfully exclude the line withdoubled origin from civilized discussion, and we can (finally) define the notion ofa variety, in a way that corresponds to the classical definition.
With our motivation from manifolds, we shouldn’t be surprised that all of ouraffine and projective schemes are separated: certainly, in the land of manifolds, theHausdorff condition comes for free for “subsets” of manifolds. (More precisely, if
Y is a manifold, and X is a subset that satisfies all the hypotheses of a manifold
except possibly Hausdorffness, then Hausdorffness comes for free. Similarly, we
will see that locally closed embeddings in something separated are also separated:combine Exercise 10.1.B and Proposition 10.1.13(a).)
As an unexpected added bonus, a separated morphism to an affine scheme
has the property that the intersection of two affine open sets in the source is affine(Proposition 10.1.8). This will make Čech cohomology work very easily on (quasicompact) schemes (Chapter 18). You might consider this an analog of the fact thatin Rn , the intersection of two convex sets is also convex. As affine schemes are
trivial from the point of view of quasicoherent cohomology, just as convex sets in
Rn have no cohomology, this metaphor is apt.
A lesson arising from the construction is the importance of the diagonal morphism. More precisely, given a morphism π : X → Y, good consequences can beleveraged from good behavior of the diagonal morphism δπ : X → X ×Y X (the279

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product of the identity morphism X → X with itself), usually through fun diagramchases. This lesson applies across many fields of geometry. (Another nice gift ofthe diagonal morphism: it will give us a good algebraic definition of differentials,
in Chapter 21.)Grothendieck taught us that one should try to define properties of morphisms,
not of objects; then we can say that an object has that property if its morphism tothe final object has that property. We discussed this briefly at the start of Chapter 7.
In this spirit, separatedness will be a property of morphisms, not schemes.
10.1.2. Defining separatedness. Before we define separatedness, we make anobservation about all diagonal morphisms.
10.1.3. Proposition. — Let π : X → Y be a morphism of schemes. Then the diagonal
morphism δ : X → X ×Y X is a locally closed embedding.

We will often use δ to denote a diagonal morphism. The corresponding locallyclosed subscheme of X ×Y X (which we also call the diagonal) will be denoted ∆.

Proof. We will describe a union of open subsets of X ×Y X covering the image of X,
such that the image of X is a closed embedding in this union.
Say Y is covered with affine open sets Vi and X is covered with affine open
sets Uij , with π : Uij → Vi . Note that Uij ×Vi Uij is an affine open subscheme ofthe product X ×Y X (basically this is how we constructed the product, by gluing
together affine building blocks). Then the diagonal is covered by these affine open
subsets Uij ×Vi Uij . (Any point p ∈ X lies in some Uij ; then δ(p) ∈ Uij ×Vi Uij .
Figure 10.1 may be helpful.) Note that δ−1 (Uij ×Vi Uij ) = Uij : clearly Uij ⊂δ−1 (Uij ×Vi Uij ), and because pr1 ◦ δ = idX (where pr1 is the first projection),δ−1 (Uij ×Vi Uij ) ⊂ Uij . Finally, we check that Uij → Uij ×Vi Uij is a closed
embedding. Say Vi = Spec B and Uij = Spec A. Then this corresponds to thenatural ring map A⊗B A → A (a1 ⊗a2 (→ a1 a2 ), which is obviously surjective. !
The open subsets we described may not cover X ×Y X, so we have not shownthat δ is a closed embedding.

10.1.4. Definition. A morphism π : X → Y is separated if the diagonal morphismδπ : X → X ×Y X is a closed embedding. An A-scheme X is said to be separated
over A if the structure morphism X → Spec A is separated. When people saythat a scheme (rather than a morphism) X is separated, they mean implicitly thatsome “structure morphism” is separated. For example, if they are talking about
A-schemes, they mean that X is separated over A.
Thanks to Proposition 10.1.3 (and once you show that a locally closed embedding whose image is closed is actually a closed embedding), a morphism isseparated if and only if the diagonal ∆ is a closed subset — a purely topologicalcondition on the diagonal. This is reminiscent of a definition of Hausdorff, as thenext exercise shows.
10.1.A. U NIMPORTANT E XERCISE ( FOR THOSE SEEKING TOPOLOGICAL MOTIVA TION ). Show that a topological space X is Hausdorff if and only if the diagonal
is a closed subset of X × X. (The reason separatedness of schemes doesn’t give
Hausdorffness — i.e., that for any two open points x and y there aren’t necessarily
disjoint open neighborhoods — is that in the category of schemes, the topological

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X ×Y XXX

F IGURE 10.1. an open neighborhood of the diagonal is coveredby Uij ×Vi Uij
space X × X is not in general the product of the topological space X with itself, see§9.1.2.)10.1.B. I MPORTANT EASY EXERCISE . Show locally closed embeddings (and inparticular open and closed embeddings) are separated. (Hint: Do this by hand.
Alternatively, show that monomorphisms are separated. Open and closed embeddings are monomorphisms, by Exercise 9.2.G.)
10.1.C. I MPORTANT EASY EXERCISE . Show that every morphism of affine schemes
is separated. (Hint: this was essentially done in the proof of Proposition 10.1.3.)
10.1.D. E XERCISE . Show that the line with doubled origin (Example 4.4.5) is notseparated, by verifying that the image of the diagonal morphism is not closed.
(Another argument is given below, in Exercise 10.2.C. A fancy argument is given
in Exercise 12.7.C.)
We next come to our first example of something separated but not affine. The
following single calculation will imply that all quasiprojective A-schemes are separated (once we know that the composition of separated morphisms are separated,
Proposition 10.1.13).
10.1.5. Proposition. — The morphism Pn
A → Spec A is separated.

We give two proofs. The first is by direct calculation. The second requires
no calculation, and just requires that you remember some classical constructions
described earlier.

n
Proof 1: Direct calculation. We cover Pn
A ×A PA with open sets of the form Ui ×A Uj ,where U0 , . . . , Un form the “usual” affine open cover. The case i = j was takencare of before, in the proof of Proposition 10.1.3. If i ̸= j then∼ Spec A[x0/i , . . . , xn/i , y0/j , . . . , yn/j ]/(xi/i − 1, yj/j − 1).
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Now the restriction of the diagonal ∆ is contained in Ui (as the diagonal morphism
composed with projection to the first factor is the identity), and similarly is contained in Uj . Thus the diagonal morphism over Ui ×A Uj is Ui ∩ Uj → Ui ×A Uj .
This is a closed embedding, as the corresponding map of rings
A[x0/i , . . . , xn/i , y0/j , . . . , yn/j ] → A[x0/i , . . . , xn/i , x−1
j/i ]/(xi/i − 1)(given by xk/i (→ xk/i , yk/j (→ xk/i /xj/i ) is clearly a surjection (as each generatorof the ring on the right is clearly in the image — note that x−1
j/i is the image of
yi/j ).
!
n
Proof 2: Classical geometry. Note that the diagonal morphism δ : Pn
A → PA × A
2
n
n
n +2n
Pn
(§9.6, a closed
A followed by the Segre embedding S : PA ×A PA → P
embedding) can also be factored as the second Veronese embedding ν2 : Pn
A →
n+2
n+2
2
−1
−1
n
+2n
(
)
(
)
P 2
(§8.2.6) followed by a linear map L : P 2→P(another closed
embedding, Exercise 8.2.D), both of which are closed embeddings.
nPn
AH ×A P▲A
▲▲▲
✉
✉
δ ✉✉
▲▲S
✉
✉✉l. (?) cl. emb. cl. emb. ▲▲▲#
✉
✉✉

2

Pn
PnI +2n
A ❍
O
❍❍
O
O
❍❍cl. emb.
cl. emb.OOO❍❍Oν2 ❍❍+OOO L
n+2
P( 2 )−1
Informally, in coordinates:([x0 , x1 , . . . , xn ], [x0 , x1 , . . . , xn ])
J
❑❑
❑❑ S
PP
❑❑
P
P
❑❑
δ PP
❑#
P
⎡
P
P
x20 ,
x0 x1 , · · ·
PP
P
⎢
P
x
x
,
x21 ,
···
1
0
⎢
[x0 , x1 , . . . , xn ]
⎢
.
.
..
..
..
❂❂
⎣
.
❂❂
❂❂
xn x0 , xn x1 , · · ·
❂❂
I
&&
ν2 ❂❂
&
&
❂❂
&
&&
❂F
&& L
[x20 , x0 x1 , . . . , xn−1 xn , x2n ]⎤x0 xn ,
x1 xn , ⎥
⎥
⎥
..
⎦
.
x2n

2
2
The composed map Pn
A may be written as [x0 , . . . , xn ] (→ [x0 , x0 x1 , x0 x2 , . . . , xn ],where the subscripts on the right run over all ordered pairs (i, j) where 0 ≤ i, j ≤n.) This forces δ to send closed sets to closed sets (or else S ◦ δ won’t, but L ◦ ν2does).
!

We note for future reference a minor result proved in the course of Proof 1.
10.1.6. Small Proposition. — If U and V are open subsets of an A-scheme X, then
∼ U ∩ V.
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Figure 10.2 may help show why this is natural. You could also interpret thisstatement as∼ U ×X V
X ×(X× X) (U ×A V) =A

which follows from the magic diagram, Exercise 1.3.S.

U×X

∆

U×V

X×V
∼ (U × V) ∩ ∆U∩V =F IGURE 10.2. Small Proposition 10.1.6

10.1.7. Important Definition (finally!). A variety over a field k, or k-variety, isa reduced, separated scheme of finite type over k. For example, a reduced finite
type affine k-scheme is a variety. (Do you see why separatedness holds?) The notion of variety generalizes our earlier notion of affine variety (§5.3.7) and projective
variety (§5.3.E, see Proposition 10.1.14). (Notational caution: In some sources, theadditional condition of irreducibility is imposed. Also, it is often assumed that k
is algebraically closed.)
Varieties over k form a category: morphisms of k-varieties are just morphisms
as k-schemes. We will shortly see (in Exercise 10.1.L) that such morphisms areautomatically finite type and separated. (Of course, the category of varieties overan algebraically closed field k long predates modern scheme theory.)
A subvariety of a variety X is a reduced locally closed subscheme of X (which
you can quickly check is a variety itself). An open subvariety of X is an open
subscheme of X. (Reducedness is automatic in this case.) A closed subvariety of X
is a reduced closed subscheme of X.
If you have read the double-starred section on group objects in a category(§6.6.4), you will automatically have the notion of a group variety. We will discuss
this a bit more in §10.3.9.
10.1.E. E XERCISE ( PRODUCTS OF IRREDUCIBLE VARIETIES OVER k ARE IRREDUCIBLE
VARIETIES ). Use Exercise 9.4.E and properties of separatedness to show that the
product of two irreducible k-varieties is an irreducible k-variety.

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10.1.F. ⋆⋆ E XERCISE ( COMPLEX ALGEBRAIC VARIETIES YIELD COMPLEX ANALYTIC
VARIETIES ; FOR THOSE WITH SUFFICIENT BACKGROUND ). Show that the analytification (Exercises 5.3.G and 6.3.N) of a complex algebraic variety is a complex
analytic variety.
Here is a very handy consequence of separatedness.
10.1.8. Proposition. — Suppose X → Spec A is a separated morphism to an affine
scheme, and U and V are affine open subsets of X. Then U ∩ V is an affine open subset ofX.
Before proving this, we state a consequence that is otherwise nonobvious. If
X = Spec A, then the intersection of any two affine open subsets of A is an affine
open subset. This is certainly not an obvious fact! We know the intersection of two
distinguished affine open sets is affine (from D(f) ∩ D(g) = D(fg)), but we havelittle handle on affine open sets in general.
Warning: this property does not characterize separatedness. For example, if
A = Spec k and X is the line with doubled origin over k, then X also has thisproperty.
∼ U ∩ V, where ∆ is the diagonal.
Proof. By Proposition 10.1.6, (U ×A V) ∩ ∆ =But U ×A V is affine (the fibered product of two affine schemes over an affine
scheme is affine, Step 1 of our construction of fibered products, Theorem 9.1.1),and (U ×A V) ∩ ∆ is a closed subscheme of the affine scheme U ×A V, and hence
U ∩ V is affine.
!
10.1.9. Redefinition: Quasiseparated morphisms.
We say a morphism π : X → Y is quasiseparated if the diagonal morphismδπ : X → X ×Y X is quasicompact.

10.1.G. E XERCISE . Show that this agrees with our earlier definition of quasiseparated (§7.3.1): show that π : X → Y is quasiseparated if and only if for any affine
open Spec A of Y, and two affine open subsets U and V of X mapping to Spec A,
U ∩ V is a finite union of affine open sets. (Possible hint: compare this to Proposition 10.1.8. Another possible hint: the magic diagram, Exercise 1.3.S.)
Here are two large classes of morphisms that are quasiseparated.
10.1.H. E ASY E XERCISE . Show that separated morphisms are quasiseparated.
(Hint: closed embeddings are affine, hence quasicompact.)
Second, any morphism from a locally Noetherian scheme is quasiseparated,
see Exercise 7.3.B(b), so (locally) Noetherian people need never worry about thisissue.
We now give four quick propositions showing that separatedness and quasiseparatedness behave well, just as many other classes of morphisms did.
10.1.10. Proposition. — Both separatedness and quasiseparatedness are preserved bybase change.

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Proof. Suppose
W

%X

"
Y

"
%Z

is a Cartesian diagram. We will show that if Y → Z is separated or quasiseparated,then so is W → X. Then you can quickly verify that
W
"
Y

δW

δY

% W ×X W"% Y ×Z Y

is a Cartesian diagram. (This is true in any category with fibered products.) As theproperty of being a closed embedding is preserved by base change (§9.2 (3)), if δY
is a closed embedding, so is δW .
The quasiseparatedness case follows in the identical manner, as quasicompactness is also preserved by base change (Exercise 9.4.B(a)).
!
10.1.11. Proposition. — The condition of being separated is local on the target. Precisely,a morphism π : X → Y is separated if and only if for any cover of Y by open subsets Ui ,π−1 (Ui ) → Ui is separated for each i.

10.1.12. Hence affine morphisms are separated, as every morphism of affine schemes
is separated (Exercise 10.1.C). In particular, finite morphisms are separated.
Proof. If π : X → Y is separated, then for any Ui "→ Y, π−1 (Ui ) → Ui is separated, as
separatedness is preserved by base change (Theorem 10.1.10). Conversely, to checkif ∆ "→ X×Y X is a closed subset, it suffices to check this on an open cover of X×Y X.
Let ρ : X ×Y X → Y be the natural map. We will use the open cover ρ−1 (Ui ), whichby construction of the fibered product is π−1 (Ui ) ×Ui π−1 (Ui ). As π−1 (Ui ) → Ui
is separated, π−1 (Ui ) → π−1 (Ui )×Ui π−1 (Ui ) is a closed embedding by definitionof separatedness.
!

10.1.I. E XERCISE . Prove that the condition of being quasiseparated is local on
the target. (Hint: the condition of being quasicompact is local on the target by
Exercise 7.3.C(a); use a similar argument as in Proposition 10.1.11.)
10.1.13. Proposition. —(a) The condition of being separated is closed under composition. In other words, if π : X →Y is separated and ρ : Y → Z is separated, then ρ ◦ π : X → Z is separated.
(b) The condition of being quasiseparated is closed under composition.
Proof. (a) Let τ = ρ ◦ π. We are given that δπ : X "→ X ×Y X and δρ : Y "→ Y ×Z Y
are closed embeddings, and we wish to show that δτ : X → X ×Z X is a closed

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embedding. Consider the diagram

X!

"

δπ

% X ×Y X
" "
Y!

ψ

δρ

% X ×Z X"% Y ×Z Y.

The square is the magic diagram (Exercise 1.3.S). As δρ is a closed embedding, ψ
is too (closed embeddings are preserved by base change, §9.2 (3)). Thus ψ ◦ δπ isa closed embedding (the composition of two closed embeddings is also a closed
embedding, Exercise 8.1.C).
(b) The identical argument (with “closed embedding” replaced by “quasicompact”) shows that the condition of being quasiseparated is closed under composition.
!
10.1.14. Corollary. — Every quasiprojective A-scheme is separated over A. In particular,
every reduced quasiprojective k-scheme is a k-variety.
Proof. Suppose X → Spec A is a quasiprojective A-scheme. The structure morphism can be factored into an open embedding composed with a closed embedding followed by Pn
A → Spec A. Open embeddings and closed embeddings are
separated (Exercise 10.1.B), and Pn
A → Spec A is separated (Proposition 10.1.5).
Compositions of separated morphisms are separated (Proposition 10.1.13), so weare done.
!
10.1.15. Proposition. — Suppose π : X → Y and π ′ : X ′ → Y ′ are separated (resp. quasiseparated) morphisms of S-schemes (where S is a scheme). Then the product morphismπ × π ′ : X ×S X ′ → Y ×S Y ′ is separated (resp. quasiseparated).
Proof. Apply Exercise 9.4.F.

!

10.1.16. Applications.
As a first application, we define the graph of a morphism.
10.1.17. Definition. Suppose π : X → Y is a morphism of Z-schemes. The morphismΓπ : X → X×Z Y given by Γπ = (id, π) is called the graph morphism. Then π factors
as pr2 ◦ Γπ , where pr2 is the second projection (see Figure 10.3). The diagram of
Figure 10.3 is often called the graph of a morphism. (We will discuss graphs ofrational maps in §10.2.4.)
10.1.18. Proposition. — The graph morphism Γ is always a locally closed embedding. If
Y is a separated Z-scheme (i.e., the structure morphism Y → Z is separated), then Γ is a
closed embedding. If Y is a quasiseparated Z-scheme, then Γ is quasicompact.
This will be generalized in Exercise 10.1.M.

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X ×Z Y
Y

pr2

Γπ

pr1

π

X

F IGURE 10.3. The graph morphism
Proof by Cartesian diagram. A special case of the magic diagram (Exercise 1.3.S) is:(10.1.18.1)

XΓπ% X ×Z Y

π

"
Y

δ

"
% Y ×Z Y.

The notions of locally closed embedding and closed embedding are preserved bybase change, so if the bottom arrow δ has one of these properties, so does the top.
The same argument establishes the last sentence of Proposition 10.1.18.
!
We next come to a strange-looking, but very useful, result. Like the magic
diagram, this result is unexpectedly ubiquitous.
10.1.19. Cancellation Theorem for a Property P of Morphisms. — Let P be a classof morphisms that is preserved by base change and composition. (Any “reasonable” classof morphisms will satisfy this, see §7.1.1.) Suppose
X❄
❄❄
❄❄
❄
τ ❄❄
8

π

Z

%Y
⑧
⑧
⑧⑧
⑧⑧ ρ
5⑧⑧

is a commuting diagram of schemes.
Suppose that the diagonal morphism δρ : Y →Y ×Z Y is in P and τ : X → Z is in P. Then π : X → Y is in P. In particular:(i) Suppose that locally closed embeddings are in P. If τ is in P, then π is in P.
(ii) Suppose that closed embeddings are in P (e.g., P could be finite morphisms, morphisms of finite type, closed embeddings, affine morphisms). If τ is in P and ρ isseparated, then π is in P.

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The Rising Sea: Foundations of Algebraic Geometry(iii) Suppose that quasicompact morphisms are in P. If τ is in P and ρ is quasiseparated, then π is in P.
The following diagram summarizes this important theorem:
∴∈P

X❄
❄❄
❄❄
❄
∈P ❄❄
8

Z

%Y
⑧
⑧
⑧⑧
⑧⑧ δ∈P⑧
5⑧

When you plug in different P, you get very different-looking (and nonobvious)
consequences, the first of which are given in Exercise 10.1.J. (Here are some facts
you can prove easily, but which can be interpreted as applications of the Cancellation Theorem in Sets, and which may thus shed light on how the Cancellation
Theorem works. If π : X → Y and ρ : Y → Z are maps of sets, and ρ ◦ π is injective,then so is π; and if ρ ◦ π is surjective and ρ is injective, then π is surjective.)
Proof. By the graph Cartesian diagram (10.1.18.1)
ΓπX% X ×Z Y

π

"
Y

δρ

"
% Y ×Z Y

we see that the graph morphism Γπ : X → X ×Z Y is in P (Definition 10.1.17), as P
is preserved by base change. By the Cartesian square
X ×Z Y

τ′

%Y
ρ

"
X

τ

"
%Z

the projection τ ′ : X ×Z Y → Y is in P as well. Thus π = τ ′ ◦ Γπ is in P!

Here now are some fun and useful exercises.

10.1.J. L ESS IMPORTANT E XERCISE . Show that an A-scheme is separated (over A)
if and only if it is separated over Z. In particular, a complex scheme is separated
over C if and only if it is separated over Z, so complex geometers and arithmetic
geometers can discuss separated schemes without confusion.
ρ
π %
%Z.
10.1.K. E ASY E XERCISE . Suppose we have morphisms X
Y(a) Show that if ρ ◦ π is a locally closed embedding (resp. locally of finite type,separated), then so is π.
(b) If ρ◦π is quasicompact, and Y is Noetherian, show that π is quasicompact. Hint:
Exercise 7.3.B(a).
(c) If ρ ◦ π is quasiseparated, show that π is quasiseparated. Hint: Exercise 7.3.B(b).

10.1.L. E ASY E XERCISE . Show that morphisms of k-varieties (i.e., morphisms as
k-schemes, see §10.1.7) are finite type and separated.

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10.1.M. E XERCISE . Suppose µ : Z → X is a morphism, and σ : X → Z is a sectionof µ, i.e., µ ◦ σ is the identity on X.
ZK
µ

σ

"
X
Show that σ is a locally closed embedding. Show that if µ is separated, then σ isa closed embedding. (This generalizes Proposition 10.1.18.) Give an example toshow that σ need not be a closed embedding if µ is not separated.
10.1.N. L ESS IMPORTANT EXERCISE . Suppose P is a class of morphisms such thatclosed embeddings are in P, and P is preserved by fibered product and composition. Show that if π : X → Y is in P then πred : Xred → Y red is in P. (Two examples
are the classes of separated morphisms and quasiseparated morphisms.) Hint:
πred% X ×Y Y red
Xred❏
❏❏
❏❏
❏❏
❏❏
❏❏ "
#
π
X

@%

Y red
"
%Y

10.1.20. ⋆⋆ Universally injective morphisms and the diagonal.
10.1.O. E ASY E XERCISE . If π : X → Y and ρ : Y → Z are morphisms, and ρ ◦ π isuniversally injective, show that π is universally injective.

10.1.P. E XERCISE .
(a) Show that π : X → Y is universally injective if and only if the diagonal morphism δπ : X → X ×Y X is surjective. (Note that δπ is always injective, as it is alocally closed embedding, by Proposition 10.1.3.)(b) Show that universally injective morphisms are separated.
(c) Show that a map between finite type schemes over an algebraically closed field
k is universally injective if and only if it is injective on closed points.

10.2 Rational maps to separated schemes
When we introduced rational maps in §6.5, we promised that in good circumstances, a rational map has a “largest domain of definition”. We are now readyto make precise what “good circumstances” means, in the Reduced-to-Separated
Theorem 10.2.2. We first introduce an important result making sense of locus
where two morphisms with the same source and target “agree”.

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10.2.A. U SEFUL EXERCISE : T HE LOCUS WHERE TWO MORPHISMS AGREE . Suppose
π : X → Y and π ′ : X → Y are two morphisms over some scheme Z.
π

:L% Y
X❄
❄❄
⑧
❄❄ π ′ ⑧⑧⑧❄❄⑧❄8 5⑧⑧⑧
Z

We can now give meaning to the phrase ’the locus where π and π ′ agree’, andthat in particular there is a largest locally closed subscheme where they agree —which is closed if Y is separated over Z. Suppose µ : W → X is some morphism(not assumed to be a locally closed embedding). We say that π and π ′ agree on µ if
π ◦ µ = π ′ ◦ µ. Show that there is a locally closed subscheme i : V "→ X on which π
and π ′ agree, such that any morphism µ : W → X on which π and π ′ agree factors
uniquely through i, i.e., there is a unique j : W → V such that µ = i ◦ j. Showfurther that if Y → Z is separated, then i : V "→ X is a closed embedding. Hint:
define V to be the following fibered product:
V

%Y

δ

" (π,π ′ )"% Y ×Z Y.
XAs δ is a locally closed embedding, V → X is too. Then if µ : W → X is any
morphism such that π ◦ µ = π ′ ◦ µ, then µ factors through V.

The fact that the locus where two maps agree can be nonreduced should notcome as a surprise: consider two maps from A1k to itself, π(x) = 0 and π ′ (x) = x2 .
They agree when x = 0, but the situation is (epsilonically) better than that — theyshould agree even on Spec k[x]/(x2 ).

10.2.1. Minor Remarks.
(i) In the previous exercise, we are describing V "→ X by way of a universal
property. Taking this as the definition, it is not a priori clear that V is a locallyclosed subscheme of X, or even that it exists.
(ii) Warning: consider two maps from Spec C to itself over Spec R, the identity
and complex conjugation. These are both maps from a point to a point, yet they
do not agree despite agreeing as maps of sets. (If you are not convinced that they
disagree as morphisms, this might help: after base change Spec C → Spec R, they
do not agree even as maps of sets.)(iii) More generally, the locus where π and π ′ agree can be interpreted as follows: π and π ′ agree at p if π(p) = π ′ (p) and the two maps of residue fields are
the same. (You may enjoying thingking this through as an exercise.)
10.2.B. E XERCISE : MAPS OF k- VARIETIES ARE DETERMINED BY THE MAPS ON CLOSEDPOINTS .
Suppose π : X → Y and π ′ : X → Y are two morphisms of k-varietiesthat are the same at the level of closed points (i.e., for each closed point p ∈ X,
π(p) = π ′ (p)). Show that π = π ′ . (This implies that the functor from the category of “classical varieties over k”, which we won’t define here, to the category of k-schemes, is fully faithful. Can you generalize this appropriately to nonalgebraically closed fields?)

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10.2.C. L ESS IMPORTANT EXERCISE . Show that the line with doubled origin X(Example 4.4.5) is not separated, by finding two morphisms π : W → X, π ′ : W →X whose domain of agreement is not a closed subscheme (cf. Proposition 10.1.3).
(Another argument was given above, in Exercise 10.1.D. A fancy argument will begiven in Exercise 12.7.C.)
We now come to the central result of this section.
10.2.2. Reduced-to-Separated Theorem. — Two S-morphisms π : U → Z, π ′ : U →Z from a reduced scheme to a separated S-scheme agreeing on a dense open subset of U are
the same.
Proof. Let V be the locus where π and π ′ agree. It is a closed subscheme of U (byExercise 10.2.A), which contains a dense open set. But the only closed subschemeof a reduced scheme U whose underlying set is dense is all of U.
!
10.2.3. Consequence 1. Hence (as X is reduced and Y is separated) if we have twomorphisms from open subsets of X to Y, say π : U → Y and π ′ : V → Y, and theyagree on a dense open subset Z of U ∩ V, then they necessarily agree on U ∩ V.
Consequence 2. A rational map has a largest domain of definition on which
π : U ""# Y is a morphism, which is the union of all the domains of definition.
In particular, a rational function on a reduced scheme has a largest domain of
definition. For example, the domain of definition of A2k ""# P1k given by (x, y) (→[x, y] has domain of definition A2k \ {(0, 0)} (cf. §6.5.3). This partially extends thedefinition of the domain of a rational function on a locally Noetherian scheme(Definition 5.5.6). The complement of the domain of definition is called the locus
of indeterminacy, and its points are sometimes called fundamental points of therational map, although we won’t use these phrases. (We will see in Exercise 22.4.Lthat a rational map to a projective scheme can be upgraded to an honest morphismby “blowing up” a scheme-theoretic version of the locus of indeterminacy.)
10.2.D. E XERCISE . Show that the Reduced-to-Separated Theorem 10.2.2 is false
if we give up reducedness of the source or separatedness of the target. Here aresome possibilities. For the first, consider the two maps from Spec k[x, y]/(y2 , xy)
to Spec k[t], where we take π given by t (→ x and π ′ given by t (→ x + y; f1 and f2
agree on the distinguished open set D(x), see Figure 10.4. For the second, consider
the two maps from Spec k[t] to the line with the doubled origin, one of whichmaps to the “upper half”, and one of which maps to the “lower half”. These twomorphisms agree on the dense open set D(t), see Figure 10.4.

F IGURE 10.4. Two different maps to a nonseparated scheme
agreeing on a dense open set (see Exercise 10.2.D)

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10.2.4. Graphs of rational maps. (Graphs of morphisms were defined in §10.1.17.) If
X is reduced and Y is separated, define the graph Γπ of a rational map π : X ""# Y
as follows. Let (U, π ′ ) be any representative of this rational map (so π ′ : U → Y
is a morphism). Let Γπ be the scheme-theoretic closure of Γπ ′ "→ U × Y "→ X × Y,
where the first map is a closed embedding (Proposition 10.1.18), and the second
is an open embedding. The product here should be taken in the category you areworking in. For example, if you are working with k-schemes, the fibered productshould be taken over k.
10.2.E. E XERCISE . Show that the graph of a rational map π : X ""# Y is independent of the choice of representative of π. Hint: Suppose ξ ′ : U → Y and ξ : V → Y
are two representatives of π. Reduce to the case where V is the domain of definition of π (§10.2.3), and ξ ′ = ξ|U . Reduce to the case V = X. Show an isomorphism∼ X, and Γξ| =∼ U. Show that the scheme-theoretic closure of U in X is all of X.
Γπ =U(Remark: the separatedness of Y is not necessary.)
In analogy with graphs of morphisms, the following diagram of a graph of arational map π can be useful (cf. Figure 10.3).
"cl. emb.%Γπ7✤ !
X × Y❊
❊❊
①①
❊❊
①
✤
①
❊❊
①
①
✤ ①①
❊❊
M
C①
π
X ❴ ❴ ❴ ❴ ❴ ❴ ❴ ❴ ❴% Y.

10.2.F. E XERCISE ( THE BLOW- UP OF THE PLANE AS THE GRAPH OF A RATIONAL
1
MAP ). Consider the rational map A2
k ""# Pk given by (x, y) (→ [x, y]. Show that
this rational map cannot be extended over the origin. (A similar argument arises in
Exercise 6.5.I on the Cremona transformation.) Show that the graph of the rational
map is the morphism (the blow-up) described in Exercise 9.3.F. (When we define
blow-ups in general, we will see that they are often graphs of rational maps, see
Exercise 22.4.M.)
10.2.5. Variations.
Variations of the short proof of Theorem 10.2.2 yield other useful results. Exercise 10.2.B is one example. The next exercise is another.
10.2.G. E XERCISE ( MAPS TO A SEPARATED SCHEME CAN BE EXTENDED OVER AN
EFFECTIVE C ARTIER DIVISOR IN AT MOST ONE WAY ). Suppose X is a Z-scheme(not necessarily reduced!), and Y is a separated Z-scheme. Suppose further that D
is an effective Cartier divisor on X. Show that any Z-morphism X \ D → Y can beextended in at most one way to a Z-morphism X → Y. (Hint: reduce to the casewhere X = Spec A, and D is the vanishing scheme of t ∈ A. Reduce to showingthat the scheme-theoretic image of D(t) in X is all of X. Show this by showing that
A → At is an inclusion.)
As noted in §6.5.2, rational maps can be defined from any X that has associatedpoints to any Y. The Reduced-to-Separated Theorem 10.2.2 can be extended to thissetting, as follows.

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10.2.H. E XERCISE ( THE “A SSOCIATED - TO -S EPARATED T HEOREM ”). Prove that
two S-morphisms π and π ′ from a locally Noetherian scheme U to a separated Sscheme Z, agreeing on an open subset containing the associated points of U, arethe same.

10.3 Proper morphisms
Recall that a map of topological spaces (also known as a continuous map!)
is said to be proper if the preimage of any compact set is compact. Properness ofmorphisms is an analogous property. For example, a variety over C will be proper
if it is compact in the classical topology (see [Se3, §7]). Alternatively, we will seethat projective A-schemes are proper over A — so this is a nice property satisfied
by projective schemes, which also is convenient to work with.
Recall (§7.3.8) that a (continuous) map of topological spaces π : X → Y is closed
if for each closed subset S ⊂ X, π(S) is also closed. A morphism of schemes isclosed if the underlying continuous map is closed. We say that a morphism ofschemes π : X → Y is universally closed if for every morphism Z → Y, the induced
morphism Z ×Y X → Z is closed. In other words, a morphism is universally closed
if it remains closed under any base change. (More generally, if P is some propertyof schemes, then a morphism of schemes is said to be universally P if it remains P
under any base change.)
To motivate the definition of properness for schemes, we remark that a continuous map π : X → Y of locally compact Hausdorff spaces which have countable bases for their topologies is universally closed if and only if it is proper (i.e.,
preimages of compact subsets are compact). You are welcome to prove this as an
exercise.
10.3.1. Definition. A morphism π : X → Y is proper if it is separated, finite type, anduniversally closed. A scheme X is often said to be proper if some implicit structure
morphism is proper. For example, a k-scheme X is often described as proper if
X → Spec k is proper. (A k-scheme is often said to be complete if it is proper. We
will not use this terminology.) If A is a ring, one often says that an A-scheme isproper over A if it is proper over Spec A.
Let’s try this idea out in practice. We expect that A1C → Spec C is not proper,
because the complex manifold corresponding to A1C is not compact. However, notethat this map is separated (it is a map of affine schemes), finite type, and (trivially)closed. So the “universally” is what matters here.
10.3.A. E ASY E XERCISE . Show that A1C → Spec C is not proper, by finding abase change that turns this into a non-closed map. Possible hint: Consider a wellchosen map such as A1C ×C A1C → A1C or A1C ×C P1C → P1C . (See Figure 19.1 foranother finite type, separated, closed morphism that is not proper. Showing thatmorphism is not proper requires more creativity.)
10.3.2. Example. As a first example: closed embeddings are proper. They areclearly separated, as affine morphisms are separated, §10.1.12. They are finite type.

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After base change, they remain closed embeddings (§9.2.2), and closed embeddings are always closed. This easily extends further as follows.
10.3.3. Proposition. — Finite morphisms are proper.
Proof. Finite morphisms are separated (as they are affine by definition, and affinemorphisms are separated, §10.1.12), and finite type (basically because finite modules over a ring are automatically finitely generated). To show that finite morphism are closed after any base change, we note that they remain finite after any
base change (finiteness is preserved by base change, Exercise 9.4.B(d)), and finite
morphisms are closed (Exercise 7.3.M).
!
10.3.4. Proposition. —(a) The notion of “proper morphism” is stable under base change.
(b) The notion of “proper morphism” is local on the target (i.e., π : X → Y is proper
if and only if for any affine open cover Ui → Y, π−1 (Ui ) → Ui is proper). Notethat the “only if” direction follows from (a) — consider base change by Ui "→ Y.
(c) The notion of “proper morphism” is closed under composition.
(d) The product of two proper morphisms is proper: if π : X → Y and π ′ : X ′ → Y ′are proper, where all morphisms are morphisms of Z-schemes, then π × π ′ : X ×Z
X ′ → Y ×Z Y ′ is proper.
(e) Suppose(10.3.4.1)

X❄
❄❄
❄❄
❄
τ ❄❄
8

π

Z

%Y
⑧
⑧
⑧⑧
⑧⑧ ρ
⑧
5⑧

is a commutative diagram, and τ is proper, and ρ is separated. Then π is proper.
A sample application of (e): a morphism (over Spec k) from a proper k-scheme
to a separated k-scheme is always proper.
Proof. (a) The notions of separatedness, finite type, and universal closedness are
all preserved by fibered product. (Notice that this is why universal closedness isbetter than closedness — it is automatically preserved by base change!)(b) We have already shown that the notions of separatedness and finite typeare local on the target. The notion of closedness is local on the target, and hence sois the notion of universal closedness.
(c) The notions of separatedness, finite type, and universal closedness are allpreserved by composition.
(d) By (a) and (c), this follows from Exercise 9.4.F.
(e) Closed embeddings are proper (Example 10.3.2), so we invoke the Cancellation Theorem 10.1.19 for proper morphisms.
!

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10.3.B. U NIMPORTANT EXERCISE (“ IMAGES OF PROPER SCHEMES ARE PROPER ”).
Suppose in the diagram
X❄
❄❄
❄❄
❄
τ ❄❄
8

π

Z

%Y
⑧⑧
⑧
⑧
⑧⑧ ρ
5⑧⑧

that τ is proper and ρ is separated and finite type. Show that the scheme-theoretic
image of X under π is a proper Z-scheme. (We won’t use this fact, but it reassures
us that properness in algebraic geometry behaves like properness in topology.)
We now come to the most important example of proper morphisms.
10.3.5. Theorem. — Projective A-schemes are proper over A.
(As finite morphisms to Spec A are projective A-schemes, Exercise 7.3.J, Theorem 10.3.5 can be used to give a second proof that finite morphisms are proper,
Proposition 10.3.3.)
Proof. The structure morphism of a projective A-scheme X → Spec A factors asa closed embedding followed by Pn
A → Spec A. Closed embeddings are proper(Example 10.3.2), and compositions of proper morphisms are proper (Proposition 10.3.4), so it suffices to show that Pn
A → Spec A is proper. We have alreadyseen that this morphism is finite type (Easy Exercise 5.3.D) and separated (Proposition 10.1.5), so it suffices to show that Pn
A → Spec A is universally closed. As
n
n
Pn
=
P
×
Spec
A,
it
suffices
to
show
that
Pn
Z
Z
A
X := PZ ×Z X → X is closed forany scheme X. But the property of being closed is local on the target on X, so
by covering X with affine open subsets, it suffices to show that Pn
B → Spec B isclosed for all rings B. This is the Fundamental Theorem of Elimination Theory(Theorem 7.4.7).
!
10.3.6. Remark: “Reasonable” proper schemes are projective. It is not easy to comeup with an example of an A-scheme that is proper but not projective! Over a field,
all proper curves are projective (see Remark 18.7.2), and all smooth surfaces areprojective (see [Ba, Thm. 1.28] for a proof of this theorem of Zariski; smoothnessof course is not yet defined). We will meet a first example of a proper but not projective variety (a singular threefold) in §16.4.10. We will later see an example ofa proper nonprojective surface in §19.11.11, and a simpler one in Exercise 20.2.G.
Once we know about flatness, we will see Hironaka’s example of a proper nonprojective irreducible smooth threefold over C (§24.7.6).
10.3.7. Functions on connected reduced proper k-schemes must be constant.
As an enlightening application of these ideas, we show that if X is a connected
reduced proper k-scheme where k = k, then Γ (X, OX ) = k. The analogous fact
for connected compact complex manifolds uses the maximum principle. We saw
this in the special case X = Pn in Exercise 4.4.E; it will be vastly generalized by
Grothendieck’s Coherence Theorem 18.9.1.
Suppose f ∈ Γ (X, OX ) (f is a function on X). This is the same as a map π : X →A1k (Exercise 6.3.F, discussed further in §6.6.1). Let π ′ be the composition of πwith the open embedding A1 "→ P1 . By Proposition 10.3.4(e), π ′ is proper, andEarly (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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in particular closed. As X is connected, the image of π ′ is as well. Thus the settheoretic image of π ′ must be either a closed point, or all of P1 . But the set-theoretic
image of π ′ lies in A1 , so it must be a closed point p (which we identify with anelement of k).
By Corollary 8.3.5, the support of the scheme-theoretic image of π is the closed
point p. By Exercise 8.3.A, the scheme-theoretic image is precisely p (with the
reduced structure). Thus π can be interpreted as the structure map to Spec k, followed by a closed embedding to A1 identifying Spec k with p. You should be able
to verify that this is the map to A1 corresponding to the constant function f = p.
(What are counterexamples if different hypotheses are relaxed?)
10.3.8. Facts (not yet proved) that may help you correctly think about finiteness.
The following facts may shed some light on the notion of finiteness. We will
prove them later.
A morphism is finite if and only if it is proper and affine, if and only if itis proper and quasifinite (Theorem 29.6.2). We have proved parts of this statement, but we will only finish the proof once we know Zariski’s Main Theorem, cf.
§7.3.15).
As an application: quasifinite morphisms from proper schemes to separated
schemes are finite. Here is why: suppose π : X → Y is a quasifinite morphism overZ, where X is proper over Z. Then by the Cancellation Theorem 10.1.19 for propermorphisms, X → Y is proper. Hence as π is quasifinite and proper, π is finite.
As an explicit example, consider the map π : P1k → P1k given by [x, y] (→[f(x, y), g(x, y)], where f and g are homogeneous polynomials of the same degreewith no common roots in P1 . The fibers are finite, and π is proper (from the Cancellation Theorem 10.1.19 for proper morphisms, as discussed after the statementof Theorem 10.3.4), so π is finite. This could be checked directly as well, but now
we can save ourselves the annoyance.
10.3.9. ⋆⋆ Group varieties. The rest of §10.3 is double-starred, and we will also
throughout work in the category of varieties over a field k. We briefly discuss groupvarieties, mainly because we can, not because we have anything profound to say.
We discuss them right now only because properness gives an unexpected consequence, that proper group varieties are abelian (thanks to the surprisingly simple
Rigidity Lemma 10.3.12).
As discussed in §10.1.7, now that we have the category of varieties over k,
we immediately have the notion of a group variety (thanks to §6.6.4 on groupobjects in any category). An algebraic group (over k) is a smooth group variety.
Examples include GLn (Exercise 6.6.N(b)), which includes Gm as a special case,and SLn (Exercise 6.6.Q).
10.3.10. Side Remarks (that we won’t prove or use). Group varieties are automatically smooth (hence algebraic groups) in characteristic 0. Group varieties are notnecessarily smooth in positive characteristic. (Let k be an imperfect field of characteristic p, and choose t ∈ k without a pth root in k. Consider the “closed sub-groupscheme” xp = typ of the “additive group” G2a = Spec k[x, y]. This group scheme
is reduced, but not geometrically reduced; and not smooth, or even regular.) Algebraic groups are automatically quasiprojective. An algebraic group G is affine(as a scheme) if and only if it admits a closed immersion into GLn for some n ≥ 0,

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such that G → GLn is a homomorphism of group schemes (Exercise 6.6.N(a)). For
further discussion on these and other issues, see [P2, §5.2].

10.3.11. Definition: Abelian varieties. We can now define one of the most importantclasses of varieties: abelian varieties. The bad news is that we have no real example
yet. We will later see elliptic curves as an example (§19.10), but we will not meetanything more exotic than that. Still, it is pleasant to know that we can make thedefinition this early.
An abelian variety over a field k is an algebraic group that is geometricallyintegral and projective (over k). It turns out that the analytification of any abelianvariety over C is a complex torus, i.e., Cn /Λ, where Λ is a lattice (of rank 2n).
(The key idea: connected compact complex Lie groups are commutative, see Remark 10.3.14; the universal cover is a simply connected commutative complex Liegroup, and thus Cn . See [BL, Lem. 1.1.1].)
We now show that abelian varieties are abelian, i.e., abelian varieties are commutative algebraic group varieties. (This is less tautological than it sounds! The
adjective “abelian” is used in two different senses; what they have in common isthat they are derived from the name of Niels Henrik Abel.) As algebraic groups
need not be commutative (witness GLn ), it is somehow surprising that commutativity could be forced by “compactness”.
The key result is the following fact, beautiful in its own right.
10.3.12. Rigidity Lemma. — Let X, Y, and Z be varieties, where X is proper andgeometrically integral, with a rational point p, and Y is irreducible. Let pr : X × Y → Y be
the projection, and let α : X × Y → Z be any morphism. Suppose q ∈ Y is such that α isconstant on X × {q}; say α(X × {q}) = r ∈ Z.
(a) Then there is a morphism ψ : Y → Z such that α = ψ ◦ pr. (In particular, for everyq ′ ∈ Y, α is constant on X × {q ′ }.)(b) If α is also constant on {p} × Y, then α is a constant function on all of X × Y.

Xq

Z
③N O
③③
③
③
③③ α
③③
% X×Y
K
ψ

pr

"
q

σp

"
%Y

The hypotheses can be relaxed in a number of ways, but this version suffices
for our purposes. If you have not read §9.5, where geometric integrality is discussed, you need not worry too much; this is automatic if k = k and X is integral(Hard Fact 9.5.2). Even if you have read §9.5, it is helpful to first read this proof
under the assumption that k = k, to avoid being distracted from the main idea bygeometric points. But of course all of §10.3.9 is double-starred, so you shouldn’t
be reading this anyway.
10.3.C. E XERCISE . Show that the properness hypothesis on X is necessary. (Canyou make this accord with your intuition?)

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Proof of the Rigidity Lemma 10.3.12. Define β : X × Y → Z by β(x, y) = α(p, y).
(More precisely, if σp is the section of Y → Y → X pulled back from Spec k (→ p,then β = α ◦ σp ◦ pr.) We will show that β(x, y) = α(x, y). This will imply (a) (takeψ = α ◦ σp ), which in turn immediately implies (b).
Proposition 9.5.17 (geometrically reduced times reduced is reduced) implies
that X × Y is reduced. Exercise 9.5.M (geometrically irreducible times irreducible isirreducible) implies that X × Y is irreducible. As Z is separated, by Exercise 10.2.A,
it suffices to show that α = β on a nonempty (hence dense) open subset of X × Y.
Let U ⊂ Z be an affine open neighborhood of r. Then α−1 (Z\U) is a closed subset of X×Y. As the projection pr is proper (using properness of X, and preservationof properness under base change, Proposition 10.3.4(a)), we have pr(α−1 (Z \ U))
is closed. Its complement is open, and contains q; let V ⊂ Y be an open neighborhood of q disjoint from pr(α−1 (Z \ U)).
10.3.D. E XERCISE . Suppose γ : Spec K → V is a geometric point of V. Show thatthe fiber XK over Spec K (or more precisely, over γ) is mapped to a point in Z by
the restriction αK of α to XK . Hint: review §10.3.7.
❦❦P Z❦❦❦❦❦❦❦❦P Q❦❦❦❦ ❦❦❦❦ ❦❦❦❦❦❦❦❦❦❦❦β❦K
❦
❦
❦❦❦❦❦
α|V
% X×V
XK ❦
αK

pr

"
Spec Kγ"%V

By Exercise 10.2.B (maps of varieties over an algebraically closed field are determined by the map of closed points), αK = βK .
10.3.E. E XERCISE . Use Exercise 10.2.A (the fact that the locus where α = β isa closed subscheme — and represents a certain functor) to show that α = β on
pr−1 (V).
!
10.3.13. Corollary. — Suppose A is an abelian variety, G is a group variety, andφ : A → G is any morphism of varieties. Then φ is the composition of a translation
and a homomorphism.
Proof. Let eA and eG be the identity points of A and G respectively. Composing
φ with a translation, we may assume that φ(eA ) = eG . Consider the morphism
α : A × A → G given by α(a1 , a2 ) = φ(a1 a2 )φ(a1 )−1 φ(a2 )−1 . Then α({eA } × A) =α(A × {eA }) = {eB }, so by the Rigidity Lemma 10.3.12, α is a constant, and sends
A × A to eB . Thus φ(a1 a2 ) = φ(a1 )φ(a2 ), so φ is a homomorphism.
!

10.3.F. E XERCISE ( ABELIAN VARIETIES ARE ABELIAN ). Show that an abelian variety is an abelian group variety. Hint: apply Corollary 10.3.13 to the inversion
morphism i : A → A.

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10.3.14. Remark. A similar idea is used to show that connected compact complex
Lie groups are abelian, see for example [BL, Lemm. 1.1.1].

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Part IV

“Geometric” properties: Dimension
and smoothness

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CHAPTER 11

Dimension

11.1 Dimension and codimension
Everyone knows what a curve is, until he has studied enough mathematics to become
confused ...
— F. Klein, [RS, p. 90]
At this point, you know a fair bit about schemes, but there are some fundamental notions you cannot yet define. In particular, you cannot use the phrase“smooth surface”, as it involves the notion of dimension and of smoothness. You
may be surprised that we have gotten so far without using these ideas. You may
also be disturbed to find that these notions can be subtle, but you should keep inmind that they are subtle in all parts of mathematics.
In this chapter, we will address the first notion, that of dimension of schemes.
This should agree with, and generalize, our geometric intuition. Although wethink of dimension as a basic notion in geometry, it is a slippery concept, as it isthroughout mathematics. Even in linear algebra, the definition of dimension of avector space is surprising the first time you see it, even though it quickly becomes
second nature. The definition of dimension for manifolds is equally nontrivial.
For example, how do we know that there isn’t an isomorphism between some 2dimensional manifold and some 3-dimensional manifold? Your answer will likely
use topology, and hence you should not be surprised that the notion of dimension
is often quite topological in nature.
A caution for those thinking over the complex numbers: our dimensions willbe algebraic, and hence half that of the “real” picture. For example, we will seevery shortly that A1C , which you may picture as the complex numbers (plus one
generic point), has dimension 1.
11.1.1. Definition(s): dimension. Surprisingly, the right definition is purely topological — it just depends on the topological space, and not on the structure sheaf. We
define the dimension of a topological space X (denoted dim X) as the supremumof lengths of chains of closed irreducible sets, starting the indexing of the closed
irreducible sets with 0. (The dimension may be infinite.) Scholars of the empty
set can take the dimension of the empty set to be −∞. (An analogy from linear
algebra: the dimension of a vector space is the supremum of the length of chainsof subspaces.) Define the dimension of a ring as the supremum of the lengths ofthe chains of nested prime ideals (where indexing starts at zero). These two definitions of dimension are sometimes called Krull dimension. (You might think a
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Noetherian ring has finite dimension because all chains of prime ideals are finite,
but this isn’t necessarily true — see Exercise 11.1.K.)
11.1.A. E ASY E XERCISE . Show that dim Spec A = dim A. (Hint: Exercise 3.7.E
gives a bijection between irreducible closed subsets of Spec A and prime ideals of
A. It is “inclusion-reversing”.)
The homeomorphism between Spec A and Spec A/N(A) (§3.4.5: the Zariski
topology disregards nilpotents) implies that dim Spec A = dim Spec A/N(A).
11.1.2. Examples. We have identified all the prime ideals of k[t] (they are 0, and(f(t)) for irreducible polynomials f(t)), Z ((0) and (p)), k (only (0)), and k[x]/(x2 )(only (x)), so we can quickly check that dim A1k = dim Spec Z = 1, dim Spec k = 0,
dim Spec k[x]/(x2 ) = 0.
11.1.3. We must be careful with the notion of dimension for reducible spaces. If Z
is the union of two closed subsets X and Y, then dim Z = max(dim X, dim Y). Thus
dimension is not a “local” characteristic of a space. This sometimes bothers us,
so we try to only talk about dimensions of irreducible topological spaces. We say
a topological space is equidimensional or pure dimensional (resp. equidimensional of dimension n or pure dimension n) if each of its irreducible components
has the same dimension (resp. they are all of dimension n). An equidimensional
dimension 1 (resp. 2, n) topological space is said to be a curve (resp. surface, nfold).
11.1.B. E XERCISE . Show that a scheme has dimension n if and only if it admits
an open cover by affine open subsets of dimension at most n, where equality isachieved for some affine open subset. Hint: You may find it helpful, here and later,
to show the following. For any topological space X and open subset U ⊂ X, there isa bijection between irreducible closed subsets of U, and irreducible closed subsets
of X that meet U.
11.1.C. E ASY E XERCISE .
finite number of points.

Show that a Noetherian scheme of dimension 0 has a

11.1.D. E XERCISE ( FIBERS OF INTEGRAL MORPHISMS , PROMISED IN §7.3.11). Suppose π : X → Y is an integral morphism. Show that every (nonempty) fiber of π
has dimension 0. Hint: As integral morphisms are preserved by base change, we
assume that Y = Spec k. Hence we must show that if φ : k → A is an integral
extension, then dim A = 0. Outline of proof: Suppose p ⊂ m are two prime idealsof A. Mod out by p, so we can assume that A is an integral domain. I claim thatany nonzero element is invertible: Say x ∈ A, and x ̸= 0. Then the minimal monicpolynomial for x has nonzero constant term. But then x is invertible — recall thecoefficients are in a field.
11.1.E. I MPORTANT E XERCISE . Show that if π : Spec A → Spec B corresponds
to an integral extension of rings, then dim Spec A = dim Spec B. Hint: show thata chain of prime ideals downstairs gives a chain upstairs of the same length, by
the Going-Up Theorem (Exercise 7.2.F). Conversely, a chain upstairs gives a chain
downstairs. Use Exercise 11.1.D to show that no two elements of the chain upstairs
go to the same element [q] ∈ Spec B of the chain downstairs.

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3 → X is the normalization of a scheme (possibly
11.1.F. E XERCISE . Show that if ν : X
3 = dim X. Feel free to assume that X isin a finite extension of fields), then dim X
integral for convenience.

11.1.G. E XERCISE . Suppose X is an affine k-scheme, and K/k is an algebraic field
extension.
(a) Suppose X has pure dimension n. Show that XK := X ×k K has pure dimension
n. (See Exercise 24.5.F for a generalization, which for example removes the affine
hypothesis. Also, see Exercise 11.2.J and Remark 11.2.16 for the fate of possible
generalizations to arbitrary field extensions.) Hint: If X = Spec A, reduce to thecase where A is an integral domain. An irreducible component of X ′ corresponds
to a minimal prime p of A ′ := A ⊗k K. Suppose a ∈ ker(A → A ′ /p). Show
that a = 0, using the fact that a lies in a minimal prime p of A ′ (and is hence a
zerodivisor, by Remark 5.5.12), and A ′ is a free A-module (so multiplication in
A ′ by a ∈ A is injective if a is nonzero). Thus A → A ′ /p is injective. Then use
Exercise 11.1.E.
(b) Prove the converse to (a): show that if XK has pure dimension n, then X has
pure dimension n.
11.1.H. E XERCISE . Show that dim Z[x] = 2. (Hint: The prime ideals of Z[x] were
implicitly determined in Exercise 3.2.Q.)
11.1.4. Codimension. Because dimension behaves oddly for disjoint unions, weneed some care when defining codimension, and in using the phrase. For example,
if Y is a closed subset of X, we might define the codimension to be dim X − dim Y,
but this behaves badly. For example, if X is the disjoint union of a point Y and acurve Z, then dim X − dim Y = 1, but this has nothing to do with the local behaviorof X near Y.
A better definition is as follows. In order to avoid excessive pathology, we define the codimension of Y in X only when Y is irreducible. (Use extreme caution whenusing this word in any other setting.) Define the codimension of an irreducible
subset Y ⊂ X of a topological space as the supremum of lengths of increasing chainsof irreducible closed subsets starting with Y (where indexing starts at 0 — recall
that the closure of an irreducible set is irreducible, Exercise 3.6.B(b)). In particular,
the codimension of a point is the codimension of its closure. The codimension of
Y in X is denoted by codimX Y.
We say that a prime ideal p in a ring has codimension equal to the supremumof lengths of the chains of decreasing prime ideals starting at p, with indexing
starting at 0. Thus in an integral domain, the ideal (0) has codimension 0; and in Z,
the ideal (23) has codimension 1. Note that the codimension of the prime ideal p in
A is dim Ap (see §3.2.8). (This notion is often called height.) Thus the codimensionof p in A is the codimension of [p] in Spec A.
(Continuing an analogy with linear algebra: the codimension of a vector subspace Y ⊂ X is the supremum of lengths of increasing chains of subspaces startingwith Y. This is a better definition than dim X − dim Y, because it works even whendim X = ∞. You might prefer to define codimX Y as dim(X/Y); that is analogous
to defining the codimension of p in A as the dimension of Ap — see the previous
paragraph.)

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11.1.I. E XERCISE . Show that if Y is an irreducible closed subset of a scheme X, andη is the generic point of Y, then the codimension of Y is the dimension of the local
ring OX,η .
Notice that Y is codimension 0 in X if it is an irreducible component of X. Similarly, Y is codimension 1 if it is not an irreducible component, and for every irreducible component Y ′ it is contained in, there is no irreducible subset strictly
between Y and Y ′ . (See Figure 11.1 for examples.) A closed subset all of whose
irreducible components are codimension 1 in some ambient space X is said to be a
hypersurface in X.
pCq

F IGURE 11.1. Behavior of codimension
11.1.J. E ASY E XERCISE . If Y is an irreducible closed subset of a scheme X, showthat
(11.1.4.1)

codimX Y + dim Y ≤ dim X.

We will soon see that equality always holds if X and Y are varieties (Theorem 11.2.9), but equality doesn’t hold in general (§11.3.9).
Warning. The notion of codimension still can behave slightly oddly. For example, consider Figure 11.1. (You should think of this as an intuitive sketch.) Herethe total space X has dimension 2, but point p is dimension 0, and codimension 1.
We also have an example of a codimension 2 subset q contained in a codimension
0 subset C with no codimension 1 subset “in between”.
Worse things can happen; we will soon see an example of a closed point in an
irreducible surface that is nonetheless codimension 1, not 2, in §11.3.9. However, forirreducible varieties this can’t happen, and inequality (11.1.4.1) must be an equality(Theorem 11.2.9).
11.1.5. In unique factorization domains, codimension 1 prime ideals are principal. For the sake of further applications, we make a short observation.

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11.1.6. Lemma. — In a unique factorization domain A, all codimension 1 prime ideals
are principal.
This is a first glimpse of the fact that codimension one is rather special — thistheme will continue in §11.3. We will see that the converse of Lemma 11.1.6 holds
as well (when A is a Noetherian integral domain, Proposition 11.3.5).
Proof. Suppose p is a codimension 1 prime. Choose any f ̸= 0 in p, and let g be
any irreducible/prime factor of f that is in p (there is at least one). Then (g) is anonzero prime ideal contained in p, so (0) ⊂ (g) ⊂ p. As p is codimension 1, wemust have p = (g), and thus p is principal.
!
11.1.7. A fun but unimportant counterexample. We end this introductory section
with a fun pathology. As a Noetherian ring has no infinite chain of prime ideals,
you may think that Noetherian rings must have finite dimension. Nagata, themaster of counterexamples, shows you otherwise with the following example.
11.1.K. ⋆⋆ E XERCISE : AN INFINITE - DIMENSIONAL N OETHERIAN RING . Let A =k[x1 , x2 , . . . ]. Choose an increasing sequence of positive integers m1 , m2 , . . . whose
differences are also increasing (mi+1 −mi > mi −mi−1 ). Let pi = (xmi +1 , . . . , xmi+1 )and S = A − ∪i pi .
(a) Show that S is a multiplicative set.
(b) Show that S−1 pi in S−1 A is the largest prime ideal in a chain of prime ideals oflength mi+1 − mi . Hence conclude that dim S−1 A = ∞.
(c) Suppose B is a ring such that (i) for every maximal ideal m, Bm is Noetherian, and (ii) every nonzero b ∈ B is contained in finitely many maximal ideals.
Show that B is Noetherian. (One possible approach: show that for any x1 , x2 , . . . ,(x1 , x2 , . . . ) is finitely generated.)(d) Use (c) to show that S−1 A is Noetherian.
11.1.8. Remark: Noetherian local rings have finite dimension. However, we shall see in
Exercise 11.3.I(a) that Noetherian local rings always have finite dimension. (Thisrequires a surprisingly hard fact, Krull’s Height Theorem 11.3.7.) Thus points oflocally Noetherian schemes always have finite codimension.

11.2 Dimension, transcendence degree, and Noether
normalization
We now give a powerful alternative interpretation for dimension for irreducible
varieties, in terms of transcendence degree. The proof will involve a classical construction, Noether normalization, which will be useful in other ways as well. In case
you haven’t seen transcendence theory, here is a lightning introduction.
11.2.A. E XERCISE / DEFINITION . Recall that an element of a field extension E/F
is algebraic over F if it is integral over F. Recall also that a field extension E/F is analgebraic extension if it is an integral extension (if all elements of E are algebraic overF). The composition of two algebraic extensions is algebraic, by Exercise 7.2.C.
If E/F is a field extension, and F ′ and F ′′ are two intermediate field extensions,

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then we write F ′ ∼ F ′′ if F ′ F ′′ is algebraic over both F ′ and F ′′ . Here F ′ F ′′ is thecompositum of F ′ and F ′′ , the smallest field extension in E containing F ′ and F ′′ . (a)Show that ∼ is an equivalence relation on subextensions of E/F. A transcendence basisof E/F is a set of elements {xi } that are algebraically independent over F (there isno nontrivial polynomial relation among the xi with coefficients in F) such that
F({xi }) ∼ E. (b) Show that if E/F has two transcendence bases, and one has cardinality n,then both have cardinality n. (Hint: show that you can substitute elements from theone basis into the other one at a time.) The size of any transcendence basis is called
the transcendence degree (which may be ∞), and is denoted tr. deg. Any finitely
generated field extension necessarily has finite transcendence degree. (Remark: A
related result was mentioned in Algebraic Fact 9.5.16.)
11.2.1. Theorem (dimension = transcendence degree). — Suppose A is a finitely
generated domain over a field k (i.e., a finitely generated k-algebra that is an integral
domain). Then dim Spec A = tr. deg K(A)/k. Hence if X is an irreducible k-variety,then dim X = tr. deg K(X)/k.
We will prove Theorem 11.2.1 shortly (§11.2.7). We first show that it is usefulby giving some immediate consequences. We seem to have immediately dim Ank =n. However, our proof of Theorem 11.2.1 will go through this fact, so it isn’t reallya consequence.
A more substantive consequence is the following. If X is an irreducible kvariety, then dim X is the transcendence degree of the function field (the stalk at
the generic point) OX,η over k. Thus (as the generic point lies in all nonempty opensets) the dimension can be computed in any open set of X. (Warning: this is false
without the finite type hypothesis, even in quite reasonable circumstances: let X
be the two-point space Spec k[x](x) , and let U consist of only the generic point, see
Exercise 3.4.K.)
Another consequence is a second proof of the Nullstellensatz 3.2.5.
11.2.B. E XERCISE : THE N ULLSTELLENSATZ FROM DIMENSION THEORY. Suppose
A = k[x1 , . . . , xn ]/I. Show that the residue field of any maximal ideal of A is a
finite extension of k. (Hint: the maximal ideals correspond to dimension 0 points,
which correspond to transcendence degree 0 finitely generated extensions of k, i.e.,
finite extensions of k.)
Yet another consequence is geometrically believable.
11.2.C. E XERCISE . If π : X → Y is a dominant morphism of irreducible k-varieties,
then dim X ≥ dim Y. (This is false more generally: consider the inclusion of thegeneric point into an irreducible curve.)
11.2.D. E XERCISE ( PRACTICE WITH THE CONCEPT ). Show that the three equations
wz − xy = 0,

wy − x2 = 0,xz − y2 = 0

cut out an integral surface S in A4k . (You may recognize these equations from Exercises 3.6.F and 8.2.A.) You might expect S to be a curve, because it is cut out by
three equations in four-space. One of many ways to proceed: cut S into pieces.
∼ Spec k[x, w]w . (You may recognize S as the affine
For example, show that D(w) =cone over the twisted cubic. The twisted cubic was defined in Exercise 8.2.A.) It

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turns out that you need three equations to cut out this surface. The first equation cuts out a threefold in A4k (by Krull’s Principal Ideal Theorem 11.3.3, which
we will meet soon). The second equation cuts out a surface: our surface, alongwith another surface. The third equation cuts out our surface, and removes the“extraneous component”. One last aside: notice once again that the cone over thequadric surface k[w, x, y, z]/(wz − xy) makes an appearance.)
11.2.2. Definition: degree of a rational map of irreducible varieties. If π : X ""# Y
is a dominant rational map of irreducible (hence integral) k-varieties of the same
dimension, the degree of the field extension is called the degree of the rational
map. This readily extends if X is reducible: we add up the degrees on each of
the components of X. If π is a rational map of integral affine k-varieties of thesame dimension that is not dominant we say the degree is 0. We will interpret
this degree in terms of counting preimages of general points of Y in §24.5.10. Notethat degree is multiplicative under composition: if ρ : Y ""# Z is a rational mapof integral k-varieties of the same dimension, then deg(ρ ◦ π) = deg(ρ) deg(π), as
degrees of field extensions are multiplicative in towers.
11.2.3. Noether normalization.
Our proof of Theorem 11.2.1 will use another important classical notion, Noether
normalization.
11.2.4. Noether Normalization Lemma. — Suppose A is an integral domain, finitely
generated over a field k. If tr. degk K(A) = n, then there are elements x1 , . . . , xn ∈ A,
algebraically independent over k, such that A is a finite (hence integral by Corollary 7.2.2)
extension of k[x1 , . . . , xn ].
The geometric content behind this result is that given any integral finite typeaffine k-scheme X, we can find a surjective finite morphism X → An
k , where n is
the transcendence degree of the function field of X (over k). Surjectivity follows
from the Lying Over Theorem 7.2.5, in particular Exercise 11.1.E. This interpretation is sometimes called geometric Noether normalization. (See Remark 30.3.7 for a
projective version of Noether normalization.)
11.2.5. Nagata’s proof of the Noether Normalization Lemma 11.2.4. Suppose we canwrite A = k[y1 , . . . , ym ]/p, i.e., that A can be chosen to have m generators. Notethat m ≥ n. We show the result by induction on m. The base case m = n isimmediate.
Assume now that m > n, and that we have proved the result for smallerm. We will find m − 1 elements z1 , . . . , zm−1 of A such that A is finite overk[z1 , . . . , zm−1 ]/q (i.e., the subring of A generated by z1 , . . . , zm−1 , where thezi satisfy the relations given by the ideal q). Then by the inductive hypothesis,k[z1 , . . . , zm−1 ]/q is finite over some k[x1 , . . . , xn ], and A is finite over k[z1 , . . . , zm−1 ],Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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so by Exercise 7.3.I, A is finite over k[x1 , . . . , xn ].
A = k[y1 , . . . , ym ]/pfinitek[z1 , . . . , zm−1 ]/q
finitek[x1 , . . . , xn ]As y1 , . . . , ym are algebraically dependent in A, there is some nonzero algebraic relation f(y1 , . . . , ym ) = 0 among them (where f is a polynomial in m variables over k).
Let z1 = y1 − yrm1 , z2 = y2 − yrm2 , . . . , zm−1 = ym−1 − yrmm−1 , where r1 , . . . ,rm−1 are positive integers to be chosen shortly. Thenf(z1 + yrm1 , z2 + yrm2 , . . . , zm−1 + yrmm−1 , ym ) = 0.
Then upon expanding this out, each monomial in f (as a polynomial in m variables)
will yield a single term that is a constant times a power of ym (with no zi factors).
By choosing the ri so that 0 ≪ r1 ≪ r2 ≪ · · · ≪ rm−1 , we can ensure that thepowers of ym appearing are all distinct, and so that in particular there is a leading
term yNm , and all other terms (including those with factors of zi ) are of smaller
degree in ym . Thus we have described an integral dependence of ym on z1 , . . . ,zm−1 as desired.
!
11.2.6. The geometry behind Nagata’s proof.
Here is the geometric intuition behind Nagata’s argument. Suppose we have an n-dimensional variety in Am
k withn < m, for example xy = 1 in A2 . One approach is to hope the projection to a
hyperplane is a finite morphism. In the case of xy = 1, if we projected to the xaxis, it wouldn’t be finite, roughly speaking because the asymptote x = 0 preventsthe map from being closed (cf. Exercise 7.3.L). If we instead projected to a random
line, we might hope that we would get rid of this problem, and indeed we usuallycan: this problem arises for only a finite number of directions. But we might havea problem if the field were finite: perhaps the finite number of directions in whichto project each have a problem. (You can show that if k is an infinite field, then thesubstitution in the above proof zi = yi − yrmi can be replaced by the linear substitution zi = yi − ai ym where ai ∈ k, and that for a nonempty Zariski-open choice ofai , we indeed obtain a finite morphism.) Nagata’s trick in general is to “jiggle” thevariables in a nonlinear way, and this jiggling kills the nonfiniteness of the map.
11.2.7. Proof of Theorem 11.2.1 on dimension and transcendence degree. Suppose X isan integral affine k-scheme of finite type. We show that dim X equals the transcendence degree n of its function field, by induction on n. (The idea is that we reducefrom X to An to a hypersurface in An to An−1 .) Assume the result is known for alltranscendence degrees less than n.
By the Noether Normalization Lemma 11.2.4, there exists a surjective finitenmorphism X → An
k . By Exercise 11.1.E, dim X = dim Ak . If n = 0, we are done, as
0dim Ak = 0.

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n
We now show that dim An
k = n for n > 0, by induction. Clearly dim Ak ≥ n,
as we can describe a chain of irreducible subsets of length n: if x1 , . . . , xn are
coordinates on An , consider the chain of ideals(0) ⊂ (x1 ) ⊂ · · · ⊂ (x1 , . . . , xn )in k[x1 , . . . , xn ]. Suppose we have a chain of prime ideals of length at least n + 1:(0) = p0 ⊂ · · · ⊂ pm .

Choose any nonzero element g of p1 , and let f be any irreducible factor of g lyingin p1 . Then replace p1 by (f). (Of course, p1 may have been (f) to begin with.) Then
K(k[x1 , . . . , xn ]/(f(x1 , . . . , xn ))) has transcendence degree n − 1, so by induction,
dim k[x1 , . . . , xn ]/(f) = n − 1,so dim k[x1 , . . . , xn ] ≤ n, completing the proof.

!

11.2.E. E XERCISE ( DIMENSION IS ADDITIVE FOR PRODUCTS OF FINITE TYPE k- SCHEMES ).
If X and Y are irreducible k-varieties, show that dim X×k Y = dim X+dim Y. (Hint:
XY
If we had surjective finite morphisms X → Adimand Y → Adim
, we could conkk
dim X+dim Y
struct a surjective finite morphism X ×k Y → Ak.)
11.2.8. Codimension is the difference of dimensions for irreducible varieties.
Noether normalization will help us show that codimension is the difference ofdimensions for irreducible varieties, i.e., that the inequality (11.1.4.1) is always anequality.
11.2.9. Theorem. — Suppose X is a pure-dimensional k scheme locally of finite type(for example, an irreducible k-variety), Y is an irreducible closed subset, and η is thegeneric point of Y. Then dim Y + dim OX,η = dim X. Hence by Exercise 11.1.I, dim Y +codimX Y = dim X — inequality (11.1.4.1) is always an equality.
Proving this will give us an excuse to introduce some useful notions, suchas the Going-Down Theorem for finite extensions of integrally closed domains(Theorem 11.2.12). Before we begin the proof, we give an algebraic translation.
11.2.F. E XERCISE . A ring A is called catenary if for every nested pair of prime
ideals p ⊂ q ⊂ A, all maximal chains of prime ideals between p and q have thesame length. (We will not use this term in any serious way later.) Show that if A isa localization of a finitely generated ring over a field k, then A is catenary.
11.2.10. Remark. Most rings arising naturally in algebraic geometry are catenary.
Important examples include: localizations of finitely generated Z-algebras; complete Noetherian local rings; Dedekind domains; and Cohen-Macaulay rings (see§26.2.13). It is hard to give an example of a noncatenary ring; see for example[Stacks, tag 02JE] or [He].
11.2.11. Proof of Theorem 11.2.9.
11.2.G. E XERCISE . Reduce the proof of Theorem 11.2.9 to the following problem.
If X is an irreducible affine k-variety and Z is a closed irreducible subset maximal
among those smaller than X (the only larger closed irreducible subset is X), thendim Z = dim X − 1.

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Let d = dim X for convenience. By Noether normalization 11.2.4, we have afinite morphism π : X → Ad corresponding to a finite extension of rings. Then π(Z)is an irreducible closed subset of Ad (finite morphisms are closed, Exercise 7.3.M).

11.2.H. E XERCISE . Show that it suffices to show that π(Z) is a hypersurface. (Hint:
the dimension of any hypersurface is d − 1 by Theorem 11.2.1 on dimension andtranscendence degree. Exercise 11.1.E implies that dim π−1 (π(Z)) = dim π(Z). But
be careful: Z is not π−1 (π(Z)) in general.)Now if π(Z) is not a hypersurface, then it is properly contained in an irreducible hypersurface H, so by the Going-Down Theorem 11.2.12 for finite extensions of integrally closed domains (which we shall now prove), there is someclosed irreducible subset Z ′ of X properly containing Z, contradicting the maximality of Z.
!
11.2.12. Theorem (Going-Down Theorem for finite extensions of integrally closed
domains). — Suppose φ : B "→ A is a finite extension of rings (so A is a finite B-module),
B is an integrally closed domain, and A is an integral domain. Then given nested prime
ideals q ⊂ q ′ of B, and a prime p ′ of A lying over q ′ (i.e., p ′ ∩ B = q ′ ), there exists a prime
p of A contained in p ′ , lying over q.
As usual, you should sketch a geometric picture of the statement. This theorem is usually stated about extending a chain of ideals, in the same way as theGoing-Up Theorem (Exercise 7.2.F), and you may want to think this through. (Another Going-Down Theorem, for flat morphisms, will be given in Exercise 24.5.E.)
This theorem is true more generally with “finite” replaced by “integral”; see[E, p. 291] (“Completion of the proof of 13.9”) for the extension of Theorem 11.2.12,
or else see [AtM, Thm. 5.16] or [Mat1, Thm. 5(v)] for completely different proofs.
See [E, Fig. 10.4] for an example (in the form of a picture) of why the “integrally
closed” hypothesis on B cannot be removed.
In the course of the proof, we will need the following fact. It is not hard, andwe could prove it now, but we leave it until Exercise 11.3.D, because the proof uses
a trick arising in Exercise 11.3.C.
11.2.13. Proposition (prime avoidance). — Suppose p1 , . . . , pn are prime ideals of aring A, and I is another ideal of A not contained in any pi . Then I is not contained in ∪pi :there is an element f ∈ I not in any of the pi .
(Can you give a geometric interpretation of this result? Can you figure out
why it is called “prime avoidance”?)

11.2.14. Proof of Theorem 11.2.12 (Going-Down Theorem for finite extensions of integrally
closed domains). The proof uses Galois theory. Let L be the normal closure of
K(A)/K(B) (the smallest subfield of K(B) containing K(A), and that is mapped toitself by any automorphism over K(B)/K(B)). Let C be the integral closure of B in
L (discussed in Exercise 9.7.I). Because A "→ C is an integral extension, there is aprime Q of C lying over q ⊂ B (by the Lying Over Theorem 7.2.5), and a prime
Q ′ of C containing Q lying over q ′ (by the Going-Up Theorem, Exercise 7.2.F).
Similarly, there is a prime P ′ of C lying over p ′ ⊂ A (and thus over q ′ ⊂ B). We
would be done if P ′ = Q ′ (just take p = Q ∩ A), but this needn’t be the case.

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However, Lemma 11.2.15 below shows there is an automorphism σ of C over B,
that sends Q ′ to P ′ , and then the image of σ(Q) in A will do the trick, completingthe proof. The following diagram, in geometric terms, may help.
Spec C/P ′ % ❱&
7
❱❱❱❱
❱❱❱❱
❱❱❱❱
σ
❱❱❱❱
❱
"
" ❱❱❱@!
% Spec C/Q !
% Spec C
Spec C/Q ′L7

E"Spec A/p ′ !

% ?! ""% Spec A

!$
K(A)
7

E "
"
Spec B/q ′ !

E "% Spec B/q ! ""% Spec B!$
K(B)11.2.15. Lemma. — Suppose B is an integrally closed domain, L/K(B) is a finite normal
field extension, and C is the integral closure of B in L. If q ′ is a prime ideal of B, thenautomorphisms of L/K(B) act transitively on the prime ideals of C lying over q ′ .
This result is often first seen in number theory, with B = Z and L a Galois
extension of Q.
Proof. Let P and Q1 be two prime ideals of C lying over q ′ , and let Q2 , . . . , Qn be
the prime ideals of C conjugate to Q1 (the image of Q1 under Aut(L/K(B))). If P
is not one of the Qi , then P is not contained in any of the Qi . (Do you see why?
Hint: Exercise 11.1.D or 11.1.E.) Hence by prime avoidance (Proposition 11.2.13),
P is not contained in their union, so there is some a ∈ P not contained in any Qi .
Thus no conjugate of a can be contained in Q1 , so the norm NL/K(B) (a) ∈ B is
not contained in Q1 ∩ B = q ′ . (Recall: If L/K(B) is separable, the norm is justthe product of the conjugates. But even if L/K(B) is not separable, the norm isthe product of conjugates to the appropriate power, because we can factor L/K(B)as a separable extension followed by a purely inseparable extension, or because
we can take an explicit basis for the extension and calculate the norm of a.) But
since a ∈ P, its norm lies in P, but also in B, and hence in P ∩ B = q ′ , yielding acontradiction.
!
We end with three exercises which may give you practice and enlightenment.
11.2.I. E XERCISE / DEFINITION . Suppose p is a closed point of a locally finite typek-scheme X. Show that the following three integers are the same:(a) the largest dimension of an irreducible component of X containing p;(b) dim OX,p ;(c) codimp X.
We call this the dimension of X at p. Use (a) to show that this function on X is anupper semicontinuous function on the closed points of X. (The Zariski topology on
the closed points of a finite type k-scheme in the obvious way: closed subsets are
cut out by equations. A case of this already arose in §7.4.5.)

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11.2.J. E XERCISE ( THE DIMENSION OF A LOCALLY FINITE TYPE k- SCHEME IS PRE SERVED BY ANY FIELD EXTENSION , CF. E XERCISE 11.1.G( A )).
Suppose X is alocally finite type k-scheme of pure dimension n, and K/k is a field extension (notnecessarily algebraic). Show that XK has pure dimension n. Hint: Reduce to thecase where X is affine, so say X = Spec A. Reduce to the case where A is an integral
domain. Show (using the axiom of choice) that K/k can be written as an algebraic
extension of a purely transcendental extension. Hence by Exercise 11.1.G(a), it
suffices to deal with the case where K/k is purely transcendental, say with transcendence basis {ei }i∈I (possibly infinite). Show that A ′ := A ⊗k K is an integral
domain, by interpreting it as a certain localization of the domain A[{ei }]. If t1 , . . . ,td is a transcendence basis for K(A)/k, show that {ei } ∪ {tj } is a transcendence basis
for K(A ′ )/k. Show that {tj } is a transcendence basis for K(A ′ )/K.
Exercise 11.2.J is conceptually very useful. For example, if X is describedby some equations with Q-coefficients, the dimension of X doesn’t depend onwhether we consider it as a Q-scheme or as a C-scheme.
11.2.16. Remark. Unlike Exercise 11.1.G, Exercise 11.2.J has finite type hypotheseson X. It is not true that if X is an arbitrary k-scheme of pure dimension n, and K/kis an arbitrary extension, then Xk necessarily has pure dimension n. For example,
you can show that dim k(x) ⊗k k(y) = 1 using the same ideas as in Exercise 9.2.K.
11.2.17. ⋆ Lines on hypersurfaces, part 1. Notice: although dimension theory is
not central to the following statement, it is essential to the proof.
11.2.K. E NLIGHTENING STRENUOUS EXERCISE : M OST SURFACES IN THREE - SPACE
OF DEGREE d > 3 HAVE NO LINES . In this exercise, we work over an algebraicallyclosed field k. For any d > 3, show that most degree d surfaces in P3 contain nolines. Here, “most” means “all closed points of a Zariski-open subset of the parameter space for degree d homogeneous polynomials in 4 variables, up to scalars”. As)*
there are d+3
such monomials, the degree d hypersurfaces are parametrized by
3
d+3
−1
(
)
3
P
(see Remark 4.5.3). Hint: Construct an incidence correspondenced+3

X = {(ℓ, H) : [ℓ] ∈ G(1, 3), [H] ∈ P( 3 )−1 , ℓ ⊂ H},

parametrizing lines in P3 contained in a hypersurface: define a closed subschemed+3
X of P( 3 )−1 × G(1, 3) that makes this notion precise. (Recall that G(1, 3) is ad+3
Grassmannian.) Show that X is a P( 3 )−1−(d+1) -bundle over G(1, 3). (Possible

hint for this: how many degree d hypersurfaces contain the line x = y = 0?)
Show that dim G(1, 3) = 4 (see §6.7: G(1, 3) has an open cover by A4 ’s). Show that)*d+3
dim X = d+3−1−(d+1)+4. Show that the image of the projection X → P( 3 )−1
3

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must lie in a proper closed subset. The following diagram may help.
)*
dim d+3− 1 − (d + 1) + 4
3❦ X ❙❙❙❙❙❙❙ (d+3❦❦❦❦❦❦❙❙P❙ 3 )−1−(d+1)
❦
❦
❦
❙❙❙
❦
❦
❦
❙❙❙
❦❦❦
❙❙0
❦
*
❦
d+3
−1
(
)
P 3
G(1, 3)dim 4(The argument readily generalizes to show that if d > 2n−3, then “most” degree dhypersurfaces in Pn have no lines. The case n = 1 and n = 2 are trivial but worth
thinking through.)
11.2.18. Side Remark. If you do the previous Exercise, your dimension count will
suggest the true facts that degree 1 hypersurfaces — i.e., hyperplanes — have 2dimensional families of lines, and that most degree 2 hypersurfaces have 1-dimensionalfamilies (rulings) of lines, as shown in Exercise 8.2.M. They will also suggest thatmost degree 3 hypersurfaces contain a finite number of lines, which reflects thecelebrated fact that regular cubic surfaces over an algebraically closed field always
contain 27 lines (Theorem 27.1.1), and we will use this “incidence correspondence”
or “incidence variety” to prove it (§27.4). The statement about quartic surfaces generalizes to the Noether-Lefschetz Theorem implying that a very general surface ofdegree d at least 4 contains no curves that are not the intersection of the surface
with a hypersurface, see [Lef, GH2]. “Very general” means that in the parameter
space (in this case, the projective space parametrizing surfaces of degree d), thestatement is true away from a countable union of proper Zariski-closed subsets.
Like “general”, (which was defined in §9.3.6), “very general” is a weaker versionof the phrase “almost every”.

11.3 Codimension one miracles: Krull’s and Hartogs’s Theorems
In this section, we will explore a number of results related to codimension
one. We introduce two results that apply in more general situations, and linkfunctions and the codimension one points where they vanish: Krull’s Principal
Ideal Theorem 11.3.3, and Algebraic Hartogs’s Lemma 11.3.11. We will find these
two theorems very useful. For example, Krull’s Principal Ideal Theorem will help
us compute codimensions, and will show us that codimension can behave oddly,
and Algebraic Hartogs’s Lemma will give us a useful characterization of uniquefactorization domains (Proposition 11.3.5). The results in this section will require(locally) Noetherian hypotheses. They are harder, in that the proofs are technical,
and don’t shed much light on the uses of the results. Thus it is more important tounderstand how to use these results than to be familiar with their proofs.
11.3.1. Krull’s Principal Ideal Theorem. In a vector space, a single linear equation
always cuts out a subspace of codimension 0 or 1 (and codimension 0 occurs only
when the equation is 0). The Principal Ideal Theorem, due to W. Krull, generalizes
this linear algebra fact.

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11.3.2. Krull’s Principal Ideal Theorem (geometric version). — Suppose X is alocally Noetherian scheme, and f is a function. The irreducible components of V(f) are
codimension 0 or 1.
This is clearly a consequence of the following algebraic statement. You know
enough to prove it for varieties (see Exercise 11.3.H), which is where we will use
it most often. The full proof is technical, and included in §11.5 (see §11.5.2) only toshow you that it isn’t excessively long.
11.3.3. Krull’s Principal Ideal Theorem (algebraic version). — Suppose A is a
Noetherian ring, and f ∈ A. Then every prime p minimal among those containing f has
codimension at most 1. If furthermore f is not a zerodivisor, then every such prime pcontaining f has codimension precisely 1.
For example, locally principal closed subschemes have “codimension 0 or 1”,
and effective Cartier divisors have “pure codimension 1”. Here is another example, that you could certainly prove directly, without the Principal Ideal Theorem.
11.3.A. E XERCISE . Show that an irreducible homogeneous polynomial in n + 1
variables over a field k describes an integral scheme of dimension n − 1 in Pnk.
11.3.B. E XERCISE . Suppose (A, m) is a Noetherian local ring, and f ∈ m. Show
that dim A/(f) ≥ dim A − 1.
11.3.C. I MPORTANT E XERCISE ( TO BE USED REPEATEDLY ). This is a cool argument.
(a) (Hypersurfaces meet everything of dimension at least 1 in projective space, unlike inaffine space.) Suppose X is a closed subset of Pn
k of dimension at least 1, and H
is a nonempty hypersurface in Pn
.
Show
that
H meets X. (Hint: note that thekaffine cone over H contains the origin in An+1.
Apply
Krull’s Principal Ideal Thek
orem 11.3.3 to the cone over X.)(b) Suppose X "→ Pn
k is a closed subset of dimension r. Show that any codimension
r linear space meets X. Hint: Refine your argument in (a). (Exercise 11.3.F generalizes this to show that any two things in projective space that you would expect tomeet for dimensional reasons do in fact meet.)(c) Show further that there is an intersection of r + 1 nonempty hypersurfaces missing X. (The key step: show that there is a hypersurface of sufficiently high degree
that doesn’t contain any generic point of X. Show this by induction on the numberof generic points. To get from m to m + 1: take a hypersurface not vanishing on
p1 , . . . , pm . If it doesn’t vanish on pm+1 , we are done. Otherwise, call this hypersurface fm+1 . Do something similar with m + 1 replaced by i for each 1 ≤ i ≤ m.
,
Then consider i f1 · · · f^i · · · fm+1 .) If k is infinite, show that there is a codimension r + 1 linear subspace missing X. (The key step: show that there is a hyperplane
not containing any generic point of a component of X.)(d) If k is an infinite field, show that there is an intersection of r hyperplanes meeting X in a finite number of points. (We will see in Exercise 12.4.C that if k = k,for “most” choices of these r hyperplanes, this intersection is reduced, and in Exercise 18.6.N that the number of points is the “degree” of X. But first of course wemust define “degree”.)

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The following exercise has nothing to do with the Principal Ideal Theorem,
but its solution is similar to that of Exercise 11.3.C(c) (and you may wish to solve
it first).
11.3.D. E XERCISE . Prove Proposition 11.2.13 (prime avoidance). Hint: by induction on n. Don’t look in the literature — you might find a much longer argument.
11.3.E. E XERCISE ( BOUND ON CODIMENSION OF INTERSECTIONS IN An
Let k
k ).
be a field. Suppose X and Y are equidimensional closed subvarieties (possibly
singular) of codimension m and n respectively in Ad
k . Show that every componentof X ∩ Y has codimension at most m + n in Ad
asfollows.
Show that the diagonalk∼ ∆ ⊂ Ad ×k Ad is a regular embedding of codimension d. (You will quickly
Ad
=
k
k
k
guess the d equations for ∆.) Figure out how to identify the intersection of X and
Y in Ad with the intersection of X × Y with ∆ in Ad ×k Ad
k . Then show that locally,X ∩ Y is cut out in X × Y by d equations. Use Krull’s Principal Ideal Theorem 11.3.3.
You will also need Exercise 11.2.E. (See Exercise 12.2.M for a generalization.)
11.3.F. F UN E XERCISE ( GENERALIZING E XERCISE 11.3.C( B )). Suppose X and Y
are equidimensional subvarieties of Pn of codimensions d and e respectively, andd + e ≤ n. Show that X and Y intersect. Hint: apply Exercise 11.3.E to the affine
cones of X and Y. Recall the argument you used in Exercise 11.3.C(a) or (b).
11.3.G. U SEFUL E XERCISE . Suppose f is an element of a Noetherian ring A, contained in no codimension zero or one prime ideals. Show that f is invertible. (Hint:
if a function vanishes nowhere, it is invertible, by Exercise 4.3.G(b).)
11.3.4. A useful characterization of unique factorization domains.
We can use Krull’s Principal Ideal Theorem to prove one of the four useful
criteria for unique factorization domains, promised in §5.4.6.
11.3.5. Proposition. — Suppose that A is a Noetherian integral domain. Then A is aunique factorization domain if and only if all codimension 1 prime ideals are principal.
This contains Lemma 11.1.6 and (in some sense) its converse. (In fact this result
is true even without Noetherian hypotheses, without much more work; see [Ka,Thm. 5, p. 4].)
Proof. One direction is Lemma 11.1.6: if A is a unique factorization domain, thenall codimension 1 prime ideals are principal. So we assume conversely that all
codimension 1 prime ideals of A are principal, and we will show that A is a uniquefactorization domain. Now each codimension one prime ideal, being principal, isgenerated by a prime element, which is unique up to multiplication by a unit. Allprime elements of a ring are irreducible, by the usual argument. (Suppose p ∈ A
is prime, and p = ab, with a, b ∈ A \ {0}. Then by primality of the ideal (p),either a or b is in (p), say without loss of generality a = kp. Then p = kpb,
from which 1 = bk, from which b is unit. Thus p is irreducible.) Conversely, if
a ∈ A is irreducible, then by the Principal Ideal Theorem 11.3.3, V(a) contains
some codimension one point [(p)], so a ∈ (p), say a = kp. But by irreducibility ofa, as p is not a unit, k must be a unit, from which (a) = (p), so a is prime. Thus wehave shown that the prime elements are the irreducible elements (and incidentally,

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The Rising Sea: Foundations of Algebraic Geometrythat they are, up to multiplicative units, in correspondence with the codimension
one prime ideals).
Finally, A is a unique factorization domain by the usual argument: each nonunit nonzero element can be factored into a finite number of irreducibles, andthe uniqueness of such a factorization can be shown using the fact that those irreducibles are also prime.
!
11.3.6. Generalizing Krull to more equations. The following generalization of
Krull’s Principal Ideal Theorem looks like it might follow by induction from Krull,
but it is more subtle. A proof is given in §11.5.3.
11.3.7. Krull’s Height Theorem. — Suppose X = Spec A where A is Noetherian,and Z is an irreducible component of V(r1 , . . . , rℓ ), where r1 , . . . , rℓ ∈ A. Then thecodimension of Z is at most ℓ.
(As mentioned in §11.1.4, height is another common word for codimension.)
11.3.H. E XERCISE . Prove Krull’s Height Theorem 11.3.7 (and hence Krull’s Principal Ideal Theorem 11.3.3) in the special case where X is an irreducible affine variety, i.e., if A is finitely generated domain over some field k. Show that dim Z ≥dim X − ℓ. Hint: Theorem 11.2.9. It can help to localize A so that Z = V(r1 , . . . , rℓ ).
11.3.I. I MPORTANT E XERCISE . Suppose (A, m) is a Noetherian local ring.
(a) (Noetherian local rings have finite dimension, promised in Remark 11.1.8) Use Krull’s
Height Theorem 11.3.7 to prove that if there are g1 , . . . , gℓ such that V(g1 , . . . , gℓ ) ={[m]}, then dim A ≤ ℓ. Hence show that A has finite dimension. (For comparison,
Noetherian rings in general may have infinite dimension, see Exercise 11.1.K.)(b) Let d = dim A. Show that there exist g1 , . . . , gd ∈ A such that V(g1 , . . . , gd ) ={[m]}. Hint: use induction on d. Find an equation gd knocking the dimension downby 1, i.e., dim A/(gd ) = dim A − 1. Suppose p1 , . . . pn correspond to the irreducible
components of Spec A of dimension d, and qi ⊃ pi are prime ideals corresponding
to irreducible closed subsets of codimension 1 and dimension d "− 1. Use primen
avoidance (Proposition 11.2.13) to find hi ∈ qi \ ∪n
p
.
Let
g
=
j
d
j=1
i=1 hi .

11.3.8. Definition. The geometric translation of the above exercise is the following.
Given a d-dimensional “germ of a reasonable space” around a point p, p can be
cut out set-theoretically by d equations, and we always need at least d equations.
These d elements of A are called a system of parameters for the Noetherian local
ring (A, m).
11.3.9. ⋆ Pathologies of the notion of “codimension”. We can use Krull’s Principal Ideal Theorem to produce the example of pathology in the notion of codimension promised earlier this chapter. Let A = k[x](x) [t]. In other words, elementsof A are polynomials in t, whose coefficients are quotients of polynomials in x,
where no factors of x appear in the denominator. (Warning: A is not k[x, t](x) .)Clearly, A is an integral domain, so xt − 1 is not a zero divisor. You can verify that∼ k[x](x) [1/x] =∼ k(x) — “in k[x](x) , we may divide by everything but x,
A/(xt − 1) =and now we are allowed to divide by x as well” — so A/(xt − 1) is a field. Thus(xt − 1) is not just prime but also maximal. By Krull’s theorem, (xt − 1) is codimension 1. Thus (0) ⊂ (xt − 1) is a maximal chain. However, A has dimension at least

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2: (0) ⊂ (t) ⊂ (x, t) is a chain of prime ideals of length 2. (In fact, A has dimension
precisely 2, although we don’t need this fact in order to observe the pathology.)
Thus we have a codimension 1 prime in a dimension 2 ring that is dimension 0.
Here is a picture of this partially ordered set of ideals.
(x, t)

(t)

❉❉
❉❉
❉❉
❉❉
(0)

(xt − 1)
✇✇
✇
✇✇
✇✇
✇✇

This example comes from geometry, and it is enlightening to draw a picture, see
Figure 11.2. Spec k[x](x) corresponds to a “germ” of A1k near the origin, and Spec k[x](x) [t]corresponds to “this × the affine line”. You may be able to see from the picturesome motivation for this pathology — V(xt − 1) doesn’t meet V(x), so it can’t haveany specialization on V(x), and there is nowhere else for V(xt − 1) to specialize.
It is disturbing that this misbehavior turns up even in a relatively benign-lookingring.

V(xt − 1)

Spec k[x](x) [t]V(x)Spec k[x](x)F IGURE 11.2. Dimension and codimension behave oddly on thesurface Spec k[x](x) [t]11.3.10. Algebraic Hartogs’s Lemma for Noetherian normal schemes.
Hartogs’s Lemma in several complex variables states (informally) that a holomorphic function defined away from a codimension two set can be extended over

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The Rising Sea: Foundations of Algebraic Geometrythat. We now describe an algebraic analog, for Noetherian normal schemes. (Itmay also be profitably compared to the Second Riemann Extension Theorem.) We
will use this repeatedly and relentlessly when connecting line bundles and divisors.
11.3.11. Algebraic Hartogs’s Lemma. — Suppose A is a integrally closed Noetherian
integral domain. ThenA = ∩p codimension 1 Ap .
The equality takes place in K(A); recall that any localization of an integral
domain A is naturally a subset of K(A) (Exercise 1.3.C). Warning: few people callthis Algebraic Hartogs’s Lemma. I call it this because it parallels the statement incomplex geometry.
One might say that if f ∈ K(A) does not lie in Ap where p has codimension
1, then f has a pole at [p], and if f ∈ K(A) lies in pAp where p has codimension 1,
then f has a zero at [p]. It is worth geometrically interpreting Algebraic Hartogs’s
Lemma as saying that a rational function on a Noetherian normal scheme with no poles
is in fact regular (an element of A). Informally: “Noetherian normal schemes havethe Hartogs property.” (We will properly define zeros and poles in §12.5.7, see also
Exercise 12.5.H.)
One can state Algebraic Hartogs’s Lemma more generally in the case thatSpec A is a Noetherian normal scheme, meaning that A is a product of Noetherian normal integral domains; the reader may wish to do so.
Another generalization (and something closer to the “right” statement) is that
if A is a subring of a field K, then the integral closure of A in K is the intersectionof all valuation rings of K containing A; see [AtM, Cor. 5.22] for explanation andproof.
11.3.12. ⋆⋆ Proof. (This proof may be stated completely algebraically, but we state
it as geometrically as possible, at the expense of making it longer. See [Stacks, tag
031T] for another proof using Serre’s criterion for normality, which we prove in§26.3.) The left side is obviously contained in the right, so assume some x lies inevery Ap but not in A. As in the proof of Proposition 5.4.2, we measure the failureof x to be a function (an element of Spec A) with the “ideal of denominators” I ofx:
I := {r ∈ A : rx ∈ A}.
(As an important remark not necessary for the proof: it is helpful to interpret theideal of denominators as scheme-theoretically measuring the failure of x to be regular, or better, giving a scheme-theoretic structure to the locus where x is not regular.) As 1 ∈/ I, we have I ̸= A. Choose a minimal prime q containing I.
Our second step in obtaining a contradiction is to focus near the point [q], i.e.,
focus attention on Aq rather than A, and as a byproduct notice that codim q >1. The construction of the ideal of denominators behaves well with respect tolocalization — if p is any prime, then the ideal of denominators of x in Ap is Ip ,and it again measures “’the failure of Algebraic Hartogs’s Lemma for x,” this timein Ap . But Algebraic Hartogs’s Lemma is vacuously true for dimension 1 rings,
so no codimension 1 prime contains I. Thus q has codimension at least 2. By
localizing at q, we can assume that A is a local ring with maximal ideal q, and thatq is the only prime containing I.

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In the third step, we construct a suitable multiple z of x that is still not a function on Spec A, such that multiplying z by anything vanishing at [q] results in afunction.
(Translation: z ∈/ A, but zq ⊂ A.) As q is the only prime containing I,√
I = q (Exercise 3.4.F), so as q is finitely generated, there is some n with I ⊃ qn (doyou see why?). Take the minimal such n, so I ̸⊃ qn−1 , and choose any y ∈ qn−1 − I.
Let z = yx. As y ∈/ I, z ∈/ A. On the other hand, yq ⊂ qn ⊂ I, so zq ⊂ Ix ⊂ A, so zq
is an ideal of A, completing this step.
Finally, we have two cases: either there is function vanishing on [q] that, when
multiplied by z, doesn’t vanish on [q]; or else every function vanishing on [q], multiplied by z, still vanishes on [q]. Translation: (i) either zq is contained in q, or (ii) it
is not.
(i) If zq ⊂ q, then we would have a finitely generated A-module (namely q)
with a faithful A[z]-action, forcing z to be integral over A (and hence in A, as A is
integrally closed) by Exercise 7.2.I, yielding a contradiction.
(ii) If zq is an ideal of A not contained in the unique maximal ideal q, then it
must be A! Thus zq = A from which q = A(1/z), from which q is principal. But
then codim q = dim A ≤ dimA/q q/q2 ≤ 1 (the inequality coming from principalityof q), contradicting codim q ≥ 2.
!
11.3.13. ⋆ Lines on hypersurfaces, part 2. (Part 1 was §11.2.17.) We now give ageometric application of Krull’s Principal Ideal Theorem 11.3.3, applied throughExercise 11.3.C(a). Throughout, we work over an algebraically closed field k.
11.3.J. E XERCISE .
(a) Supposef(x0 , . . . , xn ) = fd (x1 , . . . , xn ) + x0 fd−1 (x1 , . . . , xn ) + · · · + xd−1
f1 (x1 , . . . , xn )0
is a homogeneous degree d polynomial (so deg fi = i) cutting out a hypersurface X in Pn containing p := [1, 0, . . . , 0]. Show that there is a line through p
contained in X if and only if f1 = f2 = · · · = fd = 0 has a common zero inPn−1 = Proj k[x1 , . . . , xn ]. (Hint: given a common zero [a1 , . . . , an ] ∈ Pn−1 , showthat line joining p to [0, a1 , . . . , an ] is contained in X.)(b) If d ≤ n − 1, show that through any point p ∈ X, there is a line contained in X.
Hint: Exercise 11.3.C(a).
(c) If d ≥ n, show that for “most hypersurfaces” X of degree d in Pn (for alln+d
hypersurfaces whose corresponding point in the parameter space P( d )−1 — cf.
Remark 4.5.3 and Exercise 8.2.K — lies in some nonempty Zariski-open subset),“most points p ∈ X” (all points in a nonempty dense Zariski-open subset of X)
have no lines in X passing through them. (Hint: first show that there is a single p
in a single X contained in no line. Chevalley’s Theorem 7.4.2 may help.)
11.3.14. Remark. A projective (or proper) k-variety X is uniruled if every point p ∈X is contained in some P1 ⊂ X. (We won’t use this word beyond this remark.) Part(b) shows that all hypersurfaces of degree at most n−1 are uniruled. One can show
(using methods beyond what we know now, see [Ko1, Cor. IV.1.11]) that if char k =0, then every smooth hypersurface of degree at least n + 1 in Pn
is not uniruledk(thus making the open set in (c) explicit). Furthermore, smooth hypersurfacesof degree n are uniruled, but covered by conics rather than lines. Thus there is

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a strong difference in how hypersurfaces, behave depending on how the degree
relates to n + 1. This is true in many other ways as well. Smooth hypersurfacesof degree less than n + 1 are examples of Fano varieties; smooth hypersurfaces ofdegree n + 1 are examples of Calabi-Yau varieties (with the possible exception ofn = 1, depending on the definition); and smooth hypersurfaces of degree greaterthan n + 1 are examples of general type varieties. We define these terms in §21.5.5.

11.4 Dimensions of fibers of morphisms of varieties
In this section, we show that the dimensions of fibers of morphisms of varieties
behave in a way you might expect from our geometric intuition. The reason we
have waited until now to discuss this is because we will use Theorem 11.2.9 (forvarieties, codimension is the difference of dimensions). We discuss generalizationsin §11.4.5.
Recall that a function f from a topological space X to R is upper semicontinuous if for each x ∈ R, f−1 ((−∞, x)) is open. Informally: the function can jump“up” upon taking limits. Our upper semicontinuous functions will map to Z, so
informally functions jump “up” on closed subsets. Similarly, f is lower semicontinuous if x ∈ R, f−1 ((x, ∞)) is open.
Before we begin, let’s make sure we are on the same page with respect to ourintuition. Elimination theory (Theorem 7.4.7) tells us that the projection π : Pn
A →Spec A is closed. We can interpret this as follows. A closed subset X of Pn
A is cutout by a bunch of homogeneous equations in n + 1 variables (over A). The imageof X is the subset of Spec A where these equations have a common nontrivial solution. If we try hard enough, we can describe this by saying that the existence of a
nontrivial solution (or the existence of a preimage of a point under π : X → Spec A)
is an “upper semicontinuous” fact. More generally, your intuition might tell youthat the locus where a number of homogeneous polynomials in n + 1 variables
over A have a solution space (in Pn
A ) of dimension at least d should be a closed
subset of Spec A. (As a special case, consider linear equations. The condition form linear equations in n + 1 variables to have a solution space of dimension at leastd + 1 is a closed condition on the coefficients — do you see why, using linear algebra?) This intuition will be correct, and will use properness in a fundamental
way (Theorem 11.4.2(b)). We will also make sense of upper semicontinuity in fiber
dimension on the source (Theorem 11.4.2(a)). A useful example to think throughis the map from the xy-plane to the xz-plane (Spec k[x, y] → Spec k[x, z]), given by(x, z) (→ (x, xy). (This example also came up in §7.4.1.)
We begin our substantive discussion with an inequality that holds more generally in the locally Noetherian setting.
11.4.A. K EY E XERCISE ( CODIMENSION BEHAVES AS YOU MIGHT EXPECT FOR A
MORPHISM , OR “ FIBER DIMENSIONS CAN NEVER BE LOWER THAN EXPECTED ”).
Suppose π : X → Y is a morphism of locally Noetherian schemes, and p ∈ X andq ∈ Y are points such that q = π(p). Show that
codimX p ≤ codimY q + codimπ−1 (q) p

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November 18, 2017 draft323(see Figure 11.3). Hint: take a system of parameters (Definition 11.3.8) for q “in Y”,
and a system of parameters for p “in π−1 (q)”, and use them to find codimY q +codimπ−1 (q) p elements of OX,p cutting out p = {[m]} in Spec OX,p . Use Exercise 11.3.I.

F IGURE 11.3. Exercise 11.4.A: the codimension of a point in the
total space is bounded by the sum of the codimension of the pointin the fiber plus the codimension of the image in the target
Does Exercise 11.4.A agree with your geometric intuition? You should be able
to come up with enlightening examples where equality holds, and where equality
fails. We will see that equality always holds for sufficiently nice — flat — morphisms, see Proposition 24.5.6.
We now show that the inequality of Exercise 11.4.A is actually an equality over“most of Y” if Y is an irreducible variety.
11.4.1. Theorem. — Suppose π : X → Y is a (necessarily finite type) morphism ofirreducible k-varieties, with dim X = m and dim Y = n. Then there exists a nonempty
open subset U ⊂ Y such that for all q ∈ U, the fiber over q has pure dimension m − n, oris empty.
(Unimportant remark: the empty set technically has pure dimension m − n —every irreducible component of the empty set has dimension m−n. This makes
the statement a bit cleaner.)
Proof. We begin with three quick reductions. (i) By shrinking Y if necessary, we
may assume that Y is affine, say Spec B. (ii) We may also assume that X is affine,say Spec A. (Reason: cover X with a finite number of affine open subsets X1 , . . . , Xa ,and take the intersection of the U’s for each of the π|Xi .) (iii) If π is not dominant,then we are done, as the image misses a dense open subset U of Spec B. So we
assume now that π is dominant.
In order to motivate the rest of the argument, we describe our goal. We willproduce a nonempty distinguished open subset U of Spec B so that π−1 (U) → U

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factors through Am−n:= Am−n×k U via a finite surjective morphism:Uk(11.4.1.1)

Spec A (open emb.

$ # π−1 (U)finite surj.

"

Am−n
U

π

"
Spec B (open emb.

"$ #U

11.4.B. E XERCISE . Show that this suffices to prove the Proposition. (Hint: Use
Exercise 11.4.A, and Theorem 11.2.9 that codimension is the difference of dimensions for varieties, to show that each component of the fiber over a point of U has
dimension at least m − n. Show that any irreducible variety mapping finitely toAm−nhas dimension at most m − n.)κ
So we now work to build (11.4.1.1). We begin by noting that we have inclusions of B into both A and K(B), and from both A and K(B) into K(A). The maps
from A and K(B) into K(A) both factor through A ⊗B K(B) (whose Spec is thegeneric fiber of π), so the maps from both A and K(B) to A ⊗B K(B) must be inclusions.
(11.4.1.2)

R K(A)❥❥S❥S❥SI✆✆S❥❥❥❥❥❥ SS ✆❥❥❥❥ SSSS ✆✆✆
❥
❥
❥
S
✆
' ( " ❥❥❥❥ %A7 ! ❥A ⊗B K(B) ✆✆✆
7
✆
✆✆
✆
✆
! $ * ) ✆✆!$ !"% K(B)B

Clearly K(A) ⊗B K(B) = K(A) (as A ⊗B K(B) can be interpreted as taking A
and inverting those nonzero elements of B), and A ⊗B K(B) is a finitely generated
algebra over the field K(B).
By transcendence theory (Exercise 11.2.A), K(A) has transcendence degreem − n over K(B) (as K(A) has transcendence degree m over k, and K(B) has transcendence degree n over k). Thus by Noether normalization 11.2.4, we can findelements t1 , . . . , tm−n ∈ A ⊗B K(B), algebraically independent over K(B), suchthat A ⊗B K(B) is integral over K(B)[t1 , . . . , tm−n ].
Now, we can think of the elements ti ∈ A ⊗B K(B) as fractions, with numerators in A and (nonzero) denominators in B. If f is the product of the denominators
appearing for each ti , then by replacing B by Bf (replacing Spec B by its distinguished open subset D(f)), we may assume that the ti are all in A. Thus (after
sloppily renaming Bf as B, and Af as A) we can trim and extend (11.4.1.2) to the

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325

following.
"% A ⊗B K(B)7!$"
B[t1 , . . . , tm−n ] !
7!$% K(B)[t1 , . . . , tm−n ]
7

!$ !"
B

!$
% K(B)

A7 !

integral

Now A is finitely generated over B, and hence over B[t1 , . . . , tm−n ], say by
u1 , . . . , uq . Noether normalization implies that each ui satisfies some monic equation fi (ui ) = 0, where fi ∈ K(B)[t1 , . . . , tm−n ][t]. The coefficients of fi are a priori fractions in B, but by multiplying by all those denominators, we can assume
each fi ∈ B[t1 , . . . , tm−n ][t]. Let b ∈ B be the product of the leading coefficientsof all the fi . If U = D(b) (the locus where b is invertible), then over U, the fi(can be taken to) have leading coefficient 1, so the ui (in Ab ) are integral overBb [t1 , . . . , tm−n ]. Thus Spec Ab → Spec Bb [t1 , . . . , tm−n ] is finite and surjective(the latter by the Lying Over Theorem 7.2.5).
We have now constructed (11.4.1.1), as desired.
!
There are a couple of things worth pointing out about the proof. First, thisresult is interesting (and almost exclusively used) for classical varieties over a fieldk. But the proof of it uses the theory of varieties over another field, notably thefunction field K(B). This is an example of how the introduction of generic points
to algebraic geometry is useful even for considering more “classical” questions.
Second, the idea of the main part of the argument is that we have a result overthe generic point (Spec A⊗B K(B) finite and surjective over affine space over K(B)),
and we want to “spread it out” to an open neighborhood of the generic point ofSpec B. We do this by realizing that “finitely many denominators” appear whencorrectly describing the problem, and inverting those functions. This “spreadingout” idea is a recurring theme.
11.4.C. E XERCISE ( USEFUL CRITERION FOR IRREDUCIBILITY ). Suppose π : X → Y
is a proper morphism to an irreducible variety, and all the fibers of π are nonempty,
and irreducible of the same dimension. Show that X is irreducible.
11.4.2. Theorem (upper semicontinuity of fiber dimension). — Suppose π : X → Y
is a morphism of finite type k-schemes.
(a) (upper semicontinuity on the source) The dimension of the fiber of π at p ∈ X (the dimension of the largest component of π−1 (π(p)) containing p) is an upper semicontinuousfunction in p (i.e., on X).
(b) (upper semicontinuity on the target) If furthermore π is closed (e.g., if π is proper), thenthe dimension of the fiber of π over q ∈ Y is an upper semicontinuous function in q (i.e.,on Y).
You should be able to immediately construct a counterexample to part (b) if
the closedness hypothesis is dropped. (We also remark that Theorem 11.4.2(b) for
projective morphisms is done, in a simple way, in Exercise 18.1.B.)

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Proof. (a) Let Fn be the subset of X consisting of points where the fiber dimension
is at least n. We wish to show that Fn is a closed subset for all n. We argue by
induction on dim Y. The base case dim Y = 0 is trivial. So we fix Y, and assume theresult for all smaller-dimensional targets.
11.4.D. E XERCISE . Show that it suffices to prove the result when X and Y areintegral, and π is dominant.
Let r = dim X−dim Y be the “relative dimension” of π. If n ≤ r, then Fn = X by
Exercise 11.4.A (combined with Theorem 11.2.9, that codimension is the differenceof dimensions for varieties).
If n > r, then let U ⊂ Y be the dense open subset of Theorem 11.4.1, where“the fiber dimension is exactly r”. Then Fn does not meet the preimage of U. By
replacing Y with Y \ U (and X by X \ π−1 (U)), we are done by the inductive hypothesis.
11.4.E. E ASY EXERCISE . Prove (b) (using (a)).

!

11.4.3. Proposition (“Generically finite implies generally finite”). — Suppose
π : X → Y is a generically finite morphism of irreducible k-varieties of dimension n. Then
there is a dense open subset V ⊂ Y above which π is finite.

(If you wish, you can later relax the irreducibility hypothesis to simply requiring X and Y to be simply of pure dimension n.)

Proof. As in the proof of Theorem 11.4.1, we may assume that Y is affine, and that
π is dominant.
11.4.F. E XERCISE . Prove the result under the additional assumption that X is affine.
Hint: follow the appropriate part of the proof of Theorem 11.4.1.
For the general case, suppose that X = ∪ni=1 Ui , where the Ui are affine open
subschemes of X. By Exercise 11.4.F, there are dense open subsets Vi ⊂ Y overwhich π|Ui is finite. By replacing Y by an affine open subset of ∩Vi , we may assume
that π|Ui is finite.
11.4.G. E XERCISE . Show that π is closed. Hint: you will just use that π|Ui is closed,
and that there are a finite number of Ui .
Then X \ U1 is a closed subset, so π(X \ U1 ) is closed.
11.4.H. E XERCISE . Show that this closed subset is not all of Y.
Define V := Y \ π(X \ U1 ). Then π is finite above V: it is the restriction of thefinite morphism π|U1 : U1 → Y to the open subset V of the target Y.
!

11.4.4. Aside: Semicontinuities.
Semicontinuity is a recurring theme in algebraic geometry. It is worth keeping
an eye out for it. Other examples include the following.
(i) fiber dimension (Theorem 11.4.2 above)

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327(ii) the rank of a matrix of functions (because rank drops on closed subsets,
where various discriminants vanish)(iii) the rank of a finite type quasicoherent sheaf (Exercise 13.7.J)(iv) degree of a finite morphism, as a function of the target (§13.7.5)(v) dimension of tangent space at closed points of a variety over an algebraically closed field (Exercise 21.2.J)(vi) rank of cohomology groups of coherent sheaves, in proper flat families(Theorem 28.1.1)
All but (ii) are upper semicontinuous; (ii) is a lower semicontinuous function.
11.4.5. ⋆⋆ Generalizing results of §11.4 beyond varieties. The above arguments
can be extended to more general situations than varieties. We remain in the locally
Noetherian situation for safety, until the last sentence of §11.4.6. One fact used
repeatedly was that codimension is the difference of dimensions (Theorem 11.2.9).
This holds much more generally; see Remark 11.2.10 on catenary rings. Extensionsof Theorem 11.4.1 should require that π be finite type (which was automatic in thestatement of Theorem 11.4.1, by the Cancellation Theorem 10.1.19 for finite typemorphisms). In the proof of Theorem 11.4.1, we use that the dimension of the thegeneric fiber of the morphism π : X → Y of irreducible schemes is dim X − dim Y;
this can be proved using Proposition 24.5.6. The remaining results then readily
follow without change.
11.4.6. We make one particular generalization explicit because we will mention
it later. The ring Z (along with finitely generated algebras over it, and localizations thereof) is excellent in all possible ways, including in the ways needed in
the previous paragraph, so the argument for upper semicontinuity on the source(Theorem 11.4.2(a)) applies without change if the target is the Spec of a finitely generated Z-algebra. Any finite type morphism of locally Noetherian schemes is locally pulled back from a finite type morphism to the Spec of a finitely-generated Zalgebra (this will essentially be made precise and shown in Exercise 28.2.L), so wethen have upper semicontinuity of fiber dimension on the source for all finite typemorphisms of locally Noetherian schemes, and better yet, for all locally finitely
presented morphisms.

11.5 ⋆⋆ Proof of Krull’s Principal Ideal and Height Theorems
The details of this proof won’t matter to us, so you should probably not readit. It is included so you can glance at it and believe that the proof is fairly short,and that you could read it if you needed to.
If A is a ring, an A-module is Artinian if it satisfies the descending chain condition for submodules (any infinite descending sequence of submodules must stabilize, cf. §3.6.14). A ring is Artinian if it is Artinian over itself as a module. The notion of Artinian rings is very important, but we will get away without discussing
it much.
If m is a maximal ideal of A, then any finite-dimensional (A/m)-vector space(interpreted as an A-module) is clearly Artinian, as any descending chain
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must eventually stabilize (as dimA/m Mi is a non-increasing sequence of non-negative
integers).
11.5.A. E XERCISE . Suppose the maximal ideal m is finitely generated. Show thatfor any n, mn /mn+1 is a finite-dimensional (A/m)-vector space. (Hint: show it for
n = 0 and n = 1. Show surjectivity of Symn m/m2 → mn /mn+1 to bound thedimension for general n.) Hence mn /mn+1 is an Artinian A-module.

11.5.B. E XERCISE . Suppose A is a ring with one prime ideal√ m. Suppose m isnfinitely generated. Prove that m = (0) for some n. (Hint: As 0 is prime, it must
be m. Suppose m can be generated by r elements, each of which has kth power 0,
and show that mr(k−1)+1 = 0.)
11.5.C. E XERCISE . Show that if 0 → M ′ → M → M ′′ → 0 is an exact sequence ofmodules, then M is Artinian if and only if M ′ and M ′′ are Artinian. (Hint: given
a descending chain in M, produce descending chains in M ′ and M ′′ .)11.5.1. Lemma. — If A is a Noetherian ring with one prime ideal m, then A is Artinian,i.e., it satisfies the descending chain condition for ideals.
Proof. As we have a finite filtration
A ⊃ m ⊃ · · · ⊃ mn = (0)all of whose quotients are Artinian, A is Artinian as well.

!

11.5.2. Proof of Krull’s Principal Ideal Theorem 11.3.3. Suppose we are given f ∈ A,
with p a minimal prime containing f. By localizing at p, we may assume that A isa local ring, with maximal ideal p. Suppose q is another prime strictly containedin p.
(f) + ,
❃❃
❃❃
❃❃
❃❃
8 !"
Tp
⑦
⑦
⑦
⑦
⑦⑦
-+ ⑦⑦
q

%A

For the first part of the theorem, we must show that Aq has dimension 0. The
second part follows from our earlier work: if any minimal prime ideals are height0, f is a zerodivisor, by Remark 5.5.12 (or Theorem 5.5.10(c) and §5.5.3).
Now p is the only prime ideal containing (f), so A/(f) has one prime ideal. By
Lemma 11.5.1, A/(f) is Artinian.
For this proof only, we invoke a useful construction, the nth symbolic powerof a prime ideal: if A is any ring, and q is any prime ideal, then defineq(n) := {r ∈ A : rs ∈ qn for some s ∈ A − q}.

We return to our particular A and q. We have a descending chain of ideals inAq(1) ⊃ q(2) ⊃ · · · ,Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

November 18, 2017 draft329

so we have a descending chain of ideals in A/(f)q(1) + (f) ⊃ q(2) + (f) ⊃ · · ·which stabilizes, as A/(f) is Artinian. Say q(n) + (f) = q(n+1) + (f), soq(n) ⊂ q(n+1) + (f).

Hence for any g ∈ q(n) , we can write g = h + af with h ∈ q(n+1) . Hence af ∈ q(n) .
As p is minimal over f, f ∈/ q, so a ∈ q(n) . Thus
q(n) = q(n+1) + (f)q(n) .

As f is in the maximal ideal p, the third version of Nakayama’s Lemma (Exercise 7.2.G) gives q(n) = q(n+1) .
We now shift attention to the local ring Aq , which we are hoping has dimension 0. We have q(n) Aq = q(n+1) Aq (the symbolic power construction clearly commutes with localization). For any r ∈ qn Aq ⊂ q(n) Aq , there is some s ∈ Aq − qAqsuch that rs ∈ qn+1 Aq . As s is invertible, r ∈ qn+1 Aq as well. Thus qn Aq ⊂qn+1 Aq , but as qn+1 Aq ⊂ qn Aq , we have qn Aq = qn+1 Aq . By Nakayama’s
Lemma 7.2.9 (version 2),
qn Aq = 0.
Finally, any local ring (R, m) such that mn = 0 has dimension 0, as Spec R consistsof only one point: [m] = V(m) = V(mn ) = V(0) = Spec R.
!
11.5.3. Proof of Krull’s Height Theorem 11.3.7. We argue by induction on n. The
case n = 1 is Krull’s Principal Ideal Theorem 11.3.3. Assume n > 1. Suppose p isa minimal prime containing r1 , . . . , rn ∈ A. We wish to show that codim p ≤ n.
By localizing at p, we may assume that p is the unique maximal ideal of A. Letq ̸= p be a prime ideal of A with no prime between p and q. We shall show that q is
minimal over an ideal generated by n − 1 elements. Then codim q ≤ n − 1 by the
inductive hypothesis, so we will be done.
Now q cannot contain every ri (as V(r1 , . . . , rn ) = {[p]}), so say r1 ∈/ q. Then
V(q, r1 ) = {[p]}. As each ri ∈ p, there is some N such that rN∈(q,r1 ) (Exericise 3.4.J), so write rN
=
q
+
a
r
where
q
∈
q
(2
≤
i
≤
n)
and
a
∈
A. Note
i
i
1
i
i
i
that
(11.5.3.1)

N
V(r1 , q2 , . . . , qn ) = V(r1 , rN
2 , . . . , rn ) = V(r1 , r2 , . . . , rn ) = {[p]}.

We shall show that q is minimal among prime ideals containing q2 , . . . , qn ,completing the proof. In the ring A/(q2 , . . . , qn ), V(r1 ) = {[p]} by (11.5.3.1). By
Krull’s Principal Ideal Theorem 11.3.3, [p] is codimension at most 1, so [q] must be
codimension 0 in Spec A/(q2 , . . . , qn ), as desired.
!

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CHAPTER 12

Regularity and smoothness
One natural notion we expect to see for geometric spaces is the notion of whenan object is “smooth”. In algebraic geometry, this notion, called regularity, is easy
to define (Definition 12.2.3) but a bit subtle in practice. This will lead us to a different related notion of when a variety is smooth (Definition 12.2.6).
This chapter has many moving parts, of which §12.1–12.6 are the important
ones. In §12.1, the Zariski tangent space is motivated and defined. In §12.2, we define regularity and smoothness over a field, the central topics of this chapter, and discuss some of their important properties. In §12.3, we give a number of important
examples, mostly in the form of exercises. In §12.4, we discuss Bertini’s Theorem,
a fundamental classical result. In §12.5, we give many characterizations of discrete
valuation rings, which play a central role in algebraic geometry. Having seen clues
that “smoothness” is a “relative” notion rather than an “absolute” one, in §12.6 wedefine smooth morphisms (and in particular, étale morphisms), and give some of theirproperties. (We will revisit this definition in §21.3.1, once we know more.)
The remaining sections are less central. In §12.7, we discuss the valuative criteria for separatedness and properness. In §12.8, we mention some more sophisticated facts about regular local rings. In §12.9, we prove the Artin-Rees Lemma,
because it will have been invoked in §12.5 (and will be used later as well).

12.1 The Zariski tangent space
We begin by defining the tangent space of a scheme at a point. It behaves likethe tangent space you know and love at “smooth” points, but also makes sense atother points. In other words, geometric intuition at the “smooth” points guides
the definition, and then the definition guides the algebra at all points, which inturn lets us refine our geometric intuition.
The definition is short but surprising. The main difficulty is convincing yourself that it deserves to be called the tangent space. This is tricky to explain, because
we want to show that it agrees with our intuition, but our intuition is worse than
we realize. So you should just accept this definition for now, and later convince
yourself that it is reasonable.
12.1.1. Definition. The Zariski cotangent space of a local ring (A, m) is defined to
be m/m2 ; it is a vector space over the residue field A/m. The dual vector space is∨
the Zariski tangent space. If X is a scheme, the Zariski cotangent space TX,pat apoint p ∈ X is defined to be the Zariski cotangent space of the local ring OX,p (andsimilarly for the Zariski tangent space TX,p ). Elements of the Zariski cotangent
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space are called cotangent vectors or differentials; elements of the tangent space
are called tangent vectors.
The cotangent space is more algebraically natural than the tangent space, inthat the definition is shorter. There is a moral reason for this: the cotangent space
is more naturally determined in terms of functions on a space, and we are verymuch thinking about schemes in terms of “functions on them”. This will come up
later.
Here are two plausibility arguments that this is a reasonable definition. Hopefully one will catch your fancy.
In differential geometry, the tangent space at a point is sometimes defined as
the vector space of derivations at that point. A derivation at a point p of a manifold
is an R-linear operation that takes in functions f near p (i.e., elements of Op ), andoutputs elements f ′ (p) of R, and satisfies the Leibniz rule(fg) ′ = f ′ g + g ′ f.
(We will later define derivations in a more general setting, §21.2.17.) A derivationis the same as a map m → R, where m is the maximal ideal of Op . (The map
Op → R extends this, via the map Op → m given by f − f(p).) But m2 maps to 0, as
if f(p) = g(p) = 0, then(fg) ′ (p) = f ′ (p)g(p) + g ′ (p)f(p) = 0.

Thus a derivation induces a map m/m2 → R, i.e., an element of (m/m2 )∨ .

12.1.A. E XERCISE . Check that this is reversible, i.e., that any linear map m/m2 → R
gives a derivation. In other words, verify that the Leibniz rule holds.

Here is a second, vaguer, motivation that this definition is plausible for thecotangent space of the origin of An . (I prefer this one, as it is more primitive andelementary.) Functions on An should restrict to a linear function on the tangent
space. What (linear) function does x2 + xy + x + y restrict to “near the origin”?
You will naturally answer: x + y. Thus we “pick off the linear terms”. Hence m/m2
are the linear functionals on the tangent space, so m/m2 is the cotangent space. In
particular, you should picture functions vanishing at a point (i.e., lying in m) as
giving functions on the tangent space in this obvious way.

12.1.2. Old-fashioned example. Computing the Zariski tangent space is actuallyquite hands-on, because you can compute it just as you did when you learned
multivariable calculus. In A3 , we have a curve cut out by x + y + z2 + xyz = 0
and x − 2y + z + x2 y2 z3 = 0. (You can use Krull’s Principal Ideal Theorem 11.3.3
to check that this is a curve, but it is not important to do so.) What is the tangent
line near the origin? (Is it even smooth there?) Answer: the first surface looks likex + y = 0 and the second surface looks like x − 2y + z = 0. The curve has tangent
line cut out by x + y = 0 and x − 2y + z = 0. It is smooth (in the traditional sense).
In multivariable calculus, the students work hard to get the answer, because wearen’t allowed to tell them to just pick out the linear terms.
Let’s make explicit the fact that we are using. If A is a ring, m is a maximal
ideal, and f ∈ m is a function vanishing at the point [m] ∈ Spec A, then the Zariski
tangent space of Spec A/(f) at m is cut out in the Zariski tangent space of Spec A(at m) by the single linear equation f (mod m2 ). The next exercise will force you
to think this through.

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12.1.B. I MPORTANT E XERCISE (“K RULL’ S P RINCIPAL I DEAL T HEOREM FOR TAN GENT SPACES ” — BUT MUCH EASIER THAN K RULL’ S P RINCIPAL I DEAL T HEO REM 11.3.3!).
Suppose A is a ring, and m a maximal ideal. If f ∈ m, showthat the Zariski tangent space of A/f is cut out in the Zariski tangent space of A
by f (mod m2 ). (Note: we can quotient by f and localize at m in either order, as
quotient and localization commute, (4.3.6.1).) Hence the dimension of the Zariski
tangent space of Spec A/(f) at [m] is the dimension of the Zariski tangent space ofSpec A at [m], or one less. (That last sentence should be suitably interpreted if thedimension is infinite, although it is less interesting in this case.)
Here is another example to see this principle in action, extending Example 12.1.2:
x + y + z2 = 0 and x + y + x2 + y4 + z5 = 0 cuts out a curve, which obviously
passes through the origin. If I asked my multivariable calculus students to calculate the tangent line to the curve at the origin, they would do calculations which
would boil down (without them realizing it) to picking off the linear terms. They
would end up with the equations x + y = 0 and x + y = 0, which cut out a plane,
not a line. They would be disturbed, and I would explain that this is because thecurve isn’t smooth at a point, and their techniques don’t work. We on the other
hand bravely declare that the tangent space is cut out by x + y = 0, and (will soon)
use this pathology as definition of what makes a point singular (or non-regular).
(Intuitively, the curve near the origin is very close to lying in the plane x + y = 0.)
Notice: the cotangent space jumped up in dimension from what it was “supposed
to be”, not down. We will see that this is not a coincidence, in Theorem 12.2.1.
12.1.C. E XERCISE . Suppose Y and Z are closed subschemes of X, both containing
the point p ∈ X.
(a) Show that TZ,p is naturally a sub-κ(p)-vector space of TX,p .
(b) Show that TY∩Z,p = TY,p ∩ TZ,p , where ∩ as usual in this context is schemetheoretic intersection.
(c) Show that TY∪Z,p contains the span of TY,p and TZ,p , where ∪ as usual in thiscontext is scheme-theoretic union.
(d) Show that TY∪Z,p can be strictly larger than the span of TY,p and TZ,p . (Hint:
Figure 4.5.)
Here is a pleasant consequence of the notion of Zariski tangent space.
12.1.3. Problem. Consider the ring A = k[x, y, z]/(xy − z2 ). Show that (x, z) is nota principal ideal.
∼ k[y] has dimension 1, we see that thisAs dim A = 2 (why?), and A/(x, z) =ideal is codimension 1 (as codimension is the difference of dimensions for irreducible varieties, Theorem 11.2.9). Our geometric picture is that Spec A is a cone
(we can diagonalize the quadric as xy − z2 = ((x + y)/2)2 − ((x − y)/2)2 − z2 , atleast if char k ̸= 2 — see Exercise 5.4.J), and that (x, z) is a line on the cone. (SeeFigure 12.1 for a sketch.) This suggests that we look at the cone point.
12.1.4. Solution. Let m = (x, y, z) be the maximal ideal corresponding to theorigin. Then Spec A has Zariski tangent space of dimension 3 at the origin, andSpec A/(x, z) has Zariski tangent space of dimension 1 at the origin. But Spec A/(f)must have Zariski tangent space of dimension at least 2 at the origin by Exercise 12.1.B.

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F IGURE 12.1. V(x, z) ⊂ Spec k[x, y, z]/(xy − z2 ) is a line on a cone
12.1.5. ⋆ Remark. Another approach to solving the problem, not requiring thedefinition of the Zariski tangent space, is to use the fact that the ring is graded(where x, y, and z each have degree 1), and the ideal (x, z) is a homogeneous ideal.
(You may enjoy thinking this through.) The advantage of using the tangent spaceis that it applies to more general situations where there is no grading. For example,(a) (x, z) is not a principal ideal of k[x, y, z]/(xy − z2 − z3 ). As a different example,(b) (x, z) is not a principal ideal of the local ring (k[x, y, z]/(xy − z2 ))(x,y,z) (the“germ of the cone”). However, we remark that the graded case is still very useful.
The construction of replacing a filtered ring by its “associated graded” ring canturn more general rings into graded rings (and can be used to turn example (a)into the graded case). The construction of completion can turn local rings into
graded local rings (and can be used to turn example (b) into, essentially, the gradedcase). Filtered rings will come up in §12.9, the associated graded construction will
implicitly come up in our discussions of the blow-up in §22.3, and many aspectsof completions will be described in Chapter 29.
12.1.D. E XERCISE . Show that (x, z) ⊂ k[w, x, y, z]/(wz − xy) is a codimension 1
ideal that is not principal, using the method of Solution 12.1.4. (See Figure 12.2
for the projectivization of this situation — a line on a smooth quadric surface.)
This example was promised just after Exercise 5.4.D. An improvement is given in
Exercise 14.2.R.
12.1.E. E XERCISE . Let A = k[w, x, y, z]/(wz−xy). Show that Spec A is not factorial.
(Exercise 5.4.L shows that A is not a unique factorization domain, but this is notenough — why is the localization of A at the prime (w, x, y, z) not factorial? One
possibility is to do this “directly”, by trying to imitate the solution to Exercise 5.4.L,
but this is hard. Instead, use the intermediate result that in a unique factorization
domain, any codimension 1 prime is principal, Lemma 11.1.6, and considering
Exercise 12.1.D.) As A is integrally closed if char k ̸= 2 (Exercise 5.4.I(c)), thisyields an example of a scheme that is normal but not factorial, as promised in
Exercise 5.4.F. A slight generalization will be given in 22.4.N.
12.1.F. L ESS IMPORTANT EXERCISE (“ HIGHER - ORDER DATA”). (This exercise isfun, but won’t be used.)(a) In Exercise 3.7.B, you computed the equations cutting out the (union of the)three coordinate axes of A3k . (Call this scheme X.) Your ideal should have had
three generators. Show that the ideal cannot be generated by fewer than three

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F IGURE 12.2. The line V(x, z) on the smooth quadric surface
V(wz − xy) ⊂ P3 .
elements. (Hint: working modulo m = (x, y, z) won’t give any useful information,
so work modulo a higher power of m.)(b) Show that the coordinate axes in A3k are not a regular embedding in A3k . (Thiswas promised at the end of §8.4.)
12.1.6. Morphisms and tangent spaces. Suppose π : X → Y, and π(p) = q. Thenif we were in the category of differentiable manifolds, we would expect a tangent
map, from the tangent space of p to the tangent space at q. Indeed that is thecase; we have a map of stalks OY,q → OX,p , which sends the maximal ideal of
the former n to the maximal ideal of the latter m (we have checked that this isa “local morphism” when we briefly discussed locally ringed spaces, see §6.3.1).
Thus n2 maps to m2 , from which we see that n/n2 maps to m/m2 . If (OX,p , m)and (OY,q , n) have the same residue field κ, so n/n2 → m/m2 is a linear map of
κ-vector spaces, we have a natural map (m/m2 )∨ → (n/n2 )∨ . This is the map
from the tangent space of p to the tangent space at q that we sought. (Aside: notethat the cotangent map always exists, without requiring p and q to have the same
residue field — a sign that cotangent spaces are more natural than tangent spaces
in algebraic geometry.)
Here are some exercises to give you practice with the Zariski tangent space. If
you have some differential geometric background, the first will further convince
you that this definition correctly captures the idea of (co)tangent space.
12.1.G. I MPORTANT E XERCISE ( THE J ACOBIAN COMPUTES THE Z ARISKI COTAN GENT SPACE ). Suppose X is a finite type k-scheme. Then locally it is of the form
Spec k[x1 , . . . , xn ]/(f1 , . . . , fr ). Show that the Zariski cotangent space at a k-valued
point (a closed point with residue field k) is given by the cokernel of the Jacobian

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map kr → kn given by the Jacobian matrix⎛ ∂f1∂x1 (p) · · ·⎜..
..
(12.1.6.1)
J=⎝
.
.
∂f1
∂xn (p) · · ·∂fr∂x1 (p)..
.

∂fr∂xn (p)⎞⎟⎠.

(This makes precise our example of a curve in A3 cut out by a couple of equations,where we picked off the linear terms, see Example 12.1.2.) You might be alarmed:∂fwhat does ∂x
mean? Do you need deltas and epsilons? No! Just define deriva1
tives formally, e.g.,∂(x2 + x1 x2 + x22 ) = 2x1 + x2 .
∂x1 1
Hint: Do this first when p is the origin, and consider linear terms, just as in Example 12.1.2 and Exercise 12.1.B. For the general case, “translate p to the origin”.
12.1.7. Remark. This result can be extended to closed points of X whose residue
field is separable over k (and in particular, to all closed points if char k = 0 or if
k is finite), see Remark 21.3.10. The fact that we wait until Chapter 21 to show
this does not mean that it needs to be so complicated; a more elementary proof ispossible.
12.1.8. Warning. It is more common in mathematics (but not universal) to define
the Jacobian matrix as the transpose of this. But it will be more convenient for us
to follow this minority convention.
12.1.H. E XERCISE ( THE CORANK OF THE J ACOBIAN IS INDEPENDENT OF THE PRE SENTATION ). Suppose A is a finitely-generated k-algebra, generated by x1 , . . . ,xn , with ideal of relations I generated by f1 , . . . , fr . Let p be a point of Spec A.
(a) Suppose g ∈ I. Show that appending the column of partials of g to the Jacobianmatrix (12.1.6.1) does not change the corank at p. Hence show that the corank of
the Jacobian matrix at p does not depend on the choice of generators of I.
(b) Suppose q(x1 , . . . , xn ) ∈ k[x1 , . . . , xn ]. Let h be the polynomial y−q(x1 , . . . , xn ) ∈k[x1 , . . . , xn , y]. Show that the Jacobian matrix of (f1 , . . . , fr , h) with respect tothe variables (x1 , . . . , xn , y) has the same corank at p as the Jacobian matrix of(f1 , . . . , fr ) with respect to (x1 , . . . , xn ). Hence show that the corank of the Jacobian matrix at p is independent of the choice of generators for A.
12.1.I. E XERCISE . Suppose X is a k-scheme. Describe a natural bijection from
Mork (Spec k[ϵ]/(ϵ2 ), X) to the data of a point p with residue field k (necessarily aclosed point) and a tangent vector at p. (This is important, for example in deformation theory.)
12.1.J. E XERCISE . Find the dimension of the Zariski tangent space at the point∼ Z[x]/(x2 + 4). Find the dimension of the Zariski tangent space[(2, 2i)] of Z[2i] =√∼ Z[x]/(x2 + 2). (If you prefer geometric versions
at the point [(2, x)] of Z[ −2] =of the same examples, replace Z by C, and 2 by y: consider C[x, y]/(x2 + y2 ) andC[x, y]/(x2 + y).)Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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12.2 Regularity, and smoothness over a field
The key idea in the definition of regularity is contained in the following result,that “the dimension of the Zariski tangent space is at least the dimension of thelocal ring”.
12.2.1. Theorem. — Suppose (A, m, k) is a Noetherian local ring. Then dim A ≤dimk m/m2 .
12.2.2. Proof of Theorem 12.2.1. Note that m is finitely generated (as A is Noetherian), so m/m2 is a finitely generated (A/m = k)-module, hence finite-dimensional.
Say dimk m/m2 = n. Choose a basis of m/m2 , and lift it to elements f1 , . . . , fn of m.
Then by Nakayama’s Lemma (version 4, Exercise 7.2.H), (f1 , . . . , fn ) = m.
Then by Exercise 11.3.I (a consequence of Krull’s Height Theorem 11.3.7), dim A ≤n.
!
12.2.3. Definition. If equality holds in Theorem 12.2.1, we say that A is a regular
local ring. In particular, regular local rings are Noetherian by definition. (If a Noetherian ring A is regular at all of its prime ideals, i.e., if Ap is a regular local ringfor all prime ideals p of A, then A is said to be a regular ring. We basically won’t
use this terminology. Fact 12.8.2 reassuringly implies that regular local rings areindeed regular rings.) A locally Noetherian scheme X is regular at a point p if thelocal ring OX,p is regular. (The word nonsingular is often used as well, notably inthe case when X is finite type over a field, but for the sake of consistency we willuse “regular” throughout.) It is singular at the point otherwise, and we say thatthe point is a singularity. (The word nonregular is also used, but for the sake ofconsistency, we will use “singular”, despite the fact that this choice is inconsistent
with our choice of “regular” over “nonsingular”. Possible motivation for the inconsistency: “regular local ring” and “singularity” are both standard terminology,
so at some point we are forced to make a choice.) A scheme is regular (or nonsingular) if it is regular at all points. It is singular (or nonregular) otherwise (i.e., if it
is singular at at least one point — if it has a singularity).
12.2.A. E XERCISE . Show that a dimension 0 Noetherian local ring is regular if andonly if it is a field.
You will hopefully gradually become convinced that this is the right notion of“smoothness” of schemes. Remarkably, Krull introduced the notion of a regular
local ring for purely algebraic reasons, some time before Zariski realized that it
was a fundamental notion in geometry in 1947.
12.2.B. E XERCISE . Suppose X is a finite type k-scheme (such as a variety) of pure
dimension n, and p is a nonsingular closed point of X, so OX,p is a regular local
ring of dimension n. Suppose f ∈ OX,p . Show that OX,p /(f) is a regular local ringof dimension n − 1 if and only if f ∈ m \ m2 . (Hint: Krull’s Principal Ideal Theorem
for tangent spaces, Exercise 12.1.B.)
12.2.C. E XERCISE ( THE SLICING CRITERION FOR REGULARITY ). Suppose X is a
finite type k-scheme (such as a variety), D is an effective Cartier divisor on X
(Definition 8.4.1), and p ∈ D. Show that if p is a regular point of D then p is

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The Rising Sea: Foundations of Algebraic Geometrya regular point of X. (Hint: Krull’s Principal Ideal Theorem for tangent spaces,
Exercise 12.1.B.)
12.2.4. The Jacobian criterion.
A finite type k-scheme is locally of the form Spec k[x1 , . . . , xn ]/(f1 , . . . , fr ). The
Jacobian criterion for regularity (Exercise 12.2.D) gives a hands-on method forchecking for singularity at closed points, using the equations f1 , . . . , fr , if k = k.
12.2.D. I MPORTANT E XERCISE ( THE J ACOBIAN CRITERION — EASY, GIVEN E XER CISE 12.1.G). Suppose X = Spec k[x1 , . . . , xn ]/(f1 , . . . , fr ) has pure dimension d.
Show that a k-valued point p ∈ X is regular if and only if the corank of the Jacobianmatrix (12.1.6.1) (the dimension of the cokernel) at p is d.
12.2.E. E ASY E XERCISE . Suppose k = k. Show that the singular closed points of
the hypersurface f(x1 , . . . , xn ) = 0 in An
k are given by the equations
f=

∂f
∂f
= ··· =
= 0.
∂x1
∂xn

(Translation: the singular points of f = 0 are where the gradient of f vanishes. This
is not shocking.)
12.2.5. Smoothness over a field. There seem to be two serious drawbacks with theJacobian criterion. For finite type schemes over k, the criterion gives a necessary
condition for regularity, but it is not obviously sufficient, as we need to check regularity at non-closed points as well. We can prove sufficiency by working hard toshow Fact 12.8.2, which implies that the non-closed points must be regular as well.
A second failing is that the criterion requires k to be algebraically closed. These
problems suggest that old-fashioned ideas of using derivatives and Jacobians are
ill-suited to the fancy modern notion of regularity. But that is wrong — the fault
is with the concept of regularity. There is a better notion of smoothness over a field.
Better yet, this idea generalizes to the notion of a smooth morphism of schemes (tobe discussed in §12.6, and again in Chapter 25), which behaves well in all possible
ways (including in the sense of §7.1.1). This is another sign that some properties
we think of as of objects (“absolute notions”) should really be thought of as properties of morphisms (“relative notions”). We know enough to imperfectly (butcorrectly) define what it means for a scheme to be k-smooth, or smooth over k.
12.2.6. Definition. A k-scheme is k-smooth of dimension d, or smooth of dimension d over k, if it is of pure dimension d, and there exists a cover by affine open
sets Spec k[x1 , . . . , xn ]/(f1 , . . . , fr ) where the Jacobian matrix has corank d at all
points. (In particular, it is locally of finite type.) A k-scheme is smooth over k if it
is smooth of some dimension. The k is often omitted when it is clear from context.
12.2.F. E XERCISE ( FIRST EXAMPLES ).
(a) Show that An
k is smooth for any n and k. For which characteristics is the curvey2 z = x3 − xz2 in P2k smooth (cf. Exercise 12.3.C)?
(b) Suppose f ∈ k[x1 , . . . , xn ] is a polynomial such that the system of equations
f=

∂f
∂f
= ··· =
=0∂x1
∂xn

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has no solutions in k. Show that the hypersurface f = 0 in An
k is smooth. (Compare
this to Exercise 12.2.E, which has the additional hypothesis k = k.)
12.2.G. E XERCISE ( SMOOTHNESS OF VARIETIES IS PRESERVED BY EXTENSION OF
BASE FIELD ). Suppose X is a finite type k-scheme, and k ⊂ ℓ is a field extension.
Show that if X is smooth over k then X×Spec k Spec ℓ is smooth over ℓ. (The converse
will be proved in Exercise 21.3.C.)
The next exercise shows that we need only check closed points, thereby making a connection to classical geometry.
12.2.H. E XERCISE . Show that if the Jacobian matrix for X = Spec k[x1 , . . . , xn ]/(f1 , . . . fr )has corank d at all closed points, then it has corank d at all points. (Hint: the locus
where the Jacobian matrix has corank d can be described in terms of vanishing andnonvanishing of determinants of certain explicit matrices.)
12.2.7. You can check that any open subset of a smooth k-variety is also a smooth kvariety. With what we know now, we could show that this implies that k-smoothness
is equivalent to the Jacobian being corank d everywhere for every affine open cover(and by any choice of generators of the ring corresponding to such an open set).
Indeed, you should feel free to do this if you cannot restrain yourself. But thecokernel of the Jacobian matrix is secretly the space of differentials (which might
not be surprising if you have experience with differentials in differential geometry), so this will come for free when we give a better version of this definition in
Definition 21.3.1. The current imperfect definition will suffice for us to work out
examples. And if you don’t want to wait until Definition 21.3.1, you can use Exercise 12.2.I below to show that if k algebraically closed, then smoothness can bechecked on any open cover.
12.2.8. In defense of regularity. Having made a spirited case for smoothness, weshould be clear that regularity is still very useful. For example, it is the only concept which makes sense in mixed characteristic. In particular, Z is regular at its
points (it is a regular ring), and more generally, discrete valuation rings are incredibly useful examples of regular rings.
12.2.9. Regularity vs. smoothness.
12.2.I. E XERCISE . Suppose X is a finite type scheme of pure dimension d over analgebraically closed field k = k. Show that X is regular at its closed points if andonly if it is smooth. (We will soon learn that for finite type k-schemes, regularity atclosed points is the same as regularity everywhere, Theorem 12.8.3.) Hint to show
regularity implies smoothness: use the Jacobian criterion to show that the corankof the Jacobian is d at the closed points of X. Then use Exercise 12.2.H.
More generally, if k is perfect (e.g., if char k = 0 or k is a finite field), thensmoothness is the same as regularity at closed points (see Exercise 21.3.D). Moregenerally still, we will later prove the following fact. We mention it now because
it will make a number of statements cleaner long before we finally prove it. (There
will be no circularity.)

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12.2.10. Smoothness-Regularity Comparison Theorem. —(a) If k is perfect, every regular finite type k-scheme is smooth over k.
(b) Every smooth k-scheme is regular (with no hypotheses on perfection).
Part (a) will be proved in Exercise 21.3.D. Part (b) will be proved in §25.2.3.
The fact that Theorem 12.2.10 will be proved so far in the future does not meanthat we truly need to wait that long. We could prove both parts by the end of thischapter, with some work. We instead postpone the proof until we have machinerythat will do much of the work for us. But if you wish some insight right away, hereis the outline of the argument for (b), which you can even try to implement after
reading this chapter. We will soon show (in Exercise 12.3.O) that An
k is regular for
all fields k. Once we know the definition of étale morphism, we will realize thatevery smooth variety locally admits an étale morphism to An
k (Exercise 12.6.E).
You can then show that an if π : X → Y is an étale morphism of locally Noetherian
schemes, and p ∈ X, then p is regular if π(p) is regular. This will be shown in
Exercise 25.2.D, but you could reasonably do this after reading the definition ofétaleness.
12.2.11. Caution: Regularity does not imply smoothness. If k is not perfect, thenregularity does not imply smoothness, as demonstrated by the following example. Let k = Fp (u), and consider the hypersurface X = Spec k[x]/(xp − u). Now
k[x]/(xp − u) is a field, hence regular. But if f(x) = xp − u, then f(u1/p ) =df
1/p) = 0, so the Jacobian criterion fails — X is not smooth over k. (Neverdx (uforget that smoothness requires a choice of field — it is a “relative” notion, and wewill later define smoothness over an arbitrary scheme, in §12.6.) Technically, thisargument is not yet complete: as noted in §12.2.7, we have not shown that it suffices to check the Jacobian on any affine cover. But as mentioned in §12.2.7, this will
be rectified when we give a better definition of smoothness in Definition 21.3.1.
In case the previous example is too “small” to be enlightening (because thescheme in question is smooth over a different field, namely k[x]/(xp −u)), here is another. Let k = Fp (u) as before, with p > 2, and consider the curve Spec k[x, y]/(y2 −xp + u). Then the closed point (y, xp − u) is regular but not smooth.
Thus you should not use “regular” and “smooth” interchangeably.
12.2.12. Regular local rings are integral domains.
You might expect from geometric intuition that a scheme is “locally irreducible”
at a “smooth” point. Put algebraically:
12.2.13. Theorem. — Suppose (A, m, k) is a regular local ring of dimension n. Then A
is an integral domain.
Before proving it, we give some consequences.
12.2.J. E XERCISE . Suppose p is a regular point of a Noetherian scheme X. Show
that only one irreducible component of X passes through p.
12.2.K. E ASY E XERCISE . Show that a nonempty regular Noetherian scheme isirreducible if and only if it is connected.
12.2.L. I MPORTANT E XERCISE ( REGULAR SCHEMES IN REGULAR SCHEMES ARE
REGULAR EMBEDDINGS ).

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341(a) Suppose (A, m, k) is a regular local ring of dimension n, and I ⊂ A is an idealof A cutting out a regular local ring of dimension d. Let r = n − d. Show thatSpec A/I is a regular embedding in Spec A. Hint: show that there are elements
f1 , . . . , fr of I spanning the k-vector space I/(I ∩ m2 ). Show that the quotient of
A by both (f1 , . . . , fr ) and I yields dimension d regular local rings. Show that asurjection of integral domains of the same dimension must be an isomorphism.
(b) Suppose π : X → Y is a closed embedding of regular schemes. Show that π is aregular embedding.
Exercise 12.2.L has the following striking geometric consequence.

12.2.M. E XERCISE ( GENERALIZING E XERCISE 11.3.E). Suppose dimension d, and
X and Y are equidimensional subvarieties (possibly singular) of Ad
k codimension
m and n respectively. Show that every component of X ∩ Y has codimension at
d ∼
d
d ∼
most m + n in Ad
k as follows. Show that the diagonal Ak = ∆ ⊂ Ak ×k Ak =2d
Ak is a regular embedding of codimension d. Then follow the rest of the hint to
Exercise 11.3.E.
12.2.14. Remark. The following example shows that the regularity hypotheses in
Exercise 12.2.M cannot be (completely) dropped. Let W = Spec k[w, x, y, z]/(wz −xy) be the cone over the smooth quadric surface, which is an integral threefold.
Let X be the surface w = x = 0 and Y the surface y = z = 0; both lie in W. ThenX ∩ Y is just the origin, so we have two codimension 1 subvarieties meeting in a
codimension 3 subvariety. (It is no coincidence that X and Y are the affine cones
over two lines in the same ruling, see Exercise 8.2.M. This example will arise again
in Exercise 22.4.N.)
12.2.15. Proof of Theorem 12.2.13 (following [Liu, Prop. 2.11]). We prove the result by
induction on n. The case n = 0 follows from Exercise 12.2.A (dimension 0 regular
local rings are fields, which are certainly integral domains). We now assume n > 0,
and that we have proved the result for smaller dimension. Fix any f ∈ m \ m2 . By
Exercise 12.1.B, the Zariski tangent space of the local ring A/(f) (at its maximal
ideal m) has dimension n − 1. By Exercise 11.3.B, dim A/(f) ≥ n − 1. Thus by
Theorem 12.2.1, dim A/(f) = n − 1, and A/(f) is a regular local ring. By our
inductive hypothesis, A/(f) is an integral domain (of dimension n − 1).
Now choose any minimal prime ideal p of A with dim A/p = n. We wish toshow that p = (0). The Zariski cotangent space of A/p is a quotient of that of A,
and thus has dimension at most n. But dim A/p = n, so by Theorem 12.2.1, A/p
is a regular local ring of dimension n. By the argument of the previous paragraphwith A replaced by A/p, we see that A/(p+(f)) is an integral domain of dimension
n − 1. But it is a quotient of A/(f). The only way one integral domain can be thequotient of another of the same dimension is if the quotient is an isomorphism (asdiscussed in the hint to Exercise 12.2.L(a)).
Thus p + (f) = (f), i.e., p ⊂ fA. Hence each element u of p can be written as fv
for some v ∈ A. As f ∈/ p (as dim A/(p + (f)) = n − 1 < n = dim A/p), we havev ∈ p, and so p ⊂ fp.
Clearly fp ⊂ p, so p = fp. Then by the second version of Nakayama’s Lemma 7.2.9,
p = (0) as desired.
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12.2.16. ⋆⋆ Checking regularity of k-schemes at closed points by base changing
to k.
(We revisit these ideas using a different approach in 21.3.E, so you should readthis only if you are particularly curious.) The Jacobian criterion is a great criterion
for checking regularity of finite type k-schemes at k-valued points. The following
result extends its applicability to more general closed points.
Suppose X is a finite type k-scheme of pure dimension n, and p ∈ X is a closed
point with residue field k ′ . By the Nullstellensatz 3.2.5, k ′ /k is a finite extensionof fields; suppose that it is separable. Define π : Xk := X ×k k → X by base change
from Spec k → Spec k.

12.2.N. E XERCISE .
(a) Suppose f(x) ∈ k[x] is a separable polynomial (i.e., f has distinct roots in k),
and irreducible, so k ′′ := k[x]/(f(x)) is a field extension of k. Show that k ′′ ⊗k k is,
as a ring, k × · · · × k, where there are deg f = deg k ′′ /k factors.
(b) Show that π−1 (p) consists of deg(k ′ /k) reduced points.
12.2.O. E XERCISE . Suppose p is a closed point of X, with residue field k ′ that is
separable over k of degree d. Show that Xk is regular at all the preimages p1 , . . . ,pd of p if and only if X is regular at p as follows.
(a) Reduce to the case X = Spec A.
(b) Let m ⊂ A be the maximal ideal corresponding to p. By tensoring theexact sequence 0 → m → A → k ′ → 0 with k (field extensions preserve
exactness of sequences of vector spaces), we have the exact sequence
0 → m ⊗k k → A ⊗k k → k ′ ⊗k k → 0.

Show that m ⊗k k ⊂ A ⊗k k is the ideal corresponding to the pullback of
p to Spec A ⊗k k. Verify that (m ⊗k k)2 = m2 ⊗k k.
(c) By tensoring the short exact sequence of k-vector spaces 0 → m2 → m →m/m2 → 0 with k, show thatd.

dimk TXk ,pi = d dimk ′ TX,p .

i=1(d) Use Exercise 11.1.G(a) and the inequalities dimk TXk ,pi ≥ dim Xk anddimk ′ TX,p ≥ dim X (Theorem 12.2.1) to conclude.
12.2.17. Remark. In fact, regularity at a single pi is enough to conclude regularity
at p. You can show this by following up on Exercise 12.2.O; first deal with thecase when k ′ /k is Galois, and obtain some transitive group action of Gal(k ′ /k) on{p1 , . . . , pd }. Another approach is given in Exercise 21.3.E.
12.2.18. Remark. In Exercise 21.3.D, we will use this to prove that a variety overa perfect field is smooth if and only if it is regular at all closed points (cf. theSmoothness-Regularity Comparison Theorem 12.2.10).

12.3 Examples

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We now explore regularity in practice through a series of examples and exercises. Much of this discussion is secretly about smoothness rather than regularity.
In particular, in order to use the Jacobian criterion, we will usually work over analgebraically closed field.
12.3.1. Geometric examples.
12.3.A. E ASY E XERCISE . Suppose k is a field. Show that A1k and A2k are regular, bydirectly checking the regularity of all points. Show that P1k and P2k are regular. (Thegeneralization to arbitrary dimension is harder, so we leave it to Exercise 12.3.O.)
12.3.B. E XERCISE ( THE E ULER OR J ACOBIAN TEST FOR PROJECTIVE HYPERSUR FACES ). Suppose k = k. Show that the singular closed points of the hypersurface
f = 0 in Pn
k correspond to the locus
f=

∂f
∂f
= ··· =
= 0.
∂x0
∂xn

If the degree of the hypersurface is not divisible by char k (e.g., if char k = 0), show
,
∂f
∂f
∂f
that it suffices to check ∂x= · · · = ∂x
= 0. Hint: show that (deg f)f = i xi ∂x.
n0i(In fact, this will give the singular points in general, not just the singular closed
points, cf. §12.2.5. We won’t use this, so we won’t prove it.)12.3.C. E XERCISE . Suppose that k = k does not have characteristic 2. Show thaty2 z = x3 − xz2 in P2k is an irreducible regular curve. (Eisenstein’s Criterion gives
one way of showing irreducibility. Warning: we didn’t specify char k ̸= 3, so be
careful when using the Euler test.)
12.3.D. E XERCISE . Suppose k = k has characteristic not 2. Show that a quadrichypersurface in Pn is regular if and only if it is maximal rank. (“Maximal rank”
was defined in Exercise 5.4.J.)
12.3.E. E XERCISE . Suppose k = k has characteristic 0. Show that there exists a
regular (projective) plane curve of degree d. Hint: try a “Fermat curve” xd + yd +zd = 0. (Feel free to weaken the hypotheses. Bertini’s Theorem 12.4.2 will give
another means of showing existence.)
12.3.F. E XERCISE ( SEE F IGURE 12.3). Find all the singular closed points of thefollowing plane curves. Here we work over k = k of characteristic 0 to avoid
distractions.
(a) y2 = x2 + x3 . This is an example of a node.
(b) y2 = x3 . This is called a cusp; we met it earlier in Exercise 9.7.F.
(c) y2 = x4 . This is called a tacnode; we met it earlier in Exercise 9.7.G.
(A precise definition of a node etc. will be given in Definition 29.3.1.)
12.3.G. E XERCISE . Suppose k = k. Use the Jacobian criterion appropriately toshow that the twisted cubic Proj k[w, x, y, z]/(wz − xy, wy − x2 , xz − y2 ) is regular.
(You can do this, without any hypotheses on k, using the fact that it is isomorphic
to P1 . But do this with the explicit equations, for the sake of practice. The twisted
cubic was defined in Exercise 8.2.A.)

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F IGURE 12.3. Plane curve singularities
12.3.2. Tangent planes and tangent lines.
Suppose a scheme X ⊂ An is cut out by equations f1 , . . . , fr , and X is regularof dimension d at the k-valued point a = (a1 , . . . , an ). Then the tangent d-plane
to X at p (sometimes denoted Tp X) is given by the r equations
"
#
"
#
∂fi
∂fi
(a) (x1 − a1 ) + · · · +(a) (xn − an ) = 0,
∂x1
∂xn
∂fi
∂fi
where (as in (12.1.6.1)) ∂x(a) is the evaluation of ∂x
at a = (a1 , . . . , an ).
1
1

12.3.H. E XERCISE . Why is this independent of the choice of defining equations f1 ,. . . , fr of X?
The Jacobian criterion (Exercise 12.2.D) ensures that these r equations indeed
cut out a d-plane. (If d = 1, this is called the tangent line.) This can be readily
shown to be notion of tangent plane that we see in multivariable calculus, butnote that here this is the definition, and thus don’t have to worry about δ’s and ϵ’s.
Instead we will have to just be careful that it behaves the way we want it to.
12.3.I. E XERCISE .
(1, 1).

Compute the tangent line to the curve of Exercise 12.3.F(b) at

12.3.J. E XERCISE . Suppose X ⊂ Pnk (k as usual a field) is cut out by homogeneous
equations f1 , . . . , fr , and p ∈ X is a k-valued point that is regular of dimension
d. Define the (projective) tangent d-plane to X at p. (Definition 8.2.3 gives thedefinition of a d-plane in Pn
k , but you shouldn’t need to refer there.)
12.3.3. Side Remark to help you think cleanly. We would want the definition oftangent k-plane to be natural in the sense that for any automorphism σ of Ank (or,
in the case of the previous Exercise, Pn
),
σ(T
X)
=
T
σ(X).
You
could
verify
this
p
σ(p)
k
by hand, but you can also say this in a cleaner way, by interpreting the equations
cutting out the tangent space in a coordinate free manner. Informally speaking, weare using the canonical identification of n-space with the tangent space to n-spaceat p, and using the fact that the Jacobian “linear transformation” cuts out Tp X in
Tp An in a way independent of choice of coordinates on An or defining equationsof X. Your solution to Exercise 12.3.H will help you start to think in this way.
12.3.K. E XERCISE . Suppose X ⊂ Pn
k is a degree d hypersurface cut out by f = 0,
and L is a line not contained in X. Exercise 8.2.E (a case of Bézout’s Theorem)
showed that X and L meet at d points, counted “with multiplicity”. Suppose L
meets X “with multiplicity at least 2” at a k-valued point p ∈ L ∩ X, and that p is

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a regular point of X. Show that L is contained in the tangent plane to X at p. (Doyou have a picture of this in your mind?)12.3.4. Arithmetic examples.
12.3.L. E ASY E XERCISE . Show that Spec Z is a regular curve.
12.3.M. E XERCISE . (This tricky exercise is for those who know about the primes of
the Gaussian integers Z[i].) There are several ways of showing that Z[i] is dimension 1. (For example: (i) it is a principal ideal domain; (ii) it is the normalization of
Z in the field extension Q(i)/Q; (iii) using Krull’s Principal Ideal Theorem 11.3.3
and the fact that dim Z[x] = 2 by Exercise 11.1.H.) Show that Spec Z[i] is a regularcurve. (There are several ways to proceed. You could use Exercise 12.1.B. As an
example to work through first, consider the prime (2, 1 + i), which is cut out by theequations 2 and 1 + x in Spec Z[x]/(x2 + 1).) We will later (§12.5.10) have a simpler
approach once we discuss discrete valuation rings.
12.3.N. E XERCISE . Show that [(5, 5i)] is the unique singular point of Spec Z[5i].
∼ Z[5i]5 . Use the previous exercise.)(Hint: Z[i]5 =12.3.5. Back to geometry: An
k is regular.
The key step to showing that An
k is regular (where k is a field) is the following.
12.3.6. Proposition. — Suppose (B, n, k) is a regular local ring of dimension d. Let
φ : B → B[x]. Suppose p is a prime ideal of A := B[x] such that nB[x] ⊂ p. Then Ap is aregular local ring.

Proof. Geometrically: we have a morphism π : X = Spec B[x] → Y = Spec B, andπ([p]) = [n]. The fiber π−1 ([n]) = Spec(B[x]/nB[x]) = Spec k[x]. Thus either (i) [p] isthe fiber A1k above [n], or (ii) [p] is a closed point of the fiber A1k .
Before considering these two cases, we make two remarks. As (B, n) is a regular local ring of dimension d, n is generated by d elements of B, say f1 , . . . , fd (asdiscussed in the proof of Theorem 12.2.1), and there is a chain of prime ideals(12.3.6.1)

0 = q0 ⊂ q1 ⊂ · · · ⊂ qd = n.

Case (i): [p] is the fiber A1k . In this case, p = nB[x]. Thus by Krull’s Height
Theorem 11.3.7, the height of p = nB[x] is at most d, because p is generated by thed elements f1 , . . . , fd . But p has codimension at least d: given the chain (12.3.6.1)
ending with n, we have a corresponding chain of prime ideals of B[x] ending withnB[x]. Hence the codimension of p is precisely d, and pAp is generated by f1 , . . . ,fd , implying that it is a regular local ring.
Case (ii): [p] is a closed point of the fiber A1k . The closed point p of k[x] corresponds
to some monic irreducible polynomial g(x) ∈ k[x]. Arbitrarily lift the coefficientsof g to B; we sloppily denote the resulting polynomial in B[x] by g(x) as well.
Then p = (f1 , . . . , fd , g), so by Krull’s Height Theorem 11.3.7, the codimension of
p is at most d + 1. But p has codimension at least d + 1: given the chain (12.3.6.1)
ending with n, we have a corresponding chain in B[x] ending with nB[x], and we
can extend it by appending p. Hence the height of p is precisely d + 1, and pAp isgenerated by f1 , . . . , fd , g, implying that it is a regular local ring.
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12.3.O. E XERCISE . Use Proposition 12.3.6 to show that if X is a regular (locally
Noetherian) scheme, then so is X × A1 . In particular, show that An
k is regular.

12.4 Bertini’s Theorem
We now discuss Bertini’s Theorem, a fundamental classical result.
12.4.1. Definition: dual projective space. The dual (or dual projective space) tonPnk (with coordinates x0 , . . . , xn ), is informally the space of hyperplanes in Pk .
nSomewhat more precisely, it is a projective space Pk with coordinates a0 , . . . , an(which we denote Pn∨
with the futile intent of preventing confusion), along withk
the data of the “incidence variety” or “incidence correspondence” I ⊂ Pn × Pn∨
cut out by the equation a0 x0 +· · ·+an xn = 0. Note that the k-valued points of Pn∨
indeed correspond to hyperplanes in Pn defined over k, and this is also clearly aduality relation (there is a symmetry in the definition between the x-variables andthe a-variables). So this is concrete enough to use in practice, and extends over anarbitrary base (notably Spec Z). (But if you have a delicate and refined sensibility,
you may want to come up with a coordinate-free definition.)
12.4.2. Bertini’s Theorem. — Suppose X is a smooth subvariety of Pn
k . Then there isa nonempty (=dense) open subset U of dual projective space Pn∨
such
that
for any closedkpoint [H] ∈ U, H doesn’t contain any component of X, and the scheme H ∩ X is k-smooth.
The hypothesis that k is algebraically closed is not essential (even in the proof
below): see Exercise 12.4.A.
We remark that any theorem of this flavor is often called a “Bertini Theorem”.
One example is the Kleiman-Bertini Theorem 25.3.7, which was not proved jointly
by Kleiman and Bertini.
As an application of Bertini’s Theorem 12.4.2, a general degree d > 0 hypersurface in Pn
k intersects X in a regular subvariety of codimension 1 in X: replace
X "→ Pn with the composition% Pn ! " νd % PN
Xwhere νd is the dth Veronese embedding (8.2.8). Here “general” has its usual
meaning in algebraic geometry, see §9.3.6, except that we are considering only
closed points of U ⊂ Pn∨
A useful consequence of this, taking X = Pn : wek .
immediately see that there exists a smooth degree d hypersurface in Pn (over analgebraically closed field k), without any messing around with specific equations(as in the special case Exercise 12.3.E).
Exercise 25.3.D gives a useful improvement of Bertini’s Theorem in characteristic 0 (see Exercise 25.3.E).
Proof. In order to keep the proof as clean as possible, we assume X is irreducible,
but essentially the same proof applies in general.
The central idea of the proof is quite naive. We will describe the hyperplanes
that are “bad”, and show that they form a closed subset of dimension at most n − 1∨of Pn
k , and hence that the complement is a dense open subset. Somewhat more

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precisely, we will define a projective variety Z ⊂ X × Pn
that can informally bekdescribed as:(12.4.2.1) Z ={(p ∈ X, H ⊂ Pnk ) : p ∈ H,
and either p is a singular point of H ∩ X, or X ⊂ H}

We will see that the projection π : Z → X has fibers at closed points that are projective spaces of dimension n − 1 − dim X, and use this to show that dim Z ≤ n − 1.
∨
Thus the image of Z in Pn
will be a closed subset (Theorem 7.4.7), of dimension
kof at most n − 1, so its complement will be open and non-empty. We now put thisstrategy into action.
12.4.3. We first define Z more precisely, in terms of equations on Pn × Pn ∨ , where
the coordinates on Pn are x0 , . . . xn , and the dual coordinates on Pn ∨ are a0 , . . . ,an . Suppose X is cut out by f1 , . . . , fr . Then we take these equations as the firstof the defining equations of Z. (So far we have defined the subscheme X × Pn ∨ .)We also add the equation a0 x0 + · · · + an xn = 0. (So far we have described thesubscheme of Pn × Pn ∨ corresponding to points (p, H) where p ∈ X and p ∈ H.)
Note that the Jacobian matrix ((12.1.6.1), except with variables starting with x0
rather than x1 )
⎛ ∂f1
⎞
∂fr
∂x0 (p) · · ·∂x0 (p)
⎜
⎟
..
..
..
⎝
⎠
.
.
.
∂f1
∂fr
∂xn (p) · · ·∂xn (p)has corank equal to dim X at all closed points of X — this is precisely the Jacobian
criterion for regularity (Exercise 12.2.D). We then require that the Jacobian matrix
with a new column⎛⎞
a0⎜ .. ⎟⎝ . ⎠anappended has corank ≥ dim X (hence = dim X). This is cut out by equations (thedeterminants of certain minors). By the Jacobian description of the Zariski tangent
space, this condition encodes the requirement that the Zariski tangent space ofH ∩ X at p has dimension precisely dim X, which is dim H ∩ X + 1 (i.e., H ∩ X is
singular at p) if H does not contain X, or if H contains X. This is precisely the notionthat we wished to capture. (Remark 12.4.4 works through an example, which may
help clarify how this works.)
We next show that dim Z ≤ n − 1. For each closed point p ∈ X, let Wp be thelocus of hyperplanes containing p, such that H ∩ X is singular at p, or else contains
all of X; what is the dimension of Wp ? Suppose dim X = d. Then the restrictions
on the hyperplanes in definition of Wp correspond to d + 1 linear conditions. (Doyou see why?) This means that Wp is a codimension d + 1, or dimension n − d − 1,projective space. Thus the fiber of π : Z → X over each closed point has pure
dimension n − d − 1. By Key Exercise 11.4.A, this implies that dim Z ≤ n − 1. (Ifyou wish, you can use Exercise 11.4.C to show that dim Z = n − 1, and you canlater show that Z is a projective bundle over X, once you know what a projective
bundle is (Definition 17.2.3). But we don’t need this for the proof.)!

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12.4.4. Remark. Here is an example that may help convince you that the algebra ofparagraph 12.4.3 is describing the geometry we desire. Consider the plane conic
x20 − x21 − x22 = 0 over a field of characteristic not 2, which you might picture as thecircle x2 + y2 = 1 from the real picture in the chart U0 . Consider the point [1, 1, 0],
corresponding to (1, 0) on the circle. We expect the tangent line in the affine plane
to be x = 1, which corresponds to x0 − x1 = 0. Let’s see what the algebra gives us.
The Jacobian matrix (12.1.6.1) is
⎛
⎞ ⎛
⎞
2x0
2
⎝ −2x1 ⎠ = ⎝ −2 ⎠ ,−2x20
which indeed has rank 1 as expected. Our recipe asks that the matrix⎛⎞
2 a0⎝ −2 a1 ⎠0 a2

have rank 1 (i.e., a0 = −a1 and a2 = 0), and also that a0 x0 + a1 x1 + a2 x2 = 0,
which (you should check) is precisely what we wanted.
12.4.A. E XERCISE .
k = k hypothesis.

Reword the proof of Bertini’s Theorem so as to remove the12.4.B. E ASY E XERCISE .
(a) Prove Bertini’s Theorem with the weaker hypothesis that X has finitely many
singular points.
(b) Prove Bertini’s Theorem with the weaker hypothesis that X → Pn
k is a locallyclosed embedding.

12.4.C. E XERCISE . Continue to assume k = k. Show that if X is a projective variety of dimension n in Pm , then the intersection of X with n general hyperplanes
consists of a finite number of reduced points. More precisely: if Pm∨ is the dual
projective space, then there is a Zariski-open subset U ⊂ (Pm∨ )n such that for eachclosed point (H1 , . . . , Hn ) of U, the scheme-theoretic intersection H1 ∩ · · · ∩ Hn ∩ X
consists of a finite number of reduced points. (The number of such points, counted
correctly, is called the degree of the variety, see Exercise 18.6.N.)
12.4.5. Dual varieties. (We continue to assume k = k for convenience, although
this can be relaxed.) This gives us an excuse to mention a classical construction.
The image of Z (see (12.4.2.1)) in Pn∨ is called the dual variety of X. As dim Z =n − 1, we “expect” the dual of X to be a hypersurface of Pn∨ . It is a nonobvious
fact that this in fact is a duality: the dual of the dual of X is X itself (see [H2,
Thm. 15.24]). The following exercise will give you some sense of the dual variety.
12.4.D. E XERCISE . Show that the dual of a hyperplane in Pn is the corresponding
point of the dual space Pn∨ . In this way, the duality between Pn and Pn∨ is a
special case of duality between projective varieties.
12.4.E. E XERCISE . Suppose C ⊂ P2 is a smooth conic over an algebraically closed
field of characteristic not 2. Show that the dual variety to C is also a smooth conic.
Thus for example, through a general point in the plane (if k = k), there are two

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tangents to C. (The points on a line in the dual plane corresponds to those lines
through a point of the original plane.)
12.4.F. ⋆ E XERCISE ( THERE IS ONE SMOOTH CONIC TANGENT TO FIVE GENERALLINES , AND GENERALIZATIONS ). Continuing the notation of the previous problem, show that the number of smooth conics C containing i generally chosenpoints and tangent to 5 − i generally chosen lines is 1, 2, 4, 4, 2, 1 respectively for
i = 0, 1, 2, 3, 4, 5. You might interpret the symmetry of the sequence in terms of the
duality between the conic and the dual conic. This fact was likely known in the
paleolithic era.

12.5 Another (co)dimension one miracle: Discrete valuation rings
The case of (co)dimension 1 is important, because if you understand how
prime ideals behave that are separated by dimension 1, then you can use induction to prove facts in arbitrary dimension. This is one reason why Krull’s Principal
Ideal Theorem 11.3.3 is so useful.
A dimension 1 Noetherian regular local ring can be thought of as a “germ ofa smooth curve” (see Figure 12.4). Two examples to keep in mind are k[x](x) ={f(x)/g(x) : x ̸ | g(x)} and Z(5) = {a/b : 5 ̸ | b}. The first example is “geometric”
and the second is “arithmetic”, but hopefully it is clear that they have something
fundamental in common.

F IGURE 12.4. A germ of a curve
The purpose of this section is to give a long series of equivalent definitions ofthese rings. Before beginning, we quickly sketch these seven definitions. There
are a number of ways a Noetherian local ring can be “nice”. It can be regular, or aprincipal ideal domain, or a unique factorization domain, or normal. In dimension
1, these are the same. Also equivalent are nice properties of ideals: if m is principal;
or if all ideals are either powers of the maximal ideal, or 0. Finally, the ring canhave a discrete valuation, a measure of “size” of elements that behaves particularly
well.
12.5.1. Theorem. — Suppose (A, m) is a Noetherian local ring of dimension 1. Then thefollowing are equivalent.
(a) (A, m) is regular.
(b) m is principal.
Proof. Here is why (a) implies (b). If A is regular, then m/m2 is one-dimensional.
Choose any element t ∈ m − m2 . Then t generates m/m2 , so generates m by

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Nakayama’s Lemma 7.2.H (m is finitely generated by the Noetherian hypothesis).
We call such an element a uniformizer.
Conversely, if m is generated by one element t over A, then m/m2 is generated
by one element t over A/m = k. Since dimk m/m2 ≥ 1 by Theorem 12.2.1, we havedimk m/m2 = 1, so (A, m) is regular.
!
We will soon use a useful fact, which is geometrically motivated, and is a
special case of an important result, the Artin-Rees Lemma 12.9.3. We will prove itin §12.9.
12.5.2. Proposition. — If (A, m) is a Noetherian local ring, then ∩i mi = 0.
12.5.3. The geometric intuition for this is that any function that is analytically zeroat a point (vanishes to all orders) actually vanishes in an open neighborhood ofthat point. (Exercise 12.9.B will make this precise.) The geometric intuition alsosuggests an example showing that Noetherianness is necessary: consider the func2
tion e−1/x in the germs of C∞ -functions on R at the origin. (In particular, this
implies that the ring of C∞ -functions on R, localized at the origin, is not Noetherian!)
It is tempting to argue that
(12.5.3.1)m(∩i mi ) = ∩i mi ,and then to use Nakayama’s Lemma 7.2.9 to argue that ∩i mi = 0. Unfortunately,
it is not obvious that this first equality is true: product does not commute with
infinite descending intersections in general. (Aside: product also doesn’t commute
with finite intersections in general, as for example k[x, y, z]/(xz − yz), (z)((x) ∩(y)) ̸= (xz) ∩ (yz).) We will establish Proposition 12.5.2 in Exercise 12.9.A(b). (Wecould do it directly right now without too much effort.)
12.5.4. Theorem. — Suppose (A, m) is a Noetherian local ring of dimension 1. Then (a)and (b) are equivalent to:(c) all ideals are of the form mn (for n ≥ 0) or (0).
Proof. Assume (a): suppose (A, m, k) is a Noetherian regular local ring of dimension 1. Then I claim that mn ̸= mn+1 for any n. Otherwise, by Nakayama’s Lemma,
mn = 0, from which tn = 0. But A is an integral domain (by Theorem 12.2.13), sot = 0, from which A = A/m is a field, which doesn’t have dimension 1, contradiction.
I next claim that mn /mn+1 is dimension 1. Reason: mn = (tn ). So mn isgenerated as an a A-module by one element, and mn /(mmn ) is generated as a
(A/m = k)-module by 1 element (nonzero by the previous paragraph), so it is a
one-dimensional vector space.
So we have a chain of ideals A ⊃ m ⊃ m2 ⊃ m3 ⊃ · · · with ∩mi = (0)(Proposition 12.5.2). We want to say that there is no room for any ideal besides
these, because each pair is “separated by dimension 1”, and there is “no room atthe end”. Proof: suppose I ⊂ A is an ideal. If I ̸= (0), then there is some n such thatI ⊂ mn but I ̸⊂ mn+1 . Choose some u ∈ I − mn+1 . Then (u) ⊂ I. But u generates
mn /mn+1 , hence by Nakayama it generates mn , so we have mn ⊂ I ⊂ mn , so weare done: (c) holds.

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We now show that (c) implies (a). Assume (a) does not hold: suppose wehave a dimension 1 Noetherian local integral domain that is not regular, so m/m2
has dimension at least 2. Choose any u ∈ m − m2 . Then (u, m2 ) is an ideal, but
m2 " (u, m2 ) " m.
!
12.5.A. E ASY E XERCISE . Suppose (A, m) is a Noetherian dimension 1 local ring.
Show that (a)–(c) above are equivalent to:(d) A is a principal ideal domain.
12.5.5. Discrete valuation rings. We next define the notion of a discrete valuation
ring. Suppose K is a field. A discrete valuation on K is a surjective homomorphismv : K× → Z (in particular, v(xy) = v(x) + v(y)) satisfyingv(x + y) ≥ min(v(x), v(y))

except if x + y = 0 (in which case the left side is undefined). (Such a valuation iscalled non-archimedean, although we will not use that term.) It is often convenient
to say v(0) = ∞. More generally, a valuation is a surjective homomorphism v :K× → G to a totally ordered abelian group G, although this isn’t so important tous.
Here are three key examples.
(i) (the 5-adic valuation) K = Q, v(r) is the “power of 5 appearing in r”, e.g.,v(35/2) = 1, v(27/125) = −3.
(ii) K = k(x), v(f) is the “power of x appearing in f.”(iii) K = k(x), v(f) is the negative of the degree. This is really the same as (ii),with x replaced by 1/x.
Then 0 ∪ {x ∈ K× : v(x) ≥ 0} is a ring, which we denote Ov . It is called thevaluation ring of v. (Not every valuation is discrete. Consider the ring of Puiseux
series over a field k, K = ∪n≥1 k((x1/n )), with v : K× → Q given by v(xq ) = q.)12.5.B. E XERCISE .
Describe the valuation rings in the three examples (i)–(iii)
above. (You will notice that they are familiar-looking dimension 1 Noetherian
local rings. What a coincidence!)
12.5.C. E XERCISE . Show that {0}∪{x ∈ K× : v(x) > 0} is the unique maximal idealof the valuation ring. (Hint: show that everything in the complement is invertible.)Thus the valuation ring is a local ring.
An integral domain A is called a discrete valuation ring (or DVR) if there
exists a discrete valuation v on its fraction field K = K(A) for which Ov = A.
Similarly, A is a valuation ring if there exists a valuation v on K for which Ov = A.
Now if A is a Noetherian regular local ring of dimension 1, and t is a uniformizer (a generator of m as an ideal, or equivalently of m/m2 as a k-vector space)then any nonzero element r of A lies in some mn − mn+1 , so r = tn u where u is invertible (as tn generates mn by Nakayama, and so does r), so K(A) = At = A[1/t].
So any element of K(A)× can be written uniquely as utn where u is invertible andn ∈ Z. Thus we can define a valuation v by v(utn ) = n.
12.5.D. E XERCISE . Show that v is a discrete valuation.

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12.5.E. E XERCISE . Conversely, suppose (A, m) is a discrete valuation ring. Show
that (A, m) is a Noetherian regular local ring of dimension 1. (Hint: Show that theideals are all of the form (0) or In = {r ∈ A : v(r) ≥ n}, and (0) and I1 are the only
prime ideals. Thus we have Noetherianness, and dimension 1. Show that I1 /I2 isgenerated by the image of any element of I1 − I2 .)Hence we have proved:
12.5.6. Theorem. — An integral domain A is a Noetherian local ring of dimension 1
satisfying (a)–(d) if and only if(e) A is a discrete valuation ring.
12.5.F. E XERCISE .
valuation ring.

Show that there is only one discrete valuation on a discrete

12.5.7. Definition. Thus any Noetherian regular local ring of dimension 1 comeswith a unique valuation on its fraction field. If the valuation of an element is n > 0,
we say that the element has a zero of order n. If the valuation is −n < 0, we saythat the element has a pole of order n. We will come back to this shortly, after
dealing with (f) and (g).
12.5.8. Theorem. — Suppose (A, m) is a Noetherian local ring of dimension 1. Then(a)–(e) are equivalent to:(f) A is a unique factorization domain, and(g) A is an integral domain, integrally closed in its fraction field K = K(A).
Proof. (a)–(e) clearly imply (f), because we have the following stupid unique factorization: each nonzero element of A can be written uniquely as utn where n ∈ Z≥0
and u is invertible.
Now (f) implies (g), because unique factorization domains are integrally closed
in their fraction fields (Exercise 5.4.F).
It remains to check that (g) implies (a)–(e). We will show that (g) implies (b).
Suppose (A, m) is a Noetherian local integral domain of dimension 1, integrally closed in its fraction field K = K(A). Choose any nonzero r ∈ m. Then
S = A/(r) is a Noetherian local ring of dimension 0 — its only prime is the imageof m, which we denote n to avoid confusion. Then n is finitely generated, and eachgenerator is nilpotent (the intersection of all the prime ideals in any ring are thenilpotents, Theorem 3.2.12). Then nN = 0, where N is sufficiently large. Hence
there is some n such that nn = 0 but nn−1 ̸= 0.
Now comes the crux of the argument. Thus in A, mn ⊂ (r) but mn−1 ̸⊂ (r).
Choose s ∈ mn−1 − (r). Consider s/r ∈ K(A). As s ∈/ (r), s/r ∈/ A, so as A is
integrally closed, s/r is not integral over A.
Now (inside K(A)) sr m ̸⊂ m (or else sr m ⊂ m would imply that m is a faithfuls
A[ r ]-module, contradicting Exercise 7.2.I). But sm ⊂ mn ⊂ rA, so sr m ⊂ A. Thussr!
r m = A, from which m = s A, so m is principal.
12.5.9. Geometry of normal Noetherian schemes. We can finally make precise(and generalize) the fact that the function (x − 2)2 x/(x − 3)4 on A1C has a double
zero at x = 2 and a quadruple pole at x = 3. Furthermore, we can say that 75/34

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has a double zero at 5, and a single pole at 2. (What are the zeros and poles ofx3 (x + y)/(x2 + xy)3 on A2 ?) Suppose X is a locally Noetherian scheme. Then for
any regular codimension 1 point p (i.e., any point p where OX,p is a regular local
ring of dimension 1), we have a discrete valuation v. If f is any nonzero elementof the fraction field of OX,p (e.g., if X is integral, and f is a nonzero element of thefunction field of X), then if v(f) > 0, we say that the element has a zero of orderv(f) at p, and if v(f) < 0, we say that the element has a pole of order −v(f) at p.
(We are not yet allowed to discuss order of vanishing at a point that is not regularand codimension 1. One can make a definition, but it doesn’t behave as well as it
does when have you have a discrete valuation.)12.5.G. E XERCISE ( FINITENESS OF ZEROS AND POLES ON N OETHERIAN SCHEMES ).
Suppose X is an integral Noetherian scheme, and f ∈ K(X)× is a nonzero elementof its function field. Show that f has a finite number of zeros and poles. (Hint:
reduce to X = Spec A. If f = f1 /f2 , where fi ∈ A, prove the result for fi .)Suppose A is a Noetherian integrally closed domain. Then it is regular incodimension 1 (translation: its points of codimension at most 1 are regular). If A
is dimension 1, then obviously A is regular.

12.5.H. E XERCISE . If f is a nonzero rational function on a locally Noetherian normal scheme, and f has no poles, show that f is regular. (Hint: Algebraic Hartogs’s
Lemma 11.3.11.)
12.5.10. For example (cf. Exercise 12.3.M), Spec Z[i] is regular, because it is dimension 1, and Z[i] is a unique factorization domain. Hence Z[i] is normal, so all its
closed (codimension 1) points are regular. Its generic point is also regular, as Z[i]is an integral domain.
12.5.11. Remark. A (Noetherian) scheme can be singular in codimension 2 andstill be normal. For example, you have shown that the cone x2 + y2 = z2 in A3 incharacteristic not 2 is normal (Exercise 5.4.I(b)), but it is singular at the origin (theZariski tangent space is visibly three-dimensional).
But singularities of normal schemes are not so bad in some ways: we have
Algebraic Hartogs’s Lemma 11.3.11 for Noetherian normal schemes, which statesthat you can extend functions over codimension 2 sets.
12.5.12. Remark. We know that for Noetherian rings we have implications
unique factorization domain =⇒ integrally closed =⇒ regular in codimension 1.
Hence for locally Noetherian schemes, we have similar implications:factorial =⇒ normal =⇒ regular in codimension 1.
Here are two examples to show you that these inclusions are strict.
12.5.I. E XERCISE ( THE pinched plane). Let A be the subring k[x3 , x2 , xy, y] ofk[x, y]. (Informally, we allow all polynomials that don’t include a nonzero multipleof the monomial x.) Show that Spec k[x, y] → Spec A is a normalization. Show that
A is not integrally closed. Show that Spec A is regular in codimension 1. (Hint for
the last part: show it is dimension 2, and when you throw out the origin you get

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something regular, by inverting x2 and y respectively, and considering Ax2 andAy .)12.5.13. Example (the quadric once again). Suppose k is algebraically closed ofcharacteristic not 2. Then k[w, x, y, z]/(wz − xy) is integrally closed, but not aunique factorization domain, see Exercise 5.4.L (and Exercise 12.1.E).
12.5.14. Important remark: Finitely generated modules over a discrete valuation ring.
We record a useful fact for future reference. Recall that finitely generated modules
over a principal ideal domain are finite direct sums of cyclic modules (see for example [DF, §12.1, Thm. 5]). Hence any finitely generated module over a discrete
valuation ring A with uniformizer t is a finite direct sum of terms A and A/(tr )(for various r). See Proposition 13.7.3 for an immediate consequence.
12.5.15. Aside: Dedekind domains. A Dedekind domain is a Noetherian integral
domain of dimension at most one that is normal (integrally closed in its fraction
field). The localization of a Dedekind domain at any prime but (0) (i.e., a codimension one prime) is hence a discrete valuation ring. This is an important notion, but
we won’t use it much. Rings of integers of number fields are examples, see√ §9.7.1.
In particular, if n is a square-free integer congruent to 3 (mod 4), then Z[ n] is a
Dedekind domain, by Exercise 5.4.I(a). If you wish you can prove unique factorization of ideals in a Dedekind domain: any nonzero ideal in a Dedekind domain can be
uniquely factored into prime ideals.

12.6 Smooth (and étale) morphisms (first definition)
In §12.2.5, we defined smoothness over a field. It is an imperfect definition(see §12.2.7), and we will improve it when we know more (in §21.3.1).
We know enough to define smooth morphisms (and, as a special case, étale morphisms) in general. Our definition will be imperfect in a number of ways. For example, it will look a little surprising. As another sign, it will not be obvious that it
actually generalizes smoothness over a field. As a third sign, like Definition 12.2.6,
because it is of the form “there exist open covers satisfying some property”, it ispoorly designed to be used to show that some morphism is not smooth. But it
has a major advantage that we can give the definition right now, and the basic
properties of smooth morphisms will be straightforward to show.
12.6.1. Differential geometric motivation. The notion of a smooth morphism is motivated by the following idea from differential geometry. For the purposes of this
discussion, we say a map π : X → Y of real manifolds (or real analytic spaces, or
any variant thereof) is smooth of relative dimension n if locally (on X) it looks likeY ′ × Rn → Y ′ . Translation: it is “locally on the source a smooth fibration”. (If you
have not heard the word fibration before, don’t worry. Informally speaking, a fibration in some “category of geometric objects” is a map π : X → Y that locally on thesource is a product V × W → V; in some sort of category of real manifolds, a fibration is a “locally on the source a smooth fibration” if the W in the local description
can be taken to be Ra for some a.)

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In particular, “smooth of relative dimension 0” is the same as the importantnotion of “local isomorphism” (“isomorphism locally on the source” — not quite
the same as “covering space”). Carrying this idea too naively into algebraic geometry leads to the “wrong” definition (see Exercise 12.6.F). An explanation of how
this differential geometric intuition leads to a good definition of smoothness will
be given in §25.1, when we will be ready to give a good definition of smoothness.
For now we will settle for a satisfactory (and correct) definition.
12.6.2. Definition. A morphism π : X → Y is smooth of relative dimension n if
there exist open covers {Ui } of X and {Vi } of Y, with π(Ui ) ⊂ Vi , such that for every
i we have a commutative diagram
Ui

∼

π|Ui

"
Vi

%W
ρ|W

∼

"
% Spec B

where there is a morphism ρ : Spec B[x1 , . . . , xn+r ]/(f1 , . . . , fr ) → Spec B (inducedby the obvious map of rings in the opposite direction), and W is an open subscheme of Spec B[x1 , . . . , xn+r ]/(f1 , . . . , fr ), such that the determinant of the Jacobian matrix of the fi ’s with respect to the first r xi ’s,"#∂fj(12.6.2.1)
det,∂xi i,j≤r
is an invertible (= nowhere zero) function on W.
Étale means smooth of relative dimension 0.

12.6.3. Quick observations. From the definition, smooth morphisms are locally offinite presentation (hence locally of finite type). Also, An
B → Spec B is immediately
seen to be smooth of relative dimension n. From the definition, the locus on X
where π is smooth of relative dimension n is open. In particular, the locus where
π is étale is open.
As easy examples, open embeddings are étale, and the projection An × Y → Y
is smooth of relative dimension n. You may already have some sense of whatmakes this definition imperfect. For example, how do you know a morphism issmooth if it isn’t given to you in a form where the “right” variables xi and “right”
equations are clear?
12.6.A. M OTIVATING E XERCISE ( FOR THOSE WITH DIFFERENTIAL - GEOMETRIC BACK GROUND ). Show how Definition 12.6.2 gives the definition in differential geometry, as described in §12.6.1. (Your argument will use the implicit function theorem,
which “doesn’t work” in algebraic geometry, see Exercise 12.6.F.)
12.6.B. E XERCISE . Show that the notion of smoothness of relative dimension n is
local on both the source and target.
We can thus make sense of the phrase “π : X → Y is smooth of relative dimension n at p ∈ X”: it means that there is an open neighborhood U of p such that π|Uis smooth of relative dimension n.
The phrase smooth morphism (without reference to relative dimension n)often informally means “smooth morphism of some relative dimension n”, but

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sometimes can mean “smooth of some relative dimension in the neighborhood ofevery point”.
12.6.C. E ASY E XERCISE . Show that the notion of smoothness of relative dimension
n is preserved by base change.
12.6.D. E XERCISE . Suppose π : X → Y is smooth of relative dimension m andρ : Y → Z is smooth of relative dimension n. Show that ρ ◦ π is smooth of relative dimension m + n. (In particular, the composition of étale morphisms is étale.
Furthermore, this implies that smooth morphisms are closed under product, by
Exercise 9.4.F.)
Exercises 12.6.B–12.6.D imply that smooth morphisms (and étale morphisms)
form a “reasonable” class in the sense of §7.1.1.
12.6.E. E XERCISE . Suppose π : X → Y is smooth of relative dimension n. Show
that locally on X, π can be described as an étale cover of An
Y . More precisely, showthat for every p ∈ X, there is an open neighborhood U of p, such that π|U can befactored into
U

α

% An

β

Y

%Y

where α is étale and β is the obvious projection.
12.6.F. E XERCISE . Suppose k is a field of characteristic not 2. Let Y = Spec k[t],and X = Spec k[u, 1/u]. Show that the morphism π : X → Y induced by t (→ u2 isétale. (Sketch π!) Show that there is no nonempty open subset U of X on which π is
an isomorphism. (This shows that étale cannot be defined as “local isomorphism”,
i.e., “isomorphism locally on the source”. In particular, the naive extension of thedifferential intuition of §12.6.1 does not give the right definition of étaleness inalgebraic geometry.)
We now show that our new Definition 12.6.2 specializes to Definition 12.2.6
when the target is a field. This is the hardest fact in this section. (It is also a
consequence of Theorem 25.2.2, so you could skip the proof of Theorem 12.6.4 andtake it on faith until Chapter 25. The crux of the proof of Theorem 25.2.2 is similar
to the crux of the proof of Theorem 12.6.4.)
12.6.4. Theorem. — Suppose X is a k-scheme. Then the following are equivalent.
(i) X is smooth of relative dimension n over Spec k (Definition 12.6.2).
(ii) X has pure dimension n, and is smooth over k (in the sense of Definition 12.2.6).
This allows us to use the phrases “smooth over k” and “smooth over Spec k”
interchangeably, as we would expect. More generally, “smooth over a ring A”
means “smooth over Spec A”.
⋆ Proof. (i) implies (ii). Suppose X is smooth of relative dimension n over Spec k(Definition 12.6.2). We will show that X has pure dimension n, and is smooth overk (Definition 12.2.6). Our goal is a Zariski-local statement, so we may assume that
X is an open subset
W ⊂ Spec k[x1 , . . . , xn+r ]/(f1 , . . . , fr )as described in Definition 12.6.2.

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The central issue is showing that X has “pure dimension n”, as the latter halfof the statement would then follow because the corank of the Jacobian matrix isthen n.
Let Xk := X×k k. By Exercise 11.1.G, Xk is pure dimension n if and only if X has
pure dimension n. Let p be a closed point of the (finite type scheme) Xk . Then X
has dimension at least n at p, by Krull’s Principal Ideal Theorem 11.3.3, combinedwith Theorem 11.2.9 (codimension is the difference of dimensions for varieties).
But the Zariski tangent space at a closed point of a finite type scheme over k is
cut out by the Jacobian matrix (Exercise 12.1.G), and thus has dimension exactly n(by the corank hypothesis in the definition of k-smooth). As the dimension of the
Zariski tangent space bounds the dimension of the scheme (Theorem 12.2.1), Xk
has pure dimension n as desired.
(ii) implies (i). We will prove the result near any given closed point p of X.
(Here we use that any union of open neighborhoods of all the closed points is allof X, by Exercise 5.1.E.) We thus may reduce to the following problem. Suppose weare given X := Spec k[x1 , . . . , xn+r ]/(f1 , . . . , fN ), and a point p ∈ X, and an open
neighborhood U of p in X such that U is pure dimension n, and the Jacobian matrixof the fj ’s with respect to the xi ’s has corank n everywhere on U. We wish to showthat the structure morphism π : X → Spec k is smooth of relative dimension n at
p (in the sense of Definition 12.6.2). At p, there exists an r × r submatrix of theJacobian matrix that is invertible (by the corank condition). Rearrange the xi ’s andfj ’s so that the Jacobian of the first r of the fj ’s with respect to the first r of the xi ’s
has nonzero determinant. Let I = (f1 , . . . , fN ). We will show that I = (f1 , . . . , fr )“near p”, or more precisely, that there is some g ∈ k[x1 , . . . , xn+r ] with g(p) ̸= 0such that Ig = (f1 , . . . , fr )g (so I = (f1 , . . . , fr ) “in D(g)”).
Choose a point p ∈ Xk above p, i.e., such that if ψ is the morphism Xk → X
induced by Spec k → Spec k, then ψ(p) = p. (Why does such a p exist?) Now Xk is
k-smooth (in the sense of Definition 12.2.6, by Exercise 12.2.G) at p, hence regular(Exercise 12.2.I). Its Zariski tangent space at p is cut out in the tangent space ofAn+rby the Jacobian conditions (Exercise 12.1.G), and hence cut out by f1 , . . . , fr .
k12.6.G. E XERCISE . Suppose m is the ideal of k[x1 , . . . , xn+r ] corresponding to p.
Show that f1 , . . . , fr generate the ideal (f1 , . . . , fN ) in k[x1 , . . . , xn+r ]m . Hint: lookat your argument for Exercise 12.2.L(a).
For notational compactness, define A := k[x1 , . . . , xn+r ]/(f1 , . . . , fr ), and A :=k[x1 , . . . , xn+r ]/(f1 , . . . , fr ) = A ⊗k k. Let I (resp. I) be the ideal of A (resp. A)
generated by fr+1 , . . . , fN (so I = I ⊗k k). We wish to show that I = 0 “near p”.
Here p = [m], where m is a maximal ideal (recall p is a closed point).
Exercise 12.6.G means that Im = 0, so taking the quotient by the maximal ideal
m of A yields
I ⊗A (A/m) = 0
which implies(I ⊗A (A/m)) ⊗A/m (A/m) = 0.
But k = A/m is a field extension of ℓ := A/m, and an ℓ-vector space V is 0 if (andonly if) V ⊗ℓ k = 0, so I ⊗A (A/m) = 0, i.e., Im /mIm = 0.

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12.6.H. E XERCISE . Conclude that there is some g ∈ A \ m such that gI = 0. Hint:
Nakayama, and the finite generation of I. (Be sure you understand why the proof
is now complete.)!

12.7 ⋆ Valuative criteria for separatedness and properness(The only reason this section is placed here is that we need the theory of discrete valuation rings.)
In reasonable circumstances, it is possible to verify separatedness by checking
only maps from spectra of valuations rings. There are four reasons you might likethis (even if you never use it). First, it gives useful intuition for what separated
morphisms look like. Second, given that we understand schemes by maps to them(the Yoneda philosophy), we might expect to understand morphisms by mapping
certain maps of schemes to them, and this is how you can interpret the diagram
appearing in the valuative criterion. And the third concrete reason is that one of
the two directions in the statement is much easier (a special case of the Reducedto-Separated Theorem 10.2.2, see Exercise 12.7.A), and this is the direction we will
repeatedly use. Finally, the criterion is very useful!
Similarly, there is a valuative criterion for properness.
In this section, we will meet the valuative criteria, but aside from outlining theproof of one result (the DVR version of the valuative criterion of separatedness),
we will not give proofs, and satisfy ourselves with references. There are two reasons for this controversial decision. First, the proofs require the development ofsome commutative algebra involving valuation rings that we will not otherwise
need. Second, we will not use these results in any essential way later in this book.
We begin with a valuative criterion for separatedness that applies in a casethat will suffice for the interests of most people, that of finite type morphisms of
Noetherian schemes. We will then give a more general version for more generalreaders.
12.7.1. Theorem (Valuative criterion for separatedness, DVR version). — Suppose π : X → Y is a morphism of finite type of locally Noetherian schemes. Then π isseparated if and only if the following condition holds: for any discrete valuation ring A,
and any diagram of the form

(12.7.1.1)

Spec K(A)"#%X

open emb.

"
Spec Aπ"%Y

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November 18, 2017 draft359(where the vertical morphism on the left corresponds to the inclusion A "→ K(A)), there is
at most one morphism Spec A → X such that the diagram(12.7.1.2)

Spec K(A)"#open emb.

"
Spec A

%U X

≤1

π

"
%Y

commutes.
The idea behind the proof is explained in §12.7.3. We can show one direction
right away, in the next exercise.
12.7.A. E XERCISE ( THE EASY DIRECTION ). Use the Reduced-to-Separated Theorem 10.2.2 to prove one direction of the theorem: that if π is separated, then thevaluative criterion holds.
12.7.B. E XERCISE . Suppose X is an irreducible Noetherian separated curve. If
p ∈ X is a regular closed point, then OX,p is a discrete valuation ring, so eachregular point yields a discrete valuation on K(X). Use the previous exercise toshow that distinct points yield distinct discrete valuations.
Here is the intuition behind the valuative criterion (see Figure 12.5). We thinkof Spec of a discrete valuation ring A as a “germ of a curve”, and Spec K(A) as
the “germ minus the origin” (even though it is just a point!). Then the valuative
criterion says that if we have a map from a germ of a curve to Y, and have a lift ofthe map away from the origin to X, then there is at most one way to lift the map
from the entire germ. In the case where Y is the spectrum of a field, you can thinkof this as saying that limits of one-parameter families are unique (if they exist).

?

F IGURE 12.5. The valuative criterion for separatedness
For example, this captures the idea of what is wrong with the map of the linewith the doubled origin over k (Figure 12.6): we take Spec A to be the germ of theaffine line at the origin, and consider the map of the germ minus the origin to theline with doubled origin. Then we have two choices for how the map can extend
over the origin.

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F IGURE 12.6. The line with the doubled origin fails the valuative
criterion for separatedness. (You may notice Figure 10.4 embedded in this diagram.)
12.7.C. E XERCISE . Make this precise: show that map of the line with doubled
origin over k to Spec k fails the valuative criterion for separatedness. (Earlier arguments were given in Exercises 10.1.D and 10.2.C.)
12.7.2. ⋆⋆ Remark for experts: moduli spaces and the valuative criterion for separatedness.
If Y = Spec k, and X is a (fine) moduli space (a term we won’t define here) of some
type of object, then the question of the separatedness of X (over Spec k) has a natural interpretation: given a family of your objects parametrized by a “punctured
discrete valuation ring”, is there always at most one way of extending it over theclosed point?
12.7.3. Idea behind the proof of Theorem 12.7.1 (the valuative criterion for separatedness,
DVR version). (One direction was done in Exercise 12.7.A.) If π is not separated,
our goal is to produce a diagram (12.7.1.1) that can be completed to (12.7.1.2) inmore than one way. If π is not separated, then δ : X → X ×Y X is a locally closed
embedding that is not a closed embedding.
12.7.D. E XERCISE . Show that you can find points p not in the diagonal ∆ of X ×Y X
and q in ∆ such that p ∈ q, and there are no points “between p and q” (no points r
distinct from p and q with p ∈ r and r ∈ q). (Exercise 7.4.C may shed some light.)

Let Q be the scheme obtained by giving the induced reduced subscheme structure to q. Let B = OQ,p be the local ring of Q at p.
12.7.E. E XERCISE . Show that B is a Noetherian local integral domain of dimension
1.
If B were regular, then we would be done: composing the inclusion morphism
Q → X ×Y X with the two projections induces the same morphism q → X (i.e.,
Spec κ(q) → X) but different extensions to B precisely because p is not in the diagonal. To complete the proof, one shows that the normalization of B is Noetherian;then localizing at any prime above p (there is one by the Lying Over Theorem 7.2.5)
yields the desired discrete valuation ring A.
For an actual proof, see [Stacks, tag 0207] or [Gr-EGA, II.72.3].
With a more powerful invocation of commutative algebra, we can prove avaluative criterion with much less restrictive hypotheses.

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12.7.4. Theorem (Valuative criterion for separatedness). — Suppose π : X → Y is aquasiseparated morphism. Then π is separated if and only if for any valuation ring A withfunction field K, and any diagram of the form (12.7.1.1), there is at most one morphism
Spec A → X such that the diagram (12.7.1.2) commutes.

Because I have already failed to completely prove the DVR version, I feel nourge to prove this harder fact. The proof of one direction, that π separated implies
that the criterion holds, follows from the identical argument as in Exercise 12.7.A.
For a complete proof, see [Stacks, tags 01KY and 01KZ] or [Gr-EGA, II.7.2.3].

12.7.5. Valuative criteria for (universal closedness and) properness.
There is a valuative criterion for properness too. It is philosophically useful,and sometimes directly useful, although we won’t need it. It naturally comes fromthe valuative criterion for separatedness combined with a valuative criterion for
universal closedness.
12.7.6. Theorem (Valuative criterion for universal closedness and properness,
DVR version). — Suppose π : X → Y is a morphism of finite type of locally Noetherian
schemes. Then π is universally closed (resp. proper) if and only if for any discrete valuation
ring A and any diagram (12.7.1.1), there is at least one (resp. exactly one) morphism
Spec A → X such that the diagram (12.7.1.2) commutes.

See [Gr-EGA, II.7.3.8], [Ha1, Thm. II.4.7], or [Gr-EGA’, I.5.5] for proofs. A comparison with Theorem 12.7.1 will convince you that these three criteria naturally
form a family.
In the case where Y is a field, you can think of the valuative criterion of properness as saying that limits of one-parameter families in proper varieties always exist, and are unique. This is a useful intuition for the notion of properness.

12.7.F. E ASY E XERCISE . Use the valuative criterion for properness to prove thatPn
A → Spec A is proper if A is Noetherian. (Don’t be fooled: Because this requires
the valuative criterion, this is a difficult way to prove a fact that we already showedin Theorem 10.3.5.)
12.7.G. E XERCISE ( CF. E XERCISE 12.7.B). Suppose X is an irreducible regular (Noetherian) curve, proper over a field k (respectively, over Z). Describe a bijection
between the discrete valuations on K(X) for which the elements of k (respectivelyZ) have valuation 0, and the closed points of X.
12.7.7. Remarks for experts. There is a moduli-theoretic interpretation similar tothat for separatedness (Remark 12.7.2): X is proper if and only if there is always
precisely one way of filling in a family over a punctured spectrum of a discrete
valuation ring.
12.7.8. Finally, here is a fancier version of the valuative criterion for universal
closedness and properness.
12.7.9. Theorem (Valuative criterion for universal closedness and properness).
— Suppose π : X → Y is a quasiseparated, finite type (hence quasicompact) morphism.
Then π is universally closed (resp. proper) if and only if the following condition holds. For

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any valuation ring A and any diagram of the form (12.7.1.1), there is at least one (resp.
exactly one) morphism Spec A → X such that the diagram (12.7.1.2) commutes.

Clearly the valuative criterion for properness is a consequence of the valuative criterion for separatedness (Theorem 12.7.4) and the valuative criterion for
universal closedness. For proofs, see [Stacks, tag 01KF] or [Gr-EGA, II.7.3.8].
12.7.10. On the importance of valuation rings in general. Although we have
only discussed discrete valuation rings in depth, general valuation rings should
not be thought of as an afterthought. Serre makes the case to Grothendieck in a
letter, [GrS, p. 125]:
Tu es bien sévère pour les Valuations! Je persiste pourtant à les garder, pour plusieurs
raisons: d’abord une pratique: n rédacteurs ont sué dessus, il n’y a rien à reprocher aufourbi, on ne doit pas le vider sans raisons très sérieuses (et tu n’en as pas). . . . Même
un noethérien impénitent a besoin des valuations discrètes, et de leurs extension; en fait,Tate, Dwork, et tous les gen p-adiques te diront qu’on ne peut se limiter au cas discretet que le cas de rang 1 est indispensable; à ce moment, les méthodes noethériennes deviennent un vrai carcan, et on comprend beaucoup mieux en rédigeant le cas général que
uniquement le cas de rang 1. . . . Il n’y a pas lieu d’en faire un plat, bien sûr, et c’est
pourquoi j’avias vivement combattu le plan Weil initial qui en faisait le théorème central
de l’Algèbre Commutative, mais d’autre part il faut le garder.
You are very harsh on Valuations! I persist nonetheless in keeping them, for several
reasons, of which the first is practical: n people have sweated over them, there is nothing
wrong with the result, and it should not be thrown out without very serious reasons (which
you do not have). . . . Even an unrepentant Noetherian needs discrete valuations and theirextensions; in fact, Tate, Dwork and all the p-adic people will tell you that one cannot
restrict oneself to the discrete case and the rank 1 case is indispensable; Noetherian methods
then become a burden, and one understands much better if one considers the general caseand not only the rank 1 case. . . . It is not worth making a mountain out of it, of course,which is why I energetically fought Weil’s original plan to make it the central theorem ofCommutative Algebra, but on the other hand it must be kept.

12.8 ⋆ More sophisticated facts about regular local rings
Regular local rings have essentially every good property you could want, butsome of them require hard work. We now discuss a few fancier facts that may help
you sleep well at night.
12.8.1. Localizations of regular local rings are regular local rings.
12.8.2. Fact ([E, Cor. 19.14], [Mat2, Thm. 19.3]). — If (A, m) is a regular local ring,then any localization of A at a prime is also a regular local ring.
(We will not need this, and hence will not prove it.) This major theorem wasan open problem in commutative algebra for a long time until settled by Serre andAuslander-Buchsbaum using homological methods.

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Hence to check if Spec A is regular (A Noetherian), it suffices to check at closed
points (at maximal ideals). Assuming Fact 12.8.2 (and using Exercise 5.1.E), you
can check regularity of a Noetherian scheme by checking at closed points.
We will prove two important cases of Fact 12.8.2. The first you can do rightnow.
12.8.A. E XERCISE . Suppose X is a Noetherian dimension 1 scheme that is regular at its closed points. Show that X is reduced. Hence show (without invoking
Fact 12.8.2) that X is regular.
The second important case will be proved in §21.3.12:12.8.3. Theorem. — If X is a finite type scheme over a perfect field k that is regular at its
closed points, then X is regular.
More generally, Exercise 21.3.G will show that Fact 12.8.2 holds if A is the
localization of a finite type algebra over a perfect field.
12.8.B. E XERCISE ( GENERALIZING E XERCISE 12.3.E). Suppose k is an algebraicallyclosed field of characteristic 0. Assuming Theorem 12.8.3, show that there exists a
regular hypersurface of every positive degree d in Pn . (As in Exercise 12.3.E, feel
free to weaken the hypotheses.)
12.8.4. Regular local rings are unique factorization domains, integrally closed,
and Cohen-Macaulay.
12.8.5. Fact (Auslander-Buchsbaum Theorem). — Regular local rings are uniquefactorization domains.
(This is a hard theorem, so we will not prove it, and will therefore not use it.
For a proof, see [E, Thm. 19.19], [Mat2, Thm. 20.3] or [Gr-EGA, IV4 .21.11.1], or[Mu7, §III.7] in the special case of varieties.) Thus regular schemes are factorial,and hence normal by Exercise 5.4.F.
12.8.6. Remark: Factoriality is weaker than regularity. The implication “regular implies factorial” is strict. Here is an example showing this. Suppose k is an algebraically closed field of characteristic not 2. Let A = k[x1 , . . . , xn ]/(x21 +· · ·+x2n ) (cf.
Exercise 5.4.I). Note that Spec A is clearly singular at the origin. In Exercise 14.2.V,
we will show that A is a unique factorization domain when n ≥ 5, so Spec A
is factorial. In particular, A(x1 ,...,xn ) is a Noetherian local ring that is a uniquefactorization domain, but not regular. (More generally, it is a consequence ofGrothendieck’s proof of a conjecture of P. Samuel that a Noetherian local ring thatis a complete intersection — in particular a hypersurface — that is factorial in codimension at most 3 must be factorial, [SGA2, Exp. XI, Cor. 3.14]. For a shorter andsomewhat more elementary proof, see [CL].)
12.8.7. Regular local rings are integrally closed. The Auslander-Buchsbaum Theorem(Fact 12.8.5) implies that regular local rings are integrally closed (by Exercise 5.4.F).
We will prove this (without appealing to Fact 12.8.5) in the most important geometric cases in §26.3.5.

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12.8.8. Regular local rings are Cohen-Macaulay. In §26.2.5, we will show that regular
local rings are Cohen-Macaulay (a notion defined in Chapter 26).

12.9 ⋆ Filtered rings and modules, and the Artin-Rees Lemma
We conclude Chapter 12 by discussing the Artin-Rees Lemma 12.9.3, which
was used to prove Proposition 12.5.2. The Artin-Rees Lemma generalizes the intuition behind Proposition 12.5.2, that any function that is analytically zero at apoint actually vanishes in an open neighborhood of that point (§12.5.3). Because
we will use it later (proving the Cohomology and Base Change Theorem 28.1.6),
and because it is useful to recognize it in other contexts, we discuss it in somedetail.
12.9.1. Definitions. Suppose I is an ideal of a ring A. A descending filtration of an
A-module M(12.9.1.1)

M = M0 ⊃ M1 ⊃ M2 ⊃ · · ·is called an I-filtration if Id Mn ⊂ Mn+d for all d, n ≥ 0. An example is the I-adic
filtration where Mk = Ik M. We say an I-filtration is I-stable if for some s and all
d ≥ 0, Id Ms = Md+s . For example, the I-adic filtration is I-stable.
Let A• (I) be the graded ring ⊕n≥0 In . This is called the Rees algebra of theideal I in A, although we will not need this terminology. Define M• (I) := ⊕Mn . It
is naturally a graded module over A• (I).
12.9.2. Proposition. If A is Noetherian, M is a finitely generated A-module, and(12.9.1.1) is an I-filtration, then M• (I) is a finitely generated A• (I)-module if and only if
the filtration (12.9.1.1) is I-stable.
Proof. Note that A• (I) is Noetherian (by Exercise 4.5.D(b), as A is Noetherian, andI is a finitely generated A-module).
Assume first that M• (I) is finitely generated over the Noetherian ring A• (I),and hence Noetherian. Consider the increasing chain of A• (I)-submodules whose
kth element Lk is
M ⊕ M1 ⊕ M2 ⊕ · · · ⊕ Mk ⊕ IMk ⊕ I2 Mk ⊕ · · ·(which agrees with M• (I) up until Mk , and then “I-stabilizes”). This chain muststabilize by Noetherianness. But ∪Lk = M• (I), so for some s ∈ Z≥0 , Ls = M• (I),so Id Ms = Ms+d for all d ≥ 0 — (12.9.1.1) is I-stable.
For the other direction, assume that Md+s = Id Ms for a fixed s and all d ≥ 0.
Then M• (I) is generated over A• (I) by M ⊕ M1 ⊕ · · · ⊕ Ms . But each Mj is finitely
generated, so M• (I) is indeed a finitely generated A• (I)-module.
!
12.9.3. Artin-Rees Lemma. — Suppose A is a Noetherian ring, and (12.9.1.1) is an Istable filtration of a finitely generated A-module M. Suppose that L ⊂ M is a submodule,
and let Ln := L ∩ Mn . Thenis an I-stable filtration of L.

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Proof. Note that L• is an I-filtration, as ILn ⊂ IL ∩ IMn ⊂ L ∩ Mn+1 = Ln+1 . Also,
L• (I) is an A• (I)-submodule of the finitely generated A• (I)-module M• (I), andhence finitely generated by Exercise 3.6.X (as A• (I) is Noetherian, see the proof of
Proposition 12.9.2).
!
An important special case is the following.
12.9.4. Corollary. — Suppose I ⊂ A is an ideal of a Noetherian ring, and M is a finitely
generated A-module, and L is a submodule. Then for some integer s, Id (L ∩ Is M) =L ∩ Id+s M for all d ≥ 0.

Warning: it need not be true that Id L = L ∩ Id M for all d. (Can you think of acounterexample to this statement?)

Proof. Apply the Artin-Rees Lemma 12.9.3 to the filtration Mn = In M.

!

12.9.A. E XERCISE (K RULL I NTERSECTION T HEOREM ).
(a) Suppose I is an ideal of a Noetherian ring A, and M is a finitely generated Aj
module. Show that there is some a ≡ 1 (mod I) such that a ∩∞j=1 I M = 0. Hint:∞ jn
Apply the Artin-Rees Lemma 12.9.3 with L = ∩j=1 I M and Mn = I M. Show that
L = IL, and apply the first version of Nakayama (Lemma 7.2.8).
(b) Show that if A is a Noetherian integral domain or a Noetherian local ring, andj
I is a proper ideal, then ∩∞j=1 I = 0. In particular, you will have proved Proposition 12.5.2: if (A, m) is a Noetherian local ring, then ∩i mi = 0.
12.9.B. E XERCISE . Make the following precise, and prove it (thereby justifying
the intuition in §12.5.3): if X is a locally Noetherian scheme, and f is a function
on X that is analytically zero at a point p ∈ X, then f vanishes in a (Zariski) open
neighborhood of p.

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Part V

Quasicoherent sheaves

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CHAPTER 13

Quasicoherent and coherent sheaves
Quasicoherent and coherent sheaves generalize the notion of a vector bundle.
To motivate them, we first discuss vector bundles, and their interpretation as locally free sheaves.
Speaking very informally, a vector bundle V on a geometric space X (such asa manifold) is a family of vector spaces continuously parametrized by points of
X. In other words, for each point p of X, there is a vector space, and these vector
spaces are glued into a space V so that, as p varies, the vector space above p varies
continuously. Nontrivial examples to keep in mind are the tangent bundle to amanifold, and the Möbius strip over a circle (interpreted as a line bundle). We
will make this somewhat more precise in §13.1, but if you have not seen this idea
before, you shouldn’t be concerned; the main thing we will use this notion for ismotivation.
A free sheaf on a ringed space X is an OX -module isomorphic to OX⊕I where
the sum is over some index set I. A locally free sheaf on a ringed space X is an
OX -module locally isomorphic to a free sheaf. This corresponds to the notion ofa vector bundle (§13.1). Quasicoherent sheaves form a convenient abelian category containing the locally free sheaves that is much smaller than the full categoryof O-modules. Quasicoherent sheaves generalize free sheaves in much the waythat modules generalize free modules. Coherent sheaves are, roughly speaking, afinite rank version of quasicoherent sheaves, which form a well-behaved abeliancategory containing finite rank locally free sheaves (or equivalently, finite rank vector bundles). Just as the notion of free modules leads us to the notion of modules
in general, and finitely generated modules, the notion of free sheaves will lead us
inevitably to the notion of quasicoherent sheaves and coherent sheaves. (There isa slight fib in comparing finitely generated modules to coherent sheaves, as you
will find out in §13.6.)

13.1 Vector bundles and locally free sheaves
We recall somewhat more precisely the notion of vector bundles on manifolds.
Arithmetically-minded readers shouldn’t tune out: for example, fractional idealsof the ring of integers in a number field (defined in §9.7.1) turn out to be an example of a “line bundle on a smooth curve” (Exercise 13.1.M).
A rank n vector bundle on a manifold M is a map π : V → M with the structure of an n-dimensional real vector space on π−1 (x) for each point x ∈ M, suchthat for every x ∈ M, there is an open neighborhood U and a homeomorphism
φ : U × Rn → π−1 (U)369

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over U (so that the diagram(13.1.0.1)

π−1 (U) (❋❋❋❋❋❋π|π−1 (U) ❋❋❋
$

∼
=

U

% U × Rn①①①①①①①①projection to first factor!①①

commutes) that is an isomorphism of vector spaces over each x ∈ U. An isomorphism (13.1.0.1) is called a trivialization over U.
We call n the rank of the vector bundle. A rank 1 vector bundle is called aline bundle. (It can also be convenient to be agnostic about the rank of the vector
bundle, so it can have different ranks on different connected components. It is also
sometimes convenient to consider infinite-rank vector bundles.)13.1.1. Transition functions. Given trivializations over U1 and U2 , over theirintersection, the two trivializations must be related by an element T12 of GLn withentries consisting of functions on U1 ∩ U2 . If {Ui } is a cover of M, and we are given
trivializations over each Ui , then the {Tij } must satisfy the cocycle condition:(13.1.1.1)

Tjk |Ui ∩Uj ∩Uk ◦ Tij |Ui ∩Uj ∩Uk = Tik |Ui ∩Uj ∩UK .

−1(This implies Tij = Tji.) The data of the Tij are called transition functions (ortransition matrices) for the trivialization.
This is reversible: given the data of a cover {Ui } and transition functions Tij ,we can recover the vector bundle (up to unique isomorphism) by “gluing together
the various Ui × Rn along Ui ∩ Uj using Tij ”.

13.1.2. The sheaf of sections. Fix a rank n vector bundle V → M. The sheaf ofsections F of V (Exercise 2.2.G) is an OM -module — given any open set U, we canmultiply a section over U by a function on U and get another section.
Moreover, given a trivialization over U, the sections over U are naturally identified with n-tuples of functions of U.
n
U×R
K
π

n-tuple of functions"
U
Thus given a trivialization, over each open set Ui , we have an isomorphism∼ O ⊕n . We say that such an F is a locally free sheaf of rank n. (A sheaf F
F |Ui =Ui
∼ O ⊕n .)is free of rank n, or sometimes trivial of rank n, if F =13.1.3. Transition functions for the sheaf of sections. Suppose we have a vector
bundle on M, along with a trivialization over an open cover Ui . Suppose wehave a section of the vector bundle over M. (This discussion will apply with M
replaced by any open subset.) Then over each Ui , the section corresponds to an
n-tuple functions over Ui , say ⃗si .
13.1.A. E XERCISE . Show that over Ui ∩ Uj , the vector-valued function ⃗si is related
to ⃗sj by the (same) transition functions: Tij⃗si = ⃗sj . (Don’t do this too quickly —make sure your i’s and j’s are on the correct side.)

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371

Given a locally free sheaf F with rank n, and a trivializing open neighborhood∼ O ⊕n as O-modules), weof F (an open cover {Ui } such that over each Ui , F |Ui =Uihave transition functions Tij ∈ GLn (O(Ui ∩ Uj )) satisfying the cocycle condition(13.1.1.1). Thus the data of a locally free sheaf of rank n is equivalent to the data ofa vector bundle of rank n. This change of perspective is useful, and is similar toan earlier change of perspective when we introduced ringed spaces: understanding spaces is the same as understanding (sheaves of) functions on the spaces, andunderstanding vector bundles (a type of “space over M”) is the same as understanding functions.
13.1.4. Definition. A rank 1 locally free sheaf is called an invertible sheaf. (Unimportant aside: “invertible sheaf” is a heinous term for something that is essentiallya line bundle. The motivation is that if X is a locally ringed space, and F and G are∼ OX , then F and G are invertible sheaves [MO33489].
OX -modules with F ⊗OX G =Thus in the monoid of OX -modules under tensor product, invertible sheaves are
the invertible elements. We will never use this fact. People often informally use
the phrase “line bundle” when they mean “invertible sheaf”. The phrase line sheaf
has been proposed but has not caught on.)
13.1.5. Locally free sheaves on schemes.
We can generalize the notion of locally free sheaves to schemes (or more generally, ringed spaces) without change. A locally free sheaf of rank n on a scheme
X is defined as an OX -module F that is locally a free sheaf of rank n. Precisely,∼ O ⊕n . This open
there is an open cover {Ui } of X such that for each Ui , F |Ui =Uicover determines transition functions — the data of a cover {Ui } of X, and functions Tij ∈ GLn (O(Ui ∩ Uj )) satisfying the cocycle condition (13.1.1.1) — which inturn determine the locally free sheaf. As before, given these data, we can find thesections over any open set U. Informally, they are sections of the free sheaves overeach U ∩ Ui that agree on overlaps. More formally, for each i, they are
⎛

⎞
si1
⎜
⎟
⃗si = ⎝ ... ⎠ ∈ Γ (U ∩ Ui , OX )n ,sinsatisfying Tij⃗si = ⃗sj on U ∩ Ui ∩ Uj .
You should think of these as vector bundles, but just keep in mind that they
are not the “same”, just equivalent notions. We will later (Definition 17.1.4) define
the “total space” of the vector bundle V → X (a scheme over X) in terms of thesheaf version of Spec (or more precisely, Spec (Sym V • )). But the locally free sheaf
perspective will prove to be more useful. As one example: the definition of a
locally free sheaf is much shorter than that of a vector bundle.
As in our motivating discussion, it is sometimes convenient to let the rank vary
among connected components, or to consider infinite rank locally free sheaves.
13.1.6. Useful constructions, in the form of a series of important exercises.
We now give some useful constructions in the form of a series of exercises
about locally free sheaves on a scheme. They are useful, important, and surprisingly nontrivial! Two hints: Exercises 13.1.B–13.1.H will apply for ringed spaces
in general, so you shouldn’t use special properties of schemes. Furthermore, they

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are all local on X, so you can reduce to the case where the locally free sheaves inquestion are actually free.
13.1.B. E XERCISE . Suppose F and G are locally free sheaves on X of rank m andn respectively. Show that Hom OX (F , G ) is a locally free sheaf of rank mn.
13.1.C. E XERCISE . If E is a locally free sheaf on X of (finite) rank n, Exercise 13.1.B
implies that E ∨ := Hom (E , OX ) is also a locally free sheaf of rank n. This is called
the dual of E (cf. §2.3.4). Given transition functions for E , describe transition functions for E ∨ . (Note that if E is rank 1, i.e., invertible, the transition functionsof the dual are the inverse of the transition functions of the original.) Show that∼ E ∨∨ . (Caution: your argument showing that there is a canonical isomorE =∼ F better not also show that there is an isomorphism F ∨ =∼ F!
phism (F ∨ )∨ =We will see an example in §14.1 of a locally free F that is not isomorphic to its
dual: the invertible sheaf O(1) on Pn .)13.1.D. E XERCISE . If F and G are locally free sheaves, show that F ⊗G is a locally
free sheaf. (Here ⊗ is tensor product as OX -modules, defined in Exercise 2.6.J.) If∼ OX .
F is an invertible sheaf, show that F ⊗ F ∨ =13.1.E. E XERCISE . Recall that tensor products tend to be only right-exact in general. Show that tensoring by a locally free sheaf is exact. More precisely, if F is alocally free sheaf, and G ′ → G → G ′′ is an exact sequence of OX -modules, thenso is G ′ ⊗ F → G ⊗ F → G ′′ ⊗ F . (Possible hint: it may help to check exactness by checking exactness at stalks. Recall that the tensor product of stalks can be
identified with the stalk of the tensor product, so for example there is a “natural”∼ Gp ⊗ O
isomorphism (G ⊗OX F )p =Fp , Exercise 2.6.J(b).)
X,p13.1.F. E XERCISE . If E is a locally free sheaf of finite rank, and F and G are∼ Hom (F ⊗ E ∨ , G ). (Possible hint: firstOX -modules, show that Hom (F , G ⊗ E ) =consider the case where E is free.)13.1.G. E XERCISE AND IMPORTANT DEFINITION . Show that the invertible sheaves
on X, up to isomorphism, form an abelian group under tensor product. This iscalled the Picard group of X, and is denoted Pic X.
13.1.H. E XERCISE . If π : X → Y is a morphism of ringed spaces, and G is a locally
free sheaf of rank n on Y, figure out how to define a “pulled back” locally free
sheaf of rank n on X, denoted π∗ G . Then show that Pic is a contravariant functorfrom the category of ringed spaces to the category Ab of abelian groups.
Unlike the previous exercises, the next one is specific to schemes.
13.1.I. E XERCISE . Suppose s is a section of a locally free sheaf F on a scheme X.
Define the notion of the subscheme cut out by s = 0, denoted (for obvious reasons)V(s). Be sure to check that your definition is independent of choices! (This exercise
gives a new solution to Exercise 4.5.P.) Hint: given a trivialization over an open
set U, s corresponds to a number of functions f1 , . . . on U; on U, take the scheme
cut out by these functions. Alternate hint that avoids coordinates: figure out how
to define it as the largest closed subscheme on which s restricts to 0.

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13.1.7. Random concluding remarks.
We define rational (and regular) sections of a locally free sheaf on a scheme
X just as we did rational (and regular) functions (see for example §5.5 and §6.5).
13.1.J. E XERCISE . Show that locally free sheaves on locally Noetherian normal
schemes satisfy “Hartogs’s Lemma”: sections defined away from a set of codimension at least 2 extend over that set. (Algebraic Hartogs’s Lemma for Noetherian
normal schemes is Theorem 11.3.11.)
13.1.K. E ASY E XERCISE . Suppose s is a nonzero rational section of an invertible
sheaf on a locally Noetherian normal scheme. Show that if s has no poles, then s
is regular. (Hint: Exercise 12.5.H.)
13.1.8. Remark. Based on your intuition for line bundles on manifolds, you mighthope that every point has a “small” open neighborhood on which all invertible
sheaves (or locally free sheaves) are trivial. Sadly, this is not the case. We will
eventually see (§19.11.1) that for the curve y2 − x3 − x = 0 in A2C , every nonempty
open set has nontrivial invertible sheaves. (This will use the fact that it is an open
subset of an elliptic curve.)
13.1.L. ⋆ E XERCISE ( FOR THOSE WITH SUFFICIENT COMPLEX - ANALYTIC BACKGROUND ).
Recall the analytification functor (Exercises 6.3.N and 10.1.F), that takes a complex
finite type reduced scheme and produces a complex analytic space.
(a) If L is an invertible sheaf on a complex (algebraic) variety X, define (up tounique isomorphism) the corresponding invertible sheaf on the complex variety
Xan .
(b) Show that the induced map Pic X → Pic Xan is a group homomorphism.
(c) Show that this construction is functorial: if π : X → Y is a morphism of complex
varieties, the following diagram commutes:
Pic Y

π∗

% Pic X"
Pic Yanπ∗an"% Pic Xan

where the vertical maps are the ones you have defined.
13.1.M. ⋆ E XERCISE ( FOR THOSE WITH SUFFICIENT ARITHMETIC BACKGROUND ;SEE ALSO P ROPOSITION 14.2.10 AND §14.2.13). Recall the definition of the ringof integers OK in a number field K, Remark 9.7.1. A fractional ideal a of OK isa nonzero OK -submodule of K such that there is a nonzero a ∈ OK such thataa ⊂ OK . Products of fractional ideals are defined analogously to products ofideals in a ring (defined in Exercise 3.4.C): ab consists of (finite) OK -linear combinations of products of elements of a and elements of b. Thus fractional ideals form
a semigroup under multiplication, with OK as the identity. In fact fractional ideals
of OK form a group.
(a) Explain how a fractional ideal on a ring of integers in a number field
yields an invertible sheaf. (Although we won’t need this, it is worth noting that a fractional ideal is the same as an invertible sheaf with a trivialization at the generic point.)

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374

The Rising Sea: Foundations of Algebraic Geometry(b) A fractional ideal is principal if it is of the form rOK for some r ∈ K× .
Show that any two that differ by a principal ideal yield the same invertible sheaf.
(c) Show that two fractional ideals that yield the same invertible sheaf differ
by a principal ideal.
(d) The class group is defined to be the group of fractional ideals modulo
the principal ideals (i.e., modulo K× ). Give an isomorphism of the classgroup with the Picard group of OK .
(This discussion applies to any Dedekind domain. See Exercise 14.2.S for a followup.)
13.1.9. The problem with locally free sheaves.
Recall that OX -modules form an abelian category: we can talk about kernels,
cokernels, and so forth, and we can do homological algebra. Similarly, vector
spaces form an abelian category. But locally free sheaves (i.e., vector bundles),along with reasonably natural maps between them (those that arise as maps of
OX -modules), don’t form an abelian category. As a motivating example in the
category of differentiable manifolds, consider the map of the trivial line bundle on
R (with coordinate t) to itself, corresponding to multiplying by the coordinate t.
Then this map jumps rank, and if you try to define a kernel or cokernel you will
get confused.
This problem is resolved by enlarging our notion of nice OX -modules in anatural way, to quasicoherent sheaves.
OX -modules⊃ quasicoherent sheaves ⊃
locally free sheaves(abelian category)(abelian category)(not an abelian category)
You can turn this into two definitions of quasicoherent sheaves, equivalent tothose we will give in §13.2. We want a notion that is local on X of course. So
we ask for the smallest abelian subcategory of ModOX that is “local” and includes
vector bundles. It turns out that the main obstruction to vector bundles to be an
abelian category is the failure of cokernels of maps of locally free sheaves — as
OX -modules — to be locally free; we could define quasicoherent sheaves to be
those OX -modules that are locally cokernels, yielding a description that works
more generally on ringed spaces, as described in Exercise 13.4.B. You may wish tolater check that our future definitions are equivalent to these.
Similarly, in the locally Noetherian setting, finite rank locally free sheaves will
sit in a nice smaller abelian category, that of coherent sheaves.
quasicoherent sheaves(abelian category)⊃coherent sheaves ⊃(abelian category)finite rank locally free sheaves(not an abelian category)

13.1.10. Remark: Quasicoherent and coherent sheaves on ringed spaces. We will discussquasicoherent and coherent sheaves on schemes, but they can be defined moregenerally (see Exercise 13.4.B for quasicoherent sheaves, and [Se1, Def. 2] for coherent sheaves). Many of the results we state will hold in greater generality, but
because the proofs look slightly different, we restrict ourselves to schemes to avoid
distraction.

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13.2 Quasicoherent sheaves
We now define the notion of quasicoherent sheaf. In the same way that a scheme
is defined by “gluing together rings”, a quasicoherent sheaf over that scheme isobtained by “gluing together modules over those rings”. Given an A-module M,( on Spec A long ago (Exercise 4.1.D) — the sections
we defined an O-module M
over D(f) were Mf .

13.2.1. Theorem. — Let X be a scheme, and F an OX -module. Suppose P is the property∼M( for some A-module M. Then Pof affine open subschemes Spec A of X that F |Spec A =satisfies the two hypotheses of the Affine Communication Lemma 5.3.2.
We prove this in a moment.

13.2.2. Definition. If X is a scheme, then an OX -module F is quasicoherent if for∼ M( for some A-module M. By
every affine open subset Spec A ⊂ X, F |Spec A =Theorem 13.2.1, it suffices to check this for a collection of affine open sets covering( is a quasicoherent sheaf on Spec A, and all locally free sheaves
X. For example, M
on X are quasicoherent.
13.2.A. U NIMPORTANT E XERCISE ( NOT EVERY OX - MODULE IS A QUASICOHERENTSHEAF ).
(a) Suppose X = Spec k[t]. Let F be the skyscraper sheaf supported at the origin[(t)], with group k(t) and the usual k[t]-module structure. Show that this is an
OX -module that is not a quasicoherent sheaf. (More generally, if X is an integral
scheme, and p ∈ X is not the generic point, we could take the skyscraper sheaf at pwith group the function field of X. Except in silly circumstances, this sheaf won’t
be quasicoherent.) See Exercises 8.1.F and 13.3.I for more (pathological) examples
of OX -modules that are not quasicoherent.
(b) Suppose X = Spec k[t]. Let F be the skyscraper sheaf supported at the genericpoint [(0)], with group k(t). Give this the structure of an OX -module. Show that
this is a quasicoherent sheaf. Describe the restriction maps in the distinguishedtopology of X. (Remark: Your argument will apply more generally, for example
when X is an integral scheme with generic point η, and F is the skyscraper sheafiη,∗ K(X).)

13.2.B. U NIMPORTANT EXERCISE ( NOT EVERY QUASICOHERENT SHEAF IS LOCALLYFREE ). Use the example of Exercise 13.2.A(b) to show that not every quasicoherent
sheaf is locally free.
13.2.C. E XERCISE . Show that every (finite) rank n vector bundle on A1k is trivial ofrank n. Hint: finitely generated modules over a principal ideal domain are finite
direct sums of cyclic modules, as mentioned in Remark 12.5.14. See the aside in§14.2.8 for the difficult generalization to Ank.
13.2.3. Proof of Theorem 13.2.1. Clearly if Spec A has property P, then so does the∼M( Spec A =<f as sheaves
distinguished open Spec Af : if M is an A-module, then M|
fof OSpec Af -modules (both sides agree on the level of distinguished open sets andtheir restriction maps).

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We next show the second hypothesis of the Affine Communication Lemma 5.3.2.
Suppose we have modules M1 , . . . , Mn , where Mi is an Afi -module, along withisomorphisms φij : (Mi )fj → (Mj )fi of Afi fj -modules, satisfying the cocycle con( gives us M
(i on D(fi ) =dition (13.1.1.1). We want to construct an M such that M(
Spec Afi , or equivalently, isomorphisms ρi : Γ (D(fi ), M) → Mi , so that the bottom
triangle of(13.2.3.1)&⊗Afi &&&

&&
&&
V&&

M❏
❏❏ ⊗A
❏❏ fj
❏❏
❏❏
❏+

Mfi
Mfj
■■
❋❋
①
✉
⊗A
■
①
✉
⊗A
f
❋❋ρj
fi ✉
ρi ①①
■■ j
✉
❋
■
①
✉
■
①
✉
∼
∼ ❋❋
■■
✉
①
❋$
!①①
+
V✉✉
Mi ❋
Mfi fj
Mj
H
W❍
❍
❋❋
①
✈
①
❍
✈
❋❋
①
❍❍
✈
❋❋
①①
❍
✈✈
∼ ❍❍❍
①① ⊗Afi✈✈ ∼⊗Afj ❋❋
①
✈
M
V✈
+
C①
φij
% (Mj )f
(Mi )fji∼commutes.
13.2.D. E XERCISE . Why does this suffice to prove the result? In other words, why∼ M?
(does this imply that F |Spec A =We already know that M should be Γ (Spec A, F ), as F is a sheaf. Consider
elements of M1 × · · · × Mn that “agree on overlaps”; let this set be M. In otherwords,(13.2.3.2)%M0% M1 × · · · × Mnγ

% M12 × M13 × · · · × M(n−1)n∼ (Mj )f , and the map γ is the “differis an exact sequence (where Mij = (Mi )fj =ience” map). So M is a kernel of a morphism of A-modules, hence an A-module.
∼ Mf (and that this isomorphism satisfies (13.2.3.1)).
We are left to show that Mi =i
(At this point, we may proceed in a number of ways, and the reader may wish tofind their own route rather than reading on.)For convenience assume i = 1. Localization is exact (Exercise 1.6.F(a)), so
tensoring (13.2.3.2) by Af1 yields(13.2.3.3)

% Mf

0

% (M1 )f × (M2 )f × · · · × (Mn )f
1
1
1

1

% M12 × · · · × M1n × (M23 )f × · · · × (M(n−1)n )f
1
1

is an exact sequence of Af1 -modules.
We now identify many of the modules appearing in (13.2.3.3) in terms of M1 .
∼First of all, f1 is invertible in Af1 , so (M1 )f1 is canonically M1 . Also, (Mj )f1 =∼(M1 )fj via φ1j . Hence if i, j ̸= 1, (Mij )f1 = (M1 )fi fj via φ1i and φ1j (here the∼ (M1 )f via φ1i .
cocycle condition is implicitly used). Furthermore, (M1i )f1 =i
Thus we can write (13.2.3.3) as(13.2.3.4)

0
α

% Mf

1

% M1 × (M1 )f × · · · × (M1 )fn
2% (M1 )f × · · · × (M1 )fn × (M1 )f f × · · · × (M1 )f fn
2
2 3
n−1

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November 18, 2017 draft377∼M<1 for some M1 , so by considering the coverBy assumption, F |Spec Af1 =Spec Af1 = Spec Af1 ∪ Spec Af1 f2 ∪ Spec Af1 f3 ∪ · · · ∪ Spec Af1 fn(notice the “redundant” first term), and identifying sections of F over Spec Af1 interms of sections over the open sets in the cover and their pairwise overlaps, wehave an exact sequence of Af1 -modules
0
β

% M1

% M1 × (M1 )f × · · · × (M1 )fn
2% (M1 )f × · · · × (M1 )fn × (M1 )f f × · · · × (M1 )f fn
2
2 3
n−1

which is very similar to (13.2.3.4). Indeed, the final map β of the above sequence isthe same as the map α of (13.2.3.4), so ker α = ker β, i.e., we have an isomorphism∼ Mf .
M1 =1
13.2.E. E XERCISE . Finally, show that the bottom triangle of (13.2.3.1) is commutative, perhaps by showing that each vertex of the triangle can be identified with thesections of F over Spec Af1 f2 .
!

13.3 Characterizing quasicoherence using the distinguished affine
baseBecause quasicoherent sheaves are locally of a very special form, in order to“know” a quasicoherent sheaf, we need only know what the sections are over every
affine open set, and how to restrict sections from an affine open set U to a distinguished affine open subset of U. We make this precise by defining what I will callthe distinguished affine base of the Zariski topology — not a base in the usual sense.
The point of this discussion is to give a useful characterization of quasicoherence,but you may wish to just jump to §13.3.3.
The open sets of the distinguished affine base are the affine open subsets of
X. We have already observed that this forms a base. But forget that fact. We likedistinguished open sets Spec Af "→ Spec A, and we don’t really understand open
embeddings of one random affine open subset in another. So we just remember
the “nice” inclusions.
13.3.1. Definition. The distinguished affine base of a scheme X is the data of theaffine open sets and the distinguished inclusions.
In other words, we remember only some of the open sets (the affine open sets),and only some of the morphisms between them (the distinguished morphisms). For
experts: if you think of a topology as a category (the category of open sets), we
have described a subcategory.
We can define a sheaf on the distinguished affine base in the obvious way: wehave a set (or abelian group, or ring) for each affine open set, and we know how
to restrict to distinguished open sets. (You should think through the statement of
the identity and gluability axioms yourself.)

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Given a sheaf F on X, we get a sheaf on the distinguished affine base. You
can guess where we are going: we will show that all the information of the sheaf
is contained in the information of the sheaf on the distinguished affine base.
As a warm-up, we can recover stalks as follows. (We will be implicitly using
only the following fact. We have a collection of open subsets, and some inclusions
among these subsets, such that if we have any p ∈ U, V where U and V are in ourcollection of open sets, there is some W containing p, and contained in U and V
such that W "→ U and W "→ V are both in our collection of inclusions. In the case
we are considering here, this is the key Proposition 5.3.1 that given any two affine
open sets Spec A, Spec B in X, Spec A ∩ Spec B could be covered by affine open setsthat were simultaneously distinguished in Spec A and Spec B. In fancy language:
the category of affine open sets, and distinguished inclusions, forms a filtered set.)
The stalk Fp is the colimit lim(f ∈ F (U)) where the colimit is over all open−→
sets contained in X. We compare this to lim(f ∈ F (U)) where the colimit is over all−→affine open sets, and all distinguished inclusions. You can check that the elementsof one correspond to elements of the other. (Think carefully about this!)
13.3.A. E XERCISE . Show that a section of a sheaf on the distinguished affine baseis determined by the section’s germs.
13.3.2. Theorem. —(a) A sheaf on the distinguished affine base F b determines a unique (up to unique
isomorphism) sheaf F which when restricted to the affine base is F b . (Hence
if you start with a sheaf, and take the sheaf on the distinguished affine base, andthen take the induced sheaf, you get the sheaf you started with.)(b) A morphism of sheaves on a distinguished affine base uniquely determines amorphism of sheaves.
(c) An OX -module “on the distinguished affine base” yields an OX -module.
This proof is identical to our argument of §2.5 showing that sheaves are (essentially) the same as sheaves on a base, using the “sheaf of compatible germs”
construction. The main reason for repeating it is to let you see that all that isneeded is for the open sets to form a filtered set (or in the current case, that thecategory of open sets and distinguished inclusions is filtered).
For experts: (a) and (b) are describing an equivalence of categories between
sheaves on the Zariski topology of X and sheaves on the distinguished affine baseof X.
Proof. (a) Suppose F b is a sheaf on the distinguished affine base. Then we candefine stalks.
For any open set U of X, define the sheaf of compatible germs
F (U) :={(fp ∈ Fpb )p∈U : for all p ∈ U,

there exists Up with p ∈ Up ⊂ U, Fp ∈ F b (Up )such that Fp
q = fq for all q ∈ Up }pwhere each Up is in our base, and Fp
q means “the germ of F at q”. (As usual, those
who want to worry about the empty set are welcome to.)This really is a sheaf: convince yourself that we have restriction maps, identity,
and gluability, really quite easily.

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I next claim that if U is in our base, that F (U) = F b (U). We clearly have amap F b (U) → F (U). This is an isomorphism on stalks, and hence an isomorphism by Exercise 2.4.E.

13.3.B. E XERCISE . Prove (b) (cf. Exercise 2.5.C).
13.3.C. E XERCISE . Prove (c) (cf. Remark 2.5.3).

The proof of Theorem 13.3.2 is now complete.

!

13.3.3. A characterization of quasicoherent sheaves in terms of distinguishedinclusions. We use this perspective to give a useful characterization of quasicoherent sheaves among OX -modules. Suppose F is an OX -module, and Spec Af "→Spec A ⊂ X is a distinguished open subscheme of an affine open subscheme of
X. Let φ : Γ (Spec A, F ) → Γ (Spec Af , F ) be the restriction map. The source of φis an A-module, and the target is an Af -module, so by the universal property of
localization (Exercise 1.3.D), φ naturally factors as:φ% Γ (Spec Af , F )Γ (Spec A, F )
◗◗◗
♠X
◗◗◗
♠♠♠
♠
◗◗◗
♠
♠♠♠α
⊗A Af ◗◗◗2♠♠♠
Γ (Spec A, F )f
13.3.D. V ERY IMPORTANT EXERCISE . Show that an OX -module F is quasicoherent
if and only if for each such distinguished Spec Af "→ Spec A, α is an isomorphism.
Thus a quasicoherent sheaf is (equivalent to) the data of one module for eachaffine open subset (a module over the corresponding ring), such that the module over a distinguished open set Spec Af is given by localizing the module overSpec A. The next exercise shows that this will be an easy criterion to check.
13.3.E. I MPORTANT E XERCISE ( CF. THE Q CQS L EMMA 7.3.5). Suppose X is aquasicompact and quasiseparated scheme (i.e., covered by a finite number of affine
open sets, the pairwise intersection of which is also covered by a finite number ofaffine open sets). Suppose F is a quasicoherent sheaf on X, and let f ∈ Γ (X, OX )be a function on X. Show that the restriction map
resXf ⊂X : Γ (X, F ) → Γ (Xf , F )(here Xf is the open subset of X where f doesn’t vanish) is precisely localization. In
other words show that there is an isomorphism Γ (X, F )f → Γ (Xf , F ) making thefollowing diagram commute.
resXf ⊂X

% Γ (Xf , F )Γ (X, F )
▲▲▲
O;
▲▲▲
O
OO
O
▲▲▲
O
O ∼
⊗Γ (X,OX ) (Γ (X,OX )f ) ▲#OOOΓ (X, F )f

(Hint: Apply the exact functor ⊗Γ (X,OX ) Γ (X, OX )f to the exact sequence
0 → Γ (X, F ) → ⊕i Γ (Ui , F ) → ⊕Γ (Uijk , F )Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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where the Ui form a finite affine cover of X and Uijk form a finite affine cover ofUi ∩ Uj .)13.3.F. I MPORTANT E XERCISE ( COROLLARY TO E XERCISE 13.3.E: PUSHFORWARDS
OF QUASICOHERENT SHEAVES ARE QUASICOHERENT IN REASONABLE CIRCUM STANCES ). Suppose π : X → Y is a quasicompact quasiseparated morphism, andF is a quasicoherent sheaf on X. Show that π∗ F is a quasicoherent sheaf on Y.

13.3.G. E XERCISE ( GOOD PRACTICE : THE SHEAF OF NILPOTENTS ). If A is a ring,∼ N(A)f . Use this to define/construct the quasicoherand f ∈ A, show that N(Af ) =ent sheaf of nilpotents on any scheme X. This is an example of an ideal sheaf (ofOX ).
13.3.H. I MPORTANT E XERCISE ( TO BE USED REPEATEDLY ). Generalize Exercise 13.3.E
as follows. Suppose X is a quasicompact quasiseparated scheme, L is an invertible sheaf on X with section s, and F is a quasicoherent sheaf on X. As in Exercise 13.3.E, let Xs be the open subset of X where s doesn’t vanish. Show thatany section of F over Xs can be interpreted as the quotient of a global section ofF ⊗OX L ⊗n by sn . In other words, any section of F over Xs can be extended over
all of X, once you multiply it by a large enough power of s. More precisely: notethat ⊕n≥0 Γ (X, L ⊗n ) is a graded ring, and we interpret s as a degree 1 element of it.
Note also that ⊕n≥0 Γ (X, F ⊗OX L ⊗n ) is a graded module over this ring. Describe
a natural map))* *⊕n≥0 Γ (X, F ⊗OX L ⊗n ) s 0 → Γ (Xs , F )and show that it is an isomorphism. (Hint: after showing the existence of thenatural map, show it is an isomorphism in the affine case.)
13.3.I. L ESS IMPORTANT EXERCISE . Give a counterexample to show that Exercise 13.3.E need not hold without the quasicompactness hypothesis. (Possible hint:
take an infinite disjoint union of affine schemes. The key idea is that infinite direct
products do not commute with localization.)13.3.4. ⋆⋆ Grothendieck topologies. The distinguished affine base isn’t a topology in the usual sense — the union of two affine sets isn’t necessarily affine, forexample. It is however a first new example of a generalization of a topology — thenotion of a site or a Grothendieck topology. We give the definition to satisfy the
curious, but we certainly won’t use this notion. (For a clean statement, see [Stacks,
tag 00VH]; this is intended only as motivation.) The idea is that we should abstract
away only those notions we need to define sheaves. We need the notion of open
set, but it turns out that we won’t even need an underlying set, i.e., we won’t evenneed the notion of points! Let’s think through how little we need. For our discussion of sheaves to work, we needed to know what the open sets were, and whatthe (allowed) inclusions were, and these should “behave well”, and in particular
the data of the open sets and inclusions should form a category. (For example,
the composition of an allowed inclusion with another allowed inclusion should
be an allowed inclusion — in the distinguished affine base, a distinguished open
set of a distinguished open set is a distinguished open set.) So we just require thedata of this category. At this point, we can already define presheaf (as just a contravariant functor from this category of “open sets”). We saw this idea earlier in
Exercise 2.2.A.

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In order to extend this definition to that of a sheaf, we need to know more
information. We want two open subsets of an open set to intersect in an open
set, so we want the category to be closed under fibered products (cf. Exercise 1.3.O).
For the identity and gluability axioms, we need to know when some open sets coveranother, so we also remember this as part of the data of a Grothendieck topology.
The data of the coverings satisfy some obvious properties. Every open set covers
itself (i.e., the identity map in the category of open sets is a covering). Coverings pullback: if we have a map Y → X, then any cover of X pulls back to a cover of Y. Finally,a cover of a cover should be a cover. Such data (satisfying these axioms) is called aGrothendieck topology or a site. (There are useful variants of this definition in theliterature. Again, we are following [Stacks].) We can define the notion of a sheaf
on a Grothendieck topology in the usual way, with no change. A topos is a scaryname for a category of sheaves of sets on a Grothendieck topology.
Grothendieck topologies are used in a wide variety of contexts in and nearalgebraic geometry. Étale cohomology (using the étale topology), a generalizationof Galois cohomology, is a central tool, as are more general flat topologies, suchas the smooth topology. The definition of a Deligne-Mumford or Artin stack uses
the étale and smooth topology, respectively. Tate developed a good theory of nonarchimedean analytic geometry over totally disconnected ground fields such as Qp
using a suitable Grothendieck topology. Work in K-theory (related for example to
Voevodsky’s work) uses exotic topologies.

13.4 Quasicoherent sheaves form an abelian category
Morphisms from one quasicoherent sheaf on a scheme X to another are defined to be just morphisms as OX -modules. In this way, the quasicoherent sheaves
on a scheme X form a category, denoted QCohX . (By definition it is a full subcategory of ModOX .) We now show that quasicoherent sheaves on X form an abeliancategory.
When you show that something is an abelian category, you have to checkmany things, because the definition has many parts. However, if the objects you
are considering lie in some ambient abelian category, then it is much easier. You
have seen this idea before: there are several things you have to do to check thatsomething is a group. But if you have a subset of group elements, it is much easier
to check that it forms a subgroup.
You can look at back at the definition of an abelian category, and you will seethat in order to check that a subcategory is an abelian subcategory, it suffices tocheck only the following:(i) 0 is in the subcategory(ii) the subcategory is closed under finite sums(iii) the subcategory is closed under kernels and cokernels
In our case of QCohX ⊂ ModOX , the first two are cheap: 0 is certainly quasicoherent, and the subcategory is closed under finite sums: if F and G are sheaves∼ M∼ N,( and G =3 then F ⊕ G = M#
on X, and over Spec A, F =⊕ N (do you seewhy?), so F ⊕ G is a quasicoherent sheaf.

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We now check (iii), using the characterization of Important Exercise 13.3.D.
Suppose α : F → G is a morphism of quasicoherent sheaves. Then on any affine
open set U, where the morphism is given by β : M → N, define (ker α)(U) = ker βand (coker α)(U) = coker β. Then these behave well under inversion of a single
element: if
0→K→M→N→P→0
is exact, then so is
0 → Kf → Mf → Nf → Pf → 0,∼
∼ coker(βf ). Thus both of thesefrom which (ker β)f = ker(βf ) and (coker β)f =define quasicoherent sheaves. Moreover, by checking stalks, they are indeed thekernel and cokernel of α (exactness can be checked stalk-locally). Thus the quasicoherent sheaves indeed form an abelian category.
13.4.A. E XERCISE . Show that a sequence of quasicoherent sheaves F → G → H
on X is exact if and only if it is exact on every open set in any given affine coverof X. (In particular, taking sections over an affine open Spec A is an exact functorfrom the category of quasicoherent sheaves on X to the category of A-modules.
Recall that taking sections is only left-exact in general, see §2.6.F.) In particular, we
may check injectivity or surjectivity of a morphism of quasicoherent sheaves bychecking on an affine cover of our choice.
Caution: If 0 → F → G → H → 0 is an exact sequence of quasicoherent
sheaves, then for any open set
0 → F (U) → G (U) → H (U)is exact, and exactness on the right is guaranteed to hold only if U is affine. (To set
you up for cohomology: whenever you see left-exactness, you expect to eventuallyinterpret this as a start of a long exact sequence. So we are expecting H1 ’s on theright, and now we expect that H1 (Spec A, F ) = 0. This will indeed be the case.)
13.4.1. Warning for those already familiar with vector bundles. Morphisms of vector
bundles are more restrictive than morphisms of quasicoherent sheaves that happen to be locally free sheaves. A locally free subsheaf of a locally free sheaf does
not always yield a subvector bundle of a vector bundle. The archetypal example
is the exact sequence of quasicoherent sheaves on A1k = Spec k[t] corresponding to
the following exact sequence of k[t]-modules:(13.4.1.1)

0 → tk[t] → k[t] → k → 0.

< is a subsheaf of k[t],< but it does not correspond to a
The locally free sheaf tk[t]“subvector bundle”; the cokernel is not a vector bundle. (This example came up in§13.1.9.)
13.4.B. L ESS IMPORTANT EXERCISE ( CONNECTION TO ANOTHER DEFINITION , AND
QUASICOHERENT SHEAVES ON RINGED SPACES IN GENERAL ). Show that an OX module F on a scheme X is quasicoherent if and only if there exists an open coverby Ui such that on each Ui , F |Ui is isomorphic to the cokernel of a map of two
free sheaves:
⊕J
⊕I
OU
→ OU
→ F |Ui → 0
iiis exact. We have thus connected our definitions to the definition given at the very
start of the chapter. This is the definition of a quasicoherent sheaf on a ringed space

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in general. It is useful in many circumstances, for example in complex analytic
geometry.

13.5 Module-like constructions
In a similar way, basically any nice construction involving modules extends
to quasicoherent sheaves. (One exception: the Hom of two A-modules is an Amodule, but the Hom of two quasicoherent sheaves is quasicoherent only in “reasonable” circumstances, see Exercise 13.7.A. The failure of “niceness” is the failureof Hom to commute with localization — see the aside in Exercise 13.7.A(a) for an
example.)13.5.1. Locally free sheaves from free modules.
13.5.A. E XERCISE ( POSSIBLE HELP FOR LATER PROBLEMS ).
(a) Suppose(13.5.1.1)

0 → F ′ → F → F ′′ → 0

is a short exact sequence of quasicoherent sheaves on X. Suppose U = Spec A is
3 ⊕a , F ′′ |Spec A = A
3 ⊕b .
an affine open set where F ′ , F ′′ are free, say F ′ |Spec A = A(Here a and b are assumed to be finite for convenience, but this is not necessary,
so feel free to generalize to the infinite rank case.) Show that F is also free on
Spec A, and that 0 → F ′ → F → F ′′ → 0 can be interpreted as coming from thetautological exact sequence 0 → A⊕a → A⊕(a+b) → A⊕b → 0. (As a consequence,
given an exact sequence of quasicoherent sheaves (13.5.1.1) where F ′ and F ′′ arelocally free, F must also be locally free.)(b) In the finite rank case, show that given an open covering by trivializing affine
open sets (of the form described in (a)), the transition functions (really, matrices) of
F may be interpreted as block upper triangular matrices, where the top left a × a
blocks are transition functions for F ′ , and the bottom b × b blocks are transition
functions for F ′′ .
13.5.B. E XERCISE . Suppose (13.5.1.1) is a short exact sequence of quasicoherent
sheaves on X. By Exercise 13.5.A(a), if F ′ and F ′′ are locally free, then F is too.
(a) If F and F ′′ are locally free of finite rank, show that F ′ is too. Hint: Reduceto the case X = Spec A and F and F ′′ free. Interpret the map φ : F → F ′′ asan n × m matrix M with values in A, with m the rank of F and n the rank ofF ′′ . For each point p of X, show that there exist n columns {c1 , . . . , cn } of M thatare linearly independent at p and hence near p (as linear independence is given
by nonvanishing of the appropriate n × n determinant). Thus X can be coveredby distinguished open subsets in bijection with the choices of n columns of M.
Restricting to one subset and renaming columns, reduce to the case where thedeterminant of the first n columns of M is invertible. Then change coordinates onA⊕m = F (Spec A) so that M with respect to the new coordinates is the identity
matrix in the first n columns, and 0 thereafter. Finally, in this case interpret F ′ as# .
A⊕(m−n)(b) If F ′ and F are both locally free, show that F ′′ need not be. (Hint: consider

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The Rising Sea: Foundations of Algebraic Geometry(13.4.1.1), which we will soon interpret as the closed subscheme exact sequence(13.5.4.1) for a point on A1 .)13.5.2. Tensor products. Another important example is tensor products.
13.5.C. E XERCISE . If F and G are quasicoherent sheaves, show that F ⊗ G is aquasicoherent sheaf described by the following information: If Spec A is an affine
open, and Γ (Spec A, F ) = M and Γ (Spec A, G ) = N, then Γ (Spec A, F ⊗ G ) =M⊗A N, and the restriction map Γ (Spec A, F ⊗G ) → Γ (Spec Af , F ⊗G ) is precisely∼ Mf ⊗A Nf . (We are using thethe localization map M ⊗A N → (M ⊗A N)f =f∼ Mf ⊗A Nf . You can prove this by universal
algebraic fact that (M ⊗A N)f =fproperty if you want, or by using the explicit construction.)
Note that thanks to the machinery behind the distinguished affine base, sheafification is taken care of. This is a feature we will use often: constructions involving quasicoherent sheaves that involve sheafification for general sheaves don’t require sheafification when considered on the distinguished affine base. Along withthe fact that injectivity, surjectivity, kernels and so on may be computed on affine
opens, this is the reason that it is particularly convenient to think about quasicoherent sheaves in terms of affine open sets.
Given a section s of F and a section t of G , we have a section s ⊗ t of F ⊗ G .
If F is an invertible sheaf, this section is often denoted st.
13.5.3. Tensor algebra constructions.
For the next exercises, recall the following. If M is an A-module, then thetensor algebra T • (M) is a noncommutative algebra, graded by Z≥0 , defined asfollows. T 0 (M) = A, T n (M) = M ⊗A · · · ⊗A M (where n terms appear in the
product), and multiplication is what you expect.
The symmetric algebra Sym• M is a symmetric algebra, graded by Z≥0 , defined as the quotient of T • (M) by the (two-sided) ideal generated by all elementsof the form x⊗y−y⊗x for all x, y ∈ M. Thus Symn M is the quotient of M⊗· · ·⊗M
by the relations of the form m1 ⊗ · · · ⊗ mn − m1′ ⊗ · · · ⊗ mn′ where (m1′ , . . . , mn′ )is a rearrangement of (m1 , . . . , mn ).
The exterior algebra ∧• M is defined to be the quotient of T • M by the (twosided) ideal generated by all elements of the form x ⊗ x for all x ∈ M. Expanding(a+b)⊗(a+b), we see that a⊗b = −b⊗a in ∧2 M. This implies that if 2 is invertible
in A (e.g., if A is a field of characteristic not 2), ∧n M is the quotient of M ⊗ · · · ⊗ M
by the relations of the form m1 ⊗ · · · ⊗ mn − (−1)sgn(σ) mσ(1) ⊗ · · · ⊗ mσ(n) where
σ is a permutation of {1, . . . , n}. The exterior algebra is a “skew-commutative” Aalgebra.
Better: both Sym and ∧ can be defined by universal properties. For example, any map of A-modules M⊗n → N that is symmetric in the n entries factors
uniquely through Symn
A (M).
•
It is most correct to write TA(M), Sym•A (M), and ∧•A (M), but the “base ring”
A is usually omitted for convenience.
13.5.D. E XERCISE . Suppose F is a quasicoherent sheaf. Define the quasicoherent
sheaves T n F , Symn F , and ∧n F . (One possibility: describe them on each affine
open set, and use the characterization of Important Exercise 13.3.D.) If F is locally

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free of rank m, show that T n F , Symn F , and ∧n F are locally free, and find theirranks. (Remark: These constructions can be defined for O-modules on an arbitrary
ringed space.) We note that in this case, ∧rank F F is denoted det F , and is called
the determinant (line) bundle or (both better and worse) the determinant locally
free sheaf.
You can also define the sheaf of noncommutative algebras T • F , the sheafof commutative algebras Sym• F , and the sheaf of skew-commutative algebras∧• F .
13.5.E. E XERCISE . Suppose 0 → F ′ → F → F ′′ → 0 is an exact sequence of
locally free sheaves. Show that for any r, there is a filtration of Symr Fwith subquotients

Symr F = G 0 ⊃ G 1 ⊃ · · · ⊃ G r ⊃ G r+1 = 0∼ (Symp F ′ ) ⊗ (Symr−p F ′′ ).
G p /G p+1 =(Here are two different possible hints for this and Exercise 13.5.F: (1) Interpretthe transition matrices for F as block upper triangular, with two blocks, where
one diagonal block gives the transition matrices for F ′ , and the other gives thetransition matrices for F ′′ (cf. Exercise 13.5.1.1(b)). Then appropriately interpretthe transition matrices for Symr F as block upper triangular, with r + 1 blocks. (2)It suffices to consider a small enough affine open set Spec A, where F ′ , F , F ′′ arefree, and to show that your construction behaves well with respect to localization
at an element f ∈ A. In such an open set, the sequence is 0 → A⊕p → A⊕(p+q) →A⊕q → 0 by the Exercise 13.5.A. Let e1 , . . . , ep be the standard basis of A⊕p , andf1 , . . . , fq be the the standard basis of A⊕q . Let e1′ , . . . , ep′ be denote the images of
e1 , . . . , ep in A⊕(p+q) . Let f1′ , . . . , fq′ be any lifts of f1 , . . . , fq to A⊕(p+q) . Note thatfi′ is well-defined modulo e1′ , . . . , ep′ . Note that∼ ⊕r Symi F ′ |Spec A ⊗O
Symr F |Spec A =Symr−i F ′′ |Spec A .
i=0Spec AShow that G p := ⊕ri=p Symi F ′ |Spec A ⊗OSpec A Symr−i F ′′ |Spec A gives a well-defined(locally free) subsheaf that is independent of the choices made, e.g., of the basis e1 ,. . . , ep , f1 , . . . , fq , and the lifts f1′ , . . . , fq′ .)13.5.F. U SEFUL E XERCISE . Suppose 0 → F ′ → F → F ′′ → 0 is an exact sequenceof locally free sheaves. Show that for any r, there is a filtration of ∧r F :with subquotients∧r F = G 0 ⊃ G 1 ⊃ · · · ⊃ G r ⊃ G r+1 = 0∼ (∧p F ′ ) ⊗ (∧r−p F ′′ )G p /G p+1 =for each p. In particular, if the sheaves have finite rank, then det F = (det F ′ ) ⊗(det F ′′ ).
13.5.G. E XERCISE . Suppose F is locally free of rank n. Describe a map ∧r F ×∨∧n−r F → ∧n F that induces an isomorphism ∧r F → (∧n−r F ) ⊗∧n F . This iscalled a perfect pairing of vector bundles. (If you know about perfect pairings ofvector spaces, do you see why this is a generalization?) You might use this later inshowing duality of Hodge numbers of regular varieties over algebraically closed
fields, Exercise 21.5.L.

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13.5.H. E XERCISE ( DETERMINANT LINE BUNDLES BEHAVE WELL IN EXACT SEQUENCES ).
Suppose 0 → F1 → · · · → Fn → 0 is an exact sequence of finite rank locally freesheaves on X. Show that “the alternating product of determinant bundles is trivial”:n∼ OX .
det(F1 ) ⊗ det(F2 )∨ ⊗ det(F3 ) ⊗ det(F4 )∨ ⊗ · · · ⊗ det(Fn )(−1) =(Hint: break the exact sequence into short exact sequences. Use Exercise 13.5.B(a)to show that they are short exact sequences of finite rank locally free sheaves. Thenuse Exercise 13.5.F.)
13.5.4. Important: Quasicoherent sheaves of ideals correspond to closed subschemes. Recall that if i : X "→ Y is a closed embedding, then we have a surjection% % i∗ OX (§8.1). (The i∗ is often omitted, as we are considof sheaves on Y: OY
ering the sheaf on X as being a sheaf on Y.) The kernel IX/Y is a “sheaf of ideals”
in Y: for each open subset U of Y, the sections form an ideal in the ring of functions
on U.
Compare (hard) Exercise 8.1.H and the characterization of quasicoherent sheaves
given in (possibly hard) Exercise 13.3.D. You will see that a sheaf of ideals is quasicoherent if and only if it comes from a closed subscheme. (An example of a
non-quasicoherent sheaf of ideals was given in Exercise 8.1.F.) We call(13.5.4.1)

0 → IX/Y → OY → i∗ OX → 0

the closed subscheme exact sequence corresponding to X "→ Y. See (13.4.1.1) foran example.
13.5.5. Torsion-free sheaves (a stalk-local condition) and torsion sheaves. An Amodule M is said to be torsion-free if am = 0 implies that either a is a zerodivisor
in A or m = 0.
In the case where A is an integral domain, which is basically the only context inwhich we will use this concept, the definition of torsion-freeness can be restated as
am = 0 only if a = 0 or m = 0. In this case, the torsion submodule of M, denoted
Mtors , consists of those elements of M annihilated by some nonzero element ofA. (If A is not an integral domain, this construction needn’t yield an A-module.)
Clearly M is torsion-free if and only if Mtors = 0. We say a module M over an
integral domain A is torsion if M = Mtors ; this is equivalent to M ⊗A K(A) = 0.
If X is a scheme, then an OX -module F is said to be torsion-free if Fp is atorsion-free OX,p -module for all p. (Caution: [Gr-EGA] calls this “strictly torsionfree”.)
13.5.6. Definition: torsion quasicoherent sheaves on reduced schemes. Motivated by thedefinition of Mtors above, we say that a quasicoherent sheaf on a reduced scheme istorsion if its stalk at the generic point of every irreducible component is 0. We will
mainly use this for coherent sheaves on regular curves, where this notion is very
simple indeed (see Exercise 13.7.G(b)), but in the literature it comes up in moregeneral situations.

13.6 Finite type and coherent sheaves

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Here are three natural finiteness conditions on an A-module M. In the casewhen A is a Noetherian ring, which is the case that almost all of you will ever careabout, they are all the same.
The first is the most naive: a module could be finitely generated. In otherwords, there is a surjection A⊕p → M → 0.
The second is reasonable too. It could be finitely presented — it could have afinite number of generators with a finite number of relations: there exists a finite
presentation, i.e., an exact sequenceA⊕q → A⊕p → M → 0.

13.6.A. E XERCISE (“ FINITELY PRESENTED IMPLIES ALWAYS FINITELY PRESENTED ”).
′
Suppose M is a finitely presented A-module, and φ : A⊕p → M is any surjection.
Show that ker φ is finitely generated. Hint: Write M as the quotient of A⊕p by
a finitely generated module K. Figure out how to map the short exact sequence′
0 → K → A⊕p → M → 0 to the exact sequence 0 → ker φ → A⊕p → M → 0, anduse the Snake Lemma (Example 1.7.5).
13.6.1. The third notion is frankly a bit surprising. We say that an A-module M iscoherent if (i) it is finitely generated, and (ii) whenever we have a map A⊕p → M(not necessarily surjective!), the kernel is finitely generated.
13.6.2. Proposition. — If A is Noetherian, then these three definitions are the same.

Proof. Clearly coherent implies finitely presented, which in turn implies finitelyα% M . Then
generated. So suppose M is finitely generated. Take any A⊕pker α is a submodule of a finitely generated module over A, and is thus finitely
generated by Exercise 3.6.X. Thus M is coherent.
!
Hence most people can think of these three notions as the same thing.
13.6.3. Proposition. — The coherent A-modules form an abelian subcategory of thecategory of A-modules.
The proof in general is given in §13.8 in a series of short exercises. You should
try them only if you are particularly curious.
Proof if A is Noetherian. Recall from our discussion at the start of §13.4 that we must
check three things:(i) The 0-module is coherent.
(ii) The category of coherent modules is closed under finite sums.
(iii) The category of coherent modules is closed under kernels and cokernels.
The first two are clear. For (iii), suppose that f : M → N is a map of finitely generated modules. Then coker f is finitely generated (it is the image of N), and ker f
is too (it is a submodule of a finitely generated module over a Noetherian ring,
Exercise 3.6.X).
!
13.6.B. ⋆ E ASY E XERCISE ( ONLY IMPORTANT FOR NON -N OETHERIAN PEOPLE ). Show
A is coherent as an A-module if and only if the notion of finitely presented agrees
with the notion of coherent.

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13.6.C. E XERCISE . If f ∈ A, show that if M is a finitely generated (resp. finitely
presented, coherent) A-module, then Mf is a finitely generated (resp. finitely presented, coherent) Af -module. (The “coherent” case is the tricky one.)
13.6.D. E XERCISE .
If (f1 , . . . , fn ) = A, and Mfi is a finitely generated (resp.
finitely presented, coherent) Afi -module for all i, then M is a finitely generated(resp. finitely presented, coherent) A-module. Hint for the finitely presented case:
Exercise 13.6.A.
13.6.4. Definition. A quasicoherent sheaf F is finite type (resp. finitely presented, coherent) if for every affine open Spec A, Γ (Spec A, F ) is a finitely generated (resp. finitely presented, coherent) A-module. Note that coherent sheaves
are always finite type, and that on a locally Noetherian scheme, all three notions
are the same (by Proposition 13.6.2). Proposition 13.6.3 implies that the coherent
sheaves on X form an abelian category, which we denote CohX .
Thanks to the Affine Communication Lemma 5.3.2, and the two previous exercises 13.6.C and 13.6.D, it suffices to check “finite typeness” (resp. finite presentation, coherence) on the open sets in a single affine cover. Notice that finite rank
locally free sheaves are always finite type, and if OX is coherent, finite rank locally
free sheaves on X are coherent. (If OX is not coherent, then coherence is a pretty
useless notion on X.)
13.6.5. Associated points of coherent sheaves. Our discussion of associated points
in §5.5 immediately implies a notion of associated point for a coherent sheaf on
a locally Noetherian scheme, with all the good properties described in §5.5. (Theaffine case was done there, and the only obstacle to generalizing them to coherent sheaves was that we didn’t know what coherent sheaves were.) The phrase
associated point of a locally Noetherian scheme X (without explicit mention ofa coherent sheaf) means “associated point of OX ”, and similarly for embedded
points.
13.6.6. Why coherence? Proposition 13.6.3 is a good motivation for the definition ofcoherence: it gives a small (in a non-technical sense) abelian category in which we
can think about vector bundles.
There are two sorts of people who should care about the details of this definition, rather than living in a Noetherian world where coherent means finite type.
Complex geometers should care. They consider complex-analytic spaces with theclassical topology. One can define the notion of coherent OX -module in a way
analogous to this (see [Se1, Def. 2]). Then Oka’s Theorem states that the structure
sheaf of Cn (hence of any complex manifold) is coherent, and this is very hard (see[GR, §2.5] or [Rem, §7.2]).
The second sort of people who should care are the sort of arithmetic people
who may need to work with non-Noetherian rings, see §3.6.21, or work in nonarchimedean analytic geometry.
Warning: it is not uncommon in the later literature to incorrectly define coherent as finitely generated. Please only use the correct definition, as the wrong
definition causes confusion. Besides doing this for the reason of honesty, it will
also help you see what hypotheses are actually necessary to prove things. And

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that always helps you remember what the proofs are — and hence why things aretrue.

13.7 Pleasant properties of finite type and coherent sheavesent.

We begin with an exercise that Hom behaves reasonably if the source is coher-

13.7.A. E XERCISE .
(a) Suppose F is a coherent sheaf on X, and G is a quasicoherent sheaf on X. Show
that Hom (F , G ) is a quasicoherent sheaf. Hint: Describe it on affine open sets, andshow that it behaves well with respect to localization with respect to f. To show
∼ HomA (Mf , Nf ), use Exercise 1.6.G. Up to here, you needthat HomA (M, N)f =f
only the fact that F is finitely presented. (Aside: For an example of quasicoherent
sheaves F and G on a scheme X such that Hom (F , G ) is not quasicoherent, let A( and
be a discrete valuation ring with uniformizer t, let X = Spec A, let F = M∞∞
3
G = N with M = ⊕i=1 A and N = A. Then Mt = ⊕i=1 At , and of course Nt = At .
Consider the homomorphism φ : Mt → Nt sending 1 in the ith factor of Mt to
1/ti . Then φ is not the localization of any element of HomA (M, N).)(b) If further G is coherent and OX is coherent, show that Hom (F , G ) is also coherent.
(c) Suppose F is a coherent sheaf on X, and G is a quasicoherent sheaf on X.
Show that Hom (F , ·) is a left-exact covariant functor QCohX → QCohX , and that
Hom (·, G ) is a left-exact contravariant functor CohX → QCohX (cf. Exercise 2.6.H).
(In fact left-exactness has nothing to do with coherence or quasicoherence — it istrue even for OX -modules, as remarked in §2.6.4. But the result is easier in the
category of quasicoherent sheaves.)
13.7.1. Duals of coherent sheaves. From Exercise 13.7.A(b), assuming OX is coherent, if F is coherent, its dual F ∨ := Hom (F , OX ) (see §2.3.4) is too. This generalizes the notion of duals of vector bundles in Exercise 13.1.C. Your argument
there generalizes to show that there is always a natural “double dual” morphism
F → (F ∨ )∨ . Unlike in the vector bundle case, this is not always an isomorphism. (For an example, let F be the coherent sheaf associated to k[t]/(t) on
A1 = Spec k[t], and show that F ∨ = 0.) Coherent sheaves for which the “double dual” morphism is an isomorphism are called reflexive sheaves, but we won’t
use this notion. The canonical map F ⊗ F ∨ → OX is called the trace map — canyou see why?
13.7.B. E XERCISE . Suppose F is a finite rank locally free sheaf, and G is a quasico∼ F ∨ ⊗ G . (This holds moreherent sheaf. Describe an isomorphism Hom (F , G ) =generally if G is an O-module, but we won’t use that, so you may as well prove
the simpler result given in this exercise.)
13.7.C. E XERCISE . Suppose(13.7.1.1)

0→F →G →H →0

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is an exact sequence of quasicoherent sheaves on a scheme X, where H is a locally
free quasicoherent sheaf, and suppose E is a quasicoherent sheaf. By left-exactnessof Hom (Exercise 2.6.H),
0 → Hom (H , E ) → Hom (G , E ) → Hom (F , E ) → 0

is exact except possibly on the right. Show that it is also exact on the right. (Hint:⊕n , so$this is local, so you can assume that X is affine, say Spec A, and H = A(13.7.1.1) can be written as 0 → M → N → A⊕n → 0. Show that this exact
sequence splits, so we can write N = M ⊕ A⊕n in a way that respects the exact
sequence.) In particular, if F , G , H , and OX are all coherent, and H is locally
free, then we have an exact sequence of coherent sheaves
0 → H ∨ → G ∨ → F ∨ → 0.

13.7.D. E XERCISE ( THE SUPPORT OF A FINITE TYPE QUASICOHERENT SHEAF IS
CLOSED ).
Suppose F is a sheaf of abelian groups. Recall Definition 2.4.2 ofthe support of a section s of F , and the definition (cf. Exercise 2.7.F(b)) of the support of F . (Support is a stalk-local notion, and hence behaves well with respect
to restriction to open sets, or to stalks. Warning: Support is where the germ(s) arenonzero, not where the value(s) are nonzero.) Show that the support of a finite type
quasicoherent sheaf on a scheme X is a closed subset. (Hint: Reduce to the case X
affine. Choose a finite set of generators of the corresponding module.) Show thatthe support of a quasicoherent sheaf need not be closed. (Hint: If A = C[t], thenC[t]/(t − a) is an A-module supported at a. Consider⊕a∈C C[t]/(t − a). Be careful:"
this example won’t work if ⊕ is replaced by .)13.7.2. Remark. In particular, if X is a locally Noetherian scheme, the sheaf ofnilpotents (Exercise 13.3.G) is coherent and hence finite type, and thus has closed
support. This makes precise the statement promised in §4.2.1, that in good (Noetherian) situations, the fuzz on a scheme is supported on a closed subset. Also, as
promised in Remark 5.2.2, if X is a locally Noetherian scheme, the reduced locus
forms an open subset. (We already knew all of this as of Remark 5.5.5, but now weknow it twice as well.)We next come to a geometric interpretation of Nakayama’s Lemma, which is
why Nakayama’s Lemma should be considered a geometric fact (with an algebraic
proof).
13.7.E. U SEFUL E XERCISE : G EOMETRIC N AKAYAMA ( GENERATORS OF A FIBER
GENERATE A FINITE TYPE QUASICOHERENT SHEAF NEARBY ).
Suppose X is ascheme, and F is a finite type quasicoherent sheaf. Show that if U ⊂ X is an open
neighborhood of p ∈ X and a1 , . . . , an ∈ F (U) so that their images a1 |p , . . . , an |pgenerate the fiber F |p (defined as Fp ⊗ κ(p), §4.3.7), then there is an affine open
neighborhood p ∈ Spec A ⊂ U of p such that “a1 |Spec A , . . . , an |Spec A generate
F |Spec A ” in the following senses:(i) a1 |Spec A , . . . , an |Spec A generate F (Spec A) as an A-module;(ii) for any q ∈ Spec A, a1 , . . . , an generate the stalk Fq as an OX,q -module(and hence for any q ∈ Spec A, the fibers a1 |q , . . . , an |q generate the fiber
F |q as a κ(q)-vector space).

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In particular, if Fp ⊗ κ(p) = 0, then there exists an open neighborhood V of p suchthat F |V = 0.
13.7.F. U SEFUL E XERCISE ( LOCAL FREENESS OF A FINITELY PRESENTED SHEAF IS
A STALK - LOCAL PROPERTY; AND FREE STALKS IMPLY LOCAL FREENESS NEARBY ).
Suppose F is a finitely presented sheaf on a scheme X. Show that if Fp is a freeOX,p -module for some p ∈ X, then F is locally free in some open neighborhoodof p. Hence F is locally free if and only if Fp is a free OX,p -module for all p ∈ X.
Hint: Find an open neighborhood U of p, and n elements of F (U) that generate
Fp . Using Geometric Nakayama, Exercise 13.7.E, show that the sections generate
Fq for all q in some open neighborhood Y of p in U. Thus you have described asurjection OY⊕n → F |Y . Show that the kernel of this map is finite type, and hence⊕n
has closed support (say Z ⊂ Y), which does not contain p. Thus OY\Z
→ F |Y\Z is
an isomorphism.

This is enlightening in a number of ways. It shows that for finitely presented
sheaves, local freeness is a stalk-local condition. Furthermore, on an integral scheme,
any coherent sheaf F is automatically free over the generic point (do you seewhy?), so every finitely presented sheaf on an integral scheme is locally free overa dense open subset. And any finitely presented sheaf that is 0 at the generic pointof an irreducible scheme is necessarily 0 on a dense open subset. The last two sentences show the utility of generic points; such statements would have been moremysterious in classical algebraic geometry.

13.7.G. E XERCISE . (Torsion-free and torsion sheaves were defined in §13.5.5.)(a) Show that torsion-free coherent sheaves on a regular (hence implicitly locally
Noetherian) curve are locally free.
(b) Show that torsion coherent sheaves on a quasicompact regular integral curve
are supported at a finite number of closed points.
(c) Suppose F is a coherent sheaf on a quasicompact (for convenience) regularcurve. Describe a canonical short exact sequence 0 → Ftors → F → Flf → 0,
where Ftor is a torsion sheaf, and Flf is locally free.

To answer Exercise 13.7.G, use Useful Exercise 13.7.F (local freeness can bechecked at stalks) to reduce to the discrete valuation ring case, and recall Remark 12.5.14, the structure theorem for finitely generated modules over a principal ideal domain A: any such module can be written as the direct sum of principal
modules A/(a). For discrete valuation rings, this means that the summands are of
the form A or A/mk . Hence:

13.7.3. Proposition. — If M is a finitely generated module over a discrete valuation
ring, then M is torsion-free if and only if M is free.
(Exercise 24.2.B is closely related.)
Proposition 13.7.3 is false without the finite generation hypothesis: consider
M = K(A) for a suitably general ring A. It is also false if we give up the “dimension 1” hypothesis: consider (x, y) ⊂ C[x, y]. And it is false if we give up
the “regular” hypothesis: consider (x, y) ⊂ C[x, y]/(xy). (These examples require
some verification.) Hence Exercise 13.7.G(a) is false if we give up the “dimension
1” or “regular” hypothesis.

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13.7.4. Rank of a quasicoherent sheaf at a point.
Suppose F is a quasicoherent sheaf on a scheme X, and p is a point of X. The
vector space F |p := Fp /mFp = Fp ⊗OX,p κ(p) can be interpreted as the fiber ofthe sheaf at the point, where m is the maximal ideal of OX,p , and κ(p) is as usual
the residue field OX,p /m at p. A section of F over an open set containing p can besaid to take on a value at that point, which is an element of this vector space. The
rank of a quasicoherent sheaf F at a point p is dimκ(p) Fp /mFp (possibly infinite). More explicitly, on any affine set Spec A where p = [p] and F (Spec A) = M,then the rank is dimK(A/p) Mp /pMp . Note that this definition of rank is consistent
with the notion of rank of a locally free sheaf. In the locally free case, the rank
is a (locally) constant function of the point. The converse is sometimes true, see
Exercise 13.7.K below.
If X is irreducible, and F is a quasicoherent (usually coherent) sheaf on X,
then rank F (with no mention of a point) by convention means at the genericpoint. (For example, a rank 0 quasicoherent sheaf on an integral scheme is a torsion
quasicoherent sheaf, see Definition 13.5.6.)
13.7.H. E XERCISE . Consider the coherent sheaf F on A1k = Spec k[t] corresponding to the module k[t]/(t). Find the rank of F at every point of A1 . Don’t forget
the generic point!
13.7.I. E XERCISE . Show that at any point, rank(F ⊕ G ) = rank(F ) + rank(G ) andrank(F ⊗ G ) = rank F rank G . (Hint: Show that direct sums and tensor products∼ M/IM ⊕commute with ring quotients and localizations, i.e., (M ⊕ N) ⊗R (R/I) =∼ (M ⊗R R/I) ⊗R/I (N ⊗R R/I) =∼ M/IM ⊗R/I N/IM,
N/IN, (M ⊗R N) ⊗R (R/I) =etc.)
If F is a finite type quasicoherent sheaf, then the rank of F at p is finite, andby Nakayama’s Lemma, the rank is the minimal number of generators of Mp as
an Ap -module.
13.7.J. I MPORTANT E XERCISE . If F is a finite type quasicoherent sheaf on X, showthat rank(F ) is an upper semicontinuous function on X. Hint: generators at apoint p are generators nearby by Geometric Nakayama’s Lemma, Exercise 13.7.E.
(The example in Exercise 13.7.D shows the necessity of the finite type hypothesis.)13.7.K. I MPORTANT H ARD E XERCISE .
(a) If X is reduced, F is a finite type quasicoherent sheaf on X, and the rank isconstant, show that F is locally free. Then use upper semicontinuity of rank (Exercise 13.7.J) to show that finite type quasicoherent sheaves on an integral scheme
are locally free on a dense open set. (By examining your proof, you will see that the
integrality hypothesis can be relaxed. In fact, reducedness is all that is necessary.)
Hint: Reduce to the case where X is affine. Then show it in an open neighborhood of an arbitrary point p as follows. Suppose n = rank F . Choose n generators of the fiber F |p (a basis as an κ(p)-vector space). By Geometric Nakayama’s
Lemma 13.7.E, we can find a smaller open neighborhood p ∈ Spec A ⊂ X, with( so that the chosen generators F |p lift to generators m1 , . . . , mn of
F |Spec A = M,,
M. Let φ : A⊕n → M with (r1 , . . . , rn ) (→ri mi . If ker φ ̸= 0, then suppose(r1 , . . . , rn ) is in the kernel, with r1 ̸= 0. As r1 ̸= 0, there is some p where r1 ∈/ p—here we use the reduced hypothesis. Then r1 is invertible in Ap , so Mp has fewer

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than n generators, contradicting the constancy of rank.
(b) Show that part (a) can be false without the condition of X being reduced. (Hint:
Consider A = k[x]/(x2 ), M = k.)
You can use the notion of rank to help visualize finite type quasicoherent
sheaves, or even quasicoherent sheaves. For example, I think of a coherent sheaf
as generalizing a finite rank vector bundle as follows: to each point there is anassociated vector space, and although the ranks can jump, they fit together infamilies as well as one might hope. You might try to visualize the example of
Example 13.7.H. Nonreducedness can fit into the picture as well — how would
you picture the coherent sheaf on Spec k[ϵ]/(ϵ2 ) corresponding to k[ϵ]/(ϵ)? Howabout k[ϵ]/(ϵ2 ) ⊕ k[ϵ]/(ϵ)?
13.7.5. Degree of a finite morphism at a point. Suppose π : X → Y is a finite morphism.
Then π∗ OX is a finite type (quasicoherent) sheaf on Y, and the rank of this sheaf at
a point p is called the degree of the finite morphism at p. By Exercise 13.7.J, thedegree of π is a upper semicontinuous function on Y. The degree can jump: consider the closed embedding of a point into a line corresponding to k[t] → k given
by t (→ 0. It can also be constant in cases that you might initially find surprising —see Exercise 9.3.3, where the degree is always 2, but the 2 is obtained in a numberof different ways.
13.7.L. E XERCISE . Suppose π : X → Y is a finite morphism. By unwinding the definition, verify that the degree of π at p is the dimension of the space of functions of
the scheme-theoretic preimage of p, considered as a vector space over the residue
field κ(p). In particular, the degree is zero if and only if π−1 (p) is empty.

13.8 ⋆⋆ Coherent modules over non-Noetherian rings
This section is intended for people who might work with non-Noetherian rings,
or who otherwise might want to understand coherent sheaves in a more general
setting. Read this only if you really want to!
Suppose A is a ring. Recall the definition of when an A-module M is finitely
generated, finitely presented, and coherent. The reason we like coherence is that
coherent modules form an abelian category. Here are some accessible exercises
working out why these notions behave well. Some repeat earlier discussion inorder to keep this section self-contained.
The notion of coherence of a module is only interesting in the case that a ring iscoherent over itself. Similarly, coherent sheaves on a scheme X will be interesting
only when OX is coherent (“over itself”). In this case, coherence is clearly thesame as finite presentation. An example where non-Noetherian coherence comes
up is the ring R⟨x1 , . . . , xn ⟩ of “restricted power series” over a valuation ring R ofa non-discretely valued K (for example, a completion of the algebraic closure of
Qp ). This is relevant to Tate’s theory of non-archimedean analytic geometry overK (which you can read about in [BCDKT], for example). The importance of thecoherence of the structure sheaf underlines the importance of Oka’s Theorem incomplex geometry (stated in §13.6.5).

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13.8.A. E XERCISE . Show that coherent implies finitely presented implies finitely
generated. (This was discussed at the start of §13.6.)
13.8.B. E XERCISE . Show that 0 is coherent.
Suppose for problems 13.8.C–13.8.I that
(13.8.0.1)

0→M→N→P→0

is an exact sequence of A-modules. In this series of problems, we will show that if
two of {M, N, P} are coherent, the third is as well, which will prove very useful.
13.8.1. Hint †. The following hint applies to several of the problems: try to write
0

% A⊕p

% A⊕(p+q)

% A⊕q

%0

0

"
%M

"
%N

"
%P

%0

and possibly use the Snake Lemma 1.7.5.
13.8.C. E XERCISE . Show that N finitely generated implies P finitely generated.
(You will only need right-exactness of (13.8.0.1).)
13.8.D. E XERCISE . Show that M, P finitely generated implies N finitely generated.
Possible hint: †. (You will only need right-exactness of (13.8.0.1).)
13.8.E. E XERCISE . Show that N, P finitely generated need not imply M finitely
generated. (Hint: if I is an ideal, we have 0 → I → A → A/I → 0.)

13.8.F. E XERCISE . Show that N coherent, M finitely generated implies M coherent.
(You will only need left-exactness of (13.8.0.1).)
13.8.G. E XERCISE . Show that N, P coherent implies M coherent. Hint for (i) in thedefinition of coherence (§13.6.1):
A⊕q●
●●
●●
●●
●$

0

"
%M
"
0

A⊕p❈
❈❈
❈❈
❈❈
❈❈
"
9
%N
%P
"
0

%0
❃❃
❃❃
❃❃
❃❃
F
0

(You will only need left-exactness of (13.8.0.1).)
13.8.H. E XERCISE . Show that M finitely generated and N coherent implies P
coherent. Hint for (ii) in the definition of coherence (§13.6.1) : †.
13.8.I. E XERCISE . Show that M, P coherent implies N coherent. (Hint: †.)13.8.J. E XERCISE . Show that a finite direct sum of coherent modules is coherent.

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13.8.K. E XERCISE . Suppose M is finitely generated, N coherent. Then if φ : M →N is any map, then show that Im φ is coherent.
13.8.L. E XERCISE . Show that the kernel and cokernel of maps of coherent modules
are coherent.

At this point, we have verified that coherent A-modules form an abelian subcategory of the category of A-modules. (Things you have to check: 0 should bein this set; it should be closed under finite sums; and it should be closed undertaking kernels and cokernels.)
13.8.M. E XERCISE . Suppose M and N are coherent submodules of the coherent
module P. Show that M + N and M ∩ N are coherent. (Hint: consider the rightmap M ⊕ N → P.)

13.8.N. E XERCISE . Show that if A is coherent (as an A-module) then finitely presented modules are coherent. (Of course, if finitely presented modules are coherent, then A is coherent, as A is finitely presented!)
13.8.O. E XERCISE . If M is finitely presented and N is coherent, show that Hom(M, N)
is coherent. (Hint: Hom is left-exact in its first argument.)
13.8.P. E XERCISE . If M is finitely presented, and N is coherent, show that M ⊗A N
is coherent.
13.8.Q. E XERCISE . If f ∈ A, show that if M is a finitely generated (resp. finitely
presented, coherent) A-module, then Mf is a finitely generated (resp. finitely presented, coherent) Af -module. (Hint: localization is exact, Exercise 1.6.F(a).) This
exercise is repeated from Exercise 13.6.C to make this section self-contained.
13.8.R. E XERCISE . Suppose (f1 , . . . , fn ) = A. Show that if Mfi is finitely generated
for all i, then M is too. (Hint: Say Mfi is generated by mij ∈ M as an Afi -module.
Show that the mij generate M. To check surjectivity ⊕i,j A → M, it suffices tocheck “on D(fi )” for all i.)
13.8.S. E XERCISE . Suppose (f1 , . . . , fn ) = A. Show that if Mfi is coherent for all i,then M is too. (Hint: if φ : A⊕p → M, then (ker φ)fi = ker(φfi ), which is finitely
generated for all i. Then apply the previous exercise.)

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CHAPTER 14

Line bundles: Invertible sheaves and divisors
We next describe convenient and powerful ways of working with and classifying line bundles (invertible sheaves). We begin with a fundamental example, theline bundles O(n) on projective space, §14.1. We then introduce Weil divisors (formal sums of codimension 1 subsets), and use them to determine Pic X in a numberof circumstances, §14.2. We finally discuss sheaves of ideals that happen to be invertible (effective Cartier divisors), §14.3. A central theme is that line bundles areclosely related to “codimension 1 information”.

14.1 Some line bundles on projective space
We now describe an important family of invertible sheaves on projective spaceover a field k.
As a warm-up, we begin with the invertible sheaf OP1k (1) on P1k = Proj k[x0 , x1 ].
The subscript P1k refers to the space on which the sheaf lives, and is often omitted
when it is clear from the context. We describe the invertible sheaf O(1) using transition functions. It is trivial on the usual affine open sets U0 = D(x0 ) = Spec k[x1/0 ]and U1 = D(x1 ) = Spec k[x0/1 ]. (We continue to use the convention xi/j for describing coordinates on patches of projective space, see §4.4.9.) Thus the data ofa section over U0 is a polynomial in x1/0 . The transition function from U0 to U1
is multiplication by x0/1 = x−1
1/0 . The transition function from U1 to U0 is hence
multiplication by x1/0 = x−1
0/1 .
This information is summarized below:
open coverU0 = Spec k[x1/0 ]U1 = Spec k[x0/1 ]×x0/1 =x−1
1/0

trivialization and transition functions

k[x1/0 ] ZY

k[x0/1 ]×x1/0 =x−1
0/1

To test our understanding, let’s compute the global sections of O(1). This will
∼ k (Example 4.4.6). A global
generalize our hands-on calculation that Γ (P1k , OP1k ) =section is a polynomial f(x1/0 ) ∈ k[x1/0 ] and a polynomial g(x0/1 ) ∈ k[x0/1 ] suchthat f(1/x0/1 )x0/1 = g(x0/1 ). A little thought will show that f must be linear:f(x1/0 ) = ax1/0 + b, and hence g(x0/1 ) = a + bx0/1 . Thusdim Γ (P1k , O(1)) = 2 ̸= 1 = dim Γ (P1k , O).
397

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Thus O(1) is not isomorphic to O, and we have constructed our first (proved) example of a nontrivial line bundle!
We next define more generally OP1k (n) on P1k . It is defined in the same way,
except that the transition functions are the nth powers of those for O(1).

open covertrivialization and transition functions

U0 = Spec k[x1/0 ]U1 = Spec k[x0/1 ]−n×xn
0/1 =x1/0

k[x1/0 ] ZY

k[x0/1 ]−n×xn
1/0 =x0/1

In particular, thanks to the explicit transition functions, we see that O(n) = O(1)⊗n(with the obvious meaning if n is negative: (O(1)⊗(−n) )∨ ). Clearly also O(m) ⊗O(n) = O(m + n).
14.1.A. I MPORTANT E XERCISE . Show that dim Γ (P1 , O(n)) = n + 1 if n ≥ 0, and 0
otherwise.
14.1.1. Example. Long ago (in Exercise 2.6.J(a)), we warned that sheafification
was necessary when tensoring OX -modules: if F and G are two OX -modules on a
∼ (F ⊗ G )(X).
ringed space, then it is not necessarily true that F (X) ⊗OX (X) G (X) =1
We now have an example: let X = Pk , F = O(1), G = O(−1), and use the fact thatO(−1) has no nonzero global sections.
∼ O(n). Hence conclude that
14.1.B. E XERCISE . Show that if m ̸= n, then O(m) ̸=
we have an injection of groups Z "→ Pic P1k given by n (→ O(n).

It is useful to identify the global sections of O(n) with the homogeneous polynomials of degree n in x0 and x1 , i.e., with the degree n part of k[x0 , x1 ] (cf.
§14.1.2 for the generalization to Pm ). Can you see this from your solution to Exercise 14.1.A? We will see that this identification is natural in many ways. For
example, you can show that the definition of O(n) doesn’t depend on a choice ofaffine cover, and this polynomial description is also independent of cover. (Forthis, see Example 4.5.12; you can later compare this to Exercise 28.1.M.) As an immediate check of the usefulness of this point of view, ask yourself: where does thesection x30 − x0 x21 of O(3) vanish? The section x0 + x1 of O(1) can be multiplied by
the section x20 of O(2) to get a section of O(3). Which one? Where does the rational
section x40 (x1 + x0 )/x71 of O(−2) have zeros and poles, and to what order? (Wesaw the notion of zeros and poles in Definition 12.5.7, and will meet them again in§14.2, but you should intuitively answer these questions already.)
We now define the invertible sheaf OPm(n) on the projective space Pm
k . On thekusual affine open set Ui = Spec k[x0/i , . . . , xm/i ]/(xi/i − 1) = Spec Ai , it is trivial,
so sections (as an Ai -module) are isomorphic to Ai . The transition function from

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Ui to Uj is multiplication by xn
i/j = xj/i .

Ui = Spec k[x0/i , . . . , xm/i ]/(xi/i − 1)

Uj = Spec k[x0/j , . . . , xm/j ]/(xj/j − 1)−n×xni/j =xj/ik[x0/i , . . . , xm/i ]/(xi/i − 1) \[k[x0/j , . . . , xm/j ]/(xj/j − 1)−n×xn
j/i =xi/j

Note that these transition functions clearly satisfy the cocycle condition.
m
14.1.C. E SSENTIAL EXERCISE ( CF. E XERCISE 8.2.K). Show that dimk Γ (Pmk , OPk (n)) =
)m+n*
m .

14.1.2. As in the case of P1 , sections of O(n) on Pm
k are naturally identified withhomogeneous degree n polynomials in our m + 1 variables. (Important question:
Do you see why? Can you work out this dictionary?) Thus x + y + 2z is a sectionof O(1) on P2 . It isn’t a function, but we know where this section vanishes —precisely where x + y + 2z = 0.
)*
Also, notice that for fixed m, m+n
is a polynomial in n of degree m formn ≥ 0 (or better: for n ≥ −m). This should be telling you that this function“wants to be a polynomial,” but won’t succeed without assistance. We will latermdefine h0 (Pmk , O(n)) := dimk Γ (Pk , O(n)), and later still we will define higher
cohomology
groups, and we will define the Euler characteristic χ(Pmk , O(n)) :=
,∞
i i m
(−1)
h
(P
,
O(n))
(cohomology
willvanishindegreehigherthanm). Wek
i=0will discover the moral that the Euler characteristic is better-behaved than h0 , andso we should now suspect (and later prove, see Theorem 18.1.3, Remark r:preproof,
and Exercise 18.3.B) that this polynomial is in fact the Euler characteristic, and thereason that it agrees with h0 for n ≥ 0 is because all the other cohomology groups
should vanish.
We finally note that we can define O(n) on Pm
A for any ring A: the above
definition applies without change.
14.1.3. These are the only line bundles on Pm
k .
Suppose that k is a field. We will see in §14.2.9 that these O(n) are the only
invertible sheaves on Pm
k . The next Exercise shows this when m = 1, although thisapproach will soon be trumped.
14.1.D. E XERCISE . Show that every invertible sheaf on P1k is of the form O(n) forsome n. Hint: use the classification of finitely generated modules over a principal
ideal domain (Remark 12.5.14) to show that all invertible sheaves on A1k are trivial (a special case of Exercise 13.2.C). Reduce to determining possible transition
functions between the two open subsets in the standard cover of P1k .
Caution: there can exist invertible sheaves on P1A not of the form O(n). You
may later be able to think of examples. (Hints to find an example for when you
know more: what if Spec A is disconnected? Or if that leads to too silly an example,
what if Spec A has nontrivial invertible sheaves?)

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14.2 Line bundles and Weil divisors
The notion of Weil divisors gives a great way of understanding and classifying
line bundles, at least on Noetherian normal schemes. Some of what we discuss will
apply in more general circumstances, and the expert is invited to consider generalizations by judiciously weakening hypotheses in various statements. Before we
get started, you should be warned: this is one of those topics in algebraic geometry that is hard to digest — learning it changes the way in which you think aboutline bundles. But once you become comfortable with the imperfect dictionary todivisors, it becomes second nature.
For the rest of this section, we consider only Noetherian schemes. We do this because we will use finite decomposition into irreducible components (Exercise 5.3.B),
and Algebraic Hartogs’s Lemma 11.3.11.
Define a Weil divisor as a formal Z-linear combination of codimension 1 irreducible closed subsets of X. In other words, a Weil divisor is defined to be an
object of the form.
nY [Y]Y ⊂ X codimension 1

where the nY are integers, all but a finite number of which are zero. Weil divisors
obviously form an abelian group, denoted Weil X. For example, if X is a curve, theWeil divisors are linear combinations of closed points.
,
We say that [Y] is an irreducible (Weil) divisor. A Weil divisor D =nY [Y]
is said to be effective if nY ≥ 0 for all Y. In this case we say D ≥ 0, and by
D1 ≥ D2 we mean D1 − D2 ≥ 0. The support of a Weil divisor D, denoted
Supp D, is the subset ∪nY,
̸=0 Y. If U ⊂,X is an open set, we define the restriction
map Weil X → Weil U by nY [Y] (→ Y∩U̸=∅ nY [Y ∩ U].
Suppose now that X is regular in codimension 1. We add this hypothesis because
we will use properties of discrete valuation rings. Assume also that X is reduced.
(This is only so we can talk about rational functions without worrying about them
being defined at embedded points. Feel free to relax this hypothesis.) Supposethat L is an invertible sheaf, and s a rational section not vanishing everywhere on
any irreducible component of X. (Rational sections are given by a section over adense open subset of X, with the obvious equivalence, §13.1.7.) Then s determinesa Weil divisor.
div(s) :=valY (s)[Y]Y

where the summation runs over all irreducible divisors Y of X. We call div(s)the divisor of zeros and poles of the rational section s (cf. Definition 12.5.7). To
determine the valuation valY (s) of s along Y, take any open set U containing thegeneric point of Y where L is trivializable, along with any trivialization over U;
under this trivialization, s is a nonzero rational function on U, which thus has avaluation. Any two such trivializations differ by an invertible function (transition
functions are invertible), so this valuation is well-defined. Note that valY (s) = 0
for all but finitely many Y, by Exercise 12.5.G.
Now consider the set {(L , s)} of pairs of line bundles L with nonzero rational
sections s of L , up to isomorphism. This set (after taking quotient by isomorphism)
forms an abelian group under tensor product ⊗, with identity (OX , 1). (Tricky
question: what is the inverse of (L , s) in this group?) It is important to notice that

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401

if t is an invertible function on X, then multiplication by t gives an isomorphism∼ (L , st). Similarly, (L , s)/(L , u) = (O, s/u). The map div yields a group(L , s) =homomorphism(14.2.0.1)

div : {(L , s)}/isomorphism → Weil X.

14.2.A. E ASIER E XERCISE .
(a) (divisors of rational functions) Verify that on A1k , div(x3 /(x+1)) = 3[(x)]−[(x+1)](“= 3[0] − [−1]”).
(b) (divisor of rational sections of a nontrivial invertible sheaf) On P1k , there is a rational
section of O(1) “corresponding to” x2 /(x + y). Figure out what this means, andcalculate div(x2 /(x + y)).
The homomorphism (14.2.0.1) will be the key to determining all the line bundles on many X. (Note that any invertible sheaf will have such a rational section.
For each irreducible component, take a nonempty open set not meeting any other
irreducible component; then shrink it so that L is trivial; choose a trivialization;then take the union of all these open sets, and choose the section on this union corresponding to 1 under the trivialization.) We will see that in reasonable situations,this map div will be injective, and often an isomorphism. Thus by forgetting therational section (i.e., taking an appropriate quotient), we will have described thePicard group of all line bundles. Let’s put this strategy into action.
14.2.1. Proposition. — If X is normal and Noetherian then the map div is injective.
Proof. Suppose div(L , s) = 0. Then s has no poles. By Exercise 13.1.K, s is a regularsection. We now show that the morphism ×s : OX → L is in fact an isomorphism;∼ (L , s).
this will prove the Proposition, as it will give an isomorphism (OX , 1) =It suffices to show that ×s is an isomorphism on an open subset U of X where
L is trivial, as X is covered by trivializing open neighborhoods of L (as L is
locally trivial). Choose an isomorphism i : L |U → OU . Composing ×s with iyields a map ×s ′ : OU → OU that is multiplication by a rational function s ′ = i(s)that has no zeros and no poles. The rational function s ′ is regular because it has
no poles (Exercise 12.5.H), and 1/s ′ is regular for the same reason. Thus s ′ is aninvertible function on U, so ×s ′ is an isomorphism. Hence ×s is an isomorphism
over U.
!
Motivated by this, we try to find an inverse to div, or at least to determine theimage of div.
14.2.2. Important Definition. Assume now that X is irreducible (purely to avoid
making (14.2.2.1) look uglier — but feel free to relax this, see Exercise 14.2.B). Assume also that X is normal — this will be a standing assumption for the rest of
this section. Suppose D is a Weil divisor. Define the sheaf OX (D) by(14.2.2.1)

Γ (U, OX (D)) := {t ∈ K(X)× : div |U t + D|U ≥ 0} ∪ {0}.

Here div |U t means take the divisor of t considered as a rational function on U, i.e.,
consider just the irreducible divisors of U. (The subscript X in OX (D) is omitted
when it is clear from context.) The sections of OX (D) over U are the rational functions on U that have poles and zeros “constrained by D”: a positive coefficient

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The Rising Sea: Foundations of Algebraic Geometryin D allows a pole of that order; a negative coefficient demands a zero of that order. Away from the support of D, this is (isomorphic to) the structure sheaf (byAlgebraic Hartogs’s Lemma 11.3.11).
14.2.3. Remark. It will be helpful to note that OX (D) comes along with a canonical“rational section” corresponding to 1 ∈ K(X)× . (It is a rational section in the sensethat it is a section over a dense open set, namely the complement of Supp D.)
14.2.B. L ESS IMPORTANT EXERCISE . Generalize the definition of OX (D) to the casewhen X is not necessarily irreducible. (This is just a question of language. Onceyou have done this, feel free to drop this hypothesis in the rest of this section.)
14.2.C. E ASY E XERCISE . Verify that OX (D) is a quasicoherent sheaf. (Hint: thedistinguished affine criterion for quasicoherence of Exercise 13.3.D.)In good situations, OX (D) is an invertible sheaf. For example, let X = A1k .
ConsiderOX (−2[(x)] + [(x − 1)] + [(x − 2)]) ,often written O(−2[0] + [1] + [2]) for convenience. Then 3x3 /(x − 1) is a global
section; it has the required two zeros at x = 0 (and even one to spare), and takes
advantage of the allowed pole at x = 1, and doesn’t have a pole at x = 2, eventhough one is allowed. (Unimportant aside: the statement remains true in characteristic 2, although the explanation requires editing.)
14.2.D. E ASY E XERCISE . (This is a consequence of later discussion as well, but
you should be able to do this by hand.)(a) Show that any global section of OA1k (−2[(x)] + [(x − 1)] + [(x − 2)]) is a k[x]multiple of x2 /(x − 1)(x − 2).
(b) Extend the argument of (a) to give an isomorphism∼ O 1.
OA1k (−2[(x)] + [(x − 1)] + [(x − 2)]) =Ak
In the next several exercises, we show that in good circumstances, OX (D) is aninvertible sheaf. (In fact the OX (D) construction can be useful even if OX (D) is not
an invertible sheaf, but this won’t concern us here. An example of an OX (D) thatis not an invertible sheaf is given in Exercise 14.2.H.)
14.2.E. H ARD BUT I MPORTANT E XERCISE . Suppose L is an invertible sheaf, ands is a nonzero rational section of L .
∼ L . (You will use the normality hypoth(a) Describe an isomorphism O(div s) =∼ OU form aesis!) Hint: show that those open subsets U for which O(div s)|U =base for the Zariski topology. For each such U, define φU : O(div s)(U) → L (U)sending a rational function t (with zeros and poles “constrained by div s”) to st.
Show that φU is an isomorphism (with the obvious inverse map, division by s).
Argue that this map induces an isomorphism of sheaves φ : O(div s) → L .
(b) Let σ be the map from K(X) to the rational sections of L , where σ(t) is the ra∼ L defined via (14.2.2.1) (as described in Remark 14.2.3).
tional section of OX (D) =Show that the isomorphism of (a) can be chosen such that σ(1) = s. (Hint: the map
in part (a) sends 1 to s.)

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14.2.F. E XERCISE ( THE EXAMPLE OF §14.1). Suppose X = Pn
k , L = O(1), s is∼
the section of O(1) corresponding to x0 , and D = div s. Verify that O(mD) =m
O(m), and the canonical rational section of O(mD) is precisely s . (Watch out forpossible confusion: 1 has no pole along x0 = 0, but σ(1) = sm does have a zero if
m > 0.) For this reason, O(1) is sometimes called the hyperplane class in Pic X.
(Of course, x0 can be replaced by any linear form.)
14.2.4. Definition. If D is a Weil divisor on (Noetherian normal irreducible) X suchthat D = div f for some rational function f, we say that D is principal. Principal
divisors clearly form a subgroup of Weil X; denote this group of principal divisors
Prin X. Note that div induces a group homomorphism K(X)× → Prin X. If X can becovered with open sets Ui such that on Ui , D is principal, we say that D is locallyprincipal. Locally principal divisors form a subgroup of Weil X, which we denote
LocPrin X. (This notation is not standard.)
14.2.5. Important observation.
As a consequence of Exercise 14.2.E(a) (taking∼ O.
L = O), if D is principal (and X is normal, a standing hypothesis), then O(D) =∼(Diagram (14.2.7.1) will imply that the converse holds: if O(D) = O, then D is principal.) Thus if D is locally principal, OX (D) is locally isomorphic to OX — OX (D) is
an invertible sheaf.
14.2.G. I MPORTANT E XERCISE . Suppose OX (D) is an invertible sheaf.
(a) Show that div(σ(1)) = D, where σ was defined in Exercise 14.2.E(b).
(b) Show the converse to Observation 14.2.5: show that D is locally principal.
14.2.6. Remark. In definition (14.2.2.1), it may seem cleaner to consider those ssuch that div s ≥ D|U . The reason for the convention comes from our desire thatdiv σ(1) = D. (Taking the “opposite” convention would yield the dual bundle, inthe case where D is locally principal.)
14.2.H. L ESS IMPORTANT EXERCISE : A W EIL DIVISOR THAT IS NOT LOCALLY PRIN CIPAL . Let X = Spec k[x, y, z]/(xy − z2 ), a cone, and let D be the line z = x = 0(see Figure 12.1).
(a) Show that D is not locally principal. (Hint: consider the stalk at the origin. Use
the Zariski tangent space, see Problem 12.1.3.) In particular OX (D) is not an invertible sheaf.
(b) Show that div(x) = 2D. This corresponds to the fact that the plane x = 0 istangent to the cone X along D.
14.2.I. I MPORTANT E XERCISE . If X is Noetherian and factorial, show that for
any Weil divisor D, O(D) is an invertible sheaf. (Hint: It suffices to deal withthe case where D is irreducible, say D = [Y], and to cover X by open sets so thaton each open set U there is a function whose divisor is [Y ∩ U]. One open set
will be X − Y. Next, we find an open set U containing an arbitrary p ∈ Y, and afunction on U. As OX,p is a unique factorization domain, the prime corresponding
to Y is codimension 1 and hence principal by Lemma 11.1.6. Let f be a generatorof this prime ideal, interpreted as an element of K(X). It is regular at p, it has afinite number of zeros and poles, and through p, [Y] is the “only zero” (the only
component of the divisor of zeros). Let U be X minus all the other zeros andpoles.)

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14.2.7. The class group. We can now get a handle on the Picard group of anormal Noetherian scheme. Define the class group of X, Cl X, by Weil X/ Prin X.
By taking the quotient of the inclusion (14.2.0.1) by Prin X, we have the inclusion
Pic X "→ Cl X. This is summarized in the convenient and enlightening diagram.
(O(D),σ(1))←!D]{(L , s)}/iso.

(14.2.7.1)% LocPrin X ! "∼div% Weil X/ Prin X"{L }/iso.

Pic X

∼

^

/ Prin X"% LocPrin X/ Prin X ! ""% Cl X

O(D)←!DThis diagram is very important, and although it is short to state, it takes time todigest.
In particular, if A is a unique factorization domain, then all Weil divisors on
Spec A are principal by Lemma 11.1.6, so Cl Spec A = 0, and hence Pic Spec A = 0.
n14.2.8. As k[x1 , . . . , xn ] has unique factorization, Cl(Ank ) = 0, so Pic(Ak ) = 0 .
nGeometers might find this believable — “C is a contractible manifold, and hence
should have no nontrivial line bundles” — even if some caution is in order, as
the kinds of line bundles being considered are entirely different: holomorphic vs.
topological or C∞ . (Aside: for this reason, you might expect that An
k also has nonontrivial vector bundles. This is the Quillen-Suslin Theorem, formerly known as
Serre’s Conjecture, part of Quillen’s work leading to his 1978 Fields Medal. The
case n = 1 was Exercise 13.2.C. For a short proof by Vaserstein, see [Lan, p. 850].)
Removing a closed subset of X of codimension greater than 1 doesn’t change
the class group, as it doesn’t change the Weil divisor group or the principal divisors. (Warning: it can affect the Picard group, see Exercise 14.2.Q.)
Removing a subset of codimension 1 changes the Weil divisor group in a controllable way. Suppose Z is an irreducible codimension 1 subset of X. Then weclearly have an exact sequence:0% Z 1*→[Z] % Weil X% Weil(X − Z)

% 0.

When we take the quotient by principal divisors, taking into account the fact that
we may lose exactness on the left, we get an excision exact sequence for classgroups:(14.2.8.1)

Z

1*→[Z]% Cl X% Cl(X − Z)

% 0.

(Do you see why?)
For example, if U is an open subscheme of X = An , Pic U = {0}.
As another application, let X = Pn
k , and Z be the hyperplane x0 = 0. We haveZ

% Cl Pnk% Cl Ank%0nfrom which Cl Pn
k is generated by the class [Z], and Pic Pk is a subgroup of this.

14.2.9. By Exercise 14.2.F, [Z] (→ O(1), and as O(n) is nontrivial for n ̸= 0 (Exernncise 14.1.B), [Z] is not torsion in Cl Pn
k . Hence Pic(Pk ) "→ Cl(Pk ) is an isomorphism,

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n
∼
and Pic(Pnk ) = Z , with generator O(1). The degree of an invertible sheaf on P is
defined using this: define deg O(d) to be d.
We have gotten good mileage from the fact that the Picard group of the spectrum of a unique factorization domain is trivial. More generally, Exercise 14.2.I
gives us:14.2.10. Proposition. — If X is Noetherian and factorial, then for any Weil divisor D,
O(D) is invertible, and hence the map Pic X → Cl X is an isomorphism.
This can be used to make the connection to the class group in number theory
precise, see Exercise 13.1.M; see also §14.2.13.

14.2.11. Mild but important generalization: twisting line bundles by divisors. The
above constructions can be extended, with OX replaced by an arbitrary invertible
sheaf, as follows. Let L be an invertible sheaf on a normal Noetherian scheme X.
Then define L (D) by OX (D) ⊗ L .
14.2.J. E ASY E XERCISE .
(a) Assume for convenience that X is irreducible. Show that sections of L (D) canbe interpreted as rational sections of L with zeros and poles constrained by D,just as in (14.2.2.1):
Γ (U, L (D)) := {t nonzero rational section of L : div |U t + D|U ≥ 0} ∪ {0}.
(b) Suppose D1 and D2 are locally principal. Show that
∼ O(D1 + D2 ).
(O(D1 ))(D2 ) =14.2.12. Fun examples.
We can now actually calculate some Picard and class groups. First, a useful observation: notice that you can restrict invertible sheaves on Y to any subscheme X,
and this can be a handy way of checking that an invertible sheaf is not trivial. Effective Cartier divisors (§8.4.1) sometimes restrict too: if you have an effective Cartierdivisor on Y, then it restricts to a closed subscheme on X, locally cut out schemetheoretically by one equation. If you are fortunate and this equation doesn’t vanishon any associated point of X (§13.6.5), then you get an effective Cartier divisor on
X. You can check that the restriction of effective Cartier divisors corresponds torestriction of invertible sheaves (in the sense of Exercise 13.1.H).
14.2.K. E XERCISE : A TORSION P ICARD GROUP. Suppose that Y is a hypersurface inn∼Pnk corresponding to an irreducible degree d polynomial. Show that Pic(Pk − Y) =n∼
Z/(d). (For differential geometers: this is related to the fact that π1 (Pk − Y) =Z/(d).) Hint: (14.2.8.1).
The next two exercises explore consequences of Exercise 14.2.K, and provide
us with some examples promised in Exercise 5.4.N.
14.2.L. E XERCISE ( GENERALIZING E XERCISE 5.4.N). Keeping the same notation,
assume d > 1 (so Pic(Pn − Y) ̸= 0), and let H0 , . . . , Hn be the n + 1 coordinate
hyperplanes on Pn . Show that Pn − Y is affine, and Pn − Y − Hi is a distinguishedopen subset of it. Show that the Pn − Y − Hi form an open cover of Pn − Y. Show
that Pic(Pn − Y − Hi ) = 0. Then by Exercise 14.2.T, each Pn − Y − Hi is the Spec

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The Rising Sea: Foundations of Algebraic Geometryof a unique factorization domain, but Pn − Y is not. Thus the property of being aunique factorization domain is not an affine-local property — it satisfies only oneof the two hypotheses of the Affine Communication Lemma 5.3.2.
14.2.M. E XERCISE . Keeping the same notation as the previous exercise, show thaton Pn − Y, Hi (restricted to this open set) is an effective Cartier divisor that is not
cut out by a single equation. (Hint: Otherwise it would give a trivial element of
the class group.)
14.2.N. E XERCISE . Show that A := R[x, y]/(x2 +y2 −1) is not a unique factorization
domain, but A ⊗R C is. Hint: Exercise 14.2.L.
14.2.O. E XERCISE : P ICARD GROUP OF P1 × P1 . Consider∼ Proj k[w, x, y, z]/(wz − xy),
X = P1k ×k P1k =∼a smooth quadric surface (see Figure 8.2, and Example 9.6.2). Show that Pic X =1
1
Z ⊕ Z as follows: Show that if L = {∞} ×k P ⊂ X and M = P ×k {∞} ⊂ X, then
∼ A2 . This will give you a surjection Z ⊕ Z $ Cl X. Show that O(L)
X−L−M =
restricts to O on L and O(1) on M. Show that O(M) restricts to O on M and O(1)on L. (This exercise takes some time, but is enlightening.)
14.2.P. E XERCISE . Show that irreducible smooth projective surfaces (over k) canbe birational but not isomorphic. Hint: show P2 is not isomorphic to P1 × P1 using
the Picard group. (Aside: we will see in Exercise 20.2.D that the Picard group of
the “blown up plane” is Z2 , but in Exercise 20.2.E we will see that the blown up
plane is not isomorphic to P1 × P1 , using a little more information in the Picard
group.)
This is unlike the case for curves: birational irreducible smooth projective
curves (over k) must be isomorphic, as we will see in Theorem 17.4.3. Nonetheless,any two surfaces are related in a simple way: if X and X ′ are projective, regular,
and birational, then X can be sequentially blown up at judiciously chosen points,
and X ′ can too, such that the two results are isomorphic (see [Ha1, Thm. V.5.5];
blowing up will be discussed in Chapter 22).
14.2.Q. E XERCISE : P ICARD GROUP OF THE CONE . Let X = Spec k[x, y, z]/(xy − z2 ),a cone, where char k ̸= 2. (The characteristic hypothesis is not necessary for the
result, but is included so you can use Exercise 5.4.H to show normality of X.) Show
∼ Z/2. Hint: show that the class of Z = {x = z = 0} (thethat Pic X = 0, and Cl X =“affine cone over a line”) generates Cl X by showing that its complement D(x) is
isomorphic to an open subset of A2k . Show that 2[Z] = div(x) and hence principal,
and that Z is not principal, Exercise 14.2.H. (Remark: You know enough to showthat X−{(0, 0, 0)} is factorial. So although the class group is insensitive to removingloci of codimension greater than 1, §14.2.8, this is not true of the Picard group.)
A Weil divisor (on a normal scheme) with a nonzero multiple corresponding
to a line bundle is called Q-Cartier. (We won’t use this terminology beyond thenext exercise.) Exercise 14.2.Q gives an example of a Weil divisor that does notcorrespond to a line bundle, but is nonetheless Q-Cartier. We now give an example
of a Weil divisor that is not Q-Cartier.

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14.2.R. E XERCISE ( A NON -Q-C ARTIER DIVISOR ). On the cone over the smoothquadric surface X = Spec k[w, x, y, z]/(wz − xy), let Z be the Weil divisor cut out
by w = x = 0. Exercise 12.1.D showed that Z is not cut out scheme-theoreticallyby a single equation. Show more: that if n ̸= 0, then n[Z] is not locally principal.
Hint: show that the complement of an effective Cartier divisor on an affine scheme
is also affine, using Proposition 7.3.4. Then if some multiple of Z were locallyprincipal, then the closed subscheme of the complement of Z cut out by y = z = 0
would be affine — any closed subscheme of an affine scheme is affine. But this is
the scheme y = z = 0 (also known as the wx-plane) minus the point w = x = 0,
which we have seen is non-affine, §4.4.1.
14.2.S. ⋆ E XERCISE ( FOR THOSE WITH SUFFICIENT ARITHMETIC BACKGROUND ).
Identify the (ideal) class group of the ring of integers OK in a number field K, as defined in Exercise 13.1.M, with the class group of Spec OK , as defined in this section.
In particular, you will recover the common description of the class group as formal sums of prime ideals, modulo an equivalence relation coming from principal
fractional ideals.
14.2.13. More on class groups and unique factorization.
As mentioned in §5.4.6, there are few commonly used means of checking thata ring is a unique factorization domain. The next exercise is one of them, and
it is useful. For example, it implies the classical fact that for rings of integers in
number fields, the class group is the obstruction to unique factorization (see Exercise 13.1.M and Proposition 14.2.10).
14.2.T. E XERCISE . Suppose that A is a Noetherian integral domain. Show that A isa unique factorization domain if and only if A is integrally closed and Cl Spec A =0. (One direction is easy: we have already shown that unique factorization domains are integrally closed in their fraction fields. Also, Lemma 11.1.6 shows that
all codimension 1 prime ideals of a unique factorization domain are principal, sothat implies that Cl Spec A = 0. It remains to show that if A is integrally closed
and Cl Spec A = 0, then all codimension 1 prime ideals are principal, as this characterizes unique factorization domains (Proposition 11.3.5). Algebraic Hartogs’s
Lemma 11.3.11, may arise in your argument.) This is the third important characterization of unique factorization domains promised in §5.4.6.
My final favorite method of checking that a ring is a unique factorization domain (§5.4.6) is Nagata’s Lemma. It is also the least useful.

14.2.U. ⋆⋆ E XERCISE (N AGATA’ S L EMMA ). Suppose A is a Noetherian domain,x ∈ A an element such that (x) is prime and Ax = A[1/x] is a unique factorization
domain. Then A is a unique factorization domain. (Hint: Exercise 14.2.T. Use theshort exact sequenceZ[(x)] → Cl Spec A → Cl Spec Ax → 0(14.2.8.1) to show that Cl Spec A = 0. Prove that A[1/x] is integrally closed, thenshow that A is integrally closed as follows. Suppose T n + an−1 T n−1 + · · · + a0 = 0,
where ai ∈ A, and T ∈ K(A). Then by integral closedness of Ax , we have thatT = r/xm , where if m > 0, then r ∈/ (x). Then we quickly get a contradiction if
m > 0.)

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This leads to a fun algebra fact promised in Remark 12.8.6. Suppose k is analgebraically closed field of characteristic not 2. Let A = k[x1 , . . . , xn ]/(x21 + · · · +x2m ) where m ≤ n. When m ≤ 2, we get some special behavior. (If m = 0, we getaffine space; if m = 1, we get a nonreduced scheme; if m = 2, we get a reducible
scheme that is the union of two affine spaces.)
If m ≥ 3, we have verified that Spec A is normal, in Exercise 5.4.I(b). In fact,
if m ≥ 3, then A is a unique factorization domain unless m = 3 or m = 4 (Exercise 5.4.L; see also Exercise 12.1.E). For the case m = 3:
A = k[x, y, z, w1 , . . . , wn−3 ]/(x2 + y2 − z2 )is not a unique factorization domain, as it is has nonzero class group (by essentiallythe same argument as for Exercise 14.2.Q).
The failure at 4 comes from the geometry of the quadric surface: we havechecked that in Spec k[w, x, y, z]/(wz − xy), there is a codimension 1 irreducible
subset — the cone over a line in a ruling — that is not principal.
14.2.V. E XERCISE ( THE CASE m ≥ 5). Suppose that k is algebraically closed ofcharacteristic not 2. Show that if ℓ ≥ 3, thenA = k[a, b, x1 , . . . , xn ]/(ab − x21 − · · · − x2ℓ )is a unique factorization domain, by using Nagata’s Lemma with x = a.

14.3 ⋆ Effective Cartier divisors “=” invertible ideal sheaves
We now give a different means of describing invertible sheaves on a scheme.
One advantage of this over Weil divisors is that it can give line bundles on everywhere nonreduced schemes (such a scheme can’t be regular at any codimension 1
prime). But we won’t use this, so it is less important.
Suppose D "→ X is a closed subscheme such that the corresponding ideal sheaf
I is an invertible sheaf. Then I is locally trivial; suppose U is a trivializing affine
open set Spec A. Then the closed subscheme exact sequence (13.5.4.1)
corresponds to

0 → I → OX → OD → 0

0 → I → A → A/I → 0∼
with I = A as A-modules. Thus I is generated by a single element, say a, and thisexact sequence starts as
0

%A

×a

%A

As multiplication by a is injective, a is not a zerodivisor. We conclude that D is locally cut out by a single equation, that is not a zerodivisor. This was the definitionof effective Cartier divisor given in §8.4.1. This argument is clearly reversible, so wehave a quick new definition of effective Cartier divisor (an ideal sheaf I that is aninvertible sheaf — or equivalently, the corresponding closed subscheme).
14.3.A. E ASY E XERCISE . Show that a is unique up to multiplication by an invertible function.

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In the case where X is locally Noetherian, we can use the language of associated points (§13.6.5), so we can restate this definition as: D is locally cut out by asingle equation, not vanishing at any associated point of X.
We now define an invertible sheaf corresponding to D. The seemingly obvious
definition would be to take ID , but instead we define the invertible sheaf O(D)∨corresponding to an effective Cartier divisor to be the dual: ID. (The reason forthe dual is Exercise 14.3.C.)
14.3.B. E XERCISE .
Check that this agrees with our earlier definition of O(D),
Important Definition 14.2.2.
The ideal sheaf ID is then O(−D). We have an exact sequence

0 → O(−D) → O → OD → 0.
The invertible sheaf O(D) has a canonical section sD : Tensoring 0 → I → O
with I ∨ gives us O → I ∨ . (Easy unimportant fact: instead of tensoring I → O
with I ∨ , we could have dualized I → O, and we would get the same section.)

14.3.C. I MPORTANT AND SURPRISINGLY TRICKY EXERCISE . Recall that a section ofa locally free sheaf on X cuts out a closed subscheme of X (Exercise 13.1.I). Show
that this section sD cuts out D. (Compare this to Remark 14.2.6.)
This construction is reversible:
14.3.D. E XERCISE . Suppose L is an invertible sheaf, and s is a section that is locally not a zerodivisor. (Make sense of this! In particular, if X is locally Noetherian,this means “s does not vanish at an associated point of X”, see §13.6.5.) Show that∼ L.
s = 0 cuts out an effective Cartier divisor D, and O(D) =14.3.E. E XERCISE . Suppose I and J are invertible ideal sheaves (hence corresponding to effective Cartier divisors, say D and D ′ respectively). Show thatI J is an invertible ideal sheaf. (We define the product of two quasicoherent
ideal sheaves I J as you might expect: on each affine, we take the product of
the two corresponding ideals. To make sure this is well-defined, we need only
check that if A is a ring, and f ∈ A, and I, J ⊂ A are two ideals, then (IJ)f = If Jf
in Af .) We define the corresponding Cartier divisor to be D + D ′ . Verify that
∼ O(D) ⊗ O(D ′ ).
O(D + D ′ ) =We thus have an important correspondence between effective Cartier divisors(closed subschemes whose ideal sheaves are invertible, or equivalently locally cut
out by one non-zerodivisor, or in the locally Noetherian case, locally cut out by
one equation not vanishing at an associated point) and ordered pairs (L , s) where
L is an invertible sheaf, and s is a section that is not locally a zerodivisor (or in thelocally Noetherian case, not vanishing at an associated point). The effective Cartierdivisors form an abelian semigroup. We have a map of semigroups, from effective
Cartier divisors to invertible sheaves with sections not locally zerodivisors (andhence also to the Picard group of invertible sheaves).

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We get lots of invertible sheaves, by taking differences of two effective Cartierdivisors. In fact we “usually get them all” — it is very hard to describe an invertible sheaf on a finite type k-scheme that is not describable in such a way. For example, there are none if the scheme is regular or even factorial (basically by Proposition 14.2.10 for factoriality; and regular schemes are factorial by the AuslanderBuchsbaum Theorem 12.8.5). Exercise 16.6.F will imply that there are none if thescheme is projective. It holds in all other reasonable circumstances, see [Gr-EGA,
IV4 .21.3.4]. However, it does not always hold; the first and best example is due toKleiman, see [Kl3].

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CHAPTER 15

Quasicoherent sheaves and projective A-schemes
The first two sections of this chapter are relatively straightforward, and thelast two are trickier.

15.1 The quasicoherent sheaf corresponding to a graded module
We now describe quasicoherent sheaves on a projective A-scheme. Recall thata projective A-scheme is produced from the data of Z≥0 -graded ring S• , with S0 =A, and S+ is a finitely generated ideal (a “finitely generated graded ring over A”,§4.5.6). The resulting scheme is denoted Proj S• .
Suppose M• is a graded S• -module, graded by Z. (While reading the next section, you may wonder why we don’t grade by Z≥0 . You will see that it doesn’t
matter. A Z-grading will make things cleaner when we produce an M• from a<• as follows.
quasicoherent sheaf on Proj S• .) We define the quasicoherent sheaf M( as this might cause confusion with the affine case; but(I will avoid calling it M,<
M• is not graded in any way.) For each homogeneous f of positive degree, we define#
a quasicoherent sheaf M• (f) on the distinguished open D(f) = {p : f(p) ̸= 0} by#((M• )f )0 — note that ((M• )f )0 is an ((S• )f )0 -module, and recall that D(f) is identified with Spec((S• )f )0 (Exercise 4.5.F). As in (4.5.7.1), the subscript 0 means “the#0-graded piece”. We have obvious isomorphisms of the restriction of M
• (f) and#M
• (g) to D(fg), satisfying the cocycle conditions. (Think through this yourself, tobe sure you agree with the word “obvious”!) By Exercise 2.5.D, these sheaves glue<• on Proj S• . We then discard the temporary notation
together to a single sheaf M#M• (f).
<• is clearly quasicoherent, because it is quasicoherent on each
The O-module M
D(f), and quasicoherence is local.

<• at a point cor15.1.A. E XERCISE . Give an isomorphism between the stalk of M
responding to homogeneous prime p ⊂ S• and ((M• )p )0 . (Remark: You can use<• in terms of “compatible stalks”,
this exercise to give an alternate definition of M
cf. Exercise 4.5.M.)

Given a map of graded modules φ : M• → N• , we get an induced map of<• → N(• . Explicitly, over D(f), the map M• → N• induces (M• )f →sheaves M(N• )f , which induces φf : ((M• )f )0 → ((N• )f )0 ; and this behaves well with respect to restriction to smaller distinguished open sets, i.e., the following diagram
411

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((M• )f )0"((M• )fg )0

φf

φfg

% ((N• )f )0"% ((N• )fg )0 .

Thus ∼ is a functor from the category of graded S• -modules to the category of
quasicoherent sheaves on Proj S• .
15.1.B. E ASY E XERCISE . Show that ∼ is an exact functor. (Hint: everything in theconstruction is exact.)
15.1.C. E XERCISE . Show that if M• and M•′ agree in high enough degrees, then∼M<• =<′ . Then show that the map from graded S• -modules (up to isomorphism)
M•to quasicoherent sheaves on Proj S• (up to isomorphism) is not a bijection. (Really:show this isn’t an equivalence of categories.)
Exercise 15.1.C shows that ∼ isn’t an isomorphism (or equivalence) of categories, but it is close. The relationship is akin to that between presheaves andsheaves, and the sheafification functor (see §15.4).
<• ). (This15.1.D. E XERCISE . Describe a map of S0 -modules M0 → Γ (Proj S• , M
foreshadows the “saturation map” of §15.4.5 that takes a graded module to its
saturation, see Exercise 15.4.C.)
15.1.1. Example: Homogeneous ideals of S• give closed subschemes of Proj S• .
Recall that a homogeneous ideal I• ⊂ S• yields a closed subscheme Proj S• /I• "→∼ P3 . The ideal I• =Proj S• . For example, suppose S• = k[w, x, y, z], so Proj S• =22
(wz − xy, x − wy, y − xz) yields our old friend, the twisted cubic (defined in
Exercise 8.2.A)
15.1.E. E XERCISE . Show that if the functor ∼ is applied to the exact sequence ofgraded S• -modules
0 → I• → S• → S• /I• → 0

we obtain the closed subscheme exact sequence (13.5.4.1) for Proj S• /I• "→ Proj S• .
All closed subschemes of Proj S• arise in this way; we saw this in Exercise 8.2.C,
and will see it again in a different way in Exercise 15.4.H.

15.1.2. Remark. If M• is finitely generated (resp. finitely presented, coherent),<• , [Ro, (4.4.11)]. We will not need this fact.
then so is M

15.2 Invertible sheaves (line bundles) on projective A-schemes
Suppose M• is a graded S• -module. Define the graded module M(n)• by#• satisfies
M(n)m := Mn+m . Thus the quasicoherent sheaf M(n)#• ) = ((M• )f )n
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where here the subscript means we take the nth graded piece. (These subscripts
are admittedly confusing!)# ∼15.2.A. E XERCISE . If S• = A[x0 , . . . , xm ], so Proj S• = Pm
A , show S(n)• = O(n)using transition functions (cf. §14.1). (Recall from §14.1.2 that the global sectionsof O(n) should be identified with the homogeneous degree n polynomials in x0 ,. . . , xm . Can you see that in the context of this exercise?)
15.2.1. Definition. Motivated by this, if S• is a graded ring generated in degree 1,#• .
we define OProj S (n) (or simply O(n), where S• is implicit) by S(n)•15.2.B. I MPORTANT E XERCISE . If S• is generated in degree 1, show that OProj S• (n)is an invertible sheaf.
If S• is generated in degree 1, and F is a quasicoherent sheaf on Proj S• , define
F (n) := F ⊗ O(n). This is often called twisting F by O(n), or twisting F by n.
More generally, if L is an invertible sheaf, then F ⊗ L is often called twisting F
by L .
#• .
∼ M(n)<• (n) =15.2.C. E XERCISE . If S• is generated in degree 1, show that M(Hereafter, we can be cavalier with the placement of the “dot” in such situations.)∼ O(m) ⊗15.2.D. E XERCISE . If S• is generated in degree 1, show that O(m + n) =O(n).

15.2.2. Unimportant remark. Even if S• is not generated in degree 1, then by Exercise 6.4.G, Sd• is generated in degree 1 for some d. In this case, we may define theinvertible sheaves O(dn) for n ∈ Z. This does not mean that we can’t define O(1);this depends on S• . For example, if S• is the polynomial ring k[x, y] with the usual
grading, except without linear terms (so S• = k[x2 , xy, y2 , x3 , x2 y, xy2 , y3 ]), thenS2• and S3• are both generated in degree 1, meaning that we may define O(2) andO(3). There is good reason to call their “difference” O(1).

15.3 Globally generated and base-point-free line bundles
We now come to a topic that is harder, but that will be important. Throughout
this section, S• will be a finitely generated graded ring over A, generated in degree
1. We will prove the following result.
15.3.1. Theorem. — Any finite type quasicoherent sheaf F on Proj S• can be presented
in the form
⊕finite O(−n) → F → 0.

Because we can work with the line bundles O(−n) in a hands-on way, thisresult will give us great control over all coherent sheaves (and in particular, vector
bundles) on Proj S• . As just a first example, it will allow us to show that every
coherent sheaf on a projective k-scheme has a finite-dimensional space of global
sections (Corollary 18.1.5). (This fact will grow up to be the fact that the higher

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pushforwards of coherent sheaves under proper morphisms are also coherent, see
Theorem 18.8.1(d) and Grothendieck’s Coherence Theorem 18.9.1.)
Rather than proceeding directly to a proof, we use this as an excuse to introduce notions that are useful in wider circumstances (global generation, base-pointfreeness, ampleness), and their interrelationships. But first we use it as an excuse tomention an important classical result.
15.3.2. The Hilbert Syzygy Theorem.
Given any coherent sheaf F on Pn
k , Theorem 15.3.1 gives the existence of asurjection φ : ⊕finite O(−m) → F → 0. The kernel of the surjection is also coherent,
so iterating this construction, we can construct an infinite resolution of F by adirect sum of line bundles:· · · ⊕finite O(m2,j ) → ⊕finite O(m1,j ) → ⊕finite O(m0,j ) → F → 0.

The Hilbert Syzygy Theorem states that there is in fact a finite resolution, of length
at most n. (The Hilbert Syzygy Theorem in fact states more.) Because we won’t
use this, we don’t give a proof, but [E] (especially [E, Thm. 1.13] and the links
thereafter) has an excellent discussion. (Hilbert’s original statement, and the discussion in [E], is about graded modules. The connection to the statement here on
coherent sheaves is provided by the exactness of the functor ∼, Exercise 15.1.B.)
See the comments after Theorem 3.6.17 for the original history of this result.
15.3.3. Globally generated sheaves. Suppose X is a scheme, and F is an Omodule. The most important definition of this section is the following: F is globally generated (or generated by global sections) if it admits a surjection from afree sheaf on X:
%%F .
O ⊕I
Here I is some index set. The global sections in question are the images of the |I|
sections corresponding to 1 in the various summands of OX⊕I ; those images generate the stalks of F . We say F is finitely globally generated (or generated by afinite number of global sections) if the index set I can be taken to be finite.
More definitions in more detail: we say that F is globally generated at a pointp (or sometimes generated by global sections at p) if we can find φ : O ⊕I → F
that is surjective on stalks at p:
Op⊕I

φp

% % Fp .

(It would be more precise to say that the stalk of F at p is generated by global
sections of F .) The key insight is that global generation at p means that every
germ at p is a linear combination (over the local ring OX,p ) of germs of global
sections.
Note that F is globally generated if it is globally generated at all points p. (Reason: Exercise 2.4.E showed that isomorphisms can be checked on the level of stalks.
An easier version of the same argument shows that surjectivity can also be checkedon the level of stalks.) Notice that we can take a single index set for all of X, by
taking the union of all the index sets for each p.
15.3.A. E ASY E XERCISE ( REALITY CHECK ). Show that every quasicoherent sheaf
on every affine scheme is globally generated. Show that every finite type quasicoherent sheaf on every affine scheme is generated by a finite number of global

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sections. (Hint for both: for any A-module M, there is a surjection onto M from afree A-module.)
Clearly if F and G are globally generated, then so is F ⊕ G is as well (andsimilarly for finitely globally generated, generated at a point p, etc.). Similarly fortensor product:
15.3.B. E ASY E XERCISE ( GLOBALLY GENERATED ⊗ GLOBALLY GENERATED IS GLOB ALLY GENERATED ).
Show that if quasicoherent sheaves F and G are globally
generated at a point p, then so is F ⊗ G .
15.3.C. E ASY BUT I MPORTANT E XERCISE . Suppose F is a finite type quasicoherent
sheaf on X.
(a) Show that F is globally generated at p if and only if “the fiber of F is generated
by global sections at p”, i.e., the map from global sections to the fiber Fp /mFp issurjective, where m is the maximal ideal of OX,p . (Hint: Geometric Nakayama,
Exercise 13.7.E.)(b) Show that if F is globally generated at p, then “F is globally generated nearp”: there is an open neighborhood U of p such that F is globally generated atevery point of U.
(c) Suppose further that X is a quasicompact scheme. Show that if F is globally
generated at all closed points of X, then F is globally generated at all points of X.
(Note that nonempty quasicompact schemes have closed points, Exercise 5.1.E.)
15.3.D. E ASY E XERCISE . If F is a finite type quasicoherent sheaf on X, and X is
quasicompact, show that F is globally generated if and only if it is generated by afinite number of global sections.
15.3.E. E ASY E XERCISE . An invertible sheaf L on X is globally generated if andonly if for any point p ∈ X, there is a global section of L not vanishing at p. (SeeTheorem 16.4.1 for why we care.)15.3.4. Definitions. If L is an invertible sheaf on X, then those points where all
global sections of L vanish are called the base points of L , and the set of basepoints is called the base locus of L ; it is a closed subset of X. (We can refine this toa closed subscheme: by taking the scheme-theoretic intersection of the vanishingloci of the sections of L , we obtain the scheme-theoretic base locus.) The complement of the base locus is the base-point-free locus. If L has no base points,
it is base-point-free. By the previous discussion, (i) the base-point-free locus is
an open subset of X, and (ii) L is generated by global sections if and only if it isbase-point-free. By Exercise 15.3.B, the tensor of two base-point-free line bundles
is base-point-free.
(Remark: We will see in Exercise 18.2.I that if X is a k-scheme, and L is aninvertible sheaf on X, and K/k is any field extension, then L is base-point-free if
and only if it is “base-point-free after base change to K”. You could reasonably
prove this now.)
15.3.5. Base-point-free line bundles and maps to projective space. The main reason wecare about the definitions above is the following. Recall Exercise 6.3.M(a), whichshows that n + 1 functions on a scheme X with no common zeros yield a map to
Pn . This notion generalizes.

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15.3.F. E ASY E XERCISE ( A VITALLY IMPORTANT CONSTRUCTION ). Suppose s0 ,. . . , sn are n + 1 global sections of an invertible sheaf L on a scheme X, with nocommon zero. Define a corresponding map to Pn :X

[s0 ,...,sn ]% Pn

Hint: If U is an open subset on which L is trivial, choose a trivialization, thentranslate the si into functions using this trivialization, and use Exercise 6.3.M(a) toobtain a morphism U → Pn . Then show that all of these maps (for different U anddifferent trivializations) “agree”, using Exercise 6.3.M(b).

(In Theorem 16.4.1, we will see that this yields all maps to projective space.)
Note that this exercise as written “works over Z” (as all morphisms are “over”
the final object in the category of schemes), although many readers will just workover a particular base such as a given field k. Here is some convenient classicallanguage which is used in this case.
15.3.6. Definitions. A linear series on a k-scheme X is a k-vector space V (usuallyfinite-dimensional), an invertible sheaf L , and a linear map λ : V → Γ (X, L ). Sucha linear series is often called “V”, with the rest of the data left implicit. If the map
λ is an isomorphism, it is called a complete linear series, and is often written |L |.
The language of base points (Definition 15.3.4) readily translates to this situation.
For example, given a linear series, any point x ∈ X on which all elements of thelinear series V vanish, we say that x is a base point of V. If V has no base points,
we say that it is base-point-free. The set of base points is called the base locusof the linear series. One can similarly define the scheme-theoretic base locus (orbase scheme, although this phrase can have another meaning) of the linear series.
As a reality check, you should understand why an n + 1-dimensional linear
series on a k-scheme X with base-point-free locus U defines a morphism U → Pnk.
15.3.7. Serre’s Theorem A. We are now able to state a celebrated result of Serre.

15.3.8. Serre’s Theorem A. — Suppose S• is generated in degree 1, and finitely generated over A = S0 . Let F be any finite type quasicoherent sheaf on Proj S• . Then there
exists some n0 such that for all n ≥ n0 , F (n) is finitely globally generated.
We could now prove Serre’s Theorem A directly, but will continue to use thisas an excuse to introduce more ideas; it will be a consequence of Theorem 16.6.2.

15.3.9. Proof of Theorem 15.3.1 assuming Serre’s Theorem A (Theorem 15.3.8). Suppose
we have m global sections s1 , . . . , sm of F (n) that generate F (n). This gives amap% F (n)⊕m Ogiven by (f1 , . . . , fm ) (→ f1 s1 + · · · + fm sm on any open set. Because these global
sections generate F (n), this is a surjection. Tensoring with O(−n) (which is exact,
as tensoring with any locally free sheaf is exact, Exercise 13.1.E) gives the desired
result.
!

15.4 Quasicoherent sheaves and graded modules

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417(This section answers some fundamental questions, but it is surprisingly tricky.
You may wish to skip this section, or at least the proofs, on first reading, unless you
have a particular need for them.)Throughout this section, S• is a finitely generated graded algebra generated
in degree 1, so in particular we have the invertible sheaf O(n) for all n by Exercise 15.2.B. Also, throughout, M• is a graded S• -module, and F is a quasicoherent
sheaf on Proj S• .
We know how to get quasicoherent sheaves on Proj S• from graded S• -modules.
We will now see that we can get them all in this way. We will define a functor Γ•from (the category of) quasicoherent sheaves on Proj S• to (the category of) gradedS• -modules that will attempt to reverse the ∼ construction. They are not quite inverses, as ∼ can turn two different graded modules into the same quasicoherent
sheaf (see for example Exercise 15.1.C). But we will see a natural isomorphism∼<Γ#• (F ) = F . In fact Γ• (M• ) is a better (“saturated”) version of M• , and there is a<• ) that is akin to groupification and sheafification
saturation functor M• → Γ• (M— it is adjoint to the forgetful functor from saturated graded modules to graded
modules. And thus we come to the fundamental relationship between ∼ and Γ• :they are an adjoint pair.

QCohProj S•

graded S• -modules`❥❥❥❥∼❥❥❥ ❥❥❥❥❥❥* ❥❥❥saturateforget❚❚❚❚❚equivalence
❚❚❚❚
❚❚❚❚
Γ•
❚❚0
_
saturated graded S• -modules

We now make some of this precise, but as little as possible to move forward. In
particular, we will show that every quasicoherent sheaf on a projective A-scheme
arises from a graded module (Corollary 15.4.3), and we will see a new approach to
Exercise 8.2.C, that every closed subscheme of Proj S• arises from a homogeneous
ideal I• ⊂ S• (Exercise 15.4.H).
15.4.1. Definition of Γ• . When you do Essential Exercise 14.1.C (on global sectionsof OPm(n)), you will suspect that in good situations,k∼ Γ (Proj S• , M(n)( • ).
Mn =Motivated by this, we defineΓn (F ) := Γ (Proj S• , F (n)).
#• ),15.4.A. E XERCISE . Describe a morphism of S0 -modules Mn → Γ (Proj S• , M(n)extending the n = 0 case of Exercise 15.1.D.
15.4.B. E XERCISE . Show that Γ• (F ) is a graded S• -module. (Hint: consider Sn →
Γ (Proj S• , O(n)).)<• ) arising from the previous
15.4.C. E XERCISE . Show that the map M• → Γ• (Mtwo exercises is a map of S• -modules. We call this the saturation map.

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15.4.D. E XERCISE .
(a) Show that the saturation map need not be injective, nor need it be surjective.
(Hint: S• = k[x], M• = k[x]/(x2 ) or M• = xk[x].)(b) On the other hand, show that if S• is a finitely generated graded ring over a field
S0 = k, and M• is finitely generated, then the saturation map is an isomorphism
in large degree. In other words, show that there exists an n0 such that Mn →#• ), is an isomorphism for n ≥ n0 .
Γ (Proj S• , M(n)15.4.E. E XERCISE . Show that Γ• is a functor from QCohProj S• to the category ofgraded S• -modules. In other words, if F → G is a morphism of quasicoherent
sheaves on Proj S• , describe the natural map Γ• F → Γ• G , and show that suchmaps respect the identity and composition. Thus the saturation map can be better
called the saturation functor.
15.4.2. ⋆ The reverse map. Now that we have defined the saturation map M• →<• , we will describe a map Γ<Γ• M• F → F . While subtler to define, it will have theadvantage of being an isomorphism.

15.4.F. E XERCISE . Define the natural map Γ<• F → F as follows. First describe themap on sections over D(f). Note that sections of the left side are of the form m/fn′′
where m ∈ Γn deg f (F ), and m/fn = m ′ /fn if there is some N with fN (fn m −fn m ′ ) = 0. Sections on the right are implicitly described in Exercise 13.3.H. Show
that your map behaves well on overlaps D(f) ∩ D(g) = D(fg).
15.4.G. T RICKY E XERCISE . Show that the natural map Γ<• F → F is an isomorphism, by showing that it is an isomorphism of sections over D(f) for any f. Firstshow surjectivity, using Exercise 13.3.H to show that any section of F over D(f) isof the form m/fn where m ∈ Γn deg f (F ). Then verify that it is injective.
15.4.3. Corollary. — Every quasicoherent sheaf on a projective A-scheme arises from the∼ construction.
15.4.H. E XERCISE ( ANOTHER APPROACH TO E XERCISE 8.2.C). Show that eachclosed subscheme of Proj S• arises from a homogeneous ideal I• ⊂ S• . (Hint: Suppose Z is a closed subscheme of Proj S• . Consider the exact sequence 0 → IZ →OProj S• → OZ → 0. Apply Γ• , and then ∼. Be careful: Γ• is left-exact, but notnecessarily exact.) This fulfills promises made in Exercises 8.2.B and 15.1.E.
For the first time, we see that every closed subscheme of a projective scheme iscut out by homogeneous equations. This is the analog of the fact that every closed
subscheme of an affine scheme is cut out by equations. It is disturbing that it is so
hard to prove this fact. (For comparison, this was easy on the level of the Zariski
topology — see Exercise 4.5.H(c).)

15.4.I. ⋆⋆ E XERCISE (Γ• AND ∼ ARE ADJOINT FUNCTORS ). Describe a natural bijec∼ Hom(M<• , F ), as follows.
tion Hom(M• , Γ• F ) =(a) Show that maps M• → Γ• F are the “same” as maps ((M• )f )0 → ((Γ• F )f )0
as f varies through S+ , that are “compatible” as f varies, i.e., if D(g) ⊂Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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419

D(f), there is a commutative diagram((M• )f )0% ((Γ• F )f )0"((M• )g )0"% ((Γ• F )g )0More precisely, give a bijection between Hom(M• , Γ• F ) and the set ofcompatible maps$%
Hom((M• )f )0 → ((Γ• F )f )0.
f∈S+(b) Describe a bijection between the set of compatible maps (Hom((M• )f )0 →<• ) → Γ (D(f), F ).
((Γ• F )f )0 )f∈S+ and the set of compatible maps Γ (D(f), M(• , which is a
15.4.4. The special case M• = S• . We have a saturation map S• → Γ• S(• has the structure of a graded ring (basically because
map of S• -modules. But Γ• S
we can multiply sections of O(m) by sections of O(n) to get sections of O(m + n),
see Exercise 15.2.D).
(• induces (via the
15.4.J. E XERCISE . Show that the map of graded rings S• → Γ• S(• → Proj S• , andconstruction of Essential Exercise 6.4.A) an isomorphism Proj Γ• S
under this isomorphism, the respective O(1)’s are identified.

This addresses the following question: to what extent can we recover S• from
(Proj S• , O(1))? The answer is: we cannot recover S• , but we can recover its “saturation”. And better yet: given a projective A-scheme π : X → Spec A, along with
O(1), we obtain it as a Proj of a graded algebra in a canonical way, via∼ Proj (⊕n≥0 Γ (X, O(n))) .
X=
There is one last worry you might have, which is assuaged by the following
exercise.

15.4.K. E XERCISE . Suppose S• is a finitely generated graded ring over a field k,
so X = Proj S• is a projective k-scheme. Show that (⊕n≥0 Γ (X, O(n))) is a finitely
generated k-algebra. (Hint: S• and (⊕n≥0 Γ (X, O(n))) agree in sufficiently highdegrees, by Exercise 15.4.D.)
15.4.5. ⋆⋆ Saturated S• -modules. We end with a remark: different graded S• modules give the same quasicoherent sheaf on Proj S• , but the results of this section show that there is a “best” (= saturated) graded module for each quasicoherent sheaf, and there is a map from each graded module to its “best” version,<• . A module for which this is an isomorphism (a “best” module) is
M• → Γ• M
called saturated. We won’t use this term later.
<• is analogous to the sheafification map,
This “saturation” map M• → Γ• M
taking presheaves to sheaves. For example, the saturation of the saturation equals
the saturation.
There is a bijection between saturated quasicoherent sheaves of ideals on Proj S•
and closed subschemes of Proj S• .

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CHAPTER 16

Pushforwards and pullbacks of quasicoherent sheaves

16.1 Introduction
This chapter is devoted to pushforward and pullbacks of quasicoherent sheaves,their properties, and some applications.
Suppose B → A is a morphism of rings. Recall (from Exercise 1.5.E) that(· ⊗B A, ·B ) is an adjoint pair between the categories of A-modules and B-modules:
we have a bijection∼ HomB (N, MB )HomA (N ⊗B A, M) =functorial in both arguments. These constructions behave well with respect to localization (in an appropriate sense), and hence work (often) in the category of quasicoherent sheaves on schemes (and indeed always in the category of O-modules
on ringed spaces, see Definition 16.3.5, although we won’t particularly care). The
easier construction (M (→ MB ) will turn into our old friend pushforward. The
other (N (→ A ⊗B N) will be a relative of pullback, whom I’m reluctant to call an“old friend”.

16.2 Pushforwards of quasicoherent sheaves
We begin with the pushforwards, for which we have already done much of
the work.
The main moral of this section is that in “reasonable” situations, the pushforward of a quasicoherent sheaf is quasicoherent, and that this can be understood interms of one of the module constructions defined above. We begin with a motivating example:
16.2.A. E XERCISE . Let π : Spec A → Spec B be a morphism of affine schemes, and( is a quasicoherent sheaf on Spec A. Give an
suppose M is an A-module, so M(<
isomorphism π∗ M → MB . (Hint: There is only one reasonable way to proceed:
look at distinguished open sets.)( is quasicoherent. Perhaps more important, this implies that
In particular, π∗ M
the pushforward of a quasicoherent sheaf under an affine morphism is also quasicoherent.
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16.2.B. E XERCISE . If π : X → Y is an affine morphism, show that π∗ is an exact
functor QCohX → QCohY .

The following result, proved earlier, generalizes the fact that the pushforward
of a quasicoherent sheaf under an affine morphism is also quasicoherent.

16.2.1. Theorem (Exercise 13.3.F). — Suppose π : X → Y is a quasicompact quasiseparated morphism, and F is a quasicoherent sheaf on X. Then π∗ F is a quasicoherent sheaf
on Y.
Coherent sheaves do not always push forward to coherent sheaves. For example, consider the structure morphism π : A1k → Spec k, corresponding to k → k[t].
Then π∗ OA1k is the quasicoherent sheaf corresponding to k[t], which is not a finitely
generated k-module. But in good situations, coherent sheaves do push forward.
For example:
16.2.C. E XERCISE . Suppose π : X → Y is a finite morphism of locally Noetherian
schemes. If F is a coherent sheaf on X, show that π∗ F is a coherent sheaf. Hint:
Show first that π∗ OX is finite type.
Once we define cohomology of quasicoherent sheaves, we will quickly proventhat if F is a coherent sheaf on Pn
k , then Γ (Pk , F ) is a finite-dimensional k-module,
and more generally if F is a coherent sheaf on Proj S• , then Γ (Proj S• , F ) is a coherent A-module (where S0 = A). This is a special case of the fact that the “pushforwards of coherent sheaves by projective morphisms are also coherent sheaves”(Theorem 18.8.1(d)). (The notion of projective morphism, a relative version of
Proj S• → Spec A, will be defined in §17.3.)
More generally, given Noetherian hypotheses, pushforwards of coherent sheaves
by proper morphisms are also coherent sheaves (Grothendieck’s Coherence Theorem 18.9.1).

16.3 Pullbacks of quasicoherent sheaves
We next discuss the pullback of a quasicoherent sheaf: if π : X → Y is a morphism of schemes, π∗ is a covariant functor QCohY → QCohX . The notion of thepullback of a quasicoherent sheaf can be confusing on first (and second) glance.
(For example, it is not the inverse image sheaf, although we will see that it is related.)
Here are three contexts in which you have seen the pullback, or can understand it quickly. It may be helpful to keep these in mind, to keep you anchored in
the long discussion that follows. Suppose G is a quasicoherent sheaf on a schemeY.
(i) (restriction to open subsets) If i : U "→ Y is an open immersion, then i∗ G is
G |U , the restriction of G to U (Example 2.2.8).
(ii) (restriction to points) If i : p "→ Y is the “inclusion” of a point p in Y (forexample, the closed embedding of a closed point; see Exercise 6.3.J(b)), then i∗ G is
G |p , the fiber of G at p (Definition 4.3.7).
The similarity of the notation G |U and G |p is precisely because both are pullbacks. Pullbacks (especially to locally closed subschemes or generic points) are

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often called restriction. For this reason, if π : X → Y is some sort of inclusion (suchas a locally closed embedding, or an “inclusion of a generic point”) then π∗ G is often written as G |X and called the restriction of G to X, when π can be interpreted
as some type of “inclusion”.
(iii) (pulling back vector bundles) Suppose G is a locally free sheaf on Y, π : X → Y
is any morphism. If {Ui } are trivializing open neighborhoods for G , and Tij ∈GLr (OY (Ui ∩ Uj )) are transition matrices for G between Ui and Uj , then {π−1 Ui }are trivializing open neighborhoods for π∗ G , and π∗ Tij are transition matrices forπ∗ G . (This will be established in Theorem 16.3.7(3).)
16.3.1. Strategy. We will see three different ways of thinking about the pullback.
Each has significant disadvantages, but together they give a good understanding.
(a) Because we are understanding quasicoherent sheaves in terms of affine
open sets, and modules over the corresponding rings, we begin with an interpretation in this vein. This will be very useful for proving and understanding facts.
The disadvantage is that it is annoying to make a definition out of this (when thetarget is not affine), because gluing arguments can be tedious.
(b) As we saw with fibered product, gluing arguments can be made simpler
using universal properties, so our second “definition” will be by universal property. This is elegant, but has the disadvantage that it still needs a construction,and because it works in the larger category of O-modules, it isn’t clear from theuniversal property that it takes quasicoherent sheaves to quasicoherent sheaves.
But if the target is affine, our construction of (a) is easily seen to satisfy universal
property. Furthermore, the universal property is “local on the target”: if π : X → Y
is any morphism, i : U "→ Y is an open immersion, and G is a quasicoherent sheaf
on Y, then if π∗ G exists, then its restriction to π−1 (U) is (canonically identifiedwith) (π|U )∗ (G |U ). Thus if the pullback exists in general (even as an O-module),
affine-locally on Y it looks like the construction of (a) (and thus is quasicoherent).
(c) The third definition is one that works on ringed spaces in general. It isshort, and is easily seen to satisfy the universal property. It doesn’t obviously
take quasicoherent sheaves to quasicoherent sheaves (at least in the way that wehave defined quasicoherent sheaves) — a priori it takes quasicoherent sheaves toO-modules. But thanks to the discussion at the end of (b) above, which used (a),this shows that the pullback of a quasicoherent sheaf is indeed quasicoherent.
16.3.2. First attempt at describing the pullback, using affines. Suppose π : X → Y
is a morphism of schemes, and G is a quasicoherent sheaf on Y. We want to define
the pullback quasicoherent sheaf π∗ G on X in terms of affine open sets on X and
Y. Suppose Spec A ⊂ X, Spec B ⊂ Y are affine open sets, with π(Spec A) ⊂ Spec B.
∼ N.
3 Perhaps motivated by the fact that pullback should relate
Suppose G |Spec B =to tensor product, we want
(16.3.2.1)

Γ (Spec A, π∗ G ) = N ⊗B A.

More precisely, we would like Γ (Spec A, π∗ G ) and N ⊗B A to be identified. This
could mean that we use this to construct a definition of π∗ G , by “gluing all thisinformation together” (and showing it is well-defined). Or it could mean that we
define π∗ G in some other way, and then find a natural identification (16.3.2.1). The
first approach can be made to work (and §16.3.3 is the first step), but we will follow
the second.

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F IGURE 16.1. The pullback of a quasicoherent sheaf (module-theoretically)
16.3.3. We begin this project by fixing an affine open subset Spec B ⊂ Y. To avoid
confusion, let φ = π|π−1 (Spec B) : π−1 (Spec B) → Spec B. We will define a quasicoherent sheaf on π−1 (Spec B) that will turn out to be φ∗ (G |Spec B ) (and will also be
the restriction of π∗ G to π−1 (Spec B)).
If Spec Af ⊂ Spec A is a distinguished open set, thenΓ (Spec Af , φ∗ G ) = N ⊗B Af = (N ⊗B A)f = Γ (Spec A, φ∗ G )f

where “=” means “canonically isomorphic”. Define the restriction map Γ (Spec A, φ∗ G ) →Γ (Spec Af , φ∗ G ),(16.3.3.1)

Γ (Spec A, φ∗ G ) → Γ (Spec A, φ∗ G ) ⊗A Af ,by α (→ α ⊗ 1 (of course). Thus φ∗ G is (or: extends to) a quasicoherent sheaf on
π−1 (Spec B) (by Exercise 13.3.D).
We have now defined a quasicoherent sheaf on π−1 (Spec B), for every affine
open subset Spec B ⊂ Y. We want to show that this construction, as Spec B varies,
glues into a single quasicoherent sheaf on X.
You are welcome to do this gluing appropriately, for example using the distinguished affine base of Y (§13.3.1). This works, but can be confusing, so we takeanother approach.
16.3.4. Universal property definition of pullback. Suppose π : X → Y is a morphism of ringed spaces, and G is an OY -module. (We are of course interested in
the case where π is a morphism of schemes, and G is quasicoherent. Even once wespecialize our discussion to schemes, much of our discussion will extend withoutchange to this more general situation.) We “define” the pullback π∗ G as an OX module, using the following adjointness universal property: for any OX -module
F , there is a bijection HomOX (π∗ G , F ) ↔ HomOY (G , π∗ F ), and these bijections
are functorial in F . By universal property nonsense, this determines π∗ G up tounique isomorphism; we just need to make sure that it exists (which is why theword “define” is in quotes). Notice that we avoid worrying about when the pushforward of a quasicoherent sheaf F is quasicoherent by working in the larger category of O-modules.
16.3.A. I MPORTANT E XERCISE . If Y is affine, say Y = Spec B, show that the construction of the quasicoherent sheaf in §16.3.3 satisfies this universal property ofpullback of G . Thus calling this sheaf π∗ G is justified. (Hint: Interpret both sides

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of the alleged bijection explicitly. The adjointness in the ring/module case should
turn up.)
16.3.B. I MPORTANT E XERCISE . Suppose i : U "→ X is an open embedding of ringed
spaces, and F is an OX -module. Show that F |U satisfies the universal property ofi∗ F (and thus deserves to be called i∗ F ). In other words, for each OU -module E ,describe a bijection
HomOU (F |U , E ) (% HomO (F , i∗ E ),X

functorial in E .
We next show that if π∗ G satisfies the universal property (for the morphism
π : X → Y), then if j : V "→ Y is any open subset, and i : U = π−1 (V) "→ X (see(16.3.4.1)), then (π∗ G )|U satisfies the universal property for π|U : U → V. Thus(π∗ G )|U deserves to be called π|∗U (G |V ). You will notice that we really need towork with O-modules, not just with quasicoherent sheaves.
" i %(16.3.4.1)
π−1 (V)
U!
X
π

π|U

" "
V!

j

"
%Y

If F ′ is an OU -module, we have a series of bijections (using Important Exercise 16.3.B and adjointness of pullback and pushforward):∼ HomO (i∗ (π∗ G ), F ′ )HomOU ((π∗ G )|U , F ′ ) =U∼ HomO (π∗ G , i∗ F ′ )=XWe have thus described a bijection

∼
=
∼
=
∼
=

HomOY (G , π∗ i∗ F ′ )∼=HomOV (G |V , (π|U )∗ F ′ ).

HomOY (G , j∗ (π|U )∗ F ′ )HomOV (j∗ G , (π|U )∗ F ′ )HomOU ((π∗ G )|U , F ′ ) ↔ HomOV (G |V , (π|U )∗ F ′ ),which is clearly (by construction) functorial in F ′ . Hence the discussion in the firstparagraph of §16.3.4 is justified. For example, thanks to Important Exercise 16.3.A,
the pullback π∗ exists if Y is an open subset of an affine scheme.
At this point, we could show that the pullback exists, following the idea behind the construction of the fibered product: we would start with the definition
when Y is affine, and “glue”. We will instead take another route.
16.3.5. Third definition: pullback of O-modules via explicit construction. Suppose π : X → Y is a morphism of ringed spaces, and G is an OY -module. Of course,
our case of interest is if π is a morphism of schemes, and G is quasicoherent. Now
π−1 G is a π−1 OY -module. (Notice that we are using the ringed space (X, π−1 OY ),not (X, OX ). The inverse image construction π−1 was discussed in §2.7.) Furthermore, OX is also a π−1 OY -module, via the map π−1 OY → OX that is part of thedata of the morphism π. Define the pullback of G by π as the OX -module(16.3.5.1)

π∗ G := π−1 G ⊗π−1 OY OX .

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It is immediate that pullback is a covariant functor π∗ : ModOY → ModOX .

16.3.C. I MPORTANT E XERCISE . Show that this definition (16.3.5.1) of pullback
satisfies the universal property. Thus the pullback exists, at least as a functor
ModOY → ModOX .

16.3.D. I MPORTANT E XERCISE . Show that if π : X → Y is a morphism of schemes,
then π∗ gives a covariant functor QCohY → QCohX . (You might use §16.3.3, Exercise 16.3.B, Exercise 16.3.A, and/or §16.3.4.)
The following is then immediate from the universal property.

16.3.6. Proposition. — Suppose π : X → Y is a quasicompact, quasiseparated morphism.
Then (π∗ : QCohY → QCohX , π∗ : QCohX → QCohY ) are an adjoint pair: there is an
isomorphism(16.3.6.1)∼ HomO (G , π∗ F ),HomOX (π∗ G , F ) =Yfunctorial in both F ∈ QCohX and G ∈ QCohY .

The “quasicompact and quasiseparated” hypotheses are solely to ensure thatπ∗ indeed sends QCohX to QCohY (Exercise 13.3.F = Theorem 16.2.1).
We are now ready to show that pullback has all sorts of desired properties.
16.3.7. Theorem. — Suppose π : X → Y is a morphism of schemes, and G is a quasicoherent sheaf on Y.

∼(1) (pullback preserves the structure sheaf) There is a canonical isomorphism π∗ OY =OX .
(2) (pullback preserves finite type quasicoherent sheaves) If G is a finite type quasicoherent sheaf, so is π∗ G . Hence if X is locally Noetherian, and G is coherent,
then so is π∗ G . (It is not always true that the pullback of a coherent sheaf iscoherent, and the interested reader can think of a counterexample.)(3) (pullback preserves vector bundles, and their transition functions) If G is locally
free sheaf of rank r, then so is π∗ G . (In particular, the pullback of an invertible
sheaf is invertible.) Furthermore, if {Ui } are trivializing open neighborhoods for
G , and Tij ∈ GLr (OY (Ui ∩ Uj )) are transition matrices for G between Ui and
Uj , then {π−1 Ui } are trivializing open neighborhoods for π∗ G , and π∗ Tij aretransition matrices for π∗ G .
(4) (functoriality in the morphism) If ξ : W → X is a morphism of schemes, then∼ (π ◦ ξ)∗ G .
there is a canonical isomorphism ξ∗ π∗ G =(5) (functoriality in the quasicoherent sheaf) π∗ is a functor QCohY → QCohX .
(6) (pulling back a section) Note that a section of G is the data of a map OY → G .
By (1) and (5), if s : OY → G is a section of G then there is a natural sectionπ∗ s : OX → π∗ G of π∗ G . The pullback of the locus where s vanishes is the locus
where the pulled-back section π∗ s vanishes.
(7) (pullback on stalks) If π : X → Y, π(p) = q, then pullback induces an isomorphism(16.3.7.1)(π∗ G )p∼

% Gq ⊗O

Y,q

OX,p .

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November 18, 2017 draft427(8) (pullback on fibers of the quasicoherent sheaves) Pullback of fibers are given as
follows: if π : X → Y, where π(p) = q, then the map∼(π∗ G )|p

% G |q ⊗κ(q) κ(p)induced by (16.3.7.1) is an isomorphism.
(9) (pullback preserves tensor product) π∗ (G ⊗OY G ′ ) = π∗ G ⊗OX π∗ G ′ . (Here G ′is also a quasicoherent sheaf on Y.)(10) Pullback is a right-exact functor.
All of the above are interconnected in obvious ways that you should be able
to prove by hand. (As just one example: the germ of a pulled back section, (6), is
the expected element of the pulled back stalk, (7).) In fact much more is true, thatyou should be able to prove on a moment’s notice, such as for example that the
pullback of the symmetric power of a locally free sheaf is naturally isomorphic tothe symmetric power of the pullback, and similarly for wedge powers and tensor
powers.
16.3.E. I MPORTANT E XERCISE . Prove Theorem 16.3.7. Possible hints: You may
find it convenient to do right-exactness (10) early; it is related to right-exactness of∼ (M⊗B N)⊗B A,⊗. For the tensor product fact (9), show that (M⊗B A)⊗(N⊗B A) =and that this behaves well with respect to localization. The proof of the fiber fact(8) is as follows. Given a ring map B → A with [m] (→ [n], where m ⊂ A and n ⊂ B∼ (N ⊗B (B/n)) ⊗B/n (A/m) by
are maximal ideals, show that (N ⊗B A) ⊗A (A/m) =showing both sides are isomorphic to N ⊗B (A/m).
16.3.F. U NIMPORTANT E XERCISE . Verify that the following is an example showingthat pullback is not left-exact: consider the exact sequence of sheaves on A1k , wherei : p "→ A1k is the origin:
0 → OA1k (−p) → OA1k → i∗ O|p → 0.

(This is the closed subscheme exact sequence for p ∈ A1k , and corresponds to theexact sequence of k[t]-modules 0 → tk[t] → k[t] → k → 0. Warning: here O|p is
not the stalk Op ; it is the structure sheaf of the scheme p.) Restrict to p.

16.3.G. E XERCISE ( THE PUSH - PULL FORMULA , CF. E XERCISE 18.8.B). Supposeψ : Z → Y is any morphism, and both π : X → Y and π ′ : W → Z are quasicompact
and quasiseparated (so pushforwards send quasicoherent sheaves to quasicoherent sheaves). Suppose F is a quasicoherent sheaf on X. Suppose(16.3.7.2)

W

ψ′

%X

π′

π

"
Z

ψ

"
%Y

is a commutative diagram. Describe a natural morphism ψ∗ π∗ F → π∗′ (ψ ′ )∗ F of
quasicoherent sheaves on Z. (Possible hint: first do the special case where (16.3.7.2)
is a Cartesian diagram.)
By applying the above exercise in the special case where Z is a point q of Y,
we see that there is a natural map from the fiber of the pushforward to the sections

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over the fiber:(16.3.7.3)

π∗ F ⊗ κ(q) → Γ (π−1 (q), F |π−1 (q) ).

One might hope that (16.3.7.3) is an isomorphism, i.e., that π∗ F “glues together”
the fibers Γ (π−1 (q), F |π−1 (q) ). This is too much to ask, but at least (16.3.7.3) gives
a map. (In fact, under just the right circumstances, (16.3.7.3) is an isomorphism,see §28.1.)
16.3.H. I MPORTANT E XERCISE ( PROJECTION FORMULA , TO BE GENERALIZED IN
E XERCISE 18.8.E). Suppose π : X → Y is quasicompact and quasiseparated, and
F and G are quasicoherent sheaves on X and Y respectively.
(a) Describe a natural morphism (π∗ F ) ⊗ G → π∗ (F ⊗ π∗ G ).
(b) If G is locally free, show that this natural morphism is an isomorphism. (Hint:
what if G is free?)(c) If π is affine, then show that this natural morphism is an isomorphism.
16.3.8. Remark: Flatness. Given π : X → Y, if the functor π∗ from quasicoherent
sheaves on Y to quasicoherent sheaves on X is exact (not just right-exact as given
by Theorem 16.3.7(10)), we will say that π is a flat morphism. This is an incredibly important notion. We will return to it (and define it in a different-looking,equivalent but better way) in Chapter 24; see Exercise 24.2.L.
16.3.9. Remark: Pulling back ideal sheaves. There is one subtlety in pulling back quasicoherent ideal sheaves. Suppose i : X "→ Y is a closed embedding, and µ : Y ′ → Y
is an arbitrary morphism. Let X ′ := X ×Y Y ′ . As “closed embeddings pull back”(§9.2.2), the pulled back map i ′ : X ′ → Y ′ is a closed embedding.
X ′" #

i

% X"

′

"
Y′

#

i
µ

"
%Y

∼ OY ′ and µ∗ (i∗ OX ) =∼ (i ′ OX ′ ).
Now µ∗ induces canonical isomorphisms µ∗ OY =∗(Why is the latter isomorphism true? Hint: check affine-locally.) But it is not necessarily true that µ∗ IX/Y = IX ′ /Y ′ . (Exercise 16.3.F yields an example.) This isbecause the application of µ∗ to the closed subscheme exact sequence
0 → IX/Y → OY → i∗ OX → 0

yields something that is a priori only right-exact:µ∗ IX/Y → OY ′ → i∗′ OX ′ → 0.

Thus, as IX ′ /Y ′ is the kernel of OY ′ → i∗′ OX ′ , we see that IX ′ /Y ′ is the image ofµ∗ IX/Y in OY ′ . We can also see this explicitly from Exercise 9.2.B: affine-locally,
the ideal of the pullback is generated by the pullback of the ideal.
Note also that if µ is flat (Remark 16.3.8), then µ∗ IX/Y → IX ′ /Y ′ is an isomorphism.

16.4 Line bundles and maps to projective schemes

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Theorem 16.4.1, the converse or completion to Exercise 15.3.F, will give one
reason why line bundles are crucially important: they tell us about maps to projective space, and more generally, to quasiprojective A-schemes. Given that we have
had a hard time naming any non-quasiprojective schemes, they tell us about maps
to essentially all schemes that are interesting to us.
16.4.1. Important Theorem. — For a fixed scheme X, morphisms X → Pn are in
bijection with the data (L , s0 , . . . , sn ), where L is an invertible sheaf and s0 , . . . , sn are
sections of L with no common zeros, up to isomorphism of these data.
(This works over Z, or indeed any base.) Informally: morphisms to Pn correspond to n + 1 sections of a line bundle, not all vanishing at any point, modulo
global sections of OX∗ , as multiplication by an invertible function gives an automorphism of L . This is one of those important theorems in algebraic geometry that iseasy to prove, but quite subtle in its effect on how one should think. It takes sometime to properly digest. A “coordinate-free” version is given in Exercise 16.4.E.
16.4.2. Theorem 16.4.1 describes all morphisms to projective space, and hence by
the Yoneda philosophy, this can be taken as the definition of projective space: it
defines projective space up to unique isomorphism. Projective space Pn (over Z) isthe moduli space of line bundles L along with n + 1 sections of L with no common zeros.
(Can you give an analogous definition of projective space over X, denoted Pn
X ?)Every time you see a map to projective space, you should immediately simultaneously keep in mind the invertible sheaf and sections.
Maps to projective schemes can be described similarly. For example, if Y "→ P2k
is the curve x22 x0 = x31 − x1 x20 , then maps from a scheme X to Y are given by aninvertible sheaf on X along with three sections s0 , s1 , s2 , with no common zeros,satisfying s22 s0 − s31 + s1 s20 = 0.
Here more precisely is the correspondence of Theorem 16.4.1. Any n + 1 sections of L with no common zeros determine a morphism to Pn , by Exercise 15.3.F.
Conversely, if you have a map to projective space π : X → Pn , then we have n + 1
sections of OPn (1), corresponding to the hyperplane sections, x0 , . . . , xn . Thenπ∗ x0 , . . . , π∗ xn are sections of π∗ OPn (1), and they have no common zero. (Reminder: it is helpful to think of pulling back invertible sheaves in terms of pullingback transition functions, see Theorem 16.3.7(3).)
So to prove Theorem 16.4.1, we just need to show that these two constructions
compose to give the identity in either direction.
Proof of Important Theorem 16.4.1. Suppose we are given n + 1 sections s0 , . . . , snof an invertible sheaf L , with no common zeros, which (via Exercise 15.3.F) induce a morphism π : X → Pn . For each si , we get a trivialization of L on the
open set Xsi where si doesn’t vanish. (More precisely, we have an isomorphism∼ (O, 1), cf. Important Exercise 14.2.E(a).) The transition functions for L(L , si ) =are precisely si /sj on Xsi ∩ Xsj . As O(1) is trivial on the standard affine open setsD(xi ) of Pn , π∗ (O(1)) is trivial on Xsi = π−1 (D(xi )). Moreover, si /sj = π∗ (xi /xj )(directly from the construction of π in Exercise 15.3.F). This gives an isomorphism∼ π∗ O(1) — the two invertible sheaves have the same transition functions.
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16.4.A. E XERCISE . Show that this isomorphism can be chosen so that (L , s0 , . . . , sn ) =∗∗∗(π O(1), π x0 , . . . , π xn ), thereby completing one of the two implications of thetheorem.
For the other direction, suppose we are given a map π : X → Pn . Let si =π xi ∈ Γ (X, π∗ (O(1))). As the xi ’s have no common zeros on Pn , the si ’s haveno common zeros on X. The map [s0 , . . . , sn ] is the same as the map π. We see
this as follows. The preimage of D(xi ) (by the morphism [s0 , . . . , sn ]) is D(si ) =D(π∗ xi ) = π−1 D(xi ), so “the right open sets go to the right open sets”. To show thetwo morphisms D(si ) → D(xi ) (induced from (s1 , . . . , sn ) and π) are the same, we
use the fact that maps to an affine scheme D(xi ) are determined by their maps ofglobal sections in the opposite direction (Essential Exercise 6.3.F). Both morphisms
D(si ) → D(xi ) corresponds to the ring map π♯ : xj/i = xj /xi (→ sj /si .
!
∗16.4.3. Remark: Extending Theorem 16.4.1 to rational maps. Suppose s0 , . . . , sn are
sections of an invertible sheaf L on a scheme X. Then Theorem 16.4.1 yields amorphism X − V(s0 , . . . , sn ) → Pn . In particular, if X is integral, and the si are not
all 0, these data yield a rational map X ""# Pn .
16.4.4. Examples and applications.

16.4.B. E XERCISE ( AUTOMORPHISMS OF PROJECTIVE SPACE ). Show that all theautomorphisms of projective space Pnk (fixing k) correspond to (n + 1) × (n +1) invertible matrices over k, modulo scalars (also known as PGLn+1 (k)). (Hint:n∗∼
Suppose π : Pn
k → Pk is an automorphism. Show that π O(1) = O(1). Show that∗nn
π : Γ (P , O(1)) → Γ (P , O(1)) is an isomorphism.)
Exercise 16.4.B will be useful later, especially for the case n = 1. In this case,these automorphisms are called fractional linear transformations. (For experts:
why was Exercise 16.4.B not stated over an arbitrary base ring A? Where does theargument go wrong in that case? For what rings A does the result still work?)

16.4.C. E XERCISE .
Show that Aut(P1k ) is strictly three-transitive on k-valued
points, i.e., given two triplets (p1 , p2 , p3 ) and (q1 , q2 , q3 ) each of distinct k-valued
points of P1 , there is precisely one automorphism of P1 sending pi to qi (i =1, 2, 3).
Here are more examples of these ideas in action.
16.4.5. Example: the Veronese embedding is |OPnk (d)|.
Consider the line bundle
OPnk (d) on Pn
.
We
have
checkedthat
thesectionsof
thisline bundle form a vectork)n+d*space of dimension d , with a basis corresponding to homogeneous degree dpolynomials in the projective coordinates for Pn
k . Also, they have no common zeddros (as for example the subset of sections xd,
x
0
1 , . . . , xn have no common zeros).
Thus the complete linear series is base-point-free, and determines a morphism
n+dPn → P( d )−1 . This is the Veronese embedding (Definition 8.2.8). For example,
if n = 2 and d = 2, we get a map P2k → P5k .
In §8.2.8, we saw that this is a closed embedding. The following is a moregeneral method of checking that maps to projective space are closed embeddings.

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16.4.D. L ESS IMPORTANT EXERCISE . Suppose π : X → Pn
A corresponds to an invertible sheaf L on X, and sections s0 , . . . , sn . Show that π is a closed embedding
if and only if(i) each open set Xsi is affine, and(ii) for each i, the map of rings A[y0 , . . . , yn ] → Γ (Xsi , O) given by yj (→ sj /si
is surjective.
16.4.6. Special case of Example 16.4.5. Recall that the image of the Veronese embedding when n = 1 is called a rational normal curve of degree d (Exercise 8.2.J). Ourd d−1
map is P1k → Pdy, . . . , xyd−1 , yd ].
k given by [x, y] (→ [x , x
16.4.E. E XERCISE . Show that a (finite-dimensional) base-point-free linear series V
on X corresponding to L induces a morphism to projective space
φV : X% PV ∨ .

(This should be seen as a coordinate-free version of Theorem 16.4.1.)
16.4.F. E XERCISE . (For this exercise, we work over a field k.) Suppose we have amorphism π : X → Pn , corresponding to the base-point-free linear series Γ (Pn , O(1)) →Γ (X, L ) (so L = π∗ O(1)). If the scheme-theoretic image of X in Pn lies in a hyperplane, we say that the linear series (or X itself) is degenerate (and otherwise,
nondegenerate). Show that a base-point-free linear series V → Γ (X, L ) is nondegenerate if and only if the map V → Γ (X, L ) is an inclusion. In particular, acomplete linear series is always nondegenerate.
16.4.G. E XERCISE . Suppose we are given a map π : P1k → Pn
k where the corresponding invertible sheaf on P1k is O(d). (This can reasonably be called a degreed map, cf. Exercises 17.4.F and 18.6.I.) Show that if d < n, then the image is degenerate. Show that if d = n and the image is nondegenerate, then the image isisomorphic (via an automorphism of projective space, Exercise 16.4.B) to a rational
normal curve.

16.4.H. E XERCISE . Define the graded rings R• = k[u, v, w]/(uw − v2 ) and S• =k[x, y] (with all variables having degree 1). By Exercise 8.2.I, we have an isomor∼ Proj S• (via the Veronese embedding ν2 ). Show that this isomorphism Proj R• =phism is not induced by a map of graded rings S• → R• .
16.4.7. Remark. You may be able to show that after “regrading”, the isomorphism∼ Proj S• does arise from a map of graded rings (S2• → R• ). Exercise 19.11.B
Proj R• =gives an example where it is not possible to patch the lack of maps of graded rings
by just regrading.

16.4.I. E XERCISE : AN EARLY LOOK AT INTERSECTION THEORY, RELATED TO B ÉZOUT ’ S
T HEOREM . A classical definition of the degree of a curve in projective spaceis as follows: intersect it with a “general” hyperplane, and count the number ofpoints of intersection, with appropriate multiplicity. We interpret this in the case ofn1
π : P1k → Pn
k . Show that there is a hyperplane H of Pk not containing π(Pk ). Equiv∗
1
alently, π H ∈ Γ (P , OP1 (d)) (interpreted as the pullback of a section of OPn (1))
is not 0. Show that the number of zeros of π∗ H is precisely d. (You will have todefine “appropriate multiplicity”. What does it mean geometrically if π is a closed

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embedding, and π∗ H has a double zero? Aside: Can you make sense of this even if
π is not a closed embedding?) Thus this classical notion of degree agrees with thenotion of degree in Exercise 16.4.G. (See Exercise 8.2.E for another case of Bézout’s
Theorem. Here we intersect a degree d curve with a degree 1 hyperplane; there weintersect a degree 1 curve with a degree d hypersurface. Exercise 18.6.K will givea common generalization.)
16.4.8. Example: The Segre embedding revisited. The Segre embedding can also beinterpreted in this way. This is a useful excuse to define some notation. Suppose
F is a quasicoherent sheaf on a Z-scheme X, and G is a quasicoherent sheaf on
a Z-scheme Y. Let πX , πY be the projections from X ×Z Y to X and Y respectively.
Then F % G (pronounced “F box-times G ”) is defined to be π∗X F ⊗ π∗Y G . In
particular, OPm ×Pn (a, b) is defined to be OPm (a) % OPn (b) (over any base Z). The
Segre embedding Pm × Pn → Pmn+m+n corresponds to the complete linear series
for the invertible sheaf O(1, 1).
When we first saw the Segre embedding in §9.6, we saw (in different language)that this complete linear series is base-point-free. We also checked by hand (§9.6.1)
that it is a closed embedding, essentially by Exercise 16.4.D.
Recall that if L and M are both base-point-free invertible sheaves on a scheme
X, then L ⊗ M is also base-point-free (Exercise 15.3.B, see also Definition 15.3.4).
We may interpret this fact using the Segre embedding (under reasonable hypotheses on X). If φL : X → PM is a morphism corresponding to a (base-point-free)linear series based on L , and φM : X → PN is a morphism corresponding to alinear series on M , then the Segre embedding yields a morphism X → PM ×
PN → P(M+1)(N+1)−1 , which corresponds to a base-point-free series of sections of
L ⊗ M.
16.4.J. F UN E XERCISE . Suppose X is a quasiprojective k-scheme, and π : Pnk →X is any morphism (over k). Show that either the image of π has dimension n,or π contracts Pn
k to a point. In particular, there are no nonconstant maps fromprojective space to a smaller-dimensional quasi-projective variety. Hint: if X ⊂ PN ,∼ OPn (d). Try to show that d = 0. To do that, show thatdefine d by π∗ OPN (1) =k
if m ≤ n then m nonempty hypersurfaces in Pn have nonempty intersection. For
this, use the fact that any nonempty hypersurface in Pn
k has nonempty intersection
with any subscheme of dimension at least 1 (Exercise 11.3.C(a)).

16.4.K. E XERCISE . Explain how GLn acts (nontrivially!) on Pn−1 (over Z, or overa field of your choice). (The group scheme GLn was defined in Exercise 6.6.N.
The action of a group scheme appeared earlier in Exercise 6.6.S(a).) Hint: this ismuch more easily done with the language of functors, §6.6, using our functorialdescription of projective space (§16.4.2), than with our old description of projective
space in terms of patches. (A generalization to the Grassmannian will be given in
Exercise 16.7.L.)
16.4.9. Remark. Over an algebraically closed field k, GLn acts transitively on
the closed points of Pn, and the stabilizer of the point [1, 0, · · · , 0] consists of theksubgroup P of matrices with 0’s in the first column below the first row. (SideRemark: the point is better written as a column vector, so the GLn -action can beinterpreted as matrix multiplication in the usual way.) This suggests that Pn−1
isk

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the quotient GLn /P. This is largely true; but we first would have to make sense of
the notion of group quotient.
16.4.10. ⋆⋆ A proper nonprojective k-variety — and gluing schemes along closed
subschemes.
We conclude by using what we have developed to describe an example of ascheme that is proper but not projective (promised in Remark 10.3.6). We use aconstruction that looks so fundamental that you may be surprised to find that we
won’t use it in any meaningful way later.
∼ P3 , Zi be a line in Xi ,Fix an algebraically closed field k. For i = 1, 2, let Xi =k′and Zi be a regular conic in Xi disjoint from Zi (both Zi and Zi′ isomorphic to P1k ).
The construction of §16.4.11 will allow us to glue X1 to X2 so that Z1 is identified
with Z2′ and Z1′ is identified with Z2 , see Figure 16.2. (You will be able to make this
precise after reading §16.4.11.) The result, call it X, is proper, by Exercise 16.4.O.

F IGURE 16.2. Building a proper nonprojective variety
Then X is not projective. For if it were, then it would be embedded in projective space by some invertible sheaf L . If X is embedded, then X1 is too, so L must
restrict to an invertible sheaf on X1 of the form OX1 (n1 ), where n1 > 0. You cancheck that the restriction of L to Z1 is OZ1 (n1 ), and the restriction of L to Z1′ is
OZ1′ (2n1 ). Symmetrically, the restriction of L to Z2 is OZ2 (n2 ) for some n2 > 0,
and the restriction of L to Z2′ is OZ2′ (2n2 ). But after gluing, Z1 = Z2′ , and Z1′ = Z2 ,so we have n1 = 2n2 and 2n1 = n2 , which is impossible.
16.4.11. Gluing two schemes together along isomorphic closed subschemes.
It is straightforward to show that you can glue two schemes along isomorphic open subschemes. (More precisely, if X1 and X2 are schemes, with open sub∼ U2 , you can makeschemes U1 and U2 respectively, and an isomorphism U1 =∼sense of gluing X1 and X2 along U1 = U2 . You should think this through.) You
can similarly glue two schemes along isomorphic closed subschemes. We now
make this precise. Suppose Z1 "→ X1 and Z2 "→ X2 are closed embeddings, and∼% Z2 is an isomorphism. We will explain how to glue X1 to X2 along
φ : Z1#
φ. The result will be called X1 φ X2 .
Our motivating example is if Xi = Spec Ai and∼% A1 /I1 . Then the result
Zi = Spec Ai /Ii , and φ corresponds to φ♯ : A2 /I2will be Spec R, where R is the ring of consisting of ordered pairs (a1 , a2 ) ∈ A1 × A2

16.4.12. Motivating example.

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The Rising Sea: Foundations of Algebraic Geometrythat “agree via φ”. More precisely, this is a fibered product of rings:
R := A1 ×φ♯ : A1 /I1 →A2 /I2 A2 .
16.4.13. The general construction, as a locally ringed space. In our general situation,
we might wish to cover X1 and X2 by open charts of this form. We would thenhave
#to worry about gluing and choices, so to avoid this, we instead first construct
X1 φ X2 as a locally ringed space. As a topological space, the definition is clear:∼
we glue the underlying sets together# along the underlying sets of Z1 = Z2 , andtopologize it so that a subset of X1 φ X2 is open if and only if its restrictions to
X1 and X2#
are both open. For convenience, let Z be the image of Z1 (or equivalently
Z2 ) in X1 φ X2 . We next define the stalk of the structure sheaf at any point p ∈##X1 φ X2 . If p ∈ Xi \ Z = (X1 φ X2 ) \ X3−i (hopefully the meaning of this isclear), we define the stalk as OXi ,p . If p ∈ X1 ∩ X2 , we define the stalk to consist of∼ OZ ,p . The meaning ofelements (s1 , s2 ) ∈ OX1 ,p × OX2 ,p which agree in OZ1 ,p =2everything in this paragraph will be clear to you if you can do the following.
16.4.L. E XERCISE . Define the structure sheaf of OX1 &φ X2 in terms of compatible
germs. (What should it mean for germs to be compatible? Hint: for z ∈ Z, suppose
we have open subsets U1 of X1 and U2 of X2 , with
#U1 ∩ Z = U2 ∩ Z, so U1 andU2 glue together to give an open subset U of X1 φ X2 . Suppose we also havefunctions f1 on U1 and f2 on U2 that “agree on U ∩ Z” — what does that mean?
Then we declare that the germs of the “function on U obtained by gluing together
f1 and f2 ” are compatible.) Show that the resulting ringed space is a locally ringed
space.
#
We next want to show that the locally ringed space X1 φ X2 is a scheme.
Clearly it is a scheme away from Z. We first verify a special case.
16.4.M. E XERCISE . Show that in Example 16.4.12 the construction of §16.4.13
indeed yields Spec(A1 ×φ♯ A2 ).
16.4.N. E XERCISE . In the general case, suppose x ∈ Z. Show that there is anaffine open subset Spec Ai ⊂ X#
i such that Z ∩ Spec A1 = Z ∩ Spec A2 . Then use
Exercise 16.4.L to show that X1 φ X2 is a scheme in an open neighborhood of x,
and thus a scheme.
16.4.14. Remarks.
(a) As the notation suggests, this is a fibered coproduct in the category of schemes,
and indeed in the category of locally ringed spaces. We won’t need this fact, but
you can prove it if you wish; it isn’t hard. Unlike the situation for products, fiberedcoproducts don’t exist in general in the category of schemes. Miraculously (andfor reasons that are specific to schemes), the resulting cofibered diagram is also afibered diagram. This has pleasant ramifications. For example, this construction“behaves well with respect to” (or “commutes with”) base change; this can help
with Exercise 16.4.O(a), but if you use it, you have to prove it.
(b) Here are some interesting questions#to think through: Can we recover the gluing locus from the “glued scheme” X1 φ X2 and the two closed subschemes X1
and X2 ? (Yes.) When is a scheme X the gluing of two closed subschemes X1 and X2

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along their scheme-theoretic intersection? (Answer: When the scheme-theoretic
union of X1 and X2 is the entire scheme X.)(c) You might hope that if you have a single scheme X with two disjoint closed
subschemes W ′ and W ′′ , and an isomorphism W ′ → W ′′ , then you should be able
to glue X to itself along W ′ → W ′′ . This construction doesn’t work, and indeed it
may not be possible. You can still make sense of the quotient as an algebraic space,which we will not define here.
16.4.O. E XERCISE . We continue to use the notation Xi , φ, etc. Suppose we areworking in the category of A-schemes.
#(a) If X1 and X2 are universally closed, show that X1 φ X2 is as well.
#(b) If X1 and X2 are separated, show that X1 φ X2 is as well.
#(c) If X1 and X2 are finite type over a Noetherian ring A, show that X1 φ X2 isas well. (Hint: Reduce to the “affine” case of the Motivating Example 16.4.12.
Choose generators x1 , . . . , xn of A1 , and y1 , . . . , yn , such that xi modulo I1
agrees with yi modulo I2 via φ. Choose generators g1 , . . . , gm of I2 — here use
Noetherianness of A. Show that (xi , yi ) and (0, gi ) generate R ⊂ A1 × A2 , asfollows. Suppose (a1 , a2 ) ∈ R. Then there is some polynomial m such that′a1 = m(x1 , . . . , xn ). Hence (a1 , a2 ) − m((x,m 1 , y1 ), . . . , (xn , yn )) = (0, a2 ) for some′′′a,2m∈ I2 . Then a2 can be written as i=1 ℓi (y1 , . . . , yn )gi . But then (0, a2 ) =i=1 ℓi ((x1 , y1 ), . . . , (xn , yn ))(0, gi ).)#Thus if X1 and X2 are proper, so is X1 φ X2 .
On the# other hand, we have just seen (§16.4.10) that if X1 and X2 are projective,then X1 φ X2 needn’t be projective.

16.5 The Curve-to-Projective Extension Theorem
We now use the main theorem of the previous section, Theorem 16.4.1, toprove something useful and concrete.
16.5.1. The Curve-to-Projective Extension Theorem. — Suppose C is a pure dimension 1 Noetherian scheme over an affine base S = Spec A, and p ∈ C is a regular closed
point of it. Suppose Y is a projective S-scheme. Then any morphism C \ {p} → Y (ofS-schemes) extends to all of C.

In practice, we will use this theorem when S = Spec k, and C is a k-variety.
The only reason we assume S is affine is because we won’t know the meaning of“projective S-scheme” until we know what a projective morphism is (§17.3). But
the proof below extends immediately to general S once we know the meaning of
the statement.
Note that if such an extension exists, then it is unique: the nonreduced locus
of C is a closed subset (Exercise 5.5.E). Hence by replacing C by an open neighborhood of p that is reduced, we can use the Reduced-to-Separated Theorem 10.2.2
that maps from reduced schemes to separated schemes are determined by their
behavior on a dense open set. Alternatively, maps to a separated scheme can beextended over an effective Cartier divisor in at most one way (Exercise 10.2.G).
The following exercise shows that the hypotheses are necessary.

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16.5.A. E XERCISE . In each of the following cases, prove that the morphism C \{p} → Y cannot be extended to a morphism C → Y.

(a) Projectivity of Y is necessary. Suppose C = A1k , p = 0, Y = A1k , and C\{p} →Y is given by “t (→ 1/t”.
(b) One-dimensionality of C is necessary. Suppose C = A2k , p = (0, 0), Y = P1k ,and C \ {p} → Y is given by (x, y) (→ [x, y].
(c) Non-singularity of C is necessary. Suppose C = Spec k[x, y]/(y2 −x3 ), p = 0,
Y = P1k , and C \ {p} → Y is given by (x, y) (→ [x, y].

We remark that by combining this (easy) theorem with the (hard) valuative
criterion for properness (Theorem 12.7.6), one obtains a proof of the propernessof projective space bypassing the (tricky but not hard) Fundamental Theorem ofElimination Theory 7.4.7 (see Exercise 12.7.F). Fancier remark: the valuative criterion of properness can be used to show that Theorem 16.5.1 remains true if Y is
only required to be proper, but it requires some thought.
16.5.2. Central idea of proof. The central idea of the proof may be summarized
as “clear denominators”, and is illustrated by the following motivating example.
Suppose you have a morphism from A1 − {0} to projective space, and you wanted
to extend it to A1 . Suppose the map was given by t (→ [t4 + t−3 , t−2 + 4t]. Thenof course you would “clear the denominators”, and replace the map by t (→ [t7 +1, t + 4t4 ]. Similarly, if the map was given by t (→ [t2 + t3 , t2 + t4 ], you would
divide by t2 , to obtain the map t (→ [1 + t, 1 + t2 ].

Proof. Our plan is to maneuver ourselves into the situation where we can apply theidea of §16.5.2. We begin with some quick reductions. The nonreduced locus of C
is closed and doesn’t contain p (Exercise 5.5.E), so by replacing C by an appropriate
open neighborhood of p, we may assume that C is reduced and affine.
n
We next reduce to the case where Y = Pn
A . Choose a closed embedding Y → PA .
nn
If the result holds for P , and we have a morphism C → P with C \ {p} mapping
to Y, then C must map to Y as well. Reason: we can reduce to the case wherenthe source is an affine open subset, and the target is An
A ⊂ PA (and hence affine).
nThen the functions vanishing on Y ∩ AA pull back to functions that vanish at thegeneric points of the irreducible components of C and hence vanish everywhere
on C (using reducedness of C), i.e., C maps to Y.
Choose a uniformizer t ∈ m − m2 in the local ring of C at p. As t is a function
in some neighborhood of p, we may assume that t is a function on C (by replacing
C by this neighborhood of p). Then V(t) contains p (as t ∈ m), but V(t) does notcontain the component of C containing p (as t ∈/ m2 ), so by replacing C by an affine
open neighborhood of p in (C \ V(t)) ∪ {p} ((C \ V(t)) ∪ {p} ⊂ C is open because C(almost) has the cofinite topology (§3.4), we may assume that p is the only zero of
the function t (and of course t vanishes at p with multiplicity 1).
We have a map C \ {p} → Pn
A , which by Theorem 16.4.1 corresponds to a line
bundle L on C \ {p} and n + 1 sections of it with no common zeros in C \ {p}. Let∼ O. Then by replacing C by
U be a nonempty open set of C \ {p} on which L =nU∪p, we interpret the map to P as n+1 rational functions f0 , . . . , fn , defined away
away from p, with no common zeros away from p. Let N = mini (valp fi ). Thent−N f0 , . . . , t−N fn are n + 1 functions with no common zeros. Thus they determinena morphism C → Pn!
A extending C \ {p} → PA as desired.

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16.5.B. E XERCISE ( USEFUL PRACTICE ). Suppose X is a Noetherian k-scheme, and Z
is an irreducible codimension 1 subvariety whose generic point is a regular pointof X (so the local ring OX,Z is a discrete valuation ring). Suppose π : X ""# Y
is a rational map to a projective k-scheme. Show that the domain of definitionof the rational map includes a dense open subset of Z. In other words, rational
maps from Noetherian k-schemes to projective k-schemes can be extended overregular codimension 1 sets. (We have already seen this principle in action — see
Exercise 6.5.I on the Cremona transformation.)

16.6 Ample and very ample line bundles
Suppose π : X → Spec A is a proper morphism, and L is an invertible sheaf
on X. (The case when A is a field is the one of most immediate interest.) We
say that L is very ample over A or π-very ample, or relatively very ample if∼ Proj S• where S• is a finitely generated graded ring over A generated in degree
X=∼ OProj S (1). One often just says very ample if the
1 (Definition 4.5.6), and L =•structure morphism is clear from the context. Note that the existence of a very
ample line bundle implies that X is projective.
16.6.A. E ASY BUT I MPORTANT E XERCISE ( EQUIVALENT DEFINITION OF VERY AM PLE OVER A). Suppose π : X → Spec A is proper, and L is an invertible sheaf on
X. Show that L is very ample if and only if there exist a finite number of global
sections s0 , . . . sn of L , with no common zeros, such that the morphism

is a closed embedding.

[s0 , . . . , sn ] : X → Pn
A

16.6.B. E ASY EXERCISE ( VERY AMPLE IMPLIES BASE - POINT- FREE ). Show that avery ample invertible sheaf L on a proper A-scheme must be base-point-free.
16.6.C. E XERCISE ( VERY AMPLE ⊗ BASE - POINT- FREE IS VERY AMPLE , HENCE VERYAMPLE ⊗ VERY AMPLE IS VERY AMPLE ). Suppose L and M are invertible sheaves
on a proper A-scheme X, and L is very ample over A and M is base-point-free,then L ⊗ M is very ample. (Hint: L gives a closed embedding X "→ Pm , andM gives a morphism X → Pn . Show that the product map X → Pm × Pn is a
closed embedding, using the Cancellation Theorem 10.1.19 for closed embeddings
on X → Pm × Pn → Pm . Finally, consider the composition X "→ Pm × Pn "→Pmn+m+n , where the last closed embedding is the Segre embedding.)

16.6.D. E XERCISE ( VERY AMPLE % VERY AMPLE IS VERY AMPLE , CF. E XAMPLE 16.4.8).
Suppose X and Y are proper A-schemes, and L (resp. M ) is a very ample invertible sheaf on X (resp. Y). If πx : X ×A Y → X and πY : X ×A Y → Y are the usual
projections, show that π∗X L ⊗ π∗Y M (also known as L % M , see §16.4.8) is very
ample on X ×A Y.
16.6.1. Definition. We say an invertible sheaf L on a proper A-scheme X is ample
over A or π-ample (where π : X → Spec A is the structure morphism), or relatively
ample if one of the following equivalent conditions holds.

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16.6.2. Theorem. — Suppose π : X → Spec A is proper, and L is an invertible sheaf on
X. The following are equivalent.

(a) For some N > 0, L ⊗N is very ample over A.
(a’) For all n ≫ 0, L ⊗n is very ample over A.
(b) For all finite type quasicoherent sheaves F , there is an n0 such that for n ≥ n0 ,F ⊗ L ⊗n is globally generated.
(c) As f runs over all the global sections of L ⊗n (over all n > 0), the open subsets
Xf = {p ∈ X : f(p) ̸= 0} form a base for the topology of X.
(c’) As f runs over the global sections of L ⊗n (n > 0), those open subsets Xf which
are affine form a base for the topology of X.

(Variants of this Theorem 16.6.2 in the “absolute” and “relative” settings will
be given in Theorems 16.6.6 and 17.3.9 respectively.)
Properties (a) and (a’) relate to projective geometry, and property (b) relates toglobal generation (stalks). Properties (c) and (c’) are somehow more topological,
and while they may seem odd, they will provide the connection between (a)/(a’)
and (b). Note that (b), (c) and (c’) make no reference to the structure morphism π.
We will later (Theorem 18.7.1) meet a cohomological criterion (due, unsurprisingly,
to Serre). The Kodaira Embedding Theorem (see, for example, [GH1, p. 181]) also
gives a criterion for ampleness in the complex category: if X is a compact complex manifold, then an invertible sheaf L on X is ample if and only if it admits a
Hermitian metric with positive curvature everywhere.
The different flavor of these conditions gives some indication that amplenessis better-behaved than very ampleness in a number of ways. We mention without
proof another property: if π : X → T is a finitely presented proper morphism, and
L is an invertible sheaf on X, then those points on T where L is ample on the fiber
form an open subset of T . Furthermore, on this open subset, L is relatively ample
over the base. We won’t use these facts (proved in [Gr-EGA, IV3 .9.6.4]), but they
are good to know.
Before getting to the proof of Theorem 16.6.2, we give some sample applications. We begin by noting that the fact that (a) implies (b) gives Serre’s Theorem A(Theorem 15.3.8).
16.6.E. I MPORTANT E XERCISE . Suppose L and M are invertible sheaves on aproper A-scheme X, and L is ample. Show that L ⊗n ⊗ M is very ample forn ≫ 0. (Hint: use both (a) and (b) of Theorem 16.6.2, and Exercise 16.6.C.)
16.6.F. I MPORTANT E XERCISE . Show that every line bundle on a projective Ascheme X is the difference of two very ample line bundles. More precisely, for any
invertible sheaf L on X, we can find two very ample invertible sheaves M and N∼ M ⊗ N ∨ . (Hint: use the previous Exercise.)such that L =16.6.G. I MPORTANT E XERCISE ( USED REPEATEDLY ). Suppose π : X → Y is a finite
morphism of proper A-schemes, and L is an ample line bundle on Y. Show thatπ∗ L is ample on X. Hint: use the criterion of Theorem 16.6.2(b). Suppose F isa finite type quasicoherent sheaf on X. We wish to show that F ⊗ (π∗ L )⊗n is
globally generated for n ≫ 0. Note that (π∗ F ) ⊗ L ⊗n is globally generated forn ≫ 0 by ampleness of L on Y, i.e., there exists a surjection
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where I is some index set. Show that∼ π∗ (O ⊕I )OX⊕I =Y% π∗ (π∗ F ⊗ L ⊗n )is surjective. Pullback π∗ preserves tensor products (Theorem 16.3.7(9)), so we∼ π∗ (π∗ F ) ⊗ (π∗ L )⊗n . Show (using only
have an isomorphism π∗ (π∗ F ⊗ L ⊗n ) =∗affineness of π) that π π∗ F → F is surjective. Connect these pieces together todescribe a surjection
OX⊕I% % F ⊗ (π∗ L )⊗n .

(Remark for those who have read about ampleness in the absolute setting in §16.6.5:
the argument applies in that situation, i.e., with “proper A-schemes” changed to“schemes”, without change. The only additional thing to note is that amplenessof L on Y implies that Y is quasicompact from the definition, and separated from
Theorem 16.6.6(d). A relative version of this result appears in §17.3.8. It can be generalized even further, with “π finite” replaced by “π quasiaffine” — to be definedin §17.3.11 — see [Gr-EGA, II.5.1.12].)16.6.H. E XERCISE ( AMPLE ⊗ AMPLE IS AMPLE , AMPLE ⊗ BASE - POINT- FREE IS AM PLE . Suppose L and M are invertible sheaves on a proper A-scheme X, and L
is ample. Show that if M is ample or base-point-free, then L ⊗ M is ample.
16.6.I. L ESS IMPORTANT E XERCISE ( AMPLE % AMPLE IS AMPLE ).
cise 16.6.D with “very ample” replaced by “ample”.

Solve Exer-

16.6.3. Proof of Theorem 16.6.2 in the case X is Noetherian. Noetherian hypotheses
are used at only one point in the proof, and we explain how to remove them, andgive a reference for the details.
Obviously, (a’) implies (a).
Clearly (c’) implies (c). We now show that (c) implies (c’). Suppose we have apoint p in an open subset U of X. We seek an affine Xf containing p and containedin U. By shrinking U, we may assume that U is affine. From (c), U contains some
Xf containing p. But this Xf is affine, as it is the complement of the vanishinglocus of a section of a line bundle on an affine scheme (Exercise 7.3.F), so (c’) holds.
Note for future reference that the equivalence of (c) and (c’) did not require thehypothesis of properness.
We next show that (a) implies (c). We embed X in projective space by somepower of L . Given a closed subset Z ⊂ X, and a point p of the complement X \ Z,
we seek a section of some L ⊗N that vanishes on Z and not on p. The existence ofsuch a section follows from the fact that V(I(Z)) = Z (Exercise 4.5.H(c)): there issome element of I(Z) that does not vanish on p.
We next show that (b) implies (c). Suppose we have a point p in an open subset
U of X. We seek a section of L ⊗N that doesn’t vanish at p, but vanishes on X \ U.
Let I be the sheaf of ideals of functions vanishing on X \ U (the quasicoherent
sheaf of ideals cutting out X \ U, with reduced structure). As X is Noetherian, Iis finite type, so by (b), I ⊗ L ⊗N is generated by global sections for some N, so
there is some section of it not vanishing at p. (Noetherian note: This is the only partof the argument where we use Noetherian hypotheses. They can be removed asfollows. Show that for a quasicompact quasiseparated scheme, every ideal sheaf isgenerated by its finite type subideal sheaves. Indeed, any quasicoherent sheaf on a

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quasicompact quasiseparated scheme is the union of its finite type quasicoherent
subsheaves, see [Gr-EGA’, (6.9.9)] or [GW, Cor. 10.50]. One of these finite typeideal sheaves doesn’t vanish at p; use this as I instead.)
We now have to start working harder.
We next show that (c’) implies (b). We wish to show that F ⊗ L ⊗n is globally
generated for n ≫ 0.
We first show that (c’) implies that for some N > 0, L ⊗N is globally generated,
as follows. For each closed point p ∈ X, there is some f ∈ Γ (X, L ⊗N(p) ) notvanishing at p, so p ∈ Xf . (Don’t forget that quasicompact schemes have closed
points, Exercise 5.1.E!) As p varies, these Xf cover all of X. Use quasicompactnessof X to select a finite number of these Xf that cover X. To set notation,
say these are'⊗(Nj )/Ni

Xf1 , . . . , Xfn , where fi ∈ Γ (X, L ⊗Ni ). By replacing fi with fi j " , we may
assume that they are all sections of the same power L ⊗N of L (N = j Nj ). ThenL ⊗N is generated by these global sections.
We next show that it suffices to show that for all finite type quasicoherent
sheaves F , F ⊗ L ⊗mN is globally generated for m ≫ 0. For if we knew this, we
could apply it to F , F ⊗ L , . . . , F ⊗ L ⊗(N−1) (a finite number of times), and theresult would follow. For this reason, we can replace L by L ⊗N . In other words,
to show that (c’) implies (b), we may also assume the additional hypothesis that
L is globally generated.
For each closed point p, choose an affine open neighborhood of the form Xf ,using (c’). Then F |Xf is generated by a finite number of global sections (Easy
Exercise 15.3.A — a finite type quasicoherent sheaf on Spec A corresponds to a
finitely generated A-module). By Exercise 13.3.H, each of these generators can be
expressed as a quotient of a section (over X) of F ⊗ L ⊗M(p) by fM(p) . (Note: we
can take a single M(p) for each p.) Then F ⊗ L ⊗M(p) is globally generated at p bya finite number of global sections. By Exercise 15.3.C(b), F ⊗ L ⊗M(p) is globally
generated at all points in some open neighborhood Up of p. As L is also globally′
generated, this implies that F ⊗ L ⊗M is globally generated at all points of Up
for M ′ ≥ M(p) (cf. Easy Exercise 15.3.B). From quasicompactness of X, a finite
number of these Up cover X, so we are done (by taking the maximum of theseM(p)).
Our penultimate step is to show that (c’) implies (a). Our goal is to assume (c’),
and to find sections of some L ⊗N that embeds X into projective space. Choose acover of (quasicompact) X by n affine open subsets Xa1 , . . . , Xan , where a1 , . . . , an
are all sections of powers of L . By replacing each section with a suitable power,
we may assume that they are all sections of the same power of L , say L ⊗N . Say
Xai = Spec Ai , where (using that π is finite type) Ai = Spec A[ai1 , . . . , aiji ]/Ii .
m
By Exercise 13.3.H, each aij is of the form sij /ai ij , where sij ∈ Γ (X, L ⊗Nmij )m−mij(for some mij ). Let m = maxi,j mij . Then for each i, j, aij = (sij ai
)/am
i .
m−m
ij
For convenience, let bi = am
,
and
b
=
s
a
;
these
are
all
globalsections
of
ij
ij i
i
⊗mN
L
. Now consider the linear series generated by the bi and bij . As the D(bi ) =Xai cover X, this linear series is base-point-free, and hence (by Exercise 15.3.F)
gives a morphism to PQ (where Q = #bi + #bij − 1). Let x1 , . . . , xn , . . . , xij , . . . be
the projective coordinates on PQ , so f∗ xi = bi , and f∗ xij = bij . Then the morphismof affine schemes Xai → D(xi ) is a closed embedding, as the associated maps ofrings is a surjection (the generator aij of Ai is the image of xij /xi ).

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At this point, we note for future reference that we have shown the following.
If X → Spec A is finite type, and L satisfies (c)=(c’), then X is an open embedding
into a projective A-scheme. (We did not use separatedness.) We conclude ourproof that (c’) implies (a) by using properness to show that the image of this open
embedding into a projective A-scheme is in fact closed, so X is a projective Ascheme.
Finally, we note that (a) and (b) together imply (a’): if L ⊗N is very ample (from(a)), and L ⊗n is base-point-free for n ≥ n0 (from (b)), then L ⊗n is very ample forn ≥ n0 + N by Exercise 16.6.C.
!
16.6.4. ⋆⋆ Semiample line bundles. Just as an invertible sheaf is ample if some tensor
power of it is very ample, an invertible sheaf L is said to be semiample if sometensor power of it is base-point-free. (Translation: L is ample if some power gives
a closed embedding into projective space, and L is semiample if some power
gives just a morphism to projective space.) We won’t use this notion.
16.6.5. ⋆ Ampleness in the absolute setting. (We will not use this section inany serious way later.) Note that global generation is already an absolute notion,i.e., is defined for a quasicoherent sheaf on a scheme, with no reference to any
morphism. An examination of the proof of Theorem 16.6.2 shows that amplenessmay similarly be interpreted in an absolute setting. We make this precise. Suppose
L is an invertible sheaf on a quasicompact scheme X. We say that L is ample if as
f runs over the sections of L ⊗n (n > 0), the open subsets Xf = {p ∈ X : f(p) ̸= 0}
form a base for the topology of X. (We emphasize that quasicompactness in X ispart of the condition of ampleness of L .) For example, (i) if X is an affine scheme,
every invertible sheaf is ample, and (ii) if X is a projective A-scheme, O(1) is ample.
16.6.J. E ASY E XERCISE ( PROPERTIES OF ABSOLUTE AMPLENESS ).
(a) Fix a positive integer n. Show that L is ample if and only if L ⊗n is ample.
(b) Show that if Z "→ X is a closed embedding, and L is ample on X, then L |Z isample on Z.
The following result will give you some sense of how ampleness behaves. We
will not use it, and hence omit the proof (which is given in [Stacks, tag 01Q3]).
However, many parts of the proof are identical to (or generalize) the corresponding arguments in Theorem 16.6.2. The labeling of the statements parallels the labelling of the statements in Theorem 16.6.2.
16.6.6. Theorem (cf. Theorem 16.6.2). — Suppose L is an invertible sheaf on aquasicompact scheme X. The following are equivalent.
(b) X is quasiseparated, and for every finite type quasicoherent sheaf F , there is ann0 such that for n ≥ n0 , F ⊗ L ⊗n is globally generated.
(c) As f runs over the global sections of L ⊗n (n > 0), the open subsets Xf = {p ∈X : f(p) ̸= 0} form a base for the topology of X (i.e., L is ample).
(c’) As f runs over the global sections of L ⊗n (n > 0), those open subsets Xf which
are affine form a base for the topology of X.
(d) Let S• be the graded ring ⊕n≥0 Γ (X, L ⊗n ). Then the open sets Xf (with f ∈ S+
homogeneous of positive degree) cover X, and the associated map X → Proj S• is

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The Rising Sea: Foundations of Algebraic Geometryan open embedding. (Warning: S• need not be finitely generated, and Proj S• is
not necessarily finite type.)
Part (d) implies that X is separated (and thus quasiseparated).
16.6.7. ⋆ Transporting global generation, base-point-freeness, and ampleness to
the relative situation.
These notions can be “relativized”. We could do this right now, but we wait
until §17.3.7, when we will have defined the notion of a projective morphism, andthus a “relatively very ample” line bundle.

16.7 ⋆ The Grassmannian as a moduli space
We first defined projective space inelegantly in §4.4.9, and in §16.4.2 we gavea clean (if perhaps surprising) functorial definition. Similarly, in §6.7, we gave apreliminary description of the Grassmannian. We are now in a position to give a
better definition.
We describe the “Grassmannian functor” (which we also denote G(k, n)), thenshow that it is representable (§6.6.2). The construction works over an arbitrarybase scheme, so we work over the final object Spec Z. (You should think throughwhat to change if you wish to work with, for example, complex schemes.) The
functor is defined as follows. To a scheme B, we associate the set of locally free rank
k quotients of the rank n free sheaf,(16.7.0.1)

OB⊕n

%%Q

up to isomorphism. An isomorphism of two such quotients φ : OB⊕n → Q → 0 andφ ′ : OB⊕n → Q ′ → 0 is an isomorphism σ : Q → Q ′ such that the diagram
φ

%Q
O ⊕n❊
❊❊
❊❊
σ
❊❊
❊M "
φ′
Q′

commutes. By Exercise 13.5.B(a), ker φ is locally free of rank n − k. (Thus if you
prefer, you can extend (16.7.0.1), and instead consider the functor to take B to short
exact sequences(16.7.0.2)

0 → S → O ⊕n → Q → 0

of locally free sheaves over B, of ranks n − k, n, and k respectively.)
It may surprise you that we are considering rank k quotients of a rank n sheaf,
not rank k subobjects, given that the Grassmannian should parametrize k-dimensional
subspace of an n-dimensional space. This is done for several reasons. One is thatthe kernel of a surjective map of locally free sheaves must be locally free, while
the cokernel of an injective map of locally free sheaves need not be locally free
(Exercise 13.5.B(a) and (b) respectively). Another reason: we will see in §28.3.3
that the geometric incarnation of this problem indeed translates to this. We canalready see a key example here: if k = 1, our definition yields one-dimensional
quotients O ⊕n → L → 0. But this is precisely the data of n sections of L , with no

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common zeros, which by Theorem 16.4.1 (the functorial description of projective
space) corresponds precisely to maps to Pn , so the k = 1 case parametrizes what
we want.
We now show that the Grassmannian functor is representable for given n andk.
16.7.A. E XERCISE . Show that the Grassmannian functor is a Zariski sheaf (§9.1.7).
Hence by Key Exercise 9.1.I, to show that the Grassmannian functor is representable, we need only cover it with open subfunctors that are representable.
Throughout the rest of this section, a k-subset is a subset of {1, . . . , n} of size k.
16.7.B. E XERCISE .
(a) Suppose I is a k-subset. Make the following statement precise: there is an open
subfunctor G(k, n)I of G(k, n) where the k sections of Q corresponding to I (ofthe n sections of Q coming from the surjection φ : O ⊕n → Q) are linearly independent. Hint: in a trivializing open neighborhood of Q, where we can choose an∼ %isomorphism Q
O ⊕k , φ can be interpreted as a k × n matrix M, and thislocus is where the determinant of the k × k matrix consisting of the I columns of
M is nowhere zero. Show that this locus behaves well under transitions between
trivializations.
(b) Show that these open subfunctors G(k, n)I cover the functor G(k, n) (as I runs
through the k-subsets).
Hence by Exercise 9.1.I, to show G(k, n) is representable, we need only showthat G(k, n)I is representable for arbitrary I. After renaming the summands ofO ⊕n , without loss of generality we may assume I = {1, . . . , k}.
16.7.C. E XERCISE . Show that G(k, n){1...,k} is represented by Ak(n−k) as follows.
(You will have to make this precise.) Given a surjection φ : O ⊕n → Q, let φi : O →Q be the map from the ith summand of O ⊕n . (Really, φi is just a section of Q.)
For the open subfunctor G(k, n)I , show that
φ1 ⊕ · · · ⊕ φk : O ⊕k → Q

is an isomorphism. For a scheme B, the bijection G(k, n)I (B) ↔ Hom(B, Ak(n−k) )is given as follows. Given an element φ ∈ G(k, n)I (B), for j ∈ {k + 1, . . . , n},φj = a1j φ1 + a2j φ2 + · · · + akj φk , where aij are functions on B. But k(n − k)functions on B is the same as a map to Ak(n−k) (Exercise 6.6.E). Conversely, givenk(n − k) functions aij (1 ≤ i ≤ k < j ≤ n), define a surjection φ : O ⊕n → O ⊕kas follows: (φ1 . . . , φk ) is the identity, and φj = a1j φ1 + a2j φ2 + · · · + akj φk forj > k.
) *You have now shown that G(k, n) is representable, by covering it with nkcopies of Ak(n−k) . (You might wish to relate this to the description you gave in§6.7.) In particular, the Grassmannian over a field is smooth, and irreducible ofdimension k(n − k). (The Grassmannian over any base is smooth over that base,k(n−k)
because AB→ B is smooth, see §12.6.2.)

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16.7.1. The universal exact sequence over the Grassmannian.
tautological exact sequence

Note that we have a

0 → S → O ⊕n → Q → 0.

16.7.2. The Plücker embedding.
By applying ∧k to a surjection φ : O ⊕n → Q (over an arbitrary base B), we geta surjectionn∧k φ : O ⊕( k ) → det Q(Exercise 13.5.F). But a surjection from a rank N free sheaf to a line bundle is thesame as a map to PN−1 (Theorem 16.4.1).
n16.7.D. E XERCISE . Use this to describe a map P : G(k, n) → P( k )−1 . (This is just atautology: a natural transformation of functors induces a map of the representing
schemes. This is Yoneda’s Lemma, although if you didn’t do Exercise 1.3.Z, you
may wish to do this exercise by hand. But once you do, you may as well go backto prove Yoneda’s Lemma and do Exercise 1.3.Z, because the argument is just thesame!)
n16.7.E. E XERCISE . The projective coordinate xI on P( k )−1 corresponding to the Ithnfactor of O ⊕( k ) may be interpreted as the determinant of the map φI : O ⊕k → Q,
where the O ⊕k consists of the summands of O ⊕n corresponding to I. Make this
precise.
n
16.7.F. E XERCISE . Show that the standard open set UI of P( k )−1 corresponding to
k-subset I (i.e., where the corresponding coordinate xI doesn’t vanish) pulls backto the open subscheme G(k, n)I ⊂ G(k, n). Denote this map PI : G(k, n)I → UI .

16.7.G. E XERCISE .
Show that PI is a closed embedding as follows. We may
deal with the case I = {1, . . . , k}. Note that G(k, n)I is affine — you described it
Spec Z[aij ]1≤i≤k<j≤n in Exercise 16.7.C. Also, UI is affine, with coordinates xI ′ /I ,as I ′ varies over the other k-subsets. You want to show that the map
PI♯ : Z[xI ′ /I ]I ′ ⊂{1,...,n},|I ′ |=k} /(xI/I − 1) → Z[aij ]1≤i≤k<j≤n

is a surjection. By interpreting the map φ : O ⊕n → O ⊕k as a k × n matrix M whose
left k columns are the identity matrix and whose remaining entries are aij (1 ≤ i ≤k < j ≤ n), interpret PI♯ as taking xI ′ /I to the determinant of the k × k submatrix
corresponding to the columns in I ′ . For each (i, j) (with 1 ≤ i ≤ k < j ≤ n), findsome I ′ so that xI ′ /I (→ ±aij . (Let I ′ = {1, . . . , i − 1, i + 1, . . . , k, j}.)nZ.

Hence G(k, n) → P( k )−1 is a closed embedding, so G(k, n) is projective over

16.7.H. U NIMPORTANT E XERCISE . As an entertaining geometric consequence: if
V is a vector space over a field, show that the “pure tensors in ∧k V are pure inexactly one way”: if v1 ∧ · · · ∧ vk = w1 ∧ · · · ∧ wk ̸= 0 in ∧k V, show that there isa k × k matrix of determinant 1 relating the vi to the wi .

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n16.7.3. The Plücker equations. The equations of G(k, n) → P( k )−1 are particularly
nice. There are quadratic relations among the k × k minors of a k × (n − k) matrix,
called the Plücker equations. By our construction, they are equations satisfied by
G(k, n). It turns out that these equations cut out G(k, n), and in fact generate thehomogeneous ideal of G(k, n), but this takes more work (see [MS, §14.2]). We
explore this in one example.

16.7.I. E XERCISE . Suppose v1 , v2 , v3 , and v4 are four vectors in a two-dimensional
vector space V over some field. Show that(v1 ∧ v2 )(v3 ∧ v4 ) − (v1 ∧ v3 )(v2 ∧ v4 ) + (v1 ∧ v4 )(v2 ∧ v3 ) = 0.
The tricky part of this exercise is figuring out how to interpret multiplication (ofPlücker coordinates) here.
16.7.J. E XERCISE . Note that the Plücker embedding embeds the GrassmannianG(2, 4) into P5 .
(a) Show that G(2, 4) is cut out by the quadratic equation x12 x34 −x13 x24 +x14 x23 =0. (Hint: Use Exercise 16.7.I to show that the quadratic vanishes on G(2, 4). Butthat isn’t enough.)(b) Show that every smooth quadric hypersurface in P5 over an algebraically closed
field k of characteristic not 2 is isomorphic to the Grassmannian (over k). (For comparison, every smooth quadric hypersurface in P1k is two points; every smoothquadric hypersurface in P2k is isomorphic to P1k , §6.5.9; and every smooth quadrichypersurface in P3k is isomorphic to P1k ×k P1k , Example 9.6.2.)
16.7.K. E XERCISE ( RULINGS ON QUADRICS IN P5 ). Suppose k = k. From Remark 8.2.10, we expect two three-dimensional families of planes in G(2, 4) (interpreted as a hypersurface in P5 via the Plücker embedding, see Exercise 16.7.I).
One of them may be described as follows: for each point p ∈ P3k , we have a twodimensional family of lines through p; this is a plane in G(2, 4). There is a threedimensional family of planes corresponding to the choice of p. This is one of thetwo rulings. What is the other one? Prove (as rigorously as you can manage, given
what you know) that these are both rulings.
16.7.4. Further discussion.
16.7.L. E XERCISE . Show that the group scheme GLn acts on the GrassmannianG(k, n). (The action of a group scheme appeared earlier in Exercise 6.6.S(a).) Hint:
this is much more easily done with the language of functors, §6.6, than with thedescription of the Grassmannian in terms of patches, §6.7. (Exercise 16.4.K was the
special case of projective space.)
16.7.M. ⋆⋆ E XERCISE (G RASSMANNIAN BUNDLES ). Suppose F is a rank n locally
free sheaf on a scheme X. Define the Grassmannian bundle G(k, F ) over X. Intuitively, if F is a varying family of n-dimensional vector spaces over X, G(k, F )should parametrize k-dimensional quotients of the fibers. You may want to define
the functor first, and then show that it is representable. Your construction will
behave well under base change.

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16.7.5. (Partial) flag varieties. The discussion here extends without change to partial
flag varieties (§6.7.1), and the interested reader should think this through.

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CHAPTER 17

Relative versions of Spec and Proj, and projective
morphisms
In this chapter, we will use universal properties to define two useful constructions, Spec of a sheaf of algebras A , and Proj of a sheaf of graded algebras A• on
a scheme X. These will both generalize (globalize) our constructions of Spec ofA-algebras and Proj of graded A-algebras. We will see that affine morphisms are
precisely those of the form Spec A → X, and so we will define projective morphisms
to be those of the form Proj A• → X.
In both cases, our plan is to make a notion we know well over a ring work
more generally over a scheme. The main issue is how to glue the constructions
over each affine open subset together. The slick way we will proceed is to givea universal property, then show that the affine construction satisfies this universal property, then that the universal property behaves well with respect to open
subsets, then to use the idea that let us glue together the fibered product (or normalization) together to do all the hard gluing work. The most annoying part ofthis plan is finding the right universal property, especially in the Proj case.

17.1 Relative Spec of a (quasicoherent) sheaf of algebras

Given an A-algebra, B, we can take its Spec to get an affine scheme over Spec A:
Spec B → Spec A. We will now see a universal property description of a globalization of that notation. Consider an arbitrary scheme X, and a quasicoherent sheaf ofalgebras B on it. We will define how to take Spec of this sheaf of algebras, and wewill get a scheme Spec B → X that is “affine over X”, i.e., the structure morphism
is an affine morphism. You can think of this in two ways.

17.1.1. First, and most concretely, for any affine open set Spec A ⊂ X, Γ (Spec A, B)
is some A-algebra; call it B. Then above Spec A, Spec B will be Spec B.

17.1.2. Second, it will satisfy a universal property. We could define the A-scheme
Spec B by the fact that morphisms to Spec B (from an A-scheme W, over Spec A)
correspond to maps of A-algebras B → Γ (W, OW ) (this is our old friend Exercise 6.3.F). The universal property for β : Spec B → X generalizes this. Given a
447

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448

The Rising Sea: Foundations of Algebraic Geometrymorphism µ : W → X, the X-morphisms W → Spec B are in functorial (in W) bijection with morphisms α of sheaves of algebras making(17.1.2.1)

OX ❋
❋❋
⑥⑥
❋❋
⑥
⑥
❋❋
⑥⑥
❋❋
⑥
$
a⑥
α
% µ∗ OW
B

commute. Here the map OX → µ∗ OW is that coming from the map of ringed
spaces, and the map OX → B comes from the OX -algebra structure on B. (Forexperts: it needn’t be true that µ∗ OW is quasicoherent, but that doesn’t matter.
Non-experts should completely ignore this parenthetical comment.)
By universal property nonsense, these data determine β : Spec B → X up tounique isomorphism, assuming that it exists.
Fancy translation: in the category of X-schemes, β : Spec B → X represents thefunctor% {(α : B → µ∗ OW )}.
(µ : W → X) ✤3 where B
17.1.A. E XERCISE . Show that if X is affine, say Spec A, and B = B,
is an A-algebra, then Spec B → Spec A satisfies this universal property. (Hint:
Exercise 6.3.F.)
17.1.3. Proposition. — Suppose β : Spec B → X satisfies the universal property for(X, B), and U "→ X is an open subset. Then β|U : Spec B ×X U = (Spec B)|U → U
satisfies the universal property for (U, B|U ).

Proof. For convenience, let V = Spec B ×X U. A U-morphism W → V is the sameas an X-morphism W → Spec B (where by assumption µ : W → X factors throughU). By the universal property of Spec B, this is the same information as a map
B → µ∗ OW , which by the universal property definition of pullback (§16.3.4) is
the same as µ∗ B → OW , which is the same information as (µ|U )∗ B → OW . By
adjointness again this is the same as B|U → (µ|U )∗ OW .
!
Combining the above Exercise and Proposition, we have shown the existenceof Spec B in the case that X is an open subscheme of an affine scheme.

17.1.B. E XERCISE . Show the existence of Spec B in general, following the philosophy of our construction of the fibered product, normalization, and so forth.
We make some quick observations. First, Spec B can be “computed affinelocally on X”. We also have an isomorphism φ : B → β∗ OSpec B .

17.1.C. E XERCISE . Given an X-morphism γ:
W❄
❄❄
❄❄
µ ❄❄❄
8

γ

X

% Spec B
②②
②②
②
② β
C②②

show that the morphism α (in (17.1.2.1)) is the compositionBφ% β∗ OSpec B% β∗ γ∗ OW = µ∗ OW .

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The Spec construction gives an important way to understand affine morphisms.
Note that Spec B → X is an affine morphism. The “converse” is also true:

17.1.D. E XERCISE . Show that if µ : Z → X is an affine morphism, then we have a∼ Spec µ∗ OZ of X-schemes.
natural isomorphism Z =Hence we can recover any affine morphism using the Spec construction. Moreprecisely, a morphism is affine if and only if it is of the form µ : Spec B → X constructed above.

17.1.E. E XERCISE . Suppose µ : Spec B → X is such a morphism. Show that thecategory of quasicoherent sheaves on Spec B is equivalent to the category of quasicoherent sheaves on X with the structure of B-modules (quasicoherent B-modules
on X).
This is useful if X is quite simple but Spec B is complicated. We will use this∼ P1 , and Spec B is a more complicated curve.
before long when X =17.1.F. E XERCISE (Spec BEHAVES WELL WITH RESPECT TO BASE CHANGE ). Suppose µ : Z → X is any morphism, and B is a quasicoherent sheaf of algebras on∼ Spec µ∗ B. (If you
X. Show that there is a natural isomorphism Z ×X Spec B =think about, you will see that this generalizes the statement and possibly proof of
Proposition 17.1.3.)
17.1.4. Definition. An important example of the Spec construction is the total spaceof a finite rank locally free sheaf F , which we define to be Spec (Sym• F ∨ ).
17.1.G. E XERCISE . Suppose F is a locally free sheaf of rank n. Show that the total
space of F is a rank n vector bundle, i.e., that given any point p ∈ X, there is an
open neighborhood p ∈ U ⊂ X such that)*
∼ An .
Spec Sym• F ∨ |U =UShow that F is isomorphic to the sheaf of sections of the total space Spec (Sym• F ∨ ).
(Possible hint: use transition functions.) For this reason, the total space is also
called the vector bundle associated to a locally free sheaf F . (Caution: someauthors, e.g., [Stacks, tag 01M2], call Spec (Sym• F ), the dual of this vector bundle,
the vector bundle associated to F .)In particular, if F = OX⊕n , then Spec (Sym• F ∨ ) is called An
X , generalizing ourearlier notions of An
A . As the notion of free sheaf behaves well with respect to basen
∼ nchange, so does the notion of An
X , i.e., given X → Y, AY ×Y X = AX . (Aside: you•
may notice that the construction Spec Sym can be applied to any coherent sheaf
F (without dualizing, i.e., Spec (Sym• F )). This is sometimes called the abeliancone associated to F . This concept can be useful, but we won’t need it.)17.1.H. E XERCISE ( THE TAUTOLOGICAL BUNDLE ON Pn IS O(−1)). Suppose k is an+1field. Define the subset X ⊂ An+1
× Pn
on thek corresponding to “points of Akknn
corresponding line of Pk ”, so that the fiber of the map π : X → P corresponding
to a point l = [x0 , · · · , xn ] is the line in An+1
corresponding to l, i.e., the scalarkmultiples of (x0 , . . . , xn ). Show that π : X → Pn
k is (the line bundle correspondingto) the invertible sheaf O(−1). (Possible hint: work first over the usual affine open

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sets of Pn
k , and figure out transition functions.) For this reason, O(−1) is often
called the tautological bundle of Pnk (even over an arbitrary base, not just a field).
(Side Remark: The projection X → An+1
is the blow-up of An+1
at the “origin”,kk
see Exercise 9.3.F.)

17.2 Relative Proj of a sheaf of graded algebras
In parallel with the relative version Spec of Spec, we define a relative versionof Proj, denoted Proj (called “relative Proj” or “sheaf Proj”), of a quasicoherent
graded sheaf of algebras (satisfying some hypotheses) on a scheme X. We havealready done the case where the base X is affine, in §4.5.7, using the regular Proj
construction over a ring A. The elegant way to proceed would be to state the rightuniversal property, and then use this cleverly to glue together the constructions
over each affine, just as we did in the constructions of fibered product, normalization, and Spec . But because graded rings and graded modules make everything
confusing, we do not do this. Instead we guiltily take a more pedestrian approach.
(But the universal property can be made to work, see [Stacks, tag 01O0].)17.2.A. E XERCISE (Proj COMMUTES WITH AFFINE BASE CHANGE ). Suppose A → B
is map of rings, and S• is a Z≥0 -graded ring over A.
(a) Give a canonical isomorphism(17.2.0.1)

α : ProjB (S• ⊗A B)∼% (Proj S• ) ×Spec A Spec B
A(b) (easy) Suppose X is a projective A-scheme (§4.5.9). Show that X ×Spec A Spec B
is a projective B-scheme.
(c) Suppose S• is generated in degree 1, so OProjA S• (1) is an invertible sheaf (§15.2).
Clearly S• ⊗A B is generated in degree 1 as a B-algebra. Describe an isomorphism∼ α∗ γ∗ OProj S (1),OProj (S ⊗ B) (1) =
B

•

A

A

•

where γ is the top morphism in the pullback diagram(ProjA S• ) ×Spec A Spec B"
Spec Bγ% Proj S•
A"% Spec A

Possible hint: transition functions.
We now give a general means of constructing schemes over X (from [Stacks,
tag 01LH]), if we know what they should be over any affine open set, and how
these behave under open embeddings of one affine open set into another.
17.2.B. E XERCISE . Suppose we are given a scheme X, and the following data:(i) For each affine open subset U ⊂ X, we are given some morphism πU : ZU →U (a “scheme over U”).
(ii) For each (open) inclusion of affine open subsets V ⊂ U ⊂ X, we are given
an open embedding ρU
V : ZV "→ ZU .

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Assume this data satisfies:−1(a) for each V ⊂ U ⊂ X, ρU
V induces an isomorphism ZV → πU (V) of
schemes over V, and(b) whenever W ⊂ V ⊂ U ⊂ X are three nested affine open subsets, ρUW =VρUV ◦ ρW .

Show that there exists an X-scheme π : Z → X, and isomorphisms iU : π−1 (U) →ZU for each affine open set U, such that for nested affine open sets V ⊂ U, ρU
V
agrees with the composition
ZV

i−1
V

% π−1 (V) !

"% π−1 (U)

iU

% ZU

Hint (cf. Exercise 4.4.A): construct Z first as a set, then as a topological space, thenas a scheme. (Your construction will be independent of choices. Your solution
will work in more general situations, for example when the category of schemes isreplaced by ringed spaces, and when the affine open subsets are replaced by any
base of the topology.)
17.2.C. I MPORTANT E XERCISE AND D EFINITION (relative Proj ). Suppose S• =⊕n≥0 Sn is a quasicoherent sheaf of Z≥0 -graded algebras on a scheme X. Over∼ U ⊂ X, we have an U-scheme Proj S• (U) →each affine open subset Spec A =A
U. Show that these can be glued together to form an X-scheme, which we call
Proj X S• ; we have a “structure morphism” β : Proj X S• → X. (The structure morphism β is part of the definition.)

By the construction of Exercise 17.2.B, the preimage over any affine open set
can be computed using the original Proj construction. (You may enjoy going backand giving constructions of Xred , the normalization of X, and Spec of a quasicoherent sheaf of O-algebras using this idea. But there is a moral price to be paid bygiving up the universal property.)

17.2.1. Ongoing (reasonable) hypotheses on S• : “finite generation in degree 1”.
The Proj construction is most useful when applied to an A-algebra S• satisfying
some reasonable hypotheses (§4.5.6), notably when S• is a finitely generated Z≥0 graded A-algebra, and ideally if it is generated in degree 1. For this reason, in therest of the book, we will enforce these assumptions on S• , once we make senseof them for quasicoherent sheaves of algebras. (If you later need to relax thesehypotheses — for example, to keep the finite generation hypothesis but remove
the “generation in degree 1” hypothesis — it will not be too difficult.) Precisely,
we now always require that (i) S• is “generated in degree 1”, and (ii) S1 isfinite type. The cleanest way to make condition (i) precise is to require the naturalmap
Sym•OX S1 → S•
to be surjective. Because the Sym• construction may be computed affine-locally(§13.5.3), we can check generation in degree 1 on any affine cover. The motivation
for these hypotheses is the following construction.

17.2.D. I MPORTANT E XERCISE : O(1) ON Proj S• . If S• is finitely generated indegree 1 (Hypotheses 17.2.1), construct an invertible sheaf OProj S (1) on Proj S••that “restricts to OProjA S• (Spec A) (1) over each affine open subset Spec A ⊂ X”.

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17.2.E. E XERCISE (“Proj COMMUTES WITH BASE CHANGE ”). Suppose S• is aquasicoherent sheaf of Z≥0 -graded algebras on X. Let ρ : Z → X be any morphism.
Give a natural isomorphism∼ (Z ×X Proj S• , ψ∗ OProj S (1))(Proj ρ∗ S• , OProj ρ∗ S• (1)) =•where ψ is the “top” morphism in the base change diagram
ψ

Z ×X Proj S•% Proj S•
β

"
Z

ρ

"
% X.

17.2.2. Definition: π-very ample. Suppose π : X → Y is proper. If L is an invertible
sheaf on X, then we say that L is very ample (with respect to π), relatively very
∼ O(1),
ample, or (awkwardly) π-very ample if we can write X = Proj Y S• with L =where S• is a quasicoherent sheaf of algebras on Y satisfying Hypotheses 17.2.1(“finite generation in degree 1”). (The notion of very ampleness can be extended tomore general situations, see for example [Stacks, tag 01VM]. But this is of interest
only to people with esoteric tastes.)
17.2.F. E XERCISE . Suppose S• is finitely generated in degree 1 (Hypotheses 17.2.1).
Describe a map of graded quasicoherent sheaves φ : S• → ⊕n β∗ O(n) (β is the
structure morphism, see Exercise 17.2.C). Hint: Exercise 15.4.C.
17.2.G. E XERCISE . Suppose L is an invertible sheaf on X, and S• is a quasicoherent sheaf of graded algebras on X generated in degree 1 (Hypotheses 17.2.1).
Define S•′ = ⊕n=0 (Sn ⊗ L ⊗n ). Then S•′ has a natural algebra structure inherited from S• ; describe it. Give a natural isomorphism of “X-schemes with line
bundles”∼ (Proj S• , OProj S (1) ⊗ β∗ L ),(Proj S ′ , OProj S ′ (1)) =•••where β : Proj S• → X is the structure morphism. In other words, informallyspeaking, the Proj is the same, but the O(1) is twisted by L .

17.2.3. Definition. If F is a finite type quasicoherent sheaf on X, then Proj (Sym• F )is called its projectivization, and is denoted PF . You can check that this construc⊕(n+1)
tion behaves well with respect to base change. Define Pn). (Then
X := P(OXnnPnSpec A agrees with our earlier definition of PA , cf. Exercise 4.5.N, and PX agrees
with our earlier usage, see for example the proof of Theorem 10.3.5.) More generally, if F is locally free of rank n + 1, then PF is a projective bundle or Pn -bundle
over X. By Exercise 17.2.G, if F is a finite rank locally free sheaf on X, there is a
∼ P(L ⊗ F ).
canonical isomorphism PF =17.2.4. Example: ruled surfaces. If C is a regular curve and F is locally free ofrank 2, then PF is called a ruled surface over C. If C is further isomorphic to
P1 , PF is called a Hirzebruch surface. All vector bundles on P1 split as a direct
sum of line bundles (see §18.5.5 for a proof), so each Hirzebruch surface is of theform P(O(n1 ) ⊕ O(n2 )). By Exercise 17.2.G, this depends only on n2 − n1 . The
Hirzebruch surface P(O ⊕ O(n)) (n ≥ 0) is often denoted Fn . We will discuss the

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453

Hirzebruch surfaces in greater length in §20.2.9. We will see that the Fn are all
distinct in Exercise 20.2.Q.
17.2.H. E XERCISE . If S• is finitely generated in degree 1 (Hypotheses 17.2.1),
describe a canonical closed embedding

"
i
% PS1
❋❋
⑤
⑤
❋❋
❋❋
⑤⑤
⑤
❋
⑤
β
❋❋
M b⑤⑤
X

Proj S• !

∼ i∗ OPS (1) arising from the surjection Sym• S1 →and an isomorphism OProj S• (1) =1S• .

17.2.I. E XERCISE . Suppose F is a locally free sheaf of rank n + 1 on X. Exhibita bijection between the set of sections s : X → PF of PF → X and the set ofsurjective homomorphisms F → L → 0 of F onto invertible sheaves on X.

17.2.5. Remark (the relative version of the projective and affine cone). There is a naturalmorphism from Spec S• minus the zero-section to Proj S• (cf. Exercise 8.2.P). Just
as Proj S• [T ] contains a closed subscheme identified with Proj S• whose complement can be identified with Spec S• (Exercise 8.2.Q), Proj S• [T ] contains a closed
subscheme identified with Proj S• whose complement can be identified with Spec S• .
You are welcome to think this through.
17.2.6. Remark. If you wish, you can describe (with proof) a universal propertyof Proj S• . (You may want to describe a universal property of Proj first.) I recommend against it — a universal property should make your life easier, not harder.
One possible universal property is given in [Stacks, tag 01NS].

17.3 Projective morphisms
In §17.1, we reinterpreted affine morphisms: X → Y is an affine morphism if∼ Spec B of Y-schemes for some quasicoherent sheaf of
there is an isomorphism X =algebras B on Y. We will define the notion of a projective morphism similarly.
You might think that because projectivity is such a classical notion, there should
be some obvious definition, that is reasonably behaved. But this is not the case,and there are many possible variant definitions of projective (see [Stacks, tag
01W8]). All are imperfect, including the accepted definition we give here. Although projective morphisms are preserved by base change, we will manage toshow that they are preserved by composition only when the target is quasicompact (Exercise 17.3.B), and we will only show that the notion is local on the basewhen we add the data of a line bundle, and even then only under locally Noetherian hypotheses (§17.3.4).

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17.3.1. Definition. A morphism π : X → Y is projective if there is an isomorphism
X❂
❂❂
❂❂
π ❂❂❂
F

∼

Y

% Proj S•
①①
①①
①
①①
①
C①

for a quasicoherent sheaf of algebras S• on Y (satisfying “finite generation in degree 1”, Hypotheses 17.2.1). We say X is a projective Y-scheme, or X is projective
over Y. Using Exercise 6.4.D, this generalizes the notion of a projective A-scheme.
17.3.2. Warnings. First, notice that O(1), an important part of the concept of Proj ,is not mentioned in the definition. (I would prefer that it be part of the definition,
but this isn’t accepted practice.) As a result, the notion of affine morphism is affinelocal on the target, but the notion of projectivity of a morphism is not clearly affinelocal on the target. (In Noetherian circumstances, with the additional data of theinvertible sheaf O(1), it is, as we will see in §17.3.4. We will also later see an
example showing that the property of being projective is not local, §24.7.7.)
Second, [Ha1, p. 103] gives a different definition of projective morphism; wefollow the more general definition of Grothendieck. These definitions turn out tobe the same in nice circumstances. (But finite morphisms are not always projectivein the sense of [Ha1], while they are projective in our sense.)
17.3.A. E XERCISE .
(a) (a useful characterization of projective morphisms) Suppose π : X → Y is a morphism. Show that π is projective if and only if there exist a finite type quasicoherent sheaf S1 on Y, and a closed embedding i : X "→ PS1 (over Y, i.e., commuting
with the maps to Y). Hint: Exercise 17.2.H.
(b) (a useful characterization of projective morphisms, with line bundle) Suppose L is
an invertible sheaf on X, and π : X → Y is a morphism. Show that π is projective,∼ L , if and only if there exist a finite type quasicoherent sheaf S1 on
with O(1) =Y, a closed embedding i : X "→ PS1 (over Y, i.e., commuting with the maps to Y),∼ L.
and an isomorphism i∗ OPS1 (1) =(c) If furthermore Y admits an ample line bundle M , show that π is projective if
and only if there exists a closed embedding i : X "→ Pn
Y (over Y) for some n. (If you
wish, assume Y is proper over Spec A, so you can avoid the starred section §16.6.5.)
Hint: the harder direction is the forward implication. Use the finite type quasicoherent sheaf S1 from (a). Tensor S1 with a high enough power of M so that
it is finitely globally generated (Theorem 16.6.6, or Theorem 16.6.2 in the proper
setting), to obtain a surjection⊕(n+1)OY% % S1 ⊗ M ⊗N .

Then use Exercise 17.2.G.
17.3.3. Definition: Quasiprojective morphisms.
In analogy with projective andquasiprojective A-schemes (§4.5.9), one may define quasiprojective morphisms. If
Y is quasicompact, we say that π : X → Y is quasiprojective if π can be expressed
as a quasicompact open embedding into a scheme projective over Y. This is not a
great notion, and we will not use it. (The general definition of quasiprojective isslightly delicate — see [Gr-EGA, II.5.3] — but we won’t need it.)

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17.3.4. Properties of projective morphisms.
We start to establish a number of properties of projective morphisms. First,
the property of a morphism being projective is clearly preserved by base change,as the Proj construction behaves well with respect to base change (Exercise 17.2.E).
Also, projective morphisms are proper: properness is local on the target (Theorem 10.3.4(b)), and we saw earlier that projective A-schemes are proper over A(Theorem 10.3.5). In particular (by definition of properness), projective morphisms
are separated, finite type, and universally closed.
Exercise 17.3.G (in a future optional section) implies that if π : X → Y is aproper morphism of locally Noetherian schemes, and L is an invertible sheaf on
X, the question of whether π is a projective morphism with L as O(1) is local onY.
17.3.B. I MPORTANT C HALLENGING E XERCISE ( THE COMPOSITION OF PROJECTIVE
MORPHISMS IS PROJECTIVE , IF THE FINAL TARGET IS QUASICOMPACT ). Suppose
π : X → Y and ρ : Y → Z are projective morphisms, and Z is quasicompact. Show
that π◦ρ is projective. Hint: the criterion for projectivity given in Exercise 17.3.A(b)will be useful. (i) Deal first with the case where Z is affine. Build the following
commutative diagram, thereby finding a closed embedding X "→ PF ⊕n over Z.
In this diagram, all inclusions are closed embeddings, and all script fonts refer tofinite type quasicoherent sheaves.

"" (†) % n−1"
X ! ❘❘❘ % PE ! ❏PZ × Z Y !
❘❘❘❏❘❘❘ ❏❏❏❘❘❘ ❏❏π ❘❘❘❘ ❏❏❘❘❘❏❏ " "❘#0 !
Y

% Pn−1 ×Z PF ! "
Z

Segre

% P (F ⊕n )cf. Ex. 9.6.C, 9.6.D

"
% PF
♠♠♠
♠♠♠
♠
ρ
♠
♠♠
" c♠♠♠♠♠/
Z

Construct the closed embedding (†) as follows. Suppose M is the very ample line
bundle on Y over Z. Then M is ample, and so by Theorem 16.6.2, for m ≫ 0, E ⊗% % E ⊗ M ⊗m
M ⊗m is generated by a finite number of global sections. Suppose O ⊕n
Y

is the corresponding surjection. This induces a closed embedding P(E ⊗M ⊗m ) "→∼ PE (Exercise 17.2.G), and Pn−1 = Pn−1 ×Z Y. (ii) UnPn−1. But P(E ⊗ M ⊗m ) =YYZwind this diagram to show that (for Z affine) if L is π-very ample and M is ρ-very
ample, then for m ≫ 0, L ⊗ π∗ M ⊗m is (ρ ◦ π)-very ample. Then deal with thegeneral case by covering Z with a finite number of affines.
17.3.5. Finite morphisms are projective.

17.3.C. I MPORTANT E XERCISE : F INITE MORPHISMS ARE PROJECTIVE ( CF. E XER CISE 7.3.J). Show that finite morphisms are projective as follows. Suppose Z → X∼ Spec B where B is a finite type quasicoherent sheaf on X.
is finite, so that Z =∼ OX and Sn =∼ B for n > 0.
Describe a sheaf of graded algebras S• where S0 =∼Describe an X-isomorphism Z = Proj S• .

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In particular, closed embeddings are projective. We have the sequence of implications for morphisms
closed embedding =⇒ finite =⇒ projective =⇒ proper.
We know that finite morphisms are projective (Exercise 17.3.C), and have finite
fibers (Exercise 7.3.K). We will show the converse in Theorem 18.1.8, and state theextension to proper morphisms immediately after.
17.3.6. Caution: Consequences of projectivity not being “reasonable” in the sense of
Because the property of being projective is preserved by base change§7.1.1.
(§17.3.4), and composition to quasicompact targets (Exercise 17.3.B), the property of
being projective is “usually” preserved by products (Exercise 9.4.F): if π : X → Y
and π ′ : X ′ → Y ′ are projective, then so is π × π ′ : X × X ′ → Y × Y ′ , so long as
Y × Y ′ is quasicompact. Also, if you follow through the proof of the Cancellation
Theorem 10.1.19 for properties of morphisms, you will see that if π : X → Y is amorphism, ρ : Y → Z is a separated morphism (so the diagonal δρ is a closed embedding and hence projective), and ρ ◦ π is projective, and Y is quasicompact, then π
is projective.
17.3.D. E XERCISE .
Show that a morphism (over Spec k) from a projective kscheme to a quasicompact separated k-scheme is always projective. (Hint: theCancellation Theorem 10.1.19 for projective morphisms; see also Caution 17.3.6.)
17.3.7. ⋆⋆ Global generation and (very) ampleness in the relative setting.
We extend the discussion of §15.3 to the relative setting, in order to give ourselves the language of relatively base-point-freeness. We won’t use this discussion,
so on a first reading you should jump directly to §17.4. But these ideas come up
repeatedly in the research literature.
Suppose π : X → Y is a quasicompact quasiseparated morphism. If F isa quasicoherent sheaf on X, we say that F is relatively globally generated or
globally generated with respect to π if the natural map of quasicoherent sheavesπ∗ π∗ F → F is surjective. (Quasicompactness and quasiseparatedness are needed
ensure that π∗ F is a quasicoherent sheaf, Exercise 13.3.F.) But these hypotheses
are not very restrictive. Global generation is most useful only in the quasicompact setting, and most people won’t be bothered by quasiseparated hypotheses.
Unimportant aside: these hypotheses can be relaxed considerably. If π : X → Y
is a morphism of locally ringed spaces — not necessarily schemes — with no otherhypotheses, and F is a quasicoherent sheaf on X, then we say that F is relatively
globally generated or globally generated with respect to π if the natural mapπ∗ π∗ F → F of OX -modules is surjective.)Thanks to our hypotheses, as the natural map π∗ π∗ F → F is a morphismof quasicoherent sheaves, the condition of being relatively globally generated isaffine-local on Y.
Suppose now that L is a locally free sheaf on X, and π : X → Y is a morphism.
We say that L is relatively base-point-free or base-point-free with respect to π if
it is relatively globally generated.
17.3.E. E XERCISE . Suppose L is a finite rank locally free sheaf on X, π : X → Y
is a quasicompact separated morphism, and π∗ L is finite type on Y. (We will
later show in Grothendieck’s Coherence Theorem 18.9.1 that this latter statement

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is true if π is proper and Y is Noetherian. This is much easier if π is projective,
see Theorem 18.8.1. We could work hard and prove it now, but it isn’t worth
the trouble.) Describe a canonical morphism ψ : X → Pπ∗ L . (Possible hint: thisgeneralizes the fact that base-point-free line bundles give maps to projective space,so generalize that argument, see §15.3.5.)

We say that L is relatively ample or π-ample or relatively ample with respect to π if for every affine open subset Spec B of Y, L |π−1 (Spec B) is ample on
π−1 (Spec B) over B, or equivalently (by §16.6.5), L |π−1 (Spec B) is (absolutely) ample on π−1 (Spec B). By the discussion in §16.6.5, if L is ample then π is necessarily
quasicompact, and (by Theorem 16.6.6) separated; if π is affine, then all invertible
sheaves are ample; and if π is projective, then the corresponding O(1) is ample.
By Exercise 16.6.J, L is π-ample if and only if L ⊗n is π-ample, and if Z "→ X is a
closed embedding, then L |Z is ample over Y.
From Theorem 16.6.6(d), we have a natural open embedding X → Proj Y ⊕π∗ L ⊗d .
(Do you see what this map is? Also, be careful: ⊕π∗ L ⊗d need not be a finitely generated graded sheaf of algebras, so we are using the Proj construction where oneof the usual hypotheses doesn’t hold.)
The notions of relative global generation and relative ampleness are most useful in the proper setting, because of Theorem 16.6.2.
17.3.8. Many statements of §15.3 carry over without change. For example, we havethe following. Suppose π : X → Y is proper, F and G are quasicoherent sheaves
on X, and L and M are invertible sheaves on X. If π is affine, then F is relatively globally generated (from Easy Exercise 15.3.A). If F and G are relatively
globally generated, so is F ⊗ G (Easy Exercise 15.3.B). If L is π-very ample (Definition 17.2.2), then it is π-base-point-free (Easy Exercise 16.6.B). If L is π-very
ample, and M is π-base-point-free (if for example it is π-very ample), then L ⊗ M
is π-very ample (Exercise 16.6.C). Exercise 16.6.G extends immediately to showthat if
X❃
❃❃
❃❃
τ ❃❃❃
8

π

S

%Y
PP
P
P
PP ρ
5PP

is a finite morphism of S-schemes, and if L is a ρ-ample invertible sheaf on Y, thenπ∗ L is τ-ample.
By the nature of the statements, some of the statements of §15.3 require quasicompactness hypotheses on Y, or other patches. For example:
17.3.9. Theorem. — Suppose π : X → Y is proper, L is an invertible sheaf on X, and Y
is quasicompact. The following are equivalent.

(a) For some N > 0, L ⊗N is π-very ample.
(a’) For all n ≫ 0, L ⊗n is π-very ample.
(b) For all finite type quasicoherent sheaves F , there is an n0 such that for n ≥ n0 ,F ⊗ L ⊗n is relatively globally generated.
(c) The invertible sheaf L is π-ample.

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17.3.F. E XERCISE . Prove Theorem 17.3.9 using Theorem 16.6.2. (Unimportant
remark: The proof given of Theorem 16.6.2 used Noetherian hypotheses, but asstated there, they can be removed.)
After doing the above Exercise, it will be clear how to adjust the statement ofTheorem 17.3.9 if you need to remove the quasicompactness assumption on Y.
17.3.G. E XERCISE ( A USEFUL EQUIVALENT DEFINITION OF VERY AMPLENESS UN DER N OETHERIAN HYPOTHESES ). Suppose π : X → Y is a proper morphism, Y
is locally Noetherian (hence X is too, as π is finite type), and L is an invertible
sheaf on X. Suppose that you know that in this situation π∗ L is finite type. (Wewill later show this, as described in Exercise 17.3.E.) Show that L is very ample
if and only if (i) L is relatively base-point-free, and (ii) the canonical Y-morphism
i : X → P(π∗ L ) of Exercise 17.3.E is a closed embedding. Conclude that the notion of relative very ampleness is affine-local on Y (it may be checked on any affine
cover Y), if Y is locally Noetherian and π is proper.
As a consequence, Theorem 17.3.9 implies the notion of relative ampleness isaffine-local on Y (if π is proper and Y is locally Noetherian).
17.3.10. ⋆⋆ Ample vector bundles. The notion of an ample vector bundle is useful
in some parts of the literature, so we define it, although we won’t use the notion.
A locally free sheaf E on a proper A-scheme X is ample if OPE /X (1) is an ample invertible sheaf. In particular, using Exercise 17.2.G, you can verify that an invertible
sheaf is ample as a locally free sheaf (this definition) if and only if it is ample as aninvertible sheaf (Definition 16.6.1), preventing a notational crisis. (The proper hypotheses can be relaxed; it is included only because Definition 16.6.1 of amplenessis only for proper schemes.)
17.3.11. ⋆⋆ Quasiaffine morphisms.
Because we have introduced quasiprojective morphisms (Definition 17.3.3),
we briefly introduce quasiaffine morphisms (and quasiaffine schemes), as somereaders may have cause to use them. Many of these ideas could have been introduced long before, but because we will never use them, we deal with them all atonce.
A scheme X is quasiaffine if it admits a quasicompact open embedding into anaffine scheme. This implies that X is quasicompact and separated. Note that if X
is Noetherian (the most relevant case for most people), then any open embedding
is of course automatically quasicompact.
17.3.H. E XERCISE . Show that X is quasiaffine if and only if the canonical map
X → Spec Γ (X, OX ) (defined in Exercise 6.3.F and the paragraph following it) is aquasicompact open embedding. Thus a quasiaffine scheme comes with a canonicalquasicompact open embedding into an affine scheme. Hint: Let A = Γ (X, OX ) for
convenience. Suppose X "→ Spec R is a quasicompact open embedding. We wish
to show that X "→ Spec A is a quasicompact open embedding. Factor X "→ Spec R
through X → Spec A → Spec R. Show that X → Spec A is an open embedding in an
open neighborhood of any chosen point p ∈ X, as follows. Choose r ∈ R such thatp ∈ D(r) ⊂ X. Notice that if Xr = {q ∈ X : r(q) ̸= 0}, then Γ (Xr , OX ) = Γ (X, OX )rby Exercise 13.3.H, using the fact that X is quasicompact and quasiseparated. Use
this to show that the map Xr → Spec Ar is an isomorphism.

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It is not hard to show that X is quasiaffine if and only if OX is ample, but wewon’t use this fact.
A morphism π : X → Y is quasiaffine if the inverse image of every affine open
subset of Y is a quasiaffine scheme. By Exercise 17.3.H, this is equivalent to π
being quasicompact and separated, and the natural map X → Spec π∗ OX being aquasicompact open embedding. This implies that the notion of quasiaffineness islocal on the target (may be checked on an open cover), and also affine-local ona target (one may choose an affine cover, and check that the preimages of these
open sets are quasiaffine). Quasiaffine morphisms are preserved by base change:
if a morphism X "→ Z over Y is a quasicompact open embedding into an affine
Y-scheme, then for any W → Y, X ×Y W "→ Z ×Y W is a quasicompact open
embedding into an affine W-scheme. (Interestingly, Exercise 17.3.H is not the righttool to use to show this base change property.)
One may readily check that quasiaffine morphisms are preserved by composition, [Stacks, tag 01SN]. Thus quasicompact locally closed embeddings are quasiaffine. If X is affine, then X → Y is quasiaffine if and only if it is quasicompact (asthe preimage of any affine open subset of Y is an open subset of an affine scheme,
namely X). In particular, from the Cancellation Theorem 10.1.19 for quasicompact
morphisms, any morphism from an affine scheme to a quasiseparated scheme isquasiaffine.

17.4 Applications to curves
We now apply what we have learned to curves.
17.4.1. Theorem (every integral curve has a birational model that is regular andprojective). — If C is an integral curve of finite type over a field k, then there exists a
regular projective k-curve C ′ birational to C.

C′

P1

F IGURE 17.1. Constructing a projective regular model of a curve
C over k via a finite cover of P1
Proof. We can assume C is affine. By the Noether Normalization Lemma 11.2.4 (orusing transcendence theory, see Exercise 11.2.A), we can find some x ∈ K(C) \ kwith K(C)/k(x) a finite extension of fields. By identifying a standard open of P1kwith Spec k[x], and taking the normalization of P1 in the function field K(C) (Definition 9.7.I), we obtain a finite morphism C ′ → P1 , where C ′ is a curve (finiteness

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by Exercise 9.7.M, dim C ′ = dim P1 by Exercise 11.1.F), and regular (it is reduced
hence regular at the generic point, and regular at the closed points by the main
theorem on discrete valuation rings in §12.5). Also, C ′ is birational to C as they
have isomorphic function fields (Exercise 6.5.D).
Finally, C ′ → P1k is finite hence projective (Exercise 17.3.C), and P1k → Spec k
is projective, so as composition of projective morphisms (to a quasicompact target)
are projective (Exercise 17.3.B), C ′ → Spec k is projective.
!

17.4.2. Theorem. — If C is an irreducible separated regular curve, finite type over a fieldk, then there is an open embedding C "→ C ′ into some projective regular curve C ′ (overk).
Proof. We first prove the result in the case where C is affine. Then we have aclosed embedding C "→ An , and we consider An as a standard open subset of
Pn . Taking the scheme-theoretic closure of C in Pn , we obtain a projective integral
3 of C is a finite
curve C, containing C as an open subset. The normalization C

3 is Noetherian,morphism (finiteness of integral closure, Theorem 9.7.3(b)), so C
and regular (as normal Noetherian dimension 1 local rings are discrete valuation
rings, §12.5). Moreover, by the universal property of normalization, normalization
3of C doesn’t affect the normal open set C, so we have an open embedding C "→ C.
3 → C is finite hence projective, and C → Spec k is projective, so (byFinally, C
3 is projective.
Exercise 17.3.B) C

We next consider the case of general C. Let C1 by any nonempty affine open
subset of C. By the discussion in the previous paragraph, we have a regular pro3 . The Curve-to-Projective Extension Theorem 16.5.1 (apjective compactification C
1plied successively to the finite number of points C \ C1 ) implies that the morphism
3 extends to a birational morphism C → C
3 . Because points of a regular
C "→ C1

1

1

curve are determined by their valuation (Exercise 12.7.B), this is an inclusion ofsets. Because the topology on curves is stupid (cofinite), it expresses C as an open
3 . But why is it an open embedding of schemes?
subset of C
1

We show it is an open embedding near a point p ∈ C as follows. Let C2 be anaffine open neighborhood of p in C. We repeat the construction we used on C1 , toobtain the following diagram, with open embeddings marked.
C" # 1 .❆/
❆❆
❆❆
❆❆
❆1

01 C2" #
⑥⑥
⑥
⑥
⑥⑥
⑥
a⑥
C❄
⑧⑧ ❄❄❄⑧❄❄⑧⑧❄❄ "" 5⑧⑧⑧
8
3
3
C
C
1
2

3 extends
By the Curve-to-Projective Extension Theorem 16.5.1, the map C1 → C
2
3 → C
3 , and we similarly have a morphism π : C
3 → C
3 , extending
to π : C
12

1

2

21

2

1

3 . The composition π ◦ π is the identity morphism (as it is the identity
C2 → C
1
21
12

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rational map, see Theorem 10.2.2). The same is true for π12 ◦ π21 , so π12 and π21
are isomorphisms. The enhanced diagram
C" # 1 .❆/
❆❆
❆❆
❆❆
❆1

1 C2"
⑥0 #
⑥
⑥⑥
⑥⑥
a⑥⑥
C❄
⑧⑧ ❄❄❄⑧❄❄⑧⑧❄❄ "" 5⑧⑧⑧
8
3
3
%C
C1 (
2

commutes (by Theorem 10.2.2 again, implying that morphisms of reduced sepa3
rated schemes are determined by their behavior on dense open sets). But C → C
2

1

3 is an open embedding there
is an open embedding (in particular, at p), so C → C
1
as well.
!

17.4.A. E XERCISE . Show that all regular proper curves over k are projective.

17.4.3. Theorem (various categories of curves are the same). — The following
categories are equivalent.
(i) irreducible regular projective curves over k, and surjective k-morphisms.
(ii) irreducible regular projective curves over k, and dominant k-morphisms.
(iii) irreducible regular projective curves over k, and dominant rational maps over k.
(iv) integral curves of finite type over k, and dominant rational maps over k.
(v) the opposite category of finitely generated fields of transcendence degree 1 overk, and k-homomorphisms.
All morphisms and maps in the following discussion are assumed to be defined over k.
(Aside: The interested reader can tweak the proof below to show the following
variation of the theorem: in (i)–(iv), consider only geometrically irreducible curves,and in (v), consider only fields K such that K ∩ ks = k in K. This variation allows
us to exclude “weird” curves we may not want to consider. For example, if k = R,then we are allowing #curves such as P1C which are not geometrically irreducible,
11∼
as PC ×R Spec C = PC P1C .)Proof. Every surjective morphism is a dominant morphism, and every dominantmorphism is a dominant rational map, and each integral regular projective curve isa quasiprojective curve, so we have shown (informally speaking) how to get from(i) to (ii) to (iii) to (iv). To get from (iv) to (i), suppose we have a dominant rational
map C1 ""# C2 of integral curves. Replace C1 by a dense open set so the rational
31 → C32
map is a morphism C1 → C2 . This induces a map of normalizations C
3i be a regular projective compactification ofof regular irreducible curves. Let C
3
31 → C
32 extends toCi (for i = 1, 2), as in Theorem 17.4.2. Then the morphism C

31 → C
32 by the Curve-to-Projective Extension Theorem 16.5.1. Thisa morphism C
morphism is surjective (do you see why?), so we have produced a morphism in
category (i).

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17.4.B. E XERCISE . Put the above pieces together to describe equivalences of categories (i) through (iv).
It remains to connect (v). This is essentially the content of Exercise 6.5.D; details are left to the reader.
!
Theorem 17.4.3 has a number of implications. For example, each quasiprojective reduced curve is birational to precisely one projective regular curve. Here isanother interesting consequence, promised in Side Remark 3.1.3.
17.4.4. Degree of a projective morphism from a curve to a regular curve.
You might already have a reasonable sense that a map of compact Riemannsurfaces has a well-behaved degree, that the number of preimages of a point of
C ′ is constant, so long as the preimages are counted with appropriate multiplicity.
For example, if f locally looks like z (→ zm = y, then near y = 0 and z = 0 (butnot at z = 0), each point has precisely m preimages, but as y goes to 0, the mpreimages coalesce. Enlightening Example 9.3.3 showed this phenomenon in amore complicated context.
We now show the algebraic version of this fact. Suppose π : C → C ′ is a surjective (or equivalently, dominant) map of regular projective curves. We will showthat π has a well-behaved degree, in a sense that we will now make precise.
First we show that π is finite. Theorem 18.1.8 (finite = projective + finite fibers)
implies this, but we haven’t proved it yet. So instead we show the finiteness of
π as follows. Let C ′′ be the normalization of C ′ in the function field of C. Then
∼ K(C ′′ ) (really, equality) which leads to birational
we have an isomorphism K(C) =′′maps C (❴ ❴ ❴% C which extend to morphisms as both C and C ′′ are regular andprojective (by the Curve-to-Projective Extension Theorem 16.5.1). Thus this yields
an isomorphism of C and C ′′ . But C ′′ → C ′ is a finite morphism by the finitenessof integral closure (Theorem 9.7.3).
17.4.5. Proposition. — Suppose that π : C → C ′ is a finite morphism, where C is a (puredimension 1) curve with no embedded points (the most important case: C is reduced), andC ′ is a regular curve. Then π∗ OC is locally free of finite rank.

The “no embedded points” hypothesis is the same as requiring that every associated point of C maps to a generic point of (some component of) C ′ . Notice that
if C is reduced, then there are no restrictions on how bad the singularities of C canbe!
We will prove Proposition 17.4.5 in §17.4.10, after showing how useful it is.
The regularity hypothesis on C ′ is necessary: the normalization of a nodal curve(Figure 7.4) is an example where most points have one preimage, and one point(the “node”) has two. (We will later see, in Remark 24.4.8 and §24.4.12, that whatmatters in the hypotheses of Proposition 17.4.5 is that the morphism is finite andflat.)
17.4.6. Definition.
degree of π.

If C ′ is irreducible, the rank of this locally free sheaf is the

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17.4.C. E XERCISE . Recall that the degree of a rational map from one integral
curve to another is defined as the degree of the function field extension (Definition 11.2.2). Show that (with the notation of Proposition 17.4.5) if C and C ′ areintegral, the degree of π as a rational map is the same as the rank of π∗ OC .
17.4.7. Remark for those with complex-analytic background (algebraic degree = analytic
degree). If C → C ′ is a finite map of regular complex algebraic curves, Proposition 17.4.5 establishes that algebraic degree as defined above is the same as analytic
degree (counting preimages, with multiplicity).
17.4.D. E XERCISE . We continue the notation and hypotheses of Proposition 17.4.5.
Suppose p is a closed point of C ′ . The scheme-theoretic preimage π∗ (p) of p is adimension 0 scheme over k.
(a) Suppose C ′ is finite type over a field k, and n is the dimension of the k-vector
space of global sections of the structure sheaf of π∗ (p). Show that n = (deg π)(deg p).
(The degree of a point was defined in §5.3.8.)(b) Suppose that C is regular, and π−1 (p) = {p1 , . . . , pm }. Suppose t is a uniformizer of the discrete valuation ring OC ′ ,p . Show thatdeg π =

m.

(valpi π∗ t) deg(κ(pi )/κ(p)),

i=1where deg(κ(pi )/κ(p)) denotes the degree of the field extension of the residue
fields. (Can you extend (a) to remove the hypotheses of working over a field? If
you are a number theorist, can you recognize (b) in terms of splitting prime ideals
in extensions of rings of integers in number fields?)
17.4.E. E XERCISE . Suppose that C is an irreducible regular projective curve overk, and s is a nonzero rational function on C. Show that the number of zeros of s(counted with appropriate multiplicity) equals the number of poles. Hint: recognize this as the degree of a morphism s : C → P1k . (In the complex category, this is
an important consequence of the Residue Theorem.)
17.4.8. Remark. In Exercise 18.4.D, we will see that the number of zeros (and poles)in the previous exercise is the degree of the line bundle giving the map to P1 (via
Important Theorem 16.4.1).

17.4.F. E XERCISE . Suppose s1 and s2 are two sections of a degree d line bundle
L on an irreducible regular curve C, with no common zeros. Then s1 and s2
determine a morphism π : C → P1k . Show that the degree of π is d. (Translation:a two-dimensional base-point-free degree d linear system on C defines a degree dcover of P1 .)17.4.9. Revisiting Example 9.3.3.
Proposition 17.4.5 and Exercise 17.4.D makeprecise what general behavior we observed in Example 9.3.3. Suppose C ′ is irreducible, and that d is the rank of this allegedly locally free sheaf. Then the fiber
over any point of C with residue field K is the Spec of an algebra of dimension dover K. This means that the number of points in the fiber, counted with appropriate multiplicity, is always d.

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As a motivating example, we revisit Example 9.3.3, the map Q[y] → Q[x] given
by x (→ y2 , the projection of the parabola x = y2 to the x-axis. We observed thefollowing.
(i) The fiber over x = 1 is Q[y]/(y2 − 1), so we get 2 points.
(ii) The fiber over x = 0 is Q[y]/(y2 ) — we get one point, with multiplicity 2,
arising because of the nonreducedness.
∼ Q(i) — we get one point, with(iii) The fiber over x = −1 is Q[y]/(y2 + 1) =multiplicity 2, arising because of the field extension.
(iv) Finally, the fiber over the generic point Spec Q(x) is Spec Q(y), which is
one point, with multiplicity 2, arising again because of the field extension(as Q(y)/Q(x) is a degree 2 extension).
We thus see three sorts of behaviors ((iii) and (iv) are really the same). Note thateven if you only work with algebraically closed fields, you will still be forced todeal with this third type of behavior, because residue fields at generic points areusually not algebraically closed (witness case (iv) above).
17.4.10. Proof of Proposition 17.4.5. The key idea, useful in other circumstances, isto reduce to a fact about discrete valuation rings.
The question is local on the target, so we may assume that C ′ is affine. By
Exercise 5.4.B, we may also assume C ′ is integral.
By Important Exercise 13.7.K, if the rank of the finite type quasicoherent sheafπ∗ OC is constant, then (as C ′ is reduced) π∗ OC is locally free. We will show this byshowing the rank at any closed point p of C ′ is the same as the rank at the genericpoint.
Suppose C ′ = Spec A ′ , where A ′ is an integral domain, and p = [m]. As π isan affine morphism, C is an affine scheme as well; say C = Spec A.
We wish to show that (i) dimA ′ /m (A/m) (the rank of π∗ OC at p) equals (ii)dimK(A ′ ) (A ′× )−1 A (the rank of π∗ OC at the generic point). In other words, wetake A (considered as an A ′ -module), and (i) quotient by m, and (ii) invert allnonzero elements of A ′ , and in each case compute the result’s dimension over theappropriate field.
Both (i) and (ii) factor through localizing at m, so it suffices to show that Am is′a finite rank free Am-module, of rank d, say, as the answers to both (i) and (ii) willthen be d.
′
Now Amis a discrete valuation ring; let t be its uniformizer. We can assume′
that t ∈ A (as otherwise, we replace A ′ by a suitable localization, at the “de′nominators” of t). Then Am is a finitely generated Am-module, and hence by Re′′
mark 12.5.14 is a finite sum of principal modules, of the form Am
or Am/(tn ) (forvarious n). We wish to show that there are no summands of the latter type. But
if there were, then t (interpreted as an element of Am ) would be a zerodivisor of
Am , and thus (interpreted as an element of A) a zerodivisor of A. But then by §5.5(C), there is an associated point of C in π−1 (p), contradicting the hypotheses that
C has no embedded points.
!

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CHAPTER 18

Čech cohomology of quasicoherent sheaves
This topic is surprisingly simple and elegant. You may think cohomology
must be complicated, and that this is why it appears so late in the book. But you
will see that we need very little background. After defining schemes, we could
have immediately defined quasicoherent sheaves, and then defined cohomology,
and verified that it had many useful properties.

18.1 (Desired) properties of cohomology
Rather than immediately defining cohomology of quasicoherent sheaves, wefirst discuss why we care, and what properties it should have.
As Γ (X, ·) is a left-exact functor, if 0 → F → G → H → 0 is a short exact
sequence of quasicoherent sheaves on a scheme X, then
0 → F (X) → G (X) → H (X)is exact. We dream that this sequence continues to the right, giving a long exact
sequence. More explicitly, there should be some covariant functors Hi (i ≥ 0) fromquasicoherent sheaves on X to groups such that H0 is the global section functor Γ ,and so that there is a “long exact sequence in cohomology”.
(18.1.0.1)

0% H0 (X, F )% H0 (X, G )% H0 (X, H )% H1 (X, F )% H1 (X, G )% H1 (X, H )% ···(In general, whenever we see a left-exact or right-exact functor, we should hopefor this, and in good cases our dreams will come true. The machinery behind thisusually involves derived functors, which we will discuss in Chapter 23.)
Before defining cohomology groups of quasicoherent sheaves explicitly, we
first describe their important properties, which are in some ways more important
than the formal definition. The boxed properties will be the important ones.
Suppose X is a separated and quasicompact A-scheme. For each quasicoherent
sheaf F on X, we will define A-modules Hi (X, F ). In particular, if A = k, they are
k-vector spaces. In this case, we define hi (X, F ) = dimk Hi (X, F ) (where k is left
implicit on the left side).
(i) Each Hi is a covariant functor QCohX → ModA .
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The Rising Sea: Foundations of Algebraic Geometry(ii) The functor H0 is identified with functor Γ : H0 (X, F ) = Γ (X, F ), andthe covariance of (i) for i = 0 is just the usual covariance for Γ (F → G induces
Γ (X, F ) → Γ (X, G )).
(iii) If 0 → F → G → H → 0 is a short exact sequence of quasicoherent
sheaves on X, then we have a long exact sequence (18.1.0.1). The maps Hi (X, F ) →Hi (X, G ) come from covariance, and similarly for Hi (X, G ) → Hi (X, H ). The connecting homomorphisms Hi (X, H ) → Hi+1 (X, F ) will have to be defined.
(iv) If π : X → Y is any morphism of quasicompact separated A-schemes, andF is a quasicoherent sheaf on X, then there is a natural morphism Hi (Y, π∗ F ) → Hi (X, F )extending Γ (Y, π∗ F ) → Γ (X, F ). (Note that π is quasicompact and separated
by the Cancellation Theorem 10.1.19 for quasicompact and separated morphisms,
taking Z = Spec A in the statement of the Cancellation Theorem, so π∗ F is indeed a quasicoherent sheaf by Exercise 13.3.F.) We will later see this as part of alarger story, the Leray spectral sequence (Theorem 23.4.5). If G is a quasicoherent
sheaf on Y, then setting F := π∗ G and using the adjunction map G → π∗ π∗ G
and covariance of (ii) gives a natural pullback map Hi (Y, G ) → Hi (X, π∗ G ) (via
Hi (Y, G ) → Hi (Y, π∗ π∗ G ) → Hi (X, π∗ G )) extending Γ (Y, G ) → Γ (X, π∗ G ). In thisway, Hi is a “contravariant functor in the space”.
(v) If π : X → Y is an affine morphism, and F is a quasicoherent sheaf on X,
the natural map of (iv) is an isomorphism:

Hi (Y, π∗ F )∼% Hi (X, F ) . Whenπ is a closed embedding and Y = PN
A , this isomorphism translates calculations onarbitrary projective A-schemes to calculations on PNA.
(vi) (affine cover cohomology vanishing) If X can be covered by n affine open sets,then Hi (X, F ) = 0 for i ≥ n for all F . In particular, on affine schemes, all higher(i > 0) quasicoherent cohomology groups vanish. The vanishing of H1 in this case,along with the long exact sequence (iii) implies that Γ is an exact functor for quasicoherent sheaves on affine schemes, something we already knew (Exercise 13.4.A).
It is also true that if dim X = n, then Hi (X, F ) = 0 for all i > n and for all F (dimensional cohomology vanishing). We will prove this for projective A-schemes(Theorem 18.2.6) and even quasiprojective A-schemes (Exercise 22.4.T). See §18.2.7
for discussion of the general case.
18.1.1. Side Remark: the cohomological criterion for affineness. The converse to “affine
cover cohomology vanishing” (vi) in the case when n = 1 is Serre’s cohomologicalcriterion for affineness, [GW, Thm. 12.35]: any quasicompact quasiseparated schemeX, such that Hi (X, F ) = 0 for all i > 0 and all quasicoherent F , must be affine. (Infact, it suffices that H1 of every quasicoherent sheaf of ideals vanishes.) We will
not use this, and thus will not prove it.
18.1.2. Let’s get back to our list.
(vii) The functor Hi behaves well under direct sums, and more generally under filtered colimits: Hi (X, lim Fj ) = lim Hi (X, Fj ).
−→−→(viii) We will also identify the cohomology of all O(m) on Pn
A:

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467

)n+m*
n
• H0 (Pn
if m ≥ 0.
A , OPA (m)) is a free A-module of rank ) m
*
−m−1
n n
n
• H (PA , OPA (m)) is a free A-module of rank −n−m−1 if m ≤ −n − 1.
n
• Hi (Pn
A , OPA (m)) = 0 otherwise.
We have already shown the first statement in Essential Exercise 14.1.C.
Theorem 18.1.3 has a number of features that will be the first appearances offacts that we will prove later.
• The cohomology of these bundles vanish in degree above n ((vi) above,“affine cover cohomology vanishing”)• These cohomology groups are always finitely generated A-modules. This
will be true for all coherent sheaves on projective A-schemes (Theorem 18.1.4(i)),
and indeed (with more work) on proper A-schemes (Grothendieck’s Coherence Theorem 18.9.1).
• The top cohomology group vanishes for m > −n − 1. (We will later see
this as an example of Kodaira vanishing, see §21.5.7.)• The top cohomology group is one-dimensional for m = −n − 1 if A = k.
This is the first appearance of the dualizing sheaf.
• There is a natural duality
Hi (Pn , O(m)) × Hn−i (Pn , O(−n − 1 − m)) → Hn (Pn , O(−n − 1))This is the first appearance of Serre duality. (For the case n = 1, see Example 18.5.4.),• The alternating sum (−1)i hi (Pn , O(m)) is a polynomial in m. This is a
first example of a Hilbert polynomial.
Before proving these facts, let’s first use them to prove interesting things, as
motivation.
By Theorem 15.3.1, for any coherent sheaf F on Pn
A we can find a surjection
O(m)⊕j → F , which yields the exact sequence(18.1.3.1)

0 → G → O(m)⊕j → F → 0

for some coherent sheaf G . We can use this to prove the following.
18.1.4. Theorem. — (i) For any coherent sheaf F on a projective A-scheme X where A
is Noetherian, Hi (X, F ) is a coherent (finitely generated) A-module.
(ii) (Serre vanishing) Furthermore, for m ≫ 0, Hi (X, F (m)) = 0 for all i > 0 (evenwithout Noetherian hypotheses).
(A slightly fancier version of Serre vanishing will be given in Theorem 18.8.G.)
Proof. Because cohomology of a closed scheme can be computed on the ambient
space ((v) above), we may immediately reduce to the case X = Pn
A.

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The Rising Sea: Foundations of Algebraic Geometry(i) Consider the long exact sequence:0···% H0 (Pn , G )A% H0 (Pn , O(m)⊕j )% H0 (Pn , F )%H1 (PnA, G )% H1 (Pn , O(m)⊕j )% H1 (Pn , F )
A

% ···

% Hn−1 (Pn , G )% Hn−1 (Pn , O(m)⊕j )A% Hn−1 (Pn , F )%Hn (PnA, G )% Hn (Pn , O(m)⊕j )% Hn (Pn , F )

%0

A

A

A

A

A

A

A

The exact sequence ends here because Pn
A is covered by n + 1 affine open sets ((vi)⊕jabove, “affine cover cohomology vanishing”). Then Hn (PnA , O(m) ) is finitelyn ngenerated by Theorem 18.1.3, hence H (PA , F ) is finitely generated for all cohern−1⊕j
ent sheaves F . Hence in particular, Hn (Pn(Pn
A , G ) is finitely generated. As H
A , O(m) )
n n
n−1 n
is finitely generated, and H (PA , G ) is too, we have that H(PA , F ) is finitely
generated for all coherent sheaves F . We continue inductively downwards.
(ii) Twist (18.1.3.1) by O(N) for N ≫ 0. Then⊕jn n
Hn (PnA , O(m + N) ) = ⊕j H (PA , O(m + N)) = 0(by (vii) above), so Hn (PnA , F (N)) = 0. Translation: for any coherent sheaf, its top
cohomology vanishes once you twist by O(N) for N sufficiently large. Hence thisis true for G as well. Hence from the long exact sequence, Hn−1 (PnA , F (N)) = 0
for N ≫ 0. As in (i), we induct downwards, until we get that H1 (PnA , F (N)) = 0.
⊕j(The induction stops here, as it is not true that H0 (PnA , O(m + N) ) = 0 for large
N — quite the opposite.)!
In particular, we have proved the following, that we would have cared about
even before we knew about cohomology.
18.1.5. Corollary. — Any projective k-scheme has a finite-dimensional space of global
sections. More generally, if A is Noetherian and F is a coherent sheaf on a projective
A-scheme, then H0 (X, F ) is a coherent A-module.
(We will generalize this in Theorem 18.8.1.) I want to emphasize how remarkable this proof is. It is a question about global sections, i.e., H0 , which we thinkof as the most down to earth cohomology group, yet the proof is by downward
induction for Hn , starting with n large.
Corollary 18.1.5 is true more generally for proper k-schemes, not just projective k-schemes (Grothendieck’s Coherence Theorem 18.9.1).
Here are some important consequences. They can also be shown directly, without the use of cohomology, but with much more elbow grease.
18.1.6. As a partial converse, if h0 (X, OX ) = 1, then X is connected (why?), butneed not be reduced: witness the subscheme in P2 cut out by x2 = 0. (For experts:
the geometrically connected hypothesis is necessary, as X = Spec C is a projective
integral R-scheme, with h0 (X, OX ) = 2. Similarly, a nontrivial purely inseparable

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field extension can be used to show that the geometrically reduced hypothesis isalso necessary.)
18.1.A. C RUCIAL E XERCISE ( PUSHFORWARDS OF COHERENT SHEAVES BY PROJEC TIVE MORPHISMS ARE COHERENT ). Suppose π : X → Y is a projective morphismof locally Noetherian schemes. Show that the pushforward of a coherent sheaf
on X is a coherent sheaf on Y. (See Grothendieck’s Coherence Theorems 18.8.1
and 18.9.1 for generalizations.)
Finite morphisms are affine (from the definition) and projective (17.3.C). We
can now show that this is a characterization of finiteness.
18.1.7. Corollary. — Suppose Y is locally Noetherian. Then a morphism π : X → Y is
projective and affine if and only if π is finite.

We will see in Exercise 18.9.A that the projective hypotheses can be relaxed toproper.

Proof. We already know that finite morphisms are affine (by definition) and projective (Exercise 17.3.C), so we show the converse. Suppose π is projective and affine.
By Exercise 18.1.A, π∗ OX is coherent and hence finite type.
!
The following result was promised in §17.3.5, and has a number of useful consequences.
18.1.8. Theorem (projective + finite fibers = finite). — Suppose π : X → Y with Y
Noetherian. Then π is projective and has finite fibers if and only if it is finite. Equivalently,
π is projective and quasifinite if and only if it is finite.
(Recall that quasifinite = finite fibers + finite type. But projective includes finite
type.) It is true more generally that (with Noetherian hypotheses) proper + finite
fibers = finite, see Theorem 29.6.2.
Proof. We show π is finite near a point q ∈ Y. Fix an affine open neighborhood
Spec A of q in Y. Pick a hypersurface H in Pn
A missing the preimage of q, so H ∩ X
is closed. Let H ′ = π(H ∩ X), which is closed in Spec A, and doesn’t contain q. Let
U = Spec A − H ′ , which is an open set containing q. Then above U, π is projective
and affine, so we are done by Corollary 18.1.7.
!
A similar trick was used in the proof of the Rigidity Lemma 10.3.12.
18.1.B. E XERCISE ( UPPER SEMICONTINUITY OF FIBER DIMENSION ON THE TARGET,FOR PROJECTIVE MORPHISMS ).
Use a similar argument as in Theorem 18.1.8
to prove upper semicontinuity of fiber dimension of projective morphisms: suppose
π : X → Y is a projective morphism where Y is locally Noetherian (or more generally OY is coherent over itself). Show that {q ∈ Y : dim π−1 (q) > k} is a Zariskiclosed subset of Y. In other words, the dimension of the fiber “jumps over Zariskiclosed subsets” of the target. (You can interpret the case k = −1 as the fact thatprojective morphisms are closed, which is basically the Fundamental Theorem ofElimination Theory 7.4.7, cf. §17.3.4.) This exercise is rather important for having a
sense of how projective morphisms behave. (The case of varieties was done earlier,in Theorem 11.4.2(b). This approach is much simpler.)

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The final exercise of the section is on a different theme.
18.1.C. E XERCISE . Suppose 0 → F → G → H → 0 is an exact sequence ofquasicoherent sheaves on X, projective over A, with F coherent. Show that forn ≫ 0,
0 → H0 (X, F (n)) → H0 (X, G (n)) → H0 (X, H (n)) → 0
is also exact. (Hint: for n ≫ 0, H1 (X, F (n)) = 0.)

18.2 Definitions and proofs of key properties
This section could be read much later; the facts we will use are all stated in
the previous section. However, the arguments are not complicated, so you want
to read this right away. As you read this, you should go back and check off all thefacts in the previous section, to assure yourself that you understand everythingpromised.
18.2.1. Čech cohomology. Čech cohomology in general settings is defined using
a limit over finer and finer covers of a space. In our algebro-geometric setting, thesituation is much cleaner, and we can use a single cover.
Suppose X is quasicompact and separated, which is true for example if X is
quasiprojective over A. In particular, X may be covered by a finite number ofaffine open sets, and the intersection of any two affine open sets is also an affine
open set (by separatedness, Proposition 10.1.8). We will use quasicompactness andseparatedness only in order to ensure these two nice properties.
Suppose F is a quasicoherent sheaf, and U = {Ui }ni=1 is a finite collection ofaffine open sets covering X. For I ⊂ {1, . . . , n} define UI = ∩i∈I Ui , which is affine
by the separated hypothesis. (Here is a strong analogy for those who have seen
cohomology in other contexts: cover a topological space X with a finite numberof open sets Ui , such that all intersections ∩i∈I Ui are contractible.) Consider theČech complex&(18.2.1.1)
0→
F (UI ) → · · · →|I| = 1I ⊂ {1, . . . , n}&

|I| = iI ⊂ {1, . . . , n}

F (UI ) →&|I| = i + 1
I ⊂ {1, . . . , n}

F (UI ) → · · · .

The maps are defined as follows. The map from F (UI ) → F (UJ ) is 0 unless I ⊂ J,
i.e., J = I ∪ {j}. If j is the kth element of J, then the map is (−1)k−1 times the
restriction map resUI ,UJ .
18.2.A. E ASY E XERCISE ( FOR THOSE WHO HAVEN ’ T SEEN ANYTHING LIKE THE
Č ECH COMPLEX BEFORE ). Show that the Čech complex is indeed a complex, i.e.,that the composition of two consecutive arrows is 0.
Define HiU (X, F ) to be the ith cohomology group of the complex (18.2.1.1).
(The indexing starts with i = 0.) Note that if X is an A-scheme, then HiU (X, F ) isan A-module. We have almost succeeded in defining the Čech cohomology group

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Hi , except our definition a priori depends on a choice of a cover U . Note that
HiU (X, ·) is clearly a covariant functor QCohX → ModA .

18.2.B. E ASY E XERCISE .
axioms for F .)Identify H0U (X, F ) with Γ (X, F ). (Hint: use the sheaf

18.2.C. E XERCISE . Suppose(18.2.1.2)

0→F →G →H →0

is a short exact sequence of sheaves of abelian groups on a topological space, andU is a finite open cover such that on any intersection UI of open subsets in U , themap Γ (UI , G ) → Γ (UI , H ) is surjective. Show that we get a “long exact sequenceof cohomology for HiU ” (where we take the same definition of HiU ). In our situation, where X is a quasicompact separated A-scheme, and (18.2.1.2) is a short exact
sequence of quasicoherent sheaves on X, show that we get a long exact sequence
for the A-modules HiU .
In the proof of Theorems 18.2.4 and 18.8.1, we will make use of the fact thatyour construction of the connecting homomorphism will “commute with localization of A”. More precisely, we will need the following.
18.2.D. E XERCISE . Suppose we are given a short exact sequence (18.2.1.2) of quasicoherent sheaves on a quasicompact separated A-scheme π : X → Spec A, a cover
U of X by affine open sets, and some f ∈ A. The restriction of the sets of U
to Xf yields an affine open cover U ′ of Xf = π−1 (D(f)). Identify the long exact
sequence associated to (18.2.1.2) using HiU , localized at f, with the long exact sequence associated to the restriction of (18.2.1.2) to Xf , using the affine open coverU ′ . (First check that the maps such as HiU (F ) → HiU (G ) given by covariance“commute with localization”, and then check that the connecting homomorphisms
do as well.)
18.2.2. Theorem/Definition. — Our standing assumption is that X is quasicompact
and separated. HiU (X, F ) is independent of the choice of (finite) cover {Ui }. More precisely, for any two covers {Ui } ⊂ {Vi }, the maps Hi{Vi } (X, F ) → Hi{Ui } (X, F ) inducedby the natural map of Čech complexes (18.2.1.1) are isomorphisms. Define the Čech cohomology group Hi (X, F ) to be this group.
If you are unsure of what the “natural map of Čech complexes” is, by (18.2.3.1)
it should become clear.
18.2.3. For experts: maps of complexes inducing isomorphisms on cohomology
groups are called quasiisomorphisms. We are actually getting a finer invariant
than cohomology out of this construction; we are getting an element of the derived
category of A-modules.
Proof. We need only prove the result when |{Vi }| = |{Ui }| + 1. We will show that
if {Ui }1≤i≤n is a cover of X, and U0 is any other affine open set, then the map

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Hi{Ui }0≤i≤n (X, F ) → Hi{Ui }1≤i≤n (X, F ) is an isomorphism. Consider the exact sequence of complexes(18.2.3.1)
0
0
0
"

···

%

"

|I| = i − 1
0∈I

···

%"

|I| = i − 1···"% " |I| = i − 1 F (UI )"0"

F (UI )

0∈
/ I

"

"

|I| = i0∈I%"

|I| = i%

F (UI )"
|I| = i + 1
0∈I%"|I| = i + 1% ···F (UI )

% ···"% " |I| = i F (UI )"% " |I| = i + 1 F (UI )

% ···

"
0

"
0

"

F (UI )

0∈
/ I

%

"

F (UI )F (UI )"

0∈
/ I

Throughout, I ⊂ {0, . . . , n}. The bottom two rows are Čech complexes with respect
to two covers, and the map between them induces the desired map on cohomology.
We get a long exact sequence of cohomology from this short exact sequence ofcomplexes (Exercise 1.6.C). Thus we wish to show that the top row is exact andthus has vanishing cohomology. (Note that U0 ∩ Uj is affine by our separatedness
hypothesis, Proposition 10.1.8.) But the ith cohomology of the top row is precisely
Hi−1{Uj ∩U0 }j>0 (U0 , F ) except at step 0, where we get 0 (because the complex starts"n
off 0 → F (U0 ) →j=1 F (U0 ∩ Uj )). So it suffices to show that higher Čechgroups of affine schemes are 0. Hence we are done by the following result.
!
18.2.4. Theorem. — The higher Čech cohomology HiU (X, F ) of an affine A-scheme X
vanishes (for any affine cover U , i > 0, and quasicoherent F ).
This is another “partition of unity” argument, in the spirit of the proof of Theorem 4.1.2.
Proof. (The following argument can be made shorter using spectral sequences,
but we avoid them for the sake of clarity.) We want to show that the “extended”
complex(18.2.4.1)

0 → F (X) →

&

|I|=1

F (UI ) →

&

|I|=2

F (UI ) → · · ·(where the global sections F (X) have been appended to the start) has no cohomology, i.e., is exact. We do this with a trick.

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Suppose first that some Ui , say U0 , is X. Then the complex is the middle rowof the following short exact sequence of complexes.
(18.2.4.2)
%0
%"
%"
% ···
0
F (UI )F (UI )
|I|=1,0∈I

0"% F (X)0"% F (X)

%"

"

"

"

|I|=1 F (UI )|I|=2,0∈I%

F (UI )
|I|=1,0∈
/I

%"

"

%

"

"

|I|=2 F (UI )F (UI )
|I|=2,0∈
/I

% ···
% ···

The top row is the same as the bottom row, slid over by 1. The corresponding longexact sequence of cohomology shows that the central row has vanishing cohomology. (You should show that the “connecting homomorphism” on cohomology is
indeed an isomorphism.) This might remind you of the mapping cone construction(Exercise 1.7.E).
We next prove the general case by sleight of hand. Say X = Spec R. We wish
to show that the complex of A-modules (18.2.4.1) is exact. It is also a complex of Rmodules, so we wish to show that the complex of R-modules (18.2.4.1) is exact. To
show that it is exact, it suffices to show that for a cover of Spec R by distinguishedopen sets D(fi ) (1 ≤ i ≤ r) (i.e., (f1 , . . . , fr ) = 1 in R) the complex is exact. (Translation: exactness of a sequence of sheaves may be checked locally.) We choose acover so that each D(fi ) is contained in some Uj = Spec Aj . Consider the complex
localized at fi . As
Γ (Spec A, F )fi = Γ (Spec(Aj )fi , F )(by quasicoherence of F , Exercise 13.3.D), as Uj ∩ D(fi ) = D(fi ), we are in thesituation where one of the Ui ’s is X, so we are done.
!
We have now proved properties (i)–(iii) of the previous section. The “affine
cover cohomology vanishing” property (vi) is also straightforward: if X is covered
by n affine open sets, use these as the cover U , and notice that the Čech complex
ends by the nth step.
18.2.E. E XERCISE ( PROPERTY (v)). Suppose π : X → Y is an affine morphism,
and Y is a quasicompact and separated A-scheme (and hence X is too, as affinemorphisms are both quasicompact and separated). If F is a quasicoherent sheaf∼ Hi (X, F ). (Hint: if U is anon X, describe a natural isomorphism Hi (Y, π∗ F ) =−1
affine cover of Y, “π (U )” is an affine cover of X. Use these covers to computethe cohomology of F .)18.2.F. E XERCISE ( PROPERTY (iv)). Suppose π : X → Y is any quasicompact separated morphism, F is a quasicoherent sheaf on X, and Y is a quasicompact separated A-scheme. The hypotheses on π ensure that π∗ F is a quasicoherent sheaf on
Y. Describe a natural morphism Hi (Y, π∗ F ) → Hi (X, F ) extending Γ (Y, π∗ F ) →Γ (X, F ). (Aside: this morphism is an isomorphism for i = 0, but need not be an
isomorphism for higher i: consider i = 1, X = P1k , F = O(−2), and let Y be a point
Spec k.)

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18.2.G. E XERCISE . Prove Property (vii) of the previous section. (This can be doneby hand. Hint: in the category of modules over a ring, taking the colimit over afiltered sets is an exact functor, §1.6.12.)

We have now proved all of the properties of the previous section, except for(viii), which we will get to in §18.3.

18.2.5. Useful facts about cohomology for k-schemes.
18.2.H. E XERCISE ( COHOMOLOGY AND CHANGE OF BASE FIELD ). Suppose X is aquasicompact separated k-scheme, and F is a quasicoherent sheaf on X. Give an
isomorphism∼ Hi (X ×Spec k Spec K, F ⊗k K)
Hi (X, F ) ⊗k K =for all i, where K/k is any field extension. Here F ⊗k K means the pullback of F
to X ×Spec k Spec K. Hence hi (X, F ) = hi (X ×Spec k Spec K, F ⊗k K). If i = 0 (taking H0 = Γ ), show the result without the quasicompact and separated hypotheses.
(This is useful for relating facts about k-schemes to facts about schemes over algebraically closed fields. Your proof might use vector spaces — i.e., linear algebra— in a fundamental way. If it doesn’t, you may prove something more general, if
k → K is replaced by a flat ring map B → A. Recall that B → A is flat if ⊗B A is anexact functor ModB → ModA . A hint for this harder exercise: the FHHF Theorem,
Exercise 1.6.H. See Exercise 18.8.B(b) for the next generalization of this.)
18.2.I. E XERCISE ( BASE - POINT- FREENESS IS INDEPENDENT OF EXTENSION OF BASE
FIELD ). Suppose X is a scheme over a field k, L is an invertible sheaf on X, andK/k is a field extension. Show that L is base-point-free if and only if its pullback
to X ×Spec k Spec K is base-point-free. (Hint: Exercise 18.2.H with i = 0 implies thata basis of sections of L over k becomes, after tensoring with K, a basis of sectionsof L ⊗k K.)

18.2.6. Theorem (dimensional cohomology vanishing for quasicoherent sheaves
on projective k-schemes). — Suppose X is a projective k-scheme, and F is a quasicoherent sheaf on X. Then Hi (X, F ) = 0 for i > dim X.
In other words, cohomology vanishes above the dimension of X.
Proof. Suppose X "→ PN , and let n = dim X. We show that X may be covered by
n + 1 affine open sets. Exercise 11.3.C shows that there are n + 1 effective Cartierdivisors on PN such that their complements U0 , . . . , Un cover X. Then Ui is affine,so Ui ∩ X is affine, and thus we have covered X with n + 1 affine open sets.
!
(It turns out that n + 1 affine open sets are always necessary. One way ofproving this is by showing that the complement of a dense affine open subset isalways pure codimension 1, see for example [RV, Lem. 2.3].)

18.2.7. ⋆ Dimensional cohomology vanishing more generally. Using the theory of blowing up, Theorem 18.2.6 can be extended to quasiprojective k-schemes,see §22.4.15. Dimensional cohomology vanishing is even true in much greater
generality. To state it, we need to define cohomology with the more general machinery of derived functors (Chapter 23). If X is a Noetherian topological space(§3.6.14) and F is any sheaf of abelian groups on X, we have Hi (X, F ) = 0 for all

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475

i > dim X. (Grothendieck sketches his elegant proof in [GrS, p. 29-30]; see [Ha1,
Theorem III.2.7] for a more detailed explanation.) In particular, if X is a k-variety ofdimension n, we always have dimensional cohomology vanishing, even for crazyvarieties that can’t be covered with n + 1 affine open subsets (see §22.4.15).
18.2.8. The Künneth formula.
Suppose X and Y are quasicompact separated k-schemes, and F and G are
quasicoherent sheaves on X and Y respectively. Let πx : X ×k Y → X and πY : X ×k
Y → Y be the two projections. Recall the definition F % G := π∗X F ⊗ π∗Y G (§16.4.8).
Then we have an isomorphism∼ ⊕p+q=m Hp (X, F ) ⊗k Hq (Y, G ).
Hm (X ×k Y, F % G ) =To show this, choose affine covers of X and Y, and produce the Čech complexes
for F and G . Show that the tensor product of these two complexes (the total
complex associated to the double complex) is the Čech complex for F % G (withrespect to the products of the affine covers of X and Y). Finally, show that thecohomology of the tensor product of two complexes over k is the tensor productsof the cohomologies, a result known as the Eilenberg-Zilber Theorem.

18.2.9. The cup product. The cup product in Čech cohomology can be defined ina simple way; see [Liu, Exer. 2.17] for a particularly elegant description. We will
not need this construction.

18.3 Cohomology of line bundles on projective space
We now finally prove the last promised basic fact about cohomology, property(viii) of §18.1, Theorem 18.1.3, on the cohomology of line bundles on projective
space. More correctly, we will do one case and you will do the rest.
We begin with a warm-up that will let you (implicitly) see some of the structure that will arise in the proof. It also gives good practice in computing cohomology groups.
18.3.A. E XERCISE . Compute the cohomology groups Hi (A2k \ {(0, 0)}, O). (Hint:
the case i = 0 was done in Example 4.4.1. The case i > 1 is clear from “affine covercohomology vanishing”, property (vi) above.) In particular, show that H1 (A2k \{(0, 0)}, O) ̸= 0, and thus give another proof (see §4.4.3) of the fact that A2k \ {(0, 0)}
is not affine. (Cf. Serre’s cohomological criterion for affineness, Remark 18.1.1.)
18.3.1. Remark. Essential Exercise 14.1.C and the ensuing discussion showed thatn
H0 (PnA , OPA (m)) should be interpreted as the homogeneous degree m polynomin
als in x0 , . . . , xn (with A-coefficients). Similarly, Hn (PnA , OPA (m)) should be interpreted as the homogeneous degree m Laurent polynomials in x0 , . . . , xn , where ineach monomial, each xi appears with degree at most −1.
18.3.2. Proof of Theorem 18.1.3.
Un = D(xn ) of Pn
A.

We take the standard cover U0 = D(x0 ), . . . ,18.3.B. E XERCISE ( ESSENTIAL FOR THE PROOF OF T HEOREM 18.1.3; CF. R EMARK 18.3.1
ABOVE ). If I ⊂ {0, . . . , n}, then give an isomorphism (of A-modules) of Γ (UI , O(m))

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with the homogeneous degree m Laurent monomials (in x0 , . . . , xn , with coefficients in A) where each xi for i ∈/ I appears with non-negative degree. Your
construction should be such that the restriction map Γ (UI , O(m)) → Γ (UJ , O(m))(I ⊂ J) corresponds to the natural inclusion: a Laurent polynomial in Γ (UI , O(m))
maps to the same Laurent polynomial in Γ (UJ , O(m)).
The Čech complex for O(m) is the degree m part of(18.3.2.1)% A[x0 , x1 , x2 , x−1 ] × A[x0 , x1 , x2 , x−1 ] × A[x0 , x1 , x2 , x−1 ]
0
0

1

2

%

−1
−1 −1
−1 −1
A[x0 , x1 , x2 , x−10 , x1 ] × A[x0 , x1 , x2 , x1 , x2 ] × A[x0 , x1 , x2 , x0 , x2 ]% A[x0 , x1 , x2 , x−1 , x−1 , x−1 ]
0

1

2

% 0.

Rather than consider O(m) for each m independently, it is notationally simpler
to consider them all at once, by considering F = ⊕m∈Z O(m): the Čech complex
for F is (18.3.2.1). It is useful to write which UI corresponds to which factor (see(18.3.2.2) below). The maps (from one factor of one term to one factor of the next)
are all natural inclusions, or negative of natural inclusions, and in particular preserve degree.
We extend (18.3.2.1) by replacing the 0 → on the left by 0 → A[x0 , x1 , x2 ] →:(18.3.2.2)

0H0

U0 U1 U2

% A[x0 , x1 , x2 ]

% ···

U012

% ···

% A[x0 , x1 , x2 , x−1
, x−1
x−1
]
0
1
2

% 0.

18.3.C. E XERCISE . Show that if (18.3.2.2) is exact, except that at U012 the cohomology/cokernel is−1 −1−1 −1 −1x−1
0 x1 x2 A[x0 , x1 , x2 ],then Theorem 18.1.3 holds for n = 2. (Hint: Remark 18.3.1.)
Because the maps in (18.3.2.2) preserve multidegree (degrees of each xi independently), we can study exactness of (18.3.2.2) monomial by monomial.
0 a1 a2
The “3 negative exponents” case. Consider first the monomial xa
0 x1 x2 , where
the exponents ai are all negative. Then (18.3.2.2) in this multidegree is:

0

% 0H0

% 00 × 01 × 02

% 001 × 012 × 002% A012% 0.

Here the subscripts serve only to remind us which “Čech” terms the factors cor0 a1 a2
respond to. (For example, A012 corresponds to the coefficient of xa
0 x1 x2 in−1 −1
A[x0 , x1 , x2 , x−10 , x1 , x2 ].) Clearly this complex only has (co)homology at the
U012 spot, as desired.
The “2 negative exponents” case. Consider next the case where two of the exponents, say a0 and a1 , are negative. Then the complex in this multidegree is
0

% 0H0

% 00 × 01 × 02

% A01 × 012 × 002% A012% 0,

which is clearly exact.

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The “1 negative exponent” case. We next consider the case where one of the
exponents, say a0 , is negative. Then the complex in this multidegree is
0

% 0H0

% A0 × 01 × 02

% A01 × 012 × A02% A012%0With a little thought (paying attention to the signs on the arrows A → A), youwill see that it is exact. (The subscripts, by reminding us of the subscripts in theoriginal Čech complex, remind us what signs to take in the maps.)
The “0 negative exponent” case. Finally, consider the case where none of the
exponents are negative. Then the complex in this multidegree is
0

% AH0

% A0 × A1 × A2

% A01 × A12 × A02% A012%0

We wish to show that this is exact. We write this complex as the middle of a short
exact sequence of complexes:(18.3.2.3)%0% A2% A02 × A12% A012
%0
0
"
0

"
% AH0

"
% A0 × A1 × A2"
% A01 × A12 × A02"% A012

"
%0

"
0

"
% AH0

"
% A0 × A1

"
% A01

"
%0

"
%0

Thus we get a long exact sequence in cohomology (Theorem 1.6.6). But the top
and bottom rows are exact (basically from the “1-negative” case), i.e., cohomologyfree, so the middle row must be exact too.
18.3.D. E XERCISE . Prove Theorem 18.1.3 for general n. (I could of course justhave given you the proof for general n, but seeing the argument in action may beenlightening. In particular, your argument may be much shorter. For example, the“2-negative” case could be done in the same way as the “1-negative” case, so you
will not need n + 1 separate cases if you set things up carefully.)!
18.3.3. Remarks. (i) In fact we don’t really need the exactness of the top and bottom
rows of (18.3.2.3); we just need that they are the same, just as with (18.2.4.2).
(ii) This argument is basically the proof that the reduced homology of theboundary of a simplex S (known in some circles as a “sphere”) is 0, unless S is theempty set, in which case it is one-dimensional. The “empty set” case corresponds
to the “3-negative” case.
,i
n
j m
i−j n
Show that Hi (Pm(Pk , O(b)).
k ×k Pk , O(a, b)) =j=0 H (Pk , O(a)) ⊗k H
(Can you generalize this Künneth-type formula further?)

18.4 Riemann-Roch, degrees of coherent sheaves, and arithmetic
genus

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We have seen some powerful uses of Čech cohomology, to prove things about
spaces of global sections, and to prove Serre vanishing. We will now see someclassical constructions come out very quickly and cheaply.
In this section, we will work over a field k. Suppose F is a coherent sheaf on
a projective k-scheme X. Recall the notation (§18.1) hi (X, F ) := dimk Hi (X, F ).
By Theorem 18.1.4, hi (X, F ) is finite. (The arguments in this section will extend
without change to proper X once we have this finiteness for proper morphisms, byGrothendieck’s Coherence Theorem 18.9.1.) Define the Euler characteristic of F
by
dim
.X
χ(X, F ) :=(−1)i hi (X, F ).
i=0

We will see repeatedly here and later that Euler characteristics behave better than
individual cohomology groups. As one sign, notice that for fixed n, and m ≥ 0,"#n+m
(m + 1)(m + 2) · · · (m + n)
0 n
h (Pk , O(m)) =
=
.
m
n!

Notice that the expression on the right is a polynomial in m of degree n. (For later
reference, notice also that the leading term is mn /n!.) But it is not true that(m + 1)(m + 2) · · · (m + n)n!
for all m — it breaks down for m ≤ −n − 1. Still, you can check (using Theorem 18.1.3) thath0 (Pnk , O(m)) =(m + 1)(m + 2) · · · (m + n).
n!
So one lesson is this: if one cohomology group (usually the top or bottom) behaves
well in a certain range, and then messes up, likely it is because (i) it is actually
the Euler characteristic which behaves well always, and (ii) the other cohomology
groups vanish in that certain range.
In fact, we will see that it is often hard to calculate cohomology groups (evenh0 ), but it can be easier calculating Euler characteristics. So one important way ofgetting a hold of cohomology groups is by computing the Euler characteristic, andthen showing that all the other cohomology groups vanish. Hence the ubiquity
and importance of vanishing theorems. (A vanishing theorem usually states that acertain cohomology group vanishes under certain conditions.) We will see this inaction when discussing curves. (One of the first applications will be (19.2.5.1).)
The following exercise shows another way in which Euler characteristic behaves well: it is additive in exact sequences.
χ(Pnk , O(m)) =18.4.A. E XERCISE . Show that if 0 → F → G → H → 0 is an exact sequence ofcoherent sheaves on a projective k-scheme X, then χ(X, G ) = χ(X, F ) + χ(X, H ).
(Hint: consider the long exact sequence in cohomology.) More generally, if
0 → F1 → · · · → Fn → 0

is an exact sequence of coherent sheaves, show thatn.

(−1)i χ(X, Fi ) = 0.

i=1

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479(This exercise both generalizes the “exact” case of Exercise 1.6.B — consider thecase where X = Spec k — and uses it in the proof.)
18.4.1. The Riemann-Roch
Theorem for line bundles on a regular projective,
curve. Suppose D := p∈C ap [p] is a divisor on a regular projective curve C overa field k (where ap ∈ Z, and all but finitely many ap are 0). Define the degree of
D by.
deg D =
ap deg p.
(The degree of a point p was defined in §5.3.8, as the degree of the field extensionof the residue field over k.)

18.4.B. E SSENTIAL E XERCISE : T HE R IEMANN -R OCH T HEOREM FOR LINE BUN DLES ON A REGULAR PROJECTIVE CURVE . Show that
χ(C, OC (D)) = deg D + χ(C, OC ),by induction on |ap | (where D =ap [p] as above). Hint: to show that χ(C, OC (D)) =deg p + χ(C, OC (D − p)), tensor the closed subscheme exact sequence,

0 → OC (−p) → OC → O|p → 0(where O|p is the structure sheaf of the scheme p, not the stalk OC,p ) by OC (D),and use additivity of Euler characteristics in exact sequences (Exercise 18.4.A).
As every invertible sheaf L is of the form OC (D) for some D (see §14.2), thisexercise is very powerful.
18.4.C. I MPORTANT E ASY E XERCISE ( AND D EFINITION ). Suppose L is an invertible sheaf on a regular projective curve C over k. Define the degree of L (denoteddeg L or degC L ) as χ(C, L ) − χ(C, OC ). Let s be a nonzero rational section of L
on C. Let D be the divisor of zeros and poles of s:.
D :=vp (s)[p].
p∈CShow that deg L = deg D. In particular, the degree can be computed by counting
zeros and poles of any section not vanishing on a component of C.
18.4.D. E XERCISE . Give a new solution to Exercise 17.4.E (a nonzero rational
function on a projective curve has the same number of zeros and poles, counted
appropriately) using the ideas above.
18.4.E. E XERCISE . If L and M are two line bundles on a regular projective curve
C, show that deg L ⊗ M = deg L + deg M . (Hint: choose nonzero rational
sections of L and M .)18.4.F. E XERCISE . Suppose π : C → C ′ is a degree d morphism of integral projective regular curves, and L is an invertible sheaf on C ′ . Show that degC π∗ L =d degC ′ L . Hint: compute degC ′ L using any nonzero rational section s of L ,and compute degC π∗ L using the rational section π∗ s of π∗ L . Note that zeros pullback to zeros, and poles pull back to poles. Reduce to the case where L = O(p)for a single point p. Use Exercise 17.4.D.

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18.4.G. ⋆⋆ E XERCISE ( COMPLEX - ANALYTIC INTERPRETATION OF DEGREE ; ONLYFOR THOSE WITH SUFFICIENT ANALYTIC BACKGROUND ).
Suppose X is a connected regular projective complex curve. Show that the degree map is the composition of group homomorphisms
Pic X% Pic Xanc1

∼ Z.
% H2 (Xan , Z) ∩[Xan ]% H0 (Xan , Z) =Hint: show it for a generator O(p) of the group Pic X, using explicit transition
functions. (The first map was discussed in Exercise 13.1.L. The second map takes
a line bundle to its first Chern class, and can be interpreted as follows. The transition functions for a line bundle yield a Čech 1-cycle for OX∗ an ; this yields a
map Pic Xan → H1 (Xan , OX∗ an ). Combining this with the map H1 (Xan , OX∗ an ) →H2 (Xan , Z) from the long exact sequence in cohomology corresponding to the exponential exact sequence (2.4.10.1) yields the first Chern class map.)
18.4.2. Arithmetic genus.
Motivated by geometry (Miracle 18.4.3 below), we define the arithmetic genus
of a scheme X as 1 − χ(X, OX ). This is sometimes denoted pa (X). For integral
projective curves over an algebraically closed field, h0 (X, OX ) = 1 (§10.3.7), sopa (X) = h1 (X, OX ). (In higher dimension, this is a less natural notion.)
We can restate the Riemann-Roch formula for curves (Exercise 18.4.B) as:(18.4.2.1)

h0 (C, L ) − h1 (C, L ) = deg L − pa (C) + 1.

This is the most common formulation of the Riemann-Roch formula.
18.4.3. Miracle. If C is a regular irreducible projective complex curve, then the corresponding complex-analytic object, a compact Riemann surface, has a notion called
the genus g, which is (informally speaking) the number of holes (see Figure 18.1).
Miraculously, g = pa in this case (see Exercise 21.7.I), and for this reason, we
will often write g for pa when discussing regular (projective irreducible) curves,
over any field. We will discuss genus further in §18.6.6, when we will be able tocompute it in many interesting cases. (Warning: the arithmetic genus of P1C as an
R-variety is −1!)

F IGURE 18.1. A genus 3 Riemann surface
18.4.4. Degree and rank of a coherent sheaf.
Suppose C is an irreducible reduced curve (possibly singular), over a field k.
If F is a coherent sheaf on C, recall (from §13.7.4) that the rank of F , denoted
rank F , is its rank at the generic point of C.
18.4.H. E ASY E XERCISE . Show that the rank is additive in exact sequences: if
0→F →G →H →0

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is an exact sequence of coherent sheaves, show thatrank F − rank G + rank H = 0.
Hint: localization is exact. (Caution: your argument will use the fact that the rank
is at the generic point; the example
0<% k[t]×t<% k[t]#% k[t]/(t)%0on A1k shows that rank at a closed point is not additive in exact sequences.)
If C is projective, define the degree of F by
(18.4.4.1)

deg F = χ(C, F ) − (rank F ) · χ(C, OC ).

If F is an invertible sheaf (or if more generally the rank is the same on each irreducible component), we can drop the irreducibility hypothesis. Thus this generalizes the notion of the degree of a line bundle on a regular curve (Important
Exercise 18.4.C). We now study the behavior of this notion. (In Exercise 21.7.B,
you will show that if F is supported at a finite number of points, the degree of F
splits up into a contribution from each point.)
18.4.I. E ASY E XERCISE . Show that degree (as a function of coherent sheaves on a
fixed curve C) is additive in exact sequences.
18.4.J. E XERCISE . Show that the degree of a vector bundle is the degree of its
determinant bundle. Hint: Exercise 13.5.H.
The statement (18.4.4.1) is often called Riemann-Roch for coherent sheaves (orvector bundles) on a projective curve.
18.4.K. E XERCISE . If C is a projective curve, and L is an ample line bundle on C,
show that deg L > 0. (Hint: show it if L is very ample.)
18.4.L. E XERCISE . Suppose L is a base-point-free invertible sheaf on a proper
variety X, and hence induces some morphism φ : X → Pn . Then L is ample if andonly if φ is finite. (Hint: if φ is finite, use Exercise 16.6.G. If φ is not finite, showthat there is a curve C contracted by φ, using Theorem 18.1.8. Show that L has
degree 0 on C.)
18.4.5. Extended example: the universal plane conic has no rational sections.
We use the theory of the degree to get an interesting consequence. We workover a fixed field k. We consider the following diagram.
"cl. emb.% 2C!●
P × P5
●●
●●
●●
●● "
$
P5

% P2

If the P2 has projective coordinates x0 , x1 , x2 , then P5 has coordinates a00 , a01 ,a11 , a02 , a12 , a22 , and C is cut out by the single equation
a00 x20 + a01 x0 x1 + · · · + a22 x22 = 0.

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We interpret P5 as the parameter space of conics (in P2 ), and C as the universal
conic over P5 (parametrizing a conic C along with a point p ∈ C), which comeswith a canonical projection C → P2 .
18.4.M. E XERCISE . By interpreting C as a P4 -bundle over P2 , show that C is a
∼ Z × Z.
smooth sixfold, and that Pic C =18.4.N. E XERCISE . Fix a line ℓ ⊂ P2 and a k-valued point q ∈ P2 . Let Dℓ be thedivisor on C corresponding to (C, p) with p lying on ℓ. Let Dq be the divisor on C
corresponding to (C, p) with q ∈ C. Using your description of C as a P4 -bundle
over P2 , show that Dℓ and Dq generate Pic C .
18.4.O. E XERCISE . Suppose K is a fiber of C → P5 over a point r ∈ P5 — i.e., aconic in P2 over the field κ(r). Suppose further that neither q nor ℓ are containedin K. (This hypothesis is unnecessary, but simplifies the problem.) Show that
Dq · K = 0 and Dℓ · K = 2. Hence show that if L is any invertible sheaf on C , then
L · K is even.
18.4.P. E XERCISE . Show that there is no rational section to the projection π : C →P5 . Hint: if there were a regular section over an open subset U of P5 , it would be adivisor on π−1 (U); let D be its closure in C . Show that D meets any fiber of π over
U in multiplicity 1. Use Exercise 18.4.O to obtain a contradiction.
We can restate Exercise 18.4.P in the following dramatic way: there is no way
to write down three rational functions X0 , X1 , X2 in a00 , . . . , a22 such that
a00 X20 + a01 X0 X1 + · · · + a22 X22 = 0

without X0 = X1 = X2 = 0.
The question of a rational point on a conic is one of arithmetic. (Think: x2 +2
y = z2 .) Our solution was topological. The unification of topology and arithmetic
in this example is the beginning of a long and fruitful story in algebraic geometry.
18.4.6. Length, and Riemann-Roch for nonreduced curves.
In order to state Riemann-Roch for nonreduced curves (Exercise 18.4.S), we
need the notion of the length of a module. This is more fundamental than might besuggested by our small discussion here, which is why this section is not starred,
while the less important Riemann Roch for nonreduced curves (Exercise 18.4.S) is.
Length is a measure of a module’s size, generalizing the notion of dimension ofa vector space over a field. Modules of finite length share many properties withfinite-dimensional vector spaces. (They play an important but implicit role in the
proof of Krull’s Principal Ideal Theorem given in §11.5.)
18.4.7. Definition. The length of an A-module M, denoted ℓ(M), is the length ofthe longest strictly increasing chain of submodules of M, where the indexing (asusual) starts with 0. For example, the length of the 0-module is 0. And if A = Z
and n ̸= 0, then the length of Z/(n) is the number of prime factors of n.
18.4.8. A maximal strictly increasing chain of submodules of M is called a composition series for M. Clearly the subquotients of a composition series are all simple
modules (i.e., they contain no nontrivial submodules, and are thus isomorphic to
A/m for some maximal ideal m).

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18.4.Q. E XERCISE . Suppose M has a finite composition series, of length n. Show
that every composition series of M has length n, and in particular ℓ(M) = n. Possible hint: this parallels other composition series results you have seen in othercontexts, such as for finite groups. (If absolutely necessary, see [E, Thm. 2.13] foran argument.)
18.4.R. E XERCISE . Suppose (A, m, k) is a Noetherian local ring, and M is a finitely
generated A-module. Show that M has finite length if and only if mn M = 0 forsome n. Thus by §18.4.8, if M has finite length, then the quotients in any composition series are all isomorphic (as A-modules) to k = A/m. (Informal translation:“An A-module M has finite length if and only if it can be built out of finitely many
copies of k.”) Hint: show that mn M = mn+1 M if and only if mn M = 0.
You can use Exercise 18.4.R to show that M has finite length if and only if it
is Artinian (i.e., satisfies the descending chain condition, §11.5). (More generally,
an arbitrary module M over an arbitrary ring A has finite length if and only if it isArtinian and Noetherian, [E, Thm. 2.13].)
18.4.S. ⋆ E XERCISE (R IEMANN -R OCH FOR NONREDUCED CURVES ). Suppose C
is a projective curve over a field k, and F is a coherent sheaf on C. Show thatχ(L ⊗ F ) − χ(F ) is the sum over the irreducible components Ci of C of thedegree of L on Cred
times the length of F at the generic point ηi of Ci (the lengthiof Fηi as an Oηi -module). Hints: (1) First reduce to the case where F is schemetheoretically supported on Cred , by showing that both sides of the alleged equality
are additive in short exact sequences, and using the filtration
0 = I r F ⊂ I r−1 F ⊂ · · · ⊂ I F ⊂ F

of F , where I is the ideal sheaf cutting out Cred in C. Thus we need only consider∼ O(, nj pj )the case where C is reduced. (2) As L is projective, we can write L =where the pj are regular points distinct from the associated points of F . Use thisavatar of L , and perhaps induction on the number of pj .
In fact, all proper curves over k are projective (Remark 18.7.2), so “projective”
can be replaced by “proper” in Exercise 18.4.S. In this guise, we will use Exercise 18.4.S when discussing intersection theory in Chapter 20.
18.4.9. ⋆ Numerical equivalence, the Néron-Severi group, nef line bundles, andthe nef and ample cones.
The notion of a degree on a line bundle leads to important and useful notions.
Suppose X is a proper k-variety, and L is an invertible sheaf on X. If i : C "→ X isa one-dimensional closed reduced subscheme of X, define the degree of L on C
by degC L := degC i∗ L . If degC L = 0 for all C, we say that L is numericallytrivial.
18.4.T. E XERCISE .
(a) Show that L is numerically trivial if and only if degC L = 0 for all integral
curves C in X.
(b) Show that if π : X → Y is a proper morphism, and L is a numerically trivial
invertible sheaf on Y, then π∗ L is numerically trivial on X.
(c) Show that L is numerically trivial if and only if L is numerically trivial oneach of the irreducible components of X.

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The Rising Sea: Foundations of Algebraic Geometry(d) Show that if L and L ′ are numerically trivial, then L ⊗ L ′ and L ∨ are both
numerically trivial.
18.4.10. Numerical equivalence.
By part (d), the numerically trivial invertible
sheaves form a subgroup of Pic X, denoted Picτ X. The resulting equivalence online bundles is called numerical equivalence. Two lines bundles equivalent modulo the subgroup of numerically trivial line bundles are called numerically equivalent. A property of invertible sheaves stable under numerical equivalence is said
to be a numerical property. We will see that “nefness” and ampleness are numericalproperties (Definition 18.4.11 and Remark 20.4.2 respectively).
We will later define the Néron-Severi group NS(X) of X as Pic X modulo algebraic equivalence (Exercise 24.7.5). (We will define algebraic equivalence once wehave discussed flatness.) The highly nontrivial Néron-Severi Theorem (or Theorem of the Base) states that NS(X) is a finitely generated group. (The proof isquite difficult; see [Kl1, p. 334, Prop. 3]. For a simpler proof over C, see [GH1, p.
462].) The group Pic X/ Picτ X is denoted N1 (X). We will see (in §24.7.5) that it isa quotient of NS(X), so it is also finitely generated. As the group N1 (X) is clearly
abelian and torsion-free, it is finite free Z-module (by the classification of finitely
generated modules over a principal ideal domain, see §0.3). The rank of N1 (X)is called the Picard number, and is denoted ρ(X) (although we won’t have needof this notion, except in our discussion of the Hodge Index Theorem in §20.2.10).
For example, ρ(Pn ) = 1 and ρ((P1 )n ) = n. We define N1Q (X) := N1 (X) ⊗Z Q (soρ(X) = dimQ N1Q (X)), and call the elements of this group Q-line bundles, for lackof any common term in the literature.
18.4.U. ⋆⋆ E XERCISE ( FINITENESS OF P ICARD NUMBER IN THE COMPLEX CASE ,ONLY FOR THOSE WITH SUFFICIENT BACKGROUND ). Show (without the NéronSeveri Theorem) that if X is a complex proper variety, then ρ(X) is finite, by interpreting it as a subquotient of H2 (X, Z). Hint: show that the image of (L , C) underthe map H2 (X, Z) × H2 (X, Z) → H0 (X, Z) → Z is degC L . Hint: figure out how to
reduce to the case where C is a smooth projective curve, then use Exercise 18.4.G.
18.4.11. Definition. We say that an invertible sheaf L is numerically effective, or
nef if for all such C, degC L ≥ 0. Clearly nefness is a numerical property.
18.4.V. E ASY E XERCISE ( CF. E XERCISE 18.4.T).
(a) Show that L is nef if and only if degC L ≥ 0 for all integral curves C in X.
(b) Show that if π : X → Y is a proper morphism, and L is a nef invertible sheaf on
Y, then π∗ L is nef on X. (Hint: Exercise 18.4.F will be needed.)(c) Show that L is nef if and only if L is nef on each of the irreducible componentsof X.
(d) Show that if L and L ′ are nef, then L ⊗ L ′ is nef. Thus the nef elements of
Pic X form a semigroup.
(e) Show that ample invertible sheaves are nef.
(f) Suppose n ∈ Z>0 . Show that L is nef if and only if L ⊗n is nef.
18.4.W. E XERCISE . Define what it means for a Q-line bundle to be nef. Show thatthe nef Q-line bundles form a closed cone in N1Q (X). This is called the nef cone.

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18.4.X. E XERCISE . Describe the nef cones of P2k and P1k ×k P1k . (Notice in the lattercase that the two boundaries of the cone correspond to linear series contracting oneof the P1 ’s. This is true in general: informally speaking, linear series corresponding
to the boundaries of the cone give interesting contractions. Another example will
be given in Exercise 20.2.F.)
It is a surprising fact that whether an invertible sheaf L on X is ample depends
only on its class in N1Q (X), i.e., on how it intersects the curves in X. Because of this(as for any n ∈ Z≥0 , L is ample if and only if L ⊗n is ample, see Theorem 16.6.2), it
makes sense to define when a Q-line bundle is ample. Then by Exercise 16.6.H, theample divisors form a cone in N1Q (X), necessarily contained in the nef cone by Exercise 18.4.V(e). It turns out that if X is projective, the ample divisors are precisely
the interior of the nef cone. The new facts in this paragraph are a consequence ofKleiman’s numerical criterion for ampleness, Theorem 20.4.6.

18.5 A first glimpse of Serre duality
A common version of Riemann-Roch involves Serre duality, which unlike
Riemann-Roch is hard.
18.5.1. Theorem (Serre duality for smooth projective varieties). — Suppose X isa geometrically irreducible smooth projective k-variety, of dimension n. Then there is aninvertible sheaf ωX (or simply ω) on X such thathi (X, F ) = hn−i (X, ωX ⊗ F ∨ )for all i ∈ Z and all finite rank locally free sheaves F .

The invertible sheaf ωX is an example of a dualizing sheaf, which will be formally defined in §30.1.4. We will see in Chapter 30 that Theorem 18.5.1 is a consequence of a perfect pairing(18.5.1.1)∼ k,
Hi (X, F ) × Hn−i (X, ωX ⊗ F ∨ ) → Hn (X, ωX ) =and that smoothness can be relaxed somewhat.

18.5.2. Further miracle: the sheaf of algebraic volume forms is Serre-dualizing. The invertible sheaf ωX turns out to be the canonical sheaf (or canonical bundle) KX , which
is defined as the determinant of the cotangent bundle ΩX/k of X (see §21.5.3); we
will define the cotangent bundle in Chapter 21. We connect the dualizing sheaf tothe canonical bundle in §30.4; see Desideratum 30.1.1.
18.5.3. Back to Riemann-Roch. For the purposes of restating Riemann-Roch for a
curve C, it suffices to note that h1 (C, L ) = h0 (C, ωC ⊗ L ∨ ). Then the RiemannRoch formula can be rewritten as
h0 (C, L ) − h0 (C, ωC ⊗ L ∨ ) = deg L − pa (C) + 1.
18.5.A. E XERCISE ( ASSUMING S ERRE DUALITY ). Suppose C is a geometricallyintegral smooth projective curve over k.
(a) Show that h0 (C, ωC ) is the genus g of C. (The genus g, in the guise of the

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The Rising Sea: Foundations of Algebraic Geometryarithmetic genus pa , was defined in §18.4.2–18.4.3.)(b) Show that deg ωC = 2g − 2. (Hint: Riemann-Roch for L = ωC .)∼ O(−2). Moreover,
18.5.4. Example. If C = P1k , Exercise 18.5.A implies that ωC =we also have a natural perfect pairing (cf. (18.5.1.1))∼ H1 (P1 , O(−2)).
H0 (P1 , O(n)) × H1 (P1 , O(−2 − n)) → k =We can interpret this pairing as follows. If n < 0, both factors on the left are 0,
so we assume n > 0. Then H0 (P1 , O(n)) corresponds to homogeneous degreen polynomials in x and y, and H1 (P1 , O(−2 − n)) corresponds to homogeneousdegree −2 − n Laurent polynomials in x and y so that the degrees of x and y areboth at most −1 (see Remark 18.3.1). You can quickly check that the dimensionof both vector spaces are n + 1. The pairing is given as follows: multiply thepolynomial by the Laurent polynomial, to obtain a Laurent polynomial of degree−2. Read off the coefficient of x−1 y−1 . (This works more generally for Pn
A ; see thediscussion after the statement of Theorem 18.1.3.)
18.5.B. E XERCISE ( ASSUMING S ERRE DUALITY ): AMPLE DIVISORS ON A CONNECTEDSMOOTH PROJECTIVE VARIETY ARE CONNECTED .
Suppose X is a connectedsmooth projective k-variety of dimension at least 2, and D is an effective ample
divisor. Show that D is connected. (Hint: Suppose D = V(s), where s is a section of an ample invertible sheaf. Then V(sn ) = V(s) for all n > 0, so we may
replace L with a high power of our choosing. Use the long exact sequence for
0 → OX (−nD) → OX → OV(sn ) → 0 to show that for n ≫ 0, h0 (X, OV(sn ) ) = 1.)
Once we know that Serre duality holds for Cohen-Macaulay projective schemes(§30.3), this result will automatically extend to these schemes. (A related result is
Exercise 18.6.T, which doesn’t use Serre duality.) On the other hand, the result isfalse if X is the union of two 2-planes in P4 meeting at a point (why?), so this will
imply that this X is not Cohen-Macaulay. (We will show this in another way inCounterexample 26.2.2.)

18.5.5. ⋆ Classification of vector bundles on P1k .
As promised in Example 17.2.4, we classify the vector bundles on P1k . We
discuss this in §18.5 because you might use Serre duality at one step (although
you might not).
∼ O(a1 )⊕· · ·⊕O(ar ),18.5.6. Theorem. — If E is a rank r vector bundle on P1k , then E =for a unique nondecreasing sequence of integers a1 , . . . , ar .
This result was proved independently many times (see for example Dedekind
and Weber’s article [DW, §22]), and is a special case of a theorem of Grothendieck,[Gr2]. It is sometimes called Grothendieck’s Theorem, because Grothendieck doesn’t
have enough theorems named after him.
18.5.7. For Pn
k more generally, the case r = 1 was shown in §14.2.9, but the statement is false for r > 1 and n > 1. A counterexample is given in Exercise 21.4.G.
One true generalization is a theorem of Horrocks, which states that a finite rank
locally free sheaf E on Pn
k splits precisely when “all the middle cohomology of allof its twists is zero” — when Hi (Pnk , E (m)) = 0 for 0 < i < n and all m ∈ Z. Thishas the surprising consequence that if n ≥ 2, then a finite rank locally free sheaf on

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2
Pn
k splits if and only if its restriction to some previously chosen 2-plane (Pk ) splits— all of the additional complication turns up already in dimension 2. See [OSS,§2.3] for more.

Proof. Note that the classification makes no reference to cohomology, but the proof
uses cohomology in an essential way. It is possible to prove Theorem 18.5.6 withno cohomological machinery (see for example [HM] or [GW, p. 314-5]), or withmore cohomological machinery (see for example [Ha1, Ex. V.2.6]).
∼ O(a1 ) ⊕ · · · ⊕ O(ar ), with a1 ≤ · · · ≤ ar .
18.5.C. E XERCISE . Suppose that E =Show that the ai can be determined from the numbers dimk Hom(O(m), E ), as mranges over the integers. Use this to show the uniqueness part of Theorem 18.5.6.
We now begin the proof, by induction on r. Fix a rank r locally free sheaf E .
The case r = 1 was established in §14.2.9, so we assume r > 1.
18.5.D. E XERCISE .
Show that for m ≪ 0, Hom(O(m), E ) ̸= 0, and that form ≫ 0, Hom(O(m), E ) = 0. Hint: show that Hom (O(m), E ) = E (−m) using
Exercise 13.1.F. Use Serre vanishing (Theorem 18.1.4(ii)) for the first part. Feel free
to use Serre duality and Serre vanishing for the second part. But you may prefer
to come up with an argument without Serre duality, to avoid invoking something
we have not yet proved.
Thus there is some ar for which dim Hom(O(ar ), E ) > 0, but for which Hom(O(m), E ) =0 for all m > ar . Choose a nonzero map φ : O(ar ) → E .

18.5.E. E XERCISE . Show that φ is an injection. (Hint: O(ar ) is torsion-free, andthus the kernel is torsion-free.)18.5.F. E XERCISE . Let F be the cokernel of φ. Show that F is locally free. Hint:
Exercise 13.7.G(c) gives an exact sequence 0 → Ftors → F → Flf → 0, where Flf is
locally free. Let L be the kernel of the surjection E → Flf . Show that L is locally
free, and thus is isomorphic to O(N) for some N. Show that there is a nonzero
map φ ′ : O(ar ) → O(N). Show (using the same idea as the previous exercise) that
this nonzero map φ ′ must be an injection, so ar ≤ N. Show that N ≤ ar because
there is a nonzero map O(N) → E (recall how ar was chosen). Show that φ ′ is an
isomorphism, and thus that F = Flf .
By our inductive hypothesis, we have F = O(a1 ) ⊕ · · · ⊕ O(ar−1 ), wherea1 ≤ · · · ≤ ar−1 , so we have a short exact sequence(18.5.7.1)

0% O(ar )%E% O(a1 ) ⊕ · · · ⊕ O(ar−1 )% 0.

We next show that ar ≥ ai for i < r. We are motivated by the fact that for aquasicoherent sheaf G on P1 ,HomP1 (O(ar + 1), G ) ==HomP1 (O, G (−ar − 1)) (Exercise 13.7.B)
H0 (P1 , G (−ar − 1)).

Tensor (18.5.7.1) with O(−ar − 1) (preserving exactness, by Exercise 13.1.E), andtake the long exact sequence in cohomology. Part of the long exact sequence is
H0 (P1 , E (−ar − 1))% H0 (P1 , F (−ar − 1))% H1 (P1 , O(−1)).

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Notice that H0 (P1 , E (−ar − 1)) = 0 (as Hom(O(ar + 1), E ) = 0, by the definitionof ar ), and H1 (P1 , O(−1)) = 0 (Theorem 18.1.3). Thus0 ===

H0 (P1 , F (−ar − 1))
H0 (P1 , ⊕i O(ai − ar − 1))⊕i H0 (P1 , O(ai − ar − 1)).

Hence ai − ar − 1 < 0 (by Exercise 14.1.A), so ar ≥ ai as desired.
Finally, we wish to show that exact sequence (18.5.7.1) expresses E as a directsum (of the subsheaf and quotient sheaf). For simplicity, we focus on the caser = 2, and return to the general case in Exercise 18.5.I.
18.5.G. E XERCISE . Show that the transition functions for the vector bundle, inappropriate coordinates, are given by" −a#t 2 α(t),0t−a1where α(t) is a Laurent polynomial in t. Hint/reminder: Recall that all vector
bundles on A1k are trivial, Exercise 13.2.C. Transition matrices for extensions of
one vector bundle (with known transition matrices) by another were discussed in
Exercise 13.5.A.

18.5.H. E XERCISE . Implicit in the above 2×2 matrix is a choice of a basis of a rank 2
free module M over the ring k[t] corresponding to one of the standard affine open
subsets Spec k[t], and a rank 2 free module M ′ over the ring k[1/t] over the otherstandard open subset. Show that by an appropriate “upper-triangular” changeof basis of M, you can arrange for α(t) to have no monomials of degree ≥ −a2 .
Show that by an appropriate “upper-triangular” change of basis of M ′ , you canarrange for α(t) to have no monomials of degree ≤ −a1 . Thus by choosing basesof M and M ′ appropriately, we can take α(t) = 0. Show that this implies that
E = O(a1 ) ⊕ O(a2 ).
18.5.I. E XERCISE . Finish the proof of Theorem 18.5.6 for arbitrary r. Hint: there
are no new ideas beyond the case r = 2.
!

18.6 Hilbert functions, Hilbert polynomials, and genus
If F is a coherent sheaf on a projective k-scheme X ⊂ Pn , define the Hilbert
function of F by
hF (m) := h0 (X, F (m)).
The Hilbert function of X is the Hilbert function of the structure sheaf. The ancients were aware that the Hilbert function is “eventually polynomial”, i.e., forlarge enough m, it agrees with some polynomial. This polynomial contains lots ofinteresting geometric information, as we will soon see. In modern language, we
expect that this “eventual polynomiality” arises because the Euler characteristic
should be a polynomial, and that for m ≫ 0, the higher cohomology vanishes.
This is indeed the case, as we now verify.

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18.6.1. Theorem. — If F is a coherent sheaf on a projective k-scheme X "→ Pnk,χ(X, F (m)) is a polynomial of degree equal to dim Supp F . Hence by Serre vanishing (Theorem 18.1.4 (ii)), for m ≫ 0, h0 (X, F (m)) is a polynomial pF (m) of degree
dim Supp F . In particular, for m ≫ 0, h0 (X, OX (m)) is polynomial with degree equal
to dim X.
18.6.2. Definition. The polynomial pF (m) defined in Theorem 18.6.1 is called theHilbert polynomial. If X ⊂ Pn is a projective k-scheme, define pX (m) := pOX (m).
In Theorem 18.6.1, OX (m) is the restriction or pullback of OPnk (m). Both thedegree of the 0 polynomial and the dimension of the empty set are defined to be−1. In particular, the only coherent sheaf with Hilbert polynomial 0 is the zerosheaf.
This argument uses the notion of associated points of a coherent sheaf on a locally Noetherian scheme, §13.6.5. (The resolution given by the Hilbert Syzygy Theorem, §15.3.2, can give a shorter proof; but we haven’t proved the Hilbert Syzygy
Theorem.)
Proof. Define pF (m) = χ(X, F (m)). We will show that pF (m) is a polynomial ofthe desired degree.
We first use Exercise 18.2.H to reduce to the case where k is algebraicallyclosed, and in particular infinite. (This is one of those cases where even if you
are concerned with potentially arithmetic questions over some non-algebraicallyclosed field like Fp , you are forced to consider the “geometric” situation where thebase field is algebraically closed.)
The coherent sheaf F has a finite number of associated points. We show auseful fact that we will use again.
18.6.A. E XERCISE . Suppose X is a projective k-scheme with k infinite, and F isa coherent sheaf on X. Show that if L is a very ample invertible sheaf on X, then
∼ O(D), and where D does notthere is an effective Cartier divisor D on X with L =meet the associated points of F . (Hint: show that given any finite set of points ofPn
k , there is a hyperplane not containing any of them. This is a variant of the key
step in Exercise 11.3.C(c).)
Thus there is a hyperplane x = 0 (x ∈ Γ (X, O(1))) missing this finite numberof points. (This is where we use the infinitude of k.)×x% F is injective (on any affine open subset, F corThen the map F (−1)
responds to a module, and x is not a zerodivisor on that module, as it doesn’t
vanish at any associated point of that module, see Theorem 5.5.10(c)). Thus wehave a short exact sequence(18.6.2.1)

0% F (−1)

%F

%G

%0

where G is a coherent sheaf.
18.6.B. E XERCISE .
Show that Supp G = (Supp F ) ∩ V(x). (Hint: show that
F (−1) → F is an isomorphism away from V(x), and hence G = 0 on this locus.

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If p ∈ V(x), show that the F (−1)|p → F |p is the 0 map, and hence F |p → G |p is
an isomorphism.)Hence V(x) meets all positive-dimensional components of Supp F (Exercise 11.3.C(a)),
so dim Supp G = dim Supp F − 1 by Krull’s Principal Ideal Theorem 11.3.3 unless
F = 0 (in which case we already know the result, so assume this is not the case).
Twisting (18.6.2.1) by O(m) yields% F (m − 1)0

% F (m)% G (m)%0

Euler characteristics are additive in exact sequences (Exercise 18.4.A), from whichpF (m)−pF (m−1) = pG (m). Now pG (m) is a polynomial of degree dim Supp F −1.
The result is then a consequence from the following elementary fact about
polynomials in one variable.
18.6.C. E XERCISE . Suppose f and g are functions on the integers, f(m + 1) −f(m) = g(m) for all m, and g(m) is a polynomial of degree d ≥ 0. Show that f is apolynomial of degree d + 1.
!
)n+m*
Example 1. The Hilbert polynomial of projective space is pPn (m) =m ,where we interpret this as the polynomial (m + 1) · · · (m + n)/n!.
Example 2. Suppose H is a degree d hypersurface in Pn , with i : H "→ Pn theclosed embedding. Then from the closed subscheme exact sequence0% OPn (−d)% OPn% i∗ OH% 0,

we have

"
# "
#
n+m
m+n−d
(18.6.2.2)
pH (m) = pPn (m) − pPn (m − d) =−.
nn∼ i∗ (OH ⊗ i∗ OPn (m)).
(Implicit in this argument is the fact that (i∗ OH ) ⊗ OPn (m) =This follows from the projection formula, Exercise 16.3.H(b). You can also show
this directly with transition functions.)

18.6.D. E XERCISE . Show that the twisted cubic (P1 embedded in P3 by the complete linear series |OP1 (3)|) has Hilbert polynomial 3m + 1. (The twisted cubic wasdefined in Exercise 8.2.A.)
18.6.E. E XERCISE . More generally, find the Hilbert polynomial for the dth Veronese
embedding of Pn (i.e., the closed embedding of Pn in a bigger projective space by
way of the line bundle O(d), §8.2.6).
18.6.F. E XERCISE ( TO BE USED SEVERAL TIMES IN C HAPTER 19). Suppose
X!

"

i

% Y! "% Pnkis a sequence of closed embeddings.
(a) Show that pX (m) ≤ pY (m) for m ≫ 0. Hint: let IX/Y be the ideal sheaf of X in
Y. Consider the exact sequence
0% IX/Y (m)% OY (m)% i∗ OX (m)% 0.

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491(b) If pX (m) = pY (m) for m ≫ 0, show that X = Y. Hint: Show that if the Hilbert
polynomial of a coherent sheaf F is 0, then F = 0. (Handy trick: For m ≫ 0,
F (m) is generated by global sections.) Apply this to F = IX/Y .
From the Hilbert polynomial, we can extract many invariants, of which two
are particularly important. The first is the degree, and the second is the arithmetic
genus (§18.6.6). The degree of a projective k-scheme of dimension n is defined tobe leading coefficient of the Hilbert polynomial (the coefficient of mn ) times n!.
Using the examples above, we see that the degree of Pn in itself is 1. The
degree of the twisted cubic is 3.
18.6.G. E XERCISE . Show that the degree is always an integer. Hint: by induction,
show that any polynomial in m of degree k taking on only integer values musthave coefficient of mk an integral multiple of 1/k!. Hint for this: if f(x) takes on
only integral values and is of degree k, then f(x + 1) − f(x) takes on only integral
values and is of degree k − 1.
18.6.H. E XERCISE .
Show that the degree of a degree d hypersurface (Definition 8.2.2) is d (preventing a notational crisis).
18.6.I. E XERCISE .
Suppose a curve C is embedded in projective space via an
invertible sheaf of degree d (as defined in §18.4.4). In other words, this line bundle
determines a closed embedding. Show that the degree of C under this embedding
is d, preventing another notational crisis. Hint: Riemann-Roch, Exercise 18.4.B. (Asimilar notational crisis was averted in Exercise 17.4.F.)
18.6.J. E XERCISE . Show that the degree of the dth Veronese embedding of Pn is
dn .
18.6.K. E XERCISE (B ÉZOUT ’ S T HEOREM , GENERALIZING E XERCISES 8.2.E AND 16.4.I).
Suppose X is a projective scheme of dimension at least 1, and H is a hypersurface
not containing any associated points of X. (For example, if X is reduced and thus
has no embedded points, we are just requiring H not to contain any irreducible
components of X.) Show that deg(H ∩ X) = (deg H)(deg X). (As an example,
we have Bézout’s Theorem for plane curves: if C and D are plane curves of degrees
m and n respectively, with no common components, then C and D meet at mn
points, counted with appropriate multiplicity.)
18.6.3. This is a very handy result! For example: if two projective plane curvesof degree m and degree n share no irreducible components, then they intersect in
mn points, counted with appropriate multiplicity. (To apply Exercise 18.6.K, youneed to know that plane curves have no embedded points. You can either do thisusing Exercise 5.5.I, or save time by assuming that one of the curves is reduced.)
The notion of multiplicity of intersection is just the degree of the intersection as ak-scheme.
18.6.L. E XERCISE . Suppose C is a degree 1 curve in P3k (or more precisely, a degree
1 pure-one-dimensional closed subscheme of P3k ). Show that C is a line. Hint:
reduce to the case k = k. Suppose p and q are distinct closed points on C. Use
Bézout’s Theorem (Exercise 18.6.K) to show that any hyperplane containing p andq must contain C, and thus that C ⊂ pq.

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18.6.M. F UN E XERCISE . Let k be a field, which we assume to be algebraicallyclosed for convenience (although you are free to remove this hypothesis if youwish). Suppose C is a degree d integral curve in PN
k with N ≥ d. Show that C isN
contained in a linear Pd⊂
P.
(Exercise
18.6.Lisnot quite a special case of thiskkproblem, but the hint may still be helpful.)
18.6.N. E XERCISE ( A FORM OF B ÉZOUT ’ S T HEOREM ). Classically, the degree ofa complex projective variety of dimension n was defined as follows. We slice thevariety with n generally chosen hyperplanes. Then the intersection will be a finite
number of reduced points, by Exercise 12.4.C (a consequence of Bertini’s Theorem 12.4.2). The degree is this number of points. Use Bézout’s Theorem to makesense of this in a way that agrees with our definition of degree. You will need toassume that k is infinite.
Thus the classical definition of the degree, which involved making a choice
and then showing that the result is independent of choice, has been replaced bymaking a cohomological definition involving Euler characteristics. This should
remind you of how we got around to “correctly” understanding the degree of aline bundle. It was traditionally defined as the degree of a divisor of any nonzero
rational section (Important Exercise 18.4.C), and we found a better definition interms of Euler characteristics (§18.4.4).
18.6.4. ⋆⋆ Aside: Connection to the topological definition of degree (for those with sufficient background). Another definition of degree of a dimension d complex projective variety X ⊂ Pn
C is as the number d such that [X] is d times the “positive”
generator of H2d (PnC , Z). over C. You can show this by induction on d as follows.
Suppose X is a complex projective variety of dimension d. For a generally chosen
hyperplane, H ∩ X is a complex projective variety of (complex) dimension d − 1.
(Do you see why?) Show that [H ∩ X] = c1 (O(1)) ∪ [X] in H2(d−1) (X, Z), by suitably
generalizing the solution to Exercise 18.4.G. (A further generalization is given in§20.1.7.) For this reason, c1 (O(1)) is often called the “hyperplane class”.
18.6.5. Revisiting an earlier example. We revisit the enlightening example of Example 9.3.3 and §17.4.9: let k = Q, and consider the parabola x = y2 . We intersect itwith the four lines, x = 1, x = 0, x = −1, and x = 2, and see that we get 2 each time(counted with the same convention as with the last time we saw this example).
If we intersect it with y = 2, we only get one point — but that’s because thisisn’t a projective curve, and we really should be doing this intersection on P2k , andin this case, the conic meets the line in two points, one of which is “at ∞”.
18.6.O. E XERCISE . Show that the degree of the d-fold Veronese embedding of
Pn is dn in a different way from Exercise 18.6.J as follows. Let vd : Pn → PN be
the Veronese embedding. To find the degree of the image, we intersect it withn hyperplanes in PN (scheme-theoretically), and find the number of intersection
points (counted with multiplicity). But the pullback of a hyperplane in PN to Pn isa degree d hypersurface. Perform this intersection in Pn , and use Bézout’s theorem(Exercise 18.6.K).

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18.6.P. E XERCISE ( DEGREE IS ADDITIVE FOR UNIONS ). Suppose X and Y are twod-dimensional closed subschemes of Pn
k , with no d-dimensional irreducible components in common. Show that deg X ∪ Y = deg X + deg Y.
18.6.6. Arithmetic genus, again.
There is another central piece of information residing in the Hilbert polynomial. Notice that 1 − pX (0) = 1 − χ(X, OX ) is the arithmetic genus (§18.4.2), an
intrinsic invariant of the scheme X, independent of the projective embedding.
Imagine how amazing this must have seemed to the ancients: they defined
the Hilbert function by counting how many “functions of various degrees” thereare; then they noticed that when the degree gets large, it agrees with a polynomial;and then when they plugged 0 into the polynomial — extrapolating backwards, towhere the Hilbert function and Hilbert polynomials didn’t agree — they found a
magic invariant! Furthermore, in the case when X is a complex curve, this invariant was basically the topological genus!
We can now see a large family of curves over an algebraically closed field thatis provably not P1 ! Note that the Hilbert polynomial of P1 is (m + 1)/1 = m + 1,
so χ(OP1 ) = 1. Suppose C is a degree d curve in P2 . Then the Hilbert polynomialof C ispP2 (m) − pP2 (m − d) = (m + 1)(m + 2)/2 − (m − d + 1)(m − d + 2)/2.
Plugging in m = 0 gives us −(d2 − 3d)/2. Thus when d > 2, we have a curve thatcannot be isomorphic to P1 ! (And it is not hard to show that there exists a regulardegree d curve, Exercise 12.3.E.)
Now from 0 → OP2 (−d) → OP2 → OC → 0, using h1 (OP2 (−d)) = 0, we havethat h0 (C, OC ) = 1. As h0 − h1 = χ, we have(18.6.6.1)

h1 (C, OC ) = (d − 1)(d − 2)/2.

We now revisit an interesting question we first saw in §6.5.10. If k is an algebraically closed field, is every finitely generated transcendence degree 1 extensionof k isomorphic to k(x)? In that section, we found ad hoc (but admittedly beautiful) examples showing that the answer is “no”. But we now have a better answer.
The question initially looks like an algebraic question, but we now recognize it as afundamentally geometric one. There is an integer-valued cohomological invariantof such field extensions that is has good geometric meaning: the genus.
Equation (18.6.6.1) yields examples of curves of genus 0, 1, 3, 6, 10, . . . (corresponding
to degree 1 or 2, 3, 4, 5, . . . ). This begs some questions, such as: are there curves ofother genera? (We will see soon, in §19.5.5, that the answer is yes.) Are there othergenus 0 curves? (Not if k is algebraically closed, but sometimes yes otherwise —consider x2 +y2 +z2 = 0 in P2R , which has no R-points and hence is not isomorphic
to P1R — we will discuss this more in §19.3.) Do we have all the curves of genus 3?
(Almost all, but not quite. We will see more in §19.7.) Do we have all the curves ofgenus 6? (We are missing “most of them”, as will be suggested by §19.8.3.)
Caution: The Euler characteristic of the structure sheaf is an incomplete invariant. It doesn’t always distinguish between isomorphism classes of irreduciblesmooth projective varieties. For example, P2 and P1 ×P1 both have structure sheaf
Euler characteristic 1 (see Theorem 18.1.3 and Exercise 18.3.3), but are not isomor∼ Z (§14.2.7) while Pic(P1 × P1 ) =∼ Z ⊕ Z (Exercise 14.2.O).
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18.6.7. Complete intersections.
18.6.Q. E XERCISE .
faces in P3k .

Find the genus of the complete intersection of 2 quadric sur-

18.6.R. E XERCISE . More generally, find the genus of the complete intersection ofa degree m surface with a degree n surface in P3k . (If m = 2 and n = 3, you should
get genus 4. We will see in §19.8 that in some sense most genus 4 curves arise inthis way. Note that Bertini’s Theorem 12.4.2 ensures that there are regular curvesof this form.)
18.6.S. E XERCISE . Show that the rational normal curve of degree d in Pd is not acomplete intersection if d > 2. (Hint: If it were the complete intersection of d − 1
hypersurfaces, what would the degree of the hypersurfaces be? Why could noneof the degrees be 1?)
18.6.T. E XERCISE ( POSITIVE - DIMENSIONAL COMPLETE INTERSECTIONS ARE CON NECTED ). Show that complete intersections of positive dimension are connected.
(Hint: show that h0 (X, OX ) = 1.) For experts: this argument will even show thatthey are geometrically connected (§9.5), as h0 is preserved by field extension (Exercise 18.2.H).
18.6.U. E XERCISE . Show that the union of two planes in P4 meeting at a point is
not a complete intersection. Hint: it is connected, but you can slice with anotherhyperplane and get something not connected (see Exercise 18.6.T).

18.7 ⋆ Serre’s cohomological characterization of ampleness
Theorem 16.6.2 gave a number of characterizations of ampleness, in terms ofprojective geometry, global generation, and the Zariski topology. Here is anothercharacterization, this time cohomological, under Noetherian hypotheses. Because(somewhat surprisingly) we won’t use this result, this section is starred.
18.7.1. Theorem (Serre’s cohomological criterion for ampleness). — Suppose A
is a Noetherian ring, X is a proper A-scheme, and L is an invertible sheaf on X. Then thefollowing are equivalent.
(a-c) The invertible sheaf L is ample on X (over A).
(e) For all coherent sheaves F on X, there is an n0 such that for n ≥ n0 , Hi (X, F ⊗L ⊗n ) = 0 for all i > 0.
The label (a-c) is intended to reflect the statement of Theorem 16.6.2. We
avoid the label (d) because it appeared in Theorem 16.6.6. (Aside: the “properness” assumption cannot be removed, as can be shown by the example of Exercise 18.8.F.) Before getting to the proof, we motivate this result by giving some
applications. (As a warm-up, you can give a second solution to Exercise 16.6.G in
the Noetherian case, using the affineness of π to show that Hi (X, F ⊗(π∗ L )⊗m ) =Hi (Y, (π∗ F ) ⊗ L ⊗m ).)Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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18.7.A. E XERCISE . Suppose X is a proper A-scheme, and L is an invertible sheaf
on X. Show that L is ample on X if and only if L |Xred is ample on Xred . Hint: forthe “only if” direction, use Exercise 16.6.G. For the “if” direction, let I be the ideal
sheaf cutting out the closed subscheme Xred in X. Filter F by powers of I :0 = I r F ⊂ I r−1 F ⊂ · · · ⊂ I F ⊂ F .

(Essentially the same filtration appeared in Exercise 18.4.S, for similar reasons.)
Show that each quotient I n F /I n−1 F , twisted by a high enough power of L ,has no higher cohomology. Use descending induction on n to show each partI n F of the filtration (and hence in particular F ) has this property as well.
18.7.B. E XERCISE . Suppose X is a proper A-scheme, and L is an invertible sheaf
on X. Show that L is ample on X if and only if L is ample on each component.
Hint: follow the outline of the solution to the previous exercise, taking instead
I as the ideal sheaf of one component. Perhaps first reduce to the case where
X = Xred .
18.7.C. E XERCISE . (In Exercise 19.2.E, we will show that on a smooth projective
integral curve, an invertible sheaf is ample if and only if it has positive degree.
Use that fact in this exercise. There will be no logical circularity.) Show that a line
bundle on a projective curve is ample if and only if it has positive degree on eachcomponent.
18.7.2. Remark: Proper curves are projective. Serre’s criterion for ampleness is the
key ingredient for showing that every proper curve over a field is projective. The
steps are as follows. (i) Recall that every regular integral proper curve is projective,
Exercise 17.4.A. (ii) The hardest step is showing that every reduced integral proper
curve C is projective. This is done by choosing a regular point on each irreducible
component of C, and letting L be the corresponding invertible sheaf. Because of
Exercise 18.7.C, we hope that L will be ample. Show that a line bundle on C isample if its pullback to the normalization of C is ample (a partial converse to Exercise 16.6.G, see for example [Ha1, Ex. III.5.7(d)]). Thus our L is ample. (iii) Show
that every reduced proper curve is projective using Exercise 18.7.B. (iv) Show thatevery proper curve C is projective, using Exercise 18.7.A, after first finding an invertible sheaf on C that will be shown to be ample.
18.7.3. Very ample versus ample. The previous exercises don’t work with “ample”
replaced by “very ample”, which shows again how the notion of ampleness isbetter-behaved than very ampleness.
18.7.4. Proof of Theorem 18.7.1. For the fact that (a-c) implies (e), use the fact that
L ⊗N is very ample for some N (Theorem 16.6.2(a)), and apply Serre vanishing(Theorem 18.1.4(ii)) to F , F ⊗ L , . . . , and F ⊗ L ⊗(N−1) .
So we now assume (e), and show that L is ample by criterion (b) of Theorem 16.6.2: we will show that for any coherent sheaf F on X, F ⊗ L ⊗n is globally
generated for n ≫ 0.
We begin with a special case: we will show that L ⊗n is globally generated (i.e.,base-point-free) for n ≫ 0. To do this, it suffices to show that every closed pointp has an open neighborhood U so that there exists some Np so that n ≥ Np , L ⊗n
is globally generated for all points of Up . (Reason: by quasicompactness, every

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The Rising Sea: Foundations of Algebraic Geometryclosed subset of X contains a closed point, by Exercise 5.1.E. So as p varies over theclosed points of X, these Up cover X. By quasicompactness again, we can cover X
by a finite number of these Up . Let N be the maximum of the corresponding Np .
Then for n ≥ N, L ⊗n is globally generated in each of these Up , and hence on allof X.)
Let p be a closed point of X. For all n, mp ⊗L ⊗n is coherent (by our Noetherian
hypotheses). By (e), there exists some n0 so that for n ≥ n0 , H1 (X, mp ⊗ L ⊗n ) = 0.
By the long exact sequence arising from the closed subscheme exact sequence
0 → mp ⊗ L ⊗n → L ⊗n → L ⊗n |p → 0,

we have that L ⊗n is globally generated at p for n ≥ n0 . By Exercise 15.3.C(b),there is an open neighborhood V0 of p such that L ⊗n0 is globally generated at all
points of V0 . Thus L ⊗kn0 is globally generated at all points of V0 for all positive
integers k (using Easy Exercise 15.3.B). For each i ∈ {1, . . . , n0 − 1}, there is an open
neighborhood Vi of p such that L ⊗(n0 +i) is globally generated at all points of Vi(again by Exercise 15.3.C(b)). We may take each Vi to be contained in V0 . By Easy
Exercise 15.3.B, L ⊗(kn0 +n0 +i) is globally generated at every point of Vi (as this is
the case for L ⊗kn0 and L ⊗(n0 +i) ). Thus in the open neighborhood Up := ∩n−1
i=0 Vi ,L ⊗n is globally generated for n ≥ Np := 2n0 .
We have now shown that there exists some N such that for n ≥ N, L ⊗n is
globally generated. Now suppose F is a coherent sheaf. To conclude the proof,
we will show that F ⊗ L ⊗n is globally generated for n ≫ 0. This argument has asimilar flavor to what we have done so far, so we give it as an exercise.
18.7.D. E XERCISE . Suppose p is a closed point of X.
(a) Show that for n ≫ 0, F ⊗ L ⊗n is globally generated at p.
(b) Show that there exists an open neighborhood Up of p such that for n ≫ 0,
F ⊗ L ⊗n is globally generated at every point of Up . Caution: while it is true thatby Exercise 15.3.C(b), for each n ≫ 0, there is some open neighborhood Vn of psuch that F ⊗ L ⊗n is globally generated there, it need not be true that(18.7.4.1)∩n≫0 Vnis an open set. You may need to use the fact that L ⊗n is globally generated forn ≥ N to replace (18.7.4.1) by a finite intersection.
18.7.E. E XERCISE . Conclude the proof of Theorem 18.7.1 by showing that F ⊗L ⊗n is globally generated for n ≫ 0.
!
18.7.5. Aside: Serre’s cohomological characterization of affineness. Serre gave a characterization of affineness (Remark 18.1.1) similar in flavor to Theorem 18.7.1.
18.7.6. Theorem (Serre’s cohomological characterization of affineness). — Suppose X is a Noetherian separated scheme. Then the following are equivalent.
(a) The scheme X is affine.
(b) For any quasicoherent sheaf F on X, Hi (X, F ) = 0 for all i > 0.
(c) For any coherent sheaf of ideals I on X, H1 (X, I ) = 0.

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497

Because we won’t use it, we omit the proof. (One is given in [Ha1, Thm. III.3.7].)
Clearly (a) implies (b) implies (c) (the former from “affine cover cohomology vanishing”, Property (vi) of §18.1) without any Noetherian assumptions, so the real
substance is in the implication from (c) to (a).
Serre proved an analogous result in complex analytic geometry: Stein spaces
are also characterized by the vanishing of cohomology of coherent sheaves.

18.8 Higher pushforward (or direct image) sheaves
Cohomology groups were defined for X → Spec A where the structure morphism is quasicompact and separated; for any quasicoherent F on X, we defined
Hi (X, F ). We will now define a “relative” version of this notion, for quasicompact and separated morphisms π : X → Y: for any quasicoherent F on X, we willdefine Ri π∗ F , a quasicoherent sheaf on Y. (Now would be a good time to do
Exercise 1.6.H, the FHHF Theorem, if you haven’t done it before.)
We have many motivations for doing this. In no particular order:(1) It “globalizes” what we did before with cohomology.
(2) If 0 → F → G → H → 0 is a short exact sequence of quasicoherent
sheaves on X, then we know that 0 → π∗ F → π∗ G → π∗ H is exact, andhigher pushforwards will extend this to a long exact sequence.
(3) We will later see that this will show how cohomology groups vary in families, especially in “nice” situations. Intuitively, if we have a nice familyof varieties, and a family of sheaves on them, we could hope that the cohomology varies nicely in families, and in fact in “nice” situations, thisis true. (As always, “nice” usually means “flat”, whatever that means.
We will see that Euler characteristics are locally constant in proper flatfamilies in §24.7, and the Cohomology and Base Change Theorem 28.1.6
will show that in particularly good situations, dimensions of cohomology
groups are constant.)
All of the important properties of cohomology described in §18.1 will carry overto this more general situation. Best of all, there will be no extra work required.
In the notation Ri π∗ F for higher pushforward sheaves, the “R” stands for“right derived functor”, and corresponds to the fact that we get a long exact sequence in cohomology extending to the right (from the 0th terms). In Chapter 23,
we will see that in good circumstances, if we have a left-exact functor, there is along exact sequence going off to the right, in terms of right derived functors. Similarly, if we have a right-exact functor (e.g., if M is an A-module, then ⊗A M is aright-exact functor from the category of A-modules to itself), there may be a longexact sequence going off to the left, in terms of left derived functors.
Suppose π : X → Y, and F is a quasicoherent sheaf on X. For each Spec A ⊂ Y,
we have A-modules Hi (π−1 (Spec A), F ). We now show that these patch together
to form a quasicoherent sheaf, in the sense of §13.3.3. We need check only one
fact: that this behaves well with respect to taking distinguished open sets. In otherwords, we must check that for each f ∈ A, the natural map
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The Rising Sea: Foundations of Algebraic Geometry(induced by the map of spaces in the opposite direction — Hi is contravariant in
the space) is precisely the localization ⊗A Af . But this can be verified easily: let{Ui } be an affine open cover of π−1 (Spec A). We can compute Hi (π−1 (Spec A), F )using the Čech complex (18.2.1.1). But this induces a cover of π−1 (Spec Af ) in anatural way: if Ui = Spec Ai is an affine open subset of π−1 (Spec A), we define
Ui′ := Spec(Ai )π♯ f . The resulting Čech complex for Spec Af is the localization of
the Čech complex for Spec A. As taking cohomology of a complex commutes withlocalization (as discussed in the FHHF Theorem, Exercise 1.6.H), we have defineda quasicoherent sheaf on Y by the characterization of quasicoherent sheaves in§13.3.3.
Define the ith higher pushforward sheaf or the ith higher direct image sheaf
Ri π∗ F to be this quasicoherent sheaf.
18.8.1. Theorem. —schemes. Then:

Suppose π : X → Y is a quasicompact separated morphism of(a) Ri π∗ is a covariant functor QCohX → QCohY .
(b) We can identify R0 π∗ with π∗ F .
(c) (the long exact sequence of higher pushforward sheaves) A short exact
sequence 0 → F → G → H → 0 of sheaves on X induces a long exact
sequence(18.8.1.1)

0% R0 π∗ F% R0 π∗ G% R0 π∗ H%

R1 π∗ F% R1 π∗ G% R1 π∗ H% ···of sheaves on Y.
(d) (projective pushforwards of coherent are coherent: Grothendieck’s Coherence Theorem for projective morphisms) If π is a projective morphism and OY is coherent
on Y (this hypothesis is automatic for Y locally Noetherian), and F is a coherent
sheaf on X, then for all i, Ri π∗ F is a coherent sheaf on Y.
18.8.2. Unimportant Remark. If X and Y are Noetherian, the hypotheses “quasicompact and separated” on π can be removed; see Unimportant Remark 23.5.8 for the“separated” hypothesis.
18.8.3. Proof of Theorem 18.8.1. We first show covariance: if F → G is a morphismof quasicoherent sheaves on X, we define a map Ri π∗ F → Ri π∗ G . (It will beclear we will have shown that Ri π∗ is a functor.) It suffices to define this map on
the “distinguished affine base” of Y (Definition 13.3.1). Thus it suffices to show thefollowing: if X ′ is a quasicompact separated A-scheme, and F → G is a morphismof quasicoherent sheaves on X, then the map Hi (X ′ , F ) → Hi (X ′ , G ) constructedin §18.2 (property (i) of §18.1) “commutes with localization at f ∈ A”. But this wasshown in Exercise 18.2.D.
In a similar way, we construct the connecting homomorphism Ri π∗ H →i+1
R π∗ F in the long exact sequence (18.8.1.1), by showing that the constructionin the case where Y = Spec A “commutes with localization at f ∈ A”. Again, thiswas shown in Exercise 18.2.D.

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November 18, 2017 draft499

It suffices to check all other parts of this statement on affine open subsets of Y,
so they all follow from the analogous statements in Čech cohomology (§18.1). !
The following result is handy, and essentially immediate from our definition.
18.8.A. E ASY E XERCISE . Show that if π is affine, then for i > 0, Ri π∗ F = 0.
18.8.4. How higher pushforwards behave with respect to base change.
18.8.B. E XERCISE ( HIGHER PUSHFORWARDS AND BASE CHANGE ).
(a) (easy) Suppose ψ : Z → Y is any morphism, and π : X → Y is quasicompact andseparated. Suppose F is a quasicoherent sheaf on X. Let(18.8.4.1)

W

ψ′

%X

π′

π

"
Z

ψ

"
%Y

be a Cartesian diagram. Describe a natural morphism ψ∗ (Ri π∗ F ) → Ri π∗′ (ψ ′ )∗ Fof sheaves on Z. (Hint: the FHHF Theorem, Exercise 1.6.H. You may want to compare the i = 0 case to the push-pull formula of Exercise 16.3.G.)(b) (cohomology commutes with affine flat base change) If ψ : Z → Y is an affine morphism, and for a cover Spec Ai of Y, where ψ−1 (Spec Ai ) = Spec Bi , Bi is a flat
Ai -algebra (§1.6.11: ⊗Ai Bi is exact), and the diagram in (a) is a Cartesian diagram, show that the natural morphism of (a) is an isomorphism. (Exercise 18.2.H
was a special case of this exercise. You can likely generalize this to non-affinemorphisms — and thus show that cohomology commutes with flat base change(Theorem 24.2.8) — but we wait until Chapter 24 to discuss flatness at length.)
18.8.C. E XERCISE ( CF. E XERCISE 16.3.G). Prove Exercise 18.8.B(a) without the hypothesis that (18.8.4.1) is a Cartesian diagram, but adding the requirement that
π ′ is quasicompact and separated (just so our definition of Ri π∗′ applies). In thecourse of the proof, you will see a map arising in the Leray spectral sequence (Theorem 23.4.5). (Hint: use Exercise 18.8.B(a).)
A useful special case of Exercise 18.8.B(a) is the following.
18.8.D. E XERCISE . If q ∈ Y, describe a natural morphism (Ri π∗ F ) ⊗ κ(q) →Hi (π−1 (q), F |π−1 (q) ). (Be sure you understand the meaning of the left side!) Hint:
the FHHF Theorem, Exercise 1.6.H.
Thus the fiber of the pushforward may not be the cohomology of the fiber, but
at least it always maps to it. We will later see that in good situations this map is an
isomorphism, and thus the higher pushforward sheaf indeed “patches together”
the cohomology on fibers (the Cohomology and Base Change Theorem 28.1.6).
18.8.E. E XERCISE ( PROJECTION FORMULA , GENERALIZING E XERCISE 16.3.H). Suppose π : X → Y is quasicompact and separated, and F and G are quasicoherent

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sheaves on X and Y respectively.
F❈
π∗ G
❈❈
④
❈❈
④④
❈❈
④④
④
❈❈
④④
π∗ F❈
X
G
④
❈❈
④④
❈❈
④
❈❈
④④
❈ " ④④④Y(a) Describe a natural morphism(Ri π∗ F ) ⊗ G → Ri π∗ (F ⊗ π∗ G ).

(Hint: the FHHF Theorem, Exercise 1.6.H.)(b) If G is locally free, show that this natural morphism is an isomorphism.
The following fact uses the same trick as Theorem 18.1.8 and Exercise 18.1.B.
18.8.5. Theorem (relative dimensional cohomology vanishing). — If π : X → Y isa projective morphism and Y is locally Noetherian (or more generally OY is coherent overitself), then the higher pushforwards vanish in degree higher than the maximum dimensionof the fibers.
This is false without the projective hypothesis (see Exercise 18.8.F below). In
particular, you might hope that just as dimensional cohomology vanishing generalized from projective varieties to quasiprojective varieties or more general settings(§18.2.7) that relative dimensional cohomology vanishing would generalize fromprojective morphisms to quasiprojective morphisms, but this is not the case.
18.8.F. E XERCISE . Consider the open embedding π : An − {0} → An . By direct
calculation, show that Rn−1 π∗ OAn −{0} ̸= 0. (This calculation will remind you ofthe proof of the Hn part of Theorem 18.1.3; see also Remark 18.3.1.)
Proof of Theorem 18.8.5. Let m be the maximum dimension of all the fibers.
The question is local on Y, so we will show that the result holds near a point pof Y. We may assume that Y is affine, and hence that X "→ Pn
Y.
Let k be the residue field at p. Then π−1 (p) is a projective k-scheme of dimension at most m. By Exercise 11.3.C we can find affine open sets D(f1 ), . . . , D(fm+1 )that cover π−1 (p). In other words, the intersection of the V(fi ) does not intersect
π−1 (p).
If Y = Spec A and p = [p] (so k = Ap /pAp ), then arbitrarily lift each fi froman element of k[x0 , . . . , xn ] to an element fi′ of Ap [x0 , . . . , xn ]. Let F be the product of the denominators of the fi′ ; note that F ∈/ p, i.e., p = [p] ∈ D(F). Thenfi′ ∈ AF [x0 , . . . , xn ]. The intersection of their zero loci ∩V(fi′ ) ⊂ Pn
AF is a closed
subscheme of Pn.
Intersect
it
with
X
to
get
anotherclosedsubscheme
of Pn
AF
AF .
Take its image under π; as projective morphisms are closed, we get a closed subsetof D(F) = Spec AF . But this closed subset does not include p; hence we can find
an affine open neighborhood Spec B of p in Y missing the image. But if fi′′ are therestrictions of fi′ to B[x0 , . . . , xn ], then D(fi′′ ) cover π−1 (Spec B); in other words,
π−1 (Spec B) is covered by m + 1 affine open sets, so by affine cover cohomology
vanishing (Property (vi) of §18.1), its cohomology vanishes in degree at least m + 1.

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501

But the higher direct image sheaf is computed using these cohomology groups,
hence the higher pushforward sheaf Ri π∗ F vanishes on Spec B too.
!
18.8.G. E XERCISE ( RELATIVE S ERRE VANISHING , CF. T HEOREM 18.1.4( II )). Suppose π : X → Y is a proper morphism of Noetherian schemes, and L is a π-ample
invertible sheaf on X. Show that for any coherent sheaf F on X, for m ≫ 0,
Ri (π∗ F ⊗ L ⊗m ) = 0 for all i > 0.

18.9 ⋆ From projective to proper hypotheses: Chow’s Lemma andGrothendieck’s Coherence Theorem
The main proofs in this section are double-starred because the results aren’t
absolutely necessary in the rest of our discussions, and may not be worth reading
right now. But just knowing the statement of Grothendieck’s Coherence Theorem 18.9.1, (generalizing Theorem 18.8.1(d)) will allow you to immediately translate many of our arguments about projective schemes and morphisms to proper
schemes and morphisms, and Chow’s Lemma is a multi-purpose tool to extend
results from the projective situation to the proper situation in general.
18.9.1. Grothendieck’s Coherence Theorem. — Suppose π : X → Y is a propermorphism of locally Noetherian schemes. Then for any coherent sheaf F on X, Ri π∗ F iscoherent on Y.
The case i = 0 has already been mentioned a number of times.
18.9.A. E XERCISE .
Recall that finite morphisms are affine (by definition) andproper. Use Theorem 18.9.1 to show that if π : X → Y is proper and affine and Y
is Noetherian, then π is finite. (Hint: mimic the proof of the weaker result whereproper is replaced by projective, Corollary 18.1.7.)
The proof of Theorem 18.9.1 requires two sophisticated facts. The first is theLeray spectral sequence (Theorem 23.4.5, which applies in this situation becauseof Exercise 23.5.F). Suppose π : X → Y and ρ : Y → Z are quasicompact separated
morphisms. Then for any quasicoherent sheaf F on X, there is a spectral sequencewith E2 term given by Rp ρ∗ (Rq π∗ F ) converging to Rp+q (ρ ◦ π)∗ F . Because thiswould be a reasonable (but hard) exercise in the case we need it (where Z is affine),
we will feel comfortable using it. But because we will later prove it, we won’t
prove it now.
We will also need Chow’s Lemma.
18.9.2. Chow’s Lemma. — Suppose π : X → Spec A is a proper morphism, and A
is Noetherian. Then there exists µ : X ′ → X which is surjective and projective, such thatπ ◦ µ is also projective, and such that µ is an isomorphism on a dense open subset of X:
µ

%X
X′ ●
●●
①①
①
●●
①①
●
①① π
π◦µ ●●●①$!①
Spec A.

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In particular, if X is a proper k-variety, it admits a projective birational morphism from a projective k-variety.
Many generalizations of results from projective to proper situations go throughChow’s Lemma. We will prove this version, and state other versions of Chow’s
Lemma, in §18.9.4. Assuming these two facts, we now prove Grothendieck’s Coherence Theorem 18.9.1 in a series of exercises.
18.9.3. ⋆⋆ Proof of Grothendieck’s Coherence Theorem 18.9.1. The question is local
on Y, so we may assume Y is affine, say Y = Spec A. We work by induction on
dim Supp F , with the base case when dim Supp F = −1 (i.e., Supp F = ∅, i.e.,
F = 0), which is obvious. So fix F , and assume the result is known for all coherent
sheaves with support of smaller dimension.
18.9.B. E XERCISE . Show that we may assume that Supp F = X. (Hint: the idea isto replace X by the scheme-theoretic support of F , the smallest closed subschemeof X on which Supp F “lives”. More precisely, it is the smallest closed subscheme∼ i∗ F ′ . Show that
i : W "→ X such that there is a coherent sheaf F ′ on W, with F =this notion makes sense, using the ideas of §8.3, by defining it on each affine open
subset.)We now invoke Chow’s Lemma to construct a projective morphism µ : X ′ → X
that is an isomorphism on a dense open subset U of X (so dim X \ U < dim X), andsuch that π ◦ µ : X ′ → Spec A is projective.
Then G = µ∗ F is a coherent sheaf on X ′ , µ∗ G is a coherent sheaf on X (by the
projective case, Theorem 18.8.1(d)) and the adjunction map F → µ∗ G = µ∗ µ∗ F
is an isomorphism on U. The kernel E and cokernel H are coherent sheaves on X
that are supported in smaller dimension:
0 → E → F → µ∗ G → H → 0.

18.9.C. E XERCISE . By the inductive hypothesis, the higher pushforwards of E and
H are coherent. Show that if all the higher pushforwards of µ∗ G are coherent,
then the higher pushforwards of F are coherent.
So we are reduced to showing that the higher pushforwards of µ∗ G are coherent for any coherent G on X ′ .
µ% X π % Spec A
The Leray spectral sequence (Theorem 23.4.5) for X ′has E2 page given by Rp π∗ (Rq µ∗ G ) converging to Rp+q (π ◦ µ)∗ G . Now Rq µ∗ G iscoherent by Theorem 18.8.1(d). Furthermore, as µ is an isomorphism on a dense
open subset U of X, Rq µ∗ G is zero on U, and is thus supported on the complementof U, whose dimension is less than that of X. Hence by our inductive hypothesis,Rp π∗ (Rq µ∗ G ) is coherent for all p, and all q ≥ 1. The only possibly noncoherent
sheaves on the E2 page are in the row q = 0 — precisely the sheaves we areinterested in. Also, by Theorem 18.8.1(d) applied to π◦µ, Rp+q (π◦µ)∗ G is coherent.
18.9.D. E XERCISE . Show that Ep,qn is always coherent for any n ≥ 2, q > 0. Showp,0that En is coherent for a given n ≥ 2 if and only if Ep,0is coherent. Show that
2
p,0
Ep,q
is
coherent,
and
hence
that
E
is
coherent,
therebycompletingthe proof of∞
2Grothendieck’s Coherence Theorem 18.9.1.

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18.9.4. ⋆⋆ Proof of Chow’s Lemma.
We use the properness hypothesis on π through each of its three constituent
parts: finite type, separated, universally closed. The parts using separatedness are
particularly tricky.
As X is Noetherian, it has finitely many irreducible components. Cover X withaffine open sets U1 , . . . , Un . We may assume that each Ui meets each irreducible
component. (If some Ui does not meet an irreducible component Z, then take any
affine open subset Z ′ of Z − X − Z, and replace Ui by Ui ∪ Z ′ .) Then U := ∩i Ui isa dense open subset of X. As each Ui is finite type over A, we can choose a closediiembedding Ui ⊂ An
closure of Ui in Pn
A . Let Ui be the (scheme-theoretic)"A .
Now we have the diagonal morphism U → X ×A
Ui (where the product isover Spec A), which is a locally closed embedding (the composition of the closed"
embedding U "→ Un+1 with the open embedding Un+1 "→ X ×A Ui ). Let X ′ be"
the scheme-theoretic closure of U in X ×A Ui . Let µ be the composed morphism"
X ′ → X ×A Ui → X, so we have a diagram
X ′" # ▼▼▼▼▼▼▼µ▼cl. emb.
▼▼▼
""
▼▼d
%X
X ×A Ui
proj.

proper""proper

Ui

proj.

"
% Spec A(where the square is Cartesian). The morphism µ is projective (as it is the composition of two projective morphisms and X is quasicompact, Exercise 17.3.B). We willconclude the argument by showing that µ−1 (U) = U (or more precisely, µ is an"
isomorphism above U), and that X ′ →Ui is a closed embedding (from which
the composition&
X→
Ui → Spec A
is projective).

18.9.E. E XERCISE . Suppose T0 , . . . , Tn are separated schemes over A with isomorphic open sets, which we sloppily call V in each case. Then V is a locally closed
subscheme of T0 ×A · · ·×A Tn . Let V be the closure of this locally closed subscheme.
Show that∼ V ∩ (V ×A T1 ×A · · · ×A Tn )
V =
=
=
=

V ∩ (T0 ×A V ×A T2 ×A · · · ×A Tn )···V ∩ (T0 ×A · · · ×A Tn−1 ×A V).

(Hint for the first isomorphism: the graph of the morphism V → T1 ×A · · · ×A Tn isa closed embedding, as T1 ×A · · · ×A Tn is separated over A, by Proposition 10.1.18.
Thus the scheme-theoretic closure of V in V×A T1 ×A · · ·×A Tn is V itself. Finally, thescheme-theoretic closure can be computed locally, essentially by Theorem 8.3.4.)

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18.9.F. E XERCISE . Using (the idea behind) the previous exercise, show that µ−1 (U) =U.
""
It remains to show that X ′ →Ui is a closed embedding. Now X ′ →Ui
is closed (it is the composition of two closed maps), so it suffices to show that"
X ′ → Ui is a locally closed embedding.
18.9.G. E XERCISE . Let Ai be the closure of U in

Bi := X ×A U1 ×A · · · ×A Ui ×A · · · ×A Un(only the ith term is missing the bar), and let Ci be the closure of U in
Di := U1 ×A · · · ×A Ui ×A · · · ×A Un .

Show that there is an isomorphism Ai → Ci induced by the projection Bi → Di .
Hint: note that the section Di → Bi of the projection Bi → Di , given informally by(t1 , . . . , tn ) (→ (ti , t1 , . . . , tn ), is a closed embedding, as it can be interpreted as thegraph of a morphism to a separated A-scheme (Proposition 10.1.18). So U can beinterpreted as a locally closed subscheme of Di , which in turn can be interpreted
as a closed subscheme of Bi . Thus the closure of U in Di may be identified with
its closure in Bi .
As the Ui cover X, the µ−1 (Ui ) cover X. But µ−1 (Ui ) = Ai (closure can be becomputed locally — the closure of U in Bi is the intersection of Bi with the closure
X of U in X ×A U1 ×A · · · ×A Un ).
Hence over each Ui , we get a closed embedding of Ai "→ Di , and thus X ′ →"Ui is a locally closed embedding as desired.
!

18.9.5. Other versions of Chow’s Lemma. We won’t use these versions, but their
proofs are similar to what we have already shown.
18.9.6. Remark. Notice first that if X is reduced (resp. irreducible, integral), thenX ′ can be taken to be reduced (resp. irreducible, integral) as well.
18.9.H. E XERCISE .
By suitably crossing out lines in the proof above, weaken
the hypothesis “π is proper” to “π finite type and separated”, at the expense of
weakening the conclusion “π ◦ µ is projective” to “π ◦ µ is quasiprojective”.
18.9.7. Remark. The target Spec A can be generalized to a scheme S that is (i) Noetherian, or (ii) separated and quasicompact with finitely many irreducible components. This can be combined with Remark 18.9.6 and Exercise 18.9.H. See [Gr-EGA,
II.5.6.1] for a proof. See also [GW, Thm. 13.100] for a version that is slightly moregeneral.

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CHAPTER 19

Application: Curves
We now use what we have developed to study something explicit — curves.
Throughout this chapter, we will assume that all curves are projective, geometrically integral, regular curves over a field k. We will sometimes add the hypothesisthat k is algebraically closed. Most people are happy with working over algebraically closed fields, and those people should ignore the adverb “geometrically”.
We certainly do not need the massive machinery we have developed in orderto understand curves, but with the perspective we have gained, the development
is quite clean. The key ingredients we will need are as follows. We use a criterion for a morphism to be a closed embedding, that we prove in §19.1. We use
the “black box” of Serre duality (to be proved in Chapter 30). In §19.2, we use thisbackground to observe a very few useful facts, which we will use repeatedly. Finally, in the course of applying them to understand curves of various genera, wedevelop the theory of hyperelliptic curves in a hands-on way (§19.5), in particularproving a special case of the Riemann-Hurwitz formula.
If you are jumping into this chapter without reading much beforehand, you shouldskip §19.1, taking Theorem 19.1.1 as a black box. Depending on your background, you
may want to skip §19.2 as well, taking the “crucial tools” as a black box.

19.1 A criterion for a morphism to be a closed embedding
We will repeatedly use a criterion for when a morphism is a closed embedding,which is not special to curves. Before stating it, we recall some facts about closed
embeddings. Suppose π : X → Y is a closed embedding. Then π is projective, and it
is injective on points. This is not enough to ensure that it is a closed embedding, as
the example of the normalization of the cusp shows (Figure 9.4). Another example
is the following.
19.1.A. E XERCISE (F ROBENIUS , CF. §7.3.16). Show that π : A1Fp → A1Fp given by
x (→ xp is a bijection on points, and induces an isomorphism of residue fields on
closed points, yet is not a closed embedding.
The additional information you need to ensure that a morphism π is a closed
embedding is that the tangent map is injective at all closed points.

19.1.B. E XERCISE . Show (directly, not invoking Theorem 19.1.1) that in the two
examples described above (the normalization of a cusp and the Frobenius morphism), the tangent map is not injective at all closed points.
505

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19.1.1. Theorem. — Suppose k = k, and π : X → Y is a projective morphism of finite
type k-schemes that is injective on closed points and injective on tangent vectors at closed
points. Then π is a closed embedding.
(Remark: In this case of finite type schemes over an algebraically closed field,“injective on tangent vectors at closed points” is equivalent to “unramified”, see
Exercise 21.6.I. We will define unramified in §21.6.)
The example Spec C → Spec R shows that we need the hypothesis that k is
algebraically closed in Theorem 19.1.1. Those allergic to algebraically closed fields
should still pay attention, as we will use this to prove things about curves over kwhere k is not necessarily algebraically closed (see also Exercises 9.2.J and 19.1.E).
We need the hypothesis that the morphism be projective, as shown by theexample of Figure 19.1. It is the normalization of the node, except we erase one ofthe preimages of the node. We map A1 to the plane, so that its image is a curvewith one node. We then consider the morphism we get by discarding one of thepreimages of the node. Then this morphism is an injection on points, and is also
injective on tangent vectors, but it is not a closed embedding. (In the world ofdifferential geometry, this fails to be an embedding because the map doesn’t givea homeomorphism onto its image.)

F IGURE 19.1. The projectivity hypothesis in Theorem 19.1.1 cannot be dropped
Theorem 19.1.1 appears to be fundamentally a statement about varieties, but
it isn’t. We will reduce it to the following result.
19.1.2. Theorem. — Suppose π : X → Y is a finite morphism of Noetherian schemes
whose degree at every point of Y (§13.7.5) is 0 or 1. Then π is a closed embedding.

19.1.C. E XERCISE . Suppose π : X → Y is a finite morphism whose degree at every
point of Y is 0 or 1. Show that π is injective on points (easy). If p ∈ X is any point,
show that π induces an isomorphism of residue fields κ(π(p)) → κ(p). Show thatπ induces an injection of tangent spaces. Thus key hypotheses of Theorem 19.1.1
are implicitly in the hypotheses of Theorem 19.1.2.
19.1.3. Reduction of Theorem 19.1.1 to Theorem 19.1.2. The property of being a closed
embedding is local on the base, so we may assume that Y is affine, say Spec B.

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I next claim that π has finite fibers, not just finite fibers above closed points: thefiber dimension for projective morphisms is upper semicontinuous (Exercise 18.1.B,
or Theorem 11.4.2(b)), so the locus where the fiber dimension is at least 1 is a closed
subset, so if it is nonempty, it must contain a closed point of Y. Thus the fiber overany point is a dimension 0 finite type scheme over that point, hence a finite set.
Hence π is a projective morphism with finite fibers, thus finite by Theorem 18.1.8.
But the degree of a finite morphism is upper semicontinuous, (§13.7.5), and is
at most 1 at closed points of Y, hence is at most 1 at all points.
19.1.4. Proof of Theorem 19.1.2. The problem is local on Y, so we may assume
Y is affine, say Y = Spec B. Thus X is affine too, say Spec A, and π corresponds
to a ring morphism B → A. We wish to show that this is a surjection of rings, or(equivalently) of B-modules. Let K be the cokernel of this morphism of B-modules:(19.1.4.1)

B → A → K → 0.

We wish to show that K = 0. It suffices to show that for any maximal ideal n of B,
Kn = 0. (Do you remember why?) Localizing (19.1.4.1) at n, we obtain the exact
sequence
Bn → An → Kn → 0.
Applying the right-exact functor ⊗Bn (Bn /nBn ), we obtain(19.1.4.2)

Bn /nBnα% An /nAn% Kn /nKn% 0.

Now Spec An /nAn is the scheme theoretic preimage of [n] ∈ Spec B, so by hypothesis, it is either empty, or the map α is an isomorphism (in which case we will
see that the map α will turn out to be the map of residue fields). In the first case,
An /nAn = 0, from which Kn /nKn = 0. In the second case, Kn /nKn = 0 as well. Applying Nakayama’s Lemma 7.2.9 (noting that A is a finitely generated B-module,
hence K is too, hence Kn is a finitely generated Bn -module), Kn = 0 as desired. !
Exercise 9.2.J can be used to extend Theorem 19.1.1 to general fields k, notnecessarily algebraically closed.
19.1.D. E XERCISE . Use Theorem 19.1.1 to show that the dth Veronese embeddingnfrom Pnk , corresponding to the complete linear series |OPk (d)|, is a closed embedn
ding. Do the same for the Segre embedding from Pm×Speck Pk . (This is just forkpractice for using this criterion. It is a weaker result than we had before; we haveearlier checked both of these statements over an arbitrary base ring in Remark 8.2.8
and §9.6 respectively, and we are now checking it only over algebraically closed
fields. However, see Exercise 19.1.E below.)
19.1.E. L ESS IMPORTANT EXERCISE . Using the ideas from this section, prove thatthe dth Veronese embedding from Pn
Z (over the integers!), is a closed embedding.
(Again, we have done this before. This exercise is simply to show that these methods can easily extend to work more generally.)

19.2 A series of crucial tools

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We are now ready to start understanding curves in a hands-on way. We will
repeatedly make use of the following series of crucial remarks, and it will be important to have them at the tip of your tongue.
19.2.1. In what follows, C will be a projective, geometrically regular, geometrically integral curve over a field k, and L is an invertible sheaf on C.
19.2.2. Reminder: Serre duality. Serre duality (Theorem 18.5.1) on a geometrically irreducible regular genus g curve C over k involves an invertible sheaf ωC(of degree 2g − 2, with g sections, Exercise 18.5.A), such that for any coherent sheaf
F on C, hi (C, F ) = h1−i (C, ωC ⊗ F ∨ ) for i = 0, 1. (Better: there is a duality between the two cohomology groups.)
19.2.3. Negative degree line bundles have no nonzero section. If deg L < 0,then h0 (C, L ) = 0. Reason: deg L is the number of zeros minus the numberof poles (suitably counted) of any rational section (Important Exercise 18.4.C). If
there is a regular section (i.e., with no poles), then this is necessarily non-negative.
Refining this argument yields the following.
19.2.4. Degree 0 line bundles, and recognizing when they are trivial. If deg L =∼ OC . Reason: if there
0, then h0 (C, L ) = 0 or 1; and if h0 (C, L ) = 1 then L =is a nonzero section s, it has no poles, and hence no zeros, because deg L = 0.
∼ OC (div s) = OC . (Recall how this works, cf. Important
Then div s = 0, so L =Exercise 14.2.E: s gives a trivialization for the invertible sheaf. We have a naturalbijection for any open set Γ (U, L ) ↔ Γ (U, OU ), where the map from left to right iss ′ (→ s ′ /s, and the map from right to left is f (→ sf.) Conversely, for a geometricallyintegral projective variety, h0 (O) = 1. (§10.3.7 shows this for k algebraically closed— this is where geometric integrality is used — and Exercise 18.2.H shows that
cohomology commutes with base field extension.)
Serre duality turns these statements about line bundles of degree at most 0
into statements about line bundles of degree at least 2g − 2, as follows.
19.2.5. We know h0 (C, L ) if the degree is sufficiently high. If deg L > 2g − 2,then
(19.2.5.1)

h0 (C, L ) = deg L − g + 1.

So we know h0 (C, L ) if deg L ≫ 0. (This is important — remember this!) Reason:
h1 (C, L ) = h0 (C, ωC ⊗L ∨ ); but ωC ⊗L ∨ has negative degree (as ωC has degree
2g − 2), and thus this invertible sheaf has no sections. The result then follows fromthe Riemann-Roch Theorem in the guise of (18.4.2.1).
19.2.A. U SEFUL EXERCISE ( RECOGNIZING ωC AMONG DEGREE 2g − 2 LINE BUN DLES ). Suppose L is a degree 2g − 2 invertible sheaf. Show that it has g − 1 or g
∼ ωC .
sections, and it has g sections if and only if L =19.2.6. Twisting L by a (degree 1) point changes h0 by at most 1. Suppose p
is any closed point of degree 1 (i.e., the residue field of p is k). Then h0 (C, L ) −h0 (C, L (−p)) = 0 or 1. (The twist of L by a divisor, such as L (−p), was definedin §14.2.11.) Reason: consider 0 → OC (−p) → OC → O|p → 0, tensor with L (thisEarly (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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is exact as L is locally free) to get
0 → L (−p) → L → L |p → 0.

Then h0 (C, L |p ) = 1, so as the long exact sequence of cohomology starts off

we are done.

0 → H0 (C, L (−p)) → H0 (C, L ) → H0 (C, L |p ),19.2.7. A numerical criterion for L to be base-point-free. Suppose for thisremark that k is algebraically closed, so all closed points have degree 1 over k.
Then if h0 (C, L ) − h0 (C, L (−p)) = 1 for all closed points p, then L is base-pointfree, and hence induces a morphism from C to projective space (Theorem 16.4.1).
Reason: given any p, our equality shows that there exists a section of L that does
not vanish at p — so by definition, p is not a base point of L .
19.2.8. Next, suppose p and q are distinct (closed) points of degree 1. Then
h0 (C, L ) − h0 (C, L (−p − q)) = 0, 1, or 2 (by repeating the argument of Remark 19.2.6 twice). If h0 (C, L ) − h0 (C, L (−p − q)) = 2, then necessarily(19.2.8.1)
h0 (C, L ) = h0 (C, L (−p)) + 1 = h0 (C, L (−q)) + 1 = h0 (C, L (−p − q)) + 2.
Then the linear series L separates points p and q, i.e., the corresponding map f to
projective space satisfies f(p) ̸= f(q). Reason: there is a hyperplane of projective
space passing through q but not passing through p, or equivalently, there is asection of L vanishing at q but not vanishing at p. This is because of the lastequality in (19.2.8.1).
19.2.9. By the same argument as above, if p is a (closed) point of degree 1, thenh0 (C, L ) − h0 (C, L (−2p)) = 0, 1, or 2. I claim that if this is 2, then the map
corresponding to L (which is already seen to be base-point-free at p from theabove) separates the tangent vectors at p. To show this, we need to show that the
cotangent map is surjective. To show surjectivity onto a one-dimensional vector
space, we just need to show that the map is nonzero. So it suffices to find a function
on the target vanishing at the image of p that pulls back to a function that vanishes
at p to order 1 but not 2. In other words, we want a section of L vanishing at p
to order 1 but not 2. But that is the content of the statement h0 (C, L (−p)) −h0 (C, L (−2p)) = 1.
19.2.10. Criterion for L to be very ample. Combining some of our previous
comments: suppose C is a curve over an algebraically closed field k, and L is
an invertible sheaf such that for all closed points p and q, not necessarily distinct,
h0 (C, L ) − h0 (C, L (−p − q)) = 2, then L gives a closed embedding into projective
space, as it separates points and tangent vectors, by Theorem 19.1.1.
19.2.B. E XERCISE . Suppose that k is algebraically closed, so the previous remark
applies. Show that C \ {p} is affine. Hint: Show that if j ≫ 0, then O(jp) is basepoint-free and has at least two linearly independent sections, one of which has
divisor jp. Use these two sections to map to P1 so that the set-theoretic preimageof ∞ is p. Argue that the map is finite, and that C \ {p} is the preimage of A1 . (Atrivial variation of this argument shows that C \ {p1 , . . . , pn } is affine if n > 0.)

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19.2.11. Conclusion. We can combine much of the above discussion to give thefollowing useful fact. If k is algebraically closed, then deg L ≥ 2g implies that L
is base-point-free (and hence determines a morphism to projective space). Also,
deg L ≥ 2g + 1 implies that for any basis of sections s0 , . . . , sdeg L −g of L , themorphism given by (s0 , . . . , sdeg L −g ) is in fact a closed embedding into Pdeg L −g
(so L is very ample). Remember this!
19.2.C. E XERCISE ( ON A PROJECTIVE , REGULAR , INTEGRAL CURVE OVER k, AMPLE= POSITIVE DEGREE ). Show that an invertible sheaf L on a projective, regular,
integral curve over k is ample if and only if deg L > 0.
(This can be extended to curves over general fields using Exercise 19.2.D below.) Thus there is a blunt purely numerical criterion for ampleness of line bundles on curves. This generalizes to projective varieties of higher dimension; this iscalled the Nakai-Moishezon criterion for ampleness, Theorem 20.4.1.
19.2.D. E XERCISE ( EXTENSION TO NON - ALGEBRAICALLY CLOSED FIELDS ). Show
that the statements in §19.2.11 hold even without the hypothesis that k is algebraically closed. (Hint: to show one of the facts about some curve C and line bundle L , consider instead C ×Spec k Spec k. Then show that if the pullback of L here
has sections giving you one of the two desired properties, then there are sections
downstairs with the same properties. You may want to use facts that we have used,such as the fact that base-point-freeness is independent of extension of base field,
Exercise 18.2.I, or that the property of an affine morphism over k being a closed
embedding holds if and only if it does after an extension of k, Exercise 9.2.J.)
19.2.E. E XERCISE ( EXTENDING E XERCISE 19.2.C). Suppose L is an invertible
sheaf on a projective, geometrically regular, geometrically integral curve C (over a
field k). Show that L is ample if and only if deg L > 0. Hint: reduce to the casewhere k is algebraically closed, with the help of Exercise 9.2.J. (This was promisedin Exercise 18.7.C.)
We are now ready to take these facts and go to the races.

19.3 Curves of genus 0
We are now ready to (in some form) answer the question: what are the curvesof genus 0?
In §6.5.9, we saw a genus 0 curve (over a field k) that was not isomorphic to P1 :2x + y2 + z2 = 0 in P2R . (It has genus 0 by (18.6.6.1).) We have already observed that
this curve is not isomorphic to P1R , because it doesn’t have an R-valued point. On
the other hand, we haven’t seen a genus 0 curve over an algebraically closed fieldwith this property. This is no coincidence: the lack of an existence of a k-valued
point is the only obstruction to a genus 0 curve being P1 .
19.3.1. Proposition. — Suppose C is genus 0, and C has a k-valued (degree 1) point.
∼ P1 .
Then C =k

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Thus we see that all genus 0 (integral, regular, projective) curves over an algebraically closed field are isomorphic to P1 .
Proof. Let p be the point, and consider L = O(p). Then deg L = 1, so we canapply what we know above: first, h0 (C, L ) = 2 (Remark 19.2.5), and second,
these two sections give a closed embedding into P1k (Remark 19.2.11). But the only
closed embedding of a curve into the integral curve P1k is an isomorphism!
!
As a bonus, Proposition 19.3.1 implies that x2 +y2 +z2 = 0 in P2R has no line bundles of degree 1 over R; otherwise, we could just apply the above argument to thecorresponding line bundle. This example shows us that over a non-algebraicallyclosed field, there can be genus 0 curves that are not isomorphic to P1k . The next
result lets us get our hands on them as well.
19.3.2. Claim. — All genus 0 curves can be described as conics in P2k .
Proof. Any genus 0 curve has a degree −2 line bundle — the dualizing sheaf ωC .
Thus any genus 0 curve has a degree 2 line bundle: L = ω∨C . We apply Remark 19.2.11: deg L = 2 ≥ 2g + 1, so this line bundle gives a closed embedding
into P2 .
!
19.3.A. E XERCISE . Suppose C is a genus 0 curve (projective, geometrically integral
and regular). Show that C has a point of degree at most 2. (The degree of a point
was defined in §5.3.8.)

The geometric means of finding Pythagorean triples presented in §6.5.8 looked
quite different, but was really the same. There was a genus 0 curve C (a plane
conic) with a k-valued point p, and we proved that it was isomorphic to P1k . Theline bundle used to show the isomorphism wasn’t the degree 1 line bundle OC (p);it was the degree 1 line bundle OP2 (1)|C ⊗ OC (−p).

19.4 Classical geometry arising from curves of positive genus
We will use the following Proposition and Corollary later, and we take this as
an excuse to revisit some very classical geometry from a modern standpoint.
19.4.1. Proposition. — Recall our standing assumptions for this chapter (§19.2.1), that
C is a projective, geometrically regular, geometrically integral curve over a field k. Suppose
C is not isomorphic to P1k (with no assumptions on the genus of C), and L is an invertible
sheaf of degree 1. Then h0 (C, L ) < 2.
Proof. Otherwise, let s1 and s2 be two (independent) sections. As the divisor ofzeros of si is the degree of L , each vanishes at a single point pi (to order 1). But
p1 ̸= p2 (or else s1 /s2 has no poles or zeros, i.e., is a constant function, so s1 ands2 are linearly dependent). Thus s1 and s2 generate a base-point-free linear series,
so we have an induced morphism C → P1 . This is a finite degree 1 morphismfrom one regular curve to another (Exercise 17.4.F), which hence induces a degree
1 extension of function fields, i.e., an isomorphism of function fields, which means

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The Rising Sea: Foundations of Algebraic Geometrythat the curves are isomorphic (cf. Theorem 17.4.3). But we assumed that C is notisomorphic to P1k , so we have a contradiction.
!
19.4.2. Corollary. — If C is a projective, geometrically regular, geometrically integral∼ OC (q)curve over k, not isomorphic to P1k and p and q are degree 1 points, then OC (p) =if and only if p = q.
19.4.A. E XERCISE . Show that if k is algebraically closed, then C has genus 0 if andonly if all degree 0 line bundles are trivial.
19.4.B. E XERCISE . Suppose C is a regular plane curve of degree e > 2, and D1
and D2 are two plane curves of the same degree d not containing C. By Bézout’s
Theorem for plane curves (§18.6.3), Di meets C at de points, counted “correctly”.
Suppose D1 and D2 meet C at de − 1 of the “same points”, plus one more. Show
that the remaining points are the same as well. More precisely, suppose there isa divisor E on C of degree de − 1, and degree 1 (k-valued) points p1 and p2 suchthat Di ∩ C = E + pi (as divisors on C). Show that p1 = p2 . (The case d = e = 3
is Chasles’ Theorem, and is the first case of the Cayley-Bacharach Theorem, see[EGH].)As an entertaining application of Exercise 19.4.B, we can prove two classical results. For convenience, in this discussion we will assume k is algebraically closed,
although this assumption can be easily removed.
19.4.3. Pappus’s Theorem (Pappus of Alexandria). — (See Figure 19.2.) Supposeℓ and m are distinct lines in the plane, and α, β, γ are distinct points on ℓ, and α ′ , β ′ ,and γ ′ are distinct points on m, and all six points are distinct from ℓ ∩ m. Then points
x = αβ ′ ∩ α ′ β, y = aγ ∩ αγ ′ , and z = βγ ′ ∩ β ′ γ are collinear.

F IGURE 19.2. Pappus’ Theorem
Pascal’s “Mystical Hexagon” Theorem was discovered by Pascal at age 16.
(What were you doing at age 16?)
19.4.4. Pascal’s “Mystical Hexagon” Theorem. — (See Figure 19.3.) If a hexagon
αγ ′ βα ′ γβ ′ is inscribed in a smooth conic K, and opposite pairs of sides are extended until

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they meet, the three intersection points x = αβ ′ ∩α ′ β, y = α ′ γ∩αγ ′ , and z = βγ ′ ∩β ′ γ
are collinear.

F IGURE 19.3. Pascal’s “Mystical Hexagon” Theorem
Pappus’s Theorem can be seen as a degeneration of Pascal’s Theorem: theconic degenerates into the union of two lines, and the six points degenerate sothat α, β, and γ are on one, and α ′ , β ′ , and γ ′ are on the other. We thus prove
Pascal’s Theorem, and you should check that the proof readily applies to prove
Pappus’s Theorem.
19.4.C. E XERCISE . Suppose α, β, γ, α ′ , β ′ are five points in P2R , no three on a line.
Show that there is a unique conic C passing through the five points. Show that C isregular. Explain how to construct the tangent to C at α using only a straightedge.
(Hint for the last part: apply Pascal’s Theorem, taking γ ′ = α. You may need to
first figure out why you can apply Pascal’s Theorem in this degenerate case.)
In particular, given an ellipse C ⊂ R2 and a point α ∈ C, you will now be able
to construct the tangent to C at α using only a straightedge.
19.4.5. Proof of Pascal’s Theorem 19.4.4. We wish to show that the line xy meets βγ ′and β ′ γ at the same point. Let C be the curve that is the union of X and the linexy. (Warning, cf. §19.2.1: for one time in this chapter, we are not assuming C to be
regular!)Let D1 be the union of the three lines αβ ′ , βγ ′ , and γα ′ , and let D2 be theunion of the three lines α ′ β, β ′ γ, and γ ′ α. Note that D1 meets C at the nine points
α, β, γ, α ′ , β ′ , γ ′ , x, y, and xy ∩ βγ ′ , and D2 meets C at the same nine points,
except xy ∩ βγ ′ is replaced by xy ∩ β ′ γ. Thus if we knew Corollary 19.4.2 so that it
applied to our C (which is singular), then we would be done (cf. Exercise 19.4.B).
So we extend Corollary 19.4.2 to our situation. So to do this, we extend Proposition 19.4.1 to our situation. We have a plane cubic C which is the union of a line
∼ O(q).
L and a conic K, and two points p = xy ∩ βγ ′ and q = xy ∩ β ′ γ, with O(p) =′(Reason: both are the restriction of OP2 (3) to C, twisted by −(α + β + γ + α + β ′ +
γ ′ + x + y).) Call this invertible sheaf L . Suppose p ̸= q. Then the two sections of

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L with zeros at p and q give a base-point-free linear series, and thus a morphism
π : C → P1 , with π−1 (0) = p and π−1 (∞) = q.
By the argument in the proof of Proposition 19.4.1, π gives an isomorphism of
L with P1 . As π is proper, π(K) is closed, and as K is irreducible, π(K) is irreducible.
As π(K) does not contain 0 (or ∞), it can’t be all of P1 . Hence π(K) is a point.
The conic K meets the line L in two points with multiplicity (by Bézout’s Theorem, §18.6.3, or by simple algebra). If K ∩ L consists of two points a and b, we havea contradiction: K can’t be contracted to both π(a) and π(b). But what happens if
K meets L at one point, i.e., if the conic is tangent to the line?
19.4.D. E XERCISE . Finish the proof of Pascal’s Theorem by dealing with this case.
Hint: nilpotents will come to the rescue. The intuition is as follows: K ∩ L is a
subscheme of L of length 2, but the scheme theoretic image π(K) can be shown to
be a reduced closed point.
!
19.4.6. Remark. The key motivating fact that makes our argument work, Proposition 19.4.1, is centrally about curves not of genus 0, yet all the curves involved
in Pappus’s Theorem and Pascal’s Theorem have genus 0. The insight to keep inmind is that union of curves of genus 0 need not have genus 0. In our case, itmattered that cubic curves have genus 1, even if they are union of P1 ’s.

19.5 Hyperelliptic curves
We next discuss an important class of curves, the hyperelliptic curves. In thissection, we assume k is algebraically closed of characteristic not 2. (These hypotheses
can be relaxed, at some cost.)
A (projective regular integral) genus g curve C is hyperelliptic if it admits
a double cover of (i.e., degree 2, necessarily finite, morphism to) P1k . For convenience, when we say C is hyperelliptic, we will implicitly have in mind a choice of
double cover π : C → P1 . (We will later see that if g ≥ 2, then there is at most one
such double cover, Proposition 19.5.7, so this is not a huge assumption.) The map
π is called the hyperelliptic map.
By Exercise 17.4.D, the preimage of any closed point p of P1 consists of either
one or two points. If π−1 (p) is a single point, we say p is a branch point, and
π−1 (p) is a ramification point of π. (The notion of ramification will be defined
more generally in §21.6.)
19.5.1. Theorem (hyperelliptic Riemann-Hurwitz formula). — Suppose k = k andchar k ̸= 2, π : C → P1k is a double cover by a projective regular irreducible genus g curve
over k. Then π has 2g + 2 branch points.

This is a special case of the Riemann-Hurwitz formula, which we will stateand prove in §21.7. You may have already heard about genus 1 complex curves
double covering P1 , branched over 4 points.
To prove Theorem 19.5.1, we first prove the following.

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19.5.2. Proposition. — Assume char k ̸= 2 and k = k. Given r distinct points
p1 , . . . , pr ∈ P1 , there is precisely one double cover branched at precisely these points if r
is even, and none if r is odd.
Proof. Pick points 0 and ∞ of P1 distinct from the r branch points. All r branch
points are in P1 − ∞ = A1 = Spec k[x]. Suppose we have a double cover of
A1 , C ′ → A1 , where x is the coordinate on A1 . This induces a quadratic field
extension K over k(x). As char k ̸= 2, this extension is Galois. Let σ : K → K be
the Galois involution. Let y be a nonzero element of K such that σ(y) = −y, so
1 and y form a basis for K over the field k(x), and are eigenvectors of σ. Now
σ(y2 ) = y2 , so y2 ∈ k(x). We can replace y by an appropriate k(x)-multiple so thaty2 is a polynomial, with no repeated factors, and monic. (This is where we use thehypothesis that k is algebraically closed, to get leading coefficient 1.)
Thus y2 = xN +aN−1 xN−1 +· · ·+a0 , where the polynomial on the right (call it
f(x)) has no repeated roots. The Jacobian criterion (in the guise of Exercise 12.2.E)
implies that this curve C0′ in A2 = Spec k[x, y] is regular. Then C0′ is normal andhas the same function field as C ′ . Thus C0′ and C ′ are both normalizations of A1 in
the finite extension of fields generated by y, and hence are isomorphic. Thus wehave identified C ′ in terms of an explicit equation.
The branch points correspond to those values of x for which there is exactly
one value of y, i.e., the roots of f(x). In particular, N = r, andf(x) = (x − p1 ) · · · (x − pr ),where the pi are interpreted as elements of k.
Having mastered the situation over A1 , we return to the situation over P1 . We
will examine the branched cover over the affine open set P1 \{0} = Spec k[u], whereu = 1/x. The previous argument applied to Spec k[u] rather than Spec k[x] showsthat any such double cover must be of the form)*
C ′′ = Spec k[z, u]/ z2 − (u − 1/p1 ) · · · (u − 1/pr )
$$& %
%
r
= Spec k[z, u]/pi z2 − (−u) f (1/u)→

Spec k[u] = A1 .

So if there is a double cover over all of P1 , it must be obtained by gluing C ′′ toC , “over” the gluing of Spec k[x] to Spec k[u] to obtain P1 .
Thus in K(C), we must have′z2 = ur f(1/u) = f(x)/xr = y2 /xr
from which z2 = y2 /xr .
If r is even, considering K(C) as generated by y and x, there are two possible
values of z: z = ±y/xr/2 . After renaming z by −z if necessary, there is a single wayof gluing these two patches together (we choose the positive square root).
If r is odd, the result follows from Exercise 19.5.A below.
19.5.A. E XERCISE . Suppose char k ̸= 2. Show that x does not have a square root in
the field k(x)[y]/(y2 −√f(x)), where√ f is a polynomial with nonzero roots p1 , . . . , pr .
(Possible hint: why is 3 ∈/ Q( 2)?) Explain how this proves Proposition 19.5.2 inthe case where r is odd.
!

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For future reference, we collect here our explicit (two-affine) description of thehyperelliptic cover C → P1 .
(19.5.2.1)

Spec k[x, y]/(y2 − f(x))"
Spec k[x]z=y/xr/2

y=z/ur/2

Spec k[u, z]/(z2 − ur f(1/u))

u=1/x
x=1/u

"
Spec k[u]19.5.3. If k is not algebraically closed. If k is not algebraically closed (but of characteristic not 2), the above argument shows that if we have a double cover of A1 , thenit is of the form y2 = af(x), where f is monic (and a ̸= 0). Furthermore, if a and a ′differ (multiplicatively) by an element of (k× )2 , then y2 = af(x) is isomorphic to
y2 = a ′ f(x). You may be able to use this to show that (assuming that k× ̸= (k× )2 )a double cover is not determined by its branch points. Moreover, this failure isclassified by k× /(k× )2 . Thus we have lots of curves that are not isomorphic overk, but become isomorphic over k. These are often called twists of each other.
(In particular, once we define elliptic curves, you will be able to show that
there exist two elliptic curves over Q with the same j-invariant, that are not isomorphic, see Exercise 19.9.D.)
19.5.4. Back to proving the hyperelliptic Riemann-Hurwitz formula, Theorem 19.5.1.
Our explicit description of the unique double cover of P1 branched over r different points will allow us to compute the genus, thereby completing the proof of
Theorem 19.5.1.
We continue the notation (19.5.2.1) of the proof of Proposition 19.5.2. Suppose
P1 has affine cover by Spec k[x] and Spec k[u], with u = 1/x, as usual. Suppose
C → P1 is a double cover, given by y2 = f(x) over Spec k[x], where f has degreer, and z2 = ur f(1/u). Then C has an affine open cover by Spec k[x, y]/(y2 − f(x))
and Spec k[u, z]/(z2 − ur f(1/u)). The corresponding Čech complex for OC is0% k[x, y]/(y2 − f(x)) × k[u, z]/(z2 − ur f(1/u)))*k[x, y]/(y2 − f(x)) x

d

%

% 0.

The second (and last) nonzero part of the complex has basis consisting of monomials xn yϵ , where n ∈ Z and ϵ = 0 or 1. To compute the genus g = h1 (C, OC ),we must compute coker d. We can use the first factor k[x, y]/(y2 − f(x)) to hit themonomials xn yϵ where n ∈ Z≥0 , and ϵ = 0 or 1. The image of the second factor
is generated by elements of the form um zϵ , where m ≥ 0 and ϵ = 0 or 1. But
um zϵ = x−m (y/xr/2 )ϵ . By inspection, the cokernel has basis generated by monomials x−1 y, x−2 y, . . . , x−r/2+1 y, and thus has dimension r/2−1. Hence g = r/2−1,
from which Theorem 19.5.1 follows.
!
19.5.5. Curves of every genus. As a consequence of the hyperelliptic RiemannHurwitz formula (Theorem 19.5.1), we see that there are curves of every genus
g ≥ 0 over an algebraically closed field of characteristic not 2: to get a curve of

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genus g, consider the branched cover branched over 2g + 2 distinct points. The
unique genus 0 curve is of this form, and we will see in §19.6.2 that every genus 2
curve is of this form. We will soon see that every genus 1 curve (reminder: over analgebraically closed field!) is too (§19.9.5). But it is too much to hope that all curves
are of this form, and we will soon see (§19.7.2) that there are genus 3 curves thatare not hyperelliptic, and we will even get heuristic evidence that “most” genus 3
curves are not hyperelliptic. We will later give vague evidence (that can be madeprecise) that “most” genus g curves are not hyperelliptic if g > 2 (§19.8.3).
19.5.B. E XERCISE . Verify that a curve C of genus at least 1 admits a degree 2 coverof P1 if and only if it admits a degree 2 invertible sheaf L with h0 (C, L ) = 2.
Possibly in the course of doing this, verify that if C is a curve of genus at least
1, and C has a degree 2 invertible sheaf L with at least 2 (linearly independent)
sections, then L has precisely two sections, and that this L is base-point-free andgives a hyperelliptic map.
19.5.6. Proposition. — Assume g ≥ 2. If L corresponds to a hyperelliptic cover∼ ωC .
C → P1 , then L ⊗(g−1) =Proof. Compose the hyperelliptic map with the (g − 1)th Veronese map:
(19.5.6.1)

C

|L |

% P1

|OP1 (g−1)|% Pg−1 .

The composition corresponds to L ⊗(g−1) . This invertible sheaf has degree 2g − 2(by Exercise 18.4.F). The pullback H0 (Pg−1 , O(1)) → H0 (C, L ⊗(g−1) ) is injective
because the image of C in Pg−1 (a rational normal curve) is nondegenerate: if there
were a hyperplane s ∈ H0 (Pg−1 , O(1)) that pulled back to 0 on C, then the image ofC would lie in that hyperplane, yet a rational normal curve cannot. Thus L ⊗(g−1)
has at least g sections. But by Exercise 19.2.A, the only invertible sheaf of degree
2g − 2 with (at least) g sections is the dualizing sheaf.
!
As an added bonus, we see that the composition of (19.5.6.1) is the complete
linear series |L ⊗(g−1) | — all sections of L ⊗(g−1) come up in this way.
19.5.7. Proposition (a genus ≥ 2 curve can be hyperelliptic in only one way). —Any curve C of genus at least 2 admits at most one double cover of P1 . More precisely, if∼ M.
L and M are two degree two line bundles yielding maps C → P1 , then L =Proof. If C is hyperelliptic, then we can recover the hyperelliptic map by considering the canonical linear series given by ωC (the canonical map, which we will use
again repeatedly in the next few sections): it is a double cover of a degree g − 1 rational normal curve (by the previous proposition), which is isomorphic to P1 . This
double cover is the hyperelliptic cover (also by the proof of the previous proposition). Thus we have uniquely recovered the map C → P1 , and this map must be∼ M (using Theorem 16.4.1,
induced by both L and M , from which we have L =relating maps to projective space and line bundles).
!

19.5.8. The “space of hyperelliptic curves”. Thanks to Proposition 19.5.7, we can now
classify hyperelliptic curves of genus at least 2. Hyperelliptic curves of genus g ≥ 2
correspond to precisely 2g + 2 distinct points on P1 modulo S2g+2 , and modulo

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automorphisms of P1 . Thus “the space of hyperelliptic curves” has dimension
2g + 2 − dim Aut P1 = 2g − 1.
This is not a well-defined statement, because we haven’t rigorously defined “thespace of hyperelliptic curves” — it is an example of a moduli space. For now, takethis as a plausibility statement. It is also plausible that this space is irreducible andreduced — it is the image of something irreducible and reduced.

19.6 Curves of genus 2
19.6.1. The reason for leaving genus 1 for later. It might make most sense to jump togenus 1 at this point, but the theory of elliptic curves is especially rich and subtle,
so we will leave it for §19.9.
In general, curves have quite different behaviors (topologically, arithmetically,geometrically) depending on whether g = 0, g = 1, or g ≥ 2. This trichotomy
extends to varieties of higher dimension. We already have some inkling of it in
the case of curves. Arithmetically, genus 0 curves can have lots and lots of rational points, genus 1 curves can have lots of rational points, and by Faltings’ Theorem (Mordell’s Conjecture) any curve of genus at least 2 has at most finitely many
rational points. (Thus even before the proof of important cases of the TaniyamaShimura conjecture [Wi, TWi], and hence Fermat’s Last Theorem, we knew thatxn + yn = zn in P2 has at most finitely many rational solutions for n ≥ 4, as such)*
curves have genus n−1> 1, see (18.6.6.1).) In the language of differential geome2
try, Riemann surfaces of genus 0 are positively curved, Riemann surfaces of genus
1 are flat, and Riemann surfaces of genus at least 2 are negatively curved. It is afact that curves of genus at least 2 have finite automorphism groups (see §21.7.8),
while curves of genus 1 have one-dimensional automorphism groups, see Question 19.10.6), and the unique curve of genus 0 over an algebraically closed field
has a three-dimensional automorphism group (see Exercises 16.4.B and 16.4.C).
(See Exercise 21.5.G for more on this issue.)
19.6.2. Back to curves of genus 2.
Over an algebraically closed field, we saw in §19.3 that there is only one genus
0 curve. In §19.5 we established that there are hyperelliptic curves of genus 2. How
can we get a hold of curves of genus 2? For example, are they all hyperelliptic?
“How many” are there? We now tackle these questions.
Fix a curve C of genus g = 2. Then ωC is degree 2g − 2 = 2, and has 2 sections(Exercise 19.2.A). By Exercise 19.5.B, ωC is base-point-free, and thus induces adouble cover C → P1 (unique up to automorphisms of P1 , which we studied in
Exercises 16.4.B and 16.4.C). Conversely, any double cover C → P1 arises from adegree 2 invertible sheaf with at least 2 sections. Hence if g(C) = 2, this invertible
sheaf must be the dualizing sheaf, just because it is a degree 2g − 2 = 2 line bundle
with g = 2 sections (or alternatively, by the easiest case of Proposition 19.5.6).
Hence we have a natural bijection between genus 2 curves and genus 2 double
covers of P1 (up to automorphisms of P1 ). If the characteristic is not 2, the hyperelliptic Riemann-Hurwitz formula (Theorem 19.5.1) shows that the double cover

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is branched over 2g + 2 = 6 geometric points. In particular, we have a “threedimensional space of genus 2 curves”. This is not rigorous, but we can certainly
show that there are an infinite number of genus 2 curves. Precisely:
19.6.A. E XERCISE . Fix an algebraically closed field k of characteristic not 2. Show
that there are an infinite number of (pairwise) non-isomorphic genus 2 curves overk.
19.6.B. E XERCISE . Show that every genus 2 curve (over any field of characteristic
not 2) has finite automorphism group.

19.7 Curves of genus 3
Suppose C is a curve of genus 3. Then ωC has degree 2g−2 = 4, and has g = 3
sections.
19.7.1. Claim. — The invertible sheaf ωC is base-point-free, and hence gives a map toP2 .
Proof. We check base-point-freeness by working over the algebraic closure k (which
we can, by Exercise 18.2.I). For any point p, by Riemann-Roch,
h0 (C, ωC (−p)) − h0 (C, O(p)) = deg(ωC (−p)) − g + 1 = 3 − 3 + 1 = 1.
But h0 (C, O(p)) = 1 by Proposition 19.4.1, so
h0 (C, ωC (−p)) = 2 = h0 (C, ωC ) − 1.
Thus p is not a base point of ωC for any p, so by Criterion 19.2.7 ωC is base-point!
free.
The next natural question is: Is this a closed embedding? Again, we can checkover the algebraic closure. We use our “closed embedding test” of §19.1. If it isn’t
a closed embedding, then we can find two points p and q (possibly identical) suchthat
h0 (C, ωC ) − h0 (C, ωC (−p − q)) = 1 or 0,
i.e., h0 (C, ωC (−p − q)) = 2. But by Serre duality, this means that h0 (C, O(p +q)) = 2. We have found a degree 2 divisor with 2 sections, so C is hyperelliptic.
Conversely, if C is hyperelliptic, then we already know that ωC gives a double
cover of a regular conic in P2 , and hence ωC does not give a closed embedding.
Thus we conclude that if (and only if) C is not hyperelliptic, then the canonicalmap describes C as a degree 4 curve in P2 .
Conversely, any quartic plane curve is canonically embedded. Reason: thecurve has genus 3 (see (18.6.6.1)), and is mapped by an invertible sheaf of degree
4 with 3 sections. But by Exercise 19.2.A, the only invertible sheaf of degree 2g − 2
with g sections is ωC .
In particular, each nonhyperelliptic genus 3 curve can be described as a quartic
plane curve in only one way (up to automorphisms of P2 , which were describedin Exercise 16.4.B).
In conclusion, there is a bijection between nonhyperelliptic genus 3 curves,and plane quartic curves up to projective linear transformations.

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19.7.2. Remark. In particular, as there exist regular plane quartics (Exercise 12.3.E),
there exist nonhyperelliptic genus 3 curves.
19.7.A. E XERCISE ( CF. §19.5.8). Give a heuristic (non-rigorous) argument that thenonhyperelliptic curves of genus 3 form a family of dimension 6. (Hint: Countthe dimension of the family of regular quartic curves, and quotient by Aut P2 =PGL(3).)The genus 3 curves thus seem to come in two families: the hyperelliptic curves(a family of dimension 5), and the nonhyperelliptic curves (a family of dimension
6). This is misleading — they actually come in a single family of dimension 6.
In fact, hyperelliptic curves are naturally limits of nonhyperelliptic curves. We
can write down an explicit family. (This explanation necessarily requires somehand-waving, as it involves topics we haven’t seen yet.) Suppose we have a hyperelliptic curve branched over 2g + 2 = 8 points of P1 . Choose an isomorphism of
P1 with a conic in P2 . There is a regular quartic meeting the conic at precisely those
8 points. (This requires a short argument using Bertini’s Theorem 12.4.2, which weomit.) Then if f is the equation of the conic, and g is the equation of the quartic,
then f2 + t2 g is a family of quartics that are smooth for most t (smoothness is an
open condition in t, as we will see in Theorem 25.3.3 on generic smoothness). The
t = 0 case is a double conic. Then it is a fact that if you normalize the total space ofthe family, the central fiber (above t = 0) turns into our hyperelliptic curve. Thus
we have expressed our hyperelliptic curve as a limit of nonhyperelliptic curves.
19.7.B. U NIMPORTANT E XERCISE . A (projective) curve (over a field k) admittinga degree 3 cover of P1 is called trigonal. Show that every nonhyperelliptic genus
3 complex curve is trigonal, by taking the quartic model in P2 , and projecting toP1 from any point on the curve. Do this by choosing coordinates on P2 so that p is
at [0, 0, 1]. (The same idea, applied to cubics rather than quartics, will be used in§19.9.9.)
19.7.3. ⋆ A genus 3 curve with no nontrivial automorphisms.
We have seen that a (smooth projective integral) curve of genus at most 2 always has nontrivial automorphisms. It turns out that there are genus 3 curves
with no nontrivial automorphisms.
19.7.C. E XERCISE . Suppose C ′ ⊂ P2 is a smooth plane quartic curve (over any
field k). Show that there is bijection between automorphisms of C ′ and automorphisms of P2 preserving C ′ (as a set).
Thus to find a genus 3 curve with no nontrivial automorphisms, we need
only find a smooth quartic plane curve C ′ such that the only automorphism of
P2 fixing C ′ as a set must be the identity. Your intuition may (correctly) tell youthat most quartics have this property. But exhibiting a specific C ′ (with proof) requires rolling up our sleeves and getting to work. Poonen gives automorphism
free curves over any field in [P1]; an example in characteristic 0 is
y3 z − 3yz3 = 3x4 − 4x3 z + z4 .
19.7.D. E XERCISE . Suppose C is a smooth projective curve with no nontrivial
automorphisms. Show that no two distinct open subsets of C are isomorphic.

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19.7.4. Genus 3 curves with nontrivial automorphisms.
Certainly genus 3 curves can have automorphisms: witness hyperelliptic curves.
More impressive is the Klein quartic
x3 y + y3 z + z3 x = 0
in P2C , which has 168 automorphisms. (Can you find them all?) In fact, the automorphism group of the Klein quartic is the unique finite simple group of order
168 (the second-smallest nonabelian finite simple group). In §21.7.8, we will seethat no genus 3 curve can have more automorphisms (assuming the hard fact thatcurves of genus greater than 1 have finite automorphism groups).

19.8 Curves of genus 4 and 5
We begin with two exercises in general genus, then specialize to genus 4.
19.8.A. E XERCISE . Assume k = k (purely to avoid distraction — feel free toremove this hypothesis). Suppose C is a genus g curve, with g > 1. Show that if C
is not hyperelliptic, then the dualizing invertible sheaf gives a closed embedding
C "→ Pg−1 . (In the hyperelliptic case, we have already seen that the dualizing
sheaf gives us a double cover of a rational normal curve.) Hint: follow the genus 3
case. Such a curve is called a canonical curve, and this closed embedding is called
the canonical embedding of C.
19.8.B. E XERCISE . Suppose C is a curve of genus g > 1, over a field k that is notnecessarily algebraically closed. Show that C has a closed point of degree at most2g − 2 over the base field. (For comparison: if g = 1, for any n, there is a genus 1
curve over Q with no point of degree less than n!)
We next consider nonhyperelliptic curves C of genus 4. Note that deg ωC = 6
and h0 (C, ωC ) = 4, so the canonical map expresses C as a sextic curve in P3 . Let
i : C "→ P3 be the canonical (closed) embedding. We shall see that all such C arecomplete intersections of quadric surfaces and cubic surfaces, and conversely all
regular complete intersections of quadric surfaces and cubic surfaces are genus 4
nonhyperelliptic curves, canonically embedded.
By (19.2.5.1) (Riemann-Roch and Serre duality),⊗2h0 (C, i∗ OP3 (2)) = h0 (C, ω⊗2C ) = deg ωC − g + 1 = 12 − 4 + 1 = 9.
) *On the other hand, h0 (P3 , OP3 (2)) = 3+2= 10. Thus the restriction map
2

H0 (P3 , OP3 (2)) → H0 (C, i∗ OP3 (2))

must have a nontrivial kernel, so there is at least one quadric surface Q in P3
that contains our curve C. Now quadric surfaces are either double planes, or theunion of two planes, or cones, or regular quadrics. (They correspond to quadricforms of rank 1, 2, 3, and 4 respectively. The rank of a quadratic form was definedin Exercise 5.4.J.) But C can’t lie in a plane, so Q must be a cone or regular. In
particular, Q is irreducible.
Now C can’t lie on two (distinct) such quadrics, say Q and Q ′ . Otherwise, as
Q and Q ′ have no common components (they are irreducible and not the same!),

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The Rising Sea: Foundations of Algebraic GeometryQ ∩ Q ′ is a curve (not necessarily reduced or irreducible). By Bézout’s Theorem(Exercise 18.6.K), Q ∩ Q ′ is a curve of degree 4. Thus our curve C, being of degree
6, cannot be contained in Q ∩ Q ′ . (If you don’t see why directly, Exercise 18.6.F
might help.)
19.8.1. We next consider cubic surfaces. By (19.2.5.1) again,⊗3h0 (C, ω⊗3C ) = deg ωC − g + 1 = 18 − 4 + 1 = 15.
) *Now Sym3 Γ (C, ωC ) has dimension 4+2= 20. Thus C lies on at least a 5-dimensional
3vector space of cubics. Now a 4-dimensional subspace comes from multiplying thequadratic form Q by a linear form (?w + ?x + ?y + ?z). But hence there is still one cubic K whose underlying form is not divisible by the quadric form Q (i.e., K doesn’t
contain Q.) Then K and Q share no component, so K ∩ Q is a complete intersection
containing C as a closed subscheme. Now K ∩ Q and C are both degree 6 (the former by Bézout’s Theorem, Exercise 18.6.K, and the latter because C is embedded
by a degree 6 line bundle, Exercise 18.6.I). Also, K ∩ Q and C both have arithmetic
genus 4 (the former by Exercise 18.6.R). These two invariants determine the (linear) Hilbert polynomial, so K ∩ Q and C have the same Hilbert polynomial. Hence
C = K ∩ Q by Exercise 18.6.F.
We now show the converse, and that any regular complete intersection C of aquadric surface with a cubic surface is a canonically embedded genus 4 curve. By
Exercise 18.6.R, such a complete intersection has genus 4.

19.8.C. E XERCISE . Show that OC (1) has at least 4 sections. (Translation: C doesn’t
lie in a hyperplane.)
The only degree 2g−2 invertible sheaf with (at least) g sections is the dualizing∼ ωC , and C is indeed canonically embedded.
sheaf (Exercise 19.2.A), so OC (1) =19.8.D. E XERCISE . Give a heuristic argument suggesting that the nonhyperelliptic
curves of genus 4 “form a family of dimension 9”.
On to genus 5!
19.8.E. E XERCISE . Suppose C is a nonhyperelliptic genus 5 curve. Show that the
canonical curve is degree 8 in P4 . Show that it lies on a three-dimensional vector space of quadrics (i.e., it lies on 3 linearly independent independent quadrics).
Show that a regular complete intersection of 3 quadrics is a canonical(ly embedded) genus 5 curve.
Unfortunately, not all canonical genus 5 curves are the complete intersection of
3 quadrics in P4 . But in the same sense that most genus 3 curves can be described
as plane quartics, most canonical genus 5 curves are complete intersections of 3
quadrics, and most genus 5 curves are nonhyperelliptic. The correct way to say
this is that there is a dense Zariski-open locus in the moduli space of genus 5
curves consisting of nonhyperelliptic curves whose canonical embedding is cut
out by 3 quadrics.
(Those nonhyperelliptic genus 5 canonical curves not cut out by a three-dimensional
vector space of quadrics are precisely the trigonal curves, see Exercise 19.7.B. The
triplets of points mapping to the same point of P1 under the trigonal map turn out
to lie on a line in the canonical map. Any quadric vanishing along those 3 points

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must vanish along the line — basically, any quadratic polynomial with three zeros
must be the zero polynomial.)
19.8.F. E XERCISE . Assuming the discussion above, count complete intersectionsof three quadrics to give a heuristic argument suggesting that the curves of genus
5 “form a family of dimension 12”.
19.8.2. We have now understood curves of genus 3 through 5 by thinking of canonical curves as complete intersections. Sadly our luck has run out.
19.8.G. E XERCISE . Show that if C ⊂ Pg−1 is a canonical curve of genus g ≥ 6, thenC is not a complete intersection. (Hint: Bézout’s Theorem, Exercise 18.6.K.)
19.8.3. Some discussion on curves of general genus. However, we still have some data.
If Mg is this ill-defined “moduli space of genus g curves”, we have heuristics tofind its dimension for low g. In genus 0, over an algebraically closed field, there is
only genus 0 curve (Proposition 19.3.1), so it appears that dim M0 = 0. In genus 1,
over an algebraically closed field, we will soon see that the elliptic curves are classified by the j-invariant (Exercise 19.9.C), so it appears that dim M1 = 1. We havealso informally computed dim M2 = 3, dim M3 = 6, dim M4 = 9, dim M5 = 12.
What is the pattern? In fact in some strong sense it was known by Riemann that
dim Mg = 3g − 3 for g > 1. What goes wrong in genus 0 and genus 1? As a clue,recall our insight when discussing Hilbert functions (§18.6) that whenever somefunction is “eventually polynomial”, we should assume that it “wants to be polynomial”, and there is some better function (usually an Euler characteristic) thatis polynomial, and that cohomology-vanishing ensures that the original function
and the better function “eventually agree”. Making sense of this in the case of
Mg is far beyond the scope of our current discussion, so we will content ourselves
by observing the following facts. Every regular curve of genus greater than 1 hasa finite number of automorphisms — a zero-dimensional automorphism group.
Every regular curve of genus 1 has a one-dimensional automorphism group (seeQuestion 19.10.6). And the only regular curve of genus 0 has a three-dimensional
automorphism group (Exercise 16.4.C). (See Aside 21.5.12 for more discussion.) So
notice that for all g ≥ 0,
dim Mg − dim Aut Cg = 3g − 3

where Aut Cg means the automorphism group of any curve of genus g.
In fact, in the language of stacks (or orbifolds), it makes sense to say that thedimension of the moduli space of (smooth projective geometrically irreducible)
genus 0 curves is −3, and the dimension of the moduli space of genus 1 curves is0.

19.9 Curves of genus 1
Finally, we come to the very rich case of curves of genus 1. We will present thetheory by discussing interesting things about line bundles of steadily increasing
degree.
19.9.1. Line bundles of degree 0.

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Suppose C is a genus 1 curve. Then deg ωC = 2g − 2 = 0 and h0 (C, ωC ) =g = 1 (by Exercise 19.2.A). But the only degree 0 invertible sheaf with a section∼ OC . (If you know thatis the structure sheaf (§19.2.4), so we conclude that ωC =complex genus 1 curves are of the form C modulo a lattice, and Miracle 18.5.2 thatthe sheaf of differentials is dualizing, then you might not be surprised.)
We move on to line bundles of higher degree. Next, note that if deg L > 0,then Riemann-Roch and Serre duality (19.2.5.1) giveh0 (C, L ) = deg L − g + 1 = deg L .
19.9.2. Line bundles of degree 1.
Each degree 1 (k-valued) point q determines a line bundle O(q), and two distinct points determine two distinct line bundles (as a degree 1 line bundle has only
one section, up to scalar multiples). Conversely, any degree 1 line bundle L is of
the form O(q) (as L has a section — then just take its divisor of zeros).
Thus we have a canonical bijection between degree 1 line bundles and degree
1 (closed) points. (If k is algebraically closed, as all closed points have residue fieldk, this means that we have a canonical bijection between degree 1 line bundles andclosed points.)
Define an elliptic curve to be a genus 1 curve E with a choice of k-valued point
p. The choice of this point should always be considered part of the definition ofan elliptic curve — “elliptic curve” is not a synonym for “genus 1 curve”. (Note: agenus 1 curve need not have any k-valued points at all! For example, the genus 1
curve “compactifying” y2 = −x4 −1 in A2Q has no R-points, and hence no Q-points.
Of course, if k = k, then any closed point is k-valued, by the Nullstellensatz 3.2.4.)
We will often denote elliptic curves by E rather than C.
If (E, p) is an elliptic curve, then there is a canonical bijection between the setof degree 0 invertible sheaves (up to isomorphism) and the set of degree 1 pointsof E: simply the twist the degree 1 line bundles by O(−p). Explicitly, the bijection
is given by% div(L (p))
L ✤O(q − p) (✤q

But the degree 0 invertible sheaves form a group (under tensor product), so
we have proved:
19.9.3. Proposition (the group law on the degree 1 points of an elliptic curve). —The above bijection defines an abelian group structure on the degree 1 points of an elliptic
curve, where p is the identity.
From now on, we will identify closed points of E with degree 0 invertible
sheaves on E without comment.
For those familiar with the complex analytic picture, this is not surprising: E∼ C/Λ. (We haven’t
is isomorphic to the complex numbers modulo a lattice: E =shown this, of course.)
Proposition 19.9.3 is currently just a bijection of sets. Given that E has a muchricher structure (it has a generic point, and the structure of a variety), this is a signthat there should be a way of defining some scheme Pic0 (E), and that this should be

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an isomorphism of schemes. We will soon show (Theorem 19.10.4) that this groupstructure on the degree 1 points of E comes from a group variety structure on E.
19.9.4. Aside: The Mordell-Weil Theorem, group, and rank. This is a good excuse tomention the Mordell-Weil Theorem: for any elliptic curve E over Q, the Q-points of
E form a finitely generated abelian group, often called the Mordell-Weil group. By theclassification of finitely generated abelian groups (a special case of the classification of finitely generated modules over a principal ideal domain, Remark 12.5.14),
the Q-points are a direct sum of a torsion part, and of a free Z-module. The rankof the Z-module is called the Mordell-Weil rank.
19.9.5. Line bundles of degree 2 (see Figure 19.4).
Note that OE (2p) has 2 sections, so E admits a double cover of P1 (Exercise 19.5.B).
One of the ramification points is p: one of the sections of OE (2p) vanishes at p toorder 2, so there is a point of P1 whose preimage consists of p (with multiplicity 2).
Assume now that k = k and char k ̸= 2, so we can use the hyperelliptic RiemannHurwitz formula (Theorem 19.5.1), which implies that E has 4 ramification points(p and three others). Conversely, given 4 points in P1 , there exists a unique double
cover branched at those 4 points (Proposition 19.5.2). Thus elliptic curves correspond to 4 distinct points in P1 , where one is marked p, up to automorphisms of
P1 . Equivalently, by placing p at ∞, elliptic curves correspond to 3 points in A1 ,up to affine maps x (→ ax + b.

F IGURE 19.4. Elliptic curves yield double covers of the projective
line, branched at four points (§19.9.5)

19.9.A. E XERCISE . Show that the other three ramification points are precisely the(non-identity) 2-torsion points in the group law. (Hint: if one of the points is q,∼ O(2p), but O(q) is not congruent to O(p).)
show that O(2q) =Thus (if char k ̸= 2 and k = k) every elliptic curve has precisely four 2-torsion∼ C/Λ, this isn’t surprising.
points. If you are familiar with the complex picture E =19.9.6. Follow-up remark. An elliptic curve with full level n structure is an elliptic curve with an isomorphism of its n-torsion points with (Z/n)2 . (This notion
has problems if n is divisible by char k.) Thus an elliptic curve with full level 2
structure is the same thing as an elliptic curve with an ordering of the three other

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branch points in its degree 2 cover description. Thus (if k = k) these objects areparametrized by the λ-line, which we discuss below.
Follow-up to the follow-up. There is a notion of moduli spaces of elliptic curves
with full level n structure. Such moduli spaces are smooth curves (where this isinterpreted appropriately — they are stacks), and have smooth compactifications.
A weight k level n modular form is a section of ω⊗k
M where ωM is the dualizing (orcanonical) sheaf of this moduli space (“modular curve”).
19.9.7. The cross-ratio and the j-invariant. (We maintain the assumption that k is
algebraically closed and characteristic not 2.) If the three other points are temporarily labeled q1 , q2 , q3 , there is a unique automorphism of P1 taking p, q1 , q2
to (∞, 0, 1) respectively (as Aut P1 is three-transitive, Exercise 16.4.C). Supposethat q3 is taken to some number λ under this map, where necessarily λ ̸= 0, 1, ∞.
The value λ is called the cross-ratio of the four points (p, q1 , q2 , q3 ) of P1 (first
defined by W. Clifford, but implicitly known since the time of classical Greece).
19.9.B. E XERCISE . Show that isomorphism classes of four ordered distinct (closed)
points on P1 (over an algebraically closed field k), up to projective equivalence(automorphisms of P1 ), are classified by the cross-ratio.
We have not defined the notion of moduli space, but the previous exercise illustrates the fact that P1 − {0, 1, ∞} (the image of the cross-ratio map) is the moduli
space for four ordered distinct points of P1 up to projective equivalence.
Notice:• If we had instead sent p, q2 , q1 to (∞, 0, 1), then q3 would have been sent
to 1 − λ.
• If we had instead sent p, q1 , q3 to (∞, 0, 1), then q2 would have been sent
to 1/λ.
• If we had instead sent p, q3 , q1 to (∞, 0, 1), then q2 would have been sent
to 1 − 1/λ = (λ − 1)/λ.
• If we had instead sent p, q2 , q3 to (∞, 0, 1), then q1 would have been sent
to 1/(1 − λ).
• If we had instead sent p, q3 , q2 to (∞, 0, 1), then q1 would have been sent
to 1 − 1/(1 − λ) = λ/(λ − 1).
Thus these six values (which correspond to S3 , the symmetric group permuting q1 , q2 , and q3 ) yield the same elliptic curve, and this elliptic curve will (uponchoosing an ordering of the other 3 branch points) yield one of these six values.
This is fairly satisfactory already. To check if two elliptic curves (E, p), (E ′ , p ′ )over k = k are isomorphic, we write both as double covers of P1 ramified at p
and p ′ respectively, then order the remaining branch points, then compute theirrespective λ’s (say λ and λ ′ respectively), and see if they are related by one of thesix expressions above:(19.9.7.1)

λ ′ = λ, 1/λ, 1 − λ, 1/(1 − λ), (λ − 1)/λ, or λ/(λ − 1).

It would be far more convenient if, instead of a “six-valued invariant” λ, there
were a single invariant (let’s call it j), such that j(λ) = j(λ ′ ) if and only if one of
the equalities of (19.9.7.1) holds. This j-function should presumably be algebraic,
so it would give a map j from the λ-line A1 − {0, 1} to the A1 . By the Curve-toProjective Extension Theorem 16.5.1, this would extend to a morphism j : P1 → P1 .

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By Exercise 17.4.D, because this is (for most λ) a 6-to-1 map, the degree of thiscover is 6 (or more correctly, at least 6).
We can make this dream more precise as follows. The elliptic curves over kcorrespond to k-valued points of P1 − {0, 1, ∞}, modulo the action of S3 on λ given
above. Consider the subfield K of k(λ) fixed by S3 . Then the extension k(λ)/Kis necessarily Galois (see for example [DF, §14.2, Thm. 9]), and degree 6. We arehoping that this subfield is of the form k(j), and if so, we would obtain the j-map
P1 → P1 as described above. One could show that K is finitely generated over k,
and then invoke Lüroth’s Theorem, which we will soon prove in Example 21.7.6;
but we won’t need this.
Instead, we will just hunt for such a j. Note that λ should satisfy a sextic polynomial over k(λ) (or more precisely given what we know right now, a polynomialof degree at least six), as for each j-invariant, there are six values of λ in general.
As you are undoubtedly aware, there is such a j-invariant. Here is the formulafor the j-invariant that everyone uses:(19.9.7.2)

j = 28(λ2 − λ + 1)3.
λ2 (λ − 1)2

You can readily check that j(λ) = j(1/λ) = j(1 − λ) = · · · , and that as j (inlowest terms) has a degree 6 numerator and degree < 6 denominator, j indeeds
determines a degree 6 map from P1 (with coordinate λ) to P1 (with coordinate j).
But this complicated-looking formula begs the question: where did this formula
come from? How did someone think of it? We will largely answer this, but we will
ignore the 28 . This, as you might imagine, arises from characteristic 2 issues, andin order to invoke the results of §19.5 we have been assuming char k ̸= 2. (To see
why the 28 is forced upon us by characteristic 2, see [De, p. 64]. From a different,
complex-analytic, point of view, the 28 comes from the fact that the q-expansion of
j begins j = q−1 + 744 + · · · : the q−1 term has coefficient 1; see [Se5, p. 90, Rem. 2].
These seemingly different explanations are related by the theory of the Tate curve,see [Si2, Thm. V.3.1].)
Rather than using the formula handed to us, let’s try to guess what j is. We
won’t expect to get the same formula as (19.9.7.2), but our answer should differ by
an automorphism of the j-line (P1 ) — we will get j ′ = (aj + b)/(cj + d) for some a,b, c, d (Exercise 16.4.B).
We are looking for some j ′ (λ) such that j ′ (λ) = j ′ (1/λ) = · · · . Hence wewant some expression in λ that is invariant under this S3 -action. A first possibility
would be to take the product of the six numbers
λ · (1 − λ) ·

1 λ−1
1
λ
·
·
·
λ
λ
1−λ λ−1

This is silly, as the product is obviously 1.
A better idea is to add them all together:λ + (1 − λ) +

1 λ−1
1
λ
+
+
+
λ
λ
1−λ λ−1

This also doesn’t work, as they add to 3 — the six terms come in pairs adding to 1.
(Another reason you might realize this can’t work: if you look at the sum, youwill realize that you will get something of the form “degree at most 3” divided by“degree at most 2”. Then if j ′ = p(λ)/q(λ), then λ is a root of a cubic over j. But weEarly (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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said that λ should satisfy a sextic over j ′ . The only way we avoid a contradiction
is if j ′ ∈ k.)
But you will undoubtedly have another idea immediately. One good idea is
to take the second symmetric function in the six roots. An equivalent one that iseasier to do by hand is to add up the squares of the six terms. Even before doing
the calculation, we can see that this will work: it will clearly produce a fraction
whose numerator and denominator have degree at most 6, and it is not constant,
as when λ is some fixed small number (say 1/2), the sum of squares is some small
real number, while when λ is a large real number, the sum of squares will have to
be some large real number (different from the value when λ = 1/2).
When you add up the squares by hand (which is not hard), you will getj′ =2λ6 − 6λ5 + 9λ4 − 8λ3 + 9λ2 − 6λ + 2.
λ2 (λ − 1)2∼ k(j ′ ): you can check (again by hand) thatIndeed k(j) =2j/28 =

2λ6 − 6λ5 + 12λ4 − 14λ3 + 12λ2 − 6λ + 2.
λ2 (λ − 1)2

Thus 2j/28 − j ′ = 3.
19.9.C. E XERCISE . Explain why genus 1 curves over an algebraically closed fieldof characteristic not 2 are classified (up to isomorphism) by j-invariant. (Caution:note that the problem says “genus 1”, not “elliptic”.)
19.9.D. E XERCISE . Give (with proof) two elliptic curves over Q with the same
j-invariant that are not isomorphic. (Hint: §19.5.3.)
19.9.8. Line bundles of degree 3.
In the discussion of degree 2 line bundles 19.9.5, we assumed char k ̸= 2 andk = k, in order to invoke the Riemann-Hurwitz formula. In this section, we will
start with no assumptions, and add them as we need them. In this way, you will
see what partial results hold with weaker assumptions.
Consider the degree 3 invertible sheaf OE (3p). By Riemann-Roch (19.2.5.1),
h0 (E, OE (3p)) = deg(3p)−g+1 = 3. As deg E > 2g, this gives a closed embedding(Remark 19.2.11 and Exercise 19.2.D). Thus we have a closed embedding E "→P2k as a cubic curve. Moreover, there is a line in P2k meeting E at point p withmultiplicity 3, corresponding to the section of O(3p) vanishing precisely at p withmultiplicity 3. (A line in the plane meeting a smooth curve with multiplicity atleast 2 is a tangent line, see Definition 12.3.2. A line in the plane meeting a smooth
curve with multiplicity at least 3 is said to be a flex line, and that point is a flex
point or flex of the curve.) (See Figure 19.5.)

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F IGURE 19.5. Finding the Weierstrass cubic incarnation of an elliptic curve
Choose projective coordinates on P2k so that p maps to [0, 1, 0], and the flex lineis the line at infinity z = 0. Then the cubic is of the following form:? x3+

0 x2 y

+

0 xy2

+

+

? x2 z

+

? xyz

+

? y2 z

+

? xz2

+

? yz2

+

? z3

0 y3

=0

The coefficient of x3 is not 0 (or else this cubic is divisible by z). Dividing theentire equation by this coefficient, we can assume that the coefficient of x3 is 1.
The coefficient of y2 z is not 0 either (or else this cubic is singular at x = z = 0).
We can scale z (i.e., replace z by a suitable multiple) so that the coefficient of y2 z
is −1. If the characteristic of k is not 2, then we can then replace y by y + ?x + ?z
so that the coefficients of xyz and yz2 are 0, and if the characteristic of k is not 3,
we can replace x by x + ?z so that the coefficient of x2 z is also 0. In conclusion, if
char k ̸= 2, 3, the elliptic curve may be written as(19.9.8.1)

y2 z = x3 + axz2 + bz3 .

This is called the Weierstrass normal form of the curve.
19.9.9. From the Weierstrass cubic to the double cover. We see the hyperelliptic (double
cover) description of the curve by setting z = 1, or more precisely, by working in
the distinguished open set z ̸= 0 and using inhomogeneous coordinates. Here is
the geometric explanation of why the double cover description is visible in the
cubic description. Project the cubic from p = [0, 1, 0] (see Figure 19.6). This is a

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map E − p → P1 (given by [x, y, z] ""# [x, z]), and is basically the vertical projectionof the cubic to the x-axis. (Figure 19.7 may help you visualize this.) By the Curveto-Projective Extension Theorem 16.5.1, the morphism E − p → P1 extends over p.
If L := O(3p) is the line bundle giving the morphism E "→ P2 (via Theorem 16.4.1
describing maps to projective space in terms of line bundles), then the two sections
x and z giving the map to P1 vanish at p to order 1 and 3 respectively. To “resolve”
the rational map into an honest morphism, we interpret x and z as sections ofL (−p) = O(2p), and now they generate a base-point-free linear series, and thus amorphism to P1 . (You may be able to interpret this as implementing the proof of
Theorem 16.4.1 in a different language.) A similar idea, applied to quartics rather
than cubics, was used in Exercise 19.7.B.

F IGURE 19.6. From the Weierstrass cubic to the double cover interpretation of an elliptic curve
As a consequence, with a little sweat, we can compute the j-invariant as afunction of a and b:
28 33 a3j(a, b) =.
4a3 + 27b2
19.9.E. E XERCISE . Show that the flexes of the cubic are the 3-torsion points in thegroup E. (“Flex” was defined in §19.9.8: it is a point where the tangent line meetsthe curve with multiplicity at least 3 at that point. In fact, if k is algebraically closed
and char k ̸= 3, there are nine of them. This won’t be surprising if you are familiar∼ C/Λ.)
with the complex story, E =19.9.10. The group law, geometrically.
The group law has a beautiful classical description in terms of the Weierstrass
form. Consider Figure 19.7. In the Weierstrass coordinates, the origin p is the only
point of E meeting the line at infinity (z = 0); in fact the line at infinity corresponds
to the tautological section of O(3p). If a line meets E at three points p1 , p2 , p3 , then∼ O(3p)O(p1 + p2 + p3 ) =from which (in the group law) p1 + p2 + p3 = 0.

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531

Hence to find the inverse of a point s, we consider the intersection of E withthe line sp; −s is the third point of intersection. To find the sum of two points q
and r, we consider the intersection of E with the line qr, and call the third point s.
We then compute −s by connecting s to p, obtaining q + r.

F IGURE 19.7. The group law on the elliptic curve, geometrically
We could give this description of a group law on a cubic curve in Weierstrassnormal form to anyone familiar with the notion of projective space, and the notionof a group, but we would then have to prove that the construction we are giving
indeed defines a group. In particular, we would have to prove associativity, which
is not a priori clear. But in this case, we have already established that the degree 1
points form a group, by giving a bijection to Pic0 E, and we are merely interpreting
the group law on Pic0 E.
Note that this description works even in characteristic 2 and 3; we do not need
the cubic to be in Weierstrass normal form, and we need only that O(3p) gives aclosed embedding into P2 .
19.9.11. Degree 4 line bundles. You have probably forgotten that we began by
studying line bundles degree by degree. The story doesn’t stop in degree 3. In thesame way that we showed that a canonically embedded nonhyperelliptic curve of
genus 4 is the complete intersection in P3k of a quadric surface and a cubic surface(§19.8), we can show the following.
19.9.F. E XERCISE . Show that the complete linear series for O(4p) embeds E in
P3 as the complete intersection of two quadrics. Hint: Show the image of E iscontained in at least 2 linearly independent quadrics. Show that neither can be
reducible, so they share no components. Use Bézout’s Theorem (Exercise 18.6.K)
and Exercise 18.6.F, as in §19.8.1.

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The beautiful structure doesn’t stop with degree 4, but it gets more complicated. For example, the degree 5 embedding is not a complete intersection (ofhypersurfaces), but is the complete intersection of G(2, 5) under its Plücker embedding with five hyperplanes (or perhaps better, a codimension 5 linear space).
In seemingly different terminology, its equations are 4×4 Pfaffians of a general 5×5
skew-symmetric matrix of linear forms, although I won’t say what this means.

19.10 Elliptic curves are group varieties
Ever since 1949, I considered the construction of an algebraic theory of the Picard
variety as the task of greatest urgency in abstract algebraic geometry.
— A. Weil [Weil, p. 537]
We initially described the group law on the degree 1 points of an algebraic
curve in a rather abstract way. From that definition, it was not clear that over C thegroup operations (addition, inverse) are continuous. But the explicit description interms of the Weierstrass cubic makes this clear. In fact we can observe even more:
addition and inverse are algebraic in general. Better yet, elliptic curves are groupvarieties. (Thus they are abelian varieties, see Definition 10.3.11.)
19.10.1. (This is a clue that Pic0 (E) really wants to be a scheme, and not just agroup. Once the notion of “moduli space of line bundles on a variety” is madeprecise, this can be shown, see for example [FGIKNV, Part 5].)We begin with the “group inverse” morphism, as a warm-up.
19.10.2. Proposition. — If char k ̸= 2, 3, there is a morphism of k-varieties E → E
sending a (degree 1) point to its inverse, and this construction behaves well under field
extension of k.
In other words, the “inverse map” in the group law actually arises from a morphism of schemes — it isn’t just a set map. (You are welcome to think through thetwo remaining characteristics, and to see that essentially the same proof applies.
But the proof of Theorem 19.10.4 will give you a better sense of how to proceed.)
Proof. Consider the map (the hyperelliptic involution) y (→ −y of the Weierstrassnormal form.
!

The algebraic description of addition would be a big mess if we were to write
it down. We will be able to show algebraicity by a trick — not by writing it down
explicitly, but by thinking through how we could write it down explicitly. The
main part of the trick is the following proposition. We give it in some generality
just because it can be useful, but you may prefer to assume that k = k and C is aregular cubic.

19.10.3. Proposition. — Suppose C ⊂ P2k is a geometrically integral cubic curve (soin particular C contains no lines). Let Creg be the regular points of C. There is a uniquemorphism t : Creg × Creg → Creg such that

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533(a) if p and q are distinct regular k-valued points of C, then t(p, q) is obtained byintersecting the line pq with C, and taking the third “residual” point of intersection with C. More precisely, pq will meet C at three points with multiplicity(Exercise 8.2.E), including p and q; t(p, q) is the third point; and(b) this property remains true after extension to k.
Furthermore, if p is a k-valued point of Creg , then t(p, p) is where the tangent line ℓ to C
at p meets C again. More precisely, ℓ will meet C at three points with multiplicity, which
includes p with multiplicity 2; t(p, p) is the third point.
We will need property (b) because C may have few enough k-valued points(perhaps none!) that the morphism t can not be determined by its behavior on
them. In the course of the proof, we will see that (b) can be extended to “thisproperty remains true after any field extension of k”.
Proof. We first show (in this paragraph) that if p and q are distinct regular points,
then the third point r of intersection of pq with C is also regular. If r = p or r = q,
we are done. Otherwise, the cubic obtained by restricting C to pq has three distinct(hence reduced, i.e., multiplicity 1) roots, p, q, and r. Thus C ∩ pq is regular at r,
so r is a regular point of C by the slicing criterion for regularity, Exercise 12.2.C.
We now assume that k = k, and leave the general case to the end. Fix p, q,
and r, where p ̸= q, and r is the “third” point of intersection of pq with C. We will
describe a morphism tp,q in an open neighborhood of (p, q) ∈ Creg × Creg . By Exercise 10.2.B, showing that morphisms of varieties over k are determined by their
behavior on closed (k-valued) points, these morphisms glue together (uniquely)
to give a morphism t, completing the proof in the case k = k.
∼ Spec k[x1 , x2 ]Choose projective coordinates on P2 in such a way that U0 =contains p, q, and r, and the line pq is not “vertical”. More precisely, in Spec k[x1 , x2 ],say p = (p1 , p2 ) (in terms of “classical coordinates” — more pedantically, p =[(x1 −p1 , x2 −p2 )]), q = (q1 , q2 ), r = (r1 , r2 ), and p1 ̸= q1 . In these coordinates, thecurve C is cut out by some cubic, which we also sloppily denote C: C(x1 , x2 ) = 0.
Now suppose P = (P1 , P2 ) and Q = (Q1 , Q2 ) are two points of C ∩ U0 (notnecessarily our p and q). We attempt to compute the third point of intersection of
PQ with C, in a way that works on an open subset of C × C that includes (p, q). To
do this explicitly requires ugly high school algebra, but because we know how it
looks, we will be able to avoid dealing with any details!
2
The line PQ is given by x2 = mx1 + b, where m = PP21 −Q−Q1 and b = P2 − mP1
are both rational functions of P and Q. Then m and b are defined for all P and Q
such that P1 ̸= Q1 (and hence for an open neighborhood of (p, q), as p1 ̸= q1 , andas P1 ̸= Q1 is an open condition).
Now we solve for C∩PQ, by substituting x2 = mx1 +b into C, to get C(x1 , mx1 +b). This is a cubic in x1 , sayγ(x1 ) = Ax31 + Bx21 + Cx1 + D = 0.
The coefficients of γ are rational functions of P1 , P2 , Q1 , and Q2 . The cubic γ has 3
roots (with multiplicity) so long as A ̸= 0 , which is a Zariski-open condition on m
and b, and hence a Zariski-open condition on P1 , P2 , Q1 , Q2 . As P, Q ∈ C∩PQ∩U0 ,P1 and Q1 are two of the roots of γ(x1 ) = 0. The sum of the roots of γ(x1 ) = 0
is −B/A (by Viète’s formula), so the third root of γ is R1 := −B/A − P1 − Q1 .
Thus if we take R2 = mR1 + b, we have found the third point of intersection of

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PQ with C (which happily lies in U0 ) We have thus described a morphism fromthe open subset of (Creg ∩ U0 ) × (Creg ∩ U0 ), containing (p, q), that does what wewant. (Precisely, the open subset is defined by A ̸= 0, which can be explicitly
unwound.) We have thus completed the proof of Proposition 19.10.3 (except for
the last paragraph) for k = k. (Those who believe they are interested only inalgebraically closed fields can skip ahead.)
We extend this to Proposition 19.10.3 for every field k except F2 . Suppose[x1 ,x2 ]U0= Spec k[x1 , x2 ] is any affine open subset of P2k , along with choice of coordinates. (The awkward superscript “[x1 , x2 ]” is there to emphasize that the particular coordinates are used in the construction.) Then the construction above gives a[x ,x ]
[x ,x ]
morphism defined over k from an open subset of (Creg ∩ U0 1 2 ) × (Creg ∩ U0 1 2 )(note that all of the hypothetical algebra was done over k), that sends P and Q to
the third point of intersection of PQ with C. Note that this construction commuteswith any field extension, as the construction is insensitive to the field we are working over. Thus after base change to the algebraic closure, the map also has theproperty that it takes as input two points, and spits out the third point of intersec[x ,x ]tion of the line with the cubic. Furthermore, all of these maps (as U0 1 2 varies
over all complements U0 of lines “with k-coefficients”, and choices of coordinates
on U0 ) can be glued together: they agree on their pairwise overlaps (as after basechange to k they are the same, by our previous discussion, and two maps that are
the same after base change to k were the same to begin with by Exercise 9.2.I), andthis is what is required to glue them together (Exercise 6.2.A).
[x ,x ]
We can geometrically interpret the open subset (Creg ∩ U0 1 2 ) × (Creg ∩[x ,x ]
U0 1 2 ) by examining the construction: it is defined in the locus {P = (P1 , P2 ), Q =(Q1 , Q2 )} where (i) P1 ̸= Q1 , and (ii) the third point of intersection R of PQ with C
also lies in U0 .
So which points (P, Q) of Creg × Creg are missed? Condition (i) isn’t important;if (P, Q) satisfies (ii) but not (i), we can swap the roles of x1 and x2 , and (P, Q) willthen satisfy (i). The only way (P, Q) can not be covered by one of these open sets
is if there is no U0 (a complement of a line defined over k) that includes P, Q, andR.
19.10.A. E XERCISE . Use |k| > 2 to show that there is a linear form on P2 withcoefficients in k that misses P, Q, and R. (This is sadly not true if k = F2 — do you
see why?)
19.10.B. E XERCISE . Prove the last statement of Proposition 19.10.3.
19.10.C. ⋆⋆ U NIMPORTANT E XERCISE . Complete the proof by dealing with thecase k = F2 . Hint: first produce the morphism t over F4 . The goal is then toshow that this t is really “defined over” F2 (“descends to” F2 ). The morphism
t is initially described locally by considering the complement of a line definedover F4 (and then letting the line vary). Instead, look at the map by looking at
the complement of a line and its “conjugate”. The complement of the line andits conjugate is an affine F2 -variety. The partially-defined map t on this affine
variety is a priori defined over F4 , and is preserved by conjugation. Show that thispartially defined map is “really” defined over F2 . (If you figure out what all of thismeans, you will have an important initial insight into the theory of “descent”.)

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We can now use this to define the group variety structure on E.
19.10.4. Theorem. — Suppose (E, p) is an elliptic curve (a regular genus 1 curve overk, with a k-valued point p). Take the Weierstrass embedding of E in P2k , via the complete
linear series |OE (3p)|. Define the k-morphism e : Spec k → E by sending Spec k to p.
Define the k-morphism i : E → E via q (→ t(p, q), or more precisely, as the composition
E(id,e)

% E×E

t

% E.

Define the k-morphism m : E × E → E via (q, r) (→ t(p, t(q, r)). Then (E, e, i, m) is agroup variety over k.

By the construction of t, all of these morphisms “commute with arbitrary baseextension”.

Proof. We need to check that various pairs of morphisms described in axioms (i)–(iii) of §6.6.4 are equal. For example, in axiom (iii), we need to show that m ◦(i, id) = m ◦ (id, i); all of the axioms are clearly of this sort.
Assume first that k = k. Then each of these pairs of morphisms agree asmaps of k-points: Pic E is a group, and under the bijection between Pic E and E of
Proposition 19.9.3, the group operations translate into the maps described in thestatement of Theorem 19.10.4 by the discussion of §19.9.10.
But morphisms of k-varieties are determined by their maps on the level of
k-points (Exercise 10.2.B), so each of these pairs of morphisms are the same.
For general k, we note that from the k case, these morphisms agree after basechange to the algebraic closure. Then by Exercise 9.2.I, they must agree to begin
with.
!
19.10.5. Features of this construction. The most common derivation of the properties of an elliptic curve is to describe it as a cubic, and then to describe addition
using the explicit construction with lines. Then one has to work to prove that themultiplication described is associative.
Instead, we started with something that was patently a group (the degree 0line bundles). We interpreted the maps used in the definition of the group (addition and inverse) geometrically using our cubic interpretation of elliptic curves.
This allowed us to see that these maps were algebraic.
As a bonus, we see that in some vague sense, the Picard group of an elliptic
curve wants to be an algebraic variety.
19.10.D. E XERCISE . Suppose p and q are k-points of a genus 1 curve E. Show that
there is an automorphism of E sending p to q.
19.10.E. E XERCISE . Suppose (E, p) is an elliptic curve over an algebraically closed
field k of characteristic not 2 or 3. Show that the automorphism group of (E, p)
is isomorphic to Z/2, Z/4, or Z/6. (An automorphism of an elliptic curve (E, p)
over k = k is an automorphism of E fixing p scheme-theoretically, or equivalently, fixing the k-valued point p by Exercise 10.2.B.) Hint: reduce to the question of automorphisms of P1 fixing a point ∞ and a set of distinct three points{p1 , p2 , p3 } ∈ P1 \ {∞}. (The algebraic closure of k is not essential, so feel free to

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remove this hypothesis, using Exercise 9.2.I. What happens if the characteristic is3?)
19.10.6. Vague question. What are the possible automorphism groups of a genus 1
curve over an algebraically closed k of characteristic not 2? You should be able toconvince yourself that the group has “dimension 1”.
19.10.F. I MPORTANT E XERCISE : A DEGENERATE ELLIPTIC CURVE . Consider the
genus 1 curve C ⊂ P2k given by y2 z = x3 + x2 z, with the point p = [0, 1, 0]. Emulate
the above argument to show that C \ {[0, 0, 1]} is a group variety (where the operation is that given by tensoring line bundles). Show that it is isomorphic to Gm(the multiplicative group scheme Spec k[t, t−1 ], see Exercise 6.6.E) with coordinatet = y/x, by showing an isomorphism of schemes, and showing that multiplicationand inverse in both group varieties agree under this isomorphism.
19.10.G. E XERCISE : AN EVEN MORE DEGENERATE ELLIPTIC CURVE . Consider the
genus 1 curve C ⊂ P2k given by y2 z = x3 , with the point p = [0, 1, 0]. Emulate
the above argument to show that C \ {[0, 0, 1]} is a group variety. Show that it
is isomorphic to A1 (with additive group structure) with coordinate t = x/y, byshowing an isomorphism of schemes, and showing that multiplication/addition
and inverse in both group varieties agree under this isomorphism.
19.10.7. ⋆ Towards Poncelet’s Porism.
These ideas lead to a beautiful classical fact, Poncelet’s Porism. (A porism is anarchaic name for a type of mathematical result.)

F IGURE 19.8. Poncelet’s Porism 19.10.8
19.10.8. Poncelet’s Porism. — Suppose C and D are two ellipses in R2 , with C containing D. Choose any point p0 on C. Choose one of the two tangents ℓ1 from p0 to D. Thenℓ1 meets C at two points in total: p0 and another point p1 . From p1 , there are two tangents to D, ℓ1 and another line ℓ2 . The line ℓ2 meets C at some other point p2 . Continue
this to get a sequence of points p0 , p1 , p2 , . . . . Suppose this sequence starting with p0 isperiodic, i.e., p0 = pn for some n. Then it is periodic with any starting point p ∈ C (seeFigure 19.8).
It is possible to prove Poncelet’s Porism in an elementary manner, but a proof
involving elliptic curves is quite beautiful, and gives connections to more sophisticated ideas. Rather than proving Poncelet’s Porism, we discuss some related facts.

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19.10.H. E XERCISE . Suppose E is a smooth degree 3 projective plane curve overan algebraically closed field, q, r ∈ E, and n is a positive integer. Let F : E → E
be the morphism x (→ t(r, t(q, x)). Suppose that Fn (p) = p for some closed pointp ∈ E. Show that Fn (x) = x for all points x ∈ E.
You should feel the urge to improve this result. (Can you show that if Fn (p) =p for some p ∈ E, then Fn is the identity morphism on E? Can you extend this tothe case where the base field is not algebraically closed?)

19.10.I. E XERCISE . Suppose E is an ellipse in R2 . For each point (x, y) ∈ E, there isanother point G(x, y) := (x, y ′ ) (possibly the same), with the same first coordinate,
corresponding to the “other” intersection of the vertical line through (x, y) with E.
Similarly, there is another point H(x, y) := (x ′ , y), with the same second coordinate.
Let F = G ◦ H. Show that if F17 (p) = p for some point p ∈ E, then F17 (x) = x for all
points x ∈ E. (See Figure 19.9, where 17 is replaced by 3.) Hint: this might be best
done not using any fancy methods from algebraic geometry.
Again, you should wish to improve this. To what extent is this dependent on
the real numbers? What if the ellipse was instead a different conic section? (Andwhat does this have to do with elliptic curves? Why is the conic called E?!)F IGURE 19.9. Exercise 19.10.I (with 17 replaced by 3)

19.10.J. ⋆⋆ E XERCISE . Let k be a field, algebraically closed purely for convenience.
Suppose Q is a smooth quadric surface in P3k , and K is a cone (a rank 3 quadric) in
P3 , such that E = Q ∩ K is a smooth curve.
(a) Show that E has genus 1.
(b) If ℓ is a line on Q, show that E meets ℓ at two points (with multiplicity).
(c) Fix one of the rulings (family of lines) of Q. Show that there exists a morphism
G : E → E that takes a closed point p of E, and sends it to the other point of E on
the line ℓ in the ruling containing p.
(d) Define H : E → E similarly, using the other ruling of Q. Fix a positive integer n.

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Let F = G ◦ H. Show that if Fn (p) = p for one closed point p ∈ E, then Fn (x) = x
for all x ∈ E.

We now connect Exercise 19.10.J to Poncelet’s Porism 19.10.8. Let v be thecone point of K. Then v ∈/ Q, as otherwise v would be a singular point of Q ∩ K(do you see why?). It is a fact that projection from v gives a morphism Q → P2k
that is branched over a smooth conic D. (You may be able to make this precise,and prove it, after reading Chapter 21.) The lines on Q project to tangent lines
to D in P2k . The projection from v contracts K \ {v} to a conic C. You may beable to interpret the statement of Poncelet’s Porism 19.10.8 (for this C and D) interms of Exercise 19.10.J. (Caution: the field is not algebraically closed in Poncelet’s
Porism 19.10.8, and is algebraically closed in Exercise 19.10.J.) You may even beable to take an arbitrary C and D as in Poncelet’s Porism 19.10.8, and reverseengineer K and Q as in Exercise 19.10.J, thereby proving Poncelet’s Porism (anddiscovering why there is an elliptic curve hidden in its statement).

19.11 Counterexamples and pathologies using elliptic curves
We now give some fun counterexamples using our understanding of elliptic
curves. The main extra juice elliptic curves give us comes from the fact that elliptic
curves are the simplest varieties with “continuous Picard groups”.
19.11.1. An example of a scheme that is factorial, but such that no affine open
neighborhood of any point has ring that is a unique factorization domain.
Suppose E is an elliptic curve over C (or some other uncountable algebraicallyclosed field). Consider p ∈ E. The local ring OE,p is a discrete valuation ringand hence a unique factorization domain (Theorem 12.5.8). Then any affine open
neighborhood of E is of the form E − q1 − · · · − qn for some points q1 , . . . , qn . I
claim that its Picard group is nontrivial. Recall the exact sequence:Z⊕n(a1 ,...,an )*→a1 q1 +···+an qn% Pic E% Pic(E − q1 − · · · − qn )%0.

But the group on the left is countable, and the group in the middle is uncountable,
so the group on the right, Pic(E − q1 − · · · − qn ), is nonzero. We have shownthat every nonempty open subset of E has nonzero line bundles, as promised in
Remark 13.1.8.
As E − q1 − · · · − qn is affine, by Exercise 14.2.T the corresponding ring is not aunique factorization domain. To summarize: complex elliptic curves are factorial,
but no affine open subset has a ring that has unique factorization, as promised in§5.4.5.
19.11.A. E XERCISE . The above argument shows that over an uncountable field,
Pic E is not a finitely generated group. Show that even over the countable field Q,
Pic E is not a finitely generated group, as follows. If the elliptic curve E is generated
by q1 , . . . , qn , then there is a finite extension of fields K of Q over which all qi aredefined (the compositum of the residue fields of the qi ). Show that any point in
the subgroup of E generated by the qi must also be defined over K. Show that E

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has a point not defined over K. Use this to show that Pic E is not finitely generated.
(The same argument works with Q replaced by Fp .)19.11.2. Remark. In contrast to the above discussion, over Q, the Mordell-Weil
Theorem states that Pic E is finitely generated (see Aside 19.9.4).
19.11.3. ⋆⋆ A complex surface with infinitely many algebraic structures (for thosewith sufficient complex geometric background).
As remarked in Exercise 6.3.N, a complex manifold may have many algebraic
structures. The following example of M. Kim gives an example with infinitely
many algebraic structures. Suppose E is an elliptic curve over C with origin p,
and let L be a nontrivial line bundle on E − p. Then L is analytically trivial because E − p is a Stein space, so the analytification of the total space is independentof the choice of L . However, you know enough to show (with work) that there areinfinitely many pairwise (algebraically) nonisomorphic L , and also that their total
spaces are likewise pairwise (algebraically) nonisomorphic. This gives a complex
surface with infinitely many algebraic structures (indeed a continuum of them).
See [MO68421] for more details.
19.11.4. Counterexamples using a non-torsion point.
We next give a number of counterexamples using the existence of a non-torsion
point of a complex elliptic curve. We first show the existence of such a point.
19.11.5. We have a “multiplication by n” map [n] : E → E, which sends p to np.
If n = 0, the map [n] has degree 0 (even though [n] isn’t dominant, degree is stilldefined, see Definition 11.2.2). If n = 1, the map [n] has degree 1. Given thecomplex picture of a torus, you might not be surprised that the degree of [n] is n2 .
If n = 2, we have almost shown that it has degree 4, as we have checked that there
are precisely 4 points q such that 2p = 2q (Exercise 19.9.A). All that really shows(using Exercise 17.4.D(b)) is that the degree is at least 4. (We could check by handthat the degree is 4 is we really wanted to.)19.11.6. Proposition. — Suppose E is an elliptic curve over a field k of characteristic not
2. For each n > 0, the “multiplication by n” morphism [n] has positive degree, so there
are only a finite number of n-torsion points.
Proof. We may assume k = k, as the degree of a map of curves is independent offield extension.
We prove the result by induction; it is true for n = 1 and n = 2.
If n is odd (2k + 1, say), then assume otherwise that nq = 0 for all closed
points q. Let r be a non-trivial 2-torsion point, so 2r = 0. But nr = 0 as well, sor = (n − 2k)r = 0, contradicting r ̸= 0.
If n is even, then [n] = [2] ◦ [n/2] (degree is multiplicative under compositionof rational maps, §11.2.2), and by our inductive hypothesis both [2] and [n/2] havepositive degree.
!
In particular, the total number of torsion points on E is countable. If k is an
uncountable field, then E has an uncountable number of closed points (consider anopen subset of the curve as y2 = x3 + ax + b; there are uncountably many choices
for x, and each of them has 1 or 2 choices for y). Thus we have the following.

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19.11.7. Corollary. — If E is a curve over an uncountable algebraically closed field of
characteristic not 2 (e.g., C), then E has a non-torsion point.
Proof. For each n, there are only finitely many n-torsion points. Thus there are (atmost) countably many torsion points. The curve E has uncountably many closed
points. (One argument for this: take a double cover π : E → P1 . Then P1 has
uncountably many closed points, and π is surjective on closed points.)!

19.11.8. Remark. This argument clearly breaks down over countable fields. In fact,over Fp , all points of an elliptic curve E are torsion. (Any point x is defined oversome finite field Fpr . The points defined over Fpr form a subgroup of E, using
the explicit geometric construction of the group law, and there are finite numberof points over Fpr — certainly no more than the number of Fpr -points of P2 .) But
over Q, there are elliptic curves with non-torsion points. Even better, there areexamples over Q: [2, 1, 8] is a Q-point of the elliptic curve y2 z = x3 + 4xz2 − z3 thatis not torsion. The proof would carry us too far afield, but one method is to use
the Nagell-Lutz Theorem (see for example [Si1, Cor. 7.2]).
We now use the existence of a non-torsion point to create some interesting
pathologies.
19.11.9. A map of projective varieties not arising from a map of graded rings,
even after regrading.
19.11.B. E XERCISE . Suppose E "→ P2C is a smooth complex plane cubic (hence
genus 1), yielding a graded ring S• . Let t : E → E be translation by a non-torsion
point. Show that t does not correspond to a map of graded rings S• → S• , evenafter regrading (cf. Remark 16.4.7). (If you wish, you can show the following. If
u : E → E is a translation, show that u corresponds to some map of regraded ringsSn• → Sn• if and only if u is translation by a torsion point.)

19.11.10. An affine open subset of an affine scheme that is not a distinguishedopen set.
We can use this to construct an example of an affine scheme X and an affine
open subset Y that is not distinguished in X. Let X = E − p, which is affine (seeExercise 19.2.B, or better, note that the linear series O(3p) sends E to P2 in such a
way that the “line at infinity” meets E only at p; then E − p has a closed embedding
into the affine scheme A2 ).
Let q be another point on E so that q − p is non-torsion. Then E − p − q isaffine (Exercise 19.2.B). Assume that it is distinguished. Then there is a function f
on E − p that vanishes on q (to some positive order d). Thus f is a rational function
on E that vanishes at q to order d, and (as the total number of zeros minus polesof f is 0) has a pole at p of order d. But then d(p − q) = 0 in Pic0 E, contradicting
our assumption that p − q is non-torsion.
In particular, E − p is an affine scheme, and q is locally cut out by one equation, but it is not globally cut out even set-theoretically by one equation. This waspromised in Exercise 7.3.F and §8.4.1.
19.11.11. A proper (nonprojective) surface with no nontrivial line bundles.

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We next use a non-torsion point p on an elliptic curve E to construct a proper
nonprojective surface with no nontrivial line bundles. Let X1 be P2 , and Z1 asmooth cubic in X1 , identified with E. Let Z2 ⊂ X2 be exactly the same. Glue X1∼ Z2 given, not by the identity, but by translation
to X2 along the isomorphism Z1 =by p. (This gluing construction was described in §16.4.11.) Call the result X. By
Exercise 16.4.O (and the sentence thereafter), X is proper.
19.11.C. E XERCISE . Show that every line bundle on X is trivial. Hint: Suppose
L is an invertible sheaf on X. Then L |Xi is O(di ) for some di (for i = 1, 2). The
restriction of L |X1 to Z1 must agree with the restriction of L |X2 to Z2 . Use thisto show that d1 = d2 = 0. Explain why gluing two trivial bundles (on X1 and X2
respectively) together in this way yields a trivial bundle on X.
19.11.D. E XERCISE . Show that X is not projective.
See Exercise 20.2.G for a somewhat simpler example of a proper nonprojective
surface.
19.11.12. A Picard group that has no chance of being a scheme.
We informally observed that the Picard group of an elliptic curve “wants tobe” a scheme (see §19.10.5). This is true of projective (and even proper) varieties
in general (see [FGIKNV, Ch. 9]). On the other hand, if we work over C, the affine
scheme E − p − q (in the language of §19.11.10 above) has a Picard group that canbe interpreted as C modulo a lattice modulo a non-torsion point (e.g., C/⟨1, i, π⟩).
This has no reasonable interpretation as a manifold, let alone a variety. So the factthat the Picard group of proper varieties turns out to be a scheme should be seen
as quite remarkable.
19.11.13. A variety whose ring of functions is not finitely-generated.
We next show an example of a complex variety whose ring of global sections
is not finitely generated. (An example over Q can be constructed in the same way
using the curve of Remark 19.11.8.) This is related to Hilbert’s fourteenth problem(see [Mu5, §3]).
19.11.E. P RELIMINARY E XERCISE . Suppose X is a scheme, and L is the total spaceof a line bundle corresponding to invertible sheaf L , so L = Spec ⊕n≥0 (L ∨ )⊗n .
(This construction first appeared in Definition 17.1.4.) Show that if X is quasicompact, then H0 (L, OL ) = ⊕H0 (X, (L ∨ )⊗n ). (Possible hint: choose a trivializing
cover for L . Rhetorical question: can you figure out the more general statement if
L is a rank r locally free sheaf?)
Let E be an elliptic curve over some ground field k, N a degree 0 non-torsion
invertible sheaf on E, and P a positive-degree invertible sheaf on E. Then H0 (E, N m ⊗P n ) is nonzero if and only if either (i) n > 0, or (ii) m = n = 0 (in which case thesections are elements of k).
19.11.F. E ASY E XERCISE . Show that the ring R = ⊕m,n≥0 H0 (E, N m ⊗ P n ) is not
finitely generated.

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19.11.G. E XERCISE . Let X be the total space of the vector bundle associated to(N ⊕ P)∨ over E. Show that the ring of global sections of X is R, and hence is not
finitely generated. (Hint: interpret X as a line bundle over a line bundle over E.)
19.11.H. E XERCISE . Show that X (as in the above exercise) is a variety (and inparticular, Noetherian) whose ring of global sections is not Noetherian.

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CHAPTER 20

Application: A glimpse of intersection theory
The only reason this chapter appears after Chapter 19 is because we will use
Exercise 19.2.E in the double-starred proof of the Nakai-Moishezon criterion for
ampleness (see §20.4.5).

20.1 Intersecting n line bundles with an n-dimensional variety
Throughout this chapter, X will be a k-variety; in most applications, X will be
projective. The central tool in this chapter is the following.
20.1.1. Definition: intersection product, or intersection number (L1 , . . . , Ln ), (L1 , . . . , F ),(L1 , . . . , Y). Suppose F is a coherent sheaf on X with proper support (automatic
if X is proper) of dimension at most n, and L1 , . . . , Ln are invertible sheaves on X.
Let (L1 · L2 · · · Ln · F ) be the signed sum over the 2n subsets of {1, . . . , n}.
(20.1.1.1)(−1)m χ(X, Li∨⊗ · · · ⊗ Li∨⊗ F ).
m1{i1 ,...,im }⊂{1,...,n}We call this the intersection of L1 , . . . , Ln with F . (Never forget that wheneverwe write (L1 · · · Ln · F ), we are implicitly assuming that dim Supp F ≤ n.) The
case we will find most useful is if F is the structure sheaf of a closed subscheme
Y (of dimension at most n). In this case, we may write it (L1 · L2 · · · Ln · Y). If
the Li are all the same, say L , one often writes (L n · F ) or (L n · Y). (Be carefulwith this confusing notation: L n does not mean L ⊗n .) In some circumstances
the convention is to omit the parentheses.
We will prove many things about the intersection product in this chapter. One
fact will be left until we study flatness (Exercise 24.7.D): that it is “deformationinvariant” — that it is constant in “nice” families.
∼ OX then (L1 · L2 · · · Ln ·20.1.A. E XERCISE ( REALITY CHECK ). Show that if L1 =F ) = 0.

The following exercise suggests that the intersection product might be interesting, as it “interpolates” between two useful notions: the degree of a line bundle
on a curve, and Bézout’s Theorem.
20.1.B. E XERCISE .
(a) If X is a projective curve, and L is an invertible sheaf on X, show that (L · X) =degX L .
(b) Suppose k is an infinite field, X = PN , and Y is a dimension n subvariety of X. If
H1 , . . . , Hn are generally chosen hypersurfaces of degrees d1 , . . . , dn respectively
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The Rising Sea: Foundations of Algebraic Geometry(so dim(H1 ∩ · · · ∩ Hn ∩ Y) = 0 by Exercise 11.3.C(d)), then by Bézout’s Theorem(Exercise 18.6.K),Show thatdeg(H1 ∩ · · · ∩ Hn ∩ Y) = d1 · · · dn deg(Y).
(OX (H1 ) · · · OX (Hn ) · Y) = d1 · · · dn deg(Y).

We now describe some of the properties of the intersection product. In thecourse of proving Exercise 20.1.B(b) you may in effect solve the following exercise.
20.1.C. E XERCISE . Suppose D is an effective Cartier divisor on X that restricts toan effective Cartier divisor D|Y = D ∩ Y on Y (i.e., remains locally not a zerodivisor
on Y). Show that
(L1 · · · Ln−1 · O(D) · Y) = (L1 · · · Ln−1 · D|Y ).

More generally, if D is an effective Cartier divisor on X that does not contain any
associated point of F , show that
(L1 · · · Ln−1 · O(D) · F ) = (L1 · · · Ln−1 · F |D ).

(A similar idea came up in the proof that the Hilbert polynomial is actually polynomial; see the discussion around (18.6.2.1).)
20.1.2. Definition. For this reason, if D is an effective Cartier divisor, in the symbol
for the intersection product, we often write D instead of O(D). We interchangeably think of intersecting effective Cartier divisors rather than line bundles. For
example, we will discuss the special case of intersection theory on a surface in§20.2, and when we intersect two curves C and D, we will write the intersection as(C · D) or even C · D.
20.1.D. E XERCISE . Show that the intersection product (20.1.1.1) is preserved byfield extension of k.
20.1.3. Proposition. — Assume X is projective. For fixed F , the intersection product(L1 · · · Ln · F ) is a symmetric multilinear function of the L1 , . . . , Ln .

Proposition 20.1.3 is actually true with “projective” replaced by “proper”, see[Kl1, Prop. 2] or [Ko1, Prop. VI.2.7]. Unlike most extensions to the proper case,this is not just an application of Chow’s Lemma. It involves a different approach,involving a beautiful trick called dévissage.
Proof. Symmetry is clear. By Exercise 20.1.D, we may assume that k is infinite (e.g.,
algebraically closed). We now prove the result by induction on n.
20.1.E. E XERCISE ( BASE CASE ). Prove the result when n = 1. Hint: Exercise 18.4.S.
(In fact, you can take the base case to be n = 0, if this doesn’t confuse you.)
We now assume the result for smaller values of n. We use a trick. We wish toshow that (for arbitrary L1 , L1′ , L2 , . . . , Ln ),(20.1.3.1) (L1 · L2 · · · Ln · F ) + (L1′ · L2 · · · Ln · F ) − ((L1 ⊗ L1′ ) · L2 · · · Ln · F )is 0.

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20.1.F. E XERCISE . Rewrite (20.1.3.1) as(20.1.3.2)(L1 · L1′ · L2 · · · Ln · F ).

(There are now n + 1 line bundles appearing in the product, but this does notcontradict the definition of the intersection product, as dim Supp F ≤ n < n + 1.)
20.1.G. E XERCISE . Use the inductive hypothesis to show that (20.1.3.1) is 0 if
∼ O(D) for D an effective Cartier divisor missing the associated points of F .
Ln =In particular, if Ln is very ample, then (20.1.3.1) is 0, as Exercise 18.6.A showsthat there exists a section of Ln missing the associated points of F .
By the symmetry of its incarnation as (20.1.3.2), expression (20.1.3.1) vanishes
if L1 is very ample. Let A and B be any two very ample line bundles on X. Thenby substituting L1 = B and L1′ = A ⊗ B ∨ , using the vanishing of (20.1.3.1), wehave(20.1.3.3) (A ⊗ B ∨ · L2 · · · Ln · F ) = (A · L2 · · · Ln · F ) − (B · L2 · · · Ln · F )Both summands on the right side of (20.1.3.3) are linear in Ln , so the same is trueof the left side. But by Exercise 16.6.F, any invertible sheaf on X may be written in
the form A ⊗B ∨ (“as the difference of two very amples”), so (L1 ·L2 · · · Ln ·F ) is
linear in Ln , and thus (by symmetry) in each of the Li . (An interesting feature ofthis argument is that we intended to show linearity in L1 , and ended up showing
linearity in Ln .)!
We have an added bonus arising from the proof.
20.1.H. E XERCISE . Suppose X is projective. Show that if dim Supp F < n + 1, andL1 , L1′ , L2 , . . . , Ln are invertible sheaves on X, then (20.1.3.2) vanishes. In otherwords, the intersection product of n + 1 invertible sheaves with a coherent sheaf
F vanishes if the dim Supp F < n + 1. (In fact, the result holds with “projective”
replaced by “proper”, as the results it relies on hold in this greater generality.)
20.1.4. Proposition. — Suppose X is projective. The intersection product depends only
on the numerical equivalence classes of the Li .
(Numerical equivalence was defined in §18.4.10.) Again, the result remains
true with “projective” replaced by “proper”. But in the proof here, we use the factthat every line bundles is the difference two very ample line bundles in both theproof of Proposition 20.1.3 and in the proof of Proposition 20.1.4 itself. For the
proof of the Proposition in the proper case, see [FGIKNV, Prop. B.20].
Proof. Suppose L1 is numerically equivalent to L1′ , and L2 , . . . , Ln , and F arearbitrary. We wish to show that (L1 · L2 · · · Ln · F ) = (L1′ · L2 · · · Ln · F ).
By Exercise 20.1.D, we may assume that k is infinite (e.g., algebraically closed).
We proceed by induction on n. The case n = 1 follows from Exercise 18.4.S. We
assume that n > 1, and assume the result for “smaller n”. By multilinearity of theintersection product, and the fact that each Ln maybe written as the “difference”
of two very ample invertible sheaves (Exercise 16.6.F), it suffices to prove the resultin the case when Ln is very ample. We may then write Ln = O(D), where D is an

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The Rising Sea: Foundations of Algebraic Geometryeffective Cartier divisor missing the associated points of F (Exercise 18.6.A). Then
(L1 · L2 · · · Ln · F ) ===(L1 · L2 · · · Ln−1 · F |D ) (Exercise 20.1.C)(L1′ · L2 · · · Ln−1 · F |D ) (inductive hyp.)(L1′ · L2 · · · Ln · F ) (Exercise 20.1.C).
!

20.1.5. Asymptotic Riemann-Roch.
If Y is a proper curve, χ(Y, L ⊗m ) = m degY L + χ(Y, OY ) (see (18.4.4.1)) isa linear polynomial in m, whose leading term is an intersection product. This
generalizes as follows.
20.1.I. E XERCISE (A SYMPTOTIC R IEMANN -R OCH ). Suppose X is projective. (Asusual, the result will remain true with “projective” replaced by “proper”.) Suppose F is a coherent sheaf on X with dim Supp F ≤ n, and L is a line bundle on
X. Show that χ(X, L ⊗m ⊗ F ) is a polynomial in m of degree at most n. Show thatthe coefficient of mn in this polynomial (the “leading term”) is (L n · F )/n!. Hint:
Exercise 20.1.H implies that (L n+1 · (L ⊗i ⊗ F )) = 0. (Careful with this notation:
L n+1 doesn’t mean L ⊗(n+1) , it means L · L · · · L with n + 1 factors.) Expand
this out using (20.1.1.1) to get a recursion for χ(X, L ⊗m ⊗ F ). Your argument may
resemble the proof of polynomiality of the Hilbert polynomial, Theorem 18.6.1, so
you may find further hints there. Exercise 18.6.C in particular might help.
We know all the coefficients of this polynomial if X is a curve, by RiemannRoch (see (18.4.4.1)), or basically by definition. We will know/interpret all thecoefficients if X is a regular projective surface and F is an invertible sheaf when we
prove Riemann-Roch for surfaces (Exercise 20.2.B(b)). To understand the generalcase, we need the theory of Chern classes. The result is the Hirzebruch-RiemannRoch Theorem, which can be further generalized to the celebrated GrothendieckRiemann-Roch Theorem (see [F, §15.2]).
20.1.J. E XERCISE ( THE PROJECTION FORMULA ). Suppose π : X1 → X2 is a (projective) morphism of integral projective schemes (over a field k) of the same dimension n, and L1 , . . . , Ln are invertible sheaves on X2 . Show that(π∗ L1 · · · π∗ Ln ) = deg(X1 /X2 )(L1 · · · Ln ).
(The first intersection is on X1 , and the second is on X2 .) Hint: Let d = deg π, andassume d > 0. (Deal with the case where π is not dominant separately, so d = 0
by convention, using Chevalley’s Theorem 7.4.2.) Argue that by the multilinearityof the intersection product, it suffices to deal with the case where the Li are very
ample. Then choose sections of each Li , all of whose intersections lie in an open
subset U where π has “genuine degree d”. To find U: first use Exercise 9.3.G to finda dense open subset U ′ ⊂ X2 over which π is finite. Then use Useful Exercise 13.7.F
to show that there exists a dense open subset U ⊂ U ′ on which π∗ O is a locally
free sheaf of rank d. In the “flatness” language of Chapter 24, you are showing that
there is a dense open subset U of X2 over which π is finite and flat (and hence has“constant degree”, see Remark 24.4.8). (As usual, the result holds with “projective”
replaced with “proper”; see [Ko1, Prop. VI.2.11].)

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20.1.6. Remark: A more general projection formula. Suppose π : X1 → X2 is a propermorphism of proper varieties, and F is a coherent sheaf on X1 with dim Supp F ≤n (so dim Supp π∗ F ≤ n, using Exercise 11.2.C). Suppose also that L1 , . . . , Ln are
invertible sheaves on X2 . Then(π∗ L1 · · · π∗ Ln · F ) = (L1 · · · Ln · π∗ F ).

This is also called the projection formula (and generalizes, in a nonobvious way,
Exercise 20.1.J). Proofs are given in [FGIKNV, B.15] and [Ko1, Prop. VI.2.11]. Because we won’t use this version of the projection formula, we do not give a proof
here.
20.1.K. E XERCISE ( INTERSECTING WITH AMPLE LINE BUNDLES ). Suppose X isa projective k-variety, and L is an ample line bundle on X. Show that for any
subvariety Y of X of dimension n, (L n · Y) > 0. (Hint: use Proposition 20.1.3
and Theorem 16.6.2 to reduce to the case where L is very ample. Then show that
(L n · Y) = deg Y in the embedding into projective space induced by the linear
series |L |.)The Nakai-Moishezon criterion (Theorem 20.4.1) states that this characterizes
ampleness.
20.1.7. ⋆⋆ Cohomological interpretation in the complex projective case, generalizing Exercise 18.4.G and §18.6.4. If k = C, we can interpret (L1 · · · Ln · Y) as the degreeof(20.1.7.1)

c1 ((L1 )an ) ∪ · · · ∪ c1 ((Ln )an ) ∩ [Yan ]in H0 (Yan , Z). (Recall c1 ((Li )an ) ∈ H2 (Xan , Z), as discussed in Exercise 18.4.G.)
One way of proving this is to use multilinearity of both the intersection product∼ O(D),and (20.1.7.1) to reduce to the case where the Ln is very ample, so Ln =where D restricts to an effective Cartier divisor E on Y. Then show that if L is ananalytic line bundle on Yan with nonzero section Ean , then c1 (L ) ∩ [Yan ] = [Ean ].
Finally, use induction on n and Exercise 20.1.C.
20.1.8. Important remark: additive notation for line bundles. There is a standard, useful, but confusing convention suggested by the multilinearity of the intersection product: we write tensor product of invertible sheaves additively. This isconvenient for working with multilinearity. (Some people try to avoid confusion
by using divisors rather than line bundles, as we add divisors when we “multiply” the corresponding line bundles. This is psychologically helpful, but may add
more confusion, as one then has to worry about the whether and why and how
and when line bundles correspond to divisors.) We will use this, for example, in
Exercises 20.2.B–20.2.G, §20.2.10, and §20.4.

20.2 Intersection theory on a surface
We now apply the general machinery of §20.1 to the case of a regular projective
surface X. (What matters is that X is Noetherian and factorial, so Pic X → Cl X is
an isomorphism, Proposition 14.2.10. Recall that regular schemes are factorial by
the Auslander-Buchsbaum Theorem 12.8.5.)

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20.2.A. E XERCISE /D EFINITION .
curves) on X.
(a) Show that

Suppose C and D are effective divisors (i.e.,(20.2.0.1)

degC OX (D)|C(20.2.0.2)(20.2.0.3)==(O(C) · O(D) · X)
degD OX (C)|D .

We call this the intersection number of C and D, and denote it C · D.
(b) If C does not contain any associated point of D (so in particular, if C and D
have no components in common), show that
(20.2.0.4)

C · D = h0 (C ∩ D, OC∩D )where C ∩ D is the scheme-theoretic intersection of C and D on X.
20.2.1. Remark. In particular, if C and D have no irreducible component in common, then C · D ≥ 0. This means that we cannot hope to understand intersection
numbers using moving lemmas as in topology — if we have some irreducible
curve E with E2 < 0, we cannot “move” E to something (call if F) “equivalent” to
E, to compute E · E as E · F.
20.2.2. Important aside. The hypothesis in Exercise 20.2.A(b), that C not contain any
associated point of D, is somewhat of a red herring. In fact, D can never have any
embedded points, as we will see in §26.2.5 when discussing Cohen-Macaulay rings.
(More generally, we will see that local complete intersections have no embedded
points — this should help motivate you to learn about Cohen-Macaulayness.) The
case of A2 (and hence P2 ) can be done by hand (Exercise 5.5.I).
Thus the hypothesis in Exercise 20.2.A(b) can be replaced by the more simple“C and D have no common components”.
20.2.3. Advantages and disadvantages. We thus have four descriptions of the intersection number (20.2.0.1)–(20.2.0.4), each with advantages and disadvantages.
The Euler characteristic description (20.2.0.2) is remarkably useful (for example, inthe exercises below), but the geometry is obscured. The definition degC OX (D)|C ,(20.2.0.1), is not obviously symmetric in C and D. The definition h0 (C ∩ D, OC∩D ),(20.2.0.4), is clearly local — to each point of C ∩ D, we have a vector space. For
example, we know that in A2k , y − x2 = 0 meets the x-axis with multiplicity 2, because h0 of the scheme-theoretic intersection (k[x, y]/(y − x2 , y)) has dimension 2.
(This h0 is also the length of the dimension 0 scheme, whose definition you may beable to figure out given Definition 18.4.7 of the length of a module. But we won’t
use this terminology.)
By Proposition 20.1.3, the intersection number induces a bilinear “intersectionform”(20.2.3.1)

Pic X × Pic X → Z.

By Asymptotic Riemann-Roch (Exercise 20.1.I), χ(X, O(nD)) is a quadratic polynomial in n.

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20.2.4. You can verify that Exercise 20.2.A recovers Bézout’s Theorem for plane
curves (see Exercise 18.6.K), using χ(P2 , O(n)) = (n + 2)(n + 1)/2 (from Theorem 18.1.3), and the fact that effective Cartier divisors on P2k have no embedded
points (Exercise 5.5.I).
Before getting to a number of interesting explicit examples, we derive a coupleof fundamental theoretical facts.
20.2.B. E XERCISE . Assuming Serre duality for X (Theorem 18.5.1), prove the following for a smooth projective surface X. (We are mixing divisor and invertible
sheaf notation, as described in Remark 20.1.8, so be careful. Here KX is a divisor
corresponding to ωX .)(a) (sometimes called the adjunction formula) C · (KX + C) = 2pa (C) − 2 for any curve
C ⊂ X. Hint: compute (C · (−KX − C) instead. (See Exercise 21.5.B and §30.4 forother versions of the adjunction formula.)(b) (Riemann-Roch for surfaces)
χ(X, OX (D)) = D · (D − KX )/2 + χ(X, OX )for any Weil divisor D (cf. Riemann-Roch for curves, Exercise 18.4.B).
20.2.5. Two explicit examples: P1 × P1 and Blp P2 .
20.2.C. E XERCISE : X = P1 × P1 . Recall from Exercise 14.2.O that Pic(P1 × P1 ) =Zℓ × Zm, where ℓ is the curve P1 × {0} and m is the curve {0} × P1 . Show that theintersection form (20.2.3.1) is given by ℓ · ℓ = m · m = 0, ℓ · m = 1. (Hint: You cancompute the cohomology groups of line bundles on P1 × P1 using Exercise 18.3.3,
but it is much faster to use Exercise 20.2.A(b).) What is the class of the diagonal in
P1 × P1 in terms of these generators?
20.2.D. E XERCISE : THE BLOWN UP PROJECTIVE PLANE . (You needn’t have read
Chapter 22 to do this exercise!) Let X = Blp P2 be the blow-up of P2k at a k-valued
point (the origin, say) p — see Exercise 9.3.F, which describes the blow-up of A2k ,and “compactify”. Interpret Pic X as generated (as an abelian group) by ℓ and e,
where ℓ is a line not passing through the origin, and e is the exceptional divisor.
Show that the intersection form (20.2.3.1) is given by ℓ·ℓ = 1, e·e = −1, and ℓ·e = 0.
∼ Zℓ × Ze (as promised in the aside in Exercise 14.2.P). In
Hence show that Pic X =particular, the exceptional divisor has negative self-intersection. (This exercise will
be generalized in §22.4.13.)
20.2.6. Hint. Here is a possible hint to get the intersection form in Exercise 20.2.D.
The scheme-theoretic preimage in Blp P2 of a line through the origin is the schemetheoretic union of the exceptional divisor e and the “proper transform” m of theline through the origin. Show that ℓ = e + m in Pic(Blp P2 ) (writing the Picard
group operation additively, cf. Remark 20.1.8). Show that ℓ · m = e · m = 1 andm · m = 0.
20.2.7. Definition: (−1)-curve.
Notice that the exceptional divisor e has selfintersection −1. We will see more generally in Exercise 22.4.O that this is the casefor all exceptional divisors of blow-ups of smooth surfaces (over k) blown up at a
k-valued point. We give such curves a name. If X is a surface over k, and C ⊂ X is
∼ P1 , and C · C = −1, we say thata curve in X consisting of smooth points, with C =kEarly (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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C is a (−1)-curve. (Once we know more, we will be able to restate “C · C = −1” as“C has normal bundle O(−1)”: combine Exercise 21.2.H with Exercise 20.2.A(a).)
20.2.E. E XERCISE . Show that the blown up projective plane Blp P2 in Exercise 20.2.D
is not isomorphic to P1 ×P1 , perhaps considering their (isomorphic) Picard groups,
and identifying which classes are effective (represented by effective divisors). (Thisis an example of a pair of smooth projective birational surfaces that have isomorphic Picard groups, but which are not isomorphic. This exercise shows that F0 is
not isomorphic to F1 . (The method and result are generalized in Exercise 20.2.Q.)
20.2.F. E XERCISE ( CF. E XERCISE 18.4.X). Show that the nef cone (Exercise 18.4.W)of Blp P2 is generated by ℓ and m. Hint: show that ℓ and m are nef. By intersecting
line bundles with the curves e and ℓ, show that nothing outside the cone spannedby ℓ and m is nef. (Side Remark: note that as in Exercise 18.4.X, linear seriescorresponding to the boundaries of the cone give “interesting contractions”.)
20.2.G. E XERCISE : A PROPER NONPROJECTIVE SURFACE . Show the existence ofa proper nonprojective surface over a field as follows, paralleling the constructionof a proper nonprojective threefold in §16.4.10. Take two copies of the blown up
projective plane Blp P2 , gluing ℓ on the first to e on the second, and e on the second
to ℓ on the first. Hint: show that if L is a line bundle having positive degree on
each effective curve, then L · ℓ > L · e, using ℓ = e + m from Hint 20.2.6. (See§19.11.11 for another example of a proper nonprojective surface.)
20.2.8. Fibrations.
Suppose π : X → B is a morphism from a regular projective surface to a regularcurve and b ∈ B is a closed point. Let F = π∗ b. Then OX (F) = π∗ OB (b), which isisomorphic to O on F. Thus F · F = degF OX (F) = 0: “the self-intersection of a fiberis 0”. The same argument works without X being regular, so long as you phrase itproperly: (π∗ OX (b))2 = 0.
20.2.H. E XERCISE . Suppose E is an elliptic curve, with origin p. On E × E, let ∆ be
the diagonal. By considering the “difference” map E×E → E, for which π∗ (p) = ∆,show that ∆2 = 0. Show that N1Q (E × E) has rank at least 3. Show that in generalfor schemes X and Y, Pic X × Pic Y → Pic(X × Y) (defined by pulling back andtensoring) need not be isomorphism; the case of X = Y = P1 is misleading.

Remark: dimQ N1Q (E × E) is always 3 or 4. It is 4 if there is a nontrivial endomorphism from E to itself (i.e., not just multiplication [n] by some n, §19.11.5,
followed by a translation); the additional class comes from the graph of this endomorphism. (See [Mu3, §21, App. III] for an introduction to the tools needed toshow this.)
Our next goal is to describe the self-intersection of a curve on a ruled surface(Exercise 20.2.J). To set this up, we have a useful preliminary result.
20.2.I. E XERCISE ( THE NORMAL BUNDLE TO A SECTION OF Proj OF A RANK 2 VEC Suppose X is a scheme, and and V is a rank 2 locally free sheaf on
X. Explain how the short exact sequences

TOR BUNDLE ).

(20.2.8.1)

0→S →V →Q→0

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on X, where S and Q have rank 1, correspond to the sections σ : X → PV to theprojection PV → X. Show that the normal bundle to σ(X) in PV is Q⊗S ∨ . (A generalization is stated in §21.4.10.) Hint: (i) For simplicity, it is convenient to assume
S = OX , by replacing V by V ⊗ S ∨ , as the statement of the problem respects tensoring by an invertible sheaf (see Exercise 17.2.G). (ii) Assume now (with loss of∼ OX , and the exact sequence splits. Then describe the section
generality) that Q =1
as σ : X → P ×X, with X mapping to the 0 section. Describe an isomorphism of OX
with the normal bundle to σ(X) → P1 × X. (Do not just say that the normal bundle“is trivial”.) (iii) Now consider the case where Q is general. Choose trivializing
open neighborhoods Ui of Q, and let gij be the the transition function for Q. On
the overlap between two trivializing open neighborhoods Ui ∩ Uj , determine how
your two isomorphisms of OX with Nσ(X)/P1X from (ii) (one for Ui , one for Uj ) are
related. In particular, show that they differ by gij .
20.2.J. E XERCISE ( SELF - INTERSECTIONS OF SECTIONS OF RULED SURFACES ). Suppose C is a regular curve, and V is a rank 2 locally free sheaf on C. Then PV is a
ruled surface (Example 17.2.4). Fix a section σ of PV corresponding to a filtration(20.2.8.1). Show that σ(C) · σ(C) = degC (Q ⊗ S ∨ ).
20.2.9. The Hirzebruch surfaces Fn = Proj P1 (OP1 ⊕ OP1 (n)).
Recall the definition of the Hirzebruch surface Fn = Proj P1 (OP1 ⊕ OP1 (n))
in Example 17.2.4. It is a P1 -bundle over P1 ; let π : Fn → P1 be the structure
morphism. Using Exercise 20.2.J, corresponding to
0 → O(n) → O ⊕ O(n) → O → 0,

we have a section of π of self-intersection −n; call it E ⊂ Fn . Similarly, corresponding to
0 → O → O ⊕ O(n) → O(n) → 0,

we have a section C ⊂ Fn of self-intersection n. Let p be any k-valued point of P1 ,and let F = π∗ (p).
20.2.K. E XERCISE . Show that the line bundle O(F) is independent of the choice ofp.
20.2.L. E XERCISE . Show that Pic Fn is generated by E and F. In the course of
doing this, you will develop “local charts” for Fn , which will help you solve later
exercises.
20.2.M. E XERCISE . Compute the intersection matrix on Pic Fn . Show that E and F∼ ZE ⊕ ZF. Calculate C in terms of E and F.
are independent, and thus Pic Fn =20.2.N. E XERCISE . Show how to identify Fn \ E, along with the structure map
π, with the total space of the line bundle O(n) on P1 , with C as the 0-section.
Similarly show how to identify Fn \C with the total space of the line bundle O(−n)
on P1 ; with E as the 0-section.
20.2.O. E XERCISE . Show that h0 (Fn , OFn (C)) > 1. Hint: As OFn (C) has a section— namely C — we have that h0 (Fn , OFn (C)) ≥ 1. One way to proceed is to write
down another section using local charts for Fn .

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20.2.P. E XERCISE . Show that every effective curve on Fn is a non-negative linear
combination of E and F. (Conversely, it is clear that for every nonnegative a andb, O(aE + bF) has a section, corresponding to the effective curve “aE + bF”. The
extension of this to N1Q is called the effective cone, and this notion, extended toproper varieties more general, can be very useful. This exercise shows that E andF generate the effective cone of Fn .) Hint: show that because “F moves”, any
effective curve must intersect F nonnegatively, and similarly because “C moves”(Exercise 20.2.O), any effective curve must intersect C nonnegatively. If O(aE+bF)
has a section corresponding to an effective curve D, what does this say about a andb?
20.2.Q. E XERCISE . By comparing effective cones, and the intersection pairing,show that the Fn are pairwise nonisomorphic. (This result was promised in Example 17.2.4. Exercise 20.2.E is a special case.)
Exercise 20.2.Q is difficult to do otherwise, and foreshadows the fact that nef
and effective cones are useful tools in classifying and understanding varieties general. In particular, they are central to the minimal model program.
20.2.R. E XERCISE . If n = 0, show that there are no curves on Fn of negative selfintersection. If n ≥ 0, show that E is the unique curve on Fn with self-intersection−n, and there are no reduced curves on Fn of smaller self-intersection. This again
gives another (related) means of showing that the Fn are pairwise nonisomorphic.
20.2.S. E XERCISE . Show that the nef cone of Fn is generated by C and F. (We willsoon see that by Kleiman’s criterion for ampleness, Theorem 20.4.7, that the ample
cone is the interior of this cone, so we have now identified the ample line bundles
on Fn .)
20.2.T. E XERCISE .
We have seen earlier (Exercises 20.2.F and 18.4.X) that theboundary of the nef cone give “interesting contractions”. What are the maps givenby the two linear series corresponding to O(F) and O(C)?
After this series of exercises, you may wish to revisit Exercises 20.2.C-20.2.F,∼ P1 × P1 and F1 =∼ Blp P2 .
and interpret them as special cases: F0 =20.2.10. The Hodge Index Theorem.
We use what we have learned to prove the following celebrated result.
20.2.11. The Hodge Index Theorem. — Suppose X is an irreducible smooth projective
surface (over a field k) with L , H ∈ Pic X, with H · H > 0 and L · H = 0. Then (a)L · L ≤ 0, and (b) equality holds if and only if L is numerically trivial.
Before reading on, you should review the definitions of N1 (X) and ρ(X) in§18.4.10.

20.2.U. E XERCISE ( REASON FOR THE NAME ). (We will not use this.) By arguments
similar to the classification of quadratic forms over algebraically closed fields (Exercise 5.4.J), one can show that given a symmetric bilinear form on a real vector
space of (finite) dimension n, there are integers a+ , a0 , a− such that then given
any orthogonal basis v1 , . . . , vn , the number of vi · vi that are positive (resp. zero,
negative) is a+ (resp. a0 , a− ). The ordered pair (a+ , a− ) is often called the index,

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and is sometimes called the signature. (This result is sometimes called Sylvester’s
law of inertia. See, for example, [Lan, §XV.4] for a proof.) Show that the Hodge
Index Theorem implies that N1R (X) := N1 (X) ⊗Z R, along with the bilinear form
coming from the intersection product, has index (1, ρ(X) − 1). You should think of
the positive “term” as coming from the existence of an ample class; the “rest” isnegative.
Proof.
20.2.12. We begin with the following fact. If H is an ample invertible sheaf, M
is any invertible sheaf, and M · H > ωX · H , then h2 (X, M ) = 0. (Translation:
for all invertible sheaves “more positive than ωX ”, there is no top cohomology;
compare this to §19.2.5, the corresponding statement in dimension 1.) Reason: bySerre duality (Theorem 18.5.1), we wish to show that h0 (X, ωX ⊗ M ∨ ) = 0. But
if otherwise ωX ⊗ M ∨ had a nonzero section, then its (effective nonzero) divisor
would intersect H nonnegatively (Exercise 20.1.K), so (using additive notation,
see Remark 20.1.8) (ωX − M ) · H ≥ 0, contradicting M · H > ωX · H .
20.2.13. We next show that if H is ample, and L is any invertible sheaf with
L · L > 0 and L · H > 0, then for n ≫ 0, L ⊗n has a section. Here is why. Forn ≫ 0,
L ⊗n · H = nL · H > ωX · H ,2⊗nso h (X, L ) = 0 by §20.2.12 (taking M = L ⊗n ). Then for n ≫ 0, by RiemannRoch for surfaces (Exercise 20.2.B(b)),
h0 (X, L ⊗n ) − h1 (X, L ⊗n ) ==χ(X, L ⊗n )(n2 /2)L · L − nL · ωX /2 + χ(X, OX ).

As L · L > 0, the right side is positive for n ≫ 0, so

h0 (X, L ⊗n ) ≥ h0 (X, L ⊗n ) − h1 (X, L ⊗n ) > 0

as desired.

20.2.14. We are now ready to prove the Hodge Index Theorem. Assume first that
L · L > 0 (and L · H = 0). Then for n ≫ 0, H ′ := L ⊗ H ⊗n is ample (indeed
very ample) by Exercise 16.6.E. Then L · H ′ = L · L > 0, so by §20.2.13, L ⊗n
is effective, which implies that L · H = (L ⊗n · H )/n > 0 (using Exercise 20.1.K
again), contradicting our hypothesis.
20.2.15. Assume finally that L · H = 0, and (i) L · L = 0, but that (ii) L is notnumerically trivial. By (ii), we can find an invertible sheaf Q such that Q · L ̸= 0.
Then we can find an invertible sheaf R such that R · L ̸= 0 and R · H = 0: take
Then take L = L
some n ∈ Z so that′⊗n

R = (H · H )Q − (Q · H )H .

⊗ R, so L ′ · H = 0, but because R · L ̸= 0, we can find

L ′ · L ′ = nL · R + R · R > 0.
Then the argument of §20.2.14 applies, with L ′ in the place of L .

!

20.2.16. Generalizations and variations. The hypotheses can be weakened considerably. We used smoothness only because we need Serre duality, with an invertible

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sheaf ωX . We will see in §30.4 that we need less than smoothness for this. But wecan do much better, and a proof where X is merely required to be geometricallyirreducible and proper requires only a minor modification of our argument, see[FGIKNV, Thm. B.27].

20.3 The Grothendieck group of coherent sheaves, and an
algebraic version of homology
The construction of the intersection product (20.1.1.1) may leave you hungry for something more, especially in light of the cohomological interpretationof §20.1.7. You may want some sort of homology-like theory which is a repository
for cycles of different directions, on which (Chern classes of) line bundles can act.
We can actually do this easily, given what we know.
20.3.1. Definition. If X is a k-variety, we define the Grothendieck group of coherent sheaves, which we denote K(X) (and which is often denoted K0 (X)), as theabelian group generated symbols of the form [F ] where F is a coherent sheaf on∼ F ′ , and [F ′ ] + [F ′′ ] = [F ] if
X, subject to the relations that [F ] = [F ′ ] if F =there is a short exact sequence
0 → F ′ → F → F ′′ → 0.

(Caution: the notation K(X) is now used for two concepts, the Grothendieck groupand the function field. The meaning of the symbol should be clear from the context.) By construction, the Grothendieck group is the universal construction ofan operator on the category of coherent sheaves that “behaves well in exact sequences”. For example, if X is proper, then:(i) We have a map χ : CohX → Z, as the Euler characteristic of a coherent
sheaf is finite by Theorem 18.1.4(i) in the projective case, and more generally by Grothendieck’s Coherence Theorem 18.9.1 in the proper case. Thismap descends to χ : K(X) → Z by Exercise 18.4.A (which extends withoutchange to the proper case).
(ii) If X is integral, then the rank function
rank : CohX → Z

descends to rank : K(X) → Z. (The argument of Exercise 18.4.H applies.)

20.3.2. Definition and nonstandard notation. The Grothendieck group is filtered bydimension: let K(X)≤d be the subgroup of K(X) generated by coherent sheaves
supported in dimension at most d. Let Ad′ (X) be the dth graded piece of K(X), i.e.,Ad′ (X) := K(X)≤d /K(X)≤d−1 .
If L is an invertible sheaf on X, define L · : K(X) → K(X) by L · [F ] = [F ] −[L ∨ ⊗ F ]. (Do you see why this operator is well-defined?)

20.3.A. E XERCISE . If X is projective and k is infinite, show that L · sends K(X)≤d toK(X)≤d−1 . (Hint: for each fixed F supported on a subset of dimension at most d,
write L is a difference of two very ample invertible sheaves, and choose sectionsof those two, missing the associated points of F .)Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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20.3.3. Remark. The previous exercise holds true without k being infinite or X
being proper; see [Ko1, Prop. VI.2.5].
20.3.4. For the rest of this section, we assume X is projective and k is infinite. (Butin light of Remark 20.3.3, these hypotheses can be removed.) By Exercise 20.3.A,′
L · descends to a map Ad′ (X) → Ad−1
(X); we denote this operator c1 (L )∩. (It is
the action of the first Chern class.)
20.3.B. E XERCISE (“c1 IS ADDITIVE ”). Show that c1 (L ⊗ L ′ )∩ = (c1 (L )∩) +(c1 (L ′ )∩). Hint: show that ((L ⊗ L ′ )·) − (L ·) − (L ′ ·) = −(L ·) ◦ (L ′ ·), andthus sends K(X)≤d to K(X)≤d−2 . (This will remind you of the trick in the proof of
Proposition 20.1.3, and indeed is the motivation for that trick. Caution: the actionof Pic(X) is not additive on K(X); it only becomes additive once we pass to theassociated graded ring A•′ (X).)
If F has support of dimension at most n, and L1 , . . . , Ln are n invertible
sheaves, we can now reinterpret the intersection product (L1 · L2 · · · Ln · F ) asc1 (L1 ) ∩ · · · ∩ c1 (Ln ) ∩ [F ]where [F ] is interpreted as lying in either K(X) or An′ (X). You can now go back andread §20.1, and reprove all the results with this starting point, sometimes obtaining
interesting generalizations.
These Ad′ (X) behave like homology groups in a number of ways. If Y ⊂ X is a
closed subvariety of pure dimension d, we have a class [Y] := [OY ] ∈ Ad′ (X). The
groups Ad′ (X) have appropriate functorial properties. For example, if π : X1 → X2
is a proper morphism (even if the Xi are not themselves proper), we have a mapπ∗ : K(X1 ) → K(X2 ) given by(20.3.4.1)

π∗ [F ] = [π∗ F ] − [R1 π∗ F ] + · · ·(using the long exact sequence for higher pushforwards, Theorem 18.8.1 (c)) which
descends to a map Ad′ (X1 ) → Ad′ (X2 ), where the “later terms” on the right side of(20.3.4.1) disappear, so π∗ [F ] = [π∗ F ]. This pushforward interacts well with thefirst Chern class of line bundles, yielding the projection formula of Remark 20.1.6.
If π is instead a flat morphism (Remark 16.3.8, soon to be discussed at lengthin Chapter 24), then π∗ is exact, so we have a map π∗ : K(X2 ) → K(X1 ). If the“relative dimension” of this map is r (to be properly defined in Definition 24.5.7),′
this yields a map π∗ Ad′ (X2 ) → Ad+r
(X1 ). This interacts well with first Chernclasses and proper pushforwards.
If k = C and X is proper, there is a map Ad′ (X) → H2d (X, Q), which behaves as
you might hope (for example, in its interaction with Chern classes of line bundles).
If X is not proper (but k = C), then the map is to Borel-Moore homology rather
than usual homology.
Our groups Ad′ (X) are a good approximation of the theory of Chow groups
Ad (X), as developed in [F]. In fact, there is a surjective map(20.3.4.2)

Ad (X) → Ad′ (X)[F, Examp. 15.1.5], and this map is an isomorphism once tensored with Q [F,
Thms. 18.2 and 18.3]. This is the beginning of a long and rich story in algebraic
geometry.

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20.3.5. Side Remark. The surjection map (20.3.4.2) need not be an isomorphism,see [SGA6, Exp. XIV, §4.5-4.7] (although its kernel must be torsion, as describedabove). As a tantalizing example, if X is the group E8 , the kernel has 2-torsion,
3-torsion, and 5-torsion, see [KN, DZ].

20.4 ⋆⋆ The Nakai-Moishezon and Kleiman criteria for ampleness
Exercise 20.1.K stated that if X is projective k-variety, and L is an ample line
bundle on X, then for any subvariety Y of X of dimension n, (L n · Y) > 0. The
Nakai-Moishezon criterion states that this is a characterization:
20.4.1. Theorem (Nakai-Moishezon criterion for ampleness). — If L is an invertible sheaf on a projective k-scheme X, and for every subvariety Y of X of dimension n,(L n · Y) > 0, then L is ample.
20.4.2. Remarks. We note that X need only be proper for this result to hold ([Kl1,
Thm. III.1.1]).
Before proving the Nakai-Moishezon criterion, we point out some consequences
related to our discussion of numerical equivalence in §18.4.9. By Proposition 20.1.4,(L n · Y) depends only on the numerical equivalence class of L , so ampleness isa numerical property. As a result, the notion of ampleness makes sense on N1Q (X).
As the tensor product of two ample invertible sheaves is ample (Exercise 16.6.H),
the ample Q-line bundles in N1Q (X) form a cone, called the ample cone of X.
20.4.3. Proposition. — If X is a projective k-scheme, the ample cone is open.
20.4.4. In the rest of this section, we often use additive notation for the tensor
product of invertible sheaves, as described in Remark 20.1.8. This is because wewant to deal with intersections on the Q-vector space N1Q (X). For example by((aL1 + bL1′ ) · L2 · · · Ln · F ) (where a, b ∈ Q), we mean a(L1 · L2 · · · Ln ·F ) + b(L1′ · L2 · · · Ln · F ).
Proof. Suppose A is an ample invertible sheaf on X. We will describe a small
open neighborhood of [A ] in N1Q (X) consisting of ample Q-line bundles. Choose
invertible sheaves L1 , . . . , Ln on X whose classes form a basis of N1Q (X). By Exercise 16.6.E, there is some m such that A ⊗m ⊗ Li and A ⊗m ⊗ Li∨ are both1very ample for all n. Thus (in the additive notation of § 20.4.4), A + m
Li and1
A − m Li are both ample. As the ample ,Q-line bundles form a cone, it follows
that A + ϵ1 L1 + · · · + ϵn Ln is ample for i |ϵi | ≤ 1/m.
!

20.4.5. Proof of the Nakai-Moishezon criterion, Theorem 20.4.1. We prove the NakaiMoishezon criterion in several steps.
20.4.A. U NIMPORTANT E XERCISE . Prove the case where dim X = 0.

Step 1: initial reductions. Suppose L satisfies the hypotheses of the Theorem;
we wish to show that L is ample. By Exercises 18.7.A and 18.7.B, we may assume

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that X is integral. Moreover, we can work by induction on dimension, so we canassume that L is ample on any closed subvariety. The base case is dimension 1,
which was done in Exercise 19.2.E.
Step 2: sufficiently high powers of L have sections. We show that H0 (X, L ⊗m ) ̸= 0
for m ≫ 0.
Our plan is as follows. By Asymptotic Riemann-Roch (Exercise 20.1.I), χ(X, L ⊗m ) =nm (L n )/n! + · · · grows (as a function of m) without bound. A plausible meansof attack is to show that hi (X, L ⊗m ) = 0 for i > 0 and m ≫ 0. We won’t do that,but will do something similar.
By Exercise 16.6.C, L is the difference of two very ample line bundles, say∼ A ⊗ B −1 with A = O(A) and B = O(B). From 0 → O(−A) → O → OA → 0
L =we have(20.4.5.1)

0 → L ⊗m (−B) → L ⊗(m+1) → L ⊗(m+1) |A → 0.

From 0 → O(−B) → O → OB → 0, we have(20.4.5.2)

0 → L ⊗m (−B) → L ⊗m → L ⊗m |B → 0.

Choose m large enough so that both L ⊗(m+1) |A and L ⊗m |B have vanishinghigher cohomology (i.e., h>0 = 0 for both; use the inductive hypothesis that L isample on all proper closed subvarieties, and Serre vanishing, Theorem 18.1.4(ii)).
This implies that for i ≥ 2,
∼ Hi (X, L ⊗m (−B)) (long exact sequence for (20.4.5.2))
Hi (X, L ⊗m ) =∼=Hi (X, L ⊗m+1 )(long exact sequence for (20.4.5.1))

so the higher cohomology stabilizes (is constant) for large m. Fromχ(X, L ⊗m ) = h0 (X, L ⊗m ) − h1 (X, L ⊗m ) + constant,
H0 (L ⊗m ) ̸= 0 for m ≫ 0, completing Step 2.
So by replacing L by a suitably large multiple (ampleness is independent oftaking tensor powers, Theorem 16.6.2), we may assume L has a section D. We
now use D as a crutch.
Step 3: L ⊗m is globally generated for m ≫ 0.
As D is effective, L ⊗m is globally generated on the complement of D: wehave a section nonvanishing on that big open set. Thus any base locus must be
contained in D. Consider the short exact sequence(20.4.5.3)

0 → L ⊗(m−1) → L ⊗m → L ⊗m |D → 0

Now L |D is ample by our inductive hypothesis. Choose m so large that H1 (X, L ⊗m |D ) =0 (Serre vanishing, Theorem 18.1.4(ii)). From the exact sequence associated to(20.4.5.3),
φm : H1 (X, L ⊗(m−1) ) → H1 (X, L ⊗m )is surjective for m ≫ 0. Using the fact that the H1 (X, L ⊗m ) are finite-dimensional
vector spaces, as m grows, H1 (X, L ⊗m ) must eventually stabilize, so the φm areisomorphisms for m ≫ 0.
Thus for large m, from the long exact sequence in cohomology for (20.4.5.3),
H0 (X, L ⊗m ) → H0 (X, L ⊗m |D ) is surjective for m ≫ 0. But H0 (X, L ⊗m |D ) has nobase points by our inductive hypothesis (applied to D), i.e., for any point p of D
there is a section of L ⊗m |D not vanishing at p, so H0 (X, L ⊗m ) has no base points
on D either, completing Step 3.

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Step 4. Thus L is a base-point-free line bundle with positive degree on eachcurve (by hypothesis of Theorem 20.4.1), so by Exercise 18.4.L we are done.
!
The following result is the key to proving Kleiman’s numerical criterion of
ampleness, Theorem 20.4.7.
20.4.6. Kleiman’s Theorem. — Suppose X is a projective k-scheme. If L is a nef
invertible sheaf on X, then (L k · V) ≥ 0 for every irreducible subvariety V ⊂ X ofdimension k.
As usual, this extends to the proper case, see [Kl1, Thm. IV.2.1]. And as usual,
we postpone the proof until after we appreciate the consequences.
20.4.B. E XERCISE .
(a) (limit of amples is nef) If L and H are any two invertible sheaves such thatL + ϵH is ample for all sufficiently small ϵ > 0, show that L is nef. (Hint:limϵ→0 . This doesn’t require Kleiman’s Theorem.)(b) (nef + ample = ample) Suppose X is a projective k-scheme, H is ample, and
L is nef. Show that L + ϵH is ample for all ϵ ∈ Q≥0 . (Hint: use the NakaiMoishezon criterion: ((L + ϵH )k · V) > 0. This may help you appreciate theadditive notation.)
20.4.7. Theorem (Kleiman’s numerical criterion for ampleness). — Suppose X isa projective k-scheme.
(a) The nef cone is the closure of the ample cone.
(b) The ample cone is the interior of the nef cone.
Side remark: of course (a) is false if “projective” is relaxed to “proper” (asthe ample cone of a proper nonprojective variety is empty, but 0 is nef so the nef
cone is nonempty). However, part (b) is true if X is proper and factorial (see [Kl1,
Thm. IV.2.2] for a proof and a more general statement). Hence if X is smooth,proper, and nonprojective, then the interior of the nef cone is empty.
20.4.C. E XERCISE .
See [Kl1, p. 326, Ex. 2] for Mumford’s example of a nonample line bundle on a smooth projective surface that meets every curve positively.
Doesn’t this contradict Theorem 20.4.7?
Proof. (a) Ample invertible sheaves are nef (Exercise 18.4.V(e)), and the nef cone
is closed (Exercise 18.4.W), so the closure of the ample cone is contained in thecone. Conversely, each nef element of N1Q (X) is the limit of ample classes by Exercise 20.4.B(a), so the nef cone is contained in the closure of the ample cone.
(b) As the ample cone is open (Proposition 20.4.3), the ample cone is containedin the interior of the nef cone. Conversely, suppose L is in the interior of the nef
cone, and H is any ample class. Then L − ϵH is nef for all small enough positiveϵ. Then by Exercise 20.4.B(b), L = (L − ϵH ) + ϵH is ample.
!
Suitably motivated, we prove Kleiman’s Theorem 20.4.6.
Proof. We may immediately reduce to the case where X is irreducible and reduced.
We work by induction on n := dim X. The base case n = 1 is obvious. So we
assume that (L dim V · V) ≥ 0 for all irreducible V not equal to X. We need only
show that (L n · X) ≥ 0.

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Fix some very ample invertible sheaf H on X.
20.4.D. E XERCISE . Show that (L k · H n−k · X) ≥ 0 for all k < n. (Hint: use theinductive hypothesis).
Consider P(t) := ((L + tH )n · X) ∈ Q[t]. We wish to show that P(0) ≥ 0.
Assume otherwise that P(0) < 0. Now for t ≫ 0, L + tH is ample, so P(t) is
positive for large t. Thus P(t) has positive real roots. Let t0 be the largest positive
real root of t. (In fact there is only one positive root, as Exercise 20.4.D shows that
all the nonconstant coefficients of P(t) are nonnegative.
20.4.E. E XERCISE . Show that for (rational) t > t0 , L + tH is ample. Hint: use the
Nakai-Moishezon criterion (Theorem 20.4.1); if V ̸= X is an irreducible subvariety,show that ((L + tH )dim V · V) > 0 by expanding (L + tH )dim V .
Let Q(t) := (L · (L + tH )n−1 · X) and R(t) := (tH · (L + tH )n−1 · X), so
P(t) = Q(t) + R(t).

20.4.F. E XERCISE . Show that Q(t) ≥ 0 for all rational t > t0 . Hint (which you will
have to make sense of): (L + tH ) is ample by Exercise 20.4.E, so for N sufficiently
large, N(L + tH ) is very ample. Use the idea of the proof of Proposition 20.1.4
to intersect X with n − 1 divisors in the class of N(L + tH ) so that “((N(L +tH ))n−1 · X) is an effective curve C”. Then (L · C) ≥ 0 as L is nef. Thus
Q(t0 ) ≥ 0.
20.4.G. E XERCISE . Show that R(t0 ) > 0. (Hint: Exercise 20.4.D.)
Thus P(t0 ) > 0 as desired.

!

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CHAPTER 21

Differentials

21.1 Motivation and game plan
Differentials are an intuitive geometric notion, and we are going to figure out
the right description of them algebraically. The algebraic manifestation is somewhat non-intuitive, so it is helpful to understand differentials first in terms of geometry. Also, although the algebraic statements are odd, none of the proofs are
hard or long. You will notice that this topic could have been done as soon as we
knew about morphisms and quasicoherent sheaves. We have usually introduced
new ideas through a number of examples, but in this case we will spend a fairamount of time discussing theory, and only then get to examples.
Suppose X is a “smooth” k-variety. We would like to define a tangent bundle.
We will see that the right way to do this will easily apply in much more generalcircumstances.
• We will see that the cotangent sheaf is more “natural” for schemes than tangent
bundle/sheaf. This is similar to the fact that the Zariski cotangent space is morenatural than the tangent space (i.e., if A is a ring and m is a maximal ideal, thenm/m2 is “more natural” than (m/m2 )∨ ), as we have repeatedly discussed since§12.1. In both cases this is because we are understanding “spaces” via their (sheafof) functions on them, which is somehow dual to the geometric pictures you haveof spaces in your mind.
So we will define the cotangent sheaf first. An element of the (co)tangent space
will be called a (co)tangent vector.
• Our undergraduate intuition will continue to work — we can calculate using theJacobian matrix, the answers will be the same in cases where our undergraduate
calculations apply, and we will even have the advantage of avoiding δ’s and ϵ’s.
• Our construction will automatically apply for general X, even if X is not “smooth”(or even at all nice, e.g., finite type). The cotangent sheaf will not necessarily belocally free, but it will still be a quasicoherent sheaf.
• Better yet, this construction will naturally work “relatively”. For any π : X →Y, we will define Ωπ = ΩX/Y , a quasicoherent sheaf on X, the sheaf of relative
differentials. The fiber of this sheaf at a point will be the cotangent vectors of thefiber of the map. This will specialize to the earlier case by taking Y = Spec k. The
idea is that this glues together the cotangent sheaves of the fibers of the family.
Figure 21.1 is a sketch of the relative tangent space of a map X → Y at a pointp ∈ X — it is the tangent to the fiber. (The tangent space is easier to draw than the
cotangent space!) An element of the relative (co)tangent space is called a verticalor relative (co)tangent vector.
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F IGURE 21.1. The relative tangent space of a morphism X → Y ata point p
Thus the central concept of this chapter is the cotangent sheaf Ωπ = ΩX/Y
for a morphism π : X → Y of schemes. A good picture to have in your mind
is the following. If π : X → Y is a submersion of manifolds (a map inducing asurjection on tangent spaces), you might hope that the tangent spaces to the fibers
at each point p ∈ X might fit together to form a vector bundle. This is the relative
tangent bundle (of π), and its dual is ΩX/Y (see Figure 21.1). Even if you are notgeometrically minded, you will find this useful. (For an arithmetic example, see
Exercise 21.2.F.)

21.2 Definitions and first properties
21.2.1. The affine case: three definitions.
We first study the affine case. Suppose A is a B-algebra, so we have a morphism of rings φ : B → A and a morphism of schemes Spec A → Spec B. I willdefine an A-module ΩA/B in three ways. This is called the module of relative
differentials or the module of Kähler differentials. The module of differentials
will be defined to be this module, as well as a map d : A → ΩA/B satisfying threeproperties. (Caution: although d sends an A-module to an A-module, it is not ingeneral A-linear. A priori we take it as a homomorphism of abelian groups, but
we will momentarily make it a homomorphism of B-modules, Exercise 21.2.A.)(i) additivity: da + da ′ = d(a + a ′ ).
(ii) Leibniz: d(aa ′ ) = a da ′ + a ′ da.
(iii) triviality on pullbacks: db = 0 for b ∈ φ(B).

These properties will not be surprising if you have seen differentials in any othercontext.
21.2.A. T RIVIAL EXERCISE . Show that d is B-linear.
21.2.B. E XERCISE . Prove the quotient rule: if a ′ = as, then s2 da = s da ′ − a ′ ds.

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21.2.C. E XERCISE . State and prove the chain rule for d(f(g)) where f is a polynomial with B-coefficients, and g ∈ A. (As motivation, think of the case B = k. So for
example, dan = nan−1 da, and more generally, if f is a,polynomial in one, variable,
df(a) = f ′ (a) da, where f ′ is defined formally: if f =ci xi then f ′ =ci ixi−1 .)We will now see three definitions of the module of Kähler differentials, which
will soon “sheafify” to the sheaf of relative differentials. The first definition is aconcrete hands-on definition. The second is by universal property. And the third
will globalize well, and will allow us to define ΩX/Y conveniently in general.

21.2.2. First definition of differentials: explicit description. We define ΩA/B
to be finite A-linear combinations of symbols “da” for a ∈ A, subject to the threerules (i)–(iii) above. For example, take A = k[x, y], B = k. Then a sample differential is 3x2 dy+4 dx ∈ ΩA/B . We have identities such as d(3xy2 ) = 3y2 dx+6xy dy.
21.2.3. Key fact. Note that if A is generated over B (as an algebra) by xi ∈A (where i lies in some index set, possibly infinite), subject to some relations rj(where j lies in some index set, and each is a polynomial in the xi ), then the Amodule ΩA/B is generated by the dxi , subject to the relations drj = 0. In short,
we needn’t take every single element of A; we can take a generating set. And weneedn’t take every single relation among these generating elements; we can takegenerators of the relations.
21.2.D. E XERCISE . Verify Key fact 21.2.3. (If you wish, use the affine conormal exact sequence, Theorem 21.2.12, to verify it; different people prefer to work throughthe theory in different orders. Just take care not to make circular arguments.)
In particular:
21.2.4. Proposition. — If A is a finitely generated B-algebra, then ΩA/B is a finite type(i.e., finitely generated) A-module. If A is a finitely presented B-algebra, then ΩA/B is a
finitely presented A-module.
Recall (§7.3.17) that a ring A is finitely presented over another ring B if it can be
expressed with finite number of generators and finite number of relations:
A = B[x1 , . . . , xn ]/(r1 (x1 , . . . , xn ), . . . , rj (x1 , . . . , xn )).
If A is Noetherian, then finitely presented is the same as finite type, as the “finite
number of relations” comes for free, so most of you will not care.
Let’s now see some examples. Among these examples are three particularly
important building blocks for ring maps: adding free variables; localizing; andtaking quotients. If we know how to deal with these, we know (at least in theory)
how to deal with any ring map. (They were similarly useful in understanding thefibered product in practice, in §9.2.)
21.2.5. Example: taking a quotient. If A = B/I, then ΩA/B = 0: da = 0 for alla ∈ A, as each such a is the image of an element of B. This should be believable;in this case, there are no “vertical tangent vectors”.
21.2.6. Example: adding variables. If A = B[x1 , . . . , xn ], then ΩA/B = Adx1 ⊕· · · ⊕ Adxn . (Note that this argument applies even if we add an arbitrarily infinite

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number of indeterminates.) The intuitive geometry behind this makes the answer
very reasonable. The cotangent bundle of affine n-space should indeed be free ofrank n.
21.2.7. Explicit example: an affine plane curve. Consider the plane curve y2 =x3 − x in A2k , where the characteristic of k is not 2. Let A = k[x, y]/(y2 − x3 + x)
and B = k. By Key fact 21.2.3, the module of differentials ΩA/B is generated by dx
and dy, subject to the relation
2y dy = (3x2 − 1) dx.
Thus in the locus where y ̸= 0, dx is a generator (as dy can be expressed in terms of
3 A/B is isomorphic to the trivial line bundledx). We conclude that where y ̸= 0, Ω
(invertible sheaf). Similarly, in the locus where 3x2 −1 ̸= 0, dy is a generator. These
two loci cover the entire curve, as solving y = 0 gives x3 − x = 0, i.e., x = 0 or ±1,
3 A/B is an invertible
and in each of these cases 3x2 − 1 ̸= 0. We have shown that Ωsheaf.
We can interpret dx and dy geometrically. Where does the differential dx vanish? The previous paragraph shows that it doesn’t vanish on the patch where 2y ̸=
0. On the patch where 3x2 − 1 ̸= 0, where dy is a generator, dx = (2y/(3x2 − 1)) dy
from which we see that dx vanishes precisely where y = 0. You should find thisbelievable from the picture. We have shown that dx = 0 precisely where the curve
has a vertical tangent vector (see Figure 19.7 for a picture). Once we can pull backdifferentials (Exercise 21.2.K(a) or Theorem 21.2.27), we can interpret dx as thepullback of a differential on the x-axis to Spec A (pulling back along the projection
to the x-axis). When we do that, using the fact that dx doesn’t vanish on the x-axis,
we can interpret the locus where dx = 0 as the locus where the projection map
branches. (Can you compute where dy = 0, and interpret it geometrically?)
This discussion applies to plane curves more generally. Suppose A = k[x, y]/f(x, y),where for convenience k = k. Then the same argument as the one given above
3 A/k is free of rank 1 on the open set D(∂f/∂x), and also on D(∂f/∂y).
shows that Ω
If Spec A is a regular curve, then these two sets cover all of Spec A. (Exercise 12.2.E— basically the Jacobian criterion — gives regularity at the closed points. Furthermore, the curve must be reduced, or else as the nonreduced locus is closed, itwould be nonreduced at a closed point, contradicting regularity. Finally, reducedness at a generic point is equivalent to regularity — a scheme whose underlying
set is a point is reduced if and only if it is regular. Alternatively, we could invoke
a big result, Fact 12.8.2, to get regularity at the generic point from regularity at theclosed points.)
If, on the other hand, the plane curve is singular, then Ω is not locally free ofrank one. For example, consider the plane curve Spec A where A = C[x, y]/(y2 −x3 ), so
ΩA/C = (A dx ⊕ A dy)/(2y dy − 3x2 dx).
Then the fiber of ΩA/C over the origin (computed by setting x = y = 0) is rank 2,
as it is generated by dx and dy, with no relation.
Implicit in the above discussion is the following exercise, showing that Ω canbe computed using the Jacobian matrix.

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21.2.E. I MPORTANT BUT E ASY E XERCISE (J ACOBIAN DESCRIPTION OF ΩA/B ). Suppose A = B[x1 , . . . , xn ]/(f1 , . . . , fr ). Then ΩA/B = {⊕i Adxi }/{dfj = 0} may beinterpreted as the cokernel of the Jacobian matrix (12.1.6.1)
J : A⊕r → A⊕n .

21.2.8. Example: localization. If T is a multiplicative subset of B, and A = T −1 B,then ΩA/B = 0. Reason: by the quotient rule (Exercise 21.2.B), if a = b/t, thenda = (t db − b dt)/t2 = 0. If A = Bf , this is intuitively believable; then Spec A isan open subset of Spec B, so there should be no vertical (co)tangent vectors.
21.2.F. I MPORTANT E XERCISE ( FIELD EXTENSIONS ). This notion of relative differentials is interesting even for finite extensions of fields. In other words, even when
you map a reduced point to a reduced point, something interesting can happen
with differentials.
(a) Suppose K/k is a separable algebraic extension. Show that ΩK/k = 0. Do notassume that K/k is a finite extension! (Hint: for any α ∈ K, there is a polynomial
f(x) such that f(α) = 0 and f ′ (α) ̸= 0.)(b) Suppose k is a field of characteristic p, K = k(tp ), L = k(t). Compute ΩL/K
(where K "→ L is the “obvious” inclusion).
We now delve a little deeper, and discuss two useful and geometrically motivated exact sequences.
21.2.9. Theorem (relative cotangent sequence, affine version). — Suppose C →B → A are ring morphisms. Then there is a natural exact sequence of A-modules
A ⊗B ΩB/C

a⊗db*→a db% ΩA/Cda*→da% ΩA/B% 0.

F IGURE 21.2. A sketch of the geometry behind the relative cotangent sequence

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The proof will be quite straightforward algebraically, but the statement comes
fundamentally from geometry, and that is how best to remember it. Figure 21.2 is a
sketch of a map π : X → Y. Here X should be interpreted as Spec A, Y as Spec B, andSpec C is a point. (If you would like a picture with a higher-dimensional Spec C,
just “take the product of Figure 21.2 with a curve”.) In the Figure, Y is “smooth”,
and X is “smooth over Y” — which means roughly that all the fibers are smooth.
Suppose p is a point of X. Then the tangent space of the fiber of π at p is certainly asubspace of the tangent space of the total space of X at p. The cokernel is naturally
the pullback of the tangent space of Y at π(p). This short exact sequence for eachp should be part of a short exact sequence of “relative tangent sheaves”
0 → TX/Y → TX/Z → π∗ TY/Z → 0

on X. (We will formally define “relative tangent sheaf” in §21.2.20.) Dualizing thisyields0 → π∗ ΩY/Z → ΩX/Z → ΩX/Y → 0.

This is precisely the statement of Theorem 21.2.9, except we also have leftexactness. This discrepancy is because the statement of the theorem is more general; we will see in Exercise 21.2.S that in the “smooth” case, we indeed have leftexactness.

21.2.10. Intriguing Remark. As always, whenever you see something right-exact,
you should suspect that there should be some sort of (co)homology theory so that
this is the end of a long exact sequence. This is indeed the case, and this exact
sequence involves André-Quillen homology (see [E, p. 386] for more). You should
expect that the next term to the left should be the first homology corresponding
to A/B, and in particular shouldn’t involve C. So if you already suspect that you
have exactness on the left in the case where A/B and B/C are “smooth” (whateverthat means), and the intuition of Figure 21.2 applies, then you should expect further that all that is necessary is that A/B be “smooth”, and that this would imply
that the first André-Quillen homology should be zero. Even though you wouldn’t
precisely know what all the words meant, you would be completely correct! You
would also be developing a vague inkling about the cotangent complex. We will see
examples when left-exactness holds in a sufficiently “smooth” situation in Proposition 21.7.2 and Exercise 21.2.S. For a more general statement, see [Stacks, tag
06B6]. See also [Liu, Cor. 6.3.22] and [Liu, Prop. 6.3.11] for clean discussions of thenext two terms in the sequence: “relative cotangent sequence”, “relative conormal
sequence”, . . . .
21.2.11. Proof of Theorem 21.2.9 (the relative cotangent sequence, affine version). First,
note that surjectivity of ΩA/C → ΩA/B is clear, as this map is given by da (→ da(where a ∈ A).
Next, the composition over the middle term is clearly 0, as this composition is
given by a ⊗ db (→ adb (→ 0.
Finally, we wish to identify ΩA/B as the cokernel of A ⊗B ΩB/C → ΩA/C .
Now ΩA/B is exactly the same as ΩA/C , except we have extra relations: db = 0
for b ∈ B. These are precisely the images of 1 ⊗ db on the left.
!
21.2.12. Theorem (conormal exact sequence, affine version). — Suppose B is a
C-algebra, I is an ideal of B, and A = B/I. Then there is a natural exact sequence of

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A-modules
I/I2% A ⊗B ΩB/C a⊗db*→a db % ΩA/C

δ

% 0.

Here δ is, informally, i (→ 1 ⊗ di, or more formally, 1 ⊗ d : B/I ⊗B I → B/I ⊗B ΩB/C .

(You will recognize the map A ⊗B ΩB/C → ΩA/C from the relative cotangent
sequence, Theorem 21.2.9.) The proof is algebraic, so the geometric discussion
thereafter may help clarify how you should really think of it.

Proof. We will identify the cokernel of δ : I/I2 → A ⊗B ΩB/C with ΩA/C . Consider
A ⊗B ΩB/C . As an A-module, it is generated by db (where b ∈ B), subject tothree types of relations: dc = 0 for c ∈ φ(C) (where φ : C → B describes B asa C-algebra), additivity, and the Leibniz rule. Given any relation in B, d of that
relation is 0.
Now ΩA/C is defined similarly, except there are more relations in A; theseare precisely the elements of I ⊂ B. Thus we obtain ΩA/C by starting out withA⊗B ΩB/C , and adding the additional relations di where i ∈ I. But this is precisely
the image of δ! (Be sure that you see how the identification of the cokernel of δ with
ΩA/C is precisely via the map a ⊗ db (→ a db.)!

We now give a geometric interpretation of the conormal exact sequence, andin particular define conormal modules/sheaves/bundles.
As with the relative cotangent sequence (Theorem 21.2.9), the conormal exact
sequence is fundamentally about geometry. To motivate it, consider the sketchof Figure 21.3. In the sketch, everything is “smooth”, X is one-dimensional, Y is
two-dimensional, j is the inclusion j : X "→ Y, and Z is a point. Then at a pointp ∈ X, the tangent space TX |p clearly injects into the tangent space of j(p) in Y,
and the cokernel is the normal vector space to X in Y at p. This should give anexact sequence of bundles on X:
0 → TX → j∗ TY → NX/Y → 0.

Dualizing this should give∨
0 → NX/Y→ j∗ ΩY/Z → ΩX/Z → 0.

This is precisely what appears in the statement of the Theorem, except (i) the exact
sequence in algebraic geometry is not necessarily exact on the left, and (ii) we see∨
I/I2 instead of NSpec
A/ Spec B .
21.2.13. We resolve the first issue (i) by expecting that the sequence of Theorem 21.2.12 is exact on the left in appropriately “smooth” situations, and thisis indeed the case, see Theorem 21.3.7 and Remark 21.3.8. (If you enjoyed Remark 21.2.10, you might correctly guess several things. The next term on the leftshould be the first André-Quillen homology of A/C, so we should only need that
A/C is smooth, and B should be irrelevant. Also, if A = B/I, then we should
expect that I/I2 is the first André-Quillen homology of A/B.)
21.2.14. Conormal modules and conormal sheaves. We resolve the second issue (ii) bydeclaring I/I2 to be the conormal module, and in Definition 21.2.15 we will define
the obvious analog as the conormal sheaf.

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Y

X

Z

F IGURE 21.3. A sketch of the geometry behind the conormal exact sequence
Here is some geometric intuition as to why we might want to call (the sheaf
associated to) I/I2 the conormal sheaf, which will likely confuse you, but may offer some enlightenment. First, if Spec A is a closed point of Spec B, we expect theconormal space to be precisely the cotangent space. And indeed if A = B/m, theZariski cotangent space is m/m2 . (We made this subtle connection in §12.1.) In
particular, at some point you will develop a sense of why the conormal (=cotangent) space to the origin in A2k = Spec k[x, y] is naturally the space of linear forms
αx + βy. But then consider the z-axis in Spec k[x, y, z] = A3k , cut out by I = (x, y).
Elements of I/I2 may be written as α(z)x + β(z)y, where α(z) and β(z) are polynomial. This reasonably should be the conormal space to the z-axis: as z varies,
the coefficients of x and y vary. More generally, the same idea suggests that theconormal module/sheaf to any coordinate k-plane inside n-space corresponds toI/I2 . Now consider a k-dimensional (smooth or differentiable) real manifold X
inside an n-dimensional manifold Y, with the classical topology. We can apply
the same construction: if I is the ideal sheaf of X in Y, then I /I 2 can be identified with the conormal sheaf (essentially the conormal vector bundle), because
analytically locally X "→ Y can be identified with Rk "→ Rn . For this reason, you
might hope that in algebraic geometry, if Spec A "→ Spec B is an inclusion of something “smooth” in something “smooth”, I/I2 should be the conormal module (or,after applying the functor ∼, the conormal sheaf). Motivated by this, we define theconormal module as I/I2 always, and then notice that it has good properties (suchas Theorem 21.2.12), but take care to learn what unexpected behavior it might havewhen we are not in the “smooth” situation, by working out examples such as thatof §21.2.7.
21.2.15. Definition. Suppose i : X "→ Y is a closed embedding of schemes cut out
by ideal sheaf I . Define the conormal sheaf for a closed embedding by I /I 2 ,∨denoted by NX/Y. (The product of quasicoherent ideal sheaves was defined in

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569∨
Exercise 14.3.E.) Important: We interpret NX/Y
as a quasicoherent sheaf on X, not(just) Y — be sure you understand why we may do so. (Exercise 17.1.E is one
reason, but it may confuse the issue more than help.)∨
Define the normal sheaf as its dual NX/Y := Hom (NX/Y, OX ). This is imperfectnotation, because it suggests that the dual of N is always N ∨ . This is not always
true, as for A-modules, the natural morphism from a module to its double-dual is
not always an isomorphism. (Modules for which this is true are called reflexive,
but we won’t use this notion.)If the normal sheaf is locally free, then we call the associated vector bundle thenormal bundle of X in Y.
∨
21.2.G. E ASY E XERCISE . Define the conormal sheaf NX/Y
(and hence the normal
sheaf NX/Y ) for a locally closed embedding i : X "→ Y of schemes, a quasicoherent
sheaf on X. (Make sure your definition is well-defined!)

In the good situation of a regular embedding, the conormal sheaf (and hence
the normal sheaf) is locally free (or more informally, a vector bundle). In particular, the dual of N is indeed N ∨ . As a warm-up we deal with the important
codimension 1 case.
21.2.H. E XERCISE : N ORMAL BUNDLES TO EFFECTIVE C ARTIER DIVISORS . Suppose D ⊂ X is an effective Cartier divisor (§8.4.1). Show that the conormal sheaf∨
ND/X
is O(−D)|D (and in particular is an invertible sheaf), and hence that thenormal sheaf is O(D)|D . It may be surprising that the normal sheaf should be lo∼ A2 and D is the union of the two axes (and more generally if X
cally free if X =is regular but D is singular), because you may be used to thinking that a “tubularneighborhood” is isomorphic to the normal bundle.
We now treat the general case.
21.2.16. Proposition. —(a) Suppose A is Noetherian. If I is generated by a regular sequence x1 , . . . , xr , then themap γ : (A/I)⊕r → I/I2 given by (a1 , . . . , ar ) (→ a1 x1 + · · · + ar xr describes I/I2 as afree module of rank r over A/I with basis x1 , . . . , xr .
(b) The (co)normal sheaf of a codimension r regular embedding is locally free of rank r.
Proof. (a) Clearly γ is surjective. We now show that it is injective. It suffices to showthat it is injective upon localization to all points of Spec A/I, because injectivity
is a stalk-local condition. Then in such a localization, the regular sequence in A
remains regular (Exercise 8.4.D, plus the fact that the strict inclusion (x1 , . . . , xr ) "A remains strict upon localization at a prime containing I). We are thus reduced
to the local question. (Caution: Make sure you see why (I/I2 )m = Im /(Im )2 .)Consider the ai ’s instead as elements of A. Suppose now that (a1 , . . . , ar ) ∈ker γ; we will show that each ai is in I. Now because xr is not a zerodivisor on
A/(x1 , . . . , xr−1 ), since ar xr = 0 in A/(x1 , . . . , xr−1 ), we have ar ∈ (x1 , . . . , xr−1 ) ⊂I. Because the roles of the ai ’s are symmetric by Theorem 8.4.6, we are done.
(b) follows immediately from (a).
!

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We will soon meet a related but harder fact, that if I is the ideal sheaf of aregular embedding of codimension r, then I n /I n+1 is locally free, because it isSymn (I /I 2 ) (see Theorem 22.3.8).
21.2.17. Second definition: universal property. Here is a second definition thatis important philosophically, by universal property. Of course, it is a characterization rather than a definition: by universal property nonsense, it shows that if themodule exists (with the d map), then it is unique up to unique isomorphism, andthen one still has to construct it to make sure that it exists.
Suppose A is a B-algebra, and M is a A-module. A B-linear derivation of
A into M is a map d : A → M of B-modules (not necessarily a map of A-modules)
satisfying the Leibniz rule: d(fg) = f dg + g df. As an example, suppose B = k,
and A = k[x], and M = A. Then d/dx is a k-linear derivation. As a second
example, if B = k, A = k[x], and M = k, then (d/dx)|0 (the operator “evaluate thederivative at 0”) is a k-linear derivation.
A third example is d : A → ΩA/B , and indeed d : A → ΩA/B is the universal Blinear derivation of A. Precisely, the map d : A → ΩA/B is defined by the following
universal property: any other B-linear derivation d ′ : A → M factors uniquely
through d:
d′

%M
A❊
❊❊
②N
②
❊❊
②
❊❊
②②
②② f
d ❊❊
②
M
②
ΩA/B

Here f is a map of A-modules. (Note again that d and d ′ are not necessarily mapsof A-modules — they are only B-linear.) By universal property nonsense, if itexists, it is unique up to unique isomorphism. The map d : A → ΩA/B clearlysatisfies this universal property, essentially by definition.
The next result connects the cotangent module ΩA/B to the cotangent space ata (rational) point.
21.2.18. Proposition (the fiber of Ω at a rational point is the cotangent space). —Suppose B is a k-algebra, and m ⊂ B is a maximal ideal with residue field k. Then there is
an isomorphism of k-vector spaces δ : m/m2 → ΩB/k ⊗B k (where the k on the right is a
∼ B/m).
B-module via the isomorphism k =Corollary 21.3.9 will give a quite different proof, and generalize it to the casewhere B/m is a separable extension of k.
Proof. We instead show an isomorphism of dual vector spaces
Homk (ΩB/k ⊗B k, k) → Homk (m/m2 , k).

We have (canonical) isomorphisms

Homk (ΩB/k ⊗B k, k) ==HomB (ΩB/k ⊗B k, k)

HomB (ΩB/k , HomB (k, k))=

HomB (ΩB/k , Homk (k, k))=

HomB (ΩB/k , k),

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where in the right argument of HomB (ΩB/k , k), k is a B-module via its manifestation as B/m. By the universal property of ΩB/k (§21.2.17), HomB (ΩB/k , k) corre∼ k.
sponds to the k-derivations of B into B/m =21.2.I. E XERCISE . Show that these are precisely the elements of Homk (m/m2 , k).
(The algebra involved is essentially the same as that of Exercise 12.1.A.)!
You can verify that this δ is the one appearing in the conormal exact sequence,
Theorem 21.2.12, with I = m and A = C = k. In fact, from the conormal exact
sequence, we can immediately see that δ is a surjection, as Ωk/k = 0.
21.2.19. Remark.
Proposition 21.2.18, in combination with the Jacobian exercise 21.2.E above, gives a second proof of Exercise 12.1.G, the Jacobian method forcomputing the Zariski tangent space at a k-valued point of a finite type k-scheme.
Corollary 21.3.9 will extend this to the case of a separable closed point.
21.2.J. E XERCISE . Suppose X is a finite type scheme over an algebraically closed
field. Show that the function from the closed points of X to Z≥0 given by p (→dim Tp X is upper semicontinuous in the Zariski topology. (Be clear on what the
Zariski topology is on the set of closed points.) Corollary 21.3.9 will allow you toextend this to all fields of characteristic 0 and all finite fields.
Depending on how your brain works, you may prefer using the first (constructive) or second (universal property) definition to do the next two exercises.
21.2.K. E XERCISE .
(a) (pullback of differentials) If
A7 ′ (A7

B′ (B
is a commutative diagram, describe a natural homomorphism of A ′ -modules A ′ ⊗A
ΩA/B → ΩA ′ /B ′ . An important special case is B = B ′ .
(b) (differentials behave well with respect to base extension, affine case) If furthermore∼ B ′ ⊗B A, so the diagram is
the above diagram is a “tensor diagram” (i.e., A ′ =′“co-Cartesian”) then show that A ⊗A ΩA/B → ΩA ′ /B ′ is an isomorphism. (Depending on how your proceed, this may be trickier than you expect.)
21.2.L. E XERCISE : LOCALIZATION ( STRONGER FORM ; CF. E XAMPLE 21.2.8). Suppose φ : B → A is a map of rings, S is a multiplicative subset of A, and T is amultiplicative subset of B with φ(T ) ⊂ S, so we have the following commutative
diagram.
S−17 A (A7

T −1 B (B−1Show that the pullback of differentials S ΩA/B → ΩS−1 A/T −1 B of Exercise 21.2.K(a)is an isomorphism. (This should be believable from the intuitive picture of “vertical cotangent vectors”.) An important case is when T = {1}.

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21.2.M. E XERCISE ( FIELD EXTENSIONS CONTINUED , CF. E XERCISE 21.2.F).
(a) Compute Ωk(t)/k . (Hint: §21.2.6 followed by Exercise 21.2.L.)(b) If K/k is separably generated by t1 , . . . , tn ∈ K (i.e., t1 , . . . , tn form a transcendence basis, and K/k(t1 , . . . , tn ) is algebraic and separable), show that ΩK/k
is a free K-module (i.e., vector space) with basis dt1 , . . . , dtn . Hint: use the relative cotangent sequence (Theorem 21.2.9) for k "→ k(t1 , . . . , tn ) "→ K to showthat the dti span ΩK/k as a K-vector space. The tricky part is showing that thedti are linearly independent. Do this by showing that there exists a unique mapΩK/k → K sending dt1 to 1, and dti to 0 for i > 1. Do this first for the case whereK/k(t1 , . . . , tn ) is generated by one element; then where it is finitely generated;then the general case by defining it on all finitely generated subextensions, andusing uniqueness to show that they “all agree”.
21.2.20. Third definition: global. We now want to globalize this definition for an
arbitrary morphism of schemes π : X → Y. We could do this “affine by affine”; wejust need to make sure that the above notion behaves well with respect to “change
of affine sets”. Thus a relative differential, on X would be the data of, for every
affine U ⊂ X, a differential of the formai ,dbi , and on the intersectionof two,affine open sets U ∩ U ′ , with representatives ai dbi on U and ai′ dbi′ on thesecond, an equality on the overlap. Instead, we take a different approach. I will
give the (seemingly unintuitive) definition, then tell you how to think about it, andthen get back to the definition.
Let π : X → Y be any morphism of schemes. Recall that δ : X → X ×Y X is alocally closed embedding (Proposition 10.1.3). Define the relative cotangent sheaf∨ΩX/Y (or Ωπ ) as the conormal sheaf NX/X×of the diagonal (see §21.2.14 — andYX
if X → Y is separated you needn’t even worry about Exercise 21.2.G). (Now isalso as good a time as any to define the relative tangent sheaf TX/Y as the dual
Hom (ΩX/Y , OX ) to the relative cotangent sheaf. If we are working in the categoryof k-schemes, then ΩX/k and TX/k are often called the cotangent sheaf and tangent sheaf of X respectively.)
We now define d : OX → ΩX/Y . Let pr1 : X ×Y X → X and pr2 : X ×Y X → X be
the two projections. Then define d : OX → ΩX/Y on the open set U as follows:
df = pr∗2 f − pr∗1 f.

(Warning: this is not a morphism of quasicoherent sheaves on X, although it is
OY -linear in the only possible meaning of that phrase.) We will soon see that d isindeed a derivation of the sheaf OX (in the only possible meaning of the phrase),
and at the same time see that our new notion of differentials agrees with our old
definition on affine open sets, and hence globalizes the definition. Note that for
any open subset U ⊂ X, d induces a map(21.2.20.1)

Γ (U, OX ) → Γ (U, ΩX/Y ),which we also call d, and interpret as “taking the derivative”.
21.2.21. Motivation. Before connecting this to our other definitions, let me try
to convince you that this is a reasonable definition to make. (This discussion is
informal and nonrigorous.) Say for example that Y is a point, and X a differentiable
manifold. Then the tangent bundle TX×X on X × X is pr∗1 TX ⊕ pr∗2 TX , where pr1 andpr2 are the projections from X × X onto its two factors. Restrict this to the diagonal

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November 18, 2017 draft573∆, and look at the normal bundle exact sequence:
0 → T∆ → TX×X |∆ → N∆/X → 0.

Now the left morphism sends v to (v, v), so the cokernel can be interpreted as(v, −v). Thus N∆/X is isomorphic to TX . Thus we can turn this on its head: weknow how to find the normal bundle (or more precisely the conormal sheaf), andwe can use this to define the tangent bundle (or more precisely the cotangent
sheaf).
21.2.22. Testing this out in the affine case. Let’s now see how this works for the
special case Spec A → Spec B. Then the diagonal δ : Spec A "→ Spec A ⊗B A
corresponds to the ideal I of A ⊗B A that is the kernel of the ring map.
.
α:
xi ⊗ yi →xi yi .

21.2.23. The idealby the elements of the form 1 ⊗ a − a ⊗ 1.
, I of A ⊗B A is generated,
Reason: if α( xi ⊗ yi ) = 0, i.e., xi yi = 0, then.
.
.
xi ⊗ yi =(xi ⊗ yi − xi yi ⊗ 1) =xi (1 ⊗ yi − yi ⊗ 1).

The derivation is d : A → A ⊗B A, a (→ (1 ⊗ a − a ⊗ 1) (an element of I, taken
modulo I2 ). (We shouldn’t really call this “d” until we have verified that it agrees
with our earlier definition, but we irresponsibly will anyway.)
Let’s check that d is indeed a derivation (§21.2.17). Clearly d is B-linear, so wecheck the Leibniz rule:d(aa ′ ) − a da ′ − a ′ da

= 1 ⊗ aa ′ − aa ′ ⊗ 1 − a ⊗ a ′ + aa ′ ⊗ 1 − a ′ ⊗ a + a ′ a ⊗ 1= −a ⊗ a ′ − a ′ ⊗ a + a ′ a ⊗ 1 + 1 ⊗ aa ′= (1 ⊗ a − a ⊗ 1)(1 ⊗ a ′ − a ′ ⊗ 1)∈

I2 .

Thus by the universal property of ΩA/B , we have a natural morphism ΩA/B →I/I2 of A-modules.
21.2.24. Theorem. — The natural morphism f : ΩA/B → I/I2 induced by the universal
property of ΩA/B is an isomorphism.

Proof. We will show this as follows. (i) We will show that f is surjective, and (ii)
we will describe g : I/I2 → ΩA/B such that g ◦ f : ΩA/B → ΩA/B is the identity(showing that f is injective).
(i) The map f sends da to 1 ⊗ a − a ⊗ 1, and such elements generate I (§21.2.23),
so f is surjective.
(ii) Consider the map A⊗B A → ΩA/B defined by x⊗y (→ x dy. (This is a welldefined map, by the universal property of ⊗, see §1.3.5.) Define g : I/I2 → ΩA/B
as the restriction of this map to I. We need to check that this is well-defined, i.e.,that elements of I2 are sent to 0, i.e., we need that
$.
% $.
% .
xi ⊗ yixj′ ⊗ yj′ =xi xj′ ⊗ yi yj′ (→ 0i,jEarly (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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The Rising Sea: Foundations of Algebraic Geometry

when
.
i,j

,

,

xj′ yj′ = 0. But by the Leibniz rule,.
.
xi xj′ d(yi yj′ ) =xi xj′ yi dyj′ +xi xj′ yj′ dyi
i xi yi =

i,j

4

=

= 0.

i,j

⎞ 4
⎞
5⎛
5⎛
.
.
.
.
xi yi ⎝
xj′ dyj′ ⎠ +xi dyi ⎝xj′ yj′ ⎠
i

j

i

j

Then g ◦ f is indeed the identity, asda ✤f% 1⊗a−a⊗1 ✤g% 1 da − a d1 = da!

as desired.

21.2.N. E ASY EXERCISE . Suppose π : X → Y is a morphism of schemes, with open
subschemes Spec A ⊂ X and Spec B ⊂ Y, with Spec A ⊂ π−1 (Spec B). Identify
3 A/B , and identify d : Γ (Spec A, OX ) → Γ (Spec A, ΩX/Y ) with
ΩX/Y |Spec A with Ωd : A → ΩA/B . Thus the global construction indeed naturally “glues together” theaffine construction.
We can now use our understanding of how Ω works on affine open sets togeneralize previous statements to non-affine settings.
21.2.O. E XERCISE . If U ⊂ X is an open subset, show that the map (21.2.20.1) is a
derivation.
21.2.P. E XERCISE . Suppose π : X → Y is locally of finite type, and Y (and henceX) is locally Noetherian. Show that ΩX/Y is a coherent sheaf on X. (Feel free toweaken the Noetherian hypotheses for weaker conclusions.)
21.2.Q. E XERCISE . Suppose π : X → Y is smooth of relative dimension n. Provethat ΩX/Y is locally free of rank n. Hint: the Jacobian description of Ω, Exercise 21.2.E.
The relative cotangent exact sequence and the conormal exact sequence forschemes now directly follow.
21.2.25. Theorem. —ρπ %% Z are morphisms of(a) (relative cotangent exact sequence) Suppose X
Yschemes. Then there is an exact sequence of quasicoherent sheaves on Xπ∗ ΩY/Z% ΩX/Z% ΩX/Y% 0,

globalizing Theorem 21.2.9.
(b) (conormal exact sequence) Suppose ρ : Y → Z is a morphism of schemes, and i : X "→ Y∨
is a closed embedding, with conormal sheaf NX/Y
. Then there is an exact sequence ofsheaves on X:δ % ∗% ΩX/Z% 0,N∨
i ΩY/Z
X/Y

globalizing Theorem 21.2.12.

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21.2.R. E XERCISE . Prove Theorem 21.2.25. (What needs to be checked?)
You should expect these exact sequences to be left-exact as well in the presenceof appropriate smoothness (see Remark 21.2.10 and §21.2.13). The following is an
important special case for the relative cotangent sequence. (Theorem 21.3.7 will
give a similar statement for the conormal exact sequence, for smooth varieties.)
21.2.S. E XERCISE ( LEFT- EXACTNESS OF THE RELATIVE COTANGENT SEQUENCE WHEN
π AND ρ ARE SMOOTH ). Show that the relative cotangent exact sequence is exact
on the left if π and ρ are smooth. Hint: the (m + n) × (m + n) matrix you used in
Exercise 12.6.D is “block upper triangular”.
21.2.26. Pulling back relative differentials. Not surprisingly, the sheaf of relative
differentials pull back, and behave well under base change.
21.2.27. Theorem (pullback of differentials). —(a) If
X′

µ

%X

"
"
%Y
Y′
is a commutative diagram of schemes, there is a natural homomorphism µ∗ ΩX/Y →ΩX ′ /Y ′ of quasicoherent sheaves on X ′ . An important special case is Y = Y ′ .
(b) (Ω behaves well under base change) If furthermore the above diagram is a Cartesian∼ X ×Y Y ′ ) then µ∗ ΩX/Y → ΩX ′ /Y ′ is an isomorphism.
square (so X ′ =21.2.T. E XERCISE . Derive Theorem 21.2.27 from Exercise 21.2.K. (Why does theconstruction of Exercise 21.2.K(a) “glue well”?)

As a particular case of Theorem 21.2.27(b), the fiber of the sheaf of relative
differentials is indeed the sheaf of differentials of the fiber. Thus the sheaf of differentials notion indeed “glues together” the differentials on each fiber.
21.2.U. E XERCISE . Suppose α : X → Z and β : Y → Z are two morphisms. Describe∼ α∗ ΩX/Z ⊕ β∗ ΩY/Z .
an isomorphism ΩX×Z Y/Z =21.3 Smoothness of varieties revisited
Suppose k is a field. Since §12.2.5, we have used an awkward definition of
k-smoothness, and we now make our definition of k-smoothness more robust.
21.3.1. Redefinition. A k-scheme X is k-smooth of dimension n or smooth ofdimension n over k if it is locally of finite type, of pure dimension n, and ΩX/kis locally free of rank n. The dimension n is often omitted, and one might (possibly) want to call something smooth if it is the (scheme-theoretic) disjoint union ofthings smooth of various dimensions.
21.3.A. E XERCISE .
Definition 12.2.6.

Verify that this definition is equivalent to the one given in

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21.3.B. ⋆⋆ E XERCISE ( FOR THOSE WITH BACKGROUND IN COMPLEX GEOMETRY ).
Suppose X is a complex algebraic variety. Show that the analytification Xan of X
(defined in Exercise 5.3.G) is smooth (in the differential-geometric sense) if andonly if X is smooth (in the algebro-geometric sense, over C). In this case, showthat complex dimension of the complex manifold Xan (half the real dimension) is
dim X. Hint: the Jacobian criterion applies in both settings.
21.3.2. Important observation. As a consequence of our better definition of smoothness, we see that it can be checked on any affine cover by using the Jacobian criterion on each affine open set in the cover, as hinted in §12.2.7.
21.3.C. E XERCISE ( SMOOTHNESS IS INSENSITIVE TO EXTENSION OF BASE FIELD ).
Prove the converse to Exercise 12.2.G.
21.3.D. I MPORTANT E XERCISE . Suppose k is perfect, and X is a finite type kscheme. Show that X is smooth if and only if X is regular at all closed points. In
particular, Theorem 12.2.10(a) holds. Hint: let Xk := X ×k k. Explain why X issmooth if and only if Xk is smooth if and only if Xk is regular at all closed points
if and only if X is regular at all closed points. See Exercise 12.2.O for the last step.
21.3.3. Remark: A short definition of smoothness of varieties. Exercise 21.3.D gives
another pithy characterization of smoothness of varieties: a finite type k-scheme issmooth if it is geometrically regular.
21.3.4. Generic smoothness.
We can now verify something you may already have intuited. In positive
characteristic, this is a hard theorem, in that it uses a result from commutative
algebra that we have not proved.
21.3.5. Theorem (generic smoothness of varieties). — If X is an irreducible variety
over a perfect field k of dimension n, there is a dense open subset U of X such that U issmooth of dimension n.
Hence, by Fact 12.8.2, U is regular.Theorem 25.3.1 will generalize this to smoothmorphisms, at the expense of restricting to characteristic 0.
Proof. The n = 0 case is immediate, so we assume n > 0.
We will show that the rank at the generic point is n. Then by upper semicontinuity of the rank of a coherent sheaf (Exercise 13.7.J), it must be n in an open
neighborhood of the generic point, and we are done.
We thus have to check that if K is the fraction field of a dimension n integral
finite type k-scheme, i.e., (by Theorem 11.2.1) if K/k is a transcendence degreen extension, then ΩK/k is an n-dimensional vector space. But every extensionof transcendence degree n > 1 is separably generated (see Exercise 21.2.M(b)): we
can find n algebraically independent elements of K over k, say x1 , . . . , xn , suchthat K/k(x1 , . . . , xn ) is separable. (In characteristic 0, this is automatic from transcendence theory, see Exercise 11.2.A, as all finite extensions are separable. But it
also holds for perfect fields in positive characteristic, see [E, Cor. A1.7] or [Mat2,Thm. 26.2].) Then ΩK/k is generated by dx1 , . . . , dxn (by Exercise 21.2.M(b)). !

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21.3.6. Left-exactness of the conormal exact sequence for embeddings of smoothvarieties.
As described in §21.2.13, we expect the conormal exact sequence to be exact
on the left in appropriately “smooth” situations.
21.3.7. Theorem (conormal exact sequence for smooth varieties). — Suppose
i : X "→ Y is a closed embedding of smooth varieties over a field k, with conormal sheaf∨
NX/Y. Then the conormal exact sequence (Theorem 21.2.25(b)) is exact on the left:
0

%N∨

X/Y

δ

% i∗ ΩY/k% ΩX/k%0

is exact.
By dualizing, i.e., applying Hom (·, OX ), we obtain the normal exact sequence
0 → TX/k → TY/k |X → NX/Y → 0

which is geometrically more intuitive (see Figure 21.3 and the discussion after the
proof of Theorem 21.2.12).
21.3.8. Remark. The conormal sequence is exact on the left in even more generalcircumstances. Essentially all that is required is appropriate smoothness of i ◦g : X → Z, see [Stacks, tag 06B7], and [Stacks, tag 06BB] for related facts.

Proof. We use the fact that smooth k-varieties are regular at their closed points,
which we have only proved for perfect fields, but which we shall later prove fully(see §12.2.9).
Let I be the ideal sheaf of i : X "→ Y. By Exercise 12.2.L(b), i is a regular
embedding in the open neighborhood of all closed points, and hence everywhere.
Thus by Proposition 21.2.16(b), I /I 2 is locally free of rank r. By Exercise 5.5.L,
the associated points of ker δ are a subset of the associated points of I /I 2 , so
to show that ker δ = 0, it suffices to check this at the associated points of I /I 2 ,which are precisely the associated points of X (as I /I 2 is locally free).
As stated above, X is regular at closed points, hence (by Theorem 12.2.13) reduced (the nonreduced locus is closed, §8.3.10, and hence if nonempty would necessarily contain a closed point). Thus we need only check at the generic points of(the irreducible components of) X (as X has no embedded points, Exercise 5.5.C).
But this involves checking the left-exactness of a right-exact sequence of vector
spaces, and as the dimension of the left (the codimension of X in Y by Proposition 21.2.16(b)) is precisely the difference of the other two, we are done.
!
21.3.9. Important Corollary. — Suppose Y is a smooth k-variety, and q ∈ Y is a closed
point whose residue field κ(q) is separable over k. (This is automatic if char k = 0 or if
k is a finite field.) Then the conormal exact sequence for q "→ Y yields an isomorphism of
the Zariski cotangent space of Y at q with the fiber of ΩY/k at q.
Proof. Apply Theorem 21.3.7 with X = q. Note that q is indeed a smooth k-variety(using separability of κ(q)/k!), and that Ωq/k = 0.
!
21.3.10. Remark. This result generalizes Proposition 21.2.18 to separable closed
points. As described after the statement of Proposition 21.2.18, this has a number

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The Rising Sea: Foundations of Algebraic Geometryof consequences. For example, it extends the Jacobian description of the Zariski
tangent space to separable closed points.
21.3.E. E XERCISE ( CF. §12.2.16). Suppose p is a closed point of a k-variety X, withresidue field κ(p) that is separable over k. Define π : Xk := X ×k k → X by basechange from Spec k → Spec k. Suppose q ∈ π−1 (p). Show that Xk is regular at q if
and only if X is regular at p.

21.3.11. Proving Theorem 12.8.3, that the localization of regular local rings are
regular in the case of varieties over perfect fields. Theorems 21.3.5 and 21.3.7
allow us to prove Theorem 12.8.3 (an important case of Fact 12.8.2).
21.3.12. Proof of Theorem 12.8.3. Suppose Y is a variety over a perfect field k that isregular at its closed points (so Y is smooth, by Exercise 21.3.D), and let η be a pointof Y. We will show that Y is regular at η. Let X = η. By Theorem 21.3.5, X contains
a dense (=nonempty) open subset of smooth points. By shrinking Y by discarding
the points of X outside that open subset, we may assume X is smooth.
Then Theorem 21.3.7 (exactness of the conormal exact sequence for smoothvarieties) implies that I /I 2 is a locally free sheaf of rank codimX/Y = dim OY,η .
21.3.F. E XERCISE . Let m be the maximal ideal of OY,η . Identify the stalk of I /I 2
at the generic point η of X with m/m2 . Conclude the proof of Theorem 12.8.3. !
21.3.G. T RICKY E XERCISE ( LOCALIZATION OF REGULAR LOCAL RINGS OF VARI ETIES ARE REGULAR , PROMISED JUST AFTER T HEOREM 12.8.3). Suppose (A, m) isa regular local ring that is the localization of a finitely generated k-algebra, where
k is perfect. Show that the localization of A at a prime is also a regular local ring.

21.4 Examples
The examples below are organized by topic, not by difficulty.
21.4.1. The geometric genus of a curve. A regular irreducible projective curve
C (over a field k) has geometric genus h0 (C, ΩC/k ). (This will be generalized
to higher dimension in §21.5.3.) This is always finite, as ΩC/k is coherent (Exercise 21.2.P), and coherent sheaves on projective k-schemes have finite-dimensional
spaces of sections (Theorem 18.1.4(i)). (The geometric genus is also called thefirst algebraic de Rham cohomology group, in analogy with de Rham cohomology
in the differentiable setting.) Sadly, this isn’t really a new invariant. We will
see in Exercise 21.5.C that this agrees with our earlier definition of genus, i.e.,h0 (C, ΩC/k ) = h1 (C, OC ).
Now let’s compute some differentials!
21.4.2. The projective line. As an important first example, consider P1k , withthe usual projective coordinates x0 and x1 . As usual, the first patch corresponds
to x0 ̸= 0, and is of the form Spec k[x1/0 ] where x1/0 = x1 /x0 . The second patch
corresponds to x1 ̸= 0, and is of the form Spec k[x0/1 ] where x0/1 = x0 /x1 .

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November 18, 2017 draft579∼ O 1 . (More precisely, Ωk[x]/k =Both patches are isomorphic to A1k , and ΩA1k =Akk[x] dx.) Thus ΩP1k is an invertible sheaf (a line bundle). The invertible sheaves on
P1k are of the form O(m). So which invertible sheaf is ΩP1 /k ?
Let’s take a section, dx1/0 on the first patch. It has no zeros or poles there, so
let’s check what happens on the other patch. As x1/0 = 1/x0/1 , we have dx1/0 =−(1/x20/1 ) dx0/1 . Thus this section has a double pole where x0/1 = 0. Hence∼ O(−2).
ΩP1k /k =Note that the above argument works equally well if k were replaced by Z: ourtheory of Weil divisors and line bundles of Chapter 14 applies (P1Z is factorial),∼ O(−2).
so the previous argument essentially without change shows that ΩP1Z /Z =And because Ω behaves well with respect to base change (Theorem 21.2.27(b)),∼ O 1 (−2) for any baseand any scheme maps to Spec Z, this implies that ΩP1B /B =PB
scheme B.
(Also, as suggested by §18.5.2, this shows that ΩP1 /k is the dualizing sheaf for1
Pk ; see also Example 18.5.4. But given that we haven’t yet proved Serre duality,
this isn’t so meaningful.)
Side Remark: the fact that the degree of the tangent bundle is 2 is related to
the “Hairy Ball Theorem” (the dimension 2 case of [Hat, Thm. 2.28]).
21.4.3. Hyperelliptic curves. Throughout this discussion of hyperelliptic curves,
we suppose that k = k and char k ̸= 2, so we may apply the discussion of §19.5.
Consider a double cover π : C → P1k by a regular projective curve C, branched over
2g + 2 distinct points. We will use the explicit coordinate description of hyperelliptic curves of (19.5.2.1). By Theorem 19.5.1, C has genus g.
21.4.A. E XERCISE : DIFFERENTIALS ON HYPERELLIPTIC CURVES . What is the degree of the invertible sheaf ΩC/k ? (Hint: let x be a coordinate on one of the coordinate patches of P1k . Consider π∗ dx on C, and count poles and zeros. Use theexplicit coordinates of §19.5. You should find that π∗ dx has 2g + 2 zeros and 4
poles (counted with multiplicity), for a total of 2g − 2.) Doing this exercise will set
you up well for the Riemann-Hurwitz formula, in §21.7.
21.4.B. E XERCISE (“ THE FIRST ALGEBRAIC DE R HAM COHOMOLOGY GROUP OF A
HYPERELLIPTIC CURVE ”). Show that h0 (C, ΩC/k ) = g as follows.
2(a) Show that dx
y is a (regular) differential on Spec k[x, y]/(y − f(x)) (i.e., an element of Ω(k[x,y]/(y2 −f(x)))/k ).
(b) Show that for 0 ≤ i < g, xi (dx)/y extends to a global differential ωi on C (i.e.,with no poles).
(c) Show that the ωi (0 ≤ i < g) are linearly independent differentials. (Hint:,g−1
Show that the valuation of ωi at x = 0 is i. If ω := j=i aj ωj is a nontrivial linear
combination, with aj ∈ k, and ai ̸= 0, show that the valuation of ω at x = 0 is i,and hence ω ̸= 0.)⋆ (d) Show that the ωi form a basis for the differentials. (Hint: consider the orderof zeros of the ωi at π−1 (∞).)
21.4.C. ⋆ E XERCISE ( TOWARD S ERRE DUALITY ).
(a) Show that h1 (C, ΩC/k ) = 1. Interpret a generator of H1 (C, ΩC/k ) as x−1 dx. (Inparticular, the pullback map H1 (P1 , ΩP1 /k ) → H1 (C, ΩC/k ) is an isomorphism.)

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580

The Rising Sea: Foundations of Algebraic Geometry(b) Describe a natural perfect pairing
H0 (C, ΩC/k ) × H1 (C, OC ) → H1 (C, ΩC/k ).

In terms of our explicit coordinates, you might interpret it as follows. Recall fromthe proof of the hyperelliptic Riemann-Hurwitz formula (Theorem 19.5.1) that
H1 (C, OC ) can be interpreted as=y yy>, 2,..., g .
x xxThen the pairing
?
@ =
A
B
dx x dxdxy yy>,, . . . , xg−1
×
, 2 , . . . , g → x−1 dx
y
y
y
x x
x

is basically “multiply and read off the x−1 dx term”. Or in fancier terms: “multiply
and take the residue”. (You may want to compare this to Example 18.5.4.)
21.4.4. Discrete valuation rings. The following exercise is used in the proof of the
Riemann-Hurwitz formula, §21.7.

21.4.D. E XERCISE . Suppose that the discrete valuation ring (A, m, k) is a localization of a finitely generated k-algebra. Let t be a uniformizer of A. Show that thedifferentials are free of rank one and generated by dt, i.e., ΩA/k = A dt, as follows.
(It is also possible to show this using the ideas from §21.5.)(a) Show that ΩA/k is a finitely generated A-module.
(b) Show that ⟨dt⟩ = ΩA/k , i.e., that ×dt : A → ΩA/k is a surjection, as follows.
Let π be the projection A → A/m = k, so for a ∈ A, a − π(a) ∈ m. Defineσ(a) = (a − π(a))/t. Show that ΩA/k = ⟨dt⟩ + mΩA/k , using the fact that forevery a ∈ A,
da = σ(a) dt + t dσ(a).
Apply Nakayama’s Lemma version 3 (Exercise 7.2.G) to ⟨dt⟩ ⊂ ΩA/k . (This argument, with essentially no change, can be used to show that if (A, m, k) is a localization of a finitely generated algebra over k, and t1 , . . . , tn generate m, then dt1 , . . . ,dtn generate ΩA/k .)(c) By part (b), ΩA/k is a principal A-module. Show that ×dt is an injection as
follows. By the classification of finite generated modules over discrete valuation
rings (Remark 12.5.14), it suffices to show that tm dt ̸= 0 for all m. The surjection A → A/(tN ) induces a map ΩA/k → Ω(A/(tN ))/k , so it suffices to show that∼ k[t]/(tN ). The usual diftm dt is nonzero in Ω(A/(tN ))/k . Show that A/(tN ) =N
ferentiation rule for polynomials gives a map k[t]/(t ) → k[t]/(tN−1 ) which is a
derivation of k[t]/(tN ) over k, and tm dt will not map to 0 so long as N is sufficiently large. Put the pieces together and complete the proof. (An extension ofthese ideas can show that if (A, m, k) is a localization of a finitely generated algebra
over k that is a regular local ring, then ΩA/k is free of rank dim A.)
21.4.5. Projective space and the Euler exact sequence. We next examine thendifferentials of projective space Pn
k , or more generally PA where A is an arbitraryring. As projective space is covered by affine open sets of the form An , on which
the differentials form a rank n locally free sheaf, ΩPnA /A is also a rank n locally free
sheaf.

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21.4.6. Theorem (the Euler exact sequence). —satisfies the following exact sequence

The sheaf of differentials ΩPnA /A0% ΩPn /AA

% OPn (−1)⊕(n+1)
A

φ

% OPn

A

% 0.

This is handy, because you can get a hold of ΩPnA /A in a concrete way. See
Exercise 21.5.Q for an application. By dualizing this exact sequence, we have anexact sequence 0 → OPnA → OPnA (1)⊕(n+1) → TPnA /A → 0.

21.4.7. ⋆ Proof of Theorem 21.4.6. (What is really going on in this proof is that we
consider those differentials on An+1
\ {0} that are pullbacks of differentials on Pn
A.
A
For a different explanation, in terms of the Koszul complex, see [E, §17.5].)
We first describe a map φ : O(−1)⊕(n+1) → O, and later identify the kernel
with ΩX/Y . The map is given by
φ : (s0 , s1 , . . . , sn ) (→ x0 s0 + x1 s1 + · · · + xn sn .

(Do you see why?) You should think of this as a “degree 1” map, as each xi has
degree 1.
21.4.8. Remark. The dual φ∨ : O → O(1)⊕(n+1) of φ gives a map Pn → Pn via
Important Theorem 16.4.1 on maps to projective space. This map is the identity,
and this is one way of describing φ in a “natural” (coordinate-free) manner.
21.4.E. E ASY EXERCISE . Show that φ is surjective, by checking on the open set
D(xi ). (There is a one-line solution.)
Now we must identify the kernel of φ with the differentials, and we can dothis on each D(xi ), so long as we do it in a way that works simultaneously for each
open set. So we consider the open set U0 , where x0 ̸= 0, and we have the usual
coordinates xj/0 = xj /x0 (1 ≤ j ≤ n). Given a differential
f1 (x1/0 , . . . , xn/0 ) dx1/0 + · · · + fn (x1/0 , . . . , xn/0 ) dxn/0

we must produce n + 1 sections of O(−1). As motivation, we just look at the firstterm, and pretend that the projective coordinates are actual coordinates.
f1 dx1/0=

f1 d(x1 /x0 )x0 dx1 − x1 dx0
= f1
x20
x1
f1
= − 2 f1 dx0 +
dx1
x0
x0

Note that x0 times the “coefficient of dx0 ” plus x1 times the “coefficient of dx1 ” is0, and also both coefficients are of homogeneous degree −1. Motivated by this, we
take:
"
#
x1
xn
f1 f2
fn
(21.4.8.1) f1 dx1/0 + · · · + fn dxn/0 (→ − 2 f1 − · · · − 2 fn , , , . . . ,
x0 x0
x0
x0
x0

Note that over U0 , this indeed gives an injection of ΩPnA /A to O(−1)⊕(n+1) thatsurjects onto the kernel of O(−1)⊕(n+1) → OX (if (g0 , . . . , gn ) is in the kernel, takefi = x0 gi for i > 0).

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Let’s make sure this construction, applied to two different coordinate patches(say U0 and U1 ) gives the same answer. (This verification is best ignored on a firstreading.) Note thatf1 dx1/0 + f2 dx2/0 + · · ·

=

f1 d

1
x0/1

+ f2 d

x2/1
+ ···
x0/1

f2 x2/1
f1
f2= − 2 dx0/1 +dx2/1 − 2
dx0/1 + · · ·
x0/1
x0/1
x0/1
= −

f1 + f2 x2/1 + · · ·f2 x1
dx0/1 +
dx2/1 + · · · .
2
x0
x0/1

Under this map, the dx2/1 term goes to the second factor (where the factors areindexed 0 through n) in O(−1)⊕(n+1) , and yields f2 /x0 as desired (and similarly
for dxj/1 for j > 2). Also, the dx0/1 term goes to the “zero” factor, and yields
4 n
5
.
2
fi (xi /x1 )/(x0 /x1 ) /x1 = fi xi /x20
i=1

as desired. Finally, the “first” factor must be correct because the sum over i of xi
times the ith factor is 0.

21.4.F. E XERCISE . Finish the proof of Theorem 21.4.6, by verifying that this mapΩPnA /A → OPnA (−1)⊕(n+1) identifies ΩPnA /A with ker φ.
!

21.4.G. E XERCISE ( PROMISED IN §18.5.7). Show that h1 (P2k , ΩP2k /k ) > 0. Show
that ΩP2k /k is not the direct sum of line bundles. Show that Theorem 18.5.6 (thatall finite rank vector bundles on P1 split into line bundles) cannot be extended torank 2 vector bundles on P2k .
21.4.9. Generalizations of the Euler exact sequence. Generalizations of the Euler exact
sequence are quite useful. We won’t use them later, so no proofs will be given.
First, the argument generalizes readily if Spec A is replaced by an arbitrary basescheme. The Euler exact sequence generalizes further in a number of ways. Asa first step, suppose V is a rank n + 1 locally free sheaf (or vector bundle) on ascheme X. Then ΩPV /X sits in an Euler exact sequence:
0 → ΩPV /X → O(−1) ⊗ V ∨ → OX → 0∼ (V ∨ ⊗If π : PV → X, the map O(−1) ⊗ V ∨ → OX is induced by V ∨ ⊗ π∗ O(1) =∨V ) ⊗ OX → OX , where V ⊗ V → OX is the trace map (§13.7.1). (You may wish tocompare this to Remark 21.4.8.)
It is not obvious that this is useful, but we have already implicitly seen it in
the case of P1 -bundles over curves, in Exercise 20.2.J, where the normal bundle toa section was identified in this way.
21.4.10. ⋆⋆ Generalization to the Grassmannian. For another generalization, fix abase field k, and let G(m, n+1) be the space of sub-vector spaces of dimension m inan (n+1)-dimensional vector space V (the Grassmannian, §16.7). Over G(m, n+1)

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we have a short exact sequence of locally free sheaves
0 → S → OG(m,n+1) ⊗ V ∨ → Q → 0

where OG(m,n+1) ⊗ V ∨ is the “trivial bundle whose fibers are V ∨ ” (do you understand what that means?), and S is the “universal vector bundle” or “tautological
vector (sub)bundle”. Then there is a canonical isomorphism(21.4.10.1)∼ Hom (Q, S ).
ΩG(m,n+1)/k =21.4.H. E XERCISE . Recall that in the case of projective space, i.e., m = 1, S =O(−1) (Exercise 17.1.H). Verify (21.4.10.1) in this case using the Euler exact sequence (Theorem 21.4.6).
21.4.I. E XERCISE . Prove (21.4.10.1), and explain how it generalizes Exercise 20.2.I.
(The hint to Exercise 20.2.I may help.)
This Grassmannian fact generalizes further to Grassmannian bundles, and toflag varieties, and to flag bundles.

21.5 Understanding smooth varieties using their cotangent
bundlesIn this section, we construct birational invariants of varieties over algebraicallyclosed fields (such as the geometric genus), motivate the notion of an unramified
morphism, show that varieties are “smooth almost everywhere”, and get a firstglimpse of Hodge theory.
21.5.1. The geometric genus, and other birational invariants from i-forms ΩiX/Y .
Suppose X is a projective scheme over k. Then for each i, hi (X, ΩX/k ) is aninvariant of X, which can be useful. The first useful fact is that it, and related
invariants, are birational invariants if X is smooth, as shown in the following Exercise 21.5.A. We first define the sheaf of (relative) i-forms ΩiX/Y := ∧i ΩX/Y . Sections of ΩiX/Y (over some open set) are called (relative) i-forms (over that open
set).
21.5.2. Joke. Old Macdonald had a form; ei ∧ ei = 0.
21.5.A. E XERCISE (h0 (X, ΩiX/k ) ARE BIRATIONAL INVARIANTS ). Suppose X and
X ′ are birational irreducible smooth projective k-varieties. Show (for each i) that
∼ H0 (X ′ , Ωi ′ ). Hint: fix a birational map φ : X ""# X ′ . By ExerH0 (X, ΩiX/k ) =X /kcise 16.5.B, the complement of the domain of definition U of φ is codimension atleast 2. By pulling back i-forms from X ′ to U, we get a map φ∗ : H0 (X ′ , ΩiX ′ /k ) →H0 (U, ΩiX/k ). Use Algebraic Hartogs’s Lemma 11.3.11 and the fact that Ωi is locally free to show the map extends to a map φ∗ : H0 (X ′ , ΩiX ′ /k ) → H0 (X, ΩiX/k ). Ifψ : X ′ ""# X is the inverse rational map, we similarly get a map ψ∗ : H0 (X, ΩiX/k ) →H0 (X ′ , ΩiX ′ /k ). Show that φ∗ and ψ∗ are inverses by showing that each composition is the identity on a dense open subset of X or X ′ .

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21.5.3. The canonical bundle KX and the geometric genus pg (X). If X is a dimension
n smooth k-variety, the invertible sheaf (or line bundle) det ΩX/k = Ωn
X/k (thesheaf of “algebraic volume forms”) has particular importance, and is called thecanonical (invertible) sheaf, or the canonical (line) bundle. It is denoted KX (orKX/k ). As mentioned in §18.5.2, if X is projective, then KX is the dualizing sheafωX appearing in the statement of Serre duality, something we will establish in§30.4 (see Desideratum 30.1.1).
21.5.B. E XERCISE ( THE ADJUNCTION FORMULA FOR KX ). Suppose X is a smoothvariety, and and Z is a smooth subvariety of X. Show that
∼ KX |Z ⊗ det NZ/X .
KZ =(Hint: apply Exercise 13.5.H to Theorem 21.3.7.) In particular, by Exercise 21.2.H,
if Z is codimension 1, then
∼ (KX ⊗ OX (Z)) |Z ,KZ =which is more commonly and compactly written as KX (Z)|Z . The adjunction formula is often used inductively, for complete intersections; see Exercise 21.5.H foran example. (And see Exercise 20.2.B(a) and §30.4 for other versions of the adjunction formula.)
21.5.4. Definition. If X is a projective (or even proper) smooth k-variety, the birational invariant h0 (X, KX ) = h0 (X, ΩnX/k ) has particular importance. It is called
the geometric genus, and is denoted pg (X). We saw this in the case of curves in§21.4.1. If X is an irreducible variety that is not smooth or projective, the phrase
geometric genus refers to h0 (X ′ , KX ′ ) for some smooth projective X ′ birational toX. (By Exercise 21.5.A, this is independent of the choice of X ′ .) For example, if X
is an irreducible reduced projective curve over k, the geometric genus is the geometric genus of the normalization of X. (But in higher dimension, it is not clear
if there exists such an X ′ . It is a nontrivial fact that this is true in characteristic 0— Hironaka’s Theorem on resolution of singularities — and it is not yet known inpositive characteristic; see Remark 22.4.6.)
It is a miracle that for a complex curve the geometric genus is the same as the
topological genus and the arithmetic genus. We will connect the geometric genus
to the topological genus in our discussion of the Riemann-Hurwitz formula soon
(Exercise 21.7.I). We connect the geometric genus to the arithmetic genus in thefollowing exercise.
21.5.C. E ASY E XERCISE . Assuming Miracle 18.5.2 (that the canonical bundle is
Serre-dualizing), show that the geometric genus of an irreducible smooth projective curve over k = k equals its arithmetic genus.
21.5.D. E XERCISE . Suppose Z is a regular degree d surface in P3k . Compute thegeometric genus pg (Z) of Z. Show that no regular quartic surface in P3k is rational(i.e., birational to P2k , Definition 6.5.4). (Such quartic surfaces are examples of K3
surfaces, see Exercise 21.5.I.)
21.5.5. Important classes of varieties: Fano, Calabi-Yau, general type.

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Suppose X is a smooth projective k-variety. Then X is said to be Fano if KX∨∼ OX . (Caution: there are other definitions ofis ample, and Calabi-Yau if KX =Calabi-Yau.)
21.5.E. E XERCISE . For all j ≥ 0, the jth plurigenus of a smooth projective k-variety
is h0 (X, KX⊗j ). Show that the jth plurigenus is a birational invariant.
(By contrast, h0 (X, K ⊗j ) is not a birational invariant if j < 0. For example,
you can show that h0 (X, K ∨ ) differs for X = P2k and X = Blp P2k .)The Kodaira dimension of X, denoted κ(X), tracks the rate of growth of theplurigenera. It is the smallest k such that h0 (X, KX⊗j )/jk is bounded (as j varies
through the positive integers), except that if all the plurigenera h0 (X, KX⊗j ) are 0,
we say κ(X) = −1. It is a nontrivial fact that the Kodaira dimension always exists,
and that it is the maximum of the dimensions of images of the jth “pluricanonicalrational maps”. The latter fact implies that the Kodaira dimension is an integer between −1 and dim X inclusive. Exercise 21.5.E shows that the Kodaira dimension
is a birational invariant. If κ(X) = dim X, we say that X is of general type. (Formore information on the Kodaira dimension, see [Ii, §10.5].)
21.5.F. E XERCISE . Show that if KX is ample then X is of general type.
21.5.G. E XERCISE . Show that a smooth geometrically irreducible projective curve(over a field k) is Fano (resp. Calabi-Yau, general type) if its genus is 0 (resp. 1,
greater than 1).
The “trichotomy” of Exercise 21.5.G is morally the reason that curves behave
differently depending on which of the three classes they lie in (see §19.6.1). In somesense, important aspects of this trichotomy extend to higher dimension, which ispart of the reason for making these definitions. We now explore this trichotomy in
the case of complete intersections.
21.5.H. E XERCISE . Suppose X is a smooth complete intersection in PN
k of hypersurfaces of degree d1 , . . . , dn , where k is algebraically closed.
∼ O(−N − 1 + d1 + · · · + dn )|X .
(a) Show that KX =(b) Show that −N − 1 + d1 + · · · + dn is negative (resp. zero, positive) then X isFano (resp. Calabi-Yau, general type).
(c) Find all possible values of N and d1 , . . . , dn where X is Calabi-Yau of dimension
at most 3. Notice how small this list is.
21.5.I. E XERCISE . A K3 surface over a field k is a proper smooth geometricallyconnected Calabi-Yau surface X over k such that H1 (X, OX ) = 0. (Weil has writtenthat K3 surfaces were named in honor of Kummer, Kähler, and Kodaira, and themountain K2, [BPV, p. 288].) Prove that the dimension 2 (smooth) Calabi-Yau
complete intersections of Exercise 21.5.H(c) are all K3 surfaces.
21.5.6. Tantalizing side remark. If you compare the degrees of the hypersurfaces cutting out complete intersection K3 surfaces (in Exercise 21.5.H(c)), with the degrees
of hypersurfaces cutting out complete intersection canonical curves (see §19.8.2
and Exercise 19.8.G), you will notice a remarkable coincidence. Of course this is
not a coincidence at all.

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21.5.J. E XERCISE ( NOT EVERYTHING FITS INTO THIS TRICHOTOMY ). Suppose C isa smooth projective irreducible complex curve of genus greater than 1. Show thatC ×C P1C is neither Fano, nor Calabi-Yau, nor general type. Hint: Exercise 21.2.U.
21.5.7. ⋆ Kodaira vanishing.
The Kodaira vanishing theorem is an important result that is an important
tool in a number of different areas of algebraic geometry. We state it for the sakeof culture, but do not prove it (and hence will not use it).
21.5.8. The Kodaira Vanishing Theorem. — Suppose k is a field of characteristic 0,
and X is a smooth projective k-variety. Then for any ample invertible sheaf L , Hi (X, K ⊗L ) = 0 for i > 0.
21.5.9. The restriction on characteristic is necessary; Raynaud gave an example
where Kodaira vanishing fails in positive characteristic, [Ra]. The original proof isby proving the complex-analytic version, and then transferring it into a complex
algebraic statement using Serre’s GAGA Theorem [Se3]. The general characteristic
0 case can be reduced to C — any reduction of this sort is often called (somewhat
vaguely) an application of the Lefschetz principle. (For a precise formulation of the
Lefschetz principle, with a short proof which applies in any characteristic see [Ek].
See [FR] for a generalization. See [MO90551] for more context.) Raynaud latergave a dramatic algebraic proof of Kodaira vanishing using positive characteristic(!), see [DeI] or [Il].
21.5.K. E XERCISE . Prove the Kodaira Vanishing Theorem in the case dim X = 1,
and for Pn
k . (Neither case requires the characteristic 0 hypothesis.)21.5.10. ⋆ A first glimpse of Hodge theory.
The invariant hj (X, ΩiX/k ) is called the Hodge number hi,j (X). By Exercise 21.5.A,i,0h are birational invariants. We will soon see (in Exercise 21.5.O) that this isn’t
true for all hi,j .
21.5.L. E XERCISE . Suppose X is a smooth projective variety over k = k. Assuming Miracle 18.5.2 (that the canonical bundle is Serre-dualizing), show that Hodge
numbers satisfy the symmetry hp,q = hn−p,n−q . (Exercise 13.5.G will be useful.)
21.5.M. E XERCISE ( THE H ODGE NUMBERS OF PROJECTIVE SPACE ). Show thatp,q nhp,q (Pn(Pk ) = 0 otherwise. Hint: use the Eulerk ) = 1 if 0 ≤ p = q ≤ n and hexact sequence (Theorem 21.4.6) and apply Exercise 13.5.F.
21.5.11. Remark: The Hodge diamond. Over k = C, further miracles occur. If X isan irreducible smooth projective complex variety, then it turns out that there is adirect sum decomposition(21.5.11.1)

Hm (X, C) = ⊕i+j=m Hj (X, ΩiX/C ),,i,jfrom which hm (X, C) =i+j=m h , so the Hodge numbers (purely algebraic
objects) yield the Betti numbers (a priori topological information). Moreover, complex conjugation interchanges Hj (X, ΩiX/C ) with Hi (X, ΩjX/C ), from which(21.5.11.2)

hi,j = hj,i .

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587(Aside: This additional symmetry holds in characteristic 0 in general, but can failin positive characteristic, see for example [Ig, p. 966], [Mu1, §II], [Se4, Prop. 16].)This is the beginning of the vast and rich subject of Hodge theory (see [GH1, §0.6]
for more, or [Vo] for much more).
If we write the Hodge numbers in a diamond, with hi,j the ith entry in the(i + j)th row, then the diamond has the two symmetries coming from Serre duality
and complex conjugation. For example, the Hodge diamond of an irreduciblesmooth projective complex surface will be of the following form:
1
q

q
h1,1

pg

pg

q

q
1

where pg is the geometric genus of the surface, and q = h0,1 = h1,0 = h2,1 = h1,2
is called the irregularity of the surface. As another example, by Exercise 21.5.M,
the Hodge diamond of Pn is all 0 except for 1’s down the vertical axis of symmetry.
You won’t need the unproved statements (21.5.11.1) or (21.5.11.2) to solve thefollowing problems.
21.5.N. E XERCISE . Assuming Miracle 18.5.2 (that the canonical bundle is Serredualizing), show that the Hodge diamond of a smooth projective geometricallyirreducible genus g curve over a field k is the following.
1
g

g
1

21.5.O. E XERCISE . Show that the Hodge diamond of P1k × P1k is the following.
1

0
0

0
2

0

0
0

1
By comparing your answer to the Hodge diamond of P2k (Exercise 21.5.M), showthat h1,1 is not a birational invariant.
Notice that in both cases, h1,1 is the Picard number ρ (defined in §18.4.10). In
general, ρ ≤ h1,1 (see [GH1, §3.5, p. 456-7]).
21.5.12. ⋆ Aside: Infinitesimal deformations and automorphisms.
It is beyond the scope of this book to make this precise, but if X is a variety, H0 (X, TX ) parametrizes infinitesimal automorphisms of X, and H1 (X, TX )parametrizes infinitesimal deformations. As an example if X = P1 (over a field),∼ O(2) (§21.4.2), so h0 (P1 , TP1 ) = 3, which is precisely the dimension of theTP1 =automorphism group of P1 (Exercise 16.4.B).
n21.5.P. E XERCISE . Compute h0 (Pnk , TPk ) using the Euler exact sequence (Theorem 21.4.6). Compare this to the dimension of the automorphism group of Pnk(Exercise 16.4.B).

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21.5.Q. E XERCISE . Show that H1 (PnA , TPA ) = 0. (Thus projective space can’t
deform, and is “rigid”.)

21.5.R. E XERCISE . Assuming Miracle 18.5.2 (that the canonical bundle is Serredualizing), compute hi (C, TC ) for a genus g smooth projective geometrically irreducible curve over k, for i = 0 and 1. You should notice that h1 (C, TC ) for genus
0, 1, and g > 1 is 0, 1, and 3g − 3 respectively; after doing this, re-read §19.8.3.

21.6 Unramified morphisms
Suppose π : X → Y is a morphism of schemes. The support of the quasicoherent sheaf Ωπ = ΩX/Y is called the ramification locus, and the image of its support,π(Supp ΩX/Y ), is called the branch locus. If Ωπ = 0, we say that π is formallyunramified, and if π is also furthermore locally of finite type, we say π is unramified. (Caution: there is some lack of consensus in the definition of “unramified”;“locally of finite type” is sometimes replaced by “locally of finite presentation”,
which was the definition originally used in [Gr-EGA].)
21.6.A. E ASY E XERCISE ( EXAMPLES OF UNRAMIFIED MORPHISMS ).
(a) Show that locally closed embeddings are unramified.
(b) Show that if S is a multiplicative subset of the ring B, then Spec S−1 B → Spec B
is formally unramified. (Thus if η is the generic point of an integral scheme Y, thenSpec OY,η → Y is formally unramified.)(c) Show that finite separable field extensions (or more correctly, the corresponding
maps of schemes) are unramified.
21.6.B. E XERCISE ( PRACTICE WITH THE CONCEPT ).
(a) Show that the normalization of the node in Exercise 9.7.E (see Figure 7.4) is
unramified.
(b) Show that the normalization of the cusp in Exercise 9.7.F (see Figure 9.4) is notunramified.
21.6.C. E ASY E XERCISE ( UNRAMIFIED MORPHISMS ARE PRESERVED BY COMPOSI Suppose π : X → Y and ρ : Y → Z are unramified. Show that ρ ◦ π isunramified.

TION ).

21.6.D. E XERCISE ( CHARACTERIZATIONS OF UNRAMIFIED MORPHISMS BY THEIRFIBERS ). Suppose π : X → Y is locally of finite type.
(a) Show that π is unramified if and only if for each q ∈ Y, π−1 (q) is the (schemetheoretic) disjoint union of schemes of the form Spec K, where K is a finite separable extension of κ(q).
(b) Show that π is unramified if and only if for each geometric point q, π−1 (q) :=q ×Y X is the (scheme-theoretic) disjoint union of copies of q.
21.6.E. U NIMPORTANT E XERCISE ( FOR NUMBER THEORISTS ). Suppose φ : (A, m) →(B, n) is a local homomorphism of local rings. In algebraic number theory, such aring morphism is said to be unramified if B/φ(m)B is a finite separable extension of
A/m. Show that if φ is finite type, this agrees with our definition.

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21.6.F. E XERCISE ( FOR USE IN E XERCISE 21.6.G( B ) AND E XERCISE 25.2.G).
(a) Suppose π : X → Y is a locally finite type morphism of locally Noetherian
schemes. Show that π is unramified if and only if δπ : X → X×Y X is an open embedding. Hint: Show the following. If φ : X → Z is a closed embedding of Noetherian
schemes, and the ideal sheaf I of φ satisfies I = I 2 , then φ is also an open embedding. For that, show that if (A, m) is a Noetherian local ring, and I is a proper
ideal of A satisfying I = I2 , then I = 0. For that in turn, use Nakayama version 2(Lemma 7.2.9). Also use the fact that Supp I is closed (using Exercise 13.7.D), so
its complement is open.
⋆ (b) Adapt your proof of (a) to drop the locally Noetherian hypothesis. Hint:
show that if π is locally of finite type, then δπ is locally finitely presented.
21.6.G. E ASY E XERCISE . Suppose π : X → Y and ρ : Y → Z are locally of finite type.
Let τ = ρ ◦ π:
π
%Y
X❄
❄❄
⑧
⑧
❄❄
⑧⑧
❄
⑧⑧ ρ
τ ❄❄
⑧
8 5⑧
Z
(a) Show that if τ is unramified, then so is π.
(b) Show that if ρ is unramified, and τ is smooth of relative dimension n (e.g., étale
if n = 0), then π is smooth of relative dimension n.
(Does this agree with your geometric intuition?)21.6.H. E XERCISE . Suppose π : X → Y is locally of finite type. Show that the locus
in X where π is unramified is open.

21.6.I. E XERCISE ( THE MEANING OF UNRAMIFIEDNESS IN “ VERY GEOMETRIC ” SETShow that a morphism π : X → Y of finite type schemes over an algebraically closed field is unramified if and only if π is injective on tangent vectors at closed points. (This was mentioned in the remark immediately after Theorem 19.1.1.)

TINGS ).

21.6.1. Arithmetic side remark: the different and discriminant. If B is the ring ofintegers in a number field (§9.7.1), the different ideal of B is the annihilator ofΩB/Z . It measures how “ramified” Spec B → Spec Z is, and is a ring-theoretic
version of the ramification locus. The discriminant ideal can be interpreted as theideal of Z corresponding to effective divisor on Spec Z that is the “push forward”(not defined here, but defined as you might expect) of the divisor corresponding
to the different. It is a ring-theoretic version of the branch locus. If B/A is anextension of rings of integers of number fields, the relative different ideal (of B)
and relative discriminant ideal (of A) are defined similarly. (We won’t use theseideas.)

21.7 The Riemann-Hurwitz Formula
The Riemann-Hurwitz formula generalizes our calculation of the genus g ofa double cover of P1 branched at 2g + 2 points, Theorem 19.5.1, to higher degree
covers, and to higher genus target curves.

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21.7.1. Definition. A finite morphism between integral schemes X → Y is said to be
separable if it is dominant, and the induced extension of function fields K(X)/K(Y)is a separable extension. Similarly, a generically finite morphism is genericallyseparable if it is dominant, and the induced extension of function fields is a separable extension. Note that finite morphisms of integral schemes are automaticallyseparable in characteristic 0.
21.7.2. Proposition. — If π : X → Y is a generically separable morphism of irreduciblesmooth varieties of the same dimension n, then the relative cotangent sequence (Theorem 21.2.25) is exact on the left as well:(21.7.2.1)0% π∗ ΩY/kφ% ΩX/k% ΩX/Y% 0.

This is an example of left-exactness of the relative cotangent sequence in thepresence of appropriate “smoothness”, see Remark 21.2.10.
Proof. We must check that φ is injective. Now ΩY/k is a rank n locally free sheaf on
Y, so π∗ ΩY/k is a rank n locally free sheaf on X. A locally free sheaf on an integral
scheme (such as π∗ ΩY/k ) is torsion-free (any section over any open set is nonzero
at the generic point, see §13.5.5), so if a subsheaf of it (such as ker φ) is nonzero,
it is nonzero at the generic point. Thus to show the injectivity of φ, we need only
check that φ is an inclusion at the generic point. We thus tensor with Oη whereη is the generic point of X. Tensoring with Oη is an exact functor (localization isexact, Exercise 1.6.F), and Oη ⊗ΩX/Y = 0 (as K(X)/K(Y) is a separable extension by
hypothesis, and Ω for separable field extensions is 0 by Exercise 21.2.F(a)). Also,Oη ⊗ π∗ ΩY/k and Oη ⊗ ΩX/k are both n-dimensional Oη -vector spaces (they are
the stalks of rank n locally free sheaves at the generic point). Thus by consideringOη ⊗ π∗ ΩY/k → Oη ⊗ ΩX/k → Oη ⊗ ΩX/Y → 0(which is Oη⊕n → Oη⊕n → 0 → 0) we see that Oη ⊗ π∗ ΩY/k → Oη ⊗ ΩX/k is
injective, and thus that π∗ ΩY/k → ΩX/k is injective.
!

People not confined to characteristic 0 should note what goes wrong for nonseparable morphisms. For example, suppose k is a field of characteristic p, andconsider the map π : A1k = Spec k[t] → A1k = Spec k[u] given by u = tp . ThenΩπ is the trivial invertible sheaf generated by dt. As another (similar but different) example, if K = k(x) and K ′ = k(xp ), then the inclusion K ′ "→ K induces
π : Spec K[t] → Spec K ′ [t]. Once again, Ωπ is an invertible sheaf, generated by dx(which in this case is pulled back from ΩK/K ′ on Spec K). In both of these cases, wehave maps from one affine line to another, and there are vertical tangent vectors.
21.7.A. E XERCISE . If X and Y are smooth varieties of dimension n, and π : X → Y isgenerically separable, show that the ramification locus is pure codimension 1, andhas a natural interpretation as an effective divisor, as follows. Interpret φ as an n×n Jacobian matrix (12.1.6.1) in appropriate local coordinates, and hence interpretthe locus where φ is not an isomorphism as (locally) the vanishing scheme of thedeterminant of an n × n matrix. Hence the branch locus is also pure codimension
1. Hence we use the terms ramification divisor and branch divisor.

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591

Before getting down to our case of interest, dimension 1, we begin with something (literally) small but fun. Suppose π : X → Y is a surjective k-morphism froma smooth k-scheme that contracts a subset of codimension greater than 1. Moreprecisely, suppose π is an isomorphism over an open subset of Y, from an open
subset U of X whose complement has codimension greater than 1. Then by Exercise 21.7.A, Y cannot be smooth. (Small resolutions, to be defined in Exercise 22.4.N,
are examples of such π. In particular, you can find an example there.)Suppose now that X and Y are dimension 1. Then the ramification locus is a
finite set (ramification points) of X, and the branch locus is a finite set (branch
points) of Y. (Figure 21.4 shows a morphism with two ramification points and one
branch point.) Now assume that k = k. We examine ΩX/Y near a point p ∈ X.

F IGURE 21.4. An example where the branch divisor appears withmultiplicity 2 (see Exercise 21.7.C)
As motivation for what we will see, we note that in complex geometry, nonconstant maps from (complex) curves to curves may be written in appropriate local
coordinates as x (→ xm = y, from which we see that dy pulls back to mxm−1 dx, soΩX/Y locally looks like functions times dx modulo multiples of mxm−1 dx.
Consider now our map π : X → Y, and fix p ∈ X, and q = π(p). Because theconstruction of Ω behaves well under base change (Theorem 21.2.27(b)), we may
replace Y with Spec of the local ring OY,q at q, i.e., we may assume Y = Spec B,
where B is a discrete valuation ring (as Y is a regular curve), with residue fieldk = k corresponding to q. Then as π is finite, X is affine too. Similarly, as theconstruction of Ω behaves well with respect to localization (Exercise 21.2.8), we
may replace X by Spec OX,p , and thus assume X = Spec A, where A is a discrete
valuation ring, and π corresponds to B → A, inducing an isomorphism of residue
fields (with k).
Suppose their uniformizers are s and t respectively, with t (→ usn where u is
an invertible element of A.
X

A7

us7 n

"
Y

B

❴
t

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Recall that the differentials of a discrete valuation ring over k are generated by dof the uniformizer (Exercise 21.4.D). Thendt = d(usn ) = unsn−1 ds + sn du.
This differential on Spec A vanishes to order at least n − 1, and precisely n − 1 if
n is not divisible by the characteristic. The former case is called tame ramification,and the latter is called wild ramification. We call this order the ramification orderat this point of X.
21.7.B. E XERCISE . Show that the degree of ΩX/Y at p ∈ X is precisely the ramification order of π at p. (The degree of a coherent sheaf on a curve was definedin §18.4.4. To do this exercise, you will have to explain why a coherent sheaf F
supported on a finite set of points has a “degree” at each of these points, which
sum to the total degree of F .)21.7.C. E XERCISE : INTERPRETING THE RAMIFICATION DIVISOR IN TERMS OF NUM BER OF PREIMAGES . For this exercise, we take the base field k to be algebraicallyclosed. Suppose all the ramification above q ∈ Y is tame (which is always true incharacteristic 0). Show that the degree of the branch divisor at q is deg π − |π−1 (q)|.
Thus the multiplicity of the branch divisor counts the extent to which the numberof preimages is less than the degree (see Figure 21.4).
21.7.3. Theorem (the Riemann-Hurwitz formula). — Suppose π : X → Y is a finite
separable morphism of projective regular curves, of pure degree n. Let R be the ramification
divisor. Then
(21.7.3.1)

2g(X) − 2 = n(2g(Y) − 2) + deg R.

21.7.D. E XERCISE . Prove the Riemann-Hurwitz formula. Hint: Apply the factthat degree is additive in exact sequences (Exercise 18.4.I) to (21.7.2.1). Recall that
degrees of line bundles pull back well under finite morphisms of integral projective curves, Exercise 18.4.F. A torsion coherent sheaf on a reduced curve (such asΩπ ) is supported in dimension 0 (Exercise 13.7.G(b)), so χ(Ωπ ) = h0 (Ωπ ). Show
that the degree of R as a divisor is the same as its degree in the sense of h0 .
Here are some applications of the Riemann-Hurwitz formula.
21.7.4. Example. The degree of R is always even: any cover of a curve must be
branched over an even number of points (counted with appropriate multiplicity).
21.7.E. E ASY E XERCISE . Show that there is no nonconstant map from a smoothprojective irreducible genus 2 curve to a smooth projective irreducible genus 3
curve. (Hint: deg R ≥ 0.)
21.7.5. Example (“P1k is simply connected”). If k = k, the only connected unbranchedfinite separable cover of P1k is the isomorphism, for the following reason. Suppose
X is connected and X → P1k is unramified. Then X is a curve, and regular by
Exercise 25.2.E(a). Applying the Riemann-Hurwitz theorem, using that the ramification divisor is 0, we have 2 − 2gX = 2d with d ≥ 1 and gX ≥ 0, from which d = 1
and gX = 0.

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21.7.F. E XERCISE (“ IN CHARACTERISTIC 0, A1k IS SIMPLY CONNECTED ”. Show
that if k = k has characteristic 0, the only connected unbranched cover of A1k isitself. (Aside: in characteristic p, this needn’t hold; Spec k[x, y]/(yp − y − x) →Spec k[x] is such a map, as you can show yourself. Once the theory of the algebraic
fundamental group is developed, this translates to: “A1 is not simply connectedin characteristic p.” This cover is an example of an Artin-Schreier cover. Fun fact:
the group Z/p acts on this cover via the map y (→ y + 1. This is an example of a
Galois cover; you can check that the extension of function fields is Galois.)21.7.G. U NIMPORTANT E XERCISE . Extend Example 21.7.5 and Exercise 21.7.F,
by removing the k = k hypothesis, and changing “connected” to “geometricallyconnected”.

21.7.6. Example: Lüroth’s Theorem. Continuing the notation of Theorem 21.7.3, suppose g(X) = 0. Then from the Riemann-Hurwitz formula (21.7.3.1), g(Y) = 0.
(Otherwise, if g(Y) were at least 1, then the right side of the Riemann-Hurwitz
formula would be non-negative, and thus couldn’t be −2, which is the left side.)
Informally: the only maps from a genus 0 curve to a curve of positive genus are
the constant maps. This has a nonobvious algebraic consequence, by our identification of covers of curves with field extensions (Theorem 17.4.3): all subfields ofk(x) containing k are of the form k(y) where y = f(x) for some f ∈ k(x).
k(x)
7

P1

!$
K(C)

"
C

= P1

(It turns out that the hypotheses char k = 0 and k = k are not necessary.) This isLüroth’s Theorem.
21.7.H. E XERCISE . Use Lüroth’s Theorem to give new geometric solutions to Exercises 6.5.J and 6.5.L. (These arguments will be less ad hoc, and more suitable forgeneralization, than the algebraic solutions suggested in the hints to those exercises.)
21.7.I. ⋆ E XERCISE ( GEOMETRIC GENUS EQUALS TOPOLOGICAL GENUS ). This
exercise is intended for those with some complex background, who know that
the Riemann-Hurwitz formula holds in the complex analytic category. Suppose
C is an irreducible regular projective complex curve. Show that there is an algebraic nonconstant map π : C → P1C . Describe the corresponding map of Riemannsurfaces. Use the Riemann-Hurwitz formula to show that the algebraic notionof genus (as computed using the branched cover π) agrees with the topologicalnotion of genus (using the same branched cover). (Recall that assuming Miracle 18.5.2 — that the canonical bundle is Serre-dualizing — we know that the geometric genus equals the arithmetic genus, Exercise 21.5.C.)
21.7.J. U NIMPORTANT E XERCISE . Suppose π : X → Y is a dominant morphism of∼ π∗ ΩY .
regular curves, and R is the ramification divisor of π. Show that ΩX (−R) =(This exercise is geometrically pleasant, but we won’t use it.) Hint: This says that

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we can interpret the invertible sheaf π∗ ΩY over an open set U of X as precisely
those differentials on U vanishing along the ramification divisor.
21.7.7. Informal example: The degree of the discriminant of degree d polynomials in one
variable. You may be aware that there is a degree 2d − 2 polynomial in the coefficients ad , . . . , a0 of the degree d polynomial
f(x) = ad xd + ad−1 xd−1 + · · · + a0 = 0that vanishes precisely when f(x) has a multiple root. For example, when d = 2,
the discriminant is a21 − 4a0 a2 . We can “compute” this degree 2d − 2 using the
Riemann-Hurwitz formula as follows. (You should try to make sense of the following informal and imprecise discussion.) We work over an algebraically closed
field k of characteristic 0 for the sake of simplicity. If we take two general degree dpolynomials, g(x) and h(x), the degree of the discriminant “should be” the number of λ ∈ k for which g(x) − λh(x) has a double root. Consider the morphism
P1 → P1 given by x (→ [g(x), h(x)]. (Here we use “affine coordinates” on thesource P1 : by x we mean [x, 1].) Then this morphism has degree d. A ramification
point a mapping to the branch point [λ, 1] in the target P1 corresponds to a being
a double root of g(x) − λh(x). Thus the number of branch points should be thedesired degree of the discriminant. By the Riemann-Hurwitz formula there are
2d − 2 branch points (admittedly, with multiplicity). It is possible to turn this into
a proof, and it is interesting to do so.
21.7.8. Bounds on automorphism groups of curves.
It is a nontrivial fact that irreducible smooth projective curves of genus g ≥ 2have finite automorphism groups. (See [FK, §5.1] for a proof over C. See [Schm]for the first proof in arbitrary characteristic, although better approaches are now
available.) Granting this fact, we can show that in characteristic 0, the automorphism group has order at most 84(g − 1) (Hurwitz’s Automorphisms Theorem), as
follows.
Suppose C is an irreducible smooth projective curve over an algebraicallyclosed field k = k of characteristic 0, of genus g ≥ 2. Suppose that G is a finitegroup of automorphisms of C. We now show that |G| ≤ 84(g − 1). (The case where
k is not algebraically closed is quickly dispatched by base-changing to k.)
21.7.K. E XERCISE .
(a) Let C ′ be the smooth projective curve corresponding to the field extension
K(C)G of k (via Theorem 17.4.3). (K(C)G means the G-invariants of K(C).) Describe a morphism π : C → C ′ of degree |G|, as well as a faithful G-action on C that
commutes with π.
(b) Show that above each branch point of π, the preimages are all ramified to thesame order (as G acts transitively on them). Suppose there are n branch points
and the ith one has ramification ri (each |G|/ri times).
(c) Use the Riemann-Hurwitz formula to show that4(2g − 2) = |G| 2g(C ′ ) − 2 +n.
ri − 1i=1

ri

5

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To maximize |G|, we wish to minimize(21.7.8.1)

2g(C ′ ) − 2 +n.
ri − 1i=1ri

subject to (21.7.8.1) being positive. Note that 1/42 is possible: take g(C ′ ) = 0,n = 3, and (r1 , r2 , r3 ) = (2, 3, 7).
21.7.L. E XERCISE . Show that you can’t do better than 1/42 by considering thefollowing cases separately:(a) g(C ′ ) > 1,(b) g(C ′ ) = 1,(c) g(C ′ ) = 0 and n ≥ 5,(d) g(C ′ ) = 0 and n = 4, and(e) g(C ′ ) = 0 and n = 3.
21.7.M. E XERCISE . Use the fact that (21.7.8.1) is at least 1/42 to prove the result.
21.7.9. Remark. In positive characteristic, there can be many more automorphisms.
For example, [Stic], in characteristic p, if pn is a prime power that is not 2, thennn
the completion of the affine curve yp = x + xp +1 has genus g = pn (pn − 1)/2
and automorphism group of order p3n (p3n + 1)(p2n − 1).

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CHAPTER 22⋆ Blowing up
We next discuss an important construction in algebraic geometry, the blow-up
of a scheme along a closed subscheme (cut out by a finite type ideal sheaf). The
theory could mostly be developed immediately after Chapter 17, but the interpretation in terms of the conormal cone/bundle/sheaf of many classical examples
makes it natural to discuss blowing up after differentials.
We won’t use blowing up much in later chapters, so feel free to skip this topic
for now. But it is an important tool. For example, one can use it to resolve singularities, and more generally, indeterminacy of rational maps. In particular, blow-ups
can be used to relate birational varieties to each other.
We will start with a motivational example that will give you a picture of theconstruction in a particularly important (and the historically earliest) case, in §22.1.
We will then see a formal definition, in terms of a universal property, §22.2. The
definition won’t immediately have a clear connection to the motivational example.
We will deduce some consequences of the definition (assuming that the blow-up
actually exists). We then prove that the blow-up exists, by describing it quite explicitly, in §22.3. As a consequence, we will find that the blow-up morphism isprojective, and we will deduce more consequences from this. In §22.4, we will doa number of explicit computations, to see various sorts of applications, and to seethat many things can be computed by hand.

22.1 Motivating example: blowing up the origin in the plane
We will generalize the following notion, which will correspond to “blowing up”
the origin of A2k (Exercise 9.3.F). Our discussion will be informal. Consider thesubset of A2 ×P1 corresponding to the following. We interpret P1 as parametrizing
the lines through the origin. Consider the subvariety01
Bl(0,0) A2 := (p ∈ A2 , [ℓ] ∈ P1 ) : p ∈ ℓ) ,which is the data of a point p in the plane, and a line ℓ containing both p andthe origin. Algebraically: let x and y be coordinates on A2 , and X and Y be
projective coordinates on P1 (“corresponding” to x and y); we will consider thesubset Bl(0,0) A2 of A2 × P1 corresponding to xY − yX = 0. We have the usefuldiagram"% A2 × P1% P1
Bl(0,0) A2 !
▲▲▲
▲▲▲
▲
β ▲▲▲
▲d "
A2
597

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You can verify that it is smooth over k (Definition 12.2.6 or 21.3.1) directly (youcan now make the paragraph after Exercise 9.3.F precise), but here is an informal
argument, using the projection Bl(0,0) A2 → P1 . The projective line P1 is smooth,and for each point [ℓ] in P1 , we have a smooth choice of points on the line ℓ. Thus
we are verifying smoothness by way of a fibration over P1 .
We next consider the projection to A2 , β : Bl(0,0) A2 → A2 . This is an isomorphism away from the origin. Loosely speaking, if p is not the origin, there is
precisely one line containing p and the origin. On the other hand, if p is the origin,then there is a full P1 of lines containing p and the origin. Thus the preimage of(0, 0) is a curve, and hence a divisor (an effective Cartier divisor, as the blown-up
surface is regular). This is called the exceptional divisor of the blow-up.
If we have some curve C ⊂ A2 singular at the origin, it can be potentiallypartially desingularized, using the blow-up, by taking the closure of C \ {(0, 0)}in Bl(0,0) A2 . (A desingularization or a resolution of singularities of a variety
3 → X from a regular scheme.) For example,
X is a proper birational morphism X
2
3
2
consider the curve y = x + x , which is regular except for a node at the origin.
We can take the preimage of the curve minus the origin, and take the closure ofthis locus in the blow-up, and we will obtain a regular curve; the two branches of
the node downstairs are separated upstairs. (You can check this in Exercise 22.4.B
once we have defined things properly. The result will be called the proper transform (or strict transform) of the curve.) We are interested in desingularizations formany reasons. Because we understand regular curves quite well, we could hope tounderstand other curves through their desingularizations. This philosophy holds
true in higher dimension as well.
More generally, we can blow up An at the origin (or more informally, “blowup the origin”), getting a subvariety of An × Pn−1 . Algebraically, If x1 , . . . , xn
are coordinates on An , and X1 , . . . , Xn are projective coordinates on Pn−1 , thenthe blow-up Bl⃗0 An is given by the equations xi Xj − xj Xi = 0. Once again, this issmooth: Pn−1 is smooth, and for each point [ℓ] ∈ Pn−1 , we have a smooth choiceof p ∈ ℓ.
We can extend this further, by blowing up An+m along a coordinate m-planem
A by adding m more variables xn+1 , . . . , xn+m to the previous example; we geta subset of An+m × Pn−1 .
Because in complex geometry, submanifolds of manifolds locally “look like”
coordinate m-planes in n-space, you might imagine that we could extend this toblowing up a regular subvariety of a regular variety. In the course of making this
precise, we will accidentally generalize this notion greatly, defining the blow-up of
any finite type quasicoherent sheaf of ideals in a scheme. In general, blowing up
may not have such an intuitive description as in the case of blowing up something
regular inside something regular — it can do great violence to the scheme — buteven then, it is very useful.
Our description will depend only on the closed subscheme being blown up,
and not on coordinates. That remedies a defect was already present in the firstexample, of blowing up the plane at the origin. It is not obvious that if we pickeddifferent coordinates for the plane (preserving the origin as a closed subscheme)that we wouldn’t have two different resulting blow-ups.
As is often the case, there are two ways of understanding the notion of blowing
up, and each is useful in different circumstances. The first is by universal property,

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November 18, 2017 draft599

which lets you show some things without any work. The second is an explicit
construction, which lets you get your hands dirty and compute things (and implies
for example that the blow-up morphism is projective).
The motivating example here may seem like a very special case, but if you
understand the blow-up of the origin in n-space well enough, you will understandblowing up in general.

22.2 Blowing up, by universal property
We now define the blow-up by a universal property. The disadvantage of starting
here is that this definition won’t obviously be the same as (or even related to) the
examples of §22.1.
Suppose X "→ Y is a closed subscheme corresponding to a finite type quasicoherent sheaf of ideals. (If Y is locally Noetherian, the “finite type” hypothesis isautomatic, so Noetherian readers can ignore it.)
The blow-up of X "→ Y is a Cartesian diagram(22.2.0.1)"

EX Y !

% BlX Y
β

" "
X!

"
%Y

such that EX Y (the scheme-theoretic pullback of X by β) is an effective Cartierdivisor (defined in §8.4.1) on BlX Y, such that any other such Cartesian diagram(22.2.0.2)

D!
" "
X!

"

%W
"
% Y,

where D is an effective Cartier divisor on W, factors uniquely through it:
"

%W

" "
EX Y !

"% BlX Y

" "
X!

"
% Y.

D!

We call BlX Y the blow-up (of Y along X, or of Y with center X). (Other somewhat
archaic terms for this are monoidal transformation, σ-process, and quadratic transformation, and dilation.) We call EX Y the exceptional divisor of the blow-up. (Bl andβ stand for “blow-up”, and E stands for “exceptional”.)
By a typical universal property argument, if the blow-up exists, it is unique up
to unique isomorphism. (We can even recast this more explicitly in the language of
Yoneda’s Lemma: consider the category of diagrams of the form (22.2.0.2), where

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The Rising Sea: Foundations of Algebraic Geometrymorphisms are diagrams of the form!"D✵ 000 % W✶ ◗◗◗✵✵ 000 ✶✶ ◗◗◗
000 ✶
◗◗◗
✵✵
◗◗◗
00✶✶0
◗◗◗
0
✵✵
0
✶✶ 02
"
2% ′
✵✵
✶✶ D ′ !
W
✵✵
④
✶⑤✶ ⑤
④
④
⑤
✵✵
④
⑤ ✶
✵E ⑤⑤⑤ ✶✶3 ④④④
b
④
b! ⑤
"
% Y.
X
Then the blow-up is a final object in this category, if one exists.)
If Z "→ Y is any closed subscheme of Y, then the (scheme-theoretic) pullback
β−1 Z is called the total transform of Z. We will soon see that β is an isomorphism
away from X (Observation 22.2.2). β−1 (Z − X) is called the proper transform or
strict transform of Z. (We will use the first terminology. We will also define it in amore general situation.) We will soon see (in the Blow-up Closure Lemma 22.2.6)
that the proper transform is naturally isomorphic to BlZ∩X Z, where Z ∩ X is thescheme-theoretic intersection.
We will soon show that the blow-up always exists, and describe it explicitly.
We first make a series of observations, assuming that the blow up exists.
22.2.1. Observation. If X is the empty set, then BlX Y = Y. More generally, if
X is an effective Cartier divisor, then the blow-up is an isomorphism. (Reason:
idY : Y → Y satisfies the universal property.)∼ β−1 (U), where
22.2.A. E XERCISE . If U is an open subset of Y, then BlU∩X U =β : BlX Y → Y is the blow-up.
Thus “we can compute the blow-up locally.”

22.2.B. E XERCISE . Show that if Yα is an open cover of Y (as α runs over someindex set), and the blow-up of Yα along X ∩ Yα exists, then the blow-up of Y along
X exists.
22.2.2. Observation. Combining Observation 22.2.1 and Exercise 22.2.A, we seethat the blow-up is an isomorphism away from the locus you are blowing up:
is an isomorphism.

β|BlX Y−EX Y : BlX Y − EX Y → Y − X

22.2.3. Observation. If X = Y, then the blow-up is the empty set: the only map
W → Y such that the pullback of X is an effective Cartier divisor is ∅ "→ Y. In thiscase we have “blown Y out of existence”!

22.2.C. E XERCISE ( BLOW- UP PRESERVES IRREDUCIBILITY AND REDUCEDNESS ). Show
that if Y is irreducible, and X doesn’t contain the generic point of Y, then BlX Y isirreducible. Show that if Y is reduced, then BlX Y is reduced.
22.2.4. Existence in a first nontrivial case: blowing up a locally principal closed
subscheme.
We next see why BlX Y exists if X "→ Y is locally cut out by one equation. As thequestion is local on Y (Exercise 22.2.B), we reduce to the affine case Spec A/(t) "→Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

November 18, 2017 draft

601

Spec A. (A good example to think through is A = k[x, y]/(xy) and t = x.) Let
I = ker(A → At ) = {a ∈ A : tn a = 0 for some n > 0},and let φ : A → A/I be the projection.

22.2.D. E XERCISE . Show that φ(t) is not a zerodivisor in A/I.

22.2.E. E XERCISE . Show that β : Spec A/I → Spec A is the blow up of Spec A
along Spec A/t. In other words, show thatSpec A/(t, I)% Spec A/I"
Spec A/t"% Spec Aβ

is a “blow up diagram” (22.2.0.1). Hint: In checking the universal property reduceto the case where W (in (22.2.0.2)) is affine. Then solve the resulting problem aboutrings. Depending on how you proceed, you might find Exercise 10.2.G, about theuniqueness of extension of maps over effective Cartier divisors, helpful.
22.2.F. E XERCISE .
Spec A.

Show that Spec A/I is the scheme-theoretic closure of D(t) in

Thus you might geometrically interpret Spec A/I → Spec A as “shaving off
any fuzz supported in V(t)”. In the Noetherian case, this can be interpreted as
removing those associated points lying in V(t). This is intended to be vague, andyou should think about how to make it precise only if you want to.
22.2.5. The Blow-up Closure Lemma.
Suppose we have a fibered diagram
W!
" "
X!

"

cl. emb.

cl. emb.

%Z"%Y

where the bottom closed embedding corresponds to a finite type ideal sheaf (andhence the upper closed embedding does too). The first time you read this, it may
be helpful to consider only the special case where Z → Y is a closed embedding.
Then take the fibered product of this square by the blow-up β : BlX Y → Y, toobtain the Cartesian diagram"% Z ×Y BlX Y
W ×Y EX Y !
" "EX Y !

Cartier"% BlX Y.

(Why is the top row the base change of the bottom as claimed?) The bottom closed
embedding is locally cut out by one equation, and thus the same is true of the top
closed embedding as well. However, the local equation on Z×Y BlX Y need not be a
non-zerodivisor, and thus the top closed embedding is not necessarily an effective
Cartier divisor.

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The Rising Sea: Foundations of Algebraic Geometry

Let Z be the scheme-theoretic closure of(Z ×Y BlX Y) \ (W ×Y BlX Y)

in Z×Y BlX Y. (As W×Y BlX Y is locally principal, we are in precisely the situation of§22.2.4, so the scheme-theoretic closure is not mysterious.) Note that in the special
case where Z → Y is a closed embedding, Z is the proper transform, as definedin §22.2. For this reason, it is reasonable to call Z the proper transform of Z even if
Z isn’t a closed embedding. Similarly, it is reasonable to call Z ×Y BlX Y the totaltransform of Z even if Z isn’t a closed embedding.
Define EZ "→ Z as the pullback of EX Y to Z, i.e., by the fibered diagram
"
% Z"
EZ" !
proper transform##cl. emb.

cl. emb.

"" prin.
! loc.

"" "EX Y !

Cartier"% BlX Y.

W ×Y EX Y% Z ×Y BlX Y

total transform

Note that EZ is an effective Cartier divisor on Z. (It is locally cut out by one equation, pulled back from a local equation of EX Y on BlX Y. Can you see why this islocally not a zerodivisor?) It can be helpful to note that the top square of the diagram above is a blow-up square, by Exercises 22.2.E and 22.2.F (and the fact thatblow-ups can be computed affine-locally).
22.2.6. Blow-up Closure Lemma. — (BlW Z, EW Z) is canonically isomorphic to(Z, EZ ). More precisely: if the blow-up BlX Y exists, then (Z, EZ ) is the blow-up of Z
along W.
This will be very useful. We make a few initial comments. The first three apply
to the special case where Z → Y is a closed embedding, and the fourth comment
basically tells us we shouldn’t have concentrated on this special case.
(1) First, note that if Z → Y is a closed embedding, then the Blow-Up Closure
Lemma states that the proper transform (as defined in §22.2) is the blow-up of Z
along the scheme-theoretic intersection W = X ∩ Z.
(2) In particular, the Blow-Up Closure Lemma lets you actually compute blowups, and we will do lots of examples soon. For example, suppose C is a plane
curve, singular at a point p, and we want to blow up C at p. Then we could
instead blow up the plane at p (which we have already described how to do, evenif we haven’t yet proved that it satisfies the universal property of blowing up), andthen take the scheme-theoretic closure of C \ {p} in the blow-up.
(3) More generally, if W is some nasty subscheme of Z that we wanted to blowup, and Z were a finite type k-scheme, then the same trick would work. We could
work locally (Exercise 22.2.A), so we may assume that Z is affine. If W is cut
out by r equations f1 , . . . , fr ∈ Γ (Z, OZ ), then complete the f’s to a generating set
f1 , . . . , fn of Γ (Z, OZ ). This gives a closed embedding Y "→ An such that W is thescheme-theoretic intersection of Y with a coordinate linear space An−r .
22.2.7. (4) Most generally still, this reduces the existence of the blow-up to a specific special case. (If you prefer to work over a fixed field k, feel free to replace Z by

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November 18, 2017 draft603

k in this discussion.) Suppose that for each n, Bl(x1 ,...,xn ) Spec Z[x1 , . . . , xn ] exists.
Then I claim that the blow-up always exists. Here’s why. We may assume that Y isaffine, say Spec B, and X = Spec B/(f1 , . . . , fn ). Then we have a morphism Y → AnZ
given by xi (→ fi , such that X is the scheme-theoretic pullback of the origin. Henceby the blow-up closure lemma, BlX Y exists.
22.2.G. ⋆ T RICKY E XERCISE . Prove the Blow-up Closure Lemma 22.2.6. Hint:
obviously, construct maps in both directions, using the universal property. Constructing the following diagram may or may not help.

eff. Cartier!"%EZ 2 3✐✐ Z ✽2 ✽3
♠
♠
✾
♠
✐✐✐✐
♠
✾✾
✐
✽✽
♠
✐
♠
✐
✐
♠
✾✾
✽✽
♠♠♠
✐✐✐✐
✐
♠
✐
♠
✾
✐
✽✽
♠
✐
c! "♠♠♠ eff. Cartier]✐✐✾✐✾✐✾✽✽% BlW Z ❳❳ ✾EW ✷Z ❲❲❲❲✽✽✿✿ ❳❳❳✾✾❳❳❳❳
❲❲❲❲❲
✷✷
✽✽
✾✾ ❳❳❳❳❳❳
❲❲❲❲❲
✿✿
❳
❳
✽✽
✷✷
❲❲❲❲❲ ✿✾✾❳❳❳❳❳❳
✿
❲
❳
❲
✽,
❳
✾
✷✷
❲✿❲✿❲❲:❳❳❳❳❳f,"loc. prin.
✷✷
✿✿
!
%
Z ×Y Bl❀X Y✿ Z ×Y E✾X Y
✷✷
❀❀
✾✾
✐✐✐
❥❥❥✿✿✿
✐
❥
✷✷
✐
❥
✐
❥
❀❀
✾✾
✿✿
❥❥
✐✐✐
❥
✐
✷✷
❥
✐
❥
✐
❀❀
✾✾✐✐✐✐✿e& ! "]❥❥❥❥❥❥❥✐❀❀% Z ]✐✐✐ ✾✾✾
❀❀
W✹
✾
✾✾
❀❀
✾✾
✹✹
✾
❀❀
✾✾
✹✹
✾✾
❀❀
✾✾
✹✹
✾
e
,
✾✾
✹✹
"!
eff. Cartier% BlX Y✾✾ ✐ EX Y
✹✹
❤
✹✹
✐✐✐✾✐
❤❤❤❤
✐✐✐✐ ✾✾✾
❤❤❤❤
✐
❤
✹✹
✐
❤
✐
❤
✐
✾✾
✹&
❤❤❤❤
✐✐✐✐
,
❤❤❤❤
"]✐✐✐✐✐✐
❤
❤
!
❤
]
%Y
X

Hooked arrows indicate closed embeddings; and when morphisms are furthermore locally principal or even effective Cartier, they are so indicated. Exercise 10.2.G,on the uniqueness of extension of maps over effective Cartier divisors, may or may
not help as well. Note that if Z → Y is actually a closed embedding, then so is
Z ×Y BlX Y → BlX Y and hence also Z → BlX Y.

22.3 The blow-up exists, and is projective
22.3.1. It is now time to show that the blow-up always exists. We will see twoarguments, which are enlightening in different ways. Both will imply that theblow-up morphism is projective, and hence quasicompact, proper, finite type, andseparated. In particular, if Y → Z is quasicompact (resp. proper, finite type, separated), so is BlX Y → Z. (And if Y → Z is projective, and Z is quasicompact, thenBlX Y → Z is projective. See the solution to Exercise 17.3.B for the reason for this annoying extra hypothesis.) The blow-up of a k-variety is a k-variety (using the factthat reducedness is preserved, Exercise 22.2.C), and the blow-up of an irreducible
k-variety is a irreducible k-variety (using the fact that irreducibility is preserved,
also Exercise 22.2.C).

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Approach 1. As explained in §22.2.7, it suffices to show that BlV(x1 ,...,xn ) Spec Z[x1 , . . . , xn ]exists. But we know what it is supposed to be: the locus in Spec Z[x1 , . . . , xn ] ×
Proj Z[X1 , . . . , Xn ] cut out by the equations xi Xj − xj Xi = 0. We will show this by
the end of the section.
Approach 2. We can describe the blow-up all at once as a Proj .
22.3.2. Theorem (Proj description of the blow-up). — Suppose X "→ Y is a closed
subscheme cut out by a finite type quasicoherent sheaf of ideals I "→ OY . Then)*
Proj Y OY ⊕ I ⊕ I 2 ⊕ I 3 ⊕ · · · → Ysatisfies the universal property of blowing up.

(We made sense of products of ideal sheaves, and hence I n , in Exercise 14.3.E.)
We will prove Theorem 22.3.2 soon (§22.3.3), after seeing what it tells us. Because I is finite type, the graded sheaf of algebras has degree 1 piece that is finite
type. The graded sheaf of algebras is also clearly generated in degree 1. Thus thesheaf of algebras satisfy Hypotheses 17.2.1 (“finite generation in degree 1”).
But first, we should make sure that the preimage of X is indeed an effective
Cartier divisor. We can work affine-locally (Exercise 22.2.A), so we may assume
that Y = Spec B, and X is cut out by the finitely generated ideal I. Then)*BlX Y = ProjB B ⊕ I ⊕ I2 ⊕ · · · .

(You may recall that the ring B ⊕ I ⊕ · · · is called the Rees algebra of the ideal I inB, §12.9.1.) We are slightly abusing notation by using the notation BlX Y, as wehaven’t yet shown that this satisfies the universal property.
The preimage of X isn’t just any effective Cartier divisor; it corresponds to theinvertible sheaf O(1) on this Proj . Indeed, O(1) corresponds to taking our graded
ring, chopping off the bottom piece, and sliding all the graded pieces to the left by1 (§15.2); it is the invertible sheaf corresponding to the graded module
I ⊕ I2 ⊕ I3 ⊕ · · ·(where that first summand I has grading 0). But this can be interpreted as thescheme-theoretic pullback of X, which corresponds to the ideal I of B:)*
I B ⊕ I ⊕ I2 ⊕ · · · "→ B ⊕ I ⊕ I2 ⊕ · · · .

Thus the scheme-theoretic pullback of X "→ Y to Proj (OY ⊕ I ⊕ I 2 ⊕ · · · ), theinvertible sheaf corresponding to I ⊕I 2 ⊕I 3 ⊕· · · , is an effective Cartier divisor
in class O(1). Once we have verified that this construction is indeed the blow-up,
this divisor will be our exceptional divisor EX Y.
Moreover, we see that the exceptional divisor can be described beautifully asa Proj over X:)*(22.3.2.1)
EX Y = Proj X OY /I ⊕ I /I 2 ⊕ I 2 /I 3 ⊕ · · · .

We will later see that in good circumstances (if X is a regular embedding in Y), thisis a projectivization of a vector bundle (the “projectivized normal bundle”), see
Exercise 22.3.D(a).

22.3.3. Proof of the universal property, Theorem 22.3.2. Let’s prove that this Proj
construction satisfies the universal property. Then Approach 1 will also follow, asa special case of Approach 2.

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22.3.4. Aside: why Approach 1?. Before we begin, you may be wondering why
we bothered with Approach 1. One reason is that you may find it more comfortable to work with this one nice ring, and the picture may be geometrically clearer
to you (in the same way that thinking about the Blow-up Closure Lemma 22.2.6
in the case where Z → Y is a closed embedding is more intuitive). Another reason is that, as you will find in the exercises, you will see some facts more easily
in this explicit example, and you can then pull them back to more general examples. Perhaps most important, Approach 1 lets you actually compute blow-ups
by working affine-locally: if f1 , . . . , fn are elements of a ring A, cutting a subscheme X = Spec A/(f1 , . . . , fn ) of Y = Spec A, then BlX Y can be interpreted asa closed subscheme of Pn−1
A , by pulling back from BlV(x1 ,...,xn ) Spec Z[x1 , . . . , xn ],and taking the closure of the locus “above X” as dictated by the Blow-up Closure
Lemma 22.2.6.
Proof. Reduce to the case of affine target Spec R with ideal I ⊂ R. Reduce to thecase of affine source, with principal effective Cartier divisor t. (A principal effective Cartier divisor is locally cut out by a single non-zerodivisor.) Thus we have reduced to the case Spec S → Spec R, corresponding to f : R → S. Say (x1 , . . . , xn ) =I, with (f(x1 ), . . . , f(xn )) = (t). We will describe one map Spec S → Proj R[I] thatwill extend the map on the open set Spec St → Spec R. It is then unique, by Exercise 10.2.G. We map R[I] to S as follows: the degree one part is f : R → S, andf(Xi ) (where Xi corresponds to xi , except it is in degree 1) goes to f(xi )/t. Hencean element X of degree d goes to X/(td ). On the open set D+ (X1 ), we get the map
R[X2 /X1 , . . . , Xn /X1 ]/(x2 −X2 /X1 x1 , . . . , xi Xj −xj Xi , . . . ) → S (where there may be
many relations) which agrees with f away from V(t). Thus this map does extend
over V(I).
!
Here are some applications and observations arising from this construction of
the blow-up. First, we can verify that our initial motivational examples are indeed
blow-ups. For example, blowing up A2 (with coordinates x and y) at the origin
yields: B = k[x, y], I = (x, y), and Proj(B ⊕ I ⊕ I2 ⊕ · · · ) = Proj B[X, Y] where theelements of B have degree 0, and X and Y are degree 1 and “correspond to” x andy respectively.
22.3.5. Normal bundles to exceptional divisors. The normal bundle to an effective
Cartier divisor D is the (space associated to the) invertible sheaf O(D)|D , the invertible sheaf corresponding to the D on the total space, then restricted to D (Exercise 21.2.H). Thus in the case of the blow-up of a point in the plane, the exceptional divisor has normal bundle O(−1). (As an aside: Castelnuovo’s Criterion,∼ P1
Theorem 29.7.1, states that conversely given a smooth surface containing E =with normal bundle O(−1), E can be blown-down to a point on another smoothsurface.) In the case of the blow-up of a regular subvariety of a regular variety,the blow up turns out to be regular (see Theorem 22.3.10), the exceptional divisor
is a projective bundle over X, and the normal bundle to the exceptional divisor
restricts to O(−1) (Exercise 22.3.D).
22.3.A. H ARDER BUT ENLIGHTENING EXERCISE . If i : X "→ Pn is a projective
scheme, identify the exceptional divisor of the blow up of the affine cone over X(§8.2.12) at the origin with X itself, and that its normal bundle (§22.3.5) is isomorphic to OX (−1) := i∗ OPn (−1). (In the case X = P1 , we recover the blow-up of

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the plane at a point. In particular, we recover the important fact that the normal
bundle to the exceptional divisor is O(−1).)
22.3.6. The normal cone. Motivated by (22.3.2.1), as well as Exercise 22.3.D below,
we make the following definition. If X is a closed subscheme of Y cut out by I ,then the normal cone NX Y of X in Y is defined as)*(22.3.6.1)
NX Y := Spec X OY /I ⊕ I /I 2 ⊕ I 2 /I 3 ⊕ · · · .

This can profitably be thought of as an algebro-geometric version of a “tubularneighborhood”. But some cautions are in order. If Y is smooth, NX Y may not
be smooth. (You can work out the example of Y = A2k and X = V(xy).) And
even if X and Y are smooth, then although NX Y is smooth (as we will see shortly,
Exercise 22.3.D), it doesn’t “embed” in any way in Y (see Remark 22.3.9).
If X is a closed point p, then the normal cone is called the tangent cone to
Y at p. The projectivized tangent cone is the exceptional divisor EX Y (the Projof the same graded sheaf of algebras). Following §8.2.13, the tangent cone andthe projectivized tangent cone can be put together in the projective completionof the tangent cone, which contains the tangent cone as an open subset, and the
projectivized tangent cone as a complementary effective Cartier divisor.
In Exercise 22.3.D, we will see that at a regular point of Y, the tangent cone may
be identified with the tangent space, and the normal cone may often be identified
with the total space of the normal bundle.
22.3.B. E XERCISE . Suppose Y = Spec k[x, y]/(y2 − x2 − x3 ) (the nodal cubic, see
the bottom of Figure 7.4). Assume (to avoid distraction) that char k ̸= 2. Show thatthe tangent cone to Y at the origin is isomorphic to Spec k[x, y]/(y2 − x2 ). Thus, informally, the tangent cone “looks like” the original variety “infinitely magnified”.
22.3.C. E XERCISE . Suppose S• is a finitely generated graded algebra over a fieldk. Exercise 22.3.A gives an isomorphism of Proj S• with the exceptional divisor to
the blow-up of Spec S• at the origin. Show that the tangent cone to Spec S• at theorigin is isomorphic to Spec S• itself. (Your geometric intuition should lead you tofind these facts believable.)
22.3.7. Blowing up regular embeddings.
The case of blow-ups of regular embeddings X ⊂ Y is particularly pleasant.
For example, the exceptional divisor is a projective bundle over X. The central
reason is the following result.
22.3.8. Theorem. — If I ⊂ A is generated by a regular sequence a1 , . . . , ad , then the2
n n+1
natural map Symn
is an isomorphism.
A (I/I ) → I /I

Hence if a closed embedding i : X "→ Y is a regular embedding with ideal sheaf
I ⊂ OY , then the natural map Symn (I /I 2 ) → I n /I n+1 is an isomorphism.
Furthermore, in combination with Proposition 21.2.16, we see that I n /I n+1 is alocally free sheaf.
Before starting the proof of Theorem 22.3.8 in §22.3.11, we show its utility.
22.3.D. E XERCISE ( ASSUMING T HEOREM 22.3.8).
(a) Suppose X → Y is a regular embedding with ideal sheaf I , identify the total
space (§17.1.4) of the normal sheaf (the “normal bundle”) with the normal cone

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November 18, 2017 draft607

NX Y (22.3.6.1), and show that the exceptional divisor EX Y is a projective bundle(the “projectivized normal bundle”) over X.
(b) Show that the normal bundle to EX Y in BlX Y is O(−1) (for the projective bundle
over X).
(c) Assume further that X is a reduced closed point p. Show that p is a regularpoint of Y. Identify the total space of the tangent space to p with the tangent coneto Y at p.
22.3.9. Remark. We can now make sense of a comment made in §22.3.6, thateven if X and Y are smooth, then although NX Y is smooth, it needn’t admit anopen embedding in Y. To do this, start with a smooth complex quartic surface Y
containing a line X. (Most smooth quartic surfaces don’t contain a line, by Exercise 11.2.K, so you will have to construct one by hand.) Then NX Y is a line bundle
over X, and thus rational (i.e., birational to A2C , Definition 6.5.4). But NX Y cannot
admit an open embedding into Y, as otherwise Y would be rational, contradicting
Exercise 21.5.D.
22.3.10. Theorem. — Suppose X "→ Y is a closed embedding of smooth varieties over k.
Then BlX Y is also smooth.
Proof. (We use the fact that smooth varieties are regular, the Smoothness-Regularity
Comparison Theorem 12.2.10(b), whose proof we still have to complete.)
We need only check smoothness of BlX Y at the points of EX Y (by Observation 22.2.2). By Exercise 12.2.L(b), X "→ Y is a regular embedding. Then by Exercise 22.3.D(a), EX Y is a projective bundle over X, and thus smooth, and hence
regular at its closed points. But EX Y is an effective Cartier divisor on BlX Y. By theslicing criterion for regularity (Exercise 12.2.C), it follows that BlX Y is regular at
the closed points of EX Y, hence smooth at all points of EX Y.
!
22.3.11. ⋆ Proving Theorem 22.3.8.
The proof of Theorem 22.3.8 may reasonably be skipped on a first reading. We
prove Theorem 22.3.8 following [F, A.6.1], which in turn follows [Dav]. The proof
will be completed in §22.3.13. To begin, let α be the map of graded ringsn n+1
α : (A/I)[X1 , . . . , Xd ] → ⊕∞n=0 I /I

which takes Xi to the image of ai in I/I2 . Clearly α is surjective.
22.3.E. E XERCISE . Show that Theorem 22.3.8 would follow from the statementthat α is an isomorphism.
Because a1 is a non-zerodivisor, we can interpret A[a2 /a1 , . . . , ad /a1 ] as a subring of the total fraction ring (defined in §5.5.7). In particular, as A is a subring of
its total fraction ring, the map A → A[a2 /a1 , . . . , ad /a1 ] is an injection. Define
β : A[T2 , . . . , Td ] → A[a2 /a1 , . . . , ad /a1 ]by Ti (→ ai /a1 . Clearly, the map β is surjective, and Li := a1 Ti − ai lies in ker(β).
22.3.12. Lemma. — The kernel of β is (L2 , . . . , Ld ).

Proof. We prove the result by induction on d. We consider first the base case d = 2.
Suppose F[T2 ] ∈ ker β, so F(a2 /a1 ) = 0. Then applying the algorithm for the

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608

The Rising Sea: Foundations of Algebraic Geometrydeg F

Remainder Theorem, dividing a1(22.3.12.1)

deg F

a1

F(T2 ) by a1 T2 − a2 = L2 ,F(T2 ) = G(T2 )(a1 T2 − a2 ) + R

where G(T2 ) ∈ A[T2 ], and R ∈ A is the remainder. Substituting T2 = a2 /a1 (andusing the fact that A → A[a2 /a1 ] is injective), we have that R = 0. Then (a1 T2 −deg Fa2 )G(T2 ) ≡ 0 (mod (a1 )). Using the fact that a2 is a non-zerodivisor modulo
deg F
deg F
(a1 ) (as a1 , a2 is a regular sequence by Exercise 8.4.E), a short induction
deg F
shows that the coefficients of G(T2 ) must all be divisible by a1 . Thus F(T2 ) isdivisible by a1 T2 − a2 = L2 , so the case d = 2 is proved.
We now consider the general case d > 2, assuming the result for all smaller d.
Let A ′ = A[a2 /a1 ]. Then a1 , a3 , a4 , . . . , ad is a regular sequence in A ′ . Reason: a1
is a non-zerodivisor in A ′ . A ′ /(a1 ) = (A[T2 ]/(a1 T2 − a2 ))/(a1 ) = A[T2 ]/(a1 T2 −a2 , a1 ) = A[T2 ]/(a1 , a2 ). Then a3 is a non-zerodivisor in this ring because it isa nonzero divisor in A/(a1 , a2 ), a4 is a non-zerodivisor in A[T2 ]/(a1 , a2 , a3 ) because it is a non-zerodivisor in A/(a1 , a2 , a3 ), and so forth. Condition (ii) of theDefinition 8.4.4 of regular sequence also holds, as
A ′ /(a1 , a3 , a4 , . . . , ad ) = A[T2 ]/(a1 , a2 , . . . , ad ) ̸= 0.

Consider the composition

A[T2 , . . . , Td ] → A ′ [T3 , . . . , Td ] → A ′ [a3 /a1 , . . . , ad /a1 ] = A[a2 /a1 , . . . , ad /a1 ].

By the case d = 2, the kernel of the first map is L2 . By the inductive hypothesis,
the kernel of the second map is (L3 , . . . , Ld ). The result follows.
!
22.3.13. Proof of Theorem 22.3.8. By Exercise 22.3.E, it suffices to prove that thesurjection α is an isomorphism. Suppose F ∈ ker(α); we wish to show that F =0. We may assume that F is homogeneous, say of degree n. Consider the map
n n+1
α ′ : A[X1 , . . . , Xd ] → ⊕∞lifting α. Lift F to A[X1 , . . . , Xd ], so F ∈ ker(α ′ ).
n=0 I /I
We wish to show that F ∈ IA[X1 , . . . , Xd ]. Suppose F(a1 , . . . , ad ) = x ∈ In+1 . Then
we can write x as F ′ (a1 , . . . , ad ), where F ′ is a homogeneous polynomial of thesame degree as F, with coefficients in I. Then by replacing F by F − F ′ , we are
reduced to the following problem: suppose F ∈ A[X1 , . . . , Xd ] is homgeneous ofdegree n, and F(a1 , . . . , ad ) = 0, we wish to show that F ∈ IA[X1 , . . . , Xn ]. But
if F(a1 , . . . , ad ) = 0, then F(1, a2 /a1 , . . . , ad /a1 ) = 0 in A[a2 /a1 , . . . , ad /a1 ]. But
Lemma 22.3.12 identifies the kernel of β, so
F(1, T2 , T3 , . . . , Td ) ∈ (a1 T2 − a2 , a1 T3 − a3 , . . . , a1 Td − ad ).

Thus the coefficients of F are in (a1 , . . . , ad ) = I as desired.

!

22.4 Examples and computations
In this section we will work through a number of explicit of examples, to get asense of how blow-ups behave, how they are useful, and how one can work withthem explicitly. Throughout we work over a field k, and we assume throughout that char k = 0 to avoid distraction. The examples and exercises are loosely
arranged by topic, but not in order of importance.

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22.4.1. Example: Blowing up the plane along the origin.
Let’s first blow up the plane A2 along the origin, and see that the result agrees
with our discussion in §22.1. Let x and y be the coordinates on A2 . The blow-up is
Proj k[x, y, X, Y], where xY −yX = 0. (Here x and y have degree 0 and X and Y havedegree 1.) This is naturally a closed subscheme of A2 × P1 , cut out (in terms of the
projective coordinates X and Y on P1 ) by xY − yX = 0. We consider the two usual
patches on P1 : [X, Y] = [s, 1] and [1, t]. The first patch yields Spec k[x, y, s]/(sy − x),and the second gives Spec k[x, y, t]/(y − xt). Notice that both are smooth: the first∼ A2 , and the second is Spec k[x, t] =∼ A2 .
is Spec k[y, s] =We now describe the exceptional divisor. We first consider the first (s) patch.
The ideal is generated by (x, y), which in our ys-coordinates is (ys, y) = (y), whichis indeed principal. Thus on this patch the exceptional divisor is generated by y.
Similarly, in the second patch, the exceptional divisor is cut out by x. (This can bea little confusing, but there is no contradiction!) This explicit description will be
useful in working through some of the examples below.
22.4.A. E XERCISE . Let p be a k-valued point of P2 . Exhibit an isomorphism between Blp P2 and the Hirzebruch surface F1 = PP1 (OP1 ⊕ OP1 (1)) (Example 17.2.4).
(The map Blp P2 → P1 informally corresponds to taking a point to the line connecting it to the origin. Do not be afraid: You can do this by explicitly working withcoordinates.)
22.4.2. Resolving singularities.
22.4.3. The proper transform of a nodal curve (Figure 22.1). (You may wish to flip to
Figure 7.4 while thinking through this exercise.) Consider next the curve y2 = x3 +x2 inside the plane A2 (the nodal cubic). Let’s blow up the origin, and compute thetotal and proper transform of the curve. (By the Blow-up Closure Lemma 22.2.6,
the latter is the blow-up of the nodal curve at the origin.) In the first patch, we
get y2 − s2 y2 − s3 y3 = 0. This factors: we get the exceptional divisor y withmultiplicity two, and the curve 1 − s2 − s3 y = 0. You can easily check that the
3 meets theproper transform is regular. Also, notice that the proper transform C
exceptional divisor at two points, s = ±1. This corresponds to the two tangent
directions at the origin (as s = x/y).
22.4.B. E XERCISE (F IGURE 22.1). Describe both the total and proper transform of
the curve C given by y = x2 − x in Bl(0,0) A2 . Show that the proper transform of
C is isomorphic to C. Interpret the intersection of the proper transform of C withthe exceptional divisor E as the slope of C at the origin.
22.4.C. E XERCISE : BLOWING UP A CUSPIDAL PLANE CURVE ( CF. E XERCISE 9.7.F).
Describe the proper transform of the cuspidal curve C given by y2 = x3 in theblown-up plane Bl(0,0) A2 . Show that it is regular. Show that the proper transformof C meets the exceptional divisor E at one point, and is tangent to E there.
The previous two exercises are the first in an important sequence of singularities, which we now discuss.

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The Rising Sea: Foundations of Algebraic GeometryE
Bl(0,0) A2

E

E

C̃

C̃

C

C

A2

C̃

C

22.4.322.4.B

22.4.CF IGURE 22.1. Resolving curve singularities (§22.4.3,
cise 22.4.B, and Exercise 22.4.C)

Exer-

22.4.D. E XERCISE : RESOLVING An CURVE SINGULARITIES . Resolve the singularity
y2 = xn+1 in A2 , by first blowing up its singular point, then considering its propertransform and deciding what to do next.
22.4.4. Toward a definition of An curve singularities. You will notice that your solution to Exercise 22.4.D depends only on the “power series expansion” of the singularity at the origin, and not on the precise equation. For example, if you compare
your solution to Exercise 22.4.B with the n = 1 case of Exercise 22.4.D, you will seethat they are “basically the same”. We will make this precise in Definition 29.3.C.
22.4.E. E XERCISE ( WARM - UP TO E XERCISE 22.4.F). Blow up the cone point z2 =x2 + y2 (Figure 3.4) at the origin. Show that the resulting surface is regular. Show
that the exceptional divisor is isomorphic to P1 . (Remark: You can check that thenormal bundle to this P1 is not O(−1), as is the case when you blow up a point ona smooth surface, see §22.3.5; it is O(−2).)
22.4.F. E XERCISE ( RESOLVING An SURFACE SINGULARITIES ). Resolve the singularity z2 = y2 + xn+1 in A3 by first blowing up its singular point, then considering
its proper transform, and deciding what to do next. (A k-surface singularity analytically isomorphic to this is called an An surface singularity. For example, thecone shown in Figure 3.4 is an A1 surface singularity. We make this precise in
Exercise 29.3.C.) This exercise is a bit time consuming, but is rewarding in that it
shows that you can really resolve singularities by hand.
22.4.5. Remark: ADE-surface singularities and Dynkin diagrams (see Figure 22.2). A
k-singularity analytically isomorphic to z2 = x2 y + yn−1 (resp. z2 = x3 + y4 ,z2 = x3 + xy3 , z2 = x3 + y5 ) is a called a Dn surface singularity (resp. E6 , E7 ,E8 surface singularity). We will make this precise in Exercise 29.3.C, and you willthen be able to guess the definition of the corresponding curve singularity. If you

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611(minimally) desingularize each of these surfaces by sequentially blowing up singular points as in Exercise 22.4.F, and look at the arrangement of exceptional divisors (the various exceptional divisors and how they meet), you will discover
the corresponding Dynkin diagram. More precisely, if you create a graph, wherethe vertices correspond to exceptional divisors, and two vertices are joined by anedge if the two divisors meet, you will find the underlying graph of the corresponding Dynkin diagram. This is the start of several very beautiful stories; see
Remark 27.4.4 for a first glimpse of one of them.
An

···

···

E8

F IGURE 22.2. The exceptional divisors for resolutions of some
ADE surface singularities, and their corresponding dual graphs(see Remark 22.4.5)
22.4.6. Remark: Resolution of singularities. Hironaka’s Theorem on resolution of
singularities implies that this idea of trying to resolve singularities by blowing
up singular loci in general can succeed in characteristic 0. More precisely, if X is avariety over a field of characteristic 0, then X can be resolved by a sequence of blowups, where the nth blow-up is along a regular subvariety that lies in the singularlocus of the variety produced after the (n − 1)st stage (see [Hir], and [Ko2]). (Thecase of dimension 1 will be shown in §29.5.4, and the case of dimension 2 will be
discussed in §29.7.4.) It is not known if an analogous statement is true in positive
characteristic (except in dimension at most 2, by Lipman, see [Ar3]), but de Jong’s
Alteration Theorem [dJ] gives a result which is good enough for most applications.
3 → X from something regular, it
Rather than producing a birational proper map X
produces a proper map from something regular that is generically finite (and thecorresponding extension of function fields is separable).
Here are some other exercises related to resolution of singularities.
22.4.G. E XERCISE . Blowing up a nonreduced subscheme of a regular scheme cangive you something singular, as shown in this example. Describe the blow upof the closed subscheme V(y, x2 ) in Spec k[x, y] = A2 . Show that you get an A1
surface singularity.
22.4.H. E XERCISE .
Desingularize the tacnode (see Exercise 9.7.G and Definition 29.3.3) y2 = x4 , not in two steps (as in Exercise 22.4.D), but in a single step byblowing up (y, x2 ).
22.4.I. E XERCISE ( RESOLVING A SINGULARITY BY AN UNEXPECTED BLOW- UP ). Suppose Y is the cone x2 + y2 = z2 , and X is the line cut out by x = 0, y = z on Y. Show
that BlX Y is regular. (In this case we are blowing up a codimension 1 locus that

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is not an effective Cartier divisor, see Problem 12.1.3. But it is an effective Cartierdivisor away from the cone point, so you should expect your answer to be an
isomorphism away from the cone point.)
22.4.7. Multiplicity of a function at a regular point. In order to pose Exercise 22.4.J,
we introduce a useful concept. If f is a function on a locally Noetherian scheme
X, its multiplicity at a regular point p is the largest m such that f lies in the mth
power of the maximal ideal in the local ring OX,p . For example, if f ̸= 0, V(f) is
singular at p if and only if m > 1. (Do you see why?)
22.4.J. E XERCISE . Show that the multiplicity of the exceptional divisor in the totaltransform of a subscheme Z of An when you blow up the origin is the smallest multiplicity (at the origin) of a defining equation of Z. (For example, in the case of thenodal and cuspidal curves above, Example 22.4.3 and Exercise 22.4.C respectively,
the exceptional divisor appears with multiplicity 2.)
22.4.8. Resolving rational maps.
22.4.K. E XERCISE ( UNDERSTANDING THE BIRATIONAL MAP P2 (❴ ❴ ❴% P1 × P1 VIA
BLOW- UPS ).
Let p and q be two distinct k-points of P2 , and let r be a k-point1
1
of P × P . Describe an isomorphism Bl{p,q} P2 ↔ Blr P1 × P1 . (Possible hint:
Consider lines ℓ through p and m through q; the choice of such a pair corresponds
to the parametrized by P1 × P1 . A point s of P2 not on line pq yields a pair oflines (ps, qs) of P1 × P1 . Conversely, a choice of lines (ℓ, m) such that neither ℓand m is line pq yields a point s = ℓ ∩ m ∈ P2 . This describes a birational map
P2 (❴ ❴ ❴% P1 × P1 . Exercise 22.4.A is related.)
Exercise 22.4.K is an example of the general phenomenon explored in the next
two exercises.

22.4.L. H ARDER BUT USEFUL EXERCISE ( BLOW- UPS RESOLVE BASE LOCI OF RATIO NAL MAPS TO PROJECTIVE SPACE ). Suppose we have a scheme Y, an invertible
sheaf L , and a number of sections s0 , . . . , sn of L (a linear series, Definition 15.3.6).
Then away from the closed subscheme X cut out by s0 = · · · = sn = 0 (the baselocus of the linear series), these sections give a morphism to Pn . Show that thismorphism extends uniquely to a morphism BlX Y → Pn , where this morphism
corresponds to the invertible sheaf (β∗ L )(−EX Y), where β : BlX Y → Y is theblow-up morphism. In other words, “blowing up the base scheme resolves thisrational map”. Hint: it suffices to consider an affine open subset of Y where L istrivial. Uniqueness might use Exercise 10.2.G.
22.4.9. Remarks. (i) Exercise 22.4.L immediately implies that blow-ups can be used
to resolve rational maps to projective schemes Y ""# Z "→ Pn .
(ii) The following interpretation is enlightening. The linear series on Y pulls
back to a linear series on BlX Y, and the base locus of the linear series on Y pulls
back to the base locus on BlX Y. The base locus on BlX Y is EX Y, an effective Cartierdivisor. Because EX Y is not just locally principal, but also locally a non-zerodivisor,
it can be “divided out” from the β∗ si (yielding a section of (β∗ L )(−EX Y), thereby
removing the base locus, and leaving a base-point-free linear series. (In a sense

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613

that can be made precise through the universal property, this is the smallest “modification” of Y that can remove the base locus.) If X is already Cartier (as for example happens with any nontrivial linear system if Y is a regular pure-dimensional
curve), then we can remove a base locus by just “dividing out X”.
(iii) You may wish to revisit Exercise 19.7.B, and interpret it in terms of Exercise 22.4.L.
22.4.10. Examples. (i) The rational map Pn ""# Pn−1 given by [x0 , · · · , xn ] ""#[x1 , · · · , xn ], defined away from p = [1, 0, · · · , 0], is resolved by blowing up p.
Then by the Blow-up Closure Lemma 22.2.6, if Y is any locally closed subscheme ofPn , we can project to Pn−1 once we blow up p in Y, and the invertible sheaf giving
the map to Pn−1 is (somewhat informally speaking) β∗ (OPn (1)) ⊗ O(−Ep Y).
(ii) Consider two general cubic equations C1 and C2 in three variables, yielding two cubic curves in P2 . They are smooth (by Bertini’s Theorem 12.4.2, andmeet in 9 points p1 , . . . , p9 (using our standing assumption that we work over analgebraically closed field). Then [C1 , C2 ] gives a rational map P2 ""# P1 . To resolve the rational map, we blow up p1 , . . . , p9 . The result is (generically) an elliptic
fibration Bl{p1 ,...,p9 } P2 → P1 . (This is by no means a complete argument.)(iii) Fix six general points p1 , . . . , p6 in P2 . There is a four-dimensional vector
space of cubics vanishing at these points, and they vanish scheme-theoretically
precisely at these points. This yields a rational map P2 ""# P3 , which is resolvedby blowing up the six points. The resulting morphism turns out to be a closed
embedding, and the image in P3 is a (smooth) cubic surface. This is the famousfact that the blow up of the plane at six general points may be represented as a(smooth) cubic in P3 . (Again, this argument is not intended to be complete.) See
27.4.2 for a more precise and more complete discussion.
In reasonable circumstances, Exercise 22.4.L has an interpretation in terms ofgraphs of rational maps.
22.4.M. E XERCISE .
Suppose s0 , . . . , sn are sections of an invertible sheaf L
on an integral scheme X, not all 0. By Remark 16.4.3, these data give a rational
map φ : X ""# Pn . Give an isomorphism between the graph of φ (§10.2.4) andBlV(s0 ,...,sn ) X. (Your argument will not require working over a field k; it will
work in general.)
You may enjoy exploring the previous idea by working out how the Cremonatransformation P2 ""# P2 (Exercise 6.5.I) can be interpreted in terms of the graphof the rational map [x, y, z] ""# [1/x, 1/y, 1/z].
22.4.N. ⋆ E XERCISE . Resolve the rational map[w,x]Spec k[w, x, y, z]/(wz − xy) ❴ ❴ ❴ ❴ ❴ ❴% P1
from the cone over the quadric surface to the projective line. Let X be the resulting
variety, and π : X → Spec k[w, x, y, z]/(wz − xy) the projection to the cone over thequadric surface. Show that π is an isomorphism away from the cone point, andthat the preimage of the cone point is isomorphic to P1 (and thus has codimension
2, and therefore is different from the resolution obtained by simply blowing up thecone point). Possible hint: if Q is the quadric in P3 cut out by wz − xy = 0, then

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factor the rational map as Spec k[w, x, y, z]/(wz − xy) \ {0} → Q (cf. Exercise 8.2.P),
followed by the isomorphism Q → P1 ×P1 (Example 9.6.2), followed by projection
onto one of the factors.

This is an example of a small resolution. (A small resolution X → Y is a
resolution where the space of points of Y where the fiber has dimension r is ofcodimension greater than 2r. We will not use this notion again in any essential
way.) Notice that this resolution of the morphism involves blowing up the baselocus w = x = 0, which is a cone over one of the lines on the quadric surface wz =xy. We are blowing up an effective Weil divisor, which is necessarily not Cartier as
the blow-up is not an isomorphism. In Exercise 12.1.E, we saw that (w, x) was notprincipal, while here we see that (w, x) is not even locally principal. Essentially by
Exercise 14.2.R, V(w, x) cannot even be the support of a locally principal divisor.
22.4.11. Remark: Non-isomorphic small resolutions. If you instead resolved the map[w, y], you would obtain a similar looking small resolution
π ′ : X ′ → Spec k[w, x, y, z]/(wz − xy)(it is an isomorphism away from the origin, and the fiber over the origin is P1 ). But
it is different! More precisely, there is no morphism X → X ′ making the following
the diagram commute.
% X′
X ❘❘❘
❘❘❘
❧❧❧
❧
❧
′
❘❘❘π
π ❧❧❧❧
❘❘❘
❘❘❘
❧❧❧
❧
❧
2
c❧
Spec k[w, x, y, z]/(wz − xy)

The birational map X ""# X ′ is called the Atiyah flop, [At1].
22.4.12. Factorization of birational maps. We end our discussion of resolution of rational maps by noting that just as Hironaka’s Theorem states that one may resolve
all singularities of varieties in characteristic 0 by a sequence of blow-ups alongsmooth subvarieties, the Weak Factorization Theorem (first proved by Włodarczyk)
states that any two birational varieties X and Y in characteristic 0 may be relatedby blow-ups and blow-downs along smooth subvarieties. More precisely, there
are varieties X0 , . . . , Xn , X01 , . . . , X(n−1)n , with X0 = X and Xn = Y, with morphisms Xi(i+1) → Xi and Xi(i+1) → Xi+1 (0 ≤ i < n) which are blow-ups ofsmooth subvarieties.
22.4.13. Blow-ups and line bundles.

22.4.O. E XERCISE ( GENERALIZING E XERCISE 20.2.D). Suppose X is a regular projective surface over k, and p is a k-valued point. Let β : Blp X → X be the blow-up
morphism, and let E = Ep X be the exceptional divisor. Consider the exact sequenceZγ : 1*→[E]% Pic Blp X

α% Pic(Blp X \ E)%0(from (14.2.8.1)). Note that Blp X \ E = X \ p. Show that Pic(X \ p) = Pic X. Show∼
that β∗ : Pic X → Pic Blp X gives a section to α. Use §22.3.5 to show that OX (E)|E =OE (−1) (so in the language of Chapter 20, E is a (−1)-curve, Definition 20.2.7), andEarly (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

November 18, 2017 draft615∼ Pic X ⊕ Z. Describefrom that show that γ is an injection. Conclude that Pic Blp X =1
how to find the intersection matrix on NQ (Blp X) from that of N1Q (X).
22.4.P. E XERCISE . Suppose D is an effective Cartier divisor (a curve) on X. Let
multp D be the multiplicity of D at p (Exercise 22.4.J), and let Dpr be the propertransform of D. Show that π∗ D = Dpr + (multp D)E as effective Cartier divisors. More precisely, show that the product of the local equation for Dpr and the(multp D)th power of the local equation for E is the local equation for π∗ D, and∼ OBl X (Dpr ) ⊗hence that (i) π∗ D is an effective Cartier divisor, and (ii) π∗ OX (D) =p
OBlp X (E)⊗(multp D) . (A special case is the equation ℓ = e + m in Hint 20.2.6.)
22.4.14. Change of the canonical line bundle under blow-ups.
As motivation for how the canonical line bundle changes under blowing up,
consider π : Bl(0,0) A2 → A2 . Let X = Bl(0,0) A2 and Y = A2 for convenience. We
use Exercise 21.7.A to relate π∗ KY to KX .
We pick a generator for KY near (0, 0): dx ∧ dy. (This is in fact a generator for
KY everywhere on A2 , but for the sake of generalization, we point out that all thatmatters is that is a generator at (0, 0), and hence near (0, 0) by geometric Nakayama,
Exercise 13.7.E.) When we pull it back to X, we can interpret it as a section of KX ,which will generate KX away from the exceptional divisor E, but may contain Ewith some multiplicity µ. Recall that X can be interpreted as the data of a point in
A2 as well as the choice of a line through the origin. We consider the open subset
U where the line is not vertical, and thus can be written as y = mx. Here wehave natural coordinates: U = Spec k[x, y, m]/(y − mx), which we can interpret as
Spec k[x, m]. The exceptional divisor E meets U, at x = 0 (in the coordinates on U),
so we can calculate µ on this open set. Pulling back dx ∧ dy to U, we getdx ∧ dy = dx ∧ d(xm) = m(dx ∧ dx) + x(dx ∧ dm) = x(dx ∧ dm)
as dx ∧ dx = 0. Thus π∗ (dx ∧ dy) vanishes to order 1 along E.
22.4.Q. E XERCISE ( CF. U NIMPORTANT EXERCISE 21.7.J). Explain how this deter∼ (π∗ KY )(E).
mines a canonical isomorphism KX =22.4.R. E XERCISE . Repeat the above calculation in dimension n. Show that the
exceptional divisor appears with multiplicity (n − 1).
22.4.S. ⋆ E XERCISE . Suppose k is perfect.
(a) Suppose Y is a surface over k, and p is a regular k-valued point, and let β : X →∼ (β∗ KY )(E). Hint: to find a generator
Y be the blow-up of Y at p. Show that KX =of KX near p, choose generators x and y of m/m2 (where m is the maximal ideal of
OY,p ), and lift them to elements of OX,p . Why does dx ∧ dy generate KX at p?
(b) Repeat part (a) in arbitrary dimension (following Exercise 22.4.R).
(c) Suppose Z is a smooth m-dimensional (closed) subvariety of a smooth n-dimensional∼variety Y, and let β : X → Y be the blow-up of Y along Z. Show that KX =∗(β KY )((n − m − 1)E). (Recall from Theorem 22.3.10 that X = BlZ Y is smooth.)22.4.15. ⋆ Dimensional cohomology vanishing for quasiprojective schemes (promisedin §18.2.7).
Using the theory of blowing up, Theorem 18.2.6 (dimensional cohomology
vanishing for quasicoherent sheaves on projective k-schemes) can be extended to

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quasiprojective k-schemes. Suppose X is a quasiprojective k-variety of dimension
n. We show that X may be covered by n + 1 affine open subsets. As X is quasiprojective, there is some projective variety Y with an open embedding X "→ Y. By
replacing Y with the closure of X in Y, we may assume that dim Y = n. Put any
subscheme structure Z on the complement of X in Y (for example the reduced subscheme structure, §8.3.9). Let Y ′ = BlZ Y. Then Y ′ is a projective variety (§22.3.1),
which can be covered by n + 1 affine open subsets. The complement of X in Y ′is an effective Cartier divisor (EZ Y), so the restriction to X of each of these affine
open subsets of Y is also affine, by Exercise 7.3.F.
22.4.16. Remarks. (i) You might then hope that any dimension n variety can becovered by n + 1 affine open subsets. This is not true. For each integer m, there
is a threefold that requires at least m affine open sets to cover it, see [RV, Ex. 4.9].
By the discussion above, this example is necessarily not quasiprojective. (ii) Hereis a fact useful in invariant theory, which can be proved in the same way. Suppose
p1 , . . . , . . . , pn are closed points on a quasiprojective k-variety X. Then there is anaffine open subset of X containing all of them.
22.4.T. E XERCISE ( DIMENSIONAL COHOMOLOGY VANISHING FOR QUASIPROJEC Suppose X is a quasiprojective k-scheme of dimension d. Show
that for any quasicoherent sheaf F on X, Hi (X, F ) = 0 for i > d.

TIVE VARIETIES ).

22.4.17. ⋆⋆ Deformation to the normal cone (Figure 22.3).
The following construction is key to the modern understanding of intersection
theory in algebraic geometry, as developed by Fulton and MacPherson, [F].
22.4.U. E XERCISE : D EFORMATION TO THE NORMAL CONE . Suppose Y is a kvariety, and X "→ Y is a closed subscheme.
(a) Show that the exceptional divisor of β : BlX×0 (Y × A1 ) → Y × A1 is isomorphic
to the projective completion of the normal cone to X in Y.
(b) Let π : BlX×0 (Y × A1 ) → A1 be the composition of β with the projection to A1 .
Show that π∗ (0) is the scheme-theoretic union of BlX Y with the projective completion of the normal cone to X in Y, and the intersection of these two subschemes
may be identified with EX Y, which is a closed subscheme of BlX Y in the usual way(as the exceptional divisor of the blow-up BlX Y → Y), and a closed subscheme ofthe projective completion of the normal cone as described in Exercise 8.2.Q.
The map

BlX×0 (Y × A1 ) \ BlX Y → A1
is called the deformation to the normal cone (short for deformation of Y to the normal
cone of X in Y). Notice that the fiber above every k-point away from 0 ∈ A1 is
canonically isomorphic to Y, and the fiber over 0 is the normal cone. Because thisfamily is “nice” (more precisely, flat, the topic of Chapter 24), we can prove things
about general Y (near X) by way of this degeneration. (We will see in §24.4.10
that the deformation to the normal cone is a flat morphism, which is useful inintersection theory.)

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F IGURE 22.3. Deformation to the normal cone (§22.4.17)

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Part VI

More

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CHAPTER 23

Derived functors
Ça me semble extrêmement plaisant de ficher comme ça beaucoup de choses, pas drôlesquand on les prend séparément, sous le grand chapeau des foncteurs dérivés.
I find it very agreeable to stick all sorts of things, which are not much fun when taken
individually, together under the heading of derived functors.
— A. Grothendieck, letter to J.-P. Serre, February 18, 1955 [GrS, p. 6]
In this chapter, we discuss derived functors, introduced by Grothendieck in
his celebrated “Tôhoku article” [Gr1], and their applications to sheaves. For quasicoherent sheaves on quasicompact separated schemes, derived functor cohomology will agree with Čech cohomology (§23.5). Čech cohomology will suffice formost of our purposes, and is quite down to earth and computable, but derived
functor cohomology is worth seeing. First, it will apply much more generally inalgebraic geometry (e.g., étale cohomology) and elsewhere, although this is beyond the scope of this book. Second, it will easily provide us with some useful
notions, such as the Ext functors and the Leray spectral sequence. But derived
functors can be intimidating the first time you see them, so feel free to just skim
the main results, and to return to them later.

23.1 The Tor functors

We begin with a warm-up: the case of Tor. This is a hands-on example, but if
you understand it well, you will understand derived functors in general. Tor will
be useful to prove facts about flatness, which we will discuss in §24.3. Tor is shortfor “torsion” (see Remark 24.3.1).
If you have never seen this notion before, you may want to just remember its
properties. But I will prove everything anyway — it is surprisingly easy.
The idea behind Tor is as follows. Whenever we see a right-exact functor, wealways hope that it is the end of a long exact sequence. Informally, given a short
exact sequence(23.1.0.1)

0

% N′

%N

% N ′′% 0,

621

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we hope M ⊗A N ′ → M ⊗A N → M ⊗A N ′′ → 0 will extend to a long exact
sequence(23.1.0.2)′′′% TorA% TorA% TorA% ······i (M, N )i (M, N)i (M, N )′% TorA
1 (M, N )% TorA
1 (M, N)′′
% TorA
1 (M, N )% M ⊗A N ′% M ⊗A N% M ⊗A N ′′% 0.

More precisely, we are hoping for covariant functors TorAi (M, ·) from A-modules to
A-modules (functoriality giving 2/3 of the morphisms in (23.1.0.2)), with TorA
0 (M, N) ≡A′′′
M⊗A N, and natural connecting homomorphisms δ : TorA
(M,
N
)
→
Tor
i+1
i (M, N )for every short exact sequence (23.1.0.1) giving the long exact sequence (23.1.0.2).
(“Natural” means: given a morphism of short exact sequences, the natural square
you would write down involving the δ-morphism must commute.)
It turns out to be not too hard to make this work, and this will also motivate
derived functors. Let’s now define TorAi (M, N).
Take any resolution R of N by free modules:···% A⊕n2% A⊕n1% A⊕n0%N% 0.

More precisely, build this resolution from right to left. Start by choosing generatorsof N as an A-module, giving us A⊕n0 → N → 0. Then choose generators of thekernel, and so on. Note that we are not requiring the ni to be finite (although
we could, if N is a finitely generated module and A is Noetherian). Truncate theresolution, by stripping off the last term N (replacing → N → 0 with → 0). Thentensor with M (which does not preserve exactness). Note that M ⊗ (A⊕ni ) =M⊕ni , as tensoring with M commutes with arbitrary direct sums (Exercise 1.3.M).
Let TorAi (M, N)R be the homology of this complex at the ith stage (i ≥ 0). The
subscript R reminds us that our construction depends on the resolution, although
we will soon see that it is independent of R.
We make some quick observations.
∼• TorA
0 (M, N)R = M ⊗A N, canonically. Reason: as tensoring is right-exact,
and A⊕n1 → A⊕n0 → N → 0 is exact, we have that M⊕n1 → M⊕n0 → M ⊗A
N → 0 is exact, and hence that the homology of the truncated complex M⊕n1 →M⊕n0 → 0 is M ⊗A N.
• If M ⊗ · is exact (i.e., M is flat, §1.6.11), then TorAi (M, N)R = 0 for all i > 0.
(This characterizes flatness, see Exercise 23.1.D.)
Now given two modules N and N ′ and resolutions R and R ′ of N and N ′ , wecan “lift” any morphism N → N ′ to a morphism of the two resolutions:······% A⊕ni"% A⊕ni′

% ···
% ···

% A⊕n1% A⊕n0%N%0"% A⊕n1′"% A⊕n0′

"
% N′

%0

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We do this inductively on i. Here we use the freeness of A⊕ni : if a1 , . . . , ani are′
generators of A⊕ni , to lift the map b : A⊕ni → A⊕ni−1 (the composition of the′differential A⊕ni → A⊕ni−1 with the previously constructed A⊕ni−1 → A⊕ni−1 )′′′to c : A⊕ni → A⊕ni , we arbitrarily lift b(aj ) from A⊕ni−1 to A⊕ni , and declarethis to be c(aj ). (Warning for people who care about such things: we are using the
Axiom of Choice here.)
Denote the choice of lifts by R → R ′ . Now truncate both complexes (remove
column N → N ′ ) and tensor with M. Maps of complexes induce maps of homology (Exercise 1.6.D), so we have described maps (a priori depending on R → R ′ )

A
′
TorA
i (M, N)R → Tori (M, N )R ′ .

We say two maps of complexes f : C• → C•′ and g : C• → C•′ are homotopic if′there is a sequence of maps w : Ci → Ci+1
such that f − g = dw + wd.

23.1.A. E XERCISE . Show that two homotopic maps give the same map on homology.
23.1.B. C RUCIAL E XERCISE . Show that any two lifts R → R ′ are homotopic.

We now pull these observations together. (Be sure to digest these completely!)A′(1) We get a map of A-modules TorAi (M, N)R → Tori (M, N )R ′ , independent′of the lift R → R .
(2) Hence for any two resolutions R and R ′ of an A-module N, we get aA∼canonical isomorphism TorAi (M, N)R = Tori (M, N)R ′ . Here’s why. Choose lifts
R → R ′ and R ′ → R. The composition R → R ′ → R is homotopic to the identity (as it is a lift of the identity map N → N). Thus if fR→R ′ : TorAi (M, N)R →′′ is the map induced by R → R , and similarly fR ′ →R is the map
TorA
(M,
N)
R
i
induced by R ′ → R, then fR ′ →R ◦ fR→R ′ is the identity, and similarly fR→R ′ ◦fR ′ →R is the identity.
(3) Hence TorAi (M, ·) doesn’t depend on the choice of resolution. It is a covariant functor ModA → ModA .
23.1.1. Remark. Note that if N is a free module, then TorAi (M, N) = 0 for all M andall i > 0, as N has the trivial resolution 0 → N → N → 0 (it is “its own resolution”).
Finally, we get long exact sequences:

23.1.2. Proposition. — For any short exact sequence (23.1.0.1) we get a long exact
sequence of Tor’s (23.1.0.2).

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Proof. Given a short exact sequence (23.1.0.1), choose resolutions of N ′ and N ′′ .
Then use these to get a resolution for N as follows.
(23.1.2.1)

···
···
···

′

′′

0

0

0

"
% A⊕n1′"% A⊕n0′

"
% N′

%0

"
% A⊕(n1′ +n1′′ )"% A⊕(n0′ +n0′′ )

"
%N

%0

"
% A⊕n1′′"% A⊕n0′′"% N ′′

%0

"
0

"
0

"
0

′

′′

′

′

The map A⊕(ni+1 +ni+1 ) → A⊕(ni +ni ) is the composition A⊕ni+1 → A⊕ni "→
′
′′
′′
′′
′
′′
A⊕(ni +ni ) along with a lift of A⊕ni+1 → A⊕ni to A⊕(ni +ni ) ensuring that the
middle row is a complex.
′′′′

23.1.C. E XERCISE . Verify that it is possible to choose such a lift of A⊕ni+1 → A⊕ni′′′
to A⊕(ni +ni ) .

Hence the middle row of (23.1.2.1) is exact (not just a complex), using the longexact sequence in cohomology (Theorem 1.6.6), and the fact that the top and bottom rows are exact. Thus the middle row is a resolution, and (23.1.2.1) is a short
exact sequence of resolutions. (This is sometimes called the horseshoe construction,
as the filling in of the middle row looks like filling in the middle of a horseshoe.)
It may be helpful to notice that the columns other than the “N-column” are all “direct sum exact sequences”, and the horizontal maps in the middle row are “block
upper triangular”.
Then truncate (removing the right column 0 → N ′ → N → N ′′ → 0), tensor
with M (obtaining a short exact sequence of complexes) and take cohomology,
yielding the desired long exact sequence.
!
23.1.D. E XERCISE . Show that the following are equivalent conditions on an Amodule M.
(i) M is flat.
(ii) TorAi (M, N) = 0 for all i > 0 and all A-modules N.
(iii) TorA
1 (M, N) = 0 for all A-modules N.
23.1.3. Caution. Given that free modules are immediately seen to be flat, you
might think that Exercise 23.1.D implies Remark 23.1.1. This would follow if weA∼knew that TorAi (M, N) = Tori (N, M), which is clear for i = 0 (as ⊗ is symmetric),
but we won’t know this about Tori when i > 0 until Exercise 23.3.A.
23.1.E. E XERCISE . Show that the connecting homomorphism δ constructed above
is independent of all of choices (of resolutions, etc.). Try to do this with as little

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625

annoyance as possible. (Possible hint: given two sets of choices used to build(23.1.2.1), build a map — a three-dimensional diagram — from one version of(23.1.2.1) to the other version.)23.1.F. U NIMPORTANT E XERCISE .
Show that TorAi (M, ·) is an additive functor(Definition 1.6.1). (We won’t use this later, so feel free to skip it.)
We have thus established the foundations of Tor.

23.2 Derived functors in general
23.2.1. Projective resolutions. We used very little about free modules in the above
construction of Tor — in fact we used only that free modules are projective, i.e.,% % N , it is possible to lift any
those modules P such that for any surjection M
morphism P → N to P → M:
(23.2.1.1)

P✤ ❆
❆
✤ ❆❆❆
❆❆
exists✤"❆1%%N
M(As noted in §23.1, this uses the axiom of choice.) Equivalently, Hom(P, ·) is anexact functor (recall that Hom(Q, ·) is always left-exact for any Q). More generally, the same idea yields the definition of a projective object in any abelian category. Hence by following through our entire argument with projective modules
replacing free modules throughout, (i) we can compute TorAi (M, N) by taking any
projective resolution of N, and (ii) TorA
(M,
N)
=
0
for
any
projective A-moduleiN.
23.2.A. E XERCISE . Show that an object P is projective if and only if every short
exact sequence 0 → Q → R → P → 0 splits. Hence show that an A-module P is
projective if and only if P is a direct summand of a free module.

23.2.B. E XERCISE . Show that projective modules are flat. Hint: Exercise 23.2.A.
Be careful if you want to use Exercise 23.1.D; see Caution 23.1.3. (In fact, finitely
generated projective modules over local rings are even free, see Remark 24.4.6.)

23.2.2. Definition: Derived functors.
The above description was low-tech, but immediately generalizes drastically.
All we are using is that M ⊗A · is a right-exact functor, and that for any A-module% % N from a projective module. In general, if FN, we can find a surjection P
is any right-exact covariant functor from the category of A-modules to any abeliancategory, this construction will define a sequence of functors Li F such that L0 F = F
and the Li F’s give a long exact sequence. We can make this more general still. We
say that an abelian category has enough projectives if for any object N there is asurjection onto it from a projective object. Then if F is any right-exact covariant
functor from an abelian category with enough projectives to any abelian category,then we can define the left derived functors to F, denoted Li F (i ≥ 0). You should

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reread §23.1 and see that throughout we only use the fact we have a projective
resolution (repeatedly lifting maps as in (23.2.1.1)), as well as the fact that F sends
products to products (a consequence of additivity of the functor, see Remark 1.6.2)
to show that F applied to (23.1.2.1) preserves the exactness of the columns.
23.2.C. E XERCISE . The notion of an injective object in an abelian category is dual
to the notion of a projective object.
(a) State precisely the definition of an injective object.
(b) Define derived functors for (i) covariant left-exact functors (these are calledright derived functors), (ii) contravariant left-exact functors (also called right derived functors), and (iii) contravariant right-exact functors (these are called leftderived functors), making explicit the necessary assumptions of the category having enough injectives or projectives.
23.2.3. Notation. Recall from Definition 23.2.2 that if F is a right-exact functor, its(left) derived functors are denoted Li F (i ≥ 0, with L0 F = F). Similarly, if F is a leftexact functor, its (right-) derived functors are denoted Ri F. The i is a superscript,
to indicate that the long exact sequence is “ascending in i”.
23.2.4. The Ext functors.
23.2.D. E ASY E XERCISE ( AND DEFINITION ): Ext FUNCTORS FOR A- MODULES , FIRSTVERSION .
As Hom(·, N) is a contravariant left-exact functor in ModA , which
has enough projectives, define ExtiA (M, N) as the ith right derived functor of
Hom(·, N), applied to M. State the corresponding long exact sequence for Extmodules.
23.2.E. E ASY E XERCISE ( AND DEFINITION ): Ext FUNCTORS FOR A- MODULES , SEC OND VERSION . The category ModA has enough injectives (see §23.2.5). As Hom(M, ·)is a covariant left-exact functor in ModA , define ExtiA (M, N) as the ith right derived
functor of Hom(M, ·), applied to N. State the corresponding long exact sequencefor Ext-modules.
We seem to have a problem with the previous two exercises: we have defined
Ext (M, N) twice, and we have two different long exact sequences! Fortunately,
these two definitions of Ext agree (see Exercise 23.3.B), and two long exact sequences for Ext are better than one.
i

23.2.F. E ASY E XERCISE . Use the definition of Ext in Exercise 23.2.D to show that
if A is a Noetherian ring, and M and N are finitely generated A-modules, then
ExtiA (M, N) is a finitely generated A-module.
Ext-functors (for sheaves) will play a key role in the proof of Serre duality, see§30.2.
23.2.5. ⋆ The category of A-modules has enough injectives. We will need thefact that ModA has enough injectives, but the details of the proof won’t come up
again, so feel free to skip this discussion.

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23.2.G. E XERCISE . Suppose Q is an A-module, such that for every ideal I ⊂ A,
every homomorphism I → Q extends to A → Q. Show that Q is an injective Amodule. Hint: suppose N ⊂ M is an inclusion of A-modules, and we are given
β : N → Q. We wish to show that β extends to M → Q. Use the Axiom of Choice
to show that among those A-modules N ′ with N ⊂ N ′ ⊂ M, such that β extends
to N ′ , there is a maximal one. If this N ′ is not M, give an extension of β to N ′ +Am,
where m ∈ M \ N ′ , obtaining a contradiction.
23.2.H. E ASY E XERCISE ( USING THE A XIOM OF C HOICE , IN THE GUISE OF Z ORN ’ S
L EMMA ). Show that a Z-module (i.e., abelian group) Q is injective if and only if
it is divisible (i.e., for every q ∈ Q and n ∈ Z̸=0 , there is q ′ ∈ Q with nq ′ = q).
Hence show that any quotient of an injective Z-module is also injective.
23.2.I. E XERCISE . Show that the category of Z-modules ModZ = Ab has enoughinjectives. (Hint: if M is a Z-module, then write it as the quotient of a free Zmodule F by some K. Show that M is contained in the divisible group (F ⊗Z Q)/K.)
23.2.J. E XERCISE . Suppose Q is an injective Z-module, and A is a ring. Show that
HomZ (A, Q) is an injective A-module. Hint: First describe the A-module structureon HomZ (A, Q). You will only use the fact that Z is a ring, and that A is an algebra
over that ring.
23.2.K. E XERCISE . Show that ModA has enough injectives. Hint: suppose M is an
A-module. By Exercise 23.2.I, we can find an inclusion of Z-modules M "→ Q where
Q is an injective Z-module. Describe a sequence of inclusions of A-modules
M "→ HomZ (A, M) "→ HomZ (A, Q).
(The A-module structure on HomZ (A, M) is via the A-action on the left argument
A, not via the A-action on the right argument M.) The right term is injective by
the previous Exercise 23.2.J.
23.2.6. ⋆ Universal δ-functors.
(This discussion is best skipped on a first reading; you should move directlyto §23.3.) We now describe a more general variant of derived functors, as you may
use them in the discussion of Serre duality in Chapter 30. The advantage of thenotion of universal δ-functor is that we can apply it even in cases where A does
not have enough injectives.
Abstracting key properties of derived functors, we define the data of a (cohomological) δ-functor, from an abelian category A to another abelian category B.
A δ-functor is a collection of additive functors T i : A → B (where T i is taken to
be 0 if i < 0), along with morphisms δi : T i (A ′′ ) → T i+1 (A ′ ) for each short exact
sequence
0 → A ′ → A → A ′′ → 0(23.2.6.1)

in A , satisfying two properties:(i) (short exact sequences yield long exact sequences) For each short exact sequence (23.2.6.1), the sequence
···

i−1

% T i−1 (A ′′ ) δ% T i (A ′ )% T i (A)

% T i (A ′′ )δi

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The Rising Sea: Foundations of Algebraic Geometry(where the unlabeled maps come from the covariance of the T i ) is exact.
In particular, T 0 is left-exact.
(ii) (functoriality of (i)) For each morphism of short exact sequences in A
0

% a′

%a

% a ′′

%0

0

"
% A′

"
%A

"
% A ′′%0(where the squares commute), the δi ’s give a commutative diagram
T i (a ′′ )"T i (A ′′ )δiδi

% T i+1 (a ′ )"% T i+1 (A ′ )(where the vertical arrows come from the covariance of T i and T i+1 ).
Derived functor cohomology is clearly an example of a δ-functor; Čech cohomology of sheaves on quasicompact separated schemes is another. (You can makethese statements precise if you wish.)23.2.L. E XERCISE . Figure out the right definition of morphism of δ-functors A →B. (It should then be clear that the δ-functors from A to B form a category.)

23.2.7. Definition. A (cohomological) δ-functor (T i , δi ) is universal if for any
other δ-functor (T ′i , δ ′i ), and any natural transformation of functors α : T 0 → T ′0 ,there is a unique morphism of δ-functors (T i , δi ) → (T ′i , δ ′i ) extending α. By
universal property nonsense (and Exercise 23.2.L), given any covariant left-exact
functor F : A → B, there is at most one universal δ-functor (T i , δi ) extending F∼ F). The key fact about universal δ-functors(i.e., with a natural isomorphism T 0 =is the following.
23.2.8. Theorem. — Suppose (T i , δi ) is a covariant δ-functor from A to B, and for all
A ∈ A , there exists a monomorphism A → J with T i J = 0 for all i > 0. Then (T i , δi ) isuniversal.

23.2.M. ⋆⋆ E XERCISE . Prove Theorem 23.2.8. Partial hint: motivated by Corollary 23.2.10 below, follow our discussion of derived functors. Better hint (because
this exercise is hard): follow the hints in [Weib, Exercise 2.4.5], or follow the proofof [Lan, Ch. XX, Thm. 7.1].
23.2.9. Remark. An additive functor F : A → B is said to be effaceable if forevery A ∈ A , there is a monomorphism A → J with F(J) = 0. The hypothesesof Theorem 23.2.8 can be weakened to require only that T i is effaceable for eachi > 0, and you are welcome to prove that instead. (Indeed, [Weib, Exercise 2.4.5],[Lan, Ch. XX, Thm. 7.1], and the original source [Gr1, II.2.2.1] deal with this case.)
We give the statement of Theorem 23.2.8 for simplicity, as we will only use thisversion.
23.2.10. Corollary. — If A has enough injectives, and F is a left-exact covariant functor
A → B, then the Ri F (with the δi that accompany them) form a universal δ-functor.

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Proof. Each element of A admits a monomorphism into an injective element; this isjust the definition of “enough injectives” (Exercise 23.2.C). Higher derived functorsof an injective elements I are always 0: just compute the higher derived functor bytaking the injective resolution of I “by itself”.
!

23.3 Derived functors and spectral sequences
A number of useful facts can be easily proved using spectral sequences. By
doing these exercises, you will lose any fear of spectral sequence arguments insimilar situations, as you will realize they are all the same.
Before you read this section, you should read §1.7 on spectral sequences.
23.3.1. Symmetry of Tor.
∼
23.3.A. E XERCISE ( SYMMETRY OF Tor). Show that there is an isomorphism TorAi (M, N) =A
Tori (N, M). (Hint: take a free resolution of M and a free resolution of N. Take
their “product” to somehow produce a double complex. Use both orientations of
the obvious spectral sequence and see what you get.)On a related note:
23.3.B. E XERCISE . Show that the two definitions of Exti (M, N) given in Exercises 23.2.D and 23.2.E agree.
23.3.2. Derived functors can be computed using acyclic resolutions. Suppose
F : A → B is a right-exact additive functor of abelian categories, and that A has
enough projectives. (In other words, the hypotheses ensure the existence of leftderived functors of F. Analogous facts will hold with the other types of derived
functors, Exercise 23.2.C(b).) We say that A ∈ A is F-acyclic (or just acyclic if the
F is clear from context) if Li F A = 0 for i > 0.
The following exercise is a good opportunity to learn a useful trick (Hint 23.3.3).
23.3.C. E XERCISE . Show that you can also compute the derived functors of an
objects B of A using acyclic resolutions (not just projective resolutions), i.e., bytaking a resolution

···

% A2

% A1

% A0

%B

%0

by F-acyclic objects Ai , truncating, applying F, and taking homology. Hence Tori (M, N)
can be computed with a flat resolution of M or N.

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23.3.3. Hint for Exercise 23.3.C (and a useful trick: building a “projective resolution of acomplex”). Show that you can construct a double complex(23.3.3.1)

···
···
···

..
.

..
.

..
.

..
.

"
% P2,1"% P1,1"% P0,1

"
% P1

%0

"
% P2,0"% P1,0"% P0,0

"
% P0

%0

"
% A2

"
% A1

"
% A0

"
%B

%0

"
0

"
0

"
0

"
0

where the rows and columns are exact and the P? ’s are projective. Do this byconstructing the P? ’s inductively from the bottom right. Remove the bottom row,and the right-most nonzero column, and then apply F, to obtain a new double
complex. Use a spectral sequence argument to show that (i) the double complex
has homology equal to Li F B, and (ii) the homology of the double complex agrees
with the construction given in the statement of the exercise. If this is too confusing,read more about the Cartan-Eilenberg resolution below.
23.3.4. The Grothendieck composition-of-functors spectral sequence.
Suppose A , B, and C are abelian categories, F : A → B and G : B → C are aleft-exact additive covariant functors, and A and B have enough injectives. Thusright derived functors of F, G, and G ◦ F exist. A reasonable question is: how are
they related?
23.3.5. Theorem (Grothendieck composition-of-functors spectral sequence). —Suppose F : A → B and G : B → C are left-exact additive covariant functors, and
A and B have enough injectives. Suppose further that F sends injective elements of A
to G-acyclic elements of B. Then for each X ∈ A , there is a spectral sequence with
p,q
= Rq G(Rp F(X)) converging to Rp+q (G ◦ F)(X).
→ E2

We will soon see the Leray spectral sequence as an application (Theorem 23.4.5).

There is more one might want to extract from the proof of Theorem 23.3.5. For
example, although E0 page of the spectral sequence will depend on some choices(of injective resolutions), the E2 page will be independent of choice. For our applications, we won’t need this refinement.
We will have to work to establish Theorem 23.3.5, so the proof is possibly best
skipped on a first reading.
23.3.6. ⋆ Proving Theorem 23.3.5.
Before we give the proof (in §23.3.8), we begin with some preliminaries to motivate it. In order to discuss derived functors applied to X, we choose an injective

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resolution of X:

0

%X

% I0

% I1

% ··· .

To compute the derived functors Rp F(X), we apply F to the injective resolution I• :% F(I0 )0% F(I1 )% F(I2 )% ··· .

Note that F(Ip ) is G-acyclic, by hypothesis of Theorem 23.3.5. If we were to follow
our nose, we might take simultaneous injective resolutions I•,• of the terms in theabove complex F(I• ) (the “dual” of Hint 23.3.3 — note that only the columns are
required to be exact), and apply G, and consider the resulting double complex:

..
.7

..
.7

..
.7

0

% G(I0,2 )7% G(I1,2 )7% G(I2,2 )
7

% ···

0

% G(I0,1 )7% G(I1,1 )7% G(I2,1 )
7

% ···

0

% G(I0,0 )7% G(I1,0 )7% G(I2,0 )
7

% ···

0

0

0

(23.3.6.1)

23.3.D. E XERCISE . Consider the spectral sequence with upward orientation, starting with (23.3.6.1) as page E0 . Show that Ep,q
is Rp (G ◦ F)(X) if q = 0, and 02otherwise.
We now see half of the terms in the conclusion of Theorem 23.3.5; we are
halfway there. To complete the proof, we would want to consider another spectral
sequence, with rightward orientation, but we need to know more about (23.3.6.1);
we will build it more carefully.
23.3.7. Cartan-Eilenberg resolutions.

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Suppose · · · → Cp−1 → Cp → Cp+1 → · · · is a complex in an abelian category
B. We will build an injective resolution of C•
..
.7

..
.7

..
.7

0

% I0,2
7% I1,27% I2,27% ···0% I0,17% I1,1
7% I2,17% ···0% I0,07% I1,07% I2,0
7

% ···

0

% C0
7

% C1
7

% C2
7

% ···

0

0

0

(23.3.7.1)

satisfying some further properties.
We first define some notation for functions on a complex.
• Let Zp (K• ) be the kernel of the pth differential of a complex K• .
• Let Bp+1 (K• ) be the image of the pth differential of a complex K• . (Thesuperscript is chosen so that Bp+1 (K• ) ⊂ Kp+1 .)• As usual, let Hp (K• ) be the homology at the pth step of a complex K• .
For each p, we have complexes(23.3.7.2)

..
.7

..
.7

..
.7

Zp (I•,1 )
7

Bp (I•,1 )
7

Hp (I•,1 )
7

Zp (I•,0 )7

Bp (I•,0 )7

Hp (I•,0 )7

Zp (C• )7

Bp (C• )7

Hp (C• )
7

0

0

0

We will construct (23.3.7.1) so that the three complexes (23.3.7.2) are all injective
resolutions (of their first nonzero terms). We begin by choosing injective resolutions Bp,∗ of Bp (C• ) and Hp,∗ of Hp (C• ); these will eventually be the last two linesof (23.3.7.2).

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23.3.E. E XERCISE . Describe an injective resolution Zp,∗ of Zp (C• ) (the first line of(23.3.7.2)) making the following diagram a short exact sequence of complexes.
..
.7

..
.7

..
.7

0

% Bp,17% Zp,17% Hp,17%0

0% Bp,07% Zp,07% Hp,07%00% Bp (C• )7% Zp (C• )7% Hp (C• )
7

%0

0

0

0

(23.3.7.3)

Hint: the “dual” problem was solved in (23.1.2.1), by a “horseshoe construction”.
23.3.F. E XERCISE . Describe an injective resolution Ip,∗ of Cp making the following
diagram a short exact sequence of complexes.
..
.7

..
.7

..
.7

0

% Zp,1
7% Ip,17% Bp+1,1
7%0

0% Zp,07% Ip,07% Bp+1,07%00% Zp (C• )
7

% Cp
7

% Bp+1 (C• )
7

%0

0

0

0

(23.3.7.4)(The hint for the previous problem applies again. We remark that the first nonzero
columns of (23.3.7.3) and (23.3.7.4) appeared in (1.6.5.3).)
23.3.G. E XERCISE /D EFINITION . Build an injective resolution (23.3.7.1) of C• suchthat Zp,∗ = Zp (I•,∗ ), Bp,∗ = Bp (I•,∗ ), Hp,∗ = Hp (I•,∗ ), so the three complexes(23.3.7.2) are injective resolutions. This is called a Cartan-Eilenberg resolution ofC• .
23.3.8. Proof of the Grothendieck spectral sequence, Theorem 23.3.5. We pick up where
we left off before our digression on Cartan-Eilenberg resolutions. Choose an injective resolution I• of X. Apply the functor F, then take a Cartan-Eilenberg resolution
I•,• of FI• , and then apply G, to obtain (23.3.6.1).

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Exercise 23.3.D describes what happens when we take (23.3.6.1) as E0 in a
spectral sequence with upward orientation. So we now consider the rightward
orientation.
From our construction of the Cartan-Eilenberg resolution, we have injective
resolutions (23.3.7.2), and short exact sequences(23.3.8.1)

0% Bp (I•,q )% Zp (I•,q )(23.3.8.2)

0% Zp (I•,q )% Ip,q% Hp (I•,q )%0% Bp+1 (I•,q )%0of injective objects (from the columns of (23.3.7.3) and (23.3.7.4)). This means thatboth are split exact sequences (the central term can be expressed as a direct sum of
the outer two terms), so upon application of G, both exact sequences remain exact.
Applying the left-exact functor G to0

% Zp (I•,q )% Ip,q% Ip+1,q ,we find that GZp (I•,q ) = ker(GIp,q → GIp+1,q ). But this kernel is the definitionof Zp (GI•,q ), so we have an induced isomorphism GZp (I•,q ) = Zp (GI•,q ) (“Gand Zp commute”). From the exactness of (23.3.8.2) upon application of G, we seethat GBp+1 (I•,q ) = Bp+1 (GI•,q ) (both are coker(GZp (I•,q ) → GIp,q )). From theexactness of (23.3.8.1) upon application of G, we see that GHp (I•,q ) = Hp (GI•,q )(both are coker(GBp (I•,q ) → GZp (I•,q )) — so “G and Hp commute”).
We return to considering the rightward-oriented spectral sequence with (23.3.6.1)
as E0 . Taking cohomology in the rightward direction, we find Ep,q
= Hp (GI•,q ) =
1
p •,q
p
p •,q
GH (I ) (as G and H commute). Now H (I ) is an injective resolution of(Rp F)(X) (the last resolution of (23.3.7.2)). Thus when we compute E2 by using thevertical arrows, we find Ep,q
= Rq G(Rp F(X)).
2
You should now verify yourself that this (combined with Exercise 23.3.D) concludes the proof of Theorem 23.3.5.
!

23.4 Derived functor cohomology of O-modules
We wish to apply the machinery of derived functors to define cohomology ofquasicoherent sheaves on a scheme X. Rather than working in the category QCohX ,for a number of reasons it is simpler to work in the larger category ModOX (but see
Unimportant Remark 23.5.7).
23.4.1. Theorem. — Suppose (X, OX ) is a ringed space. Then the category of OX modules ModOX has enough injectives.
As a side benefit (of use to others more than us), taking OX = Z, we see thatthe category of sheaves of abelian groups on a fixed topological space have enoughinjectives.

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23.4.2. Proof. We prove Theorem 23.4.1 in a series of exercises, following Godement, [GrS, p. 27-28]. Suppose F is an OX -module. We will exhibit an injection F "→ Q ′ into an injective OX -module. For each p ∈ X, choose an inclusion Fp "→ Qp into an injective OX,p -module (possible as the category of OX,p modules has enough injectives, Exercise 23.2.K).
23.4.A. E XERCISE . Show that the skyscraper sheaf Qp := ip,∗ Qp , with module
Qp at point p ∈ X, is an injective OX -module. (You can cleverly do this by abstract
nonsense, using Exercise 23.5.B, but it is just as quick to do it by hand.)23.4.B. E ASY E XERCISE . Show that the direct product (possibly infinite) of injective objects in an abelian category is also injective.
"
By the previous two exercises, Q ′ := p∈X Qp is an injective OX -module.
23.4.C. E ASY E XERCISE . By considering stalks, show that the natural map F →Q ′ is an injection.
This completes the proof of Theorem 23.4.1.

!

We can now make a number of definitions.
23.4.3. Definition. If (X, OX ) is a ringed space, and F is an OX -module, define
Hi (X, F ) as Ri Γ (X, F ). If furthermore π : (X, OX ) → (Y, OY ) is a map of ringed
spaces, we have higher pushforwards (or derived pushforwards) Ri π∗ : ModOX →ModOY .
We have defined these notions earlier in special cases, for quasicoherent sheaves
on quasicompact separated schemes (for Hi ), or for quasicompact separated morphisms of schemes (for Ri π∗ ), in Chapter 18. We will soon (§23.5) show that theseolder definitions agree with Definition 23.4.3. Thus the derived functor definition
applies much more generally than our Čech definition. But it is worthwhile to notethat almost everything we use will come out of the Čech definition. A notable exception is the Leray spectral sequence, which we now discuss.
23.4.4. The Leray spectral sequence.
23.4.5. Theorem (Leray spectral sequence). — Suppose π : (X, OX ) → (Y, OY ) is amorphism of ringed spaces. Show that for any OX -module F , there is a spectral sequencewith E2 term given by Hq (Y, Rp π∗ F ) abutting to Hp+q (X, F ).
This is an immediate consequence of the Grothendieck composition-of-functors
spectral sequence (Theorem 23.3.5) once we prove that the pushforward of an injective O-module is an acyclic O-module. We do this now.
23.4.6. Definition. We make an intermediate definition that is independently important. A sheaf F on a topological space is flasque (also sometimes called flabby)
if all restriction maps are surjective, i.e., if resU⊂V : F (V) → F (U) is surjective for
all U ⊂ V.

23.4.D. E XERCISE . Suppose (X, OX ) is a ringed space.
(a) Show that if(23.4.6.1)

0 → F ′ → F → F ′′ → 0

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is an exact sequence of OX -modules, and F ′ is flasque, then (23.4.6.1) is exact on
sections over any open set U, i.e. 0 → F ′ (U) → F (U) → F ′′ (U) → 0 is exact.
(b) Given an exact sequence (23.4.6.1), if F ′ is flasque, show that F is flasque if
and only if F ′′ is flasque.
23.4.E. E ASY E XERCISE ( PUSHFORWARD OF FLASQUES ARE FLASQUE ).
(a) Suppose π : X → Y is a continuous map of topological spaces, and F is a flasquesheaf of sets on X. Show that π∗ F is a flasque sheaf on Y.
(b) Suppose π : (X, OX ) → (Y, OY ) is a morphism of ringed spaces, and F is a
flasque OX -module. Show that π∗ F is a flasque OY -module.

23.4.F. E XERCISE ( INJECTIVE SHEAVES ARE FLASQUE ). Suppose (X, OX ) is a ringed
space, and Q is an injective OX -module. Show that Q is flasque. Hint: If U ⊂V ⊂ X, then describe an injection of OX -modules 0 → (iV )! OV → (iU )! OU , whereiU : U "→ X and iV : V "→ X are the obvious open embeddings. Apply the exact
contravariant functor Hom(·, Q). (The morphisms (iV )! and (iU )! are extensions
by zero, see Exercise 2.7.G. Now might be a good time to do that Exercise.)

23.4.G. E XERCISE ( FLASQUE IMPLIES Γ - ACYCLIC ). Suppose F is a flasque OX module. Show that F is Γ -acyclic (that Hi (X, F ) = 0 for i > 0, §23.3.2) as follows.
As ModOX has enough injectives, choose an inclusion of F into some injective I ,and call its cokernel G :0 → F → I → G → 0.

Then I is flasque by Exercise 23.4.F, so G is flasque by Exercise 23.4.D(b). Take thelong exact sequence in (derived functor) cohomology, and show that H1 (X, F ) = 0.
Your argument works for any flasque sheaf F , so H1 (X, G ) = 0 as well. Show that
H2 (X, F ) = 0. Turn this into an induction.
Thus if π : X → Y is a morphism of ringed spaces, and I is an injective OX module, then I is flasque (Exercise 23.4.F), so π∗ I is flasque (Exercise 23.4.E(b)),
so π∗ I is acyclic for the functor Γ (Exercise 23.4.G), so this completes the proof of
the Leray spectral sequence (Theorem 23.4.5).
!

23.4.H. E XERCISE . Extend the Leray spectral sequence (Theorem 23.4.5) to deal
with a composition of higher pushforwards for(X, OX )π% (Y, OY )ρ

% (Z, OZ ).

23.4.7. ⋆⋆ The category of OX -modules need not have enough projectives. In
contrast to Theorem 23.4.1, the category of OX -modules need not have enoughprojectives. For example, let X be P1k with the Zariski-topology (in fact we willneed very little about X — only that it is not an Alexandrov space), but take OX to
be the constant sheaf Z. We will see that ModOX — i.e., the category of sheaves ofabelian groups on X — does not have enough projectives. If ModOX had enoughprojectives, then there would be a surjection ψ : P → Z from a projective sheaf. Fix
a closed point q ∈ X. We will show that the map on stalks ψq : Pq → Zq is the zeromap, contradicting the surjectivity of ψ. For each open subset U of X, denote by ZU
the extension to X of the constant sheaf associated to Z on U by 0 (Exercise 2.7.G— ZU (V) = Z if V ⊂ U, and ZU (V) = 0 otherwise). For each open neighborhood

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V of q, let W be a strictly smaller open neighborhood. Consider the surjection
ZX−q ⊕ ZW → Z. By projectivity of P, the surjection ψ lifts to P → ZX−q ⊕ ZW .
The map P(V) → Z(V) factors through ZX−q (V) ⊕ ZW (V) = 0, and hence must be
the zero map. Thus the map ψq : Pq → Zq map is zero as well (do you see why?)
as desired.

23.5 Čech cohomology and derived functor cohomology agree
We next prove that Čech cohomology and derived functor cohomology agree,where the former is defined.
23.5.1. Theorem. — Suppose X is a quasicompact separated scheme, and F is a quasicoherent sheaf. Then the Čech cohomology of F agrees with the derived functor cohomology
of F .
This statement is not as precise as it should be. We would want to know that
this isomorphism is functorial in F , and that it respects long exact sequences (sothe connecting homomorphism defined for Čech cohomology agrees with that forderived functor cohomology). There is also an important extension to higher pushforwards. We leave these issues for the end of this section, §23.5.5
In case you are curious: so long as it is defined appropriately, Čech cohomology agrees with derived functor cohomology in a wide variety of circumstances
outside of scheme theory (if the underlying topological space is paracompact), but
not always (see [Gr1, §3.8] for a counterexample). Remark 23.5.6 is also related.
The central idea in the proof (albeit with a twist) is a spectral sequence argument in the same style as those of §23.3, and uses two “cohomology-vanishing”
ingredients, one for each orientation of the spectral sequence.
(A) If (X, OX ) is a ringed space, Q is an injective OX -module, and X = ∪i Ui isa finite open cover, then Q has no ith Čech cohomology with respect to this coverfor i > 0.
(B) If X is an affine scheme, and F is a quasicoherent sheaf on X, then (Ri Γ )(F ) =0 for i > 0.
Translation: (A) says that building blocks of derived functor cohomology haveno Čech cohomology, and (B) says that building blocks of Čech cohomology haveno derived functor cohomology.
23.5.A. P RELIMINARY E XERCISE .
Suppose (X, OX ) is a ringed space, Q is aninjective O-module, and i : U "→ X is an open subset. Show that Q|U is injective
on U. Hint: use the fact that i−1 has an exact left adjoint i! (extension by zero), see
Exercise 2.7.G, and the following diagrams.
0

%A
" b
Q|U

%B
?

0

% i! A" C
Q% i! B?

In the course of Exercise 23.5.A, you will have proved the following fact, which
we shall use again in Exercise 30.3.C. (You can also use it to solve Exercise 23.4.A.)

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23.5.B. E XERCISE ( RIGHT ADJOINTS TO EXACT FUNCTORS PRESERVE INJECTIVES ).
Show that if (F, G) is an adjoint pair of additive functors between abelian categories, and F is exact, then G sends injective elements to injective elements.
23.5.2. Proof of Theorem 23.5.1, assuming (A) and (B). As with the facts proved in§23.3, we take the only approach that is reasonable: we choose an injective resolution 0 → F → Q• of F and a Čech cover of X, mix these two types of information
in a double complex, and toss it into our spectral sequence machine (§1.7). Moreprecisely, choose a finite affine open cover X = ∪i Ui and an injective resolution
0 → F → Q0 → Q1 → · · · .

Consider the double complex(23.5.2.1)

..
.7

..
.7

..
.7

..
.7

07

% ⊕i Q2 (Ui )7% ⊕i,j Q2 (Uij )7% ⊕i,j,k Q2 (Uijk )
7

% ···

07

% ⊕i Q1 (Ui )7% ⊕i,j Q1 (Uij )7% ⊕i,j,k Q1 (Uijk )
7

% ···

07

% ⊕i Q0 (Ui )7% ⊕i,j Q0 (Uij )7% ⊕i,j,k Q0 (Uijk )
7

% ···

0

%0

%0

%0

% ···

We take this as the E0 term in a spectral sequence. First, we use the rightward
filtration. As higher Čech cohomology of injective O-modules is 0 (assumption(A)), we get 0’s everywhere except in “column 0”, where we get Qi (X) in row i:
..
.7

..
.7

..
.7

..
.7

07

Q2 (X)
7

07

07

···

07

Q1 (X)
7

07

07

···

07

Q0 (X)
7

07

07

···

0

0

0

0

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639

Then we take cohomology in the vertical direction, and we get derived functor
cohomology of F on X on the E2 page:
..
.

..
.

..
.

..
.

0 =✺
R2 Γ (X, F )
0K✲
0
=✺✺
✺✺
✲✲
✺✺
✺✺
✲✲
✺✺
✺✺
✲✲
✺✺
✺✺
✲
✺
1
✺
0 =✺
R Γ (X, F ) ✺✺0K✲ ✲✲ 0
✺
✺
=
✺
✺✺
✺✺
✲✲ ✲✲
✺✺
✺✺
✺✺
✺✺
✲
✲
✺
✺
✺
✺
✺✺
✺✺ ✲✲ ✲✲
✺✺
✺
✲
✺✺
✲✲
✺✺ Γ (X, F ) ✺✺✺
0
✲✲ 0
✺✺ 0
✺✺
✲✲
✺✺
✺✺
✲✲
✺✺
✺✺
0
0
0

···
···
···

We then start over on the E0 page, and this time use the filtration corresponding
to choosing the upward arrow first. By Proposition 23.5.A, Qi |UJ is injective on
UJ , so we are computing the derived functor cohomology of F on UJ . Then thehigher derived functor cohomology of F on UJ is 0 (assumption (B)), so all entries
are 0 except possibly on row 0. Thus the E1 term is:(23.5.2.2)
%0
%0
%0
% ···
0
0

%0

%0

%0

% ···

0

% ⊕i Γ (Ui , F )% ⊕i,j Γ (Uij , F )% ⊕i,j,k Γ (Uijk , F )

% ···

0

%0

%0

%0

% ···

Row 0 is precisely the Čech complex of F , so the spectral sequence converges at
the E2 term, yielding the Čech cohomology. Since one orientation yields derived
functor cohomology and one yields Čech cohomology, we are done.
!
So it remains to show (A) and (B).
23.5.3. Ingredient (A): injectives have no Čech cohomology.
23.5.C. E XERCISE . Suppose X = ∪j Uj is a finite cover of X by open sets, andF is a flasque sheaf (Definition 23.4.6) on X. Show that the Čech complex for F
with respect to ∪n
j=1 Uj has no cohomology in positive degree, i.e., that it is exact
except in degree 0 (where it has cohomology F (X), by the sheaf axioms). Hint:
use induction on n. Consider the short exact sequence of complexes (18.2.4.2) (seeEarly (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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The Rising Sea: Foundations of Algebraic Geometry

also (18.2.3.1)). The corresponding long exact sequence will immediately give thedesired result for i > 1, and flasqueness will be used for i = 1.
Thus flasque sheaves have no Čech cohomology, so injective O-modules inparticular (Exercise 23.4.F) have none. This is all we need for our algebro-geometricapplications, but to show you how general this machinery is, we give an entertaining application.
23.5.D. U NIMPORTANT E XERCISE ( PERVERSE PROOF OF INCLUSION - EXCLUSION
The inclusion-exclusion principle is(equivalent to) the following: suppose that X is a finite set, and Ui (1 ≤ i ≤ n)
are finite sets covering X. As usual, define UI = ∩i∈I Ui for I ⊂ {1, . . . , n}. Then
.
.
.
.
|X| =
|Ui | −
|U|I| | +|U|I| | −|U|I| | + · · · .
THROUGH COHOMOLOGY OF SHEAVES ).

|I|=2

|I|=3

|I|=4

Prove this by endowing X with the discrete topology, showing that the constantsheaf Q is flasque, considering the Čech complex computing Hi (X, Q) using thecover {Ui }, and using Exercise 1.6.B.

23.5.4. ⋆ Ingredient (B): quasicoherent sheaves on affine schemes have no derived functor cohomology.
We show the following statement by induction on k. Suppose X is an affine
scheme, and F is a quasicoherent sheaf on X. Then Ri Γ (X, F ) = 0 for 0 < i ≤ k.
The result is vacuously true for k = 0; so suppose we know the result for all
0 < k ′ < k (with X replaced by any affine scheme). Suppose α ∈ Rk Γ (X, F ). We
wish to show that α = 0. Choose an injective resolution by OX -modules
..
.7
d1

Q7 1
d0

Q7 0
F7
0
Then α has a representative α in Qk (X) = Γ (X, Qk ), such that dα ′ = 0. Because
the injective resolution is exact, α ′ is locally a boundary. In other words, in the
open neighborhood of any point p ∈ X, there is an open set Vp such that α|Vp =dα ′ for some α ′ ∈ Qk−1 (Vp ). (Be sure you see why this is true! Recall that taking
cokernel of a map of sheaves requires sheafification, see Proposition 2.6.1.) By
shrinking Vp if necessary, we can assume Vp is affine. By the quasicompactnessof X, we can choose a finite number of the Vp ’s that cover X. Rename these as Ui ,so we have an affine cover X. Consider the Čech cover of X with respect to this′

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affine cover (not the affine cover you might have thought we would use — that of
X by itself — but instead an affine cover tailored for our particular α). Consider
the double complex (23.5.2.1), as the E0 term in a spectral sequence.
First consider the rightward orientation. As in the argument in §23.5.2, thespectral sequence converges at E2 , where we get 0 everywhere, except that thederived functor cohomology appears in the 0th column.
Next, start over again, choosing the upward filtration. On the E1 page, row
0 is the Čech complex, as in (23.5.2.2). All the rows between 1 and k − 1 are 0
by our inductive hypothesis, but we don’t yet know anything about the higher
rows. Because we are interested in the kth derived functor, we focus on the kth
antidiagonal (Ep,k−p). The only possibly nonzero terms in this antidiagonal
are
•
"
0,k
k,0
Ek,0
and
E
.
We
look
first
at
the
term
on
the
bottom
row
E
=
Γ
(U
,
F
),
I
1
1
1
|I|=k
which is part of the Čech complex:
&
&
&
··· →
Γ (UI , F ) →Γ (UI , F ) →Γ (UI , F ) → · · · .
|I|=k−1|I|=k|I|=k+1

But we have already verified that the Čech cohomology of a quasicoherent sheaf
on an affine scheme vanishes — this is the one spot where we use the quasicoherence of F . Thus this term vanishes by the E2 page (i.e., Ek,0
= 0 for i ≥ 2).
i
So the only term of interest in the kth antidiagonal of E1 is E0,k
1 , which is thehomology of&&&(23.5.4.1)
Qk−1 (Ui ) →Qk (Ui ) →Qk+1 (Ui ),
"

i

i

i

which is i R Γ (Ui , F ) (using Preliminary Exercise 23.5.A which stated that the
Qj |Ui are injective on Ui , so they can be used to compute Rk (Γ (Ui , F )). So E0,k
2 is
the homology of&&0→
Rk Γ (Ui , F ) →Rk Γ (Uij , F )
k

i

i,j

and thereafter all differentials to and from the E0,kterms will be 0, as the sources•
and targets of those arrows
will
be
0.
Considernow
our lift of α ′ of our original" kkclass α ∈ R Γ (X, F ) to i R Γ (Ui , F ). Its image in the homology of (23.5.4.1) iszero — this was how we chose our cover Ui to begin with! Thus α = 0 as desired,completing our proof.
!
23.5.E. ⋆⋆ E XERCISE .
The proof is not quite complete. We have a class α ∈Rk Γ (X, F ), and we have interpreted Rk Γ (X, F ) as
⎛
⎞
&
&
ker ⎝
Rk Γ (Ui , F ) →Rk Γ (Uij , F )⎠ .
ii,j

We have two maps Rk Γ (X, F ) → Rk Γ (Ui , F ), one coming from the natural restriction (under which we can see that the image of α is zero), and one coming from theactual spectral sequence machinery. Verify that they are the same map. (Possible
hint: with the filtration used, the E0,k
∞ term is indeed the quotient of the homologyof the double complex, so the map goes the right way.)
23.5.5. ⋆ Tying up loose ends.

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23.5.F. I MPORTANT E XERCISE . State and prove the generalization of Theorem 23.5.1
to higher pushforwards Ri π∗ , where π : X → Y is a quasicompact separated morphism of schemes.
23.5.G. E XERCISE . Show that the isomorphism of Theorem 23.5.1 is functorial in
F , i.e., given a morphism F → G , the diagram
Hi (X, F ) (% Ri Γ (X, F )"Hi (X, G ) ("% Ri Γ (X, G )commutes, where the horizontal arrows are the isomorphisms of Theorem 23.5.1,
and the vertical arrows come from functoriality of Hi and Ri Γ . (Hint: “spectral
sequences are functorial in E0 ”, which can be easily seen from the construction,although we haven’t said it explicitly. See Remark 1.7.8, on the functoriality ofspectral sequences.)
23.5.H. E XERCISE . Show that the isomorphisms of Theorem 23.5.1 induce isomorphisms of long exact sequences.
23.5.6. Remark. If you wish, you can use the above argument to prove the following theorem of Leray. Suppose we have a sheaf of abelian groups F on a
topological space X, and some covering {Ui } of X such that the (derived functor)
cohomology of F in positive degree vanishes on every finite intersection of the Ui .
Then the cohomology of F can be calculated by the Čech cohomology of the cover{Ui }; there is no need to pass to the inductive limit of all covers, as is the case forČech cohomology in general.
23.5.7. Unimportant Remark: Working in QCohX rather than ModOX . In our definitionof derived functors of quasicoherent sheaves on X, we could have tried to work in
the category of quasicoherent sheaves QCohX itself, rather than in the larger category ModOX . There are several reasons why this would require more effort. It
is not hard to show that QCohX has enough injectives if X is Noetherian (see forexample [Ha1, Exer. III.3.6(a)]). Because we don’t have “extension by zero” (Exercise 2.7.G) in QCoh, the proofs that injective quasicoherent sheaves on an open set
U restrict to injective quasicoherent sheaves on smaller open subsets V (the analog
of Exercise 23.5.A) and that injective quasicoherent sheaves are flasque (the analog
of Exercise 23.4.F) are harder. You can use this to show that Hi and Ri π∗ computed
in QCoh are the same as those computed in ModO (once you make the statements
precise). It is true that injective elements of QCohX (X Noetherian) are injective in
ModOX , but this requires work (see [Mur, Prop. 68]).
It is true that QCohX has enough injectives for any scheme X, but this is muchharder, see [EE]. And as is clear from the previous paragraph, “enough injectives”
is only the beginning of what we want.
23.5.8. Unimportant Remark. Theorem 23.5.1 implies that if π : X → Y is quasicompact and separated, then Ri π∗ sends QCohX to QCohY (by showing an isomorphismwith Čech cohomology). If X and Y are Noetherian, the hypothesis “separated” canbe relaxed to “quasiseparated”, using the ideas of Unimportant Remark 23.5.7. But

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because morphisms from a Noetherian scheme are automatically quasiseparated(Easy Exercise 7.3.B(b)), we can then remove the redundant “quasiseparated” assumption. This sounds exciting, but it is not nearly as useful as the separated case,because without Čech cohomology, it is hard to compute anything.

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CHAPTER 24

Flatness
The concept of flatness is a riddle that comes out of algebra, but which technically is
the answer to many prayers.
— D. Mumford [Mu7, III.10]
It is a riddle, wrapped in a mystery, inside an enigma; but perhaps there is a key.
— W. Churchill

24.1 Introduction
We come next to the important concept of flatness (first introduced in §16.3.8).
We could have discussed flatness at length as soon as we had discussed quasicoherent sheaves and morphisms. But it is an unexpected idea, and the algebra andgeometry are not obviously connected, so we have left it for relatively late. “Exact” seems more descriptive terminology than “flat” (so we would have “exact
modules” and “exact morphisms”), but this has not caught on.
Serre has stated that he introduced flatness purely for reasons of algebra in his
landmark “GAGA” paper [Se3], and that it was Grothendieck who recognized its
geometric significance.
A flat morphism π : X → Y is the right notion of a “nice”, or “nicely varying”
family over Y. For example, if π is a projective flat family over a connected Noetherian base (translation: π : X → Y is a projective flat morphism, with Y connected
and Noetherian), we will see that various numerical invariants of fibers are constant, including the dimension (§24.5.5), and numbers interpretable in terms of an
Euler characteristic (see §24.7):(a) the Hilbert polynomial (Corollary 24.7.2),(b) the degree (in projective space) (Exercise 24.7.B(a)),(b’) (as a special case of (b)) if π is finite, the degree of π (recovering andextending the fact that the degree of a projective map between regularcurves is constant, §17.4.4, see Exercise 24.4.G and §24.4.12),(c) the arithmetic genus (Exercise 24.7.B(b)),(d) the degree of a line bundle if the fiber is a curve (Corollary 24.7.3), and(e) intersections of divisors and line bundles (Exercise 24.7.D).
One might think that the right hypothesis might be smoothness, or more generally some sort of equisingularity, but we only need something weaker. And thisis a good thing: branched covers are not fibrations in any traditional sense, yetthey still behave well — the double cover A1 → A1 given by y (→ x2 has constantdegree 2 (§9.3.3, revisited in §17.4.9). Another key example is that of a family ofsmooth curves degenerating to a nodal curve (Figure 24.1) — the topology of the
645

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The Rising Sea: Foundations of Algebraic Geometry(underlying analytic) curve changes, but the arithmetic genus remains constant.
One can prove things about regular curves by first proving them about a nodal
degeneration, and then showing that the result behaves well in flat families. Degeneration techniques such as this are ubiquitous in algebraic geometry.

F IGURE 24.1. A flat family of smooth curves degenerating to a
nodal curve: y2 = x3 + x2 + tx.
Given the cohomological nature of the constancy of Euler characteristic result,
you should not be surprised that the hypothesis needed (flatness) is cohomologicalin nature — it can be characterized by vanishing of Tor (Exercise 23.1.D), which we
use to great effect in §24.3.
But flatness is important for other reasons too. As a start: as this is the rightnotion of a “nice family”, it allows us to correctly define the notion of moduli space.
For example, the Hilbert scheme of Pn “parametrizes closed subschemes of Pn ”.
Maps from a scheme B to the Hilbert scheme correspond to (finitely presented)
closed subschemes of Pn
B flat over B. By universal property nonsense, this defines
the Hilbert scheme up to unique isomorphism (although we of course must showthat it exists, which takes some effort — see Theorem 28.3.1 for more). The moduli
space of smooth projective curves is defined by the universal property that maps tothe moduli space correspond to projective flat (finitely presented) families whosegeometric fibers are smooth curves. (Sadly, this moduli space does not exist...) On
a related note, flatness is central in deformation theory: it is key to understanding
how schemes (and other geometric objects, such as vector bundles) can deform(cf. §21.5.12). Finally, the notion of Galois descent generalizes to (faithfully) “flatdescent”, which allows us to “glue” in more exotic Grothendieck topologies in thesame way we do in the Zariski topology (or more classical topologies); but this isbeyond the scope of our current discussion.
Happy families are all alike; every unhappy family is unhappy in its own way. (See,for example, Figure 24.2, and Exercises 24.2.G(d), 24.2.K, and 24.4.H.)— L. Tolstoy, Anna Karenina, [To, l. 1]
24.1.1. Structure of the chapter.
Flatness has many aspects of different flavors, and it is easy to lose sight of
the forest for the trees. Because the algebra of flatness seems so unrelated to thegeometry, it can be nonintuitive. We will necessarily begin with algebraic foundations, but you should focus on the following points: methods of showing things
are flat (both general criteria and explicit examples), and classification of flat modules over particular kinds of rings. You should try every exercise dealing with
explicit examples such as these.

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Here is an outline of the chapter, to help focus your attention.
• In §24.2, we discuss some of the easier facts, which are algebraic in nature.
• §24.3, §24.4, and §24.6 give ideal-theoretic criteria for flatness. §24.3 and§24.4 should be read together. The first uses Tor to understand flatness,and the second uses these insights to develop ideal-theoretic criteria forflatness. §24.6, on local criteria for flatness, is harder.
• §24.5 is relatively free-standing, and could be read immediately after §24.2.
It deals with topological aspects of flatness, such as the fact that flat morphisms are open in good situations.
• In §24.7, we explain fact that “the Euler characteristic of quasicoherent
sheaves is constant in flat families” (with appropriate hypotheses), andits many happy consequences. This section is surprisingly easy given its
utility.
You should focus on what flatness implies and how to “picture” it, but also on
explicit criteria for flatness in different situations, such as for integral domains(Observation 24.2.2), principal ideal domains (Exercise 24.4.B), discrete valuation
rings (Exercise 24.4.C), the dual numbers (Exercise 24.4.D), and local rings (Theorem 24.4.5).

24.2 Easier facts
Many facts about flatness are easy or immediate, although a number are tricky.
As always, I will try to make clear which is which, to help you remember the easyfacts and the key ideas of proofs of the harder facts. We will pick the low-hanging
fruit first.
We recall the definition of a flat A-module (§1.6.11). If M ∈ ModA , M ⊗A · isalways right-exact (Exercise 1.3.H). We say that M is a flat A-module (or flat over
A or A-flat) if M ⊗A · is an exact functor. We say that a ring morphism B → A is flat
if A is flat as a B-module. (In particular, the algebra structure of A is irrelevant.)
24.2.1. Two key examples.
(i) Free A-modules (even of infinite rank) are clearly flat. More generally, projective modules are flat (Exercise 23.2.B).
(ii) Localizations are flat: Suppose S is a multiplicative subset of B. Then B →S−1 B is a flat ring morphism because localization is exact (Exercise 1.6.F(a)).

24.2.A. E ASY E XERCISE : FIRST EXAMPLES .
(a) (trick question) Classify flat modules over a field k.
(b) Show that A[x1 , . . . , xn ] is a flat A-module.
(c) Show that the ring morphism Q[x] → Q[y], with x (→ y2 , is flat. (This will help
us understand Example 9.3.3 better, see §24.4.12.)
We make some quick but important observations.

24.2.2. Important Observation.
module, thenIf x is a non-zerodivisor of A, and M is a flat AM

×x%M

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The Rising Sea: Foundations of Algebraic Geometry%A
is injective. (Reason: apply the exact functor M⊗A to the exact sequence 0
In particular, flat modules are torsion-free. (Torsion-freeness was defined in §13.5.5.)
This observation gives an easy way of recognizing when a module is not flat. We
will use it many times.

×x% A .)

24.2.B. E XERCISE : ANOTHER EXAMPLE . Show that a finitely generated module
over a discrete valuation ring is flat if and only if it is torsion-free if and only if it isfree. Hint: Remark 12.5.14 classifies finitely generated modules over a discrete valuation ring. (Exercise 24.4.B sheds more light on flatness over a discrete valuation
ring. Proposition 13.7.3 is also related.)
24.2.C. E XERCISE ( FLATNESS IS PRESERVED BY CHANGE OF BASE RING ). Show
that if M is a flat B-module, B → A is a homomorphism, then M ⊗B A is a flat
A-module.

24.2.D. E XERCISE ( TRANSITIVITY OF FLATNESS ). Show that if A is a flat B-algebra,
and M is A-flat, then M is also B-flat.
24.2.3. Proposition (flatness is a stalk/prime-local property). — An A-module M
is flat if and only if Mp is a flat Ap -module for all prime ideals p.
Proof. Suppose first that M is a flat A-module. Given any exact sequence of Ap modules
0 → N ′ → N → N ′′ → 0,

(24.2.3.1)

0 → M ⊗A N ′ → M ⊗A N → M ⊗A N ′′ → 0

is exact too. But M⊗A N is canonically isomorphic to Mp ⊗Ap N (do you see why?),so Mp is a flat Ap -module.
Suppose next that Mp is a flat Ap -module for all p. Given any short exact
sequence (24.2.3.1), tensoring with M yields(24.2.3.2)

0%K% M ⊗A N ′% M ⊗A N% M ⊗A N ′′%0(using right-exactness of ⊗, Exercise 1.3.H) where K is the kernel of M ⊗A N ′ →M ⊗A N. We wish to show that K = 0. It suffices to show that Kp = 0 for every prime p ⊂ A (see the comment after Exercise 4.3.F). Given any p, localizing(24.2.3.1) at p and tensoring with the exact Ap -module Mp yields(24.2.3.3)

0% Mp ⊗Ap Np′% Mp ⊗Ap Np% Mp ⊗Ap Np′′

% 0.

∼ (M⊗A N)p ,But localizing (24.2.3.2) at p and using the isomorphisms Mp ⊗Ap Np =we obtain the exact sequence0% Kp% Mp ⊗Ap Np′% Mp ⊗Ap Np% Mp ⊗Ap Np′′

% 0,which is the same as the exact sequence (24.2.3.3) except for the Kp . Hence Kp = 0
as desired.
!
24.2.4. Flatness for schemes.
Motivated by Proposition 24.2.3, the extension of the notion of flatness toschemes is straightforward.

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24.2.5. Definition: flat quasicoherent sheaves. We say that a quasicoherent sheaf
F on a scheme X is flat at p ∈ X if Fp is a flat OX,p -module. We say that aquasicoherent sheaf F on a scheme X is flat (over X) if it is flat at all p ∈ X. In lightof Proposition 24.2.3, we can check this notion on an affine open cover of X.
24.2.6. Definition: flat morphism. Similarly, we say that a morphism of schemes
π : X → Y is flat at p ∈ X if OX,p is a flat OY,π(p) -module. We say that a morphismof schemes π : X → Y is flat if it is flat at all p ∈ X. We can check flatness locally on
the source and target.
We can combine these two definitions into a single fancy definition.

24.2.7. Definition: flat quasicoherent sheaf over a base. Suppose π : X → Y is a morphism of schemes, and F is a quasicoherent sheaf on X. We say that F is flat (overY) at p ∈ X if Fp is a flat OY,π(p) -module. We say that F is flat (over Y) if it is flatat all p ∈ X.
Definitions 24.2.5 and 24.2.6 correspond to the cases X = Y and F = OX
respectively. (Definition 24.2.7 applies without change to the category of ringed
spaces, but we won’t use this.)24.2.E. E ASY E XERCISE ( REALITY CHECK ). Show that open embeddings are flat.
Our results about flatness over rings above carry over easily to schemes.
24.2.F. E XERCISE . Show that a map of rings B → A is flat if and only if thecorresponding morphism of schemes Spec A → Spec B is flat. More generally, if
B → A is a map of rings, and M is an A-module, show that M is B-flat if and( is flat over Spec B. (Be careful: this requires more than merely invoking
only if M
Proposition 24.2.3.)

Thus if π : X → Y is an affine morphism, and F is a quasicoherent sheaf on X,
then F is flat over Y if and only if π∗ F is flat over Y.

24.2.G. E ASY E XERCISE ( EXAMPLES AND REALITY CHECKS ).
(a) If X is a scheme, and p ∈ X, show that the natural morphism Spec OX,p → X isflat. (Hint: localization is flat, §24.2.1.)(b) Show that An
A → Spec A is flat.
(c) If F is a locally free sheaf on a scheme X, show that PF → X (Definition 17.2.3)
is flat.
(d) Show that Spec k → Spec k[t]/(t2 ) is not flat. (Draw a picture to try to see what
is not “nice” about this morphism. Some more insight about flatness of the dual
numbers will be given in the criterion of Exercise 24.4.D.)
24.2.H. E XERCISE ( TRANSITIVITY OF FLATNESS ). Suppose π : X → Y and F isa quasicoherent sheaf on X, flat over Y. Suppose also that ψ : Y → Z is a flat
morphism. Show that F is flat over Z.
24.2.I. E XERCISE ( FLATNESS IS PRESERVED BY BASE CHANGE ). Suppose π : X → Y
is a morphism, and F is a quasicoherent sheaf on X, flat over Y. If ρ : Y ′ → Y is
any morphism, and ρ ′ : X ×Y Y ′ → X is the induced morphism, show that (ρ ′ )∗ F

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650

is flat over Y ′ .

The Rising Sea: Foundations of Algebraic Geometry(ρ ′ )∗ F

X ×Y Y ′

F
ρ′

%X
π

"
"
ρ
%Y
Y′
In particular, using Exercise 24.2.A(a), if X and Y ′ are k-schemes, and F is any
quasicoherent sheaf on X, then (ρ ′ )∗ F is flat over Y ′ . For example, X ×k Y ′ isalways flat over Y ′ — “products over a field are flat over their factors”. (Feel free toimmediately generalize this further; for example, F can be a quasicoherent sheaf
on a scheme Z over X, flat over Y.)
The following exercise is very useful for visualizing flatness and non-flatness(see for example Figure 24.2).
24.2.J. E XERCISE ( FLAT MAPS SEND ASSOCIATED POINTS TO ASSOCIATED POINTS ).
Suppose π : X → Y is a flat morphism of locally Noetherian schemes. Show thatany associated point of X must map to an associated point of Y. (Feel free to immediately generalize this to a coherent sheaf F on X, flat over Y, without π itself
needing to be flat.) Hint: suppose π♯ : (B, n) → (A, m) is a local morphism of Noetherian local rings (i.e., π♯ (n) ⊂ m, §6.3.1). Suppose n is not an associated primeof B. Show that there is an element f ∈ n not in any associated prime of B (perhaps using prime avoidance, Proposition 11.2.13), and hence is a non-zerodivisor.
Show that π♯ f ∈ m is a non-zerodivisor of A using Observation 24.2.2, and thus
show that m is not an associated prime of A.
24.2.K. E XERCISE . Use Exercise 24.2.J to show that the following morphisms are
not flat (see Figure 24.2):(a) Spec k[x, y]/(xy) → Spec k[x],(b) Spec k[x, y]/(y2 , xy) → Spec k[x],(c) Bl(0,0) A2k → A2k .
Hint for (c): first pull back to a line through the origin to obtain a something akinto (a). (This foreshadows the statement and proof of Proposition 24.5.6, which saysthat for flat morphisms “there is no jumping of fiber dimension”.)
24.2.L. E XERCISE . Show that π : X → Y is flat if and only if π∗ : QCohY → QCohX
is an exact functor. (Translation: the earlier definition of flat morphism in Remark 16.3.8 agrees with Definition 24.2.6. But Definition 24.2.6 is “better”!)
24.2.8. Theorem (cohomology commutes with flat base change). — Suppose
X′

ρ′

%X

π′

"
Y′

π
ρ

"
%Y

is a Cartesian diagram, and π (and thus π ′ ) is quasicompact and separated (so higher
pushforwards of quasicoherent sheaves by π and π ′ exist, as described in §18.8). Suppose

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F IGURE 24.2. Morphisms that are not flat (Exercise 24.2.K)

also that ρ is flat, and F is a quasicoherent sheaf on X. Then the natural morphisms∗(Exercise 18.8.B(a)) ρ∗ (Ri π∗ F ) → Ri π∗′ (ρ ′ F ) are isomorphisms.

24.2.M. E XERCISE . Prove Theorem 24.2.8. Hint: Exercise 18.8.B(b) is the special
case where ρ is affine.

A useful special case is where Y ′ is the generic point of a reduced componentof Y. In other words, in light of Exercise 24.2.G(a), the stalk of the higher pushforward of F at the generic point is the cohomology of F on the fiber over the genericpoint. This is a first example of something important: understanding cohomologyof (quasicoherent sheaves on) fibers in terms of higher pushforwards. (We would
certainly hope that higher pushforwards would tell us something about higher cohomology of fibers, but this is certainly not a priori clear!) In comparison to thisresult, which shows that cohomology of any quasicoherent sheaf commutes withflat base change, §24.7 and Chapter 28 deal with when and how cohomology of aflat quasicoherent sheaf commutes with any base change.
24.2.9. Pulling back closed subschemes (and ideal sheaves) by flat morphisms.
Closed subschemes pull back particularly well under flat morphisms, and thiscan be helpful to keep in mind. As pointed out in Remarks 16.3.8 and 16.3.9, inthe case of flat morphisms, pullback of ideal sheaves as quasicoherent sheaves agrees
with pullback in terms of the pullback of the corresponding closed subschemes.
In other words, closed subscheme exact sequences pull back (remain exact) under flat pullbacks. This is in fact not just a necessary condition for flatness; it isalso sufficient, which can be shown using the ideal-theoretic criterion for flatness(Theorem 24.4.1). There is an analogous fact about pulling ideal sheaves of flat
subschemes by arbitrary pullbacks, see §24.3.2.
24.2.N. E XERCISE .
(a) Suppose D is an effective Cartier divisor on Y and π : X → Y is a flat morphism.
Show that the pullback of D to X (by π) is also an effective Cartier divisor.
(b) Use part (a) to show that under a flat morphism, regular embeddings pull backto regular embeddings.

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24.2.O. U NIMPORTANT E XERCISE .
(a) Suppose π : X → Y is a morphism, and Z "→ Y is a closed embedding cut out by
an ideal sheaf I ⊂ OY . Show that (π∗ I )n = π∗ (I n ). (Products of quasicoherent
ideal sheaves were defined in Exercise 14.3.E.)∗(b) Suppose further that π is flat, Y = Ank , and Z is the origin. Let J = π I be thequasicoherent sheaf of algebras on X cutting out the pullback W of Z. Prove thatthe graded sheaf of algebras ⊕n≥0 J n /J n+1 (do you understand the multiplication?) is isomorphic to OW [x1 , . . . , xn ] (interpreted as a graded sheaf of algebras).
∼ Symn (J /J 2 ).)(Hint: first show that J n /J n+1 =24.2.P. U NIMPORTANT E XERCISE .
(a) Show that blowing up commutes with flat base change. More precisely, if
π : X → Y is any flat morphism, and Z "→ Y is any closed embedding, give a∼ BlZ× X X. (You can proceed by universal
canonical isomorphism (BlZ Y) ×Y X =Yproperty, using Exercise 24.2.N(a), or by using the Proj construction of the blow
up and Exercise 24.2.O.)(b) Give an example to show that blowing up does not commute with base change
in general.

24.3 Flatness through Tor
We defined the Tor (bi-)functor in §23.1: TorAi (M, N) is obtained by taking afree resolution of N, removing the N, tensoring it with M, and taking homology.
Exercise 23.1.D characterized flatness in terms of Tor: M is A-flat if TorA
1 (M, N) =0 for all N. In this section, we reap the easier benefits of this characterization,
recalling key properties of Tor when needed. In §24.4, we work harder to extract
more from Tor.
It is sometimes possible to compute Tor from its definition, as shown in thefollowing exercise that we will use repeatedly.
24.3.A. E XERCISE . Define (M : x) as {m ∈ M : xm = 0} ⊂ M — it consists of theelements of M annihilated by x. If x is not a zerodivisor, show that⎧⎪
⎨M/xM if i = 0;A
Tori (M, A/(x)) = (M : x) if i = 1;⎪⎩0
if i > 1.
Hint: use the resolution
0

%A

×x

%A

% A/(x)%0of A/(x).
24.3.1. Remark. As a corollary of Exercise 24.3.A, we see again that flat modules
over an integral domain are torsion-free (and more generally, Observation 24.2.2).
Also, Exercise 24.3.A gives the reason for the notation Tor — it is short for torsion.
24.3.B. E XERCISE . If B is A-flat, use the FHHF Theorem (Exercise 1.6.H(c)) to giveB∼
an isomorphism B ⊗A TorAi (M, N) = Tori (B ⊗ M, B ⊗ N).

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Recall that the Tor functor is symmetric in its entries (there is an isomorphismA
TorA
i (M, N) ↔ Tori (N, M), Exercise 23.3.A). This gives us a quick but very useful
result.
24.3.C. E ASY E XERCISE . If 0 → N ′ → N → N ′′ → 0 is an exact sequence ofA-modules, and N ′′ is flat (e.g., free), show that 0 → M ⊗A N ′ → M ⊗A N →M ⊗A N ′′ → 0 is exact for any A-module M.

We would have cared about this result long before seeing Tor, so it gives somemotivation for learning about Tor. (Unimportant side question: Can you prove
this without Tor, using a diagram chase?)
24.3.D. E XERCISE ( IMPORTANT CONSEQUENCE OF E XERCISE 24.3.C). Suppose
0 → F ′ → F → F ′′ → 0 is a short exact sequence of quasicoherent sheaves on ascheme Y, and F ′′ is flat (e.g., locally free). Show that if π : X → Y is any morphismof schemes, the pulled back sequence 0 → π∗ F ′ → π∗ F → π∗ F ′′ → 0 remains
exact.

24.3.E. E XERCISE ( CF. E XERCISE 13.5.B FOR THE ANALOGOUS FACTS ABOUT VEC Suppose 0 → M ′ → M → M ′′ → 0 is an exact sequence ofA-modules.
(a) If M and M ′′ are both flat, show that M ′ is too. (Hint: Recall the long exact sequence for Tor, Proposition 23.1.2. Also, use that N is flat if and only if
Tori (N, N ′ ) = 0 for all i > 0 and all N ′ , Exercise 23.1.D.)(b) If M ′ and M ′′ are both flat, show that M is too. (Same hint.)(c) If M ′ and M are both flat, show that M ′′ need not be flat.
TOR BUNDLES ).

24.3.F. E XERCISE . If 0 → M0 → M1 → · · · → Mn → 0 is an exact sequence of flat
A-modules, show that it remains exact upon tensoring with any other A-module.
(Hint: as always, break the exact sequence into short exact sequences.)
24.3.G. E ASY E XERCISE . If 0 → M0 → M1 → · · · → Mn → 0 is an exact sequence,and Mi is flat for i > 0, show that M0 is flat too. (Hint: as always, break the exact
sequence into short exact sequences.)
We will use the Exercises 24.3.F and 24.3.G later in this chapter.
24.3.2. Pulling back quasicoherent ideal sheaves of flat closed subschemes byarbitrary morphisms (promised in §24.2.9). Suppose
W

α

%X

"
Y

β

"
%Z

is a fibered product, and V "→ X is a closed subscheme. Then Y ×Z V is a closed subscheme of W (§9.2.2). There are two possible senses in which IV/X can be “pulledback” to W: as a quasicoherent sheaf α∗ IV/X , and as the ideal of the “pulled back”
closed subscheme IY×Z V/W . As pointed out in Remark 16.3.9, these are not necessarily the same, but they are the same if β is flat. We now give another importantcase in which they are the same.

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24.3.H. E XERCISE .
If V is flat over Z (with no hypotheses on β), show that∼ IY× V/W . Hint: Easy Exercise 24.3.C.
α∗ IV/X =Z

24.4 Ideal-theoretic criteria for flatness
The following theorem will allow us to classify flat modules over a number ofrings. It is a refined version of Exercise 23.1.D, that M is a flat A-module if andonly if TorA
1 (M, N) = 0 for all A-modules N.
24.4.1. Theorem (ideal-theoretic criterion for flatness). — The A-module M is flat
if and only if TorA
1 (M, A/I) = 0 for every ideal I.
24.4.2. Remarks. Before getting to the proof, we make some side remarks that may
give some insight into how to think about flatness. Theorem 24.4.1 is profitably
stated without the theory of Tor. It is equivalent to the statement that M is flat if
and only if for all ideals I ⊂ A, I⊗A M → M is an injection, and you can reinterpretthe proof in this guise. Perhaps better, M is flat if and only if I ⊗A M → IM is an
isomorphism for every ideal I.
Flatness is often informally described as “continuously varying fibers”, andthis can be made more precise as follows. An A-module M is flat if and only if
it restricts nicely to closed subschemes of Spec A. More precisely, what we losein this restriction, the submodule IM of elements which “vanish on Z”, is easy
to understand: it consists of formal linear combinations of elements i ⊗ m, withno surprise relations among them — i.e., the tensor product I ⊗A M. This is thecontent of the following exercise, in which you may use Theorem 24.4.1.
24.4.A. ⋆ E XERCISE ( THE EQUATIONAL CRITERION FOR,FLATNESS ). Show that
an A-module M is flat if and only if for every relation ,ai mi = 0 with ai ∈ A
and mi ∈ M, there exist mj′ ∈ M and aij ∈ A such that j aij mj′ = mi for all i,
and i ai aij = 0 in A for all j. (Translation: whenever elements of M satisfy an
A-linear relation, this is “because” of linear equations holding in A.)
24.4.3. Unimportant remark. In the statement of Theorem 24.4.1, it suffices to check
only finitely generated ideals. This is essentially the content of the following statement, which you can prove if you wish: Show that an A-module M is flat if andonly if for all finitely generated ideals I, the natural map I ⊗A M → M is an injection.
Hint: if there is a counterexample for an ideal J that is not finitely generated, use
it to find another counterexample for an ideal I that is finitely generated.
24.4.4. Proof of the ideal-theoretic criterion for flatness, Theorem 24.4.1. By Exercise 23.1.D,A
we need only show that TorA
1 (M, A/I) = 0 for all I implies Tor1 (M, N) = 0 for all
A-modules N, and hence that M is flat.
We first prove that TorA
1 (M, N) = 0 for all finitely generated modules N, by
induction on the number n of generators a1 , . . . , an of N. The base case (if n = 1,
∼ A/ Ann(a1 )) is our assumption. If n > 1, then Aan =∼ A/ Ann(an ) is aso N =
submodule of N, and the quotient Q is generated by the images of a1 , . . . , an−1 ,so the result follows by considering the Tor1 portion of the Tor long exact sequence

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for
0 → A/ Ann(an ) → N → Q → 0.

We deal with the case of general N by abstract nonsense. Notice that N is theunion of its finitely generated submodules {Nα }. In fancy language, this union
is a filtered colimit — any two finitely generated submodules are contained in a
finitely generated submodule (specifically, the submodule they generate). Filteredcolimits of modules commute with cohomology (Exercise 1.6.L), so Tor1 (M, N) is
the colimit over α of Tor1 (M, Nα ) = 0, and is thus 0.
!
We now use Theorem 24.4.1 to get explicit characterizations of flat modules
over three (types of) rings: principal ideal domains, dual numbers, and some local
rings.
Recall Observation 24.2.2, that flatness implies torsion-free. The converse istrue for principal ideal domains:
24.4.B. E XERCISE ( FLAT = TORSION - FREE FOR A PID). Show that a module over a
principal ideal domain is flat if and only if it is torsion-free.
24.4.C. E XERCISE ( FLATNESS OVER A DVR). Suppose M is a module over a discrete valuation ring A with uniformizer t. Show that M is flat if and only if t is nota zerodivisor on M, i.e., (using the notation defined in Exercise 24.3.A) (M : t) = 0.
(See Exercise 24.2.B for the case of finitely generated modules.) This yields a simple and very important geometric interpretation of flatness over a regular curve,
which we discuss in §24.4.9.
24.4.D. E XERCISE ( FLATNESS OVER THE DUAL NUMBERS ). Show that M is flatover k[t]/(t2 ) if and only if the “multiplication by t” map M/tM → tM is an
isomorphism. (This fact is important in deformation theory and elsewhere.) Hint:
k[t]/(t2 ) has only three ideals.
24.4.5. Important Theorem (flat = free = projective for finitely presented modules over local rings). — Suppose (A, m) is a local ring (not necessarily Noetherian),and M is a finitely presented A-module. Then M is flat if and only if it is free if and only
if it is projective.
24.4.6. Remarks. Warning: modules over local rings can be flat without being free:
Q is a flat Z(p) -algebra (Z(p) is the localization of Z at p, not the p-adics), as all
localizations are flat (§24.2.1), but it is not free (do you see why?).
Also, non-Noetherian people may be pleased to know that with a little work,“finitely presented” can be weakened to “finitely generated”: use [Mat2, Thm. 7.10]
in the proof below, where finite presentation comes up.
Proof. For any ring, free modules are projective (§23.2.1), and projective modules
are flat (Exercise 23.2.B), so we need only show that flat modules are free for a local
ring.
(At this point, you should see Nakayama coming from a mile away.) Now
M/mM is a finite-dimensional vector space over the field A/m. Choose a basis of
M/mM, and lift it to elements m1 , . . . , mn ∈ M. Consider A⊕n → M given byei (→ mi . We will show this is an isomorphism. It is surjective by Nakayama’s
Lemma (see Exercise 7.2.H): the image is all of M modulo the maximal ideal,

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hence is everything. As M is finitely presented, by Exercise 13.6.A (“finitely presented implies always finitely presented”), the kernel K is finitely generated. Tensor 0 → K → A⊕n → M → 0 with A/m. As M is flat, the result is still exact(Exercise 24.3.C):
0 → K/mK → (A/m)⊕n → M/mM → 0.

But (A/m)⊕n → M/mM is an isomorphism by construction, so K/mK = 0. As K is
finitely generated, K = 0 by Nakayama’s Lemma 7.2.9.
!

Here is an immediate and useful corollary — really just a geometric interpretation.
24.4.7. Corollary (flat = locally free for finitely presented sheaves). — A finitely
presented sheaf F on X is flat (over X) if and only if it is locally free.
Proof. Local freeness of a finitely presented sheaf can be checked at the stalks,
Exercise 13.7.F.
!
24.4.E. E XERCISE . Suppose π : X → Y is a finite flat morphism of locally Noetherian schemes, and F is a finite rank locally free sheaf on X.
(a) Show that π∗ F is a finite rank locally free sheaf on Y.
(b) If Y is irreducible with generic point η, the degree of π above η is n, andF is locally free of rank r, show that π∗ F is locally free of rank nr. (Thismay be trickier than it looks.)

24.4.8. Remark. In particular, taking F = OX , we see that in the locally Noetheriansetting, finite flat morphisms have locally constant degree.
24.4.F. ⋆ E XERCISE ( INTERESTING VARIANT OF T HEOREM 24.4.5, BUT UNIMPOR TANT FOR US ).
Suppose A is a ring (not necessarily local), and M is a finitely

presented A-module. Show that M is flat if and only if it is projective. Hint: showthat M is projective if and only if Mm is free for every maximal ideal m. The harderdirection of this implication uses the fact that HomAm (Mm , Nm ) = HomA (M, N)m ,which follows from Exercise 1.6.G. (Remark: There exist finitely generated flat modules that are not projective.
"∞ They are necessarily not finitely presented. Example
without proof: let A = i=1 F2 , interpreted as functions Z≥0 → Z/2, and let M
be the module of functions modulo those of proper support, i.e., those vanishingat almost all points of Z≥0 .)24.4.G. E XERCISE .
Prove the following useful criterion for flatness: Suppose
π : X → Y is a finite morphism, and Y is reduced and locally Noetherian. Then
π is flat if and only if π∗ OX is locally free, if and only if the rank of π∗ OX is locallyconstant (dimκ(q) (π∗ OX )q ⊗ κ(q) is a locally constant function of q ∈ Y). Partial
hint: Exercise 13.7.K.
24.4.H. E XERCISE . Show that the normalization of the node (see Figure 7.4) is
not flat. Hint: use Exercise 24.4.G. (This exercise can be strengthened to show that
nontrivial normalizations are never flat.)

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24.4.I. E XERCISE . In A4k = Spec k[w, x, y, z], let X be the union of the wx-planewith the yz-plane:(24.4.8.1)

X = Spec k[w, x, y, z]/(wy, wz, xy, xz).

The projection A4k → A2k given by k[a, b] → k[w, x, y, z] with a (→ w − y, b (→ x − z
restricts to a morphism X → A2k . Show that this morphism is not flat.

24.4.9. Flat families over regular curves. Exercise 24.4.C gives an elegant geometric criterion for when morphisms to regular curves are flat.
24.4.J. E XERCISE ( CRITERION FOR FLATNESS OVER A REGULAR CURVE ). Suppose
π : X → Y is a morphism from a locally Noetherian scheme to a regular (locally
Noetherian) curve. (The local Noetherian hypothesis on X is so we can discuss its
associated points.) Show that π is flat if and only if all associated points of X map
to a generic point of Y. (This is a partial converse to Exercise 24.2.J, that flat maps
always send associated points to associated points. As with Exercise 24.2.J, feel
free to immediately generalize your argument to a coherent sheaf F on X.)
24.4.10. For example, a nonconstant map from an integral (locally Noetherian)
scheme to a regular curve must be flat. (As another example, the deformation
to the normal cone, discussed in the double-starred section §22.4.17 and depicted
in Figure 22.3, is flat.) Exercise 24.4.H (and the comment after it) shows that theregular condition is necessary. The example of two planes meeting at a point in
Exercise 24.4.I shows that the dimension 1 condition is necessary.

24.4.11. ⋆ Unimportant remark: A valuative criterion for flatness. Exercise 24.4.J showsthat flatness over a regular curves is geometrically intuitive (and is “visualizable”).
It gives a criterion for flatness in general: suppose π : X → Y is finitely presented
morphism. If π is flat, then for every morphism Y ′ → Y where Y ′ is the Spec of adiscrete valuation ring, π ′ : X ×Y Y ′ → Y ′ is flat, so no associated points of X ×Y Y ′map to the closed point of Y ′ . If Y is reduced and locally Noetherian, then this isa sufficient condition; this can reasonably be called a valuative criterion for flatness.
(Reducedness is necessary: consider Exercise 24.2.G(d).) This gives an excellent
way to visualize flatness, which you should try to put into words (perhaps after
learning about flat limits below). See [Gr-EGA, IV3 .3.11.8] for a proof (and an
extension without Noetherian hypotheses).
24.4.12. Revisiting the degree of a projective morphism from a curve to a regular curve.
As hinted after the statement of Proposition 17.4.5, we can now better understandwhy nonconstant projective morphisms from a curve to a regular curve have awell-defined degree, which can be determined by taking the preimage of any point(§17.4.4). (Example 9.3.3 was particularly enlightening.) This is because such maps
are flat by Exercise 24.4.J, and then the degree is constant by Remark 24.4.8 (see also
Exercise 24.4.G). Also, Exercise 24.4.G yields a new proof of Proposition 17.4.5.
24.4.13. Flat limits. Here is an important consequence of Exercise 24.4.J, which we
can informally state as: we can take flat limits over one-parameter families. Moreprecisely: suppose A is a discrete valuation ring, and let 0 be the closed point ofSpec A and η the generic point. Suppose X is a locally Noetherian scheme over A,
and Y is a closed subscheme of X|η . Let Y ′ be the scheme-theoretic closure of Y in

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X. Then Y ′ is flat over A. Similarly, suppose Z is a one-dimensional Noetherian
scheme, 0 is a regular point of Z, and π : X → Z is a morphism from a locally
Noetherian scheme to Z. If Y is a closed subscheme of π−1 (Z − {0}), and Y ′ isthe scheme-theoretic closure of Y in X, then Y ′ is flat over Z. In both cases, theclosure Y ′ |0 is often called the flat limit of Y. (Feel free to weaken the Noetherian
hypotheses on X.)
24.4.K. E XERCISE . Suppose (with the language of the previous paragraph) that A
is a discrete valuation ring, X is a locally Noetherian A-scheme, and Y is a closed
subscheme of the generic fiber X|η . Show that there is only one closed subscheme
Y ′ of X such that Y ′ |η = Y, and Y ′ is flat over A.
24.4.L. H ARDER E XERCISE ( AN EXPLICIT FLAT LIMIT ). Let X = A3 × A1 → Z = A1
over a field k, where the coordinates on A3 are x, y, and z, and the coordinates on
A1 are t. Define Y away from t = 0 as the union of the two lines y = z = 0 (thex-axis) and x = z − t = 0 (the y-axis translated by t). Find the flat limit at t = 0.
(Hints: (i) it is not the union of the two axes, although it includes this union. The
flat limit is nonreduced at the node, and the “fuzz” points out of the plane they
are contained in. (ii) (y, z)(x, z) ̸= (xy, z). (iii) Once you have a candidate flat limit,
be sure to check that it is the flat limit. (iv) If you get stuck, read Example 24.4.14
below.)
Consider a projective version of the previous example, where two lines in P3
degenerate to meet. The limit consists of two lines meeting at a node, with somenonreduced structure at the node. Before the two lines come together, their spaceof global sections is two-dimensional. When they come together, it is not immediately obvious that their flat limit also has two-dimensional space of global sections
as well. The reduced version (the union of the two lines meeting at a point) has
a one-dimensional space of global sections, but the effect of the nonreduced structure on the space of global sections may not be immediately clear. However, we
will see that “cohomology groups can only jump up in flat limits”, as a consequence (indeed the main moral) of the Semicontinuity Theorem 28.1.1.
24.4.14. ⋆⋆ Example of variation of cohomology groups in flat families. We canuse a variant of Exercise 24.4.L to see an example of a cohomology group actually
jumping. We work over an algebraically closed field to avoid distractions. Before
we get down to explicit algebra, here is the general idea. Consider a twisted cubic
C in P3 . A projection prp from a random point p ∈ P3 will take C to a nodal
plane cubic. Picture this projection “dynamically”, by choosing coordinates so p
is at [1, 0, 0, 0], and considering the map φt : [w, x, y, z] (→ [tw, x, y, z]; φ1 is theidentity on P3 , φt is an automorphism of P3 for t ̸= 0, and φ0 is the projection.
The limit of φt (C) as t → 0 will be a nodal cubic, with nonreduced structure at
the node “analytically the same” as what we saw when two lines came together(Exercise 24.4.L). (The phrase “analytically the same” can be made precise once wedefine completions in §29.)
Let’s now see this in practice. Rather than working directly with the twisted
cubic, we use another example where we saw a similar picture. Consider the nodal(affine) plane cubic y2 = x3 + x2 . Its normalization (see Figure 7.4, Example (3) of§7.3.8, Exercise 9.7.E, . . . ) was obtained by adding an extra variable m corresponding to y/x (which can be interpreted as blowing up the origin, see §22.4.3). We use

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the variable m rather than t (used in §7.3.8) in order to reserve t for the parameterfor the flat family.
We picture the nodal cubic C as lying in the xy-plane in A3 = Spec k[x, y, m],
3 projecting to it, with m = y/x. What are the equations
and the normalization C
3
for C? Clearly, they include the equations y2 = x3 + x2 and y = mx, but these
are not enough — the m-axis (i.e., x = y = 0) is also in V(y2 − x3 − x2 , y − mx).
A little thought (and the algebra we have seen earlier in this example) will makeclear that we have a third equation m2 = (x+1), which along with y = mx impliesy2 = x2 +x3 . Now we have enough equations: k[x, y, m]/(m2 −(x+1), y−mx) is an
integral domain, as it is clearly isomorphic to k[m]. Indeed, you should recognize
this as the algebra appearing in Exercise 9.7.E.
Next, we want to formalize our intuition of the dynamic projection to the xy3 ⊂ A3 . We picture it as follows. Given a point (x, y, m) at time 1, at time
plane of C
3 and at time t = 0
t we want it to be at (x, y, mt). At time t = 1, we “start with” C,
3
we have (set-theoretically) C. Thus at time t ̸= 0, the curve C is sent to the curve
cut out by equations
k[x, y, m]/(m2 − t2 (x + 1), ty − mx).

The family over Spec k[t, t−1 ] is thusk[x, y, m, t, t−1 ]/(m2 − t2 (x + 1), ty − mx).
Notice that we have inverted t because we are so far dealing only with nonzero t.
For t ̸= 0, this is certainly a “nice” family, and so surely flat. Let’s make sure thisis true.
24.4.M. E XERCISE . Check this, as painlessly as possible! Hint: by a clever changeof coordinates, show that the family is constant “over Spec k[t, t−1 ]”, and hence
pulled back (in some way you must figure out) via Spec k[t, t−1 ] → Spec k fromSpec k[X, Y, M]/(M2 − (X + 1), Y − MX) → Spec k,which is flat by Trick Question 24.2.A(a).

We now figure out the flat limit of this family over t = 0, in Spec k[x, y, m, t] →A1 = Spec k[t]. We first hope that our flat family is given by the equations we havealready written down:Spec k[x, y, m, t]/(m2 − t2 (x + 1), ty − mx).
But this is not flat over A1 = Spec k[t], as the fiber dimension jumps (§24.5.5):
substituting t = 0 into the equations (obtaining the fiber over 0 ∈ A1 ), we findSpec k[x, y, m]/(m2 , mx). This is set-theoretically the xy-plane (m = 0), which ofcourse has dimension 2. Notice for later reference that this “false limit” is schemetheoretically the xy-plane, with some nonreduced structure along the y-axis. (This may
remind you of Figure 4.4.)
So we are missing at least one equation. One clue as to what equation is missing: the equation y2 = x3 + x2 clearly holds for t ̸= 0, and does not hold for our
naive attempt at a limit scheme m2 = mx = 0. So we put this equation back in,and have a second hope for describing the flat family over A1 :Spec k[x, y, m, t]/(m2 − t2 (x + 1), ty − mx, y2 − x3 − x2 ) → Spec k[t].

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Let A = k[x, y, m, t]/(m2 − t2 (x + 1), ty − mx, y2 − x3 − x2 ) for convenience.
The morphism Spec A → A1 is flat at t = 0. How can we show it? We could hope
to show that A is an integral domain, and thus invoke Exercise 24.4.J. Instead weuse Exercise 24.4.B, and show that t is not a zerodivisor on A. We do this by givinga “normal form” for elements of A.
24.4.N. E XERCISE . Show that each element of A can be written uniquely as apolynomial in x, y, m, and t such that no monomial in it is divisible by m2 , mx, or
y2 . Then show that t is not a zerodivisor on A, and conclude that Spec A → A1 is
indeed flat.
Thus the flat limit when t = 0 is given bySpec k[x, y, m]/(m2 , mx, y2 − x2 − x3 ).
24.4.O. E XERCISE . Show that the flat limit is nonreduced, and the “nonreducedness has length 1 and supported at the origin”. More precisely, if X = Spec A/(t),show that IXred is a skyscraper sheaf, with value k, supported at the origin. Sketch
this flat limit X.
24.4.15. Note that we have a nonzero global function on X, given by m, which is
supported at the origin (i.e., 0 away from the origin).
We now use this example to get a projective example with interesting behavior.
We take the projective completion of this example, to get a family of cubic curvesin P3 degenerating to a nodal cubic C with a nonreduced point.
24.4.P. E XERCISE . Do this: describe this family (in P3 × A1 ) precisely.
Take the long exact sequence corresponding to0% ICred% OC% OCred% 0,

to get0% H0 (C, ICred )% H1 (C, ICred )α% H0 (C, OC )% H0 (C, OCred )% H1 (C, OC )% H1 (C, OCred )% H2 (C, ICred )We have H1 (C, ICred ) = H2 (C, ICred ) = 0 as ICred is supported in dimension 0(by dimensional cohomology vanishing, Theorem 18.2.6). Also, Hi (Cred , OCred ) =Hi (C, OCred ) (property (v) of cohomology, see §18.1). The (reduced) nodal cubic
Cred has h0 (O) = 1 (§10.3.7) and h1 (O) = 1 (cubic plane curves have genus 1,(18.6.6.1)). Also, h0 (C, ICred ) = 1 as observed above. Finally, α is not 0, as there
exists a nonzero function on C vanishing on Cred (§24.4.15 — convince yourself
that this function extends from the affine patch Spec A to the projective completion).
Using the long exact sequence, we conclude h0 (C, OC ) = 2 and h1 (C, OC ) = 1.
Thus in this example we see that (h0 (O), h1 (O)) = (1, 0) for the general member

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of the family (twisted cubics are isomorphic to P1 ), and the special member (theflat limit) has (h0 (O), h1 (O)) = (2, 1). Notice that both cohomology groups have
jumped, yet the Euler characteristic has remained the same. The first behavior, as
stated after Exercise 24.4.L, is an example of the Semicontinuity Theorem 28.1.1.
The second, constancy of Euler characteristics in flat families, is what we turn tonext. (It is no coincidence that the example had a singular limit, see §28.1.3.)

24.5 Topological aspects of flatness
We now discuss some topological aspects and consequences of flatness, thatboil down to the Going-Down Theorem for flat morphisms (§24.5.2), which in turncomes from faithful flatness. Because dimension in algebraic geometry is a topological notion, we will show that dimensions of fibers behave well in flat families(§24.5.5).
24.5.1. Faithful flatness. The notion of faithful flatness is handy for many reasons,and we describe only a few. A B-module M is faithfully flat if for all complexesof B-modules(24.5.1.1)

N ′ → N → N ′′ ,(24.5.1.1) is exact if and only if (24.5.1.1)⊗B M is exact. A B-algebra A is faithfully
flat if it is faithfully flat as a B-module.
24.5.A. E XERCISE . Show that a flat B-module M is faithfully flat if and only if for
all B-modules N, M ⊗B N = 0 implies that N = 0.
24.5.B. E XERCISE .
equivalent.

Suppose M is a flat B-module. Show that the following are(a) M is faithfully flat;(b) for all prime ideals p ⊂ B, M ⊗B κ(p) is nonzero;(c) for all maximal ideals m ⊂ B, M ⊗B κ(m) = M/mM is nonzero.
Suppose π : X → Y is a morphism of schemes. We say that π is faithfully
flat if it is flat and surjective. (Unlike flatness, faithful flatness is not that useful anotion for quasicoherent sheaves, so we do not define faithfully flat quasicoherent
sheaves over a base.)
24.5.C. E XERCISE ( CF. E XERCISE 24.5.B( B )). Suppose B → A is a ring morphism.
Show that A is faithfully flat over B if and only if Spec A → Spec B is faithfully flat.
Faithful flatness is preserved by base change, as both surjectivity and flatness
are (Exercises 9.4.D and 24.2.I respectively). Similarly, faithful flatness is presevedby composition, as both surjectivity and flatness are .
24.5.D. E XERCISE . Suppose π : Spec A → Spec B is flat.
(a) Show that π is faithfully flat if and only if every closed point q ∈ Spec B is in theimage of π. (Hint: Exercise 24.5.B(c).)(b) Hence show that every flat (local) morphism of local rings (Definition 6.3.1) is

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faithfully flat. (Morphisms of local rings are assumed to be local, i.e., the maximal
ideal pulls back to the maximal ideal.)
24.5.2. Going-Down for flat morphisms. A consequence of Exercise 24.5.D is thefollowing useful result, whose statement makes no mention of faithful flatness.
(The statement is not coincidentally reminiscent of the Going-Down Theorem for
finite extensions of integrally closed domains, Theorem 11.2.12.)
24.5.E. E XERCISE (G OING -D OWN T HEOREM FOR FLAT MORPHISMS ).
(a) Suppose that B → A is a flat morphism of rings, corresponding to a map
π : Spec A → Spec B. Suppose q ⊂ q ′ are prime ideals of B, and p ′ is a prime
ideal of A with π([p ′ ]) = [q ′ ]. Show that there exists a prime p ⊂ p ′ of A with
π([p]) = [q]. Hint: show that Bq ′ → Ap ′ is a flat local ring homomorphism, andhence faithfully flat by the Exercise 24.5.D(b).
(b) Part (a) gives a geometric consequence of flatness. Draw a picture illustrating
this.
(c) Recall the Going-Up Theorem, described in §7.2.4. State the Going-Down Theorem for flat morphisms in a way parallel to Exercise 7.2.F, and prove it.
24.5.F. E XERCISE . Suppose π : X → Y is an integral (e.g., finite) flat morphism, andY has pure dimension n. Show that X has pure dimension n. (This generalizes
Exercise 11.1.G(a).) Hint: π satisfies both Going-Up (see Exercise 7.2.F) and GoingDown.
24.5.G. I MPORTANT E XERCISE : FLAT MORPHISMS ARE OPEN ( IN REASONABLE SITUATIONS ). Suppose π : X → Y is locally of finite type and flat, and Y (and henceX) is locally Noetherian. Show that π is an open map (i.e., sends open sets to opensets). Hint: reduce to showing that π(X) is open for all such π. Reduce to the casewhere X andY are both affine. Use Chevalley’s Theorem 7.4.2 to show that π(X) is
constructible. Use the Going-Down Theorem for flat morphisms, Exercise 24.5.E,
to show that π(X) is closed under generization. Conclude using Exercise 7.4.C.
24.5.3. Remark (proof of Proposition 9.5.4). Suppose A and B are finite type kalgebras. Then by Exercise 24.5.G, Spec A ×Spec k Spec B → Spec B is an open map.

24.5.4. Follow-ups to Exercise 24.5.G.
(i) Of course, not all open morphisms are flat: witness Spec k[t]/(t) → Spec k[t]/(t2 ).
(ii) Also, in quite reasonable circumstances, flat morphisms are not open: witness
Spec k(t) → Spec k[t] (flat by Example 24.2.1(ii)).
(iii) On the other hand, you can weaken the hypotheses of “locally of finite type”
and “locally Noetherian” to just “locally finitely presented” [Gr-EGA, IV2 .2.4.6]— as with the similar generalization in Exercise 9.3.I of Chevalley’s Theorem 7.4.2,
use the fact that any such morphisms is “locally” pulled back from a Noetheriansituation. We won’t use this, and hence omit the details.
24.5.5. Dimensions of fibers are well-behaved for flat morphisms.

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24.5.6. Proposition. — Suppose π : X → Y is a flat morphism of locally Noetherian
schemes, with p ∈ X and q ∈ Y such that π(p) = q. Then
codimX p = codimY q + codimπ−1 (q) p(see Figure 11.3).
Informal translation: the dimension of the fibers is the difference of the dimensions of X and Y (at least locally). Compare this to Exercise 11.4.A, which statedthat without the flatness hypothesis, we would only have inequality (≤).
24.5.H. E XERCISE . Prove Proposition 24.5.6 as follows. As just mentioned, Exercise 11.4.A gives one inequality, so show the other. Given a chain of irreducible
closed subsets in Y containing q, and a chain of irreducible closed subsets in
π−1 (q) ⊂ X containing p, construct a chain of irreducible closed subsets in X containing p, using the Going-Down Theorem for flat morphisms (Exercise 24.5.E).
As a consequence of Proposition 24.5.6, if π : X → Y is a flat map of irreducible
varieties, then the fibers of π all have pure dimension dim X − dim Y. (Warning:
Spec k[t]/(t) → Spec k[t]/(t2 ) does not exhibit dimensional jumping of fibers, isopen, and sends associated points to associated points, cf. Exercise 24.2.J, but is not
flat. If you prefer a reduced example, the normalization of the cuspidal plane cubicSpec k[t] → Spec k[x, y]/(y2 − x3 ), shown in Figure 9.4, also has these properties.)
This leads us to the following useful definition.

24.5.7. Definition. If π : X → Y is a flat morphism that is locally of finite type, andall fibers of π have pure dimension n, we say that π is flat of relative dimension n.
(In particular, when one says a morphism is flat of relative dimension n, the locallyfinite type hypotheses are implied. Remark 24.5.8 motivates this hypothesis.)24.5.I. E XERCISE . Suppose π : X → Y is a flat morphism of finite type k-schemes,
and Y is equidimensional (so “codimension is the difference of dimensions”, cf.
Theorem 11.2.9). Show that the following are equivalent.
(i) The scheme X has pure dimension dim Y + n.
(ii) The morphism π is flat of relative dimension n.
24.5.J. E XERCISE . Suppose π : X → Y and ρ : Y → Z are morphisms of locally Noetherian schemes, flat of relative dimension m and n respectively (hence locally offinite type). Show that ρ◦π is flat of relative dimension m+n. Hint: Exercise 24.5.I.
24.5.K. E XERCISE . Show that the notion of a morphism being “flat of relative
dimension n” is preserved by arbitrary base change. Hint to show that fiber dimension is preserved: Exercise 11.2.J.
24.5.L. E XERCISE ( DIMENSION IS ADDITIVE FOR PRODUCTS OF VARIETIES ). If X
and Y are k-varieties of pure dimension m and n respectively, show that X ×k Y
has pure dimension m + n. (We have waited egregiously long to prove this basic
fact!)
24.5.8. Remark. The reason for the “locally finite type” assumption in the definition of “flat of relative dimension n” is that we want any class of morphism to be

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behave “reasonably” (in the sense of §7.1.1). In particular, we want our notion of“flatness of relative dimension n” to be preserved by base change. Consider thefibered diagram% Spec k(y)Spec k(x) ⊗k k(y)π′"Spec k(x)π"% Spec k.

Both π and π ′ are trivially flat, because they are morphisms to spectra of fields(Exercise 24.2.A(a)). But the dimension of the fiber of π is 0, while (as described in
Remark 11.2.16) the dimension of the fiber of π ′ is 1.
24.5.9. Generic flatness. (This would be better called “general flatness”.)
24.5.M. E ASY E XERCISE ( GENERIC FLATNESS ). Suppose π : X → Y is a finite typemorphism to a Noetherian integral scheme, and F is a coherent sheaf on X. Show
that there is a dense open subset U ⊂ Y over which F is flat. Translation: F isflat at all points of X mapping to U. (An important special case is if F = OX , inwhich case this shows there is a dense open subset U over which π is flat.) Hint:
Grothendieck’s Generic Freeness Lemma 7.4.4.
24.5.10. Interpretation of the degree of a generically finite morphism. We use this tointerpret the degree of a rational map of varieties in terms of counting preimages,
fulfilling a promise made in §11.2.2. Suppose π : X ""# Y is a generically finite
rational map of k-varieties of degree d. For simplicity we assume that X and Y are
irreducible. (But feel free to relax this.) Replace X by an open subset on which π
is a morphism. By Proposition 11.4.3 (“generically finite implies generally finite”),
there is a dense open subset V of Y over which π is finite. By “generic flatness”(Exercise 24.5.M, there is a dense open subset V ′ of V over which π is finite andflat. Then by Remark 24.4.8 (“finite flat morphisms have locally constant degree”),
over this open subset V ′ , π has “locally constant degree”, which is necessarily d .
24.5.11. Exercise 24.5.M can be improved:
24.5.12. Theorem (generic flatness, improved version, [Stacks, tag 052B]). — If
π : X → Y is a finite type morphism of schemes, Y is reduced, and F is a finite type quasicoherent sheaf on X, then there is an open dense subscheme U ⊂ Y such that F |π−1 (U) is
of finite presentation over π−1 (U), and flat over U.
We won’t use this result, so we omit the proof.
24.5.13. Flatness is an open condition. Generic flatness can be used to show that inreasonable circumstances, the locus where a quasicoherent sheaf is flat over a baseis an open subset. More precisely:
24.5.14. Theorem (flatness is an open condition). — Suppose π : X → Y is a locallyfinite type morphism of locally Noetherian schemes, and F is a finite type quasicoherent
sheaf on X.
(a) The locus of points of X at which F is Y-flat is an open subset of X.
(b) If π is closed (e.g., proper), then the locus of points of Y over which F is flat (i.e.
those q ∈ Y for which F is flat at all p ∈ π−1 (q)) is an open subset of Y.

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665

Part (b) follows immediately from part (a). Part (a) reduces to a nontrivial
statement in commutative algebra, see for example [Mat2, Thm. 24.3] or [Gr-EGA,
IV3 .11.1.1]. As is often the case, Noetherian hypotheses can be dropped in exchange for local finite presentation hypotheses on the morphism π, see [Gr-EGA,
IV3 .11.3.1] or [Stacks, tag 00RC].

24.6 Local criteria for flatness(This is the hardest section on ideal-theoretic criteria for flatness, and could
profitably be postponed to a second reading.)
In the case of a Noetherian local ring, there is a greatly improved version of
the ideal-theoretic criterion of Theorem 24.4.1: we need check only one ideal —the maximal ideal. The price we pay for the simplicity of this “local criterion for
flatness” is that it is harder to prove.
24.6.1. Theorem (local criterion for flatness). — Suppose (A, m) is a Noetherian local ring, and M is a finitely generated A-module. Then M is flat if and only if
TorA
1 (M, A/m) = 0.
This is a miracle: flatness over all of Spec A is determined by what happens
over the closed point. (Caution: the finite generation hypothesis is necessary. Let
A = k[x, y](x,y) and M = k(x), with y acting as 0. Then M is not flat by Observation 24.2.2, but it turns out that it satisfies the local criterion otherwise.)
Theorem 24.6.1 is an immediate consequence of the following more generalstatement.
24.6.2. Theorem (local criterion for flatness, more general version). — Suppose(B, n) → (A, m) is a local morphism of Noetherian local rings (§6.3.1), and that M is a
finitely generated A-module. Then M is B-flat if and only if TorB
1 (M, B/n) = 0.

24.6.3. ⋆ Proof of Theorem 24.6.2. A sign of the difficulty of this result is that theArtin-Rees Lemma 12.9.3 (or a consequence thereof) is used twice — once for the
local ring (A, m) (in the guise of the Krull Intersection Theorem, Exercise 12.9.A),
and once for the local ring (B, n).
Recall from Exercise 18.4.R that a B-module N has finite length if there is a
∼ B/n forfinite sequence 0 = N0 ⊂ N1 ⊂ · · · ⊂ Nn = N with Nm /Nm−1 =1 ≤ m ≤ n.
24.6.A. E ASY P RELIMINARY E XERCISE . With the same hypotheses as Theorem 24.6.2,
B
suppose that TorB
1 (M, B/n) = 0. Show that Tor1 (M, N) = 0 for all B-modules N of
finite length, by induction on the length of N.
By Exercise 23.1.D, if M is B-flat, then TorB
1 (M, B/n) = 0, so it remains toassume that TorB
(M,
B/n)
=
0
and
show
that
M
is B-flat.
1
By the ideal theoretic criterion for flatness (Theorem 24.4.1, see §24.4.2), wewish to show that φ : I ⊗B M → M is an injection for all ideals I of B, i.e., thatker φ = 0.
Note that I ⊗B M inherits an A-module structure (as M is an A-module). It
is furthermore a finitely generated A-module (do you see why?), so by the Krull

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Intersection Theorem (Exercise 12.9.A), ∩t mt (I ⊗B M) = 0. Thus it suffices toshow that ker φ ⊂ mt (I ⊗B M) for all t.
As n ⊂ m (or more correctly, the image of n is contained in m), it suffices toshow that ker φ ⊂ nt (I ⊗B M) for all t. Notice that nt (I ⊗B M) is (the image inI ⊗B M of) (nt I) ⊗B M.
24.6.B. E XERCISE . Show that for each s, nt ∩ I ⊂ ns I for t ≫ 0, so it suffices toshow that ker φ ⊂ (nt ∩ I) ⊗B M for all t. Hint: Use the Artin-Rees Lemma 12.9.3,
taking A there to be B here, Mt = nt , I = n, and L = I.
Consider the short exact sequence
0 → nt ∩ I → I → I/(nt ∩ I) → 0.

Applying (·) ⊗B M, and using the fact that I/(nt ∩ I) is finite length, we have that)*
0 → (nt ∩ I) ⊗B M → I ⊗B M → I/(nt ∩ I) ⊗B M → 0
is exact using Exercise 24.6.A. Our goal is thus to show that ker φ maps to 0 in)*)*
(24.6.3.1)
I/(nt ∩ I) ⊗B M = (I + nt )/nt ⊗B M.
Applying (·) ⊗B M to the short exact sequence(24.6.3.2)

0 → (I + nt )/nt → B/nt → B/(I + nt ) → 0,

and using Exercise 24.6.A (as B/(I+nt ) is finite length), the top row of the diagram(24.6.3.3)% ((I + nt )/nt ) ⊗B M α% (B/nt ) ⊗B M
% (B/(I + nt )) ⊗B M
%00
7
7
I ⊗B Mφ% B ⊗B M

is exact, and the square clearly commutes. But then any element of I ⊗B M mapping to 0 in B ⊗B M = M must map to 0 (under the right vertical arrow) in(B/nt ) ⊗B M, and hence must have mapped to 0 in ((I + nt )/nt ) ⊗B M by the
injectivity of α, as desired.
!
This argument basically shows that flatness is an “infinitesimal” property, depending only on the completion of the scheme at the point in question. This is
made precise as follows.
Suppose (B, n) → (A, m) is a (local) homomorphism of Noetherian local rings,
and M is an A-module. If M is flat over B, then for each t ∈ Z≥0 , M/(nt M) is
flat over B/nt (flatness is preserved by base change, 24.2.I). (You should of course
restate this in your mind in the language of schemes and quasicoherent sheaves.)
The infinitesimal criterion for flatness states that this necessary criterion for flatness
is actually sufficient.
24.6.C. ⋆ E XERCISE ( THE INFINITESIMAL CRITERION FOR FLATNESS ). Suppose(B, n) → (A, m) is a (local) homomorphism of local rings, and M is a finitely generated A-module. Suppose further that for each t ∈ Z≥0 , M/(nt M) is flat overB/nt . Show that M is flat over B. (In combination with Exercise 24.6.A, this gives
another proof of the local criterion of flatness, Theorem 24.6.2.) Hint: follow the
proof of Theorem 24.6.2. Given the hypothesis, then for each t, we wish to showthat ker φ maps to 0 in (24.6.3.1). We wish to apply (·)⊗B M to (24.6.3.2) and obtain

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the (exactness of the) top row of (24.6.3.3). To do this, show that applying (·) ⊗B M
to (24.6.3.2) is the same as applying (·) ⊗B/nt (M/nt M). Then proceed as in therest of the proof of Theorem 24.6.2.
Exercise 24.6.C implies the useful fact that if A is a Noetherian local ring, thenits completion (at its maximal ideal) is flat over A.
24.6.4. The slicing criterion for flatness.
A useful variant of the local criterion is the following. Suppose t is a nonzerodivisor of B in m (geometrically: an effective Cartier divisor on the target passing through the closed point). If M is flat over B, then t is not a zerodivisor of M
(Observation 24.2.2). Also, M/tM is a flat B/tB-module (flatness commutes withbase change, Exercise 24.2.I). The next result says that this is a characterization offlatness, at least when M is finitely generated, or somewhat more generally.
24.6.5. Theorem (slicing criterion for flatness). — Suppose (B, n) → (A, m) is alocal morphism of Noetherian local rings, M is a finitely generated A-module, and t ∈ n
is a non-zerodivisor on B. Then M is B-flat if and only if(i) t is not a zerodivisor on M, and(ii) M/tM is flat over B/(t).
(For slicing criteria for other properties, see Exercise 12.2.C and Theorem 26.2.3.)
Proof. Assume that t is not a zerodivisor on M, and M/tM is flat over B/(t). We
will show that M is B-flat. (As stated at the start of §24.6.4, the other implication isa consequence of what we have already shown.)
By the local criterion for flatness, Theorem 24.6.2, we know
B/(t)

Tor1

(M/tM, (B/(t))/n) = 0,

and we wish to show that TorB
1 (M, B/n) = 0.
result then follows from the following lemma.

(Note that (B/(t))/n = B/n.) The!

24.6.6. Lemma. — Suppose M is a B-module, and t ∈ n is not a zerodivisor on M. Thenfor any B/(t)-module N, we have(24.6.6.1)

B/(t)

TorB
i (M, N) = Tori

(M/tM, N).

Proof. We calculate the left side of (24.6.6.1) by taking a free resolution of M:(24.6.6.2)· · · → F2 → F1 → F0 → M → 0.

By Exercise 24.3.A, TorBi (M, B/(t)) = 0 for i > 0 (here we use that t is not a zerodivisor on M, to show that TorB
1 (M, B/(t)) = 0). But this Tor module is computedby tensoring the free resolution (24.6.6.2) of M with B/(t). Thus the complex(24.6.6.3)· · · → F2 /tF2 → F1 /tF1 → F0 /tF0 → M/tM → 0

is exact (exactness except at the last term comes from the vanishing of Tori ). This isa free resolution of M/tM over the ring B/(t)! The left side of (24.6.6.1) is obtainedby tensoring (24.6.6.2) by N and truncating and taking homology, and the rightside is obtained by tensoring (24.6.6.3) by N and truncating and taking homology.
As (·) ⊗B N = (· ⊗B (B/t)) ⊗B/t N, we have established (24.6.6.1) as desired. !

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24.6.D. E XERCISE . Give a second (admittedly less direct) proof of the criterion for
flatness over a discrete valuation ring of Exercise 24.4.J, using the slicing criterion
for flatness (Theorem 24.6.5).
24.6.E. E XERCISE .
Use the slicing criterion to give a second solution to Exercise 24.4.I, on two planes in P4 meeting at a point.
The following exercise gives a sort of slicing criterion for flatness in the source.
24.6.F. E XERCISE . Suppose A is a B-algebra, A and B are Noetherian, M is a finitely
generated A-module, and f ∈ A has the property that for all maximal ideals n ⊂ B,
multiplication by f is injective on M/nM. Show that if M is B-flat, then M/fM isalso B-flat. (Hint: Use the local criterion for flatness, Theorem 24.6.2. Notice that
0

%M

×f

%M

% M/fM%0

is a flat resolution of M/fM.)
Exercise 24.6.F has an immediate geometric interpretation: “Suppose π : X →Y is a morphism of Noetherian schemes, F is a coherent sheaf on X, and Z "→ X is alocally principal subscheme ...” In the special case where F = OX , this leads to thenotion of a relative effective Cartier divisor: a locally principal subscheme of X
that is an effective Cartier divisor on all the fibers of π. Exercise 24.6.F implies that
if π is flat, then any relative effective Cartier divisor is also flat. (See Exercise 28.3.F
for an important application.)24.6.7. Remark: Local slicing criterion for flatness in the source, without Noetherian
assumptions.
The Noetherian hypotheses in Exercise 24.6.F can be removed. Suppose (B, n) →(A, m) is a flat (local) homomorphism of local rings (not necessarily Noetherian).
Suppose further that A is the localization of a finitely presented B-algebra. (Thismeans that A is essentially finitely presented, or essentially of finite presentation, overB, but we won’t need this language.) If f ∈ A is a nonzero divisor in B/mB, thenA/f is flat over B. See [Stacks, tag 046Z] for a proof.
24.6.8. ⋆⋆ Fibral flatness. We conclude by mentioning a criterion for flatness thatis useful enough to be worth recognizing, but not so useful as to merit proof here.
24.6.G. E XERCISE . Suppose we have a commuting diagram(24.6.8.1)

X❄
❄❄
❄❄
❄
τ ❄❄
8

π

Z

%Y
⑧⑧
⑧
⑧
⑧⑧ ρ
5⑧⑧

and a quasicoherent sheaf F on X, and points p ∈ X, q = π(p) ∈ Y, r = τ(p) ∈ Z.
Suppose ρ is flat at q, and F is π-flat at p. Show that F is τ-flat at p, and F |r is
πr -flat at p. Here F |r is the restriction of F to the fiber above r, and πr : τ−1 (r) →ρ−1 (r) is the restriction of π above r.
The Fibral Flatness Theorem states that in good circumstances the converse istrue.

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24.6.9. The Fibral Flatness Theorem ([Gr-EGA, IV3 .11.3.10], see also [Stacks, tag
039A]). — Suppose we have a commuting diagram (24.6.8.1) and a finitely presented
quasicoherent sheaf F on X, and points p ∈ X, q = π(p) ∈ Y, r = τ(p) ∈ Z, with
Fp ̸= 0. Suppose either X and Y are locally Noetherian, or ρ and τ are locally of finite
presentation. Then the following are equivalent.
(a) F is τ-flat at p, and F |r is πr -flat at p.
(b) ρ is flat at q, and F is π-flat at p.
This is a useful way of showing that a F is π-flat. The architecture of the argument is as follows. First reduce to the case where X, Y, and Z are affine. Cleverly
reduce to the Noetherian case (see [Gr-EGA, IV.11.2.7]), then solve the resulting
nontrivial problem in commutative algebra (see [Gr-EGA, IV.11.3.10.1]).

24.7 Flatness implies constant Euler characteristic
We come to an important consequence of flatness promised in §24.1. We will
see that this result implies many answers and examples to questions that we would
have asked before we even knew about flatness.
24.7.1. Important Theorem (Euler characteristic is constant in flat families). —Suppose π : X → Y is a projective morphism of locally Noetherian
schemes, and F is a,i icoherent sheaf on X, flat over Y. Then χ(Xq , F |Xq ) =(−1)
h (Xq , F |Xq ) is ai≥0−1
locally constant function of q ∈ Y (where Xq = π (q)).
This is first sign that “cohomology behaves well in flat families.” (We willsoon see a second: the Semicontinuity Theorem 28.1.1. A different proof of Theorem 24.7.1, giving an extension to the proper case, will be given in §28.2.5.) The
Noetherian hypotheses are used to ensure that π∗ F (m) is a coherent sheaf.
Theorem 24.7.1 gives a necessary condition for flatness. Converses (yielding asufficient condition) are given in Exercise 24.7.A(b)–(d).

Proof. We make three quick reductions. (i) The question is local on the target Y, so
we may reduce to case Y is affine, say Y = Spec B, so π factors through a closednembedding X "→ Pn
B for some n. (ii) We may reduce to the case X = PB , bynconsidering F as a sheaf on PB . (iii) We may reduce to showing that for m ≫ 0,
h0 (Xq , F (m)|Xq ) is a locally constant function of q ∈ Y (by Serre vanishing form ≫ 0, Theorem 18.1.4(ii), h0 agrees with the Euler characteristic).
Twist by O(m) for m ≫ 0, so that all the higher cohomology of F vanishes.
Now consider the Čech complex C • for F (m). Note that all the terms in theČech complex C • are flat, because F is flat. (Do you see why?) As all higher
cohomology groups (higher pushforwards) vanish, C • is exact except at the firstterm, where the cohomology is Γ (π∗ F (m)). We add the module Γ (π∗ F (m)) to thefront of the complex, so it is once again exact:(24.7.1.1)% Γ (π∗ F (m))
% C1
% C2
% ···
% C n+1% 0.
0(We have done this trick of tacking on a module before, for example in (18.2.4.1).)
Thus by Exercise 24.3.G, as we have an exact sequence in which all but the first

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The Rising Sea: Foundations of Algebraic Geometryterms are flat, the first term is flat as well. Thus π∗ F (m) is a flat coherent sheaf on
Y, and hence locally free (Corollary 24.4.7), and thus has locally constant rank.
Suppose q ∈ Y. We wish to show that the Hilbert function hF |Xq (m) is alocally constant function of q. To compute hF |Xq (m), we tensor the Čech resolution with κ(q) and take cohomology. Now the extended Čech resolution (withΓ (π∗ F (m)) tacked on the front), (24.7.1.1), is an exact sequence of flat modules,
and hence remains exact upon tensoring with κ(q) (Exercise 24.3.F). Thus∼ Γ (π∗ F (m)|q ),Γ (π∗ F (m)) ⊗ κ(q) =so the Hilbert function hF |Xq (m) is the rank at q of a locally free sheaf, which is alocally constant function of q ∈ Y.
!
Before we get to the interesting consequences of Theorem 24.7.1, we mention
some converses.

24.7.A. U NIMPORTANT E XERCISE ( CONVERSES TO T HEOREM 24.7.1). (We won’t
use this exercise for anything.)(a) Suppose A is a ring, and S• is a finitely generated A-algebra that is flat over A.
Show that Proj S• is flat over A.
(b) Suppose π : X → Y is a projective morphism of locally Noetherian schemes(which as always includes the data of an invertible sheaf OX (1) on X), such thatπ∗ OX (m) is locally free for all m ≥ m0 for some m0 . Show that π is flat. Hint:
describe X as
$ C%
Proj OY(⊕m≥m0 π∗ OX (m)) .
(c) More generally, suppose π : X → Y is a projective morphism of locally Noetherian schemes, and F is a coherent sheaf on X, such that π∗ F (m) is locally free forall m ≥ m0 for some m0 . Show that F is flat over Y.
(d) Suppose π : X → Y is a projective morphismof locally Noetherian schemes,,
and F is a coherent sheaf on X, such that (−1)i hi (Xq , F |q ) is a locally constantfunction of q ∈ Y. If Y is reduced, show that F must be flat over Y. (Hint: Exercise 13.7.K shows that constant rank implies local freeness in particularly nicecircumstances.)
We now give some ridiculously useful consequences of Theorem 24.7.1.
24.7.2. Corollary. — Assume the same hypotheses and notation as in Theorem 24.7.1.
Then the Hilbert polynomial of F |Xq is locally constant as a function of q ∈ Y.
24.7.B. C RUCIAL E XERCISE . Suppose X → Y is a projective flat morphism, where
Y is connected. Show that the following functions of q ∈ Y are constant: (a) the
degree of the fiber, (b) the dimension of the fiber, (c) the arithmetic genus of thefiber.
24.7.C. E XERCISE . Use §24.4.9 and Exercise 24.7.B(a) to give another solution to
Exercise 17.4.D(a) (“the degree of a finite map from a curve to a regular curve isconstant”).
Another consequence of Corollary 24.7.2 is something remarkably useful.
24.7.3. Corollary. — An invertible sheaf on a flat projective family of curves has locallyconstant degree on the fibers.

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671(Recall that the degree of a line bundle on a projective curve requires no hypotheses on the curve such as regularity, see (18.4.4.1).)
Proof. An invertible sheaf L on a flat family of curves is always flat (as locally it isisomorphic to the structure sheaf). Hence χ(Xq , Lq ) is a constant function of q. By
the definition of degree given in (18.4.4.1), deg(Xq , Lq ) = χ(Xq , Lq ) − χ(Xq , OXq ).
The result follows from the local constancy of χ(Xq , Lq ) and χ(Xq , OXq ) (Theorem 24.7.1).
!
The following exercise is a serious generalization of Corollary 24.7.3.
24.7.D. ⋆ E XERCISE FOR THOSE WHO HAVE READ STARRED C HAPTER 20: INTER SECTION NUMBERS ARE LOCALLY CONSTANT IN FLAT FAMILIES .
Supposeπ : X → Y is a flat projective morphism to a connected scheme; L1 , . . . , Ln areline bundles on X; and F is a coherent sheaf on X, flat over Y, such that the support of F when restricted to any fiber of π has dimension at most n. If q is any
point of Y, define (the temporary notation) (L1 · L2 · · · Ln · F )q to be the intersection on the fiber Xq of L1 , . . . , Ln with F |Xq (Definition 20.1.1). Show that
(L1 · L2 · · · Ln · F )q is independent of q.
Corollary 24.7.3 motivates the following definition.

24.7.4. Definition. Suppose L1 and L2 are line bundles on a k-variety X. We saythat L1 and L2 are algebraically equivalent if there exists a connected (but notnecessarily irreducible) k-variety Y with two k-valued points q1 and q2 , and a line
bundle L on X × Y such that the restriction of L to the fibers over q1 and q2 are
isomorphic to L1 and L2 respectively.
24.7.E. E XERCISE . Show that “algebraic equivalence” really is an equivalence
relation. Show that the line bundles algebraically equivalent to O form a subgroupof Pic X. This subgroup is denoted Pic0 X.
24.7.5. Identify the group of line bundles Pic X modulo algebraic equivalence with
Pic X/ Pic0 X. This quotient is called the Néron-Severi group. (This definition waspromised in §18.4.10.) By Proposition 20.1.4, Pic0 X ⊂ Picτ X: algebraic equivalence implies numerical equivalence. (Side Remark: a line bundle on a properk-scheme X is numerically trivial if and only if there exists an integer m ̸= 0 withL⊗m algebraically trivial. Thus Picτ X/Pic0 X is torsion. See [SGA6, XIII, Thm. 4.6]
for a proof, or [Laz, Cor. 1.4.38] for the projective case.)
24.7.6. ⋆ Hironaka’s example of a proper nonprojective smooth threefold.
In §16.4.10, we produced a proper nonprojective variety, but it was singular.
We can use Corollary 24.7.3 to give a smooth example, due to Hironaka.
Inside P3k , fix two conics C1 and C2 , which meet in two (k-valued) points, p1
and p2 . We construct a proper map π : X → P3k as follows. Away from pi , we blow
up Ci and then blow up the proper transform of C3−i (see Figure 24.3). This iswell-defined, as away from p1 and p2 , C1 and C2 are disjoint, blowing up one andthen the other is the same as blowing up their union, and thus the order doesn’t
matter.
Note that π is proper, as it is proper away from p1 , and proper away from p2 ,and the notion of properness is local on the base (Proposition 10.3.4(b)). As P3k is

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F IGURE 24.3. Hironaka’s example of a proper nonprojective
smooth threefoldproper (over k, Theorem 7.4.7), and compositions of proper morphisms are proper(Proposition 10.3.4(c)), X is proper.
24.7.F. E XERCISE . Show that X is smooth. (Hint: Theorem 22.3.10.) Let Ei be thepreimage of Ci \ {p1 , p2 }. Show that π|Ei : Ei → Ci \ {p1 , p2 } is a P1 -bundle (andflat).

24.7.G. E XERCISE . Let Ei be the closure of Ei in X. Show that Ei → Ci is flat.
(Hint: Exercise 24.4.J.)
24.7.H. E XERCISE .
Show that π∗ (pi ) is the union of two P1 ’s, say Yi and Zi ,meeting at a point, such that Yi , Y3−i , Z3−i ∈ Ei but Zi ∈/ Ei .

24.7.I. E XERCISE . Show that X is not projective as follows. Suppose otherwise
L is a very ample line bundle on X, so L has positive degree on every curve(including the Yi and Zi ). Using flatness of Ei → Ci , and constancy of degree in flatfamilies (Exercise 24.7.D), show that degYi L = degY3−i L + degZ3−i L . Obtaina contradiction. (This argument will remind you of the argument of §16.4.10.)
24.7.7. The notion of “projective morphism” is not local on the target.
Note that
π : X → P3 is not projective, as otherwise X would be projective (as the composition of projective morphisms is projective if the final target is quasicompact, Exercise 17.3.B). But away from each pi , π is projective (as it is a composition of
blow-ups, which are projective by construction, and the final target is quasicompact, so Exercise 17.3.B applies). Thus the notion of “projective morphism” is not
local on the target.

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CHAPTER 25

Smooth and étale morphisms, and flatness
We have defined smooth and étale morphisms earlier (Definition 12.6.2), but
we are now in a position to understand them much better. We will see that thenotion of unramified morphism (§21.6) is a natural companion to them.
Our three algebro-geometric definitions won’t be so obviously a natural triplet,but we will discuss the definitions given in [Gr-EGA] (§25.2.6), and in this context
the three types of morphisms look very similar. (We briefly mention other approaches and definitions in §25.2.7.)

25.1 Some motivation
The three classes of morphisms we will discuss in this chapter are the analogsof the following types of maps of manifolds, in differential geometry.
• Submersions are maps inducing surjections of tangent spaces everywhere.
They are useful in the notion of a fibration. (Perhaps a more relevant
notion from differential geometry, allowing singularities, is: “locally on
the source a smooth fibration”.)• Isomorphisms locally on the source (or local isomorphisms) are maps inducing
isomorphisms of tangent spaces.
• Immersions are maps inducing injections of tangent spaces.
(Recall our warning from §8.1.2: “immersion” is often used in algebraic geometry
with a different meaning.)[picture to be made later]
F IGURE 25.1. Sketches of notions from differential geometry: (a)maps “locally on the source a smooth fibration”, (b) local isomorphisms, and (c) immersions. (In algebraic geometry: (a) smoothmorphisms, (b) étale morphisms, and (c) unramified morphisms.)
In order to better understand smooth and étale morphisms, we temporarily
forget our earlier definitions, and consider some examples of things we want tobe analogs of “local isomorphism” (or “locally on the source an isomorphism”)
and “locally on the source a smooth fibration”, and see if they help us make good
definitions.
25.1.1. “Local isomorphisms” (étale morphisms). Consider the parabola x = y2
projecting to the x-axis, over the complex numbers. (This example has come up
673

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repeatedly, in one form or another, including in Exercise 12.6.F.) We might reasonably want this to be a local isomorphism (on the source) away from the origin.
We might also want the notion of local isomorphism space to be an open condition: the locus where a morphism is a local isomorphism should be open on thesource. This is true for the differential geometric definition. But then this morphism should be a local isomorphism over the generic point, and here we get a
non-trivial residue field extension (C(y)/C(y2 )), not an isomorphism. Thus weare forced to consider (the Spec’s of) certain finite extensions of fields to be “isomorphisms locally on the source” — in this case, even “covering spaces”. (We will
see in Exercise 25.2.C that we want precisely the finite separable extensions; you
could have shown this earlier.)
Note also in this example there are no (nonempty) Zariski-open subsets U ⊂Spec C[x] \ {0} and V ⊂ Spec C[x, y]/(x − y2 ) \ {(0, 0)} where the map sends U into
V isomorphically (Exercise 12.6.F), so this is not a local isomorphism in a way you
may have seen before. This leads to the notion of the étale topology, which is noteven a topology in the usual sense, but a “Grothendieck topology” (§13.3.4). Theétale topology is beyond the scope of this book.
25.1.2. Submersions (smooth morphisms).
25.1.3. Fibers are smooth varieties. As a first approximation of the algebro-geometricversion of submersion, we will want the fibers to be smooth varieties (over the
residue field). So the very first thing we need is to generalize the notion of “variety” over a base. It is reasonable to do this by having a locally finite type hypothesis. For somewhat subtle reasons, we will require the stronger condition that themorphism be locally of finite presentation. (If you really care, you can see where
it comes up in our discussion. But of course there is no difference for Noetherian
readers.)
The fibers are not just varieties; they should be smooth varieties (of dimensionn, say). From the case of smoothness over a field, or from our intuition of whatsmooth varieties should look like, we expect the sheaf of differentials on the fibers
to be locally free of rank n, or even better, that the sheaf of relative differentials be
locally free of rank n.
25.1.4. Flatness. At this point, our first approximation of “smooth morphism” issome version of “locally finitely presented, and fibers are smooth varieties”. Butthat isn’t quite enough. For example, a horrible map from a scheme X to a curve
Y that maps a different regular variety to each point of Y (X is the infinite disjoint
union of these) should not be considered a smooth fibration in any reasonable
sense. Also, we might not want to consider Spec k → Spec k[ϵ]/(ϵ2 ) to be a submersion; for example, this isn’t surjective on tangent spaces, and more generally
the picture “doesn’t look like a fibration”.
Both problems are failures of π : X → Y to be a nice, “’continuous” family.
Whenever we are looking for some vague notion of “niceness” we know that “flatness” will be in the definition.
For comparison, note that “unramified” has no flatness hypothesis, and indeed we didn’t expect it, as we would want the inclusion (closed embedding) of
the origin into A1 to be unramified. But then weird things may be unramified.

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F IGURE 25.2. We don’t want this to be a “smooth morphism”#
For example, if X = z∈C Spec C, then the morphism X → A1C sending the point
corresponding to z to the point z ∈ A1C is unramified. Such is life.

25.1.5. Desired Alternate Definitions. We might hope that a morphism π : X → Y issmooth of relative dimension n if and only if(i) π is locally of finite presentation,(ii) π is flat of relative dimension n, and(iii) ΩX/Y is locally free of rank n.
We might similarly hope that a morphism π : X → Y is étale if and only if(i) π is locally of finite presentation,(ii) π is flat, and(iii) ΩX/Y = 0.
We will shortly show (in Theorem 25.2.2 and Exercise 25.2.B respectively) thatthese desired definitions are equivalent to Definition 12.6.2.

25.2 Different characterizations of smooth and étale morphisms
The main result in this section is a description of equivalent characterizationsof smooth morphisms, Theorem 25.2.2. We will state the theorem, and then giveconsequences, and then finally give a proof.
25.2.1. But first, we discuss a preliminary fact. It requires something from §26.2.7:
if k is a field, and f1 , . . . , fi ∈ k[x1 , . . . , xN ] cut out a dimension N − i subscheme
X of ANk , and fi+1 ∈ k[x1 , . . . , xN ] does not vanish on any irreducible componentof X, then fi+1 is not a zerodivisor on X. For now, take this fact for granted; youshouldn’t read all of Chapter 26 in order to do this exercise. There are other ways
around this issue, but it is not worth the effort.
25.2.A. E XERCISE . Suppose π : X → Y is smooth of relative dimension n. Either (i)assume Y (and hence X) is locally Noetherian, or (ii) accept Remark 24.6.7 (the slicing criterion for flatness in the source without Noetherian hypotheses, which had a

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The Rising Sea: Foundations of Algebraic Geometryreference rather than a proof). Show that π is flat. Hint: suppose π is the morphism
Spec B[x1 , . . . xn+r ]/(f1 , . . . , fr ) → Spec B. Consider B[x1 , . . . , xn+r ]/(f1 , . . . , fi ) asi runs from 0 to r, and use induction on i and the slicing criterion for flatness inthe source ((i) Exercise 24.6.F or (ii) Remark 24.6.7).
We now come to the central result of this section.
25.2.2. Theorem. — Suppose π : X → Y is a morphism of schemes. Then the following
are equivalent.

(i) The morphism π is smooth of relative dimension n (Definition 12.6.2).
(ii) The morphism π is locally finitely presented and flat of relative dimension n; andΩπ is locally free of rank n (Desired Alternate Definition 25.1.5).
(iii) The morphism π is locally finitely presented and flat, and the fibers are smoothk-schemes of pure dimension n.
(iv) The morphism π is locally finitely presented and flat, and the geometric fibers aresmooth k-schemes of pure dimension n.

Before proving Theorem 25.2.2, we motivate it, by giving a number of exercises
and results assuming it.
25.2.B. E XERCISE . Show that a morphism π is étale (Definition 12.6.2) if and only
if π is locally of finite presentation, flat, and Ωπ = 0 (Desired Alternate Definition 25.1.5).
25.2.C. E ASY E XERCISE . Suppose π : X → Y is a morphism. Show that the following are equivalent.
(a) π is étale.
(b) π is smooth and unramified.
(c) π is locally finitely presented, flat, and unramified.
(d) π is locally finitely presented, flat, and for each q ∈ Y, π−1 (q) is the disjoint union of schemes of the form Spec K, where K is a finite separable
extension of κ(q).
(e) π is locally finitely presented, flat, and for each geometric point q of Y,
π−1 (q) is the (scheme-theoretic) disjoint union of copies of q.

25.2.D. E XERCISE . If π : X → Y is étale, show that any preimage p ∈ X of any regular point q ∈ Y whose local ring has dimension n is also a regular point whose local
ring has dimension n. Hint: Prove the result by induction on dim OY,q . “Slice” byan element of mY,q \ m2Y,q . Use the slicing criterion for regularity (Exercise 12.2.C).
25.2.3. Proof of the Smoothness-Regularity Comparison Theorem 12.2.10(b) (every smoothk-scheme is regular). By Exercise 12.6.E, any dimension n smooth k-scheme X can
locally be expressed as an étale cover of An
k , which is regular by Exercise 12.3.O.
Then by Exercise 25.2.D, X is regular.
!
25.2.E. E XERCISE .
Suppose π : X → Y is a morphism of equidimensional kvarieties, and Y is smooth. Use the conormal exact sequence (Theorem 21.2.12) toshow the following.
(a) Suppose that dim X = dim Y = n, and π is unramified. Show that X is smooth.

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677⋆ (b) Suppose dim X = m > dim Y = n, and the (scheme-theoretic) fibers of π overclosed points are smooth of dimension m − n. Show that X is smooth.
25.2.F. E XERCISE . Suppose π : X → Y is a morphism (over k) of smooth equidimensional k-varieties, where dim X = m and dim Y = n. (These two parts areclearly essentially the same geometric situation.)(a) Show that if the fibers of π are all smooth of dimension m − n, then π is smoothof relative dimension m − n.
(b) Show that if π∗ ΩY → ΩX is injective (“Tπ is surjective”, or “the relative cotangent sequence is exact on the left”), then π is smooth of relative dimension m − n.
Hint: for each point q ∈ Y, OY,q is a regular local ring. Work by induction on
dim OY,q , and “slice” by an element of mY,q \ m2Y,q .
25.2.G. E XERCISE . Suppose π : X → Y is locally finitely presented, ρ : Y → Z isétale, and τ = ρ ◦ π. Show that π is smooth of dimension n (e.g., étale, taking
n = 0) if and only if τ is. (We showed earlier that π is unramified if and only if τ
is. Do you see why and where?) Hint: use the Cancellation Theorem 10.1.19 forflat morphisms and Exercise 21.6.F. For the Ω part of the problem, use the relative
cotangent sequence, Theorem 21.2.25. (If you only solved Exercise 21.6.F(a), then
you will still be able to use it to prove this in the Noetherian case.)
25.2.4. ⋆⋆ Proof of Theorem 25.2.2.
(i) implies (ii). As noted in §12.6.2, local finite presentation is clear. Flatness
follows from Exercise 25.2.A. Finally, the sheaf of relative differentials is locally
free of rank n by Exercise 21.2.Q.
(ii) implies (iii). Suppose π satisfies (ii). We wish to show that the fibers aresmooth. The hypotheses of (ii) are preserved by base change, so we may assume
that Y = Spec k. Then X is indeed smooth (see the equivalent Definition 21.3.1).
(iii) implies (iv) by Exercise 12.2.G.
We now come to the main part of the argument: (iv) implies (i). (Part of thisargument will be reminiscent of the proof of Theorem 12.6.4, in which we also had
to prove smoothness.)Fix a point p ∈ X. Let q = π(p) ∈ Y, say q = Spec k. We will show that there isan open neighborhood Ui of p and Vi of q of the form stated in Definition 12.6.2.
It suffices to deal with the case that p is a closed point in π−1 (q), because closed
points are dense in finite type schemes (Exercise 5.3.F). (We cannot and do notassume that q is a closed point of Y.)
Because π is finitely presented, there are affine open neighborhoods U of p and
V of q, with π(U) ⊂ V, such that the morphism π|U : U → V can be written as
Spec B[x1 , . . . , xN ]/I → Spec B,
where I is finitely generated.
Fix a geometric point q = Spec k of Y mapping to q. Fix a point p of thegeometric fiber over q, which maps to p ∈ X. (Why is there such a p?) We then

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have a diagram
Uk% Uk ! "%U

Spec k[x1 , . . . , xN ]/Ik

% Spec k[x1 , . . . , xN ]/Ik ! "% Spec B[x1 , . . . , xN ]/I"
Spec k"% Spec k ! ""% Spec Bq% q! "%V

π|U

where Uk (resp. Uk ) are the fibers of U over q (resp. q), and Ik (resp. Ik ) is the ideal
cutting out Uk (resp. Uk ) in AN(resp. ANk ).
k
We make an important observation, using §24.3.2 (for a flat closed subscheme,
the pullback of the defining ideal as a quasicoherent sheaf is the same as the idealof the pulled back closed subscheme). Because B[x1 , . . . , xN ]/I is flat over B, wehave that Ik = I ⊗B k and Ik = I ⊗B k.
By our hypothesis (iv), the geometric fiber Uk ⊂ AN
is smooth, and hencekregular at its closed points by Exercise 12.2.D. (Here we use the fact that k is algebraically closed.)
Let r = N − n be the codimension of Uk in ANk (and the codimension of Uk in
NAk ). By the Jacobian criterion for smoothness over a field (§21.3.2), the cotangent
space for Uk at p is cut out in ΩAN |p by r linear equations (over k). By the conormalkexact sequence for Uk ⊂ AN
(Theorem 21.2.25(b)), these equations are spanned bykthe elements of Ik , which in turn are spanned by elements of I (as Ik = I ⊗B k, soelements of Ik are finite sums of pure tensors i ⊗ a with i ∈ I and a ∈ k). Thus
we can choose f1 , . . . , fr in I so that the images of f1 , . . . , fr in k[x1 , . . . , xN ] cut out
the Zariski tangent space of Uk at p (as the Zariski tangent space at a closed pointof a variety over an algebraically closed field is computed by the Jacobian matrix,
Exercise 12.2.D).
25.2.H. E XERCISE . Show that f1 , . . . , fr generate the ideal (Ik )p . Hint: look at yourargument for Exercise 12.2.L(a).
Rearrange the xi ’s so that the Jacobian matrix of the fi with respect to the first rof the xi is invertible at p, and hence at p, and hence in an open neighborhood of p.
Then (an open neighborhood of p in) Spec B[x1 , . . . , xN ]/(f1 , . . . , fr ) is of the form
we are looking for. We will show that U is the same as Spec B[x1 , . . . , xN ]/(f1 , . . . , fr )(i.e., I = (f1 , . . . , fr )) near p, thereby completing the proof.
For notational compactness (at the cost of having subscripts later on with confusingly different meanings), define
A :=
Ak :=
Ak

:=

B[x1 , . . . , xN ]/(f1 , . . . , fr ),A ⊗B k = k[x1 , . . . , xN ]/(f1 , . . . , fr ),A ⊗B k = k[x1 , . . . , xN ]/(f1 , . . . , fr ).

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We have the diagram
A/J (
g❆❆
❆❆
❆❆
❆❆

JA
PP
P
P
PP
PP

B
where J denotes the image of I in A. We wish to show that “J is 0 near p”.
Let m be the maximal ideal of A corresponding to p, and let m be the maximal
ideal of Ak corresponding to p.
Let Jk be the image of Ik in Ak . We use §24.3.2 a second time in this proof.
Because A/J = B[x1 , . . . , xN ]/I is flat over B, Jk can be interpreted both as thepullback of J as a quasicoherent sheaf, or as the ideal sheaf of the “pulled backclosed subscheme corresponding to J”. More precisely,(25.2.4.1)
Similarly,(25.2.4.2)

J ⊗B k = JAk .
J ⊗B k = JAk .

Now Exercise 25.2.H means that
) *J · Ak m = (0),so taking the quotient by the maximal ideal m of (Ak )m yields)*)*
J · Ak ⊗Ak Ak /m = (0),which by (25.2.4.1) can be rewritten as)*)*
J ⊗B k ⊗A⊗B k Ak /m = (0),which in turn can be readily rewritten as
and then as)*(J ⊗B k) ⊗A⊗B k Ak /m = (0))*((J ⊗B k) ⊗A⊗B k (Ak /m)) ⊗Ak /m Ak /m = (0)But Ak /m = k is a field extension of Ak /m = k, and a k-vector space V is 0 if (andonly if) V ⊗k k = 0, so we have shown that((J ⊗B k) ⊗A⊗B k (Ak /m)) = (0).

Using (25.2.4.2) (essentially reversing our steps, with k replaced by k), we have
(J · Ak ) ⊗Ak (Ak /m) = 0,

which means that J ⊗A (A/m) = 0.

25.2.I. E XERCISE . Show that Jm = 0, and from this conclude that there is anelement f ∈ A \ m such that fJ = 0. (Hint: Nakayama, and finite generation of J.)

Then J = 0 in the open neighborhood D(f) ⊂ Spec A of p, implying that (i)holds.
!
This proof can be made notably easier by working in characteristic 0, where
we can take (iii) rather than (iv) as our starting point to prove (i), using the factthat smooth schemes over perfect fields k are regular at closed points (see Exercise 21.3.D).

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25.2.5. ⋆⋆ Formally unramified, smooth, and étale.
[Gr-EGA] takes a different starting point for the definition of unramified, smooth,and étale. The definitions there make clear that these three definitions form a family.
The cost of these definitions are that they are perhaps less immediately motivated by geometry, and it is harder to show some basic properties. The benefit isthat it is possible to show more (for example, left-exactness of the relative cotangent and conormal exact sequences, and good interpretations in terms of completions of local rings). But we simply introduce these ideas here, and do not explore
them. See [BLR, §2.2] for an excellent discussion. (You should largely ignore what
follows, unless you find later in life that you really care.)25.2.6. Definition. We say that π : X → Y is formally smooth (resp. formallyétale, formally unramified) if for all affine schemes Z, and every closed subscheme
Z0 ⊂ Z defined by a nilpotent ideal, and every morphism Z → Y, the canonicalmap HomY (Z, X) → HomY (Z0 , X) is surjective (resp. bijective, injective). Thisis summarized in the following diagram, which is reminiscent of the valuative
criteria for separatedness and properness.
Z0" #nilpotent ideal

"
Z

?

%X
A
π

"
%Y

(You can check that this is the same as the definition we would get by replacing“nilpotent” by “square-zero”. This is sometimes an easier formulation to workwith.)
Then [Gr-EGA] defines smooth as morphisms that are formally smooth andlocally of finite presentation, and similarly for unramified and étale.
One can show that [Gr-EGA]’s definitions of formally unramified, and smoothagree with the definitions we give. For “formally unramified” (where our definition given in §21.6 is not obviously the same as the definition of [Gr-EGA] givenhere), see [Gr-EGA, IV4 .17.2.1] or [Stacks, tag 00UO]. For “smooth”, see [Gr-EGA,
IV4 .17.5.1] or [Stacks, tag 00TN]. (Our characterization of étale as smooth of relative dimension 0 then agrees with [Gr-EGA]. Our definition of unramified asformally unramified plus locally of finite type disagrees with [Gr-EGA], as mentioned in §21.6.)
25.2.7. Other starting points. Unlike many other definitions in algebraic geometry, there are a number of quite different ways of defining smooth, étale, andunramified morphisms, and it is nontrivial to relate them. We have just described
the approach of [Gr-EGA]. Another common approach is the characterization ofsmooth morphisms as locally finitely presented, flat, and with regular geometric fibers (see Theorem 25.2.2). Yet another definition is via a naive version of
the cotangent complex; this is the approach taken by [Stacks, tag 00T2], and isless frightening than it sounds. Finally, the different characterizations of Exercise 25.2.C give a number of alternate initial definitions of étaleness.

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25.3 Generic smoothness and the Kleiman-Bertini Theorem
We will now discuss a number of important results that fall under the rubricof “generic smoothness”, an idea we first met in §21.3.4. All will require working
over a field of characteristic 0 in an essential way.
25.3.1. Theorem (generic smoothness on the source). — Let k be a field of characteristic 0, and let π : X → Y be a dominant morphism of integral k-varieties. Then there is anonempty (=dense) open set U ⊂ X such that π|U is smooth of dimension dim X − dim Y.
The key idea already appeared in Theorem 21.3.5, when we showed that every
variety has an open subset that is regular.
Proof. Define n = dim X − dim Y (the “relative dimension”). Now K(X)/K(Y) is a
finitely generated field extension of transcendence degree n (from transcendence
theory, see for example Exercise 11.2.A), so ΩX/Y has rank n at the generic point,
by Exercise 21.2.M(b). (Here we use the hypothesis char k = 0, as we are using
the fact that K(X)/K(Y) is separably generated by any transcendence basis.) By
upper semicontinuity of fiber rank of a coherent sheaf (Exercise 13.7.J), it is rank nfor every point in a dense open set. On a reduced scheme, constant rank implies
locally free of that rank (Exercise 13.7.K), so ΩX/Y is locally free of rank n on that
set. Also, by generic flatness (Exercise 24.5.M), it is flat on a dense open set. Let U
be the intersection of these two open sets.
!
25.3.2. Example. An examination of the proof yields an example showing that thisresult fails in positive characteristic: consider the purely inseparable extension
Fp (t)/Fp (tp ). The same problem can arise even over an algebraically closed field of
characteristic p: consider A1k = Spec k[t] → Spec k[u] = A1k , given by u (→ tp .
If furthermore X is smooth, the situation is even better.

25.3.3. Theorem (generic smoothness on the target). — Suppose π : X → Y is amorphism of k-varieties, where char k = 0, and X is smooth (over k). Then there is adense open subset U of Y such that π|π−1 (U) is a smooth morphism.

Note that π−1 (U) may be empty! Indeed, if π is not dominant, we will have totake such a U.
To prove Theorem 25.3.3, we use a neat trick. Suppose π : X → Y is a morphismof schemes that are finite type over k, where char k = 0. Define
67)*D
Xr = p ∈ X : rank π∗ ΩY/k → ΩX/k Dp ≤ r .

25.3.A. E XERCISE . Show that Xr is a closed subset of X. Hint: If X and Y are bothsmooth, show that the rank condition implies that Xr is cut out by “determinantal
equations”, or use an appropriate form of uppersemicontinuity.
By Chevalley’s Theorem 7.4.2, π(Xr ) is constructible, so we can make sense ofdim π(Xr ).

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25.3.4. Lemma. — If π : X → Y is a morphism of schemes that are finite type over k,where char k = 0, then dim π(Xr ) ≤ r.

Proof. In this proof, we make repeated use of the identification of Zariski cotangent
spaces at closed points with the fibers of cotangent sheaves, using Corollary 21.3.9.
We can replace X by an irreducible component of Xr , and Y by the closure of thatcomponent’s image in Y (with reduced subscheme structure, see §8.3.9). The resulting map will have all of X contained in Xr , so we may as well assume X = Xr . Thisboils down to the following linear algebra observation: if a linear map ρ : V1 → V2
has rank at most r, and Vi′ is a subspace of Vi , with ρ sending V1′ to V2′ , thenthe restriction of ρ to V1′ has rank at most that of ρ itself.) Thus we have a dominant morphism π : X → Y, and we wish to show that dim Y ≤ r. Using genericsmoothness on the source (Theorem 25.3.1) for Y → Spec k, we can shrink Y further so as to assume that it is smooth. By generic smoothness on the source forπ : X → Y, there is a nonempty open subset U ⊂ X such that π : U → Y is smooth.
But then for any point p ∈ U, the cotangent map ΩY |π(p) → ΩX |p is injective(Exercise 21.2.S), and has rank at most r. Taking p to be a closed point, we havedim Y = dimπ(p) Y ≤ dim ΩY |π(p) ≤ r.
!
There is not much left to do to prove the theorem.

25.3.5. Proof of Theorem 25.3.3. Reduce to the case Y smooth over k (by restricting toa smaller open set, using generic smoothness of Y, Theorem 25.3.1). Say n = dim Y.
Now dim π(Xn−1 ) ≤ n − 1 by Lemma 25.3.4 , so remove π(Xn−1 ) from Y as well.
Then the rank of Ωπ is at least n for each closed point of X. But as Y is regular of
dimension n, we have that Ωπ = (π∗ ΩY/k → ΩX/k ) is injective for every closed
point of X, and hence π∗ ΩY/k → ΩX/k is an injective map of sheaves (do you seewhy?). Thus π is smooth by Exercise 25.2.F(b).
!

25.3.6. ⋆⋆ The Kleiman-Bertini Theorem. The same idea of bounding the dimension of some “bad locus” can be used to prove the Kleiman-Bertini Theorem 25.3.7 (due to Kleiman), which is useful in (for example) enumerative geometry. Throughout this discussion k = k, although the definitions and results can begeneralized. Suppose G is a group variety (over k = k), and we have a G-action ona variety X. We say that the action is transitive if it is transitive on closed points.
A better definition (that you can show is equivalent) is that the action morphism
G × X → X restricted to a fiber above any closed point of X is surjective. A variety
X with a transitive G-action is said to be a homogeneous space for G. For example,
G acts on itself by left-translation, and via this action G is a homogeneous spacefor itself.
25.3.B. E ASY E XERCISE . Suppose G is a group variety over an algebraically closed
field k of characteristic 0. Show that every homogeneous space X for G is smooth.
(In particular, taking X = G, we see that G is smooth.) Hint: X has a dense open set
U that is smooth by Theorem 25.3.1, and G acts transitively on the closed points ofX, so we can cover X with translates of U.
25.3.7. The Kleiman-Bertini Theorem, [Kl2, Thm. 2]. — Suppose X is homogeneous
space for a group variety G (over a field k = k of characteristic 0). Suppose α : Y → X andβ : Z → X are morphisms from smooth k-varieties Y and Z.

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683(a) Then there is a nonempty open subset V ⊂ G such that for every σ ∈ V(k), Y ×X Z
defined by
Y ×X Z

%Z

"
Y

"
%X

β
σ◦α

(Y is “translated by σ”) is smooth of dimension dim Y + dim Z − dim X.
(b) Furthermore, there is a nonempty open subset V ⊂ G such that(25.3.7.1)(G ×k Y) ×X Z → G

is a smooth morphism of relative dimension dim Y + dim Z − dim X over V.
The first time you hear this, you should think of the special case where Y → X
and Z → X are locally closed embeddings (Y and Z are smooth subvarieties of X).
In this case, the Kleiman-Bertini theorem says that the second subvariety will meeta “general translate” of the first “transversely”.
Proof. It is more pleasant to describe this proof “backwards”, by considering how
we would prove it ourselves. We will use generic smoothness twice.
Clearly (b) implies (a), so we prove (b).
In order to show that the morphism (25.3.7.1) is smooth over a nonempty
open set V ⊂ G, it would suffice to apply generic smoothness on the target (Theorem 25.3.3) to (25.3.7.1). Hence it suffices to show that (G ×k Y) ×X Z is a smooth
k-variety. Now Z is smooth over k, so it suffices to show that (G ×k Y) ×X Z → Z
is a smooth morphism (as the composition of two smooth morphisms is smooth,
Exercise 12.6.D). But this is obtained by base change from G×k Y → X, so it suffices
to show that this latter morphism is smooth (as smoothness is preserved by basechange).
Now G×k Y → X is a G-equivariant morphism. (By “G-equivariant”, we meanthat the G-action on both sides respects the morphism.) By generic smoothness ofthe target (Theorem 25.3.3), this is smooth over a dense open subset X. But thenby transitivity of the G-action on X, this morphism is smooth everywhere.
!
25.3.C. E XERCISE ( POOR MAN ’ S K LEIMAN -B ERTINI ). Prove Theorem 25.3.7(a)without the hypotheses on k (on algebraic closure or characteristic), and without
the smoothness in the conclusion. Hint: This is a question about dimensions offibers of morphisms, so you could have solved this after reading §11.4.
25.3.D. E XERCISE ( IMPROVED CHARACTERISTIC 0 B ERTINI ).
Suppose Z is asmooth k-variety, where char k = 0 and k = k. Let V be a finite-dimensional
base-point-free linear series on Z, i.e., a finite vector space of sections of some invertible sheaf L on Z. Show that a general element of V, considered as a closed
subscheme of Z, is regular. (More explicitly: each element s ∈ V gives a closed subscheme of Z. Then for a general s, considered as a point of PV, the corresponding
closed subscheme is smooth over k.) Hint: figure out what this has to do with the
Kleiman-Bertini Theorem 25.3.7. Let n = dim V, G = GL(V), X = P|V ∨ |, take Z in
Kleiman-Bertini to be the Z of the problem, and let Y be the “universal hyperplane”
over P|V ∨ | (the incidence variety I ⊂ PV × PV ∨ of Definition 12.4.1).

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25.3.E. E ASY E XERCISE . Interpret Bertini’s Theorem 12.4.2 over a characteristic 0
field as a corollary of Exercise 25.3.D.
In characteristic 0, Exercise 25.3.D is a good improvement on Bertini’s Theorem. For example, we don’t need L to be very ample, or X to be projective. But
unlike Bertini’s Theorem, Exercise 25.3.D fails in positive characteristic, as shownby the one-dimensional linear series {pQ : Q ∈ P1 }. This is essentially Example 25.3.2. (Do you see why this does not contradict Bertini’s Theorem 12.4.2?)

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CHAPTER 26

Depth and Cohen-Macaulayness
We now introduce the notion of depth. Depth is an algebraic rather than geometric concept, so we concentrate on developing some geometric sense of what
it means. Most important is the geometric bound on depth by dimension of (closures of) associated points (Theorem 26.1.2). A central tool to understanding depth
is the Koszul complex, but we avoid this approach, as we can prove what we needdirectly.
When the depth of a local ring equals its dimension, the ring is said to be
Cohen-Macaulay, and this is an important way in which schemes can be “nice”.
For example, regular local rings are Cohen-Macaulay (§26.2.5), as are regular embeddings in smooth varieties (Proposition 26.2.6). Cohen-Macaulayness will be
key to the proof of Serre duality in Chapter 30, through the Miracle Flatness Theorem 26.2.11.
Another application of depth is Serre’s R1 + S2 criterion for normality (Theorem 26.3.2), which we will use to prove that regular schemes are normal (§26.3.5)
without having to show that regular local rings are factorial (the Auslander-Buchsbaum
Theorem, Fact 12.8.5), and to prove that regular embeddings in smooth schemes
are normal if they are regular in codimension 1 (§26.3.3).

26.1 Depth
Recall the theory of regular sequences from §8.4.3.
26.1.1. Definition. Suppose (A, m) is a Noetherian local ring, and M is a finitely
generated A-module. The depth of M (denoted depth M) is the length of thelongest M-regular sequence with elements in m. (More generally, if R is a Noetherian ring, I ⊂ R is an ideal, and M is a finitely generated R-module, then the
I-depth of M, denoted depthI M, is the length of the longest M-regular sequencewith elements in I. We won’t need this.)26.1.A. I MPORTANT E XERCISE . Suppose M is a nonzero finitely generated module
over a Noetherian local ring (A, m). Show that depth M = 0 if and only if every
element of m is a zerodivisor of M if and only if m is an associated prime of M.
26.1.B. E XERCISE . Suppose M is a finitely generated module over a Noetherian
local ring A. Show that depth M ≤ dim Supp M. In particular,(26.1.1.1)

depth A ≤ dim A.

Hint: Krull’s Principal Ideal Theorem 11.3.3. (We will improve this result in Theorem 26.1.6.)
685

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At this point, it is hard to determine the depth of an A-module M. You canstart trying to build an M-regular sequence by successively choosing x1 , x2 , . . . ,but how do you know you have made the right choices to find the longest one?
The happy answer is that you can’t go wrong; this is the content of the next result.
(We then describe how to find the depth of a module M in practice, in §26.1.4.)
26.1.2. Theorem. — Suppose M is a finitely generated module over a Noetherian local
ring (A, m). Then all maximal M-regular sequences contained in m (M-regular sequences
x1 , . . . , xn in m that cannot be extended to x1 , . . . , xn+1 ) have the same length. Thus thedepth of M is the length of any maximal M-regular sequence.
We prove Theorem 26.1.2 by giving a cohomological criterion, Theorem 26.1.3,
for a regular sequence to be maximal. (You will then prove Theorem 26.1.2 in Exercise 26.1.D.) We will also use this criterion to give a better bound on the depth inTheorem 26.1.6. Theorem 26.1.2 is the key technical result of this Chapter. An important moral is that depth should be understood as a “cohomological” property.
26.1.3. Theorem (cohomological criterion for existence of regular sequences). —Suppose (A, m) is a Noetherian local ring, and M is a finitely generated A-module. The
following are equivalent.
(i) For every finitely generated A-module N with Supp N = {[m]} (i.e., N has finite
length), ExtiA (N, M) = 0 for all i < n.
(ii) ExtiA (A/m, M) = 0 for all i < n.
(iii) There exists an M-regular sequence in m of length n.
This result can be extended in various ways, see for example [Mat1, Thm. 28].
Proof. Clearly (i) implies (ii).
26.1.C. E XERCISE . Prove that (ii) implies (i). Hint: apply Exercise 5.5.M to N,
so that it admits a filtration with each subquotient isomorphic to A/m, or see Exercise 18.4.R. Use induction on the length of the filtration, and the long exact sequence for ExtiA (·, M).
Proof that (iii) implies (ii). The case n = 0 is vacuous. We inductively prove theresult for all n. Suppose (iii) is satisfied, where n ≥ 1, and assume that we know(ii) for “all smaller n”. Choose a regular sequence x1 , . . . , xn of length n. Then x1
is a non-zerodivisor on M, so we have an exact sequence(26.1.3.1)

0

%M

×x1

%M

% M/x1 M% 0.

Then M/x1 M has a regular sequence x2 , . . . , xn−1 of length n − 1, so by the inductive hypothesis, ExtiA (A/m, M/x1 M) = 0 for i < n − 1. Taking the Ext long exact
sequence for ExtiA (A/m, ·) for (26.1.3.1), we find that
ExtiA (A/m, M)

×x1% ExtiA (A/m, M)

is an injection for i < n. Now ExtiA (A/m, M) can be computed by taking an injective resolution of M, and applying HomA (A/m, ·). Hence as x1 lies in m (andthus annihilates A/m), multiplication by x1 is the zero map. Thus (ii) holds for n
as well.

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687

Proof that (ii) implies (iii). The case n = 0 is vacuous.
We deal next with the case n = 1, by showing the contrapositive. Assume
that there are no non-zerodivisors in m on M, so by Exercise 26.1.A, m is an associated prime of M. Thus from §5.5.9 we have an injection A/m "→ M, yielding
HomA (A/m, M) ̸= 0 as desired.
We now inductively prove the result for all n > 1. Suppose (ii) is satisfied,
where n ≥ 2, and assume that we know (iii) for “all smaller n” . Then by the casen = 1, there exists a non-zerodivisor x1 on M, so we have a short exact sequence(26.1.3.1). A portion of the Ext long exact sequence for ExtiA (A/m, ·) for (26.1.3.1)
isExtiA (A/m, M)% ExtiA (A/m, M/x1 M)% Exti+1 (A/m, M).
A

By assumption, both ExtiA (A/m, M) and Exti+1
A (A/m, M) are 0 for i < n − 1, so
ExtiA (A/m, M/x1 M) = 0 for i < n − 1, so by the inductive hypothesis, we have an(M/x1 M)-regular sequence x2 , . . . , xn of length n − 1 in m. Adding x1 to the frontof this sequence, we are done.
!
26.1.D. E XERCISE . Prove Theorem 26.1.2. Hint: by Theorem 26.1.3 (notably, theequivalence of (ii) and (iii)), you have control of how long an M-regular sequencein m can be. Use the criterion, and the long exact sequence used in the proof of
Theorem 26.1.3, to show that any M-regular sequence in m can be extended to thislength.
26.1.E. E XERCISE . Suppose M is a finitely generated module over a Noetherian local ring (A, m). If x is a non-zerodivisor in m, show that depth(M/xM) =depth M − 1. Hint: Theorem 26.1.2.
26.1.4. Finding the depth of a module. We can now compute the depth of M bysuccessively finding non-zerodivisors, as follows. Is there a non-zerodivisor x on
M in m?
(a) If not, then depth M = 0 (Exercise 26.1.A).
(b) If so, then choose any such x, and (using the previous exercise) repeat theprocess with M/xM.
The process must terminate by Exercise 26.1.B.
26.1.F. I MPORTANT E XERCISE . Suppose (A, m) is a dimension d regular local ring.
Show that depth A = d. Hint: Exercise 12.2.B.
26.1.G. E XERCISE ( CF. E XERCISE 24.4.I). Suppose X = Spec R, whereR = k[w, x, y, z]/(wy, wz, xy, xz),
the union of two co-ordinate two-planes in A4k meeting at the origin. Show thatthe depth of the local ring of X at the origin is 1. Hint: Show that w − y is not a
zerodivisor, and that R/(w − y) has an embedded point at the origin.
26.1.5. Depth is bounded by the dimension of associated prime ideals.
Theorem 26.1.3 can be used to give an important improvement of the bound(26.1.1.1) on depth by the dimension:

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26.1.6. Theorem. — The depth of a module M is at most the smallest dim A/p as p runs
over the associated prime ideals of M.
(The example of two planes meeting at a point in Exercise 26.1.G shows that
this bound is not sharp.) The key step in the proof of Theorem 26.1.6 is the following result of Ischebeck.
26.1.7. Lemma. — Suppose (A, m) is a Noetherian local ring, and M and N are nonzero
finitely generated A-modules. Then ExtiA (N, M) = 0 for i < depth M − dim Supp N.
Proof. Consider the following statements.
(generalr ) Lemma 26.1.7 holds for dim Supp N ≤ r.
(primer ) Lemma 26.1.7 holds for dim Supp N ≤ r and N = A/p for some prime p.

Note that (general0 ) (and hence (prime0 )) is true, as in this case Supp N = {[m]},and the result follows from Theorem 26.1.3 (from “(iii) implies (i)”).
26.1.H. E XERCISE . Use Exercise 5.5.M, and Exercise 5.5.L (or the easier fact that if
0 → M ′ → M → M ′′ → 0 is a short exact sequence of A-modules, then Supp M =Supp M ′ ∪ Supp M ′′ ), to show that (primer ) implies (generalr ).

We conclude the proof by showing that (generalr−1 ) implies (primer ) for r ≥ 1.
Fix a prime p ⊂ A with dim A/p = r. Since dim A/p > 0, we can choose x ∈ m \ p.
Consider the exact sequence(26.1.7.1)

0

% A/p

×x

% A/p

% A/(p + (x))% 0,noting that dim Supp A/(p+(x)) ≤ r−1 (do you see why?). Then the Ext long exact
sequence obtained by applying HomA (·, M) to (26.1.7.1), along with the vanishingof ExtiA (A/(p + (x))), M) for i < depth M − r + 1 (by (generalr−1 )), implies that
ExtiA (A/p, M)

×x% ExtiA (A/p, M)

is an isomorphism for i < depth M − r. But ExtiA (A/p, M) is a finitely generated
A-module (Exercise 23.2.F), so by Nakayama’s Lemma 7.2.8, ExtiA (A/p, M) = 0 for
i < depth M − r.
!
26.1.I. E ASY E XERCISE . Prove Theorem 26.1.6. Hint: from §5.5.9, if p ∈ Ass(M),then HomA (A/p, M) ̸= 0.

26.2 Cohen-Macaulay rings and schemes
26.2.1. Definition. A Noetherian local ring (A, m) is Cohen-Macaulay (or often
CM for short) if depth A = dim A, i.e., if equality holds in (26.1.1.1). (One may
define Cohen-Macaulay module similarly, but we won’t need this concept.) A
locally Noetherian scheme is Cohen-Macaulay if all of its local rings are CohenMacaulay.

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26.2.A. E XERCISE . Show that every locally Noetherian scheme of dimension 0 is
Cohen-Macaulay. Show that a locally Noetherian scheme of dimension 1 is CohenMacaulay if and only if it has no embedded points.
26.2.2. (Counter)example. Let X be the example of Exercise 26.1.G — two planes
meeting at a point. By Exercise 26.1.G, X is not Cohen-Macaulay.
26.2.B. E XERCISE . Suppose A is Cohen-Macaulay Noetherian local ring. Use
Theorem 26.1.6 to show that Spec A is equidimensional, and has no embedded
points. (It is not true that Noetherian local rings of pure dimension having noembedded prime ideals are Cohen-Macaulay, see Example 26.2.2.)
26.2.3. Theorem (slicing criterion for Cohen-Macaulayness). — Suppose (A, m) isa Noetherian local ring, and x ∈ m is a non-zerodivisor. Then (A, m) is Cohen-Macaulay
if and only if (A/x, m) is Cohen-Macaulay.
Compare this to the slicing criteria for regularity and flatness (Exercise 12.2.C
and Theorem 24.6.5 respectively).
26.2.4. Geometric interpretation of the slicing criterion. Suppose X is a locally Noetherian scheme, and D is an effective Cartier divisor. If X is Cohen-Macaulay, thenso is D. If D is Cohen-Macaulay, then X is Cohen-Macaulay at the points of D.
26.2.C. E XERCISE . Prove Theorem 26.2.3, using Theorem 26.1.2, the fact that maximal regular sequences (in m) all have the same length.
26.2.D. E XERCISE . Show that if (A, m) is Cohen-Macaulay, then a set of elements
x1 , . . . , xr ∈ m is a regular sequence (for A) if and only if dim A/(x1 , . . . , xr ) =dim A − r.
26.2.5. By Exercise 26.1.F, regular local rings (Noetherian by definition) are CohenMacaulay. In particular, as smooth schemes over a field k are regular (Theorem 12.2.10(b)),
we see that smooth k-schemes are Cohen-Macaulay. (In combination with Exercise 26.2.B, this explains why effective Cartier divisors on smooth varieties haveno embedded points, which justifies a comment made in Aside 20.2.2.) Combining
this with Exercise 26.2.D or §26.2.4, we have the following.
26.2.6. Proposition. — Regular embeddings in smooth k-schemes are Cohen-Macaulay.
26.2.7. As a consequence of Proposition 26.2.6 and the fact that Cohen-Macaulay
schemes have no embedded points (Exercise 26.2.B), we see that regular embedn
dings in smooth k-schemes (in An
k or Pk for example) have no embedded points,
generalizing Exercise 5.5.I (the hypersurface in An
k case). This is not clear without the theory of Cohen-Macaulayness! (This fact was used in Exercise 25.2.A, see§25.2.1.)
26.2.8. Alternate definition of Cohen-Macaulayness.
The slicing criterion (Theorem 26.2.3) gives an enlightening alternative inductive definition of Cohen-Macaulayness
in terms of effective Cartier divisors, in the spirit of the method of §26.1.4 for computing depth. Suppose as before that (A, m) is a Noetherian local ring.
(i) If dim A = 0, then A is Cohen-Macaulay (Exercise 26.2.A).

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The Rising Sea: Foundations of Algebraic Geometry(iia) If dim A > 0, and every element of m is a zerodivisor, then A is not CohenMacaulay (by Exercise 26.1.A).
(iib) Otherwise, choose any non-zerodivisor x in m. Then A is Cohen-Macaulay
if and only if A/(x) (necessarily of dimension dim A − 1 by Krull’s Principal Ideal Theorem 11.3.3) is Cohen-Macaulay.
The following example could have been stated (but not proved) before we
knew any algebraic geometry at all. (We work over C rather than over an arbitrary
field simply to ensure that the statement requires as little background as possible.)
26.2.E. F UN E XERCISE (M AX N OETHER ’ S AF + BG T HEOREM ). Suppose f, g ∈C[x0 , x1 , x2 ] are two homogeneous polynomials, cutting out two curves in P2C thatmeet “transversely”, i.e., at a finite number of reduced points. Suppose h ∈C[x0 , x1 , x2 ] is a homogeneous polynomial vanishing at these points. Show thath ∈ (f, g). Hint: show that the intersection of the affine cones V(f) and V(g) in
A3 has no embedded points. (This problem is quite nontrivial to do without thetheory developed in this chapter! As a sign that this is subtle: you can easily construct three quadratics e, f, g ∈ C[x0 , x1 , x2 ] cutting out precisely the two points[1, 0, 0] and [0, 1, 0] ∈ P2C , yet the line z = 0 is not in the ideal (e, f, g) for degree
reasons.)
26.2.9. Miracle flatness.
We conclude with a remarkably simple and useful criterion for flatness, which
we shall use in the proof of Serre duality. The main content is the following algebraic result.
26.2.10. Miracle Flatness Theorem (algebraic version). — Suppose φ : (B, n) →(A, m) is a (local) homomorphism of Noetherian local rings, such that A is Cohen-Macaulay,
and B is regular, and A/nA = A ⊗B (B/n) (the ring corresponding to the fiber) has pure
dimension dim A − dim B. Then φ is flat.
Proof. We prove Theorem 26.2.10 by induction on dim B. If dim B = 0, then B is afield (Exercise 12.2.A), so the result is immediate, as everything is flat over a field(Exercise 24.2.A(a)). Assume next that dim B > 0, and we have proved the result
for all “B of smaller dimension”. Choose x ∈ n − n2 , so B/x is a regular local ringof dimension dim B − 1 . Thendim A/xA ≤==dim B/(x) + dim A/nA (Key Exercise 11.4.A)
dim B − 1 + dim A/nA (Krull’s Principal Ideal Theorem 11.3.3)
dim A − 1(by hypothesis of Theorem 26.2.10).

By Krull’s Principal Ideal Theorem 11.3.3, dim A/xA ≥ dim A − 1, so we have
dim A/xA = dim B/(x) + dim A/nA = dim A − 1. By Exercise 26.2.D, A/xA is a
Cohen-Macaulay ring, and x is a non-zerodivisor on A. The inductive hypothesis
then applies to (B/(x), n) → (A/xA, m), so A/xA is flat over B/(x). Then by the
local slicing criterion for flatness (Theorem 24.6.5), B → A is flat, as desired.
!

26.2.11. Miracle Flatness Theorem. — Suppose π : X → Y is a morphism of equidimensional finite type k-schemes, where X is Cohen-Macaulay, Y is regular, and the fibersof π have dimension dim X − dim Y. Then π is flat.

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26.2.F. E XERCISE . Prove the Miracle Flatness Theorem 26.2.11. (Do not forget that
schemes usually have non-closed points!)
The geometric situation in the Miracle Flatness Theorem 26.2.11 is part of thefollowing pretty package.
26.2.G. E XERCISE . Suppose π : X → Y is a map of locally finite type k-schemes,
where both X and Y are equidimensional, and Y is regular. Show that if any two of
the following hold, then the third does as well:(i) π is flat of relative dimension dim X − dim Y.
(ii) X is Cohen-Macaulay.
(iii) Every fiber Xy is Cohen-Macaulay of pure dimension dim X − dim Y.
Hint: if φ : B → A is a flat ring map, then φ sends non-zerodivisors to nonzerodivisors (Observation 24.2.2).

The statement of Exercise 26.2.G can be improved, at the expense of killing thesymmetry. In the implication (ii) and (iii) imply (i), the Cohen-Macaulay hypotheses in (iii) are not needed.
26.2.12. Example. As an example of Exercise 26.2.G in action, we consider theexample of two planes meeting a point, continuing the notation of Exercise 26.1.G.
Consider the morphism π : X → Y := A2 given by (w − y, x − z). Then Y is regular (by the Smoothness-Regularity Comparison Theorem 12.2.10(b)). But X is not
Cohen-Macaulay (Exercise 26.1.G) and π is not flat (Exercise 24.4.I), so by Exercise 26.2.G, you can use either of these to prove the other.
26.2.13. Fancy properties of Cohen-Macaulayness.
We mention a few additional properties of Cohen-Macaulayness without proof.
They are worth seeing, but we will not use them.
If X is a locally Noetherian scheme, the locus of Cohen-Macaulay points is
open, see [Mat2, Thm 24.5] (and also [Stacks, tag 00RD]). In particular:
26.2.14. Fact (cf. Fact 12.8.5). — Any localization of a Cohen-Macaulay Noetherian
local ring at a prime ideal is also Cohen-Macaulay.
(See [Stacks, tag 00NB], [E, Prop. 18.8], [Mat2, Thm. 17.3(iii)], or [Mat1, Thm. 30]
for a direct proof.) In particular, for quasicompact schemes, Cohen-Macaulayness
may be checked at closed points. The fact that Cohen-Macaulayness is preserved
by localization can be used to quickly show that Cohen-Macaulay local rings arecatenary (see [E, Cor. 18.10], [Mat1, Thm. 31(ii)], or [Mat2, Thm. 17.9]). A ring A
is Cohen-Macaulay if and only if A[x] is Cohen-Macaulay ([E, Prop 18.9], [Mat2,
Thm 17.7], [Mat1, Thm. 33]), if and only A[[x]] is [Mat2, p. 137]. A local ring is
Cohen-Macaulay if and only if its completion is Cohen-Macaulay ([E, Prop. 18.8],[Mat2, Thm 17.5]).

26.3 ⋆ Serre’s R1 + S2 criterion for normality
The notion of depth yields a useful criterion for normality, due to Serre.

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26.3.1. Definition. Suppose A is a Noetherian ring, and k ∈ Z≥0 . We say A has
property Rk (A is regular in codimension ≤ k, or more sloppily, A is regular incodimension k) if for every prime p ⊂ A of codimension at most k, Ap is regular.
(In light of Fact 12.8.2, a Noetherian local ring is regular and only if it has property
Rk for all k.)
We say A has property Sk if for every prime p ⊂ A, the local ring Ap has
depth at least min(k, dim Ap ) — “the local rings are Cohen-Macaulay up until
codimension k, and have depth at least k thereafter”. (In light of Fact 26.2.14, a
Noetherian local ring is Cohen-Macaulay if and only if it has property Sk for allk.)
26.3.A. E ASY EXERCISE . Note that a Noetherian ring A trivially always has property S0 . Show that a Noetherian ring A has property R0 if Spec A is “genericallyreduced”: it is reduced at the generic point of each of its irreducible components.
Show that a Noetherian ring A has property S1 if and only if Spec A has no embedded points. (Possible hint: Exercise 26.1.A.)
26.3.B. E XERCISE . Show that a Noetherian ring A has no nilpotents (Spec A is
reduced) if and only if it has properties R0 and S1 . (Hint: show that a Noetherian
scheme is reduced if and only if each irreducible component is generically reduced,
and it has no embedded points. Another hint: Exercise 5.5.K.)
Incrementing the subscripts in Exercise 26.3.B yields Serre’s criterion.
26.3.2. Theorem (Serre’s criterion for normality). — A Noetherian ring A is normal
if and only if it has properties R1 and S2 .
(Recall that a Noetherian ring A is normal if its localizations Ap at all primes
p are integrally closed domains.) Thus failure of normality can have two possible
causes: it can be failure of R1 (something we already knew, from the equivalenceof (a) and (g) in Theorem 12.5.8), or it can be the more subtle failure of S2 . Examples of varieties satisfying R1 but not S2 are given in Example 26.3.4 (two planes
meeting at a point) and Exercise 26.3.D (the pinched plane).
26.3.3. Applications. As usual, before giving a proof, we give some applications
to motivate the result. First, it implies that Cohen-Macaulay schemes are normal
if and only if they are regular in codimension 1. Thus to check normality of hypersurfaces (or more generally, regular embeddings) in Pnk (or more generally, in
any smooth variety), it suffices to check that their singular locus has codimension greater than one. (You should think through the details of why these statements are true.) In particular, this gives a new (more complicated) proof of Exercise 5.4.I(b) and (c).
26.3.C. E XERCISE ( PRACTICE WITH THE CONCEPT ). Show that two-dimensional
normal varieties are Cohen-Macaulay.
26.3.4. Example: two planes meeting at a point. The variety X of Exercise 26.1.G (twoplanes meeting at a point) is not normal (why?), but it is regular away from theorigin. This implies that X does not have property S2 (and hence is not CohenMacaulay), without the algebraic manipulations of Exercise 26.1.G.

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We already knew that this example was not Cohen-Macaulay, but the sameidea can show that the pinched plane Spec k[x3 , x2 , xy, y] (appearing in Exercise 12.5.I)
is not Cohen-Macaulay. Because of the “extrinsic” description of the ring, it is difficult to do this in another way.
26.3.D. E XERCISE : THE PINCHED PLANE IS NOT C OHEN -M ACAULAY. Let A be
the subring k[x3 , x2 , xy, y] of k[x, y]. Show that A is not Cohen-Macaulay. Hint:
Exercise 12.5.I showed that A is not integrally closed.
26.3.5. Regular local rings are integrally closed. Serre’s criterion can be used toshow that regular local rings are integrally closed without going through the hard
Fact 12.8.5 (the Auslander-Buchsbaum Theorem) that they are unique factorization domains. Regular local rings are Cohen-Macaulay (§26.2.5), and (by Exercise 26.3.B) regular in codimension 0 as they are integral domains (Theorem 12.2.13),
so we need only show that regular local rings are regular in codimension 1. We
can invoke a different hard Fact 12.8.2 that localizations of regular local rings areagain regular, but at least we have shown this for localizations of finitely generated
algebras over a perfect field (see Theorem 12.8.3 and Exercise 21.3.G).
26.3.6. Caution. As is made clear by the following exercise, the condition S2 is acondition on all prime ideals, not just those of codimension at most 2.
26.3.E. E XERCISE . Give an example of a variety satisfying R1 , and Cohen-Macaulay
at all points of codimension at most 2, which is not normal.
26.3.7. Proof of Theorem 26.3.2. The proof of Serre’s criterion will take us until
the end of this section.
26.3.8. Normal implies R1 + S2 .
Suppose first that A is normal. It is reduced, and thus satisfies (R0 and) S1 by
Exercise 26.3.B. It satisfies R1 from the equivalence of (a) and (g) in Theorem 12.5.8.
All that is left is to verify property S2 . We thus must show that if (A, m) is a normal
Noetherian local ring of dimension greater than 1, then A has depth at least 2.
Choose any nonzero x ∈ m. (Don’t forget that normality implies that the local
ring A is an integral domain.) We wish to show that depth A/(x) > 0. Assume
otherwise that depth A/(x) = 0, so (by Exercise 26.1.A) m is an associated prime ofA/(x). Thus there is some nonzero y ∈ A/(x) (i.e., y ∈/ (x)) with ym = 0 in A/(x),i.e., multiplying any element z ∈ m by y gives you a multiple of x. Let w = y/x ∈ K(A).
We will show that w ∈ A, and thus that y ∈ (x), yielding a contradiction.
For each prime ideal p ⊂ A of codimension 1, choose any z ∈ m \ p. Thenzw = zy/x ∈ (y/x)m ⊂ A by the italicized statement in the previous paragraph,so w ∈ Ap . Thus w ∈ Ap for all codimension 1 prime ideals p ⊂ A, so by Algebraic
Hartogs’s Lemma 11.3.11 (using the hypothesis that A is integrally closed), w ∈ A
as desired.
26.3.9. R1 + S2 implies normal: the integral domain case. Normality is a stalk-local
condition by definition, so it suffices to show:(†) if (A, m) is a dimension d Noetherian local ring satisfying R1 and S2 thenit is an integrally closed integral domain.
(As A satisfies R0 and S1 , A is reduced by Exercise 26.3.B.)

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The case in which A is an integral domain is notably easier, and may help
motivate the statement of the criterion, so we deal with this case first, and leave thegeneral case as something to read when you have too much time on your hands.
Note that this case suffices for many of the consequences we discuss, notably thenormality of regular schemes (§26.3.5).
Suppose now that A is an integral domain. Suppose x ∈ K(A) is integral over
A. We must prove that x ∈ A. Write x = f/g, where f, g ∈ A. We wish to show thatf ∈ (g), or equivalently that f restricts to 0 on the closed subscheme W := V(g) ofSpec A.
26.3.F. E XERCISE . Show that W := V(g) has no embedded points. Translation:
A/(g) satisfies S1 .
Let η1 , . . . , ηm be the generic points of the irreducible components of W, corresponding to prime ideals p1 , . . . , pm of A. By Exercise 26.3.F, η1 ,. . . , ηm are all
the associated points of W, so the natural map(26.3.9.1)

A/(g) →m&

OW,ηi =i=1

m&

(A/(g))pi

i=1

is an injection, by Theorem 5.5.10(b).
The ring Api is regular in codimension 1, so by parts (a) and (g) of Theorem 12.5.8, Api is integrally closed. Now x is integral over A and thus over Api , soby the integral closedness of Api , x ∈ Api . Hence f is 0 in each Api /(g) = (A/(g))pi .
From the injectivity of (26.3.9.1), we have that f is 0 on W as desired.
26.3.10. ⋆⋆ R1 + S2 implies normal: the general case. We prove (†) by induction on d.
26.3.G. E XERCISE . Prove the desired result if d = 0 and d = 1.
Our next step is to show that A is integrally closed in its total fraction ring,
the localization of A at the multiplicative subset of non-zerodivisors (defined in§5.5.7). Suppose we have f, g ∈ A, with g a non-zerodivisor, such that f/g satisfies
the monic equation(26.3.10.1)(f/g)n +n.

ai (f/g)n−i = 0

i=1

with ai ∈ A. We prove f ∈ (g) by induction on d. The cases d = 0 and 1 follow
from Exercise 26.3.G. Assume now that d > 1, and that the result is known for“smaller d”.
As depth A ≥ 2, depth A/(g) ≥ 1, so there is a non-zerodivisor t on A/(g)with t ∈ m. Consider the localization of (26.3.10.1) at all the prime ideals corresponding to points of D(t) (those prime ideals not containing t), each of which
has codimension < d. By the inductive hypothesis, f ∈ (g) in each of these localizations, so f is zero in (A/(g))t . But as t is a non-zerodivisor on A/(g), the map
A/(g) → (A/(g))t is an injection (Exercise 1.3.C), so f is zero in A/(g), as desired.
We have thus shown that A is integrally closed in its total fraction ring.
26.3.11. Lemma. — Suppose R is a reduced ring with finitely many minimal prime ideals
p1 , . . . , pn . If R is integrally closed in its total fraction ring, then R is a finite product ofintegral domains.

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26.3.12. Proof of Lemma 26.3.11. The pi are the"nassociated prime ideals of R, by
Exercise 5.5.C, so the natural map φ : R →i=1 Rpi is an inclusion (by Theorem 5.5.10(b)).

26.3.H.
E XERCISE . Show that the map φ identifies the total fraction ring of R with
"n
R
.
p
i
i=1
"
Suppose ei is the ith idempotent of the total fraction ring Rpi (i.e., (0, . . . , 0, 1, 0, . . . , 0),
where the 1 is in the ith spot). Since R is integrally closed in its total fraction ring,it contains each of the ei .
26.3.I.
Conclude the proof of Lemma 26.3.11, by describing R ="n E XERCISE .
i=1 Rei , and showing that each Rei is an integrally closed integral domain. Possible hint: Remark 3.6.3.
!
We now return to our proof of Serre’s criterion. By Lemma 26.3.11, we now
know that A is a product of integral domains. But Spec A is connected (A is a local
ring!), so A is an integral domain.
!

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CHAPTER 27

Twenty-seven lines
This topic is placed in Chapter 27 for purely numerological reasons. As aresult, it invokes one fact that we will only meet in a later chapter: Castelnuovo’s
Criterion (Theorem 29.7.1) is used in the proof of Proposition 27.4.1. This is afeature rather than a bug: you needn’t read the details of the later chapters in orderto enjoy this one. In fact, you should probably read this many chapters earlier,taking key facts as “black boxes”, to appreciate the value of seemingly technicaldefinitions and theorems before you learn about them in detail. This topic is afitting conclusion to an energetic introduction to algebraic geometry, even if you
do not earlier cover everything needed to justify all details.

27.1 Introduction
Wake an algebraic geometer in the dead of night, whispering: “27”. Chances are, hewill respond: “lines on a cubic surface”.
— R. Donagi and R. Smith, [DS] (on page 27, of course)
Since the middle of the nineteenth century, geometers have been entranced by
the fact that there are 27 lines on every smooth cubic surface, and by the remarkable structure of the configuration of the lines. Their discovery by Cayley andSalmon in 1849 has been called the beginning of modern algebraic geometry, [Do,p. 55].
The reason so many people are bewitched by this fact is because it requires
some magic, and this magic connects to many other things, including fundamental
ideas we have discussed, other beautiful classical constructions (such as Pascal’s
Mystical Hexagon Theorem 19.4.4, the fact that most smooth quartic plane curves
have 28 bitangents, exceptional Lie groups, . . . ), and many themes in modern algebraic geometry (deformation theory, intersection theory, enumerative geometry,arithmetic and diophantine questions, ...). It will be particularly pleasant for us, as
it takes advantage of many of the things we have learned.
You are now ready to be initiated into the secret fellowship of the twenty-sevenlines.
27.1.1. Theorem. — Every smooth cubic surface in P3k has exactly 27 lines.
Theorem 27.1.1 is closely related to the following.
27.1.2. Theorem. — Every smooth cubic surface over k is isomorphic to P2 blown up at
6 points.
697

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F IGURE 27.1. Twenty-seven lines on a cubic surface

There are many reasons why people consider these facts magical. First, thereis the fact that there are always 27 lines. Unlike most questions in enumerative geometry, there are no weasel words such as “a general cubic surface” or “most cubic
surfaces” or “counted correctly” — as in, “every monic degree d polynomial has
d roots — counted correctly”. And somehow (and we will see how) it is precisely
the smoothness of the surface that makes it work.
Second, there is the magic that you always get the blow-up of the plane at six
points (§27.4).
Third, there is the magical incidence structure of the 27 lines, which relates to
E6 in Lie theory. The Weyl group of E6 is the symmetry group of the incidence
structure (see Remark 27.3.5). In a natural way, the 27 lines form a basis of the
27-dimensional fundamental representation of E6 .
27.1.3. Structure of this chapter. Throughout this chapter, X will be a smooth cubic surface over an algebraically closed field k. In §27.2, we establish some preliminary facts. In §27.3, we prove Theorem 27.1.1. In §27.4, we prove Theorem 27.1.2.
We remark here that the only input that §27.4 needs from §27.3 is Exercise 27.3.J.
This can be done directly by hand (see for example [Rei, §7], [Be, Thm. IV.13], or[Sh, p. 246-7]), and Theorem 27.1.2 readily implies Theorem 27.1.1, using Exercise 27.4.E. We would thus have another, shorter, proof of Theorem 27.1.1. The
reason for giving the argument of §27.3 is that it is natural given what we have

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done so far, it gives you some glimpse of some ideas used more broadly in the subject (the key idea is that a map from one moduli space to another is finite and flat),and it may help you further appreciate and digest the tools we have developed.

27.2 Preliminary facts
By Essential Exercise 14.1.C, there is a 20-dimensional vector space of cubic
forms in four variables, so the cubic surfaces in P3 are parametrized by P19 .
27.2.A. E XERCISE . Show that the any smooth cubic surface X is “anticanonicallyembedded” — it is embedded by the complete anticanonical linear series KX∨ .
Hint: the adjunction formula, Exercise 21.5.B.
27.2.B. E XERCISE . Suppose X ⊂ P3k is a smooth cubic surface. Suppose C is a
curve on X. Show that C is a line if and only if C is a (−1)-curve (Definition 20.2.7).
Hint: the adjunction formula again, perhaps in the guise of Exercise 20.2.B(a); andalso Exercise 18.6.L.
It will be useful to find a single cubic surface with exactly 27 lines:
27.2.C. E XERCISE . Show that the Fermat cubic surface(27.2.0.1)

x30 + x31 + x32 + x33 = 0

in P3C has precisely 27 lines, each of the form
x0 + ωxi = xj + ω ′ xk = 0,
where {1, 2, 3} = {i, j, k}, j < k, and ω and ω ′ are cube roots of 1 (possibly thesame). Hint: up to a permutation of coordinates, show that every line in P3 can be
written x0 = ax2 + bx3 , x1 = cx2 + dx3 . Show that this line is on (27.2.0.1) if andonly if(27.2.0.2)

a3 + c3 + 1 = b3 + d3 + 1 = a2 b + c2 d = ab2 + cd2 = 0

Show that if a, b, c, and d are all nonzero, then (27.2.0.2) has no solutions.
27.2.1. ⋆ The singular cubic surfaces give an irreducible divisor in the parameter
space P19 of all cubic surfaces. This discussion is starred not because it is hard,
but because it is not needed in the rest of the chapter. Nonetheless, it is a pretty
application of what we have learned, and it foreshadows key parts of the argumentin §27.3.
Fix a field k. Bertini’s Theorem 12.4.2 shows that the locus of smooth cubics
in the parameter space P19 of all cubics is dense and open. The proof suggests
more: that the complement ∆ of this locus is of pure codimension 1. We now
show that this is the case, and even that this “discriminant hypersurface” ∆ is also
irreducible.
27.2.D. E XERCISE . (Hint for both: recall the solution to Exercise 11.2.K.)(a) Define the incidence correspondence Y ⊂ P19 × P3 corresponding to the data ofa cubic surface X, along with a singular point p ∈ X. (This is part of the problem!

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We need Y as a scheme, not just as a set.) Let µ be the projection Y → P19 .
⑥⑥
⑥⑥
⑥
a⑥
⑥
µ

P19

Y❅
❅❅ 15
❅❅P -bundle❅❅
8P3(b) Show that Y is an irreducible smooth variety of dimension 18, by describing it
as a P15 -bundle over P3 .
27.2.E. E XERCISE . Show that there exists a cubic surface with a single singularpoint. Feel free to assume k is your favorite field; the main point is to familiarize
yourself with geometric ideas, not the peculiarities of positive characteristic fields.
(Hint if char k ̸= 2: Exercise 12.3.C.)
27.2.F. E XERCISE . Show that µ(Y) = ∆ is a closed irreducible subset of codimension 1. Hint for the codimension 1 statement: Exercise 11.4.A or Theorem 11.4.1,
in combination with Exercise 27.2.E.
Your argument will generalize with essentially no change to deal with degreed hypersurfaces in Pn .

27.3 Every smooth cubic surface (over k) has 27 lines
We are now ready to prove Theorem 27.1.1. Until Exercise 27.3.K, to avoid
distraction, we assume char k = 0. (However, the following argument carries
through without change if char k ̸= 3. The one required check — and the reason for
the restriction on the characteristic — is that Exercise 27.2.C works with C replacedby any such k.)
27.3.A. E XERCISE . (Hint for both: recall the solution to Exercise 11.2.K.)(a) Define the incidence correspondence Z ⊂ P19 × G(1, 3) corresponding to thedata of a line ℓ in P3 contained in a cubic surface X. (As in Exercise 27.2.D, this ispart of the problem.) Let π be the projection Z → P19 .
⑥⑥

π ⑥⑥

P19

⑥⑥
a⑥⑥

Z❊
❊❊
❊❊P15 -bundle❊❊❊❊M
G(1, 3)(b) Show that Z is an irreducible smooth variety of dimension 19, by describing it
as a P15 -bundle over the Grassmannian G(1, 3).
27.3.B. E XERCISE . Use the fact that there exists a cubic surface with a finite numberof lines (Exercise 27.2.C), and the behavior of dimensions of fibers of morphisms(Exercise 11.4.A or Theorem 11.4.1) to show the following.
(a) Every cubic surface contains a line, i.e., π is surjective. (Hint: show that
π is projective.)

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701(b) “Most cubic surfaces have a finite number of lines”: there is a dense
open subset U ⊂ P19 such that the cubic surfaces parametrized by closed
points of U have a positive finite number of lines. (Hint: upper semicontinuity of fiber dimension, Theorem 11.4.2.)
The following fact is the key result in the proof of Theorem 27.1.1, and oneof the main miracles of the 27 lines, that ensures that the lines stay distinct on asmooth surface. It states, informally, that two lines can’t come together without
damaging the surface. This is really a result in deformation theory: we are explicitly showing that a line in a smooth cubic surface has no first-order deformations.
27.3.1. Theorem. — If ℓ is a line in a regular cubic surface X, then {ℓ ⊂ X} is a reduced
isolated point of the fiber of π : Z → P19 .
Before proving Theorem 27.3.1, we use it to prove Theorem 27.1.1.

27.3.2. Proof of Theorem 27.1.1. Now π is a projective morphism, and over P19 \ ∆,π has relative dimension 0, and hence has finite fibers. Hence by Theorem 18.1.8,
π is finite over P19 \ ∆.
Furthermore, as Z is regular (hence Cohen-Macaulay, §26.2.5) and P19 is regular, the Miracle Flatness Theorem 26.2.11 implies that π is flat over P19 \ ∆.
Thus, over P19 \ ∆, π is a finite flat morphism, and so the fibers of π (again,
away from ∆) always have the same number of points, “counted correctly” (Remark 24.4.8). But by Theorem 27.3.1, above each closed point of P19 \ ∆, each pointof the fiber of π counts with multiplicity one (here using k = k). Finally, by Exercise 27.2.C, the Fermat cubic surface gives an example of one regular cubic surface
with precisely 27 lines, so (as P19 \ ∆ is connected) we are done.
!
We have actually shown that away from ∆, Z → P19 is a finite étale morphismof degree 27.

27.3.3. ⋆ Proof of Theorem 27.3.1. Choose projective coordinates so that the line ℓ
is given, in a distinguished affine subset (with coordinates named x, y, z), by thez-axis. (We use affine coordinates to help visualize what we are doing, although
this argument is better done in projective coordinates. On a second reading, you
should translate this to a fully projective argument.)
27.3.C. E XERCISE . Consider the lines of the form (x, y, z) = (a, b, 0) + t(a ′ , b ′ , 1)(where (a, b, a ′ , b ′ ) ∈ A4 is fixed, and t varies in A1 ). Show that a, b, a ′ , b ′ can beinterpreted as the “usual” coordinates on one of the standard open subsets of theGrassmannian (see §6.7), with [ℓ] as the origin.
Having set up local coordinates on the moduli space, we can get down tobusiness. Suppose f(x, y, z) is the (affine version) of the equation for the cubic
surface X. Because X contains the z-axis ℓ, f(x, y, z) ∈ (x, y). More generally, theline(27.3.3.1)(x, y, z) = (a, b, 0) + t(a ′ , b ′ , 1)lies in X precisely when f(a + ta ′ , b + tb ′ , t) is 0 as a cubic polynomial in t. This isequivalent to four equations in a, a ′ , b, and b ′ , corresponding to the coefficientsof t3 , t2 , t, and 1. This is better than just a set-theoretic statement:

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F IGURE 27.2. Parameters for the space of “lines near ℓ”, in termsof where they meet the z = 0 plane and the z = 1 plane
27.3.D. E XERCISE . Verify that these four equations are local equations for thescheme-theoretic fiber π−1 ([X]).
Now we come to the crux of the argument, where we use the regularity of X
(along ℓ). We have a specific question in algebra. We have a cubic surface X given
by f = 0, containing ℓ, and we know that X is regular (including “at ∞”, i.e., in P3 ).
To show that [ℓ] = V(a, a ′ , b, b ′ ) is a reduced isolated point in the fiber, we workin the ring k[a, a ′ , b, b ′ ]/(a, a ′ , b, b ′ )2 , i.e., we impose the equations(27.3.3.2)

a2 = aa ′ = · · · = (b ′ )2 = 0,

and try to show that a = a ′ = b = b ′ = 0. (It is essential that you understand
why we are setting (a, a ′ , b, b ′ )2 = 0. You can also interpret this argument interms of the derivatives of the functions involved — which after all can be interpreted as forgetting higher-order information and remembering only linear terms
in the relevant variables, cf. Exercise 12.1.G. See [Mu6, §8D] for a description ofthis calculation in terms of derivatives.)
Suppose f(x, y, z) = cx3 x3 + cx2 y x2 y + · · · + c1 1 = 0, where cx3 , cx2 y , · · · ∈ k.
Because ℓ ⊂ X, i.e., f ∈ (x, y), we have c1 = cz = cz2 = cz3 = 0. We now substitute(27.3.3.1) into f, and then apply (27.3.3.2). Only the coefficients of f of monomials
involving precisely one x or y survive:cx (a + a ′ t) + cxz (a + a ′ t)t + cxz2 (a + a ′ t)t2
+cy (b + b ′ t) + cyz (b + b ′ t)t + cyz2 (b + b ′ t)t2
= (a + a ′ t)(cx + cxz t + cxz2 t2 ) + (b + b ′ t)(cy + cyz t + cyz2 t2 )is required to be 0 as a polynomial in t. (Recall that cx , . . . , cyz2 are fixed elementsof k.) Let Cx (t) = cx +cxz t+cxz2 t2 and Cy (t) = cy +cyz t+cyz2 t2 for convenience.

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Now X is regular at (0, 0, 0) precisely when cx and cy are not both 0 (as cz = 0).
More generally, X is regular at (0, 0, t0 ) precisely if cx + cxz t0 + cxz2 t20 = Cx (t0 )and cy + cyz t0 + cyz2 t20 = Cy (t0 ) are not both zero. You should be able to quickly
check that X is regular at the point of ℓ “at ∞” precisely if cxz2 and cyz2 are notboth zero. We summarize this as follows: X is regular at every point of ℓ precisely
if the two quadratics Cx (t) and Cy (t) have no common roots, including “at ∞”.
We now use this to force a = a ′ = b = b ′ = 0 using (a + a ′ t)Cx (t) + (b +′b t)Cy (t) ≡ 0.
We deal first with the special case where Cx and Cy have two distinct roots,
both finite (i.e., cxz2 and cyz2 are nonzero). If t0 and t1 are the roots of Cx (t), thensubstituting t0 and t1 into (a + a ′ t)Cx (t) + (b + b ′ t)Cy (t), we obtain b + b ′ t0 = 0,
and b + b ′ t1 = 0, from which b = b ′ = 0. Similarly, a = a ′ = 0.
27.3.E. E XERCISE . Deal with the remaining cases to conclude the proof of Theorem 27.3.1. (It is possible to do this quite cleverly. For example, you may be able
to re-choose coordinates to ensure that Cx and Cy have finite roots.)!
27.3.4. The configuration of lines.
By the “configuration of lines” on a cubic surface, we mean the data of whichpairs of the 27 lines intersect. We can readily work this out in the special caseof the Fermat cubic surface (Exercise 27.2.C). (It can be more enlightening to use
the description of X as a blow-up of P2 , see Exercise 27.4.E.) We now show thatthe configuration is the “same” (interpreted appropriately) for all smooth cubic
surfaces.
27.3.F. E XERCISE . Construct a degree 27! finite étale map W → P19 \ ∆, thatparametrizes a cubic surface along with an ordered list of 27 distinct lines. Hint: let
W ′ be the 27th fibered power of Z over P19 \ ∆, interpreted as parametrizing a
cubic surface with an ordered list of 27 lines, not necessarily distinct. Let W be thesubset corresponding to where the lines are distinct, and show that W is open andclosed in W ′ , and thus a union of connected components of W ′ .
We now make sense of the statement that the configuration of lines on the
Fermat cubic surface (call it X0 ) is the “same” as the configuration on some othersmooth cubic surface (call it X1 ). Lift the point [X0 ] to a point w0 ∈ W. Let W ′′ bethe connected component of W containing w0 .
27.3.G. E XERCISE . Show that the morphism W ′′ → P19 \ ∆ is finite and étale (andsurjective).

Choose a point w1 ∈ W ′′ mapping to [X1 ]. Because W parametrizes a “labeling” or ordering of the 27 lines on a surface, we now have chosen an identificationof the lines on X0 with those of X1 . Let the lines be ℓ1 , . . . , ℓ27 on X0 , and let thecorresponding lines on X1 be m1 , . . . , m27 .
27.3.H. E XERCISE ( USING STARRED E XERCISE 24.7.D). Show that ℓi · ℓj = mi · mj
for all i and j.

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27.3.I. E XERCISE . Show that for each smooth cubic surface X ⊂ P3k , each line on X
meets exactly 10 other lines ℓ1 , ℓ1′ , . . . , ℓ5 , ℓ5′ on X, where ℓi and ℓi′ meet for each i,and no other pair of the lines meet.
27.3.J. E XERCISE . Show that every smooth cubic surface contains two disjointlines ℓ and ℓ ′ , such that there are precisely five other lines ℓ1 , . . . , ℓ5 meeting both ℓ
and ℓ ′ . (In fact, for any two disjoint lines ℓ and ℓ ′ in a smooth cubic surface, there
are precise five other lines meeting them; but this is strictly harder than what the
problem asks.)27.3.5. Remark: The Weyl group W(E6 ). The symmetry group of the configurationof lines — i.e., the subgroup of the permutations of the 27 lines preserving theintersection data — magically turns out to be the Weyl group of E6 , a group oforder 51840. (You know enough to at least verify that the size of the group is51840, using the Fermat surface of Exercise 27.2.C, but this takes some work.) It is
no coincidence that the degree of W ′′ over P19 \ ∆ is 51840, and the Galois groupof the Galois closure of K(Z)/K(P19 \ ∆) is isomorphic to W(E6 ) (see [H1, III.3]).
27.3.K. ⋆⋆ E XERCISE . Prove Theorem 27.3.1 in arbitrary characteristic. Begin by
figuring out the right statement of Exercise 27.3.A over Z, and proving it. Thenfollow the argument given in this section, making changes when necessary.
27.3.6. ⋆ Fano varieties of lines, and Hilbert schemes.
In Exercises 27.3.A
and 27.3.K, you constructed a moduli space of lines contained in a X, as a scheme.
Your argument can be generalized to any X ⊂ PN . This construction is called
the Fano variety of lines of X (not to be confused with the notion of a Fano variety,§21.5.5), and is an example of a Hilbert scheme.

27.4 Every smooth cubic surface (over k) is a blown up plane
We now prove Theorem 27.1.2. As stated in §27.1.3, this section is remarkably
independent from the previous one; all we will need is Exercise 27.3.J, and it ispossible to prove this in other ways.
Suppose X is a smooth cubic surface (over k). Suppose ℓ is a line on X, andchoose coordinates on the ambient P3 so that ℓ is cut out by x0 and x1 . Projection
from ℓ gives a rational map P3 ""# P1 (given by [x0 , x1 , x2 , x3 ] (→ [x0 , x1 ]), which
extends to a morphism on X. The reason is that this rational map is resolved byblowing up the closed subscheme V(x0 , x1 ) (Exercise 22.4.L). But (x0 , x1 ) cuts out
the effective Cartier divisor ℓ on X, and blowing up an effective Cartier divisor
does not change X (Observation 22.2.1).
Now choose two disjoint lines ℓ and ℓ ′ as in Exercise 27.3.J, and consider themorphism ρ : X → P1 × P1 , where the map to the first P1 is projection from ℓ,and the map to the second P1 is the projection from ℓ ′ . The first P1 can then be
identified with ℓ ′ , and the second with ℓ.
27.4.A. E XERCISE . Show that the morphism ρ is birational. Hint: given a generalpoint of (p, q) ∈ ℓ ′ × ℓ, we obtain a point of X as follows: the line pq in P3 meetsthe cubic X at three points by Bézout’s Theorem 8.2.E: p, q, and some third point

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November 18, 2017 draft705

x ∈ X; send (p, q) to x. (This idea appeared earlier in the development of the grouplaw on the cubic curve, see Proposition 19.10.3.) Given a general point x ∈ X, weobtain a point (p, q) ∈ ℓ ′ × ℓ by projecting from ℓ ′ and ℓ.
In particular, we have shown for the first time that every smooth cubic surface
over k is rational.

27.4.B. E XERCISE : ρ CONTRACTS PRECISELY ℓ1 , . . . , ℓ5 . Show that ρ is an isomorphism away from the ℓi mentioned in Exercise 27.3.J, and that each ρ(ℓi ) is a pointpi ∈ ℓ ′ × ℓ.
27.4.1. Proposition. — The morphism ρ : X → P1 × P1 is the blow-up at P1 × P1 at
the five pi .

Proof. By Castelnuovo’s Criterion (Theorem 29.7.1), as the lines ℓi are (−1)-curves(Exercise 27.2.B), they can be contracted. More precisely, there is a morphism
β : X → X ′ that is the blow-up of X ′ at five closed points p1′ , . . . , p5′ , such thatℓi is the exceptional divisor at pi′ . We wish to show that X ′ is P1 × P1 .
The morphism ρ : X → P1 ×P1 yields a morphism ρ ′ : X ′ \{p1′ , . . . p5′ } → P1 ×P1 .
We now show that ρ ′ extends over pi′ for each i, sending pi′ to pi . Choose an open
neighborhood of pi ∈ P1 × P1 isomorphic to A2 , with coordinates x and y. Thenboth x and y pull back to functions on a punctured open neighborhoods of pi′ (i.e.,there is some open neighborhood U of pi′ such that x and y are functions on U\{pi′ }).
By Algebraic Hartogs’s Lemma 11.3.11, they extend over pi′ , and this extension is
unique as P1 × P1 is separated — use the Reduced-to-Separated Theorem 10.2.2 if
you really need to. Thus ρ ′ extends over pi′ . (Do you see why ρ ′ (pi′ ) = pi ?)27.4.C. E XERCISE . Show that the birational morphism ρ ′ : X ′ → P1 × P1 is invertible. Hint: you can use Zariski’s Main Theorem (in the guise of Exercise 29.6.D),
but you needn’t use something so powerful. Instead, note that the birational map
ρ ′−1 is a morphism away from p1 , . . . , p5 . Use essentially the same argument as
in the last paragraph to extend ρ ′−1 over each pi .
!
As a consequence we see that X is the blow-up of P1 × P1 at 5 points. Because
the blow-up of P1 × P1 at one point is isomorphic to the blow-up of P2 at two
points (Exercise 22.4.K), Theorem 27.1.2 then follows.
!
27.4.2. Reversing the process.
(This is a more precise version of 22.4.10(iii).) The process can be reversed: wecan blow-up P2 at six points, and embed it in P3 . We first explain why we can’t
blow up P2 at just any six points and hope to embed the result in P3 . Because the
cubic surface is embedded anticanonically (Exercise 27.2.A), we see that any curve
C in X must satisfy (KX · C) < 0.
27.4.D. E XERCISE . Suppose P2 is sequentially blown up at p1 , . . . , p6 , resulting ina smooth surface X.
(a) If pi lies on the exceptional divisor of the blow-up at pj (i > j), then show that
there is a curve C ⊂ X isomorphic to P1 , with (KX · C) ≥ 0.
(b) If the pi are distinct points on P2 , and three of them are collinear, show that

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there is a curve C ⊂ X isomorphic to P1 , with (KX · C) ≥ 0.
(c) If the six pi are distinct points on a smooth conic, show that there is a curve
C ⊂ X isomorphic to P1 , with (KX · C) ≥ 0.

Thus the only chance we have of obtaining a smooth cubic surface by blowing
up six points on P2 is by blowing up six distinct points, no three on a line and not
all on a conic.
27.4.3. Proposition. — The anticanonical map of P2 blown up at six distinct points, nothree on a line and not all on a conic, gives a closed embedding into P3 , as a cubic surface.

Because we won’t use this, we only describe the main steps of the proof: first∼
count sections of the anticanonical bundle KP∨
2 = OP2 (3) (there is a 4-dimensional
2vector space of cubics on P vanishing at the six points, and these correspond tosections of the anticanonical bundle of the blow-up via Exercise 22.4.S(a)). Thenshow that these sections separate points and tangent vectors of X, thus showingthat the anticanonical linear series gives a closed embedding, Theorem 19.1.1. Judicious use of the Cremona transformation (Exercise 6.5.I) can reduce the amountof tedious case-checking in this step.
27.4.E. E XERCISE . Suppose X is the blow-up of P2k at six distinct points p1 , . . . ,p6 , no three on a line and not all on a conic. Verify that the only (−1)-curves on X
are the six exceptional divisors, the proper transforms of the 15 lines pi pj , and theproper transforms of the six conics through five of the six points, for a total of 27.
27.4.F. E XERCISE . Solve Exercises 27.3.I and 27.3.J again, this time using the description of X as a blow-up of P2 .
27.4.4. Remark. If you blow-up 4 ≤ n ≤ 8 points on P2 , with no three on a line
and no six on a conic, then the symmetry group of the configuration of lines is a
Weyl group, as shown in the following table.
n

4
5
6
7
8
W(A4 ) W(D5 ) W(E6 ) W(E7 ) W(E8 )(If you know about Dynkin diagrams, you may see the pattern, and may be able
to interpret what happens for n = 3 and n = 9.) This generalizes part of Remark 27.3.5, and the rest of it can similarly be generalized.

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CHAPTER 28

Cohomology and base change theorems

28.1 Statements and applications
Higher pushforwards are easy to define, but it is hard to get a geometric senseof what they are, or how they behave. For example, given a morphism π : X → Y,
and a quasicoherent sheaf F on X, you might reasonably hope that the fibers of
Ri π∗ F are the cohomologies of F along the fibers. More precisely, given ψ : q → Y
corresponding to the inclusion of a point (better: ψ : Spec κ(q) → Y), yielding thefibered diagram(28.1.0.1)

Xq

ψ′

%X

π′

π

"
q

ψ

"
% Y,

one might hope that the morphism
∗ p
p
′ ∗
φp
q : ψ (R π∗ F ) → H (Xq , (ψ ) F )(given in Exercise 18.8.C) is an isomorphism. (Note: F |Xq and (ψ ′ )∗ F are symbols
for the same thing. The first is often preferred, but we sometimes use the second
because we will consider more general ψ and ψ ′ .) We could then picture Ri π∗ F
as somehow fitting together the cohomology groups of fibers into a coherent sheaf.
(Warning: this is too much to hope for, see Exercise 28.1.A.)
It would also be nice if hp (Xq , (ψ ′ )∗ F ) was constant, and φp
q put them together into a nice locally free sheaf (vector bundle) π∗ F .
There is no reason to imagine that the particular choice of base change ψ : q →Y should be special. As long as we are dreaming, we may as well hope that ingood circumstances, given a Cartesian diagram (18.8.4.1)(28.1.0.2)

W

ψ′

π′

%X
π

"
Z

ψ

"
% Y,

the natural morphism(28.1.0.3)

∗ p
p ′
′ ∗
φp
Z : ψ (R π∗ F ) → R π∗ (ψ ) F
707

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The Rising Sea: Foundations of Algebraic Geometryof sheaves on Z (Exercise 18.8.B(a)) is an isomorphism. (In some cases, we canalready address this question. For example, cohomology commutes with flat basechange, Theorem 24.2.8, so the result holds if ψ is flat. Also related: if F is flat overY, then the Euler characteristic of F on fibers is locally constant, Theorem 24.7.1.)
There is no point in dreaming if we are not going to try to make our dreams
come true. So let’s formalize them. Suppose F is a coherent sheaf on X, π : X → Y
is proper, Y (hence X) is Noetherian, and F is flat over Y. We formalize our dreams
into three nice properties that we might wish in this situation. We will see that they
are closely related.
pp(a) Given a Cartesian square (28.1.0.1), is φp
q : R π∗ F ⊗κ(q) → H (Xq , F |Xq )an isomorphism?
∗ p
p ′
′ ∗
(b) Given a Cartesian square (28.1.0.2), is φp
Z : ψ (R π∗ F ) → R π∗ (ψ ) F
an isomorphism?
(c) Is Rp π∗ F locally free?

We turn first to property (a). The dimension of the left side Rp π∗ F ⊗ κ(q) is anupper semicontinuous function of q ∈ Y by upper semicontinuity of rank of finite
type quasicoherent sheaves (Exercise 13.7.J). The Semicontinuity Theorem statesthat the dimension of the right is also upper semicontinuous. More formally:
28.1.1. Semicontinuity Theorem. — Suppose π : X → Y is a proper morphism of
Noetherian schemes, and F is a coherent sheaf on X flat over Y. Then for each p ≥ 0,
the function Y → Z given by q (→ dimκ(q) Hp (Xq , F |Xq ) is an upper semicontinuousfunction of q ∈ Y.
Translation: ranks of cohomology groups are upper semicontinuous in proper
flat families. (A proof will be given in §28.2.4.)

28.1.2. Example. You may already have seen an example of cohomology groups
jumping, in §24.4.14. Here is a simpler example, albeit not of the structure sheaf.
Let (E, p0 ) be an elliptic curve over a field k, and consider the second projection
π : E × E → E. Let L be the invertible sheaf (line bundle) on E × E corresponding
to the divisor that is the diagonal, minus the section of p0 × E of π (where p0 ∈ E).
Then L |p0 (i.e., L |p0 ×E ) is trivial, but L |p is non-trivial for any p ̸= p0 (as we
showed in our study of genus 1 curves, in §19.9). Thus h0 (E, L |p ) is 0 in general,
but jumps to 1 for p = p0 .
28.1.A. E XERCISE . Show that π∗ L = 0. Thus we cannot picture π∗ L as “gluing
together” h0 of the fibers; in this example, cohomology does not commute with“base change” or “taking fibers”.
28.1.3. Side Remark. In characteristic 0, the cohomology of O doesn’t jump insmooth families. Over C, this is because Betti numbers are constant in connectedfamilies, and (21.5.11.1) (from Hodge theory) expresses the Betti constants hkBetti assums (over i + j = k) of upper semicontinuous functions hj (Ωi ), so the Hodge
numbers hj (Ωi ) must in fact be constant. The general characteristic 0 case canbe reduced to C by an application of the Lefschetz principle (which also arose in§21.5.9). But ranks of cohomology groups of O for smooth families of varieties canjump in positive characteristic (see for example [MO70920]). Also, the exampleof §24.4.14 shows that the “smoothness” hypothesis cannot be removed.

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28.1.4. Grauert’s Theorem. If Rp π∗ F is locally free (property (c)) and φp
q is
an isomorphism (property (a)), then hp (Xq , F |Xq ) is clearly locally constant. The
following is a partial converse.
28.1.5. Grauert’s Theorem. — If π : X → Y is proper, Y is reduced and locally Noetherian, F is a coherent sheaf on X flat over Y, and hp (Xq , F |Xq ) is a locally constant function of q ∈ Y, then Rp π∗ F is locally free, and φp
Z is an isomorphism for all ψ : Z → Y.

In other words, if cohomology groups of fibers have locally constant dimension (over a reduced base), then they can be fit together to form a vector bundle,
and the fiber of the pushforward is identified with the cohomology of the fiber.
(No Noetherian hypotheses are needed.)
We further note that if Y is integral, π is proper, and F is a coherent sheaf
on X flat over Y, then by the Semicontinuity Theorem 28.1.1 there is a dense open
subset of Y on which Rp π∗ F is locally free (and on which the fiber of the pth higher
pushforward is the pth cohomology of the fiber).
The following statement is even more magical than Grauert’s Theorem 28.1.5.
28.1.6. Cohomology and Base Change Theorem. — Suppose π is proper, Y is locally
Noetherian, F is coherent over X and flat over Y, and φp
q is surjective. Then the followinghold.
(i) There is an open neighborhood U of q such that for any ψ : Z → U, φp
Z is anisomorphism. In particular, φpisanisomorphism.
q(ii) Furthermore, φp−1
is surjective (hence an isomorphism by (i)) if and only ifq
Rp π∗ F is locally free in some open neighborhood of q (or equivalently, (Rp π∗ F )q
is a free OY,q -module, Exercise 13.7.F). This in turn implies that hp is constantin an open neighborhood of q.
(Proofs of Theorems 28.1.5 and 28.1.6 will be given in §28.2. Note in (ii) that if−1
p = 0, φp−1
is automatically surjective, as the codomain of φ−1(Xq , F |Xq ),qq is H
which is 0 by definition.)
This is amazing: the hypothesis that φp
q is surjective involves what happens
only over reduced points, and it has implications over the (possibly nonreduced)
scheme as a whole! This might remind you of the local criterion for flatness (Theorem 24.6.2), and indeed that is the key technical ingredient of the proof.
Here are some consequences.
28.1.B. E XERCISE . Use Theorem 28.1.6 to give a second solution to Exercise 24.4.E.
(This is a big weapon to bring to bear on this problem, but it is still enlightening;
the original solution to Exercise 24.4.E foreshadowed the proof of the Cohomology
and Base Change Theorem 28.1.6.)
28.1.C. E XERCISE . Suppose π : X → Y is proper, Y is locally Noetherian, and F is acoherent sheaf on X, flat over Y. Suppose further that Hp (Xq , F |Xq ) = 0 for someq ∈ Y. Show that there is an open neighborhood U of q such that (Rp π∗ F )|U = 0.
28.1.D. E XERCISE . Suppose π : X → Y is proper, Y is locally Noetherian, and F isa coherent sheaf on X, flat over Y. Suppose further that Hp (Xq , F |Xq ) = 0 for allq ∈ Y. Show that the (p − 1)st cohomology commutes with arbitrary base change:
φp−1
is an isomorphism for all ψ : Z → Y.
Z

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28.1.E. E XERCISE . Suppose π is proper, Y is locally Noetherian, and F is a coherent sheaf on X flat over Y. Suppose further that Rp π∗ F = 0 for p ≥ p0 . Show that
Hp (Xq , F |Xq ) = 0 for all q ∈ Y, p ≥ p0 .
28.1.F. E XERCISE . Suppose π is proper, Y is locally Noetherian, and F is a coherent
sheaf on X, flat over Y. Suppose further that Y is reduced. Show that there exists
a dense open subset U of Y such that φp
Z is an isomorphism for all ψ : Z → U
and all p. (Hint: find suitable open neighborhoods of the generic points of Y. See
Exercise 24.2.M and the paragraph following it.)28.1.7. An important class of morphisms: Proper, O-connected morphisms π : X →Y of locally Noetherian schemes.
If a morphism π : X → Y satisfies the property that the natural map OY →π∗ OX is an isomorphism, we say that π is O-connected.
28.1.G. E ASY E XERCISE . Show that proper O-connected morphisms of locally
Noetherian schemes are surjective.

28.1.8. We will soon meet Zariski’s Connectedness Lemma 29.5.1, which showsthat proper, O-connected morphisms of locally Noetherian schemes have connected
fibers. In some sense, this class of morphisms is really the right class of morphisms
capturing what we might want by “connected fibers”; this is the motivation for theterminology. The following result gives some evidence for this point of view, inthe flat context.
28.1.H. I MPORTANT E XERCISE . Suppose π : X → Y is a proper flat morphism of
locally Noetherian schemes, whose fibers satisfy h0 (Xq , OXq ) = 1. (Important remark: this is satisfied if π has geometrically connected and geometrically reduced
fibers, by §10.3.7.) Show that π is O-connected. Hint: consider
OY ⊗ κ(q)% (π∗ OX ) ⊗ κ(q)φ0q

∼ κ(q) .
% H0 (Xq , OXq ) =The composition is surjective, hence φ0q is surjective, hence it is an isomorphismby the Cohomology and Base Change Theorem 28.1.6(i). Then by the Cohomology
and Base Change Theorem 28.1.6(ii), π∗ OX is locally free, thus of rank 1. Perhaps
use Nakayama’s Lemma to show that a map of invertible sheaves OY → π∗ OXthat is an isomorphism on fibers over closed points points is necessarily an isomorphism of sheaves.
28.1.9. ⋆ Unimportant remark. This class of proper, O-connected morphisms is
not preserved by arbitrary base change, and thus is not “reasonable” in the senseof §7.1.1. But you can show that they are preserved by flat base change, using
the fact that cohomology commutes with flat base change, Theorem 24.2.8. Furthermore, the conditions of Exercise 28.1.H behave well under base change, andNoetherian hypotheses can be removed from the Cohomology and Base Change
Theorem 28.1.6 (at the expense of finitely presented hypotheses, see §28.2.9), so
the class of morphisms π : X → Y that are proper, finitely presented, and flat, withgeometrically connected and geometrically reduced fibers, is “reasonable” (anduseful).

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28.1.10. We next address the following question. Suppose π : X → Y is a morphismof schemes. Given an invertible sheaf L on X, we ask when it is the pullback of
an invertible sheaf M on Y. For this to be true, we certainly need that L is trivial
on the fibers. We will see that if π is a proper O-connected morphism of locally
Noetherian schemes, then this often suffices. Given L , we recover M as π∗ L ; thefibers of M are one-dimensional, and glue together to form a line bundle. We now
begin to make this precise.
28.1.I. E XERCISE . Suppose π : X → Y is a proper, O-connected morphism of locally
Noetherian schemes. Show that if M is any invertible sheaf on Y, then the naturalmorphism M → π∗ π∗ M is an isomorphism. In particular, we can recover M fromπ∗ M by applying the pushforward π∗ .

28.1.11. Proposition. — Suppose π : X → Y is a flat, proper, O-connected morphism of
locally Noetherian schemes. Suppose also that Y is reduced, and L is an invertible sheaf on
X that is trivial on the fibers of π (i.e., Lq is a trivial invertible sheaf on Xq for all q ∈ Y).
Then π∗ L is an invertible sheaf on Y (call it M ), and the natural map π∗ M → L is an
isomorphism.

Proof. By Grauert’s Theorem 28.1.5, π∗ L is locally free of rank 1 (again, call itM ), and M ⊗OY κ(q) → H0 (Xq , Lq ) is an isomorphism. We have a natural mapof invertible sheaves π∗ M = π∗ π∗ L → L . To show that it is an isomorphism,
we need only show that it is surjective. (Do you see why? If A is a ring, andφ : A → A is a surjection of A-modules, why is φ an isomorphism?) For this, itsuffices to show that it is surjective on the fibers of π. (Do you see why? Hint: if
the cokernel of the map is not 0, then it is not 0 above some point of Y.) But thisfollows from the hypotheses.
!
Proposition 28.1.11 has some pleasant consequences. For example, if you have
two invertible sheaves A and B on X that are isomorphic on every fiber of π, thenthey differ by a pullback of an invertible sheaf on Y: just apply Proposition 28.1.11
to A ⊗ B ∨ . But Proposition 28.1.11 has the unpleasant hypothesis that the target must be reduced. We can get rid of this hypothesis by replacing the use of
Grauert’s Theorem with the Cohomology and Base Change Theorem 28.1.6. We
do this now.
28.1.12. Projective bundles.
28.1.J. E XERCISE . Let X be a locally Noetherian scheme, and let pr1 : X × Pn → X
be the projection onto the first factor. Suppose L is a line bundle on X × Pn ,whose degree on every fiber of pr1 is zero. Use the Cohomology and Base Change
Theorem 28.1.6 to show that (pr1 )∗ L is an invertible sheaf on X. Use Nakayama’s
Lemma (in some guise) to show that the natural map pr∗1 ((pr1 )∗ L ) → L is an
isomorphism.
Your argument will apply just as well to the situation where pr1 : X × Pn → X
is replaced by a Pn -bundle over X, pr1 : Z → X; or by pr1 : Z → X which is asmooth morphism whose geometric fibers are integral curves of genus 0.
Furthermore, the locally Noetherian hypotheses can be removed, see §28.2.9.

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28.1.K. E XERCISE .
Suppose X is a connected Noetherian scheme. Show that
∼ Pic X × Z. Hint: the map Pic X × Pic Pn → Pic(X × Pn ) is given by
Pic(X × Pn ) =(L , O(m)) (→ pr∗1 L ⊗ pr∗2 O(m), where pr1 : X × Pn → X and pr2 : X × Pn → Pn
are the projections from X × Pn to its factors. (The notation % is often used for thisconstruction, see §16.4.8.)∼
A very similar argument will show that if Z is a Pn -bundle over X, then Pic Z =Pic X × Z. You will undoubtedly also be able to figure out the right statement if X
is not connected.

28.1.13. Remark. As mentioned in §19.10.1, the Picard group of a scheme often“wants to be a scheme”. You may be able to make this precise in the case of Pic Pn
Z.
In this case, the scheme Pic Pn
is“Z
copies
of
Spec
Z”,
with
the
“obvious”
groupZ
scheme structure. Can you figure out what functor it represents? Can you showthat it represents this functor? This will require extending Exercise 28.1.K out of
the Noetherian setting, using §28.2.9.
28.1.L. E XERCISE . Suppose π : X → Y is a projective flat morphism over a Noetherian integral scheme, all of whose geometric fibers are isomorphic to Pn (overthe appropriate field). Show that π is a projective bundle if and only if there isan invertible sheaf L on X that restricts to O(1) on all the geometric fibers. (Onedirection is clear: if it is a projective bundle, then it has a O(1) which comes fromthe projectivization, see Exercise 17.2.D. In the other direction, the candidate vector bundle is π∗ L . Show that it is indeed a locally free sheaf of the desired rank.
Show that its projectivization is indeed π : X → Y.)

Caution: the map π : Proj R[x, y, z]/(x2 + y2 + z2 ) → Spec R shows that not every projective flat morphism over a Noetherian integral scheme, all of whose geometric fibers are isomorphic to Pn , is necessarily a Pn -bundle. However, Tsen’s Theorem implies that if the target is a smooth curve over an algebraically closed field, thenthe morphism is a Pn -bundle (see [GS, Thm. 6.2.8]). Example 18.4.5 shows that“curve” cannot be replaced by “5-fold” in this statement — the “universal smoothplane conic” is not a P1 -bundle over the parameter space U ⊂ P5 of smooth plane
conics. If you wish, you can extend Example 18.4.5 to show that “curve” cannot
even be replaced by “surface”. (Just replace the P5 of all conics with a generally
chosen P2 of conics — but then figure out what goes wrong if you try to replace itwith a generally chosen P1 of conics.)
28.1.M. E XERCISE . Suppose π : X → Y is the projectivization of a vector bundle F∼ Proj Sym• F ). Recall from §17.2.3 thatover a locally Noetherian scheme (i.e., X =∼ Proj Sym• (F ⊗ L )). Show that these are thefor any invertible sheaf L on Y, X =only ways in which it is the projectivization of a vector bundle. (Hint: recover F
by pushing forward O(1).)
28.1.14. The Hodge bundle.

28.1.N. E XERCISE ( THE H ODGE BUNDLE ). Suppose π : X → Y is a flat proper morphism of locally Noetherian schemes, and the fibers of π are regular irreducible
curves of genus g. Show that π∗ ΩX/Y is a locally free sheaf on Y of rank g, andEarly (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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that the construction of π commutes with base change: given a Cartesian square(28.1.14.1)

X′

ψ′

π′

"
Y′

%X
π

ψ

"
% Y,

there is a canonical isomorphism
) ′
*
)
*
∼ ψ∗ π∗ ΩX/Y .
π∗ ΩX ′ /Y ′ =(The locally free sheaf π∗ ΩX/Y is called the Hodge bundle.) Hint: use the Cohomology and Base Change Theorem 28.1.6 twice, once with p = 2, and once with
p = 1.

28.2 ⋆ Proofs of cohomology and base change theorems
The key to proving the Semicontinuity Theorem 28.1.1, Grauert’s Theorem 28.1.5,
and the Cohomology and Base Change Theorem 28.1.6 is the following wonderfulidea of Mumford (see [Mu3, p. 47 Lem. 1]). It turns questions of pushforwards(and how they behave under arbitrary base change) into something computablewith vector bundles (hence questions of linear algebra). After stating it, we willinterpret it.
28.2.1. Key Theorem. — Suppose π : X → Spec B is a proper morphism, and F is acoherent sheaf on X, flat over Spec B. Then there is a complex(28.2.1.1)

···

% K−1

% K0

% K1

% ···

% Kn

%0

of finitely generated free B-modules and an isomorphism of functors∼ Hp (K• ⊗B A)(28.2.1.2)
Hp (X ×B A, F ⊗B A) =for all p, for all ring maps B → A.

Because (28.2.1.1) is a complex of free B-modules, all of the information is contained in the maps, which are matrices with entries in B. This will turn questions
about cohomology (and base change) into questions about linear algebra. For example, semicontinuity will turn into the fact that ranks of matrices (with functions
as entries) drop on closed subsets (§11.4.4(ii)).
Although the complex (28.2.1.1) is infinite, by (28.2.1.2) it has no cohomology
in negative degree, even after any ring extension B → A (as the left side of (28.2.1.2)
is 0 for p < 0).
The idea behind the proof is as follows: take the Čech complex, produce acomplex of finite rank free modules mapping to it “with the same cohomology” (aquasiisomorphic complex, §18.2.3). We first construct the complex so that (28.2.1.2)
holds for B = A, and then show the same complex works for general A. We begin
with a lemma.
28.2.2. Lemma. — Let B be a Noetherian ring. Suppose C• is a complex of B-modules
such that Hi (C• ) are finitely generated B-modules, and such that Cp = 0 for p > n. Then
there exists a complex K• of finite rank free B-modules such that Kp = 0 for p > n, and a

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homomorphism of complexes α : K• → C• such that α induces isomorphisms Hi (K• ) →Hi (C• ) for all i.

Proof. We build this complex inductively. (This may remind you of Hint 23.3.3.)
Assume we have defined (Kp , αp , δp ) for p ≥ m + 1 (as in (28.2.2.1)) such that thesquares commute, and the top row is a complex, and αq defines an isomorphismof cohomology Hq (K• ) → Hq (C• ) for q ≥ m + 2 and a surjection ker δm+1 →Hm+1 (C• ), and the Kp are finite rank free B-modules. (Our base case is m = p:take Kn = 0 for n > p.)(28.2.2.1)

Km+1

···

% Cm−1

% Cm

Km+2

αm+1

"

% Cm+1δm

δm+1 %

"

δm+1

% ···

αm+2

% Cm+2% ··· .

(We sloppily use δq for the horizontal morphisms in both rows.)
We construct (Km , δm , αm ). Choose generators of Hm (C• ), say c1 , . . . , cM . Let
"
#
m+1
m+1
m+1 α
m+1
•
D
:= ker ker(δ) −→ H(C )(where the δm+1 is the differential of the top complex K• ). Choose generators of
Dm+1 , say d1 , . . . , dN . Let Km = B⊕(M+N) . Define δm : Km → Km+1 by sending
the last N generators to d1 , . . . , dN , and the first M generators to 0. Define αmby sending the first M generators of B⊕(M+N) to (lifts of) c1 , . . . , cM , and sending
the last N generators to arbitrarily chosen lifts of the αm+1 (di ) (as the αm+1 (di )are 0 in Hm+1 (C• ), and thus lie in the image of δm ), so the square (with upper left
corner Km ) commutes. Then by construction, we have completed our inductive
step:
Km

δm

m+1

% Km+1 δαm···% Cm−1

"

"

% Cm

δm

% Km+2

αm+1% Cm+1

δm+1

"

% ···

αm+2

% Cm+2% ··· .
!

28.2.3. Lemma. — Suppose α : K• → C• is a morphism of complexes of flat B-modules,
bounded on the right (i.e., Kn = Cn = 0 for n ≫ 0), inducing isomorphisms of cohomology (a quasiisomorphism, §18.2.3). Then “this quasiisomorphism commutes with arbitrary change of base ring”: for every B-algebra A, the maps Hp (K• ⊗B A) → Hp (C• ⊗B A)
are isomorphisms.

Proof. The mapping cone M• of α : K• → C• is exact by Exercise 1.7.E. Then M• ⊗B
A is still exact, by Exercise 24.3.F. But M• ⊗B A is the mapping cone ofα ⊗B A : K• ⊗B A → C• ⊗B A,

so by Exercise 1.7.E, α ⊗B A induces an isomorphism of cohomology (i.e., is aquasiisomorphism) too.
!

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Proof of Key Theorem 28.2.1. Choose a finite affine covering of X. Take the Čech complex C• for F with respect to this cover. Recall that Grothendieck’s Coherence Theorem 18.9.1 (which had Noetherian hypotheses) showed that the cohomology of
F is coherent. (Theorem 18.9.1 required serious work. If you need Theorem 28.2.1
only in the projective case, the analogous statement with projective hypotheses,
Theorem 18.8.1(d), was much easier.) Apply Lemma 28.2.2 to get the nicer variant K• of the same complex C• . By Lemma 28.2.3, if we tensor with A and takecohomology, we get the same answer whether we use K• or C• .
!
We now use Theorem 28.2.1 to prove some of the fundamental results statedearlier: the Semicontinuity Theorem 28.1.1, Grauert’s Theorem 28.1.5, and the Cohomology and Base Change Theorem 28.1.6. In the course of proving Semicontinuity, we will give a new proof of Theorem 24.7.1, that Euler characteristics are
locally constant in flat families (that applies more generally in proper situations).
28.2.4. Proof of the Semicontinuity Theorem 28.1.1. The result is local on Y, so
we may assume Y is affine. Let K• be a complex as in Key Theorem 28.2.1.
Then for q ∈ Y,dimκ(q) Hp (Xq , F |Xq ) ==(28.2.4.1)

dimκ(q) ker(δp ⊗B κ(q)) − dimκ(q) im(δp−1 ⊗B κ(q))
dimκ(q) (Kp ⊗B κ(q)) − dimκ(q) im(δp ⊗B κ(q))− dimκ(q) im(δp−1 ⊗B κ(q))

Now dimκ(q) im(δp ⊗B κ(q)) is a lower semicontinuous function on Y. (Reason:
the locus where the dimension is less than some number N is obtained by setting
all N × N minors of the matrix Kp → Kp+1 to 0; cf. §11.4.4(ii)).) The same is true
for dimκ(q) im(δp−1 ⊗B κ(q)). The result follows.
!

28.2.5. A new proof (and extension to the proper case) of Theorem 24.7.1 that
Euler characteristics of flat sheaves are locally constant.
If K• were finite “on the left” as well — if Kp = 0 for p ≪ 0 — then wewould have a short proof of Theorem 24.7.1. By taking alternating sums (over p)of (28.2.4.1), we would have that.
.
χ(Xq , F |Xq ) =(−1)p hp (Xq , F |Xq ) =(−1)p rank Kp ,which is locally constant. The only problem is that the sums are infinite. We patch
this problem as follows. Define J• by Jp = Kp for p ≥ 0, Jp = 0 for p < −1, andJ−1 := ker(K0 → K1 ). Combine the J• into a complex, by defining Jp → Jp+1 as theobvious map induced by K• . We have a map of complexes J• → K• . Clearly thisinduces an isomorphism on cohomology (as J• patently has the same cohomology
as K• at step p ≥ 0, and both have 0 cohomology for p < 0). Thus the compositionβ : J• → K• → C• induces an isomorphism on cohomology as well.
Now J−1 is coherent (as it is the kernel of a map of coherent modules). Consider the mapping cone M• of β : J• → C• :0 → J−1 → C−1 ⊕ J0 → C0 ⊕ J1 → · · · → Cn−1 ⊕ Jn → Cn → 0.

From Exercise 1.7.E, as J• → C• induces an isomorphism on cohomology, the mapping cone has no cohomology — it is exact. All terms in it are flat except possibly
J−1 (the Cp are flat by assumption, and Ji is free for i ̸= −1). Hence J−1 is flat

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too, by Exercise 24.3.G. But flat coherent sheaves are locally free (Theorem 24.4.7).
Then Theorem 24.7.1 follows from.
.
χ(Xq , F |Xq ) =(−1)p hp (Xq , F |Xq ) =(−1)p rank Jp .
!

28.2.6. ⋆⋆ Proof of Grauert’s Theorem 28.1.5 and the Cohomology and Base
Change Theorem 28.1.6.
Thanks to Theorem 28.2.1, Theorems 28.1.5 and 28.1.6 are now statements
about complexes of free modules over a Noetherian ring. We begin with some
general comments on dealing with the cohomology of a complex
% Kp

···

δp

% Kp+1% ··· .

We define some notation for functions on a complex (most of which already appeared in §23.3.7).
• Let Zp be the kernel of the pth differential of a complex, so for example
Zp K• = ker δp .
• Let Bp+1 be the image of the pth differential, so for example Bp+1 K• =im δp .
• Let W p+1 be the cokernel of the pth differential, so for example W p+1 K• =coker δp .
• As usual, let Hp be the homology at the pth step.
We have exact sequences (cf. (1.6.5.3) and (1.6.5.4))(28.2.6.1)

% Zp

0

% Kp

% Kp+1% W p+1(28.2.6.2)

0

% Zp

% Kp

% Bp+1%0(28.2.6.3)

0

% Bp

% Zp

% Hp

%0

(28.2.6.4)

0

% Bp

% Kp

% Wp

%0

% Hp

% Wp

% Bp+1% 0.

(28.2.6.5)

0%0

We proceed by a series of exercises, some of which were involved in the proofof the FHHF Theorem (Exercise 1.6.H). Suppose C• is any complex in an abeliancategory A with enough projectives, and suppose F is any right-exact functor fromA.
28.2.A. E XERCISE ( COKERNELS COMMUTE WITH RIGHT- EXACT FUNCTORS ). De∼% W p FC• . (Hint: consider Cp−1 → Cp →scribe an isomorphism γp : FW p C•
p •W C → 0.)28.2.B. E XERCISE .
(a) Describe a map βp : FBp C• → Bp FC• . Hint: (28.2.6.4) induces
R1 FW p C•

% FBp C•
βp

=

0

"
% Bp FC•

"
% FCp

% FCp

% FW p C•

%0

γp ∼

"
% W p FC•

% 0.

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717(b) Show that βp is surjective. Possible hint: use Exercise 1.7.B, a weaker versionof the Snake Lemma, to get an exact sequence% ker βp%0

% ker γp

% coker βp%0% coker γp% R1 FW p C•

R1 FCp% 0.

28.2.C. E XERCISE .
(a) Describe a map αp : FZp C• → Zp FC• . Hint: use (28.2.6.2) to induce
R1 FBp+1 C•% FZp C•
αp

=

0

"
% Zp FC•"% FCp% FBp+1 C•

% FCp

%0

βp+1

"

% Bp+1 FC•%0(b) Use Exercise 1.7.B to get an exact sequence
R1 FCp% ker αp%0

% ker βp+1

% coker αp%0% coker βp+1

% R1 FBp+1 C•% 0.

28.2.D. E XERCISE .
(a) Describe a map ϵp : FHKp → HFKp . (This is the FHHF Theorem for right-exact
functors, Exercise 1.6.H(a).) Hint: (28.2.6.3) induces
R1 FHp C•% FBp C•

% FZp C•

% FHp C•

βp

αp

ϵp

0

"
% Bp FC•"% Zp FC•"% Hp FC•%0%0(b) Use Exercise 1.7.B to get an exact sequence:
R1 FZp C•% R1 FHp C•% ker βp

% ker αp

% ker ϵp% coker βp% coker αp% coker ϵp% 0.

28.2.7. Back to the theorems we want to prove. Recall the properties we discussed at
the start of §28.1.

pp(a) Given a Cartesian square (28.1.0.1), is φp
q : R π∗ F ⊗κ(q) → H (Xq , F |Xq )an isomorphism?
∗ p
p ′
′ ∗
(b) Given a Cartesian square (28.1.0.2), is φp
Z : ψ (R π∗ F ) → R π∗ (ψ ) F
an isomorphism?
(c) Is Rp π∗ F locally free?

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We reduce to the case Y and Z are both affine, say Y = Spec B. We apply ourgeneral results of §28.2.6 to the complex (28.2.1.1) of Theorem 28.2.1.
28.2.E. E XERCISE . Suppose W p K• and W p+1 K• are flat. Show that the answer to(b), and hence (a), is yes. Show that the answer to (c) is yes if Y is reduced or locally
Noetherian. Hint: (You will take F to be the functor (·) ⊗B A, where A is some Balgebra.) Use (28.2.6.4) (shifted) to show that Bp+1 K• is flat, and then (28.2.6.5) toshow that Hp K• is flat. By Exercise 28.2.A, the construction of the cokernel W •behaves well under base change. The flatness of Bp+1 and Hp imply that theirconstructions behave well under base change as well — apply F to (28.2.6.4) and(28.2.6.5) respectively. (If you care, you can check that Zp K• is also locally free, andbehaves well under base change.)
28.2.F. E XERCISE . Prove Grauert’s Theorem 28.1.5. (Reminder: you won’t need
Noetherian hypotheses.) Hint: By (28.2.4.1), W p K• and W p+1 K• have constantrank. But finite type quasicoherent sheaves having constant rank on a reduced
scheme are locally free (Exercise 13.7.K), so we can invoke Exercise 28.2.E. Conclude that Hp K• is flat of constant rank, and hence locally free.
28.2.8. Proof of the Cohomology and Base Change Theorem 28.1.6. Keep in mind that
we now have locally Noetherian hypotheses. We have reduced to the case Y and Z
are both affine, say Y = Spec B. Let F be the functor · ⊗B κ(q). The key input is the
local criterion for flatness (Theorem 24.6.2): R1 FW p K• = 0 if and only if W p K• isflat at q (and similarly with W replaced by other letters). In particular, R1 FKp = 0
for all p. Also keep in mind that if a coherent sheaf on a locally Noetherian scheme(such as Spec B) is flat at a point q, then it is flat in an open neighborhood of thatpoint, by Corollary 24.4.7 (flat = locally free for such sheaves).
28.2.G. E XERCISE . Look at the boxed snakes in §28.2.6 (with C• = K• ), and show
the following in order, starting from the assumption that coker ϵp = 0:
• coker αp = 0, ker βp+1 = 0, R1 FW p+1 K• = 0;
• W p+1 K• is flat at q, Bp+1 K• is flat at q (use (28.2.6.4) with the indexing
shifted by one), Zp K• is flat at q (use (28.2.6.3));• R1 FBp+1 K• = 0;
• ker αp = 0, ker ϵp = 0.
It might be useful for later to note that∼ ker βp =∼ R1 FHp K•
R1 FW p K• =At this point, we have shown that φp
q is an isomorphism — part of part (i) of
the theorem.
28.2.H. E XERCISE .
rem 28.1.6.

Prove part (i) of the Cohomology and Base Change Theo-

Also, φp−1
surjective implies W p K• is flat (in the same way that you showedqpφq surjective implies W p+1 K• is flat), so we get Hp is free by Exercise 28.2.E, yielding half of (ii).

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November 18, 2017 draft719

28.2.I. E XERCISE . For the other direction of (ii), shift the grading of the last two
boxed snakes down by one, to obtain further isomorphisms∼ coker αp−1 =∼ coker ϵp−1 .
ker βp =For the other direction of (ii), note that if the stalks W p K• and W p+1 K• at y are
flat, then they are locally free (by Corollary 24.4.7), and hence W p K• and W p+1 K•
are locally free in an open neighborhood of q by Exercise 13.7.F. Thus the stalks of
W p K• and W p+1 K• are flat in an open neighborhood of q, and the same argument
applies for any point in this open neighborhood to show that W p+1 K• , Bp+1 K• ,and Zp K• are all flat.
28.2.J. E XERCISE . Use this to show the following, possibly in order:• R1 FCp+1 = R1 FBp+1 = R1 FZp = 0.
• ker βp+1 = 0, coker αp = 0, coker ϵp = 0.
28.2.K. E XERCISE . Put all the pieces together and finish the proof of part (ii) of the
Cohomology and Base Change Theorem 28.1.6.
!
28.2.9. ⋆ Removing Noetherian conditions.
It can be helpful to have versions of the theorems of §28.1 without Noetherian
conditions; important examples come from moduli theory, and will be discussed
in the next section. Noetherian conditions can often be exchanged for finite presentation conditions. We begin with an extension of Exercise 9.3.H.
28.2.L. E XERCISE . Suppose π : X → Spec B is a finitely presented morphism, andF is a finitely presented quasicoherent sheaf on X. Show that there exists a basechange diagram of the form
(28.2.9.1)

F′

F
σ

X

π′

π

"
Spec B% X′

ρ"% Spec Z[x1 , . . . , xN ]/I

where N is some integer, I ⊂ Z[x1 , . . . , xN ], and π ′ is finitely presented (= finite
type as the target is Noetherian, see §7.3.17), and a finitely presented (= coherent)∼ σ∗ F ′ .
quasicoherent sheaf F ′ on X ′ with F =28.2.10. Properties of π ′ . (The ideal I appears in the statement of Exercise 28.2.L
not because it is needed there, but to make the statement of this remark correct.)
If π is proper, then diagram (28.2.9.1) can be constructed so that π ′ is also proper(using [Gr-EGA, IV3 .8.10.5]). Furthermore, if F is flat over Spec B, then (28.2.9.1)
can be constructed so that F ′ is flat over Spec Z[x1 , . . . , xN ]/I (using [Gr-EGA,
IV3 .11.2.6]). This requires significantly more work.
28.2.M. E XERCISE . Assuming the results stated in §28.2.10, prove the following results, with the “locally Noetherian” hypotheses removed, and “finite presentation”
hypotheses added:

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The Rising Sea: Foundations of Algebraic Geometry(a) the constancy of Euler characteristic in flat families (Theorem 24.7.1, extended to the proper case as in §28.2.5);(b) the Semicontinuity Theorem 28.1.1;(c) Grauert’s Theorem 28.1.5 (you will have to show that Z[x1 , . . . , xN ]/I in(28.2.9.1) can be taken to be reduced); and(d) the Cohomology and Base Change Theorem 28.1.6.
28.2.11. Necessity of finite presentation conditions. The finite presentation conditions
are necessary. There is a projective flat morphism to a connected target where thefiber dimension jumps. There is a finite flat morphism where the degree of thefiber is not locally constant. There is a projective flat morphism to a connected
target where the fibers are curves, and the arithmetic genus is not constant. See[Stacks, tag 05LB] for the first example; the other two use the same idea.

28.3 Applying cohomology and base change to moduli problems
The theory of moduli relies on ideas of cohomology and base change. We
explore this by examining two special cases of one of the primordial moduli spaces,
the Hilbert scheme: the Grassmannian, and the fact that degree d hypersurfaces inprojective space are “parametrized” by another projective space (corresponding todegree d polynomials, see Remark 4.5.3).
As suggested in §24.1, the Hilbert functor HilbY Pn of Pn
Y parametrizes finitely
presented closed subschemes of Pn
Y , where Y is an arbitrary scheme. More precisely, it is a contravariant functor sending the Y-scheme X to the set of finitely prensented closed subschemes of X ×Y Pn
Y = PX flat over X (and sending morphisms
X1 → X2 to pullbacks of flat families). An early achievement of Grothendieck
was the construction of the Hilbert scheme, which can then be cleverly used toconstruct many other moduli spaces.
28.3.1. Theorem (Grothendieck). — HilbZ Pn is representable by a scheme locally offinite type.
(Grothendieck’s original argument is in [Gr5]. A readable construction is
given in [Mu2], and in [FGIKNV, Ch. 5].)28.3.A. E ASY E XERCISE . Assuming Theorem 28.3.1, show that HilbY Pn is representable, by showing that it is represented by HilbZ Pn ×Z Y. Thus the general casefollows from the “universal” case of Y = Z.
28.3.B. E XERCISE . Assuming Theorem 28.3.1, show that HilbZ Pn is the disjoint
p(m) nunion of schemes HilbZ
P , each one corresponding to finitely presented closednsubschemes of PZ whose fibers have fixed Hilbert polynomial p(m). Hint: Corollary 24.7.2.
p(m)28.3.2. Theorem (Grothendieck). — Each HilbZ

Pn is projective over Z.

In order to get some feeling for the Hilbert scheme, we discuss two important
examples, without relying on Theorem 28.3.1.

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28.3.3. The Grassmannian.
We have defined the Grassmannian G(k, n) twice before, in §6.7 and §16.7.
The second time involved showing the representability of a (contravariant) functor(from Sheaves to Sets), of rank k locally free quotient sheaves of a rank n free sheaf.
We now consider a parameter space for a more geometric problem. The space
will again be G(k, n), but because we won’t immediately know this, we invent
some temporary notation. Let G ′ (k, n) be the contravariant functor (from Schemes
to Sets) which assigns to a scheme B the set of finitely presented closed subschemesof Pn−1, flat over B, whose fiber over any point b ∈ B is a (linearly embedded)
B
Pk−1
in
Pn−1
κ(b)
κ(b) :" cl. subscheme % n−1
PB
♣♣
♣
♣
♣
flat, f. pr.
♣♣♣π" D♣♣♣♣♣
B(28.3.3.1)

X!

(This describes the map to Sets; you should think through how pullback makes
this into a contravariant functor.)
28.3.4. Theorem. — The functor G ′ (k, n) is represented by G(k, n).
Translation: there is a natural bijection between diagrams of the form (28.3.3.1)(where the fibers are Pk−1 ’s) and diagrams of the form (16.7.0.1) (the diagrams that
G(k, n) parametrizes, or represents).
One direction is notably easier. Suppose we are given a diagram of the form
(16.7.0.1) over a scheme B,OB⊕n(28.3.4.1)% % Q,

where Q is locally free of rank k. Applying Proj B to the Sym• construction on bothOB⊕n and Q, we obtain a closed embedding(28.3.4.2)

Proj B (Sym• Q) !

"

▲▲▲
▲▲▲
▲▲▲
▲▲d

B

)
*
% Proj Sym• O ⊕n
B
B
♣♣
♣♣♣
♣
♣
♣♣
h♣♣♣

= Pn−1 × B(as, for example, in Exercise 17.2.H).
The fibers are linearly embedded Pk−1 ’s (as base change, in this case to a pointof B, commutes with the Proj construction, Exercise 17.2.E). Note that Proj (Sym• Q)
is flat and finitely presented over B, as it is a projective bundle. We have constructed a diagram of the form (28.3.3.1).
We now need to reverse this. The trick is to produce (28.3.4.1) from our geometric situation (28.3.3.1), and this is where cohomology and base change will beused.
Given a diagram of the form (28.3.3.1) (where the fibers are Pk−1 ’s), consider
the closed subscheme exact sequence for X:
0 → IX → OPn−1 → OX → 0.
B

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Tensor this with OPn−1 (1):B

(28.3.4.3)

0 → IX (1) → OPn−1 (1) → OX (1) → 0.
B

Note that OX (1) restricted to each fiber of π is O(1) on Pk−1 (over the residue
field), for which all higher cohomology vanishes (§18.3).
28.3.C. E XERCISE . Show that Ri π∗ OX (1) = 0 for i > 0, and π∗ OX (1) is locally freeof rank k. Hint: use the Cohomology and Base Change Theorem 28.1.6. Either use
the non-Noetherian discussion of §28.2.9 (which we haven’t proved), or else justassume B is locally Noetherian.
28.3.D. E XERCISE . Show that the long exact sequence obtained by applying π∗ to(28.3.4.3) is just a short exact sequence of locally free sheaves
0 → π∗ IX (1) → π∗ OPn−1 (1) → π∗ OX (1) → 0.
Bof ranks n − k, n, and k respectively, where the middle term is canonically identified with OB⊕n .
The surjection OB⊕n → π∗ OX (1) is precisely a diagram of the sort we wished
to construct, (16.7.0.1).

28.3.E. E XERCISE . Close the loop, by using these two “inverse” constructions toshow that G(k, n) represents the functor G ′ (k, n).
28.3.5. Hypersurfaces.
Ages ago (in Remark 4.5.3), we informally said that hypersurfaces of degree dn+dn
in P are parametrized by a P( d )−1 . We now make this precise. We work overa base Z for suitable generality. You are welcome to replace Z by a field of yourchoice, but by the same argument as in Easy Exercise 28.3.A, all other cases are
obtained from this one by base change.
Define the contravariant functor Hd,n : Sch → Sets from schemes to sets asfollows. To a scheme B, we associated the set of all closed subschemes X "→ Pn
B,flat and finitely presented over B, all of whose fibers are degree d hypersurfaces in
Pn (over the appropriate residue field). To a morphism B1 → B2 , we obtain a map
Hd,n (B2 ) → Hd,n (B1 ) by pullback.
n+d28.3.6. Proposition. — The functor Hd,n is represented by P( d )−1 .

As with the case of the Grassmannian, one direction is easy, and the otherrequires cohomology and base change.
n+d28.3.F. E ASY E XERCISE . Over P( d )−1 , described a closed subscheme X "→n+dPn × P( d )−1 that will be the universal hypersurface. Show that X is flat andn+dfinitely presented over P( d )−1 . (For flatness, you can use the local criterion offlatness on the source, Exercise 24.6.F, but it is possible to deal with it easily byworking by hand.)n+dThus given any morphism B → P( d )−1 , by pullback, we have a degree dhypersurface X over B (an element of Hd,n (B)).

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November 18, 2017 draft723

Our goal is to reverse this process: from a degree d hypersurface π : X → PnBn+dover B (an element of Hd,n (B)), we want to describe a morphism B → P( d )−1 .
nConsider the closed subscheme exact sequence for X "→ Pn
B , twisted by OPB (d):(28.3.6.1)

0 → IX (d) → OPnB (d) → OX (d) → 0.

28.3.G. E XERCISE ( CF. E XERCISE 28.3.C). Show that the higher pushforwards (byπ) of each term of (28.3.6.1) is 0, and that the long exact sequence of pushforwardsof (28.3.6.1) is
0 → π∗ IX (d) → π∗ OPnB (d) → π∗ OX (d) → 0.
)*
where the middle term is free of rank n+d(whose summands can be identifieddwith degree d monomials in the projective variables x1 , . . . , xn (see Exercise 8.2.K),and the left term π∗ IX (d) is locally free of rank 1 (basically, a line bundle).
⊕(n+d)(It is helpful to interpret the middle term OB d as parametrizing homogeneous degree d polynomials in n+1 variables, and the rank 1 subsheaf of π∗ IX (d)as “the equation of X”. This will motivate what comes next.)⊕(n+d)Taking the dual of the injection π∗ IX (d) "→ OB d , we have a surjection
⊕(n+d
d )

OB

%%L

from a free sheaf onto an invertible sheaf L = (π∗ IX (d))∨ , which (by the univern+d
sal property of projective space) yields a morphism B → P( d )−1 .
28.3.H. E XERCISE . Close the loop: show that these two constructions are inverses,
thereby proving Proposition 28.3.6.
28.3.7. Remark. The proof of the representability of the Hilbert scheme shares anumber of features of our arguments about the Grassmannian and the parameter
space of hypersurfaces.

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CHAPTER 29

Power series and the Theorem on Formal Functions

29.1 Introduction
Power series are a central tool in analytic geometry. Their analog in algebraic
geometry, completion, is similarly useful. We will only touch on some aspects ofthe subject.
In §29.2, we deal with some algebraic preliminaries. In §29.3, we use completions to (finally) give a definition of various singularities, such as nodes. (We won’t
use these definitions in what follows.) In §29.4, we state the main technical resultof the chapter, the Theorem on Formal Functions 29.4.2. The subsequent three sections give applications. In §29.5, we prove Zariski’s Connectedness Lemma 29.5.1
and the Stein Factorization Theorem 29.5.3. In §29.6, we prove a commonly usedversion of (Grothendieck’s version of) Zariski’s Main Theorem 29.6.1; we rely on§29.5. In §29.7, we prove Castelnuovo’s Criterion for contracting (−1)-curves,which we used in Chapter 27. The proof of Castelnuovo’s Criterion also uses §29.5.
Finally, in §29.8, we prove the Theorem on Formal Functions 29.4.2.
There are deliberately many small sections in this chapter, so you can see thatthey tend not to be as hard at they look, with the exception of the proof of the Theorem on Formal Functions itself, and possibly Theorem 29.2.6 relating completion
to exactness and flatness.

29.2 Algebraic preliminaries
Suppose A is a ring, and J1 ⊃ J2 ⊃ · · · is a decreasing sequence of ideals.
^ and is called
Then we may take the limit lim A/Jn . This ring is often denoted A,←−
the completion of A (for this sequence of ideals); the sequence of ideals is left
implicit. The most important case of completion is if Jn = In , where I is an idealof A. The limit lim A/In is called the I-adic completion of A, or the completion←−of A along I or at I.
29.2.1. Example. We define k[[x1 , . . . , xn ]] as the completion of k[x1 , . . . , xn ] at the
maximal ideal (x1 , . . . , xn ). This is the ring of formal power series in n variables
over k.
The p-adic numbers (Example 1.4.3) are another example.
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29.2.A. E XERCISE . Suppose that J1′ ⊃ J2′ ⊃ · · · is a decreasing series of ideals that′′is cofinal with Jn . In other words, for every Jn , there is some JN
with Jn ⊃ JN,′′
and for every Jn , there is some JN with Jn ⊃ JN . Show that there is a canonical∼ lim A/Jn . Thus what matters is less the specific sequence
isomorphism lim A/Jn′ =←−←−of ideals than the “cofinal” equivalence class.

^29.2.2. Preliminary remarks. We have an obvious morphism A → A.
^
If we put the discrete topology on A/Jn , then A naturally has the structure of
a topological ring (a ring that is a topological space, where all the ring operations
are continuous, or equivalently, a ring object in the category of topological spaces).
We then can have the notion of a topological module M over a topological ring A ′— a module over the underlying ring A ′ , with a topology, such that the action of
A ′ on M is continuous.
^ is an injection, we say that A is
In the case of completion at an ideal I, if A → A
I-adically separated or complete with respect to I, if the map is an isomorphism?
although we won’t use this phrase. (For example, the Krull Intersection Theorem
implies that if I is a proper ideal of a Noetherian integral domain or a Noetherian
local ring, then A is I-adically separated, see Exercise 12.9.A(b).)
29.2.B. E XERCISE . Suppose m is a maximal ideal of a ring A. Show that thecompletion of A at m is canonically isomorphic to the completion of Am at m.
^ is an injecIf (A, m) is a Noetherian local ring, then the natural map A → A
tion: anything in the kernel must lie in ∩mi , which is 0 by the Krull Intersection
Theorem (Exercise 12.9.A(b)). Thus “no information is lost by completing”, just as
analytic functions are (locally) determined by their power series expansion.
29.2.C. E XERCISE . Suppose that (A, m) is Noetherian local ring containing its
residue field k (i.e., it is a k-algebra), of dimension d. Let x1 , . . . , xn be elements of
A whose images are a basis for m/m2 . Show that the map of k-algebras(29.2.2.1)^
k[[t1 , . . . , tn ]] → A

defined by ti (→ xi is a surjection. (First explain why there is such a map!) As
usual, for local rings, the completion is assumed to be at the maximal ideal.

Exercise 29.2.C is a special case of the Cohen Structure Theorem. (See [E, §7.4]
for more on this topic.)
29.2.D. E XERCISE . Let X be a locally Noetherian scheme over k, let p ∈ X bea rational (k-valued) point. Suppose p ∈ X is regular of codimension d (i.e.,∼ k[[x1 , . . . , xd ]] as topologicaldim OX,p = d). Describe an isomorphism O^X,p =rings. (Hint: As in Exercise 29.2.C, choose d elements of m that restrict to a basisof m/m2 ; these will be your x1 , . . . , xd . Show that the map (29.2.2.1) has no kernel.
It may help to identify mn /mn+1 with Symn (m/m2 ) using Theorem 22.3.8.)
The converse also holds: if (A, m, k) is a Noetherian local ring that is a k∼ k[[x1 , . . . , xd ]], then A is a regular local ring of dimension d;^ =algebra, and A
see [AtM, Prop. 11.24] for the key step.
29.2.3. Remark. Suppose p is a smooth rational (k-valued) point of a k-variety ofdimension d, and f ∈ OX,p (f is a “local function”). By way of the isomorphism of

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727

Exercise 29.2.D, we can interpret f as an element of k[[x1 , . . . , xd ]]. This should beinterpreted as the “power series expansion of f at p”, in the “local coordinates x1 ,. . . , xd ”.
29.2.4. Aside: Some more geometric motivation.
Exercise 29.2.D may give some motivation for completion: “in the completion,
regular schemes look like affine space”. This is often stated in the suggestive language of “formally locally, regular schemes are isomorphic to affine space”; thiswill be made somewhat more precise in §29.4.1.
We now give a little more geometric motivation for completion, that we will
not use later on. Recall from §12.6.1 that étale morphisms are designed to look like“local isomorphisms” in differential geometry. But Exercise 12.6.F showed that thismetaphor badly fails in one important way. More precisely, let Y = Spec k[t], and
X = Spec k[u, 1/u], and let p ∈ X be given by u = 1, and q ∈ Y be given by t = 1.
The morphism π : X → Y induced by t (→ u2 is étale, and π(p) = q. But there is no
open neighborhood of p that π maps isomorphically onto an open neighborhoodof q. However, the following Exercise shows that π induces an isomorphism ofcompletions (an isomorphism of “formal neighborhoods”).
29.2.E. E XERCISE . Continuing the notation of the previous paragraph, show thatπ induces an isomorphism of completions O^Y,q → O^X,p .

29.2.F. E XERCISE . Suppose Y = Spec A, and X = Spec A[t]/(f(t)), and π : X → Y is
the morphism induced by A → A[t]/(f(t)). Suppose p ∈ X is a closed point, q ∈ Y
is a closed point, π(p) = q, and π is étale at p and induces an isomorphism ofresidue fields. Show that π induces an isomorphism of completions O^Y,q → O^X,p .
(If you are not familiar with Hensel’s Lemma, you will rediscover its central idea
in the course of solving this exercise.)With a little more care, you can show more generally that if π : X → Y is anétale morphism, π(p) = q, and π induces an isomorphism of residue fields at p,then π induces an isomorphism of completions O^Y,q → O^X,p . (You may even wish
to think about how to remove the hypothesis of isomorphism of residue fields.)
These ideas are close to the definition of “formal étaleness” discussed in §25.2.6.
You can interpret these results as a statement that “the implicit function theorem works formally locally”, even though it doesn’t work Zariski-locally. (Theétale topology is somewhere between these two; it is partially designed so that the
implicit function theorem in this sense always holds, essentially by fiat. But this is
not the place to discuss the étale topology.)
29.2.5. Completion and exactness.
We conclude this section with an interesting and useful statement.
29.2.6. Theorem. — Suppose A is a Noetherian ring, and I ⊂ A is an ideal. For any^ = lim M/Ij M be the completion of M with respect to I.
A-module M, let M←−^ (of A with respect to I) is flat over A.
(a) The completion A^ ⊗A M → M^ is an isomorphism.
(b) If M is finitely generated, then the natural map A(c) If 0 → M → N → P → 0 is a short exact sequence of finitely generated A-modules,
^ → N
^ → P
^ → 0 is exact. (Thus completion preserves exact sequences ofthen 0 → M
finitely generated modules, by Exercise 1.6.E.)

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We will use (a) in §29.3, and (b) in §29.8.
29.2.7. Remark. Before proving Theorem 29.2.6, we make some remarks. Parts(a) and (b) together clearly imply part (c), but we will use (c) to prove (a) and(b). Also, note a delicate distinction (which helps me remember the statement): if
0 → M → N → P → 0 is an exact sequence of A-modules, not necessarily finitely
generated, then
(29.2.7.1)

is always exact, but(29.2.7.2)^ ⊗A M → A^ ⊗A N → A^ ⊗A P → 0
0→A
^ →N
^ →P
^→0
0→M

need not be exact — and when it is exact, it is often because the modules are finitely
generated, and thus (29.2.7.2) is really (29.2.7.1).
Caution: completion is not always exact. Consider the exact sequence of k[t]modules
203% ⊕∞ k[t] ×(t,t ,t ,... ) % ⊕∞ k[t]n=1n=1% ⊕∞ k[t]/(tn )
n=1

% 0.

After completion with respect to the ideal ((t, t, t, . . . )), the sequence is no longer
exact in the middle: (t2 , t3 , t4 , . . .) maps to 0, but is not in the image of the completion of the previous term.
⋆⋆ Proof. The key step is to prove (c), which we do through a series of exercises.
Suppose that 0 → M → N → P → 0 is a short exact sequence of finitely generated
A-modules.
^ →P^ is surjective. Hint: consider an element of
29.2.G. E XERCISE . Show that Nj^
P as a sequence (pj ∈ P/I P)j≥0 , where the image of pj+1 is pj , cf. Exercise 1.4.B.
Build a preimage (nj ∈ N/Ij N)j≥0 by induction on j.
^ → P)^ with M.
^
We now wish to identify ker(N29.2.H. E XERCISE . Show that for each j ≥ 0,
(29.2.7.3)

0 → M/(M ∩ Ij N) → N/Ij N → P/Ij P → 0

is exact. (Possible hint: show that 0 → M ∩ Ij N → M → N/Ij N → P/Ij P → 0 isexact.)The short exact sequences (29.2.7.3) form an inverse system as j varies. Its limit
is left-exact (because limits always are), but it is also right-exact by Exercise 1.6.M,
as the “transition maps on the left” M/(M ∩ Ij+1 N) → M/(M ∩ Ij N) are clearlysurjective. Thus
^ →P^→0
0 → lim M/(M ∩ Ij N) → N←−
is exact. To complete the proof of (c), it suffices (by Exercise 29.2.A) to show thatthe sequence of submodules Ij M is cofinal with the sequence M ∩ Ij N, so that the^term lim M/(M ∩ Ij N) on the left of (29.2.7.4) is naturally identified with M.
←−(29.2.7.4)

29.2.I. E XERCISE . Prove this. Hint: clearly Ij M ⊂ M ∩ Ij N. By Corollary 12.9.4 to
the Artin-Rees Lemma 12.9.3, for some integer s, M ∩ Ij+s N = Ij (M ∩ Is N) for allj ≥ 0, and clearly Ij (M ∩ Is N) ⊂ Ij M.

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729

This completes the proof of part (c) of Theorem 29.2.6.
For part (b), present M as
(29.2.7.5)

A⊕m

α

% A⊕n

%M

%0

where α is an m × n matrix with coefficients in A. Completion is exact in this caseby part (c), and clearly commutes with finite direct sums, so
^ ⊕m
A

%A
^ ⊕n

%M
^

%0

is exact. Tensor product is right-exact, and commutes with direct sums (Exercise 1.3.M), so
^ ⊕m
A

%A
^ ⊕n

%A
^ ⊗A M%0^ ⊕m to A^ ⊕n in both right-exact seis exact as well. Notice that the maps from A
quences are the same; they are both α. Thus their cokernels are identified, and (b)follows.
Finally, to prove (a), we need to extend the ideal-theoretic criterion for flatness(Theorem 24.4.1) slightly. Recall (§24.4.2) that it is equivalent to the fact that an
A-module M is flat if and only if for all ideals I, the natural map I ⊗A M → M isan injection.

29.2.J. E XERCISE ( STRONGER FORM OF THE IDEAL - THEORETIC CRITERION FOR FLATNESS ). Show that an A-module M is flat if and only if for all finitely generated ideals
I, the natural map I ⊗A M → M is an injection. (Hint: if there is a counterexample
for an ideal J that is not finitely generated, use it to find another counterexample
for an ideal I that is finitely generated.)
By this criterion, to prove (a) it suffices to prove that the multiplication map^ →A^ is an injection for all finitely generated ideals I. But by part (b), this is
I ⊗A A^ is an injection; and this follows from part (c).
the same showing that ^I → A!

29.3 Defining types of singularities
Singularities are best defined in terms of completions. As an important firstexample, we finally define “node”.
29.3.1. Definition. Suppose X is a dimension 1 variety over k, and p ∈ X is a
closed point. We say that X has a node at p if the completion of OX,p at mX,p isisomorphic (as topological rings) to k[[x, y]]/(xy).
29.3.A. E XERCISE . Suppose k = k and char k ̸= 2. Show that the curve in A2k cut
out by (y2 − x2 − x3 ), (which we have been studying repeatedly since Figure 7.4)
has a node at the origin.
29.3.B. E XERCISE . Suppose k = k and char k ̸= 2, and we have f(x, y) ∈ k[x, y].
Show that Spec k[x, y]/(f(x, y)) has a node at the origin if and only if f has no termsof degree 0 or 1, and the degree 2 terms are not a perfect square. (This generalizes
Exercise 29.3.A.)

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The definition of node outside the case of varieties over algebraically closed
fields is more problematic, and we give some possible ways forward. For varieties
over a non-algebraically closed field k, one can always base change to the closurek. As an alternative approach, if p is a k-valued point of a variety over k (notnecessarily algebraically closed), then we could take the same Definition 29.3.1;
this might reasonably be called a split node, because the branches (or more precisely, the tangent directions) are distinguished. Those singularities that are notsplit nodes, but which become nodes after base change to k (such as the origin inSpec R[x, y]/(x2 + y2 )) might reasonably be called non-split nodes.
29.3.2. Other singularities. We may define other singularities similarly. To avoid
complications, we do so over an algebraically closed field.
29.3.3. Definition. Suppose X is a variety over k, and p is a closed point of X,
where char k ̸= 2, 3. We say that X has a cusp (resp. tacnode, triple point) at p
if O^X,p is isomorphic to the completion of the curve Spec k[x, y]/(y2 − x3 ) (resp.
Spec k[x, y]/(y2 − x4 ), Spec k[x, y]/(y3 − x3 )). We say that X has an ordinary multiple point of multiplicity m, or ordinary m-fold point, if O^X,p is isomorphic tothe completion of the curve Spec k[x, y]/(f(x, y)) where f is a homogeneous polynomial of degree m with no repeated roots. (You can see quickly that an ordinary
2-fold point — or ordinary double point — is precisely a node, and an ordinary 3-foldpoint is a triple point.)
29.3.C. T RIVIAL E XERCISE . (For this exercise, work over an algebraically closed
field for simplicity.) Define An curve singularity (see §22.4.4). Define An surface
singularity (see Exercise 22.4.F). Define Dn , E6 , E7 , and E8 surface and curve singularities (see Remark 22.4.5).
29.3.4. Using this definition. We now give an example of how this definition canbe used.
29.3.D. E XERCISE . Suppose X is a k-variety with a node at a closed point p. Show
that the blow up of X at p yields a morphism β : X̃ → X, where the exceptional
divisor β−1 (p) consists of two reduced smooth points. Hint: use the fact that completion is flat (Theorem 29.2.6(a)), and that blowing up commutes with flat basechange (Exercise 24.2.P(a)), to turn this into a calculation on the “formal model” ofthe node, Spec k[[x, y]]/(xy).
29.3.E. E XERCISE . Continuing the terminology of the previous exercise, describe
an exact sequence(29.3.4.1)

0 → OX → β∗ OX̃ → Op → 0.

Hint: cohomology commutes with flat base change, Theorem 24.2.8 (or more simply, the affine case version Exercise 18.8.B(b)).
29.3.F. E XERCISE . We continue the terminology of the previous two exercises. If X
is a pure-dimensional reduced projective curve, show that pa (X) = pa (X̃) + 1.
Thus “resolving a node” of a curve reduces the arithmetic genus of the curveby 1. If you wish, you can readily show that “resolving a cusp” reduces the genus
by 1, and “resolving a tacnode” reduces the genus by 2. In general, for each type of

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curve singularity, the contribution it makes to the genus — the difference between
the genus of the curve and that of its normalization, when the curve is projective— is called the δ-invariant. Thus δ for a node or cusp is 1, and δ for a tacnode is 2.
29.3.G. E XERCISE . Show that for any singularity type (other than a smooth point),
δ > 0.
29.3.H. E)XERCISE. Show that a reduced irreducible degree d plane curve can have*at most d−1
singularities.
2

29.3.I. E XERCISE . Here is a trick for speedily working out the genus of a nodal(projective reduced) curve when it comes up in examples. (The description will
be informal.) Assuming it is possible, draw a sketch on a piece of paper (i.e., thereal plane) where nodes “look like nodes”, and a curve of genus g is drawn withg holes, as in Figure 29.1. (In Figure 29.1, we see three irreducible components,of geometric genus 0, 2, and 3 respectively. The first two components are smooth,while the third meets itself twice. There are also three additional nodes, joining
each pair of the irreducible components.) Then if the curve is connected, count the
number of holes visible in the picture. Prove that this recipe works. (Feel free tofigure out what to do if the curve is not connected.)

F IGURE 29.1. This curve has genus 8 (count the “holes”!)
This can be quite useful. For example, the sketches) in *Figure 29.2 can remind
1
1
you that a degree d plane curve has arithmetic genus d−1
2 , the curve on P × P
coming from a section of O(a, b) (§16.4.8) has genus (a − 1)(b − 1), and a curve inclass 2C + 2F on the Hirzebruch surface F2 (using the language of Exercise 20.2.M)
has arithmetic genus 3. In each case, we are taking a suitable curve in the class, taking advantage of the fact that arithmetic genus is constant in flat families (Crucial
Exercise 24.7.B).
29.3.5. Other definitions. If k = C, then definitions of this sort agree with theanalytic definitions (see [Ar1, §1]). For example, a complex algebraic curve singularity is a node if and only if it is analytically isomorphic to an open neighborhoodof xy = 0 in C2 . There is also a notion of isomorphism “étale-locally”, which we

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F IGURE 29.2. These curves have arithmetic genus 6, 6, and 3 respectively
do not define here. Once again, this leads to the same definition of these types of
singularities (see [Ar2, §2]).

29.4 The Theorem on Formal Functions
Suppose π : X → Y is proper morphism of locally Noetherian schemes, and F isa coherent sheaf on X, so Ri π∗ F is a coherent sheaf (Grothendieck’s Coherence
Theorem 18.9.1). Fix a point q ∈ Y. We already have a sense that there is animperfect relationship between the fiber of Ri π∗ F at q and the cohomology of F
“on the fiber Xq ”, and much of Chapter 28 was devoted to making this precise.
The Theorem on Formal Functions deals with this issue in a different way.
Rather than comparing the fiber of Ri π∗ F with the cohomology of F restricted
to the fiber, it gives a precise isomorphism between information on “infinitesimal
thickenings” of the situation.
29.4.1. We now make this more precise. If Z is a closed subscheme of W cut out byideal sheaf I , we say that the closed subscheme cut out by I n+1 is the nth orderformal (or infinitesimal) neighborhood (or thickening) of Z (in W). The phrase“formal neighborhood” without mention of an “order” refers to the information
contained in all of these neighborhoods at once, often in the form of a limit of thesort we will soon describe. (This also leads us to the notion of formal schemes, an
important notion we will nonetheless not need.)
We turn now to our situation of interest. Rather than the fiber of Ri π∗ F at q,
we consider its completion: we take the stalk at q, and complete it at mY,q . Rather
than the cohomology of F along the fiber, we consider the cohomology of F whenrestricted to the nth formal neighborhood Xn of the fiber, and take the limit. The
Theorem of Formal Functions says that these are canonically identified.
Before we state the Theorem on Formal Functions, we need to be precise about
what these limits are, and how they are defined. To concentrate on the essential,
we do this in the special case where q is a closed point of Y, and leave the (mild)
extension to the general case to you (Exercise 29.4.A).
We deal first with Ri π∗ F . It is helpful to note that by Exercise 29.2.B, we
can compute the restriction to the nth formal neighborhood either by restricting/tensoring to the stalk Spec OY,q , or in any affine open neighborhood of q ∈ Y.

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For convenience, we pick an affine open neighborhood Spec B ⊂ Y of q. (Again,
by Exercise 29.2.B, it won’t matter which affine open subset we take.) Define
XB := π−1 (Spec B) for convenience. Let m ⊂ B be the maximal ideal corresponding to q. Then Ri π∗ (F )^q is canonically the completion of the coherent B-module
Hi (XB , F |XB ) at m:
Ri π∗ (F )^q = lim Hi (XB , F |XB )/mn Hi (XB , F |XB ).
←−We turn next to the cohomology of F on thickenings of the fiber. We haveclosed embeddings Xn "→ Xn+1 , and thus maps of cohomology groups Hi (Xn+1 , F |Xn+1 ) →Hi (Xn , F |Xn ). We have base change maps
(29.4.1.1)

Hi (XB , F |XB )/mn Hi (XB , F |XB ) → Hi (Xn , F |Xn )(see (28.1.0.3)) such that the square

Hi (XB , F |XB )/mn+1 Hi (XB , F |XB )"Hi (XB , F |XB )/mn Hi (XB , F |XB )% Hi (Xn+1 , F |X

n+1

)

"
% Hi (Xn , F |Xn )commutes. (Do you see why? Basically, this is again because they are base change
maps.) Thus we have an induced map of limits:(29.4.1.2)
29.4.A. E XERCISE .
closed.

Ri π∗ (F )^q → lim Hi (Xn , F |Xn ).
←−Extend the previous discussion to the case where q is not

The Theorem on Formal Functions states that this is an isomorphism.
29.4.2. Theorem on Formal Functions. — Suppose π : X → Y is a proper morphismof locally Noetherian schemes, F is a coherent sheaf on X, and q ∈ Y. Then (29.4.1.2) is
an isomorphism for all i ≥ 0.

Warning: the Theorem on Formal Functions does not imply anything about
the maps “at finite level”, i.e., about (29.4.1.1) for “finite n”.
The proof of Theorem 29.4.2 is quite subtle, and is postponed to the doublestarred §29.8. We first give some important applications.

29.5 Zariski’s Connectedness Lemma and Stein Factorization
We now state and prove Zariski’s Connectedness Lemma, which was mentioned in §28.1.8.
29.5.1. Zariski’s Connectedness Lemma. — If a proper morphism π : X → Y of
locally Noetherian schemes is O-connected (see §28.1.7), then π−1 (q) is connected forevery q ∈ Y.
The proof requires the following result.

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29.5.A. E ASY E XERCISE ( THE COMPLETION OF A LOCAL RING IS A LOCAL RING ).
Suppose (A, m) is a Noetherian local ring. Show that the completion of A along^ := lim A/mn , is a local ring, with maximal ideal mA.
^ Hint: show that any
m, A
←−
^
^
element of A not in mA is invertible.

Proof. Assume otherwise#that there is some q ∈ Y such that π−1 (q) is not con−1
nected, say π (q) = X1 X2 (where X1 and X2 are nonempty open subsets ofthe fiber, see Figure 29.3). Then the nth order formal neighborhood of π−1 (q), having the same topological space, is also disconnected. We use the useful trick ofidempotents (Remark 3.6.3). Let e1 be the function 1 on X1 and 0 on X2 , and let
e2 be the function 1 on X2 and 0 on X1 . These functions make sense for any orderformal neighborhood, and they have natural images in the inverse limit. Thus we^ with e1 + e2 = 1, e1 e2 = 0. Now, OY,q
^ is a local
get nonzero elements e1 , e2 ∈ OY,qring (Exercise 29.5.A). But e1 e2 = 0 implies that neither e1 nor e2 is invertible (orelse the other one would be 0), so both e1 and e2 are in the maximal ideal, andhence e1 + e2 can’t be 1. We thus have a contradiction.
!

F IGURE 29.3. “Formal neighborhoods” of X1 and X2 in the proofof Zariski’s Connectedness Lemma 29.5.1
29.5.2. Stein factorization. We next show the existence of a Stein factorization. We
could have given this construction long before, but now Zariski’s Connectedness
Lemma 29.5.1 will give us some impressive consequences.
29.5.3. Stein Factorization Theorem. — Any proper morphism π : X → Y of locally
Noetherian schemes can be factored into β◦α, where α : X → Y ′ is proper and O-connected(hence has connected fibers, by Zariski’s Connectedness Lemma 29.5.1), and β : Y ′ → Y isa finite morphism.
(29.5.3.1)

O-conn.
% Y′
X❃
α
❃❃
⑦
β ⑦⑦
❃❃
⑦
❃
π ❃❃
⑦⑦
8 5⑦⑦ finiteY

F IGURE 29.4. Stein Factorization

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We note that by the Cancellation Theorem 10.1.19 for projective morphisms, if
π is projective, then so is α (as δβ , being a closed embedding, is projective).
Although it is not in the statement of the theorem, the proof produces a specificfactorization, which is called the Stein factorization of π. The picture to have inmind is that the Stein factorization (roughly) contracts the continuous parts of thefibers, in some canonical way; see Figure 29.4.
As usual, the Noetherian hypotheses can be removed (see [Stacks, tag 03H2]).
Proof of the Stein Factorization Theorem 29.5.3. By Grothendieck’s Coherence Theorem 18.9.1, π∗ OX is a coherent sheaf (of algebras) on Y. Define Y ′ := Spec π∗ OX ,so (as π∗ OX is finite type) the structure morphism β : Y ′ → Y is finite. We have afactorization
% Y′
X❃
α
❃❃
⑦
β ⑦⑦
❃❃
⑦
❃
π ❃❃
⑦⑦
8 5⑦⑦ finite
Y
Finally, we show that α is O-connected. By the quasicoherence of both OY ′ andα∗ OX , it suffices to check on an affine cover of Y ′ . We choose as our affine coverof Y ′ the pullback of an affine cover of Y. If U is an affine open subset of Y ′ ,then on β−1 (U), on the level of rings, OY ′ → α∗ OX is precisely the isomorphismΓ (U, π∗ OX ) → Γ (U, π∗ OX ).
!

29.5.B. E XERCISE . Suppose π : X → Y is a proper morphism of locally Noetherian
schemes, that is an isomorphism over a dense open subset of Y. Suppose further
that Y is normal. Show that π is O-connected. (This applies, for example, to blowups of smooth varieties under smooth centers.) Hence, by Zariski’s Connectedness
Lemma 29.5.1, for every q ∈ Y, π−1 (q) is connected. This is called Zariski’s Main
Theorem (the version for birational morphisms). Hint: Exercise 9.7.P.
29.5.C. E XERCISE . Suppose π : X → Y and π ′ : X → Y ′ are two proper morphismsof locally Noetherian schemes, each contracting the same connected set of X (andan isomorphism elsewhere), and suppose Y and Y ′ are normal. Show that π and
π ′ are the same (or more precisely, that there is an isomorphism i : Y → Y ′ suchthat π ′ = i ◦ π). Informal translation: If X → Y has connected fibers, then Y isdetermined by just knowing the locus contracted on X. Hint: Identify Y and Y ′ astopological spaces, and then identify π and π ′ as maps of topological spaces. Use
Stein factorization to recover the structure sheaf as π∗ OX .
29.5.D. E XERCISE .
Show that the construction of the Stein factorization of amorphism π : X → Y of locally Noetherian schemes commutes with any flat basechange Z → Y of locally Noetherian schemes.

29.5.4. Resolution of singularities of curves.
If C is a reduced projective curve over a field k, then we can resolve its singularities by normalization: ν : C̃ → C. (See Remark 22.4.6 for some discussion ofresolution of singularities, and Theorem 12.5.8 for the reason why normal curves
are nonsingular.) But as we have seen from examples, it requires luck and insightto figure out how to take an integral closure, and I often have neither. We now take

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The Rising Sea: Foundations of Algebraic Geometrythe (inspired) guesswork out of desingularization by explaining how to desingularize by blowing up. The algorithm is simple: we find a singular point, then blow
it up, then look for more singular points to repeat the process. Clearly if there areno singular points to be found, then we are done; the problem is to show that thisis guaranteed to terminate. We do this by making use of an integer invariant: thearithmetic genus. We must show (i) that if pa (C) = pa (C̃), then ν is an isomorphism — C is already nonsingular; (ii) if C ′ → C is a birational morphism, thenpa (C ′ ) ≤ pa (C) (so pa (C̃) will be a lower bound for the arithmetic genus throughout the process); and (iii) that blowing up a singular point decreases the arithmetic
genus.
To set up these results, we consider the following situation. Suppose π : C ′ →C is a finite morphism (where C is as described in the previous paragraph, and C ′is a reduced curve), that is an isomorphism away from a finite closed subset of C.
Then the pullback map OC → π∗ OC ′ has no kernel, as C is reduced, and the map
is an isomorphism on the generic points of the components of C. The cokernel G
is supported on a finite set of closed points, and thus Hi (C, G ) = 0 for i > 0, and if
H0 (C, G ) = 0, then G = 0. From the exact sequence
0 → OC → π∗ OC ′ → G → 0,

and the fact that π is affine, χ(C ′ , OC ′ ) = χ(C, OC ) + χ(C, G ), from which pa (C) ≥pa (C ′ ), with equality if and only if π is O-connected (i.e., G = 0).
29.5.E. E XERCISE . Show that if π is O-connected, then π is an isomorphism. Hint:
use Zariski’s Connectedness Lemma 29.5.1 to show that the fibers are connected,
and hence that π is bijective. Identify C and C ′ as topological spaces. Then use theisomorphism OC → π∗ OC to identify C and C ′ as ringed spaces.

To complete our strategy, it remains to show that if β : C ′ → C is the blowup of C at a point p, and β is an isomorphism, then C is already nonsingularat that point. But if β is an isomorphism, then p is the exceptional divisor, andhence effective Cartier, and hence cut out locally by a singular equation. But thenmC,p ⊂ OC,p is principal, so OC,p is a regular local ring of dimension 1.
(You should verify that we have completed our strategy. Can you extend thisargument to the case where C is not projective?)
Resolution of singularities for surfaces will be briefly discussed in §29.7.4.

29.6 Zariski’s Main Theorem
Zariski’s Main Theorem is misnamed in all possible ways (although not horribly so). It is not a single Theorem, because there are many results that go by this
name, and they often seem quite unrelated. What they have in common is thatthey are intellectual descendants of a particular result of Zariski. This result is
not the Main Theorem of Zariski’s prolific career; the name comes because it was
the Main Theorem of a particular paper, [Z]. And finally, it is not Zariski’s in any
stronger sense than this; the modern versions are due to Grothendieck.
We have already seen one version of Zariski’s Main Theorem: the “birational”
form of Exercise 29.5.B. Our goal in this section is to prove a strong statement thatis close to the optimal version of the one most often used.

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Before we start, we note that the fiber of a proper morphism over a point q ∈ Y
is a proper (hence finite type) scheme over the residue field κ(q), and we say thata point is isolated in its fiber if it forms a component of the fiber of dimension 0.
29.6.1. Zariski’s Main Theorem (Grothendieck’s form). — Suppose π : X → Y is aproper morphism of locally Noetherian schemes.
(a) The set of points of X that are isolated in their fiber forms an open subset X0 ⊂ X.
(b) The morphism π|X0 : X0 → Y factors into an open embedding X0 "→ Y ′ followed by afinite morphism Y ′ → Y. Moreover, the morphism π : X → Y factors through Y ′ .
(29.6.1.1)

TX❄⑦⑦ ❄❄❄ α
⑦
❄❄
⑦
❄❄
⑦⑦
8
-+ ⑦⑦
5
π open L Y ′X0 4 ❅
⑧
❅❅
⑧⑧
❅❅
⑧
❅
⑧⑧ β finite
π|X0 ❅❅ _
8 5⑧⑧
Y
open

As with Stein factorization, the proof of the theorem yields a specific factorization.
If π is a morphism of varieties, we already know part (a), by upper semicontinuity of fiber dimension Theorem 11.4.2(a). For this reason, the proof in the caseof varieties is easier, and (even if you are interested in the more general case) you
are advised to read the proof restricting to this simpler case, to concentrate on themain ideas. But even in the case of varieties, we will need some of the ideas arisingin the proof of (a), so you should read it until advised to skip ahead to (b).
Proof. (a) Take the Stein factorization (29.5.3.1) of π:X❅❅❅❅❅O-conn.
α ❅❅
1
π
Y′
⑦
β ⑦⑦
⑦
⑦⑦
⑦
" a⑦ finiteY.
By the finiteness of β, a point of X is an isolated point in its fiber of π if and only if it
is an isolated point in its fiber of α. Thus we may replace by π by α. (Basically, we
are reducing to the case where π is O-connected, but we prefer to carefully call themorphism under consideration α so we can return to considering π in the proof of(b).)
Because fibers of proper O-connected morphisms of locally Noetherian schemes
are connected (Zariski’s Connectedness Lemma 29.5.1), a point is isolated in its
fiber of α if and only if it is its fiber.
If you are considering only the variety case, at this point you should jump to
the proof of part (b).
To show that X0 is open, we will identify it with the locus where Ωα = 0 (theunramified locus of α — not of π!), which is open by Exercise 21.6.H.

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29.6.A. E XERCISE . If Ωα |p = 0, show that p is isolated in its fiber of α (and henceof π).
We now show the other direction. Suppose p is isolated in its fiber, and letr ∈ Y ′ be α(p). We will show that α induces an isomorphism α♯ : OY ′ ,r → OX,p , so
ΩOX,p /OY ′ ,r = 0. As the construction of Ω behaves well with respect to localizationon the source and target (Exercise 21.2.L), this implies that (Ωα )p = 0, so Ωα is 0
at p.
The stalk OY ′ ,r is obtained by taking the limit (of sections of O) over all open
subsets of Y ′ containing r. For any such open subset V ⊂ Y ′ , the condition∼ Γ (α−1 (V), OX ) (compatible withα∗ OX = OY ′ gives an isomorphism Γ (V, OY ′ ) =inclusions). We will show that the system of open subsets α−1 (V) (as V varies
through open neighborhoods of r), which each necessarily contain p, is cofinal
with the system of open neighborhoods of p in X. To do this, we must show thatfor any open neighborhood U of p in X, there is an open neighborhood V of r in
Y such that α−1 (V) ⊂ U. To do this, note that X \ U is closed in X, so α(X \ U) is
closed in Y (as α is proper), so its complement V := Y \ α(X \ U) is open. Simply
note that α−1 (V) ⊂ U (implicitly using that α, being proper and O-connected, issurjective, see Exercise 28.1.G), and we are done.
(b) In our proof of (a), we have established that α gives a bijection between
X0 and an open subset of Y ′ . Notice that this is furthermore a homeomorphism:
any open subset of Y ′ pulls back to an open subset of X0 by the continuity ofα; and if U is an open subset of X0 , then α(X \ U) is a closed subset of Y ′ , so
its complement (which is α(U), using that proper O-connected morphisms are
surjective, Exercise 28.1.G) is open.
Using this isomorphism of topological spaces, the condition of O-connectednessof α (restricted to α(X0 )) shows that α gives an isomorphism of ringed spaces (i.e.,of schemes) X0 → α(X0 ). We have found our desired factorization.
!
29.6.B. E XERCISE . Prove Zariski’s Main Theorem with Y affine, say Spec A, andX a quasiprojective A-scheme. (Hint: by the Definition 4.5.9 of quasiprojective Ascheme, we can find an open embedding X "→ X ′ into a projective A-scheme. Apply Theorem 29.6.1 to X ′ → Spec A.)

For the next two applications (Exercises 29.6.D and 29.6.E), we will need an annoying hypothesis, which is foreshadowed by the previous exercise. We say thata morphism π : X → Y of locally Noetherian schemes satisfies (†) if for all affines
Ui in some open cover of Y, the restriction π|π−1 (Ui ) : π−1 (Ui ) → Ui factors as
an open embedding into a scheme proper over Ui . A morphism satisfying (†) isnecessarily separated and finite type. It turns out that this is sufficient: Nagata’s
Compactification Theorem states that every separated finite type morphism of Noetherian schemes π : X → Y can be factored into an open embedding into a scheme
proper over Y. (See [Lü] for a proof. The Noetherian hypotheses can be replacedby the condition that Y is quasicompact and quasiseparated, see [Co].)29.6.C. E XERCISE . Show that if π : X → Y is a morphism of varieties over k, and
X is an open subset of a proper variety Z, then π satisfies (†). (Hint: X is an open
subscheme of its closure in Z ×k Y.)

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29.6.D. E XERCISE . Suppose π : X → Y is a quasifinite birational morphism of integral locally Noetherian schemes, and that Y is normal. Suppose further π satisfies(†). Show that π is an open embedding. (Hence if furthermore π is a bijection, thenπ must be an isomorphism.)
29.6.E. E XERCISE . Suppose π : X → Y is a quasifinite morphism of Noetherian
schemes satisfying (†). Show that π factors as an open embedding into a finite
morphism.
"
open
% Y′
X! ❃
❃❃
⑦
❃❃
⑦⑦
❃
⑦⑦
π ❃❃
⑦
finite
8 5⑦⑦
YThis shows that it is reasonable to think of quasifinite morphisms as “open subsets
of finite morphisms”, as promised in §7.3.15.

As a final application of Zariski’s Main Theorem, we can finally prove a characterization of finite morphisms that we have mentioned a number of times.

29.6.2. Theorem. — Suppose π : X → Y is a morphism of locally Noetherian schemes.
The following are equivalent.
(a) π is finite.
(b) π is affine and proper.
(c) π is proper and quasifinite.

As usual, Noetherian hypotheses can be removed: [Gr-EGA, IV3 .8.11.1] showsthat proper, quasifinite, locally finitely presented morphisms are finite.
Proof. Finite morphisms are affine by definition, and proper by Proposition 10.3.3,
so (a) implies (b).
To show that (b) implies (c), we need only show that affine proper morphisms
have finite fibers (as the “finite type” part of the definition of quasifiniteness istaken care of by properness). This was shown in Exercise 18.9.A.
Finally, we assume (c), that π is proper and quasifinite, and show (a), that π isfinite. By Zariski’s Main Theorem 29.6.1, we have a factorization (29.6.1.1).
"
open
% Y′
X0 ! ❅
γ
⑧
❅❅
⑧
❅❅
⑧⑧
π ❅❅❅
⑧⑧ finite⑧
8 5⑧YNow X = X0 . (Do you see why? Hint: Exercise 7.4.D followed by Exercise 7.3.H.)
By the Cancellation Theorem 10.1.19 for proper morphisms (using that Y ′ → Y is
finite hence separated), γ is proper.
29.6.F. E XERCISE . Show that proper quasicompact open embeddings are closed
embeddings. Hint: use Corollary 8.3.5 to show that the image is open and closed
in the target.
Applying this to our situation, we see that X → Y is the composition of two
finite morphisms γ : X "→ Y ′ and Y ′ → Y, and is thus finite itself.
!

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29.6.3. ⋆⋆ Other versions of Zariski’s Main Theorem.
The Noetherian conditions in Theorem 29.6.1 can be relaxed (see [Gr-EGA,
IV3 .8.12.6] or [GW, Thm. 12.73]): if π : X → Y is a separated morphism of finite
type, and Y is quasicompact and quasiseparated, then the set of points X0 isolated
in their fiber is open in X, and for every quasicompact open subset U of X0 , there
is a factorization
% Y′
U❄
γ
❄❄
⑦
⑦
β ⑦
❄❄
⑦
❄
⑦⑦
π|U ❄❄⑦
8 5⑦ finiteYsuch that β is finite, γ|U is a quasicompact open embedding, and γ−1 (γ(U)) = U.
There are also topological and power series forms of Zariski’s Main Theorem,see [Mu7, §III.9].

29.7 Castelnuovo’s Criterion for contracting (−1)-curves
We now prove a result used in Chapter 27. We showed in Exercise 22.4.O that if weblow up a regular point of a surface at a (reduced) point, the exceptional divisor isa (−1)-curve (see Definition 20.2.7: it is isomorphic to P1 , and has normal bundle
OC (−1)). Castelnuovo’s Criterion is the converse: if we have a quasiprojective
surface containing a (−1)-curve, that surface is obtained by blowing up anothersurface at a reduced regular point. We say that we can blow down the (−1)-curve.
29.7.1. Theorem (Castelnuovo’s Criterion). — Let C ⊂ X be a (−1)-curve on asmooth projective surface over k. Then there exists a birational morphism π : X → Y suchthat Y is a smooth projective surface, π(C) is a k-valued point, π is the blow-up of Y atthat point, and C is the exceptional divisor of the blow-up.
By Exercise 29.5.C, there is only one way to “blow down C”: this contractionis unique.
Proof. The proof is in three steps. Step 1. We construct π : X → Y that contracts C toa point q (and is otherwise an isomorphism). At this point we will know that Y is a
projective variety. Step 2. Then we show that Y is smooth; this is the hard step, andrequires the Theorem on Formal Functions 29.4.2. Step 3. Finally, we recognize π
as the blow-up Blq Y → Y.

Step 1. We do this by finding an invertible sheaf L whose complete linear
series will do the job. We start with a very ample invertible sheaf H on X suchthat H1 (X, H ) = 0. We can find such an invertible sheaf H by choosing any
very ample invertible sheaf, and then taking a suitably large multiple, invoking
Exercise 16.6.C (very ample ⊗ very ample = very ample) and Serre vanishing (Theorem 18.1.4(ii)). Let k = H ·C; k > 0 by the very ampleness of H (Exercise 20.1.K).
29.7.A. E XERCISE . Show that H1 (X, H (iC)) = 0 for 0 ≤ i ≤ k. Hint: use induction
on i, using(29.7.1.1)

0 → H ((i − 1)C) → H (iC) → H (iC)|C → 0.
∼ P1 ; which O(n) is it?
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Define L := H (kC). We will use the sections of H to obtain sections of L ,via the map φ : H0 (X, H ) → H0 (X, H (kC)).
29.7.B. E XERCISE . Show that L is base-point-free. Hint: To show that L has
no base points away from C, consider the image of φ. To show that L has nobase points on C, use that from (29.7.1.1) for i = k, we have that H0 (X, L ) →∼ OC .
H0 (C, L |C ) is surjective, and L |C =Thus the complete linear series |L | yields a morphism π ′ from X to some projective space PN .
29.7.C. E XERCISE . Show that π ′ precisely contracts C to a point q ′ . More explicitly,
show that π ′ sends C to a point q ′ , and π ′ |X\C : X \ C → PN \ {q ′ } is a closed embedding. Hint: To show that π ′ gives a closed embedding away from C: consider
the image of φ. To show that C is contracted by π ′ , use the fact that deg L |C = 0.
Let Y ′ be the image of π ′ , so we have a morphism X → Y ′ (which we also
call π ′ ). Let ν : Y → Y ′ be the normalization of Y ′ , so π ′ lifts to π : X → Y by theuniversal property of normalization (§9.7). (We are normalizing in order to use
Exercise 29.5.B in Step 2.)
29.7.D. E ASY EXERCISE . Show that ν−1 (q ′ ) ⊂ Y is a single point, which we call q;
and that π(C) = q. (Hint: the image of π is closed, and π(C) is connected.)
29.7.E. E XERCISE . Let X be a smooth projective surface, containing a curve C ⊂ X
isomorphic to P1k . If C·C < 0, show that there is a morphism π : X → X ′ contracting
C to a point, and leaving the rest of X unchanged. (This will not be used, and isincluded to give practice with the argument of Step 1.)
Step 2. We now show that Y is a smooth surface. As Y is normal, we have
∼ OY (π : X → Y is O-connected), by Exercise 29.5.B. We will showthat π∗ OX =∼ k[[x, y]] as topological rings, asthat Y is smooth at q by showing that O^Y,p =∼then dimk (m/m2 ) = 2. By the Theorem on Formal Functions 29.4.2, O^Y,p =lim H0 (Cn′ , OCn′ ), where Cn′ is the closed subscheme of X that is the scheme-theoretic←−
pullback of the nth order neighborhood of q.
We are not precisely sure what the nth order neighborhood is, but that is fine;nthis inverse system is cofinal with O(−nC) = IC/X
(do you see why?), so we cantake this inverse limit instead (by Exercise 29.2.A). It suffices to show that for all∼ k[[x, y]]/(x, y)n where Cn is defined as 0 → O(−nC) →n ≥ 0, H0 (Cn , OCn ) =OX → OCn → 0.
We do this by induction on n. As a k-vector space, this is easy, using
0 → I n /I n+1 → OCn+1 → OCn → 0(the closed subscheme exact sequence (13.5.4.1)) for Cn in Cn+1* ), using the canonical isomorphism H0 (P1 , OP1 (n)) = Symn H0 (P1 , OP1 (1)) . The tricky thing isthat we want this isomorphism as rings.
29.7.F. E XERCISE . Prove this. (This may remind you of how we found the ringof functions on the total space of an invertible sheaf in Exercise 19.11.E. This is no
coincidence.)
Hence we have smoothness, completing Step 2.

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Step 3. We now must recognize π as the blow-up of Y at q. As C is an effective
Cartier divisor on X, by the universal property of blowing up means that there isa unique morphism α making the following diagram commute
X❂
❂❂
❂❂
π ❂❂❂
F

α

Y

% Blq Y
④
④
④
④
④④ β
b④④

By the Cancellation Theorem 10.1.19 for projective morphisms, α is projective.
29.7.G. E XERCISE . Show that α has finite fibers.
As projective morphisms with finite fibers are finite (Theorem 18.1.8), α is a
finite morphism. But then α is a birational finite morphism to a normal scheme,
and thus an isomorphism by Exercise 9.7.P.
!
So far we have only used the Theorem on Formal Functions 29.4.2 for H0 . The
next exercise will give you a chance to see it in action for other cohomology groups.
29.7.H. E XERCISE .
(a) Suppose β : X = Blq Y → Y is the blow-up of a smooth surface over k at asmooth k-valued point. Show that R1 β∗ OX = R2 β∗ OX = 0.
(b) Suppose further that Y is projective. Show that the natural maps β∗ : H1 (Y, OY ) →H1 (X, OX ) and β∗ : H2 (Y, OY ) → H2 (X, OX ) are isomorphisms.
29.7.2. Elementary transformations.
Recall the definition of the Hirzebruch surfaces Fn (over a field k) from Example 17.2.4. If n > 0, then on Fn , we have a unique curve E of maximally negative
self-intersection (Exercise 20.2.R). It has self-intersection E · E = −n. It is a section∼ P1 × P1 , E is usually defined asof the projection π : Fn → P1 . (If n = 0, so Fn =1
1
1
any “constant” section of π : P × P → P — it is not unique.)Suppose q is a k-valued point of Fn . Let F be the fiber of π containing q.
29.7.I. E XERCISE . Show that F · F = 0.

Let β : Blq Fn → Fn be the blow-up at q, and let Z be the exceptional divisor(as the name “E” is taken). Let F ′ be the proper transform of F, and let E ′ be theproper transform of E.
29.7.J. E XERCISE . Show that F ′ · F ′ = −1.

Thus by Castelnuovo’s Criterion, we can blow down F ′ , to obtain a new surface Y.
∼ Fn−1 . If q ∈ E,
29.7.K. E XERCISE . Suppose n > 0. If q ∈/ E, show that Y =∼ Fn+1 . Possible hint: If you knew that Y was a Hirzebruch surface
show that Y =Fm , you could recognize it by the self-intersection of the unique curve of negative
self-intersection if m > 0.
∼ F1 . Hint: Exercise 22.4.K.
29.7.L. E XERCISE . Suppose n = 0. Show that Y =This discussion can be generalized to P1 -bundles over more general curves.
The following is just a first step in the story.

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29.7.M. E XERCISE . Suppose C is a smooth curve over k, π : X → C is a P1 -bundle
over C. If q is a k-valued point of C, then we can blow up q and blow down theproper transform of the fiber through q. Show that the result is another P1 -bundle
X ′ over C. (This is called an elementary transformation of the ruled surface.)
29.7.N. E XERCISE . Suppose X = Proj C (L ⊕ M ), and q lies on the section corresponding to
0 → L → L ⊕ M → M → 0.
′ ∼Show that X = Proj (L (π(q)) ⊕ M )).
C∼ Proj V ,Can you speculate (or show) what happens if you are just told that X =C
where V is a locally free sheaf of rank 2, and q lies on a section corresponding to(20.2.8.1)?
29.7.3. Minimal models of surfaces.
A surface over an algebraically closed field is called minimal if it has no (−1)curves. If a surface X is not minimal, then we can choose a (−1)-curve, andblow it down. If the resulting surface is not minimal, we can again blow downa (−1)-curve, and so on. By the finiteness of the Picard number ρ (which we haveadmittedly not proved, see §18.4.10), as blowing down reduces ρ by 1 (by Exercise 22.4.O), this process must terminate. By this means we construct a minimalmodel of X — a minimal surface birational to X. If minimal models were unique,this would provide a means of classifying surfaces up to birationality: if X and X ′were smooth projective surfaces, to see if they were birational, we would find theirminimal models, and see if they were isomorphic.
Sadly, minimal models are not unique. The example of P1 × P1 and P2 showthat there can be more than one minimal model in a birational equivalence class.
(Why are these two surfaces minimal?) Furthermore, the isomorphism between
the blow-up of P2 at two points with the blow-up of P1 × P1 at one point (Exercise 22.4.K) shows that a smooth surface can have more than one minimal model.
Nonetheless, the failure of uniqueness is well-understood: it was shown byZariski that the only surfaces with more than one minimal model are those birational to C×P1 for some curve C. This is part of the Enriques-Kodaira classificationof surfaces. For more on this, see [Be], [Ba], or [BPV, §6].
29.7.4. Here are two more facts worth mentioning but not proving here. First, singular surfaces can be desingularized by normalizing and then repeatedly blowing
up points; this explicit desingularization is analogous to the version for curvesin §29.5.4, and is generalized by Hironaka’s Theorem on resolution of singularities(Remark 22.4.6). Second, birational maps X (❴ ❴ ❴% Y between smooth surfaces canbe factored into a finite number of blow-ups followed by a finite number of blowdowns. As a consequence, by Exercise 29.7.H, birational projective surfaces havethe same Hi (O) for i = 0, 1, 2. (Using facts stated in §21.5.10, you can show thatover C, this implies that the entire Hodge diamond except for h1,1 is a birational
invariant of surfaces.)

29.8 ⋆⋆ Proof of the Theorem on Formal Functions 29.4.2

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In this section, we will prove the following.
29.8.1. Theorem. — Suppose X → Spec B is a proper morphism, with B Noetherian,^ is the I-adic completionand F is a coherent sheaf on X. Suppose I is an ideal of B, and B
lim B/In of B. Let Xn := X ×B (B/In+1 ) (the nth order formal neighborhood of the fiber←−X0 ) and let Fn be the pullback of F to Xn . Then for all i, the natural map(29.8.1.1)

is an isomorphism.

^ → lim Hi (Xn , Fn )Hi (X, F ) ⊗B B←−

29.8.A. E XERCISE . Prove that Theorem 29.8.1 implies Theorem 29.4.2. (This isessentially immediate if q is a closed point, and might require a little thought if q
is not closed; cf. Exercise 29.4.A.)
On the right side of (29.8.1.1), we have morphisms X1 "→ X2 "→ · · · "→ X, fromwhich we get restriction maps Hi (X1 , F1 ) ← Hi (X2 , F2 ) ← · · · ← Hi (X, F ). But
notice that the Xi are all supported on the same underlying set, and it is helpful tokeep in mind that the Fi ’s are all sheaves on the same topological space.
We will most often apply this in the case where I is a maximal ideal. But wemay as well prove the result in this generality. The result can be relativized (with B
and I replaced by a scheme and a closed subscheme), but we won’t bother givinga precise statement.
Our proof will only use properness through the fact that the pushforward ofcoherent sheaves under proper morphisms are coherent (Grothendieck’s Coherence Theorem 18.9.1), which required hard work. If you haven’t read that proof,
you can retreat to the projective case with little loss.
^
We will show that both sides of (29.8.1.1) are B-modules,
and that the map of^(29.8.1.1) is an isomorphism of topological B-modules.
Write In F for the coherent sheaf on X defined as you might expect: if I is
the pullback of 3I from Spec B to X, In F is the image of I n ⊗ F → F . Then
Hi (Xn , Fn ) can be interpreted as Hi (X, F /In F ) (by §18.1 property (v)), so theterms in the limit on the right side of (29.8.1.1) all live on the single space X. From
now on we work on X.
The map in the statement of the theorem arises from the restriction Hi (X, F ) →Hi (X, F /In F ). This appears in the long exact sequence associated to
0 → In F → F → F /In F → 0,

and in particular in the portion
Hi (X, In F )un% Hi (X, F )We shrink this further to:(29.8.1.2)% Hi (X, F )/ im(un )0% Hi (X, F /In F )% Hi (X, F /In F )% Hi+1 (X, In F )% ker(vn )vn% Hi+1 (X, F ).

%0and then apply lim to both sides. Limits are left-exact (§1.6.12), so the result is←−
left-exact. But the situation is even better than that. The transition maps of the left
term for our exact sequences,
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are clearly surjective, so by Exercise 1.6.M we have an exact sequence(29.8.1.3)% lim Hi (X, F )/ im(un ) µ % lim Hi (X, F /In F )% lim ker(vn )
0
←−
←−
←−

745

% 0.

Two key facts will imply our result. (i) We will show that {im(un )} is cofinal
with the I-adic topology on Hi (X, F ). Thus by Exercise 29.2.A, we have a natural
isomorphism of the left side of (29.8.1.3) with
∼ Hi (X, F )^.
lim Hi (X, F )/In Hi (X, F ) =←−Because Hi (X, F ) is a finitely generated B-module (by Grothendieck’s Coherence
Theorem 18.9.1, or more simply Exercise 18.1.A in the projective case), this is iso^ by Theorem 29.2.6(b), the left side of (29.8.1.1). (ii) We
morphic to Hi (X, F ) ⊗B B
will show that there is some d such that the transition map ker(vn+d ) → ker(vn )is the zero map for n ≫ 0. This implies that the right term lim ker(vn ) in (29.8.1.3)←−
is 0. (Do you see why?) As a result, the map µ in (29.8.1.3) becomes our desired
isomorphism. We now begin these two tasks.
(i) We will show that there is some n0 such that for n ≥ n0 ,In Hi (X, F ) ⊂ im(un ) ⊂ In−n0 Hi (X, F ).

The first inclusion is straightforward: note that Hi (X, F ) → Hi (X, F /In F ) is Blinear with kernel im(un ), and target annihilated by In .
We will make use of the ring R• := B ⊕ I ⊕ I2 ⊕ · · · (the Rees algebra, defined in§12.9.1). It is a graded ring generated in degree 1. Because I is a finitely generated
ideal of B, R• is a finitely generated B-algebra, and thus Noetherian.
For each a ∈ Im , we have a commutative diagram of quasicoherent sheaves
In F
"#
"
F

×a

×a

% In+m" F#"%F

which induces a commutative diagram of B-modules
Hi (X, In F )×a% Hi (X, In+m F )"Hi (X, F )×a"% Hi (X, F ).

In this way, ⊕n≥0 Hi (X, In F ) and ⊕n≥0 Hi (X, F ) are graded R• -modules (wherea is considered as an element of Rm ). Then ⊕n≥0 im(un ) is the image of the mapof R• -modules(29.8.1.4)⊕ Hi (X, In F ) → ⊕Hi (X, F ),and is thus itself an R• -module.
We will show that ⊕n≥0 Hi (X, In F ) is a finitely generated R• -module. This will
imply that its image ⊕n≥0 im(un ) is also finitely generated, and thus by Proposition 12.9.2, for some n0 im(un+1 ) = I im(un ) for n ≥ n0 , which in turn implies (byinduction) that for n ≥ n0 , we have im(un ) ⊂ In−n0 Hi (X, F ), completing task (i).

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Now cohomology of quasicoherent sheaves commutes with direct sums (§18.1(vii)), so ⊕n≥0 Hi (X, In F ) = Hi (X, ⊕n≥0 In F ).
29.8.B. E XERCISE . Consider the fibered diagram
X ×B R•

σ

%X

ρ

"
Spec R•"% Spec B.

Interpret ⊕n Hi (X, In F ) as ρ∗ σ∗ F . (Be sure to check that the R• -action determinedby the pushforward agrees with the R• -action on ⊕n Hi (X, In F ) described above!)
But σ∗ F is coherent (pullback takes finite type quasicoherent sheaves to finite type quasicoherent sheaves, and all schemes here are locally Noetherian),so ρ∗ σ∗ F is coherent (proper pushforward of coherent sheaves on Noetherian
schemes are coherent, by Grothendieck’s Coherence Theorem 18.9.1). We have
now completed task (i).
(ii) For the same reason that ⊕n≥0 Hi (X, In F ) is a coherent R• -module, ⊕n≥0 Hi+1 (X, In F )is a coherent R• -module. (Just replace i with i + 1 throughout the discussion in (i).)
Notice that ⊕n≥0 ker vn is the kernel of the map of graded R• -modules⊕n≥0 Hi+1 (X, In F )⊕n≥0 vn% ⊕n≥0 Hi+1 (X, F )(which is just (29.8.1.4) with i replaced by i + 1) and thus itself has the structure of a graded R• -module. It is a submodule of the finitely generated module⊕Hi+1 (X, In F ), so ⊕n≥0 ker vn is finitely generated as well. We invoke Proposition 12.9.2 a second time, to show that there is some d such that ker(vn+1 ) =I ker(vn ) for n ≥ d, from which(29.8.1.5)

ker(vA+B ) = IA ker vBforA ≥ 0 and B ≥ d.

For any a ∈ Id , the map
×a : In F → In+d F

induces a map Hi+1 (X, In F ) → Hi+1 (X, In+d F ), which sends ker(vn ) to ker(vn+d ).
(This can be interpreted as coming from the action of R• on the module ⊕ ker(vn ),taking a as an element of Rd .) As a runs through Id , the B-linear spans of theimages ofker(vn )% Hi+1 (X, In F )×a% Hi+1 (X, In+d F )for n ≥ d is Id ker(vn ) = ker(vn+d ) by (29.8.1.5). The composition
Hi+1 (X, In F )×a% Hi+1 (X, In+d F )% Hi+1 (X, In F )(the latter map coming from inclusion of sheaves) is exactly the map
×a : Hi+1 (X, In F ) → Hi+1 (X, In F ).

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Thus the image of ker(vn+d ) → ker(vn ) is exactly the B-linear span of the imagesof the composition maps"
% Hi+1 (X, In F ) ×a % Hi+1 (X, In F )ker(vn ) !
as a runs through Id .
It thus suffices to show that the multiplication by a ∈ Id on Hi+1 (X, In F ) annihilates ker(vn ) for n ≥ d. Because ker(vn ) = In−d ker(vd ) for all n ≥ d ((29.8.1.5)
again), it suffices to show that any element a of Id annihilates ker(vd ). But ker(vd )is the image of Hi (X, F /Id F ) (from the long exact sequence, see (29.8.1.2)), andthus is annihilated by any element of Id , as desired.
!

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CHAPTER 30⋆ Proof of Serre duality
Réfléchissant un peu à ton théorème de dualité, je m’aperçois que sa formulationgénérale est à peu près évidente, et d’ailleurs je viens de vérifier qu’elle se trouve implicitement (dans le cas de l’espace projectif) dans ton théorème donnant les T q (M) par des Ext.
(J’ai bien l’impression, salaud, que tes §3 et 4 du Chap. 3 peuvent se faire aussi sans aucun
calcul).
Thinking a bit about your duality theorem, I notice that its general form is almost
obvious, and in fact I just checked that (for a projective space) it is implicitly contained inyour theorem giving the T q (M) in terms of Exts. (I have the impression, you bastard, that
§3 and 4 in your Chap. 3 could be done without any computation).
— A. Grothendieck, letter to J.-P. Serre, December 15, 1955 [GrS, p. 19]

30.1 Introduction
We first met Serre duality in §18.4 (Theorem 18.5.1), and we have repeatedly
seen how useful it is. We will prove Theorem 18.5.1 (Corollary 30.3.10, combinedwith Exercise 30.4.I, see Remark 30.4.9), as well as stronger versions, and we willbe left with a desire to prove even more. We give three statements (Serre duality
for vector bundles; Serre duality for Hom; and Serre duality for Ext), in two versions(functorial and trace). (These names are idiosyncratic and nonstandard.) We give
several variants for a number of reasons. First, the easier statements will be easierto prove, and the hardest statements we won’t be able to prove here. Second, they
may help give you experience in knowing how to know what to hope for, andwhat to try to prove.
Throughout this chapter, X will be a projective k-scheme of pure dimension
n. We will want a coherent sheaf ω (or with more precision, ωX , or even better,ωX/k ) on X, the dualizing sheaf, which will play a role in the statements of duality.
For the best statements, we will want a trace morphism(30.1.0.1)

t : Hn (X, ωX ) → k.

30.1.1. Desideratum: the determinant of the cotangent bundle is dualizing forsmooth varieties. If X is smooth, we will want ωX = KX in this case (recall
KX = det ΩX ) — the miracle that the canonical bundle is Serre-dualizing (§18.5.2).
In particular, ωX is an invertible sheaf. This will be disturbingly harder to prove749

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than the basic duality statements we show; we will only get to it later (Exercise 30.4.I). But we will prove more, for example that ωX is an invertible sheaf
if X is a regular embedding in a smooth variety (Exercise 30.4.H).
30.1.2. Desideratum: Serre duality for vector bundles. The first version of duality, which (along with Desideratum 30.1.1) gives Theorem 18.5.1, is the following:
if F is locally free of finite rank, then we have a functorial isomorphism∼ Hn−i (X, F ∨ ⊗ ωX )∨ .
(30.1.2.1)
Hi (X, F ) =More precisely, we want to construct a particular isomorphism (30.1.2.1), or equivalently, a particular perfect pairing Hi (X, F ) × Hn−i (X, F ∨ ⊗ ωX ) → k. This isomorphism will be functorial in F , i.e., it gives a natural isomorphism of covariant
functors∼ %(30.1.2.2)
Hi (X, ·)Hn−i (X, ·∨ ⊗ ωX )∨ .

We call this functorial Serre duality for vector bundles.
Better still, there should be a cup product in cohomology, which can be used
to construct a map Hi (X, F ) × Hn−i (X, F ∨ ⊗ ωX ) → Hn (X, ωX ). This should
be functorial in F (in the sense that we get a natural transformation of functors Hi (X, ·) → Hn−i (X, ·∨ ⊗ ωX )∨ ⊗ Hn (X, ωX )). Combined with the trace map(30.1.0.1), we get a map Hi (X, F ) × Hn−i (X, F ∨ ⊗ ωX ) → k, which should yield(30.1.2.2). We call this the trace version of Serre duality for vector bundles. (Thiswas hinted at after the statement of Theorem 18.5.1.)
In fact, the trace version of Serre duality for vector bundles (and hence thefunctorial version) is true when X is Cohen-Macaulay, and in particular when X issmooth. We will prove the functorial version (see Corollary 30.3.10). We will givean indication of how the trace version can be proved in Remark 30.3.15.
30.1.3. Desideratum: duality for more general X and F .
A weaker sort ofduality will hold with weaker hypotheses. We will show that (without CohenMacaulay hypotheses) for any coherent sheaf F on a pure n-dimensional projective k-scheme X, there is a functorial isomorphism(30.1.3.1)

Hom(F , ωX )∼% Hn (X, F )∨ .

We call this functorial Serre duality for Hom.
In parallel with Serre duality for vector bundles, we have a natural candidate
for the perfect pairing:(30.1.3.2)

Hom(F , ωX ) × Hn (X, F )% Hn (X, ωX )

t

% k,

where the trace map t is some linear functional on Hn (X, ωX ). If this composition
is a perfect pairing, we say that X (with the additional data of (ωX , t)) satisfies thetrace version of Serre duality for Hom. Unlike the trace version of Serre duality
for vector bundles, we already know what the “cup product” map Hom(F , ωX ) ×
Hn (X, F ) → Hn (X, ωX ) is: an element [σ : F → ωX ] of Hom(F , ωX ) induces— by covariance of Hn (X, ·), see §18.1 — a map Hn (X, F ) → Hn (X, ωX ). The
resulting pairing is clearly functorial in F , so the trace version of Serre duality
for Hom implies the functorial version. Unlike the case of Serre duality for vector
bundles, the functorial version of Serre duality for Hom also implies the traceversion:

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30.1.A. E XERCISE . Show that the functorial version of Serre duality for Hom implies the trace version. In other words, the trace map is already implicit in thefunctorial isomorphism Hom(F , ωX ) → Hn (X, F )∨ . Hint: consider the commuting diagram (coming from functoriality)
Hom(F , ωX )7

% Hn (X, F )∨7

Hom(ωX , ωX )% Hn (X, ωX )∨ .

30.1.4. Definition. Suppose X is a projective k-scheme of pure dimension n. A
coherent sheaf ω = ωX = ωX/k along with a map t : Hn (X, ωX ) → k is called
dualizing if the natural map (cf. (30.1.3.1))(30.1.4.1)

Hom(F , ωX ) × Hn (X, F )% Hn (X, ωX )t%kis a perfect pairing for all coherent sheaves F on X. We call ωX the dualizing sheaf
and t the trace map. (The earlier discussion of ωX and t was aspirational. This
now is a definition.) If X has such (ωX , t), we say that X satisfies Serre duality (forHom). The following proposition justifies the use of the word “the” (as opposed
to “a”) in the phrase “the dualizing sheaf”.
30.1.5. Proposition. — If a dualizing sheaf and trace (ωX , t) exists for X, this data is
unique up to unique isomorphism.
Proof. Suppose we have two such (ωX , t) and (ωX′ , t ′ ). From the two morphisms(30.1.5.1)

Hom(F , ωX ) × Hn (X, F )% Hn (X, ωX )t

Hom(F , ωX′ ) × Hn (X, F )% Hn (X, ω ′ )

t′

X

%k

% k,

∼ Hom(F , ω ′ ), which is functorial inwe get a natural bijection Hom(F , ωX ) =X
F . By the typical universal property argument (Exercise 1.3.Z), this induces a∼ ω ′ . From (30.1.5.1), under this isomorphism, the(unique) isomorphism ωX =X′
two trace maps t and t must be the same too.
!
We will prove the functorial and trace versions of Serre duality for Hom in
Corollary 30.3.12. The special case of projective space will be a key ingredient; weprove this case now.
30.1.6. Serre duality (for Hom) for projective space. Define ωPnk (or just ω for connvenience) as OPnk (−n − 1). Let t be any isomorphism Hn (Pnk , ωPk ) → k (Theorem 18.1.3). As the notation suggests, (ωPnk , t) will be dualizing for projective
space Pnk.
30.1.B. E XERCISE . Suppose F = OPnk (m). Show that the natural map (30.1.4.1)
is a perfect perfect pairing. (Hint: do this by hand. See the discussion after Theorem 18.1.3.) Hence show that if F is a direct sum of line bundles on Pn
k , the naturalmap (30.1.4.1) is a perfect pairing.

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30.1.7. Proposition. — The functorial version (and hence the trace version by Exercise 30.1.A) of Serre duality for Hom holds for Pnk.
Proof. We wish to show that the (functorial) natural map
Hom(F , ωPnk )∼% Hn (Pn , F )∨ ,kcoming from the cup pairing and the choice of isomorphism t, is an isomorphism(cf. (30.1.3.1)). Fix a coherent sheaf F on Pn
k . By Theorem 15.3.1, we can present
F as
%G
%E
%F
%0
(30.1.7.1)
0where E is a finite direct sum of line bundles, and G is coherent. Applying theleft-exact functor Hom(·, ωPnk ) to (30.1.7.1), we have the exact sequence(30.1.7.2)% Hom(F , ωPn )0k% Hom(E , ωPn )k% Hom(G , ωPn ).
kTaking the long exact sequence in cohomology for (30.1.7.1) and dualizing, wehave the exact sequence(30.1.7.3)

0% Hn (Pn , F )∨k

% Hn (Pn , E )∨k

% Hn (Pn , G )∨k

The (functorial) pairing (30.1.3.2) gives a map from (30.1.7.2) to (30.1.7.3):(30.1.7.4)%0% Hn (Pn , F )∨% Hn (Pn , E )∨% Hn (Pn , G )∨
07
k7
k7
k7
7
α

0

β

γ

%0

% Hom(F , ωPn )

ϵ

δ

k

% Hom(E , ωPn )k% Hom(G , ωPn )kMaps α and β are obviously isomorphisms, and Exercise 30.1.B shows that δ is an
isomorphism. Thus by the subtle version of the Five Lemma (Exercise 1.7.D, as βand δ are injective and α is surjective), γ is injective. This shows that the natural′ ∨′map Hom(F ′ , ωPnk ) → Hn (Pnk , F ) is injective for all coherent sheaves F , and′in particular for F = G . Thus ϵ is injective. Then by the dual of the subtle versionof the Five Lemma (as β and δ are surjective, and ϵ is injective), γ is surjective. !
30.1.8. Mathematical puzzle.
Here is a puzzle to force you to confront a potentially confusing point. We will see that Desideratum 30.1.1 holds for P1 , so
∼ ΩP1 . What then is the trace map t : H1 (P1 , ΩP1 ) → k? The Čech complex
ωP1 =for H1 (P1 , ΩP1 ) (with the usual cover of P1 ) is given by(30.1.8.1)

0% ΩP1 (U0 ) × ΩP1 (U1 )α% ΩP1 (U0 ∩ U1 )% 0.

If U0 = Spec k[x], and U0 ∩ U1 = Spec k[x, 1/x], then the differentials on U0 ∩ U1
are those of the form f(x) dx where f(x) is a Laurent polynomial (for example:(x−3 + x−1 + 3 + 17x4 ) dx). To compute H1 (P1 , ΩP1 ), we need to find which suchdifferentials on U0 ∩ U1 are in the image of α in (30.1.8.1). Clearly any term of theform xi dx (for i ≥ 0) extends to a differential on U0 , and thus is in the image of
α. A short calculation shows that any term of the form xi dx (i < −1) extends to adifferential on U1 . Thus the cokernel of α can be described as the one-dimensional
k-vector space generated by x−1 dx. We have an obvious isomorphism to k: take

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the coefficient of x−1 dx, which can be interpreted as “take the residue at 0”. But
there is another choice, which is equally good: take the residue at ∞ — certainly
there is no reason to privilege 0 over ∞ (or U0 over U1 )! But these two residues
are not the same — they add to 0 (as you can quickly calculate — you may also
believe it because of the Residue Theorem in the theory of Riemann surfaces). So:
which one is the trace?
30.1.9. Desideratum: a stronger version of duality, involving Ext.
Ta formule:n
Hn−p (X, F ) ′ = Extp
O (X, F, Ω )m’excite beaucoup, car je suis bien convaincu que c’est la bonne façon d’énoncer le théorèmede dualité à la fois dans le cas analytique et dans le cas algébrique (le plus important —pour moi!).
I find your formulan
Hn−p (X, F ) ′ = Extp
O (X, F, Ω )very exciting, as I am quite convinced that it is the right way to state the duality theorem
in both the analytic case and the algebraic case (the more important one — for me!).
— J.-P. Serre, letter to A. Grothendieck, December 22, 1955 [GrS, p. 21]
The vector bundle and Hom versions of Serre duality have a common extension. If we have an isomorphism of functors(30.1.9.1)

Exti (·, ωX )∼% Hn−i (X, ·)∨ ,we say that X satisfies functorial Serre duality for Ext. (The case i = 0 is functorial
Serre duality for Hom, or by Exercise 30.1.A, the trace version.) Here Exti is thederived functor of Hom; see the start of §30.2 below.
In Exercise 30.2.I, we will find that the functorial version of Serre duality for
Ext (resp. the trace version) implies the functorial version of Serre duality or vector
bundles (resp. the trace version). Thus to prove Theorem 18.5.1, it suffices to provefunctorial Serre duality for Ext, and Desideratum 30.1.1 (see Remark 30.4.9). We
will prove the for functorial version of Serre duality for Ext when X is CohenMacaulay in Corollary 30.3.14.
30.1.10. Functorial Serre duality for Ext holds for projective space.
We now prove that functorial Serre duality for Ext holds for projective space.
We will use the machinery of universal δ-functors (§23.2.6), so you may wish toeither quickly skim that section, or else ignore this discussion.
30.1.C. E XERCISE .
Show that (ExtiPnk (·, ωPnk )) is a (contravariant) universal δfunctor. Hint: Ext is not a derived functor in its first argument, so you can’t use
the “projective” version of Corollary 23.2.10. Instead, use Theorem 23.2.8, and theexistence of a surjection O(m)⊕N → F for each F , for some m < 0.
∨30.1.D. E XERCISE . Show that (Hn−i (Pnk , ·) ) is a universal δ-functor. (What are
the δ-maps?) Hint: try the same idea as in the previous exercise.

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The Rising Sea: Foundations of Algebraic Geometry∼ Hn (Pn , ·)∨ ,Proposition 30.1.7 gives an isomorphism of functors Ext0Pnk (·, ωPnk ) =kso by the Definition 23.2.7 of universal δ-functor, we have an isomorphism of δ∼ (Hn−i (Pn , ·)∨ ), thereby proving functorial Serre dualfunctors (ExtiPnk (·, ωPnk )) =knity for Ext for Pk .
30.1.11. Trace version of Serre duality for Ext. As with the previous versions, thefunctoriality of functorial Serre duality for Ext should come from somewhere. We
should expect a natural cup product Exti (F , ωX ) × Hn−i (X, F ) → Hn (X, ωX ),which coupled with the trace map (30.1.0.1) should yield the isomorphism (30.1.9.1).
We call this the trace version of Serre duality for Ext. We will not be able to prove
the trace version, as we will not define this cup product. But we will give someindication of how it works in §30.2.3.
30.1.12. Necessity of Cohen-Macaulay hypotheses. We remark that the CohenMacaulay hypotheses are necessary everywhere they are stated. The following
example applies in all cases. Let X be the union of two 2-planes in P4k , meeting ata point p. If there were a coherent dualizing sheaf ωX on X, then for d ≫ 0, we
would have h1 (X, OX (−d)) = h1 (X, ωX (d)) by Serre duality, which must be 0 by
Serre vanishing (Theorem 18.1.4(ii)). We show that this is not the case.
30.1.E. E XERCISE . Let H be a hyperplane in P4 not passing through p; say H is
hyperplane x0 = 0. Let Z be the intersection of dH (the divisor xd
0 = 0) with X.
Show (cheaply) that for d ≥ 0, h0 (Z, OZ ) > 1. Show that h1 (X, OX ) = 0 (perhaps
using an easy Čech cover, or perhaps by comparing the cohomology of X to thatof its normalization, as we did in with curves in (29.3.4.1)). Use the exact sequence
0 → OX (−Z) → OX → OZ → 0

to show that h1 (X, OX (−d)) > 0.

30.2 Ext groups and Ext sheaves for O-modules
Recall that for any ringed space X, the category ModOX has enough injectives(Theorem 23.4.1). Thus for any OX -module F on X, we may define
ExtiX (F , ·) : ModOX → ModΓ (X,OX )as the ith right derived functor of HomX (F , ·), and we have a corresponding longexact sequence for ExtiX (F , ·). We similarly define a sheaf version of thisExt iX (F , ·) : ModOX → ModOX

as the ith right derived functor of Hom X (F , ·). In both cases, the subscript X isoften omitted when it is clear from the context, although this can be dangerous
when more than one space is relevant to the discussion. (We saw Ext functors for
A-modules in §23.2.4.)
Warning: it is not clear (and in fact not true, see §23.4.7) that ModOX has
enough projectives, so we cannot define Exti as a derived functor in its left argument. Nonetheless, we will see that it behaves as though it is a derived functor— it is “computable by acyclics”, and has a long exact sequence (Remark 30.2.1).

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Another warning: with this definition, it is not clear that if F and G are quasicoherent sheaves on a scheme, then the Ext i (F , G ) are quasicoherent, and indeed
the aside in Exercise 13.7.A(a) points out this need not be true even for i = 0. But
Exercise 30.2.F will reassure you.
Exercise 23.5.A (an injective OX -module, when restricted to an open subset
U ⊂ X, is injective on U) has a number of useful consequences.
30.2.A. E XERCISE . Suppose I is an injective OX -module. Show that Hom OX (·, I )is an exact contravariant functor. (A related fact: HomOX (·, I ) is exact, by thedefinition of injectivity, Exercise 23.2.C(a).)
30.2.B. E XERCISE . Suppose X is a ringed space, F and G are OX -modules, and U is
∼ Ext i (F |U , G |U ).
an open subset. Describe a canonical isomorphism Ext iX (F , G )|U =U
Hint: take an injective resolution of G on X, and restrict it to U. Use Exercise 30.2.A
to show that the result is an injective resolution of G |U .
30.2.C. E XERCISE . Suppose X is a ringed space, and G is an OX -module.
(a) Show that%G
Ext (OX , G ) =0i

if i = 0, andotherwise.

∼ Hi (X, G ).
(b) Describe a canonical isomorphism Exti (OX , G ) =30.2.D. E XERCISE . Use Exercise 30.2.C(a) to show that if E is a locally free sheaf
on X, then Ext i (E , G ) = 0 for i > 0.
In the category of modules over a ring, we like projectives more than injectives,because free modules are easy to work with. It would be wonderful if locally freesheaves on schemes were always projective, but sadly this is not true. For if OX
were projective, then Exti (OX , ·) would be zero (as a functor) for i > 0, but by
Exercise 30.2.C(b), it is the functor Hi (X, ·), which is often nonzero. Nonetheless,
we can still compute Ext using a locally free resolution, as shown by the following
exercise.
30.2.E. I MPORTANT E XERCISE . Suppose X is a ringed space, and(30.2.0.1)

···

% E1

% E0

%F

%0

is a resolution of an OX -module F by locally free sheaves. (Of course we are mostinterested in the case where X is a scheme, and F is quasicoherent, or even coherent.) Let E• denote the truncation of (30.2.0.1), where F is removed. Describe an∼ Hi (Hom (E• , G )). In other words, Ext • (F , G )) can be
isomorphism Ext i (F , G ) =computed by taking a locally free resolution of F , truncating, applying Hom (·, G ),and taking homology. Hint: choose an injective resolution
0

%G

% I0

% I1

% ···

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and consider the spectral sequence whose E0 term is
..
.7

..
.7

Hom (E0 , I1 )7

% Hom (E1 , I1 )
7

% ···

Hom (E0 , I0 )% Hom (E1 , I0 )% ··· .

This result is important: to compute Ext , we can compute it using finite rank
locally free resolutions. You can work affine by affine (by Exercise 30.2.B), and oneach affine you can use a free resolution of the left argument. As another consequence of Exercise 30.2.E:
30.2.F. E XERCISE . Suppose F and G are coherent sheaves on a quasiprojective
k-scheme X. Show that Ext iX (F , G ) is a coherent sheaf as well. (Your argument
will work on any scheme for which there always exist resolutions by finite rank
locally free sheaves.)
30.2.1. Remark. The statement “Exti (F , G ) behaves like a derived functor in the firstargument” is true in a number of ways. We can compute it using a resolution of F
by locally frees, which are acyclic for Exti (·, G ). And we even have a corresponding long exact sequence, as shown in the next exercise.
30.2.G. E XERCISE . Suppose 0 → F ′′ → F → F ′ → 0 is an exact sequence of
OX -modules on a ringed space X. For any OX -module G , describe a long exact
sequence0% Hom(F ′′ , G )% Hom(F , G )% Hom(F ′ , G )% Ext1 (F ′′ , G )% Ext1 (F , G )% Ext1 (F ′ , G )% ··· .

Hint: take an injective resolution 0 → G → I 0 → · · · . Use the fact that if I is
injective, then Hom(·, I ) is exact (the definition of injectivity, Exercise 23.2.C(a)).
Hence get a short exact sequence of complexes
0 → Hom(F ′′ , I • ) → Hom(F , I • ) → Hom(F ′ , I • ) → 0

and take the long exact sequence in cohomology.
Here are two useful exercises.

30.2.H. E XERCISE . Suppose X is a ringed space, F and G are OX -modules, and E
is a locally free sheaf on X. Describe isomorphisms∼ Ext i (F , G ⊗ E ) =∼ Ext i (F , G ) ⊗ E
Ext i (F ⊗ E ∨ , G ) =and∼ Exti (F , G ⊗ E ).
Exti (F ⊗ E ∨ , G ) =Hint: show that if I is injective then I ⊗ E is injective.

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30.2.I. E XERCISE . If F is a locally free sheaf on a scheme X, and ωX is any coherent∼ Hi (X, F ∨ ⊗ ωX ) (functorial
sheaf on X, describe an isomorphism Exti (F , ωX ) =in F ). Show that functorial Serre duality for Ext implies functorial Serre duality for vector bundles. (You may wish to ponder the trace versions as well.) As
a consequence, by §30.1.10, functorial Serre duality for vector bundles holds for
projective space. Hint: Exercises 30.2.H and 30.2.C(b).
30.2.2. The local-to-global spectral sequence for Ext.
“Sheaf” Ext and “global” Ext are related by a spectral sequence. This is astraightforward application of the Grothendieck composition-of-functors spectral
sequence, once we show that Hom (F , I ) is acyclic for the functor Γ .
30.2.J. E XERCISE . Suppose I is an injective OX -module. Show that Hom (F , I )is flasque (and thus Γ -acyclic by Exercise 23.4.G). Hint: suppose j : U "→ V is an inclusion of open subsets. We wish to show that Hom (F , I )(V) → Hom (F , I )(U)is surjective. Note that I |V is injective on V (Exercise 23.5.A). Apply the exact
functor HomV (·, I |V ) to the inclusion j! (F |U ) "→ F |V of sheaves on V (Exercise 2.7.G).
30.2.K. E XERCISE ( THE LOCAL - TO - GLOBAL SPECTRAL SEQUENCE FOR Ext). Suppose X is a ringed space, and F and G are OX -modules. Describe a spectral sequence with E2 -term Hj (X, Ext i (F , G )) abutting to Exti+j (F , G ). (Hint: use theGrothendieck composition-of-functors spectral sequence, Theorem 23.3.5. Recall
that Hom(F , ·) = Γ (Hom (F , ·)), Exercise 2.3.C.)
30.2.3. ⋆⋆ Composing Ext’s (and Hi ’s): the Yoneda cup product.
It is useful and reassuring to know that Ext’s can be composed, in a reasonable
sense. We won’t need this, and so just outline the ideas, so you can recognize themin the future should you need them. For more detail, see [Gr3, §2] or [Cart].
If C is an abelian category, and A• and B• are complexes in C , then define
Hom• (A• , B• ) as the integer-graded group of graded homomorphisms: the elementsof Homn (A• , B• ) are the maps from the complex A• to B• shifted “to the right byn”. Define δ : Hom• (A• , B• ) byδ(u) = du + (−1)n+1 ud
for each u ∈ Homn (A• , B• ) (where d sloppily denotes the differential in both A•
and B• ). Then δ2 = 0, turning Hom• (A• , B• ) into a complex. Let H• (A• , B• ) be thecohomology of this complex. If C• is another complex in C , then composition ofmaps of complexes yields a map Hom• (A• , B• ) × Hom• (B• , C• ) → Hom• (A• , C• )which induces a map on cohomology:H• (A• , B• ) × H• (B• , C• ) → H• (A• , C• )(30.2.3.1)

which can be readily checked to be associative. In particular, H• (A• , A• ) has the
structure of a graded associative non-commutative ring (with unit), and H• (A• , B• )(resp. H• (B• , A• )) has a natural graded left-module (resp. right-module) structure
over this ring. The cohomology groups H• (A• , B• ) are functorial in both A• andB• . A short exact sequence of complexes 0 → A•′ → A• → A•′′ → 0 induces longexact sequences···% Hi (A•′ , B• )% Hi (A• , B• )% Hi (A•′′ , B• )% Hi+1 (A•′ , B• )% ···Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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and
···

% H• (B• , A•′′ )% H• (B• , A• )% H• (B• , A•′ )% ··· .

Suppose now that C has enough injectives. Suppose A, B ∈ C , and let IA• be
any injective resolution of A (more precisely: take an injective resolution of A, andremove the “leading” A), and similarly for IB• . Then it is a reasonable exercise todescribe canonical isomorphisms
•
B ∼
•
B ∼
H• (IA• , I• ) = H (A, I• ) = Ext (A, B)

where in the middle term, the “A” is interpreted as a complex that is zero, except
the 0th piece is A.
Then the map (30.2.3.1) induces a (graded) map(30.2.3.2)

Ext• (A, B) × Ext• (B, C) → Ext• (A, C)

extending the natural map Hom(A, B)×Hom(B, C) → Hom(A, C). (Of course, onemust show that the map (30.2.3.2) is independent of choice of injective resolutionsof B and C.)
In particular, in the category of O-modules on a ringed space X, we have (using
Exercise 30.2.C(b)) a natural map
Hi (X, F ) × Extj (F , G ) → Hi+j (X, G ).

This is the source of the natural map in the trace version of Serre duality for Ext,
discussed at the end of §30.1.

30.3 Serre duality for projective k-schemes
We now prove various versions of Serre duality for projective k-schemes, byleveraging what we know about Serre duality for projective space.
30.3.1. The weird functor π!sh .
The key construction is a right adjoint π!sh to the pushforward π∗ , when π isan affine morphism. (Caution: the notation π!sh is nonstandard, and is introduced
only for the purposes of the arguments we will give.) This is surprising, as we
usually think of π∗ as a right-adjoint (to the pullback π∗ ), not a left adjoint. Thiscan be seen as very roughly analogous to the surprising occasional left-adjoint to
the pullback: extension by zero (Exercise 2.7.G).
30.3.2. We begin with the ring-theoretic version. Suppose B → A is a ring morphism, M is an A-module, and N is a B-module. Note that HomB (A, N) naturally
has the structure of an A-module. Also, M naturally carries the structure of a Bmodule; when we wish to emphasize its structure as a B-module, we sometimes
call it MB (see Exercise 1.5.E).
Consider the map(30.3.2.1)

HomA (M, HomB (A, N)) → HomB (MB , N)

defined as follows. Given m ∈ M, and an element φ of HomA (M, HomB (A, N)),send m to φm (1).

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30.3.A. E XERCISE .
(a) Show that (30.3.2.1) is a homomorphism of B-modules.
(b) Show that (30.3.2.1) is a bijection. Thus (M (→ MB , N (→ HomB (A, N)) is anadjoint pair ModA ↔ ModB .
(c) Show that this bijection (30.3.2.1) behaves well with respect to localization at anelement of B.
Exercise 30.3.A(c) implies that this naturally “sheafifies” to a construction for
an affine morphism π : X → Y.

30.3.B. E XERCISE . Suppose π : X → Y is an affine morphism.
(a) Explain how the map
π!sh : QCohY% QCohG ✤% π−1 Hom Y (π∗ OX , G )X(where Hom Y (π∗ OX , G ) is interpreted as a quasicoherent sheaf on X via Exercise 17.1.E,
and in this guise is denoted π−1 Hom Y (π∗ OX , G )) globalizes the construction of Exercise 30.3.A, yielding a covariant functor QCohY → QCohX .
(b) Describe a natural isomorphism of quasicoherent sheaves on Y
∼ Hom Y (π∗ OX , G ).
π∗ π!sh G =(c) Show that (π∗ , π!sh ) is an adjoint pair between QCohX and QCohY .
Caution: we have defined π!sh only for categories of quasicoherent sheaves.
If π is a finite morphism, and Y (and hence X) is locally Noetherian (the case thatwill be relevant for us), then π!sh is a covariant functor from the category of coherent
sheaves on Y to coherent sheaves on X. We may show this affine-locally, using thenotation of §30.3.2. As A and N are both coherent B-modules, HomB (A, N) is acoherent B-module (cf. Exercise 13.7.A(b)), hence a finitely generated B-module,
and hence a finitely generated A-module, hence a coherent A-module.
Thus if π is a finite morphism of locally Noetherian schemes, (π∗ , π!sh ) is anadjoint pair between CohX and CohY .
30.3.3. If F ∈ QCohX and G ∈ QCohY , then there is a natural isomorphism(30.3.3.1)

π∗ Hom X (F , π!sh G ) → Hom Y (π∗ F , G ),which affine-locally is the isomorphism (30.3.2.1) described in Exercise 30.3.A.
If G is an OX -module (not necessarily quasicoherent), it is in general not clear
how to make sense of this construction to define an OY -module π!sh G . (Try it andsee!) However, in the special case where π is a closed embedding, we can makesense of π!sh G , as discussed in the next exercise.
30.3.C. E XERCISE ( USED IN §30.4). Suppose π : X → Y is a closed embedding ofschemes and G is an OY -module.
(a) Explain why π!sh (G ) := Hom Y (π∗ OX , G ) naturally has the structure of an OX module. Hint: if I is the ideal sheaf of X, explain how π!sh (G ) (over some open
subset U ⊂ Y) is annihilated by “functions vanishing on X” (elements of I (U)).

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Hence we have defined a map π!sh : ModOY → ModOX extending the map of Exercise 30.3.B(a).
(b) Show that (π∗ , π!sh ) is an adjoint pair between ModOX and ModOY .
(c) Show that π!sh sends injective OY -modules to injective OX -modules. (Hint: π∗ isexact; use Exercise 23.5.B.)
30.3.4. Remark. If π is finite and flat, which is the case most of interest to us, π!shagrees (in the only possible sense of the word) with π! , one of Grothendieck’s “sixoperations”, and indeed this motivates our notation (“a variant of π! for sheaves”).
But π! naturally lives in the world of derived categories, so we will not discuss it
here. (For more on π! , see [KS1, Ch. III].)
We now apply the machinery of π!sh to Serre duality.
30.3.5. Projective n-dimensional schemes are covers of Pn .
30.3.6. Proposition. — Suppose X is a projective k-scheme of pure dimension n.
(a) There exists a finite morphism π : X → Pn .
(b) If furthermore X is Cohen-Macaulay, then π is flat.

Proof. Part (b) follows from part (a) by the Miracle Flatness Theorem 26.2.11, so wewill prove part (a).
Choose a closed embedding j : X "→ PN . For simplicity of exposition, first
assume that k is an infinite field. By Exercise 11.3.C(d), there is a linear space
L of codimension n + 1 (one less than complementary dimension) disjoint fromX. Projection from L yields a morphism π : X → Pn . The morphism π is affine(PN \ L → Pn is affine, and the closed embedding X "→ PN \ L is, like all closed
embeddings, affine) and projective (by the Cancellation Theorem 10.1.19 for projective morphisms), so π is finite (projective affine morphisms of locally Noetherian
schemes are finite, Corollary 18.1.7).
30.3.D. E XERCISE . Prove Proposition 30.3.6(a), without the assumption that k isinfinite. Hint: using Exercise 11.3.C(c), show that there is some d such that there isan intersection of n + 1 degree d hypersurfaces missing X. Then apply the above
argument to the dth Veronese embedding of PN (§8.2.6).
!
30.3.7. Remark. More generally, the above argument shows that any projective
k-scheme of dimension n (not necessarily equidimensional) can be expressed as afinite cover of Pn . This might be seen as a projective version of Noether normalization.
30.3.8. Serre duality on X via π!sh .
Suppose π : X → Y is a finite morphism of projective k-schemes of pure dimension n, and we have a coherent sheaf ωY on Y. (We will soon apply this in the casewhere Y = Pn
k , but we may as well avoid distraction and needless specificity.)
30.3.9. Proposition. — Suppose π is flat. If functorial Serre duality for Ext holds for Y,
with dualizing sheaf ωY , then functorial Serre duality for vector bundles holds for X, withdualizing sheaf π!sh (ωY ).

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761

Note the mismatch of the hypotheses and the conclusion: we use functorial
Serre duality for Ext in the hypothesis, and get functorial Serre duality only forvector bundles in the conclusion.
Proof. For each i, and each finite rank locally free sheaf F on X, we have isomorphisms (functorial in F ):∼Hn−i (X, F ∨ ⊗ π!sh ωY ) =
∼
=
∼
=
∼
=
∼
=
∼
=

Hn−i (Y, π∗ (F ∨ ⊗ π!sh ωY )) (affineness of π, §18.1 (v))
Hn−i (Y, π∗ (Hom X (F , π!sh ωY ))) (Exercise 13.7.B)
Hn−i (Y, Hom Y (π∗ F , ωY )) (equ. (30.3.3.1))

Hn−i (Y, (π∗ F )∨ ⊗ ωY ) (Exercises 24.4.E and 13.7.B)
Hi (Y, π∗ F )∨i

H (X, F )∨(functorial Serre duality for Ext on Y)(affineness of π, §18.1 (v))(The hypothesis of flatness is used to show that π∗ F is locally free.)!

30.3.10. Corollary. — Functorial Serre duality for vector bundles holds for every CohenMacaulay equidimensional projective k-scheme.
Proof. Combine Proposition 30.3.9 with Proposition 30.3.6(b) and §30.1.10.

!

30.3.11. Proposition. — Suppose π : X → Y is a finite morphism of projective k-schemes.
If the trace version of Serre duality for Hom holds for Y with dualizing sheaf (ωY , tY ),then functorial Serre duality for Hom holds for X with dualizing sheaf π!sh ωY .
Recall from Exercise 30.1.A that for Serre duality for Hom, the trace version isequivalent to the functorial version. We state the Proposition in this awkward way
because we use tY , and conclude functoriality for Serre duality for Hom for X. But
the trace map for X will come cheaply as the diagonal arrow in (30.3.11.1).
Note that we have no flatness hypotheses on π (unlike the corresponding
propositions for Serre duality for vector bundles and, soon, for Ext).
Proof. The following diagram commutes, and is functorial in F .
(30.3.11.1)% Hn (X, π! ωY )HomX (F , π!sh ωY ) × Hn (X, F )
sh❉
❉❉
❉❉
❉❉
❉❉
"
❉ tX
Hn (Y, π∗ π!sh ωY ) ❉❉❉
❉❉
❉❉
❉❉
❉❉
❉9
"
"
tY
n
n
% H (Y, ωY )%k
HomY (π∗ F , ωY ) × H (Y, π∗ F )(The left vertical arrow is an isomorphism; it is an isomorphism on each factor.)
The commutativity of the diagram relies on the adjointness of (π∗ , π!sh ), and theaffineness of π.
!
30.3.12. Corollary. — The functorial and trace versions of Serre duality for Hom holdfor all equidimensional projective k-schemes (with no Cohen-Macaulay hypotheses).

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Proof. Combine Proposition 30.3.11 with Proposition 30.3.6(a) and Proposition 30.1.7.
!
30.3.13. Proposition. — If X is a Cohen-Macaulay projective k-scheme of pure dimension n, and π : X → Pn is a finite flat morphism (as in Proposition 30.3.6(b)), thenfunctorial Serre duality for Ext holds for X with dualizing sheaf ωX := π!sh (ωPn ).
Unlike Proposition 30.3.9 and 30.3.11, we did not state this for general π : X →Y; we will use the fact that the target is Pn .

Proof. The argument will parallel that of §30.1.10: we will show an isomorphismof δ-functors(30.3.13.1)

ExtiX (F , π!sh (ωPn )) → Hn−i (X, F )∨ .

We already have an isomorphism for i = 0 (Corollary 30.3.12), so it suffices to showthat both sides of (30.3.13.1) are universal δ-functors. We do this by verifying thecriterion of Theorem 23.2.8. For any coherent sheaf F on X we can find a surjection
O(−m)⊕N → F for m ≫ 0 (and some N), so it suffices to show that for m ≫ 0,
ExtiX (O(−m), π!sh (ωPn )) = 0

and Hn−i (X, O(−m))∨ = 0

for i > 0 and m ≫ 0. For the first, we have (for i > 0 and m ≫ 0)∼
ExtiX (O(−m), π!sh (ωPn )) ==Hi (X, π!sh (ωPn )(m)) (Exercise 30.2.H)
0(Serre vanishing, Thm. 18.1.4(ii))

For the second, we haveHn−i (X, OX (−m)) =∼=Hn−i (X, OX ⊗ π∗ O(−m))

Hn−i (Pn , (π∗ OX ) ⊗ O(−m))

by the projection formula (Exercise 18.8.E(b)). Now π∗ OX is locally free by Exercise 24.4.E, so by functorial Serre duality for vector bundles on Pn , we have (form ≫ 0)∼
Hn−i (Pn , (π∗ OX ) ⊗ O(−m)) =∼==Hi (Pn , (π∗ OX )∨ ⊗ O(m) ⊗ ωPn )Hi (Pn , ((π∗ OX )∨ ⊗ ωPn ) ⊗ O(m))
0(Serre vanishing, Thm. 18.1.4(ii))!

30.3.14. Corollary. — Functorial Serre duality for Ext holds for all equidimensional
Cohen-Macaulay projective k-schemes.
Proof. Combine Proposition 30.3.13 with Proposition 30.3.6(b) and the projective
space case of §30.1.10.
!
30.3.15. Remark: the trace version of Serre duality for vector bundles. Here is an outlineof how the above proof can be extended to prove the trace version of Serre duality
for vector bundles on all equidimensional Cohen-Macaulay projective k-schemes.
Continuing the notation of Proposition 30.3.13, the Yoneda cup product of §30.2.3

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followed by the trace map gives a pairing
ExtiX (F , ωX ) × Hn−i (X, F )∪% Hn (X, ωX )t%k,which is functorial in F . This induces a functorial map (30.3.13.1). One showsthat this particular map is a map of δ-functors. Then by the above argument, thisparticular map is an isomorphism.

30.4 The adjunction formula for the dualizing sheaf, and ωX = KX
The dualizing sheaf behaves well with respect to slicing by effective Cartierdivisors, and this will be useful to obtaining Desideratum 30.1.1, i.e., the miracle
that the canonical bundle is Serre-dualizing (§18.5.2). In order to show this, wegive a description for the dualizing sheaf for a subvariety of a variety satisfying
Serre duality for Ext.
30.4.1. Preliminary Observation. Suppose π : X → Y is a closed embedding. Thenfor any OY -module F , Ext iY (π∗ OX , F ) naturally has the structure of an OX -module.
(Reason: we compute this by taking an injective resolution of F by OY -modules,
then truncating, then applying Hom Y (π∗ OX , ·). But for any OY -module G , Hom Y (π∗ OX , G )has the structure of an OX -module.) To emphasize its structure as an OX -module,
we write it as π−1 Ext iY (π∗ OX , F ).
30.4.A. E XERCISE . Show that if Y (and hence X) is locally Noetherian, and G isa coherent sheaf on Y, then π−1 Ext iY (π∗ OX , G ) is a coherent sheaf on X. (Hint:
Ext iY (π∗ OX , G ) is a coherent sheaf on Y, Exercise 30.2.F.)
30.4.2. The dualizing sheaf for a subvariety in terms of the dualizing sheaf of
the ambient variety.
For the rest of this section, π : X → Y will be a closed embedding of equidimensional projective k-schemes of dimension n and N respectively, where Y satisfies
functorial Serre duality for Ext. Let r = N − n (the codimension of X in Y). By
Corollary 30.3.12, X satisfies functorial Serre duality for Hom. We now identify
ωX in terms of ωY .
30.4.3. Theorem. — We have ωX = π−1 Ext rY (π∗ OX , ωY ).
Before we prove Theorem 30.4.3, we explain what happened to the “earlier”

Ext iY (π∗ OX , ωY )’s.
30.4.4. Proposition. — Suppose that π : X "→ Y is a closed embedding of equidimensional projective k-schemes of dimension n and N respectively, and Y satisfies functorial
Serre duality for Ext. Then for all i < r := N − n, Ext iY (π∗ OX , ωY ) = 0.
Proof. As Ext iY (π∗ OX , ωY ) is coherent (Exercise 30.2.F), it suffices to show that
Ext iY (π∗ OX , ωY ) ⊗ O(m) has no nonzero global sections for m ≫ 0 (as for any
coherent sheaf G on Y, G (m) is generated by global sections for m ≫ 0 by Serre’s

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Theorem A, Theorem 15.3.8). By Exercise 30.2.H,
Hj (Y, Ext iY (π∗ OX , ωY )(m)) = Hj (Y, Ext iY (π∗ OX , ωY (m))).

If j > 0, then for m ≫ 0, by Serre vanishing, Hj (Y, Ext iY (π∗ OX , ωY )(m)) = 0. Thusby the local-to-global spectral sequence for Ext (Exercise 30.2.K),
H0 (Y, Ext iY (π∗ OX , ωY (m))) = ExtiY (π∗ OX , ωY (m))).

By Exercise 30.2.H again, then functorial Serre duality for Ext on Y,
ExtiY (π∗ OX , ωY (m))) = ExtiY (π∗ OX (−m), ωY ) = HN−i (Y, π∗ OX (−m))

which is 0 if N − i > n, as the cohomology of a quasicoherent sheaf on a projective scheme vanishes in degree higher than the dimension of the sheaf’s support(dimensional cohomology vanishing, Theorem 18.2.6).
!
The most difficult step in the proof of Theorem 30.4.3 is the following.
30.4.5. Lemma. — Suppose that π : X → Y is a closed embedding of codimension r ofequidimensional projective k-schemes, and Y satisfies functorial Serre duality for Ext. We
have an isomorphism, functorial in F ∈ CohX :∼ Extr (π∗ F , ωY ).
HomX (F , π−1 Ext rY (π∗ OX , ωY )) =Y

(30.4.5.1)

Proof. Choose an injective resolution
0 → ωY → I 0 → I 1 → I 2 → · · ·of ωY . To compute the right side of (30.4.5.1), we drop the ωY from this resolution,
and apply HomY (π∗ F , ·), to obtain0 → HomY (π∗ F , I 0 ) → HomY (π∗ F , I 1 ) → HomY (π∗ F , I 2 ) → · · · ,or (by adjointness of π∗ and π!sh for O-modules when π is a closed embedding,
Exercise 30.3.C(b)):
0 → HomX (F , π!sh I 0 ) → HomX (F , π!sh I 1 ) → HomX (F , π!sh I 2 ) → · · · .

Motivated by this, consider the complex(30.4.5.2)

0 → π!sh I 0 → π!sh I 1 → π!sh I 2 → · · · .

Note first that (30.4.5.2) is indeed a complex, and second that π!sh I i are injective
OX -modules (Exercise 30.3.C(c)).
30.4.B. E XERCISE .
Ext iY (π∗ OX , ωY ).

Show that the cohomology of (30.4.5.2) at the ith step isThus by Proposition 30.4.4, the complex (30.4.5.2) is exact before the rth step.
30.4.C. E XERCISE .
J ⊕ K , so thatShow that there exists a direct sum decomposition π!sh I r =0 → π!sh I 0 → π!sh I 1 → · · · → π!sh I r−1 → J → 0

is exact. Hint: work out the case r = 1 first. Another hint: Notice that if
0

% I0

α

%F

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

November 18, 2017 draft765

is exact and I 0 is injective, then there is a map β : F → I 0 “splitting” α, allowing
us to write F as a direct sum I 0 ⊕ J 1 . If F is furthermore injective, then J 1 isinjective too. This is the beginning of an induction.
30.4.D. E XERCISE . (We continue the notation of the previous exercise.) Identify
ker(K → π!sh I r+1 ) with π−1 Ext rY (π∗ OX , ωY ).

30.4.E. E XERCISE . Put together the pieces above to complete the proof of Lemma 30.4.5.
!
We are now ready to prove Theorem 30.4.3.
30.4.6. Proof of Theorem 30.4.3. Suppose F is a coherent sheaf on X. We wish tofind an isomorphism∼ Hn (X, F )∨ ,HomX (F , π−1 Ext rY (π∗ OX , ωY )) =functorial in F . By Lemma 30.4.5, we have a functorial isomorphism∼ Extr (π∗ F , ωY ).
HomX (F , π−1 Ext rY (π∗ OX , ωY )) =Y
By functorial Serre duality for Ext on Y, we have a functorial isomorphism∼ HN−r (Y, π∗ F )∨ .
ExtrY (π∗ F , ωY ) =As F is a coherent sheaf on X, and N − r = n, the right side is is precisely
Hn (X, F )∨ .
!
30.4.7. Applying Theorem 30.4.3. We first apply Theorem 30.4.3 in the special
case where X is an effective Cartier divisor on Y. We can compute the dualizing
sheaf π−1 Ext 1Y (π∗ OX , ωY ) by computing Ext 1Y (π∗ OX , ωY ) using any locally freeresolution (on Y) of π∗ OX (Exercise 30.2.E). But π∗ OX has a particularly simple
resolution, the closed subscheme exact sequence (13.5.4.1) for X:(30.4.7.1)

0 → OY (−X) → OY → π∗ OX → 0.

We compute Ext (π∗ OX , ωY ) by truncating (removing the π∗ OX ), and applying
Hom (·, ωY ): Ext • (π∗ OX , ωY ) is the cohomology of•

0 → Hom Y (OY , ωY ) → Hom (OY (−X), ωY ) → 0,i.e.,

0 → ωY → ωY ⊗ OY (X) → 0.

We immediately see that Ext (π∗ OX , ωY ) = 0 if i ̸= 0, 1. Furthermore, Ext 0 (π∗ OX , ωY ) =0 by Proposition 30.4.4. (Unimportant aside: you can use this to show that ωY has
no embedded points.)
We now consider ωX = coker(ωY → ωY ⊗ OY (X)). Tensoring (30.4.7.1) with
the invertible sheaf OY (X), and then tensoring with ωY , yields
iωY → ωY ⊗ OY (X) → ωY ⊗ OY (X)|X → 0

The right term is often (somewhat informally) written as ωY (X)|X . Thus
ωX = coker(ωY → ωY ⊗ OY (X)) = ωY (X)|X ,and this identification is canonical.
We have shown the following.

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The Rising Sea: Foundations of Algebraic Geometry

30.4.8. Proposition (the adjunction formula). — Suppose that Y is a Cohen-Macaulay
projective scheme of pure dimension n (which satisfies functorial Serre duality for Ext by
Corollary 30.3.14), and X is an effective Cartier divisor on Y. (Hence X is Cohen-Macaulayby §26.2.4, and thus satisfies functorial Serre duality for Ext by Corollary 30.3.14.) Then
ωX = ωY (X)|X . In particular, if ωY is an invertible sheaf on Y, then ωX is an invertible
sheaf on X.
As an immediate application, we have the following.
30.4.F. E XERCISE . Suppose X is a complete intersection in Pn , of hypersurfacesof degrees d1 , . . . , dr . (Note that X is Cohen-Macaulay by Proposition 26.2.6, andthus satisfies functorial
Serre duality for Ext, by Corollary 30.3.14.) Show that
∼ OX (−n − 1 + , di ). If furthermore X is smooth, show that ωX =∼ det ΩX .
ωX =(Hint for the last sentence: use the adjunction formula for K , Exercise 21.5.B).
But we can say more.

30.4.G. E XERCISE . Suppose Y is a smooth equidimensional variety, and π : X "→ Y
is a codimension r regular embedding with normal sheaf NX/Y (which is locally
free, by Proposition 21.2.16(b)). Suppose L is an invertible sheaf on Y.
(a) Show that Ext iY (π∗ OX , L ) = 0 if i ̸= r.
∼ (det NX/Y ) ⊗O L |X .
(b) Describe a canonical isomorphism π−1 Ext rY (π∗ OX , L ) =X(Note that because Exercise 30.4.G has nothing explicitly to do with duality,
we have no projectivity assumptions on Y; it is a completely local question.) From
Exercise 30.4.G we deduce the following.
30.4.H. I MPORTANT E XERCISE . Suppose X is a codimension r regular embeddingin Pn
k . (Then X is Cohen-Macaulay by Proposition 26.2.6, and thus satisfies functorial Serre duality for Ext by Corollary 30.3.14.) Show that
∼ (det NX/Y ) ⊗ ωPn |X .
ωX = π−1 Ext r n (π∗ OX , ωPn ) =P

In particular, ωX is an invertible sheaf.

30.4.I. I MPORTANT E XERCISE . Suppose X is a smooth pure codimension r subvariety of Pnk (and hence a regular embedding, by Exercise 12.2.L(b)). Show that
∼ KX . Hint: both sides satisfy adjunction (see Exercise 21.5.B for adjunction
ωX =∼ (det NX/Y ) ⊗ (KPn )|X .
for Ω): they are isomorphic to (det NX/Y ) ⊗ ωPn |X =30.4.9. Remark. As long promised, the version of Serre duality given in Theorem 18.5.1 now follows by combining Corollary 30.3.10 with Exercise 30.4.I.
Aux yeux de ces amateurs d’inquiétude et de perfection, un ouvrage n’est jamais
achevé, mot qui pour eux n’a aucun sens, mais abandonné; et cet abandon, qui le livre
aux flammes ou au public (et qu’il soit l’effet de la lassitude ou de l’obligation de livrer),leur est une sorte d’accident, comparable à la rupture d’une réflexion . . .
In the eyes of those lovers of anxiety and perfection, a work is never finished — aword that for them has no meaning — but abandoned; and this abandonment, whether tothe flames or to the public (and which is the result of weariness or an obligation to deliver)is a kind of accident to them, like the breaking off of a reflection . . .

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November 18, 2017 draft767— P. Valéry, [Val, p. 1497]

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

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K. Schwede, Gluing schemes and a scheme without closed points, in Recent Progress in Arithmetic and Algebraic Geometry, 157–172, Contemp. Math. 386, Amer. Math. Soc., Providence, RI, 2005.
J.-P. Serre, Faisceaux algébriques cohérents, Ann. of Math. (2) 61 (1955), 197–278.
, Sur la dimension cohomologique des anneaux et des modules noethériens, in Proc. Intern. Symp. on Alg. Number Theory, 176–189, Tokyo-Nikko, 1955.
, Géométrie algébrique et géométrie analytique, Ann. de l’institut Fourier 6 (1956),
1–42.
, Sur la topologie des variétés algébriques en caractéristique p, in Symposium Internacional de Topologı́a Algebraica, International Symposium on Algebraic Topology, 24–53, Univ.
Nacional Autónoma de México and UNESCO, Mexico City, 1958.
, A Course in Arithmetic, Grad. Texts in Math. 7, Springer-Verlag, New York, 1973.
A. Grothendieck, Séminaire de Géométrie Algébrique du Bois Marie 1962: Cohomologie locale
des faisceaux cohérents et Théorèmes de Lefschetz locaux et globaux (SGA2), North-Holland
Publ. Co., Amsterdam, 1968.
P. Berthelot, A. Grothendieck, and L. Illusie eds., Séminaire de Géométrie Algébrique duBois Marie 1966–67: Théorie des intersections et théorème de Riemann Roch (SGA 6), Lect.
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I. Shafarevich, Basic Algebraic Geometry 1: Varieties in Projective Space, 2nd ed., M. Reid
trans., Springer-Verlag, Berlin, 1994.
J. Silverman, The Arithmetic of Elliptic Curves, 2nd ed., Grad. Texts in Math. 206, SpringerVerlag, New York-Berlin, 2009.
, Advanced Topics in the Arithmetic of Elliptic Curves, Grad. Texts in Math. 151,
Springer-Verlag, New York-Berlin, 1994.
The Stacks Project Authors, Stacks Project, http://stacks.math.columbia.edu.
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http://stacks.math.columbia.edu/tags. To search:
http://stacks.math.columbia.edu/search.)
H. Stichtenoth, Über die Automorphismengruppe eines algebraischen Funktionenkörpers vonPrimzahlcharakteristik. I. Eine Abschätzung der Ordnung der Automorphismengruppe, Arch.
Math. (Basel) 24 (1973), 527–544.
A. Tarski, A decision method for elementary algebra and geometry, RAND Corp., Santa Monica, CA, 1948.
R. Taylor and A. Wiles, Ring-theoretic properties of certain Hecke algebras, Ann. of Math. (2)141 (3) (1995), 553–572.
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R. Vakil, The Rising Sea: Foundations of Algebraic Geometry, preprint 2013.
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La Pléiade, Paris, 1957.
B. L. van der Waerden, Algebra vol. II, J. R. Schulenberger trans., Springer-Verlag, New
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C. Voisin, Hodge Theory and Complex Algebraic Geometry I and II, L. Schneps trans., Cambridge Stud. in Adv. Math. 76 and 77, Cambridge U.P., Cambridge, 2007.
C. Weibel, An Introduction to Homological Algebra, Cambridge Stud. in Adv. Math. 38,
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A. Weil, Scientific works — Collected papers Vol. II (1951–1964), Springer-Verlag, New YorkHeidelberg, 1979.
H. Weyl, Philosophy of Mathematics and Natural Science, Princeton U.P., Princeton, NJ,
2009.
H. Whitney, Geometric Integration Theory, Princeton U.P., Princeton, NJ, 1957.
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[Z]

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O. Zariski, Foundations of a general theory of birational correspondences, Trans. Amer. Math.
Soc. 53 (1943), 490–542.

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Index(−1)-curve, 549, 614, 699, 705, 706, 725, see also
Castelnuovo’s Criterion, 740, 743
blowing down, 740(M : x), 652(L1 · L2 · · · Ln · F ) (intersection of L1 , . . . ,Ln with F ), 543
A-scheme, 159
ADE singularities, 610
An singularity, 730
An -singularities, 610
A• (I), 364
BlX Y (blow-up of Y along X), 599
Creg , 533
D(f), 149
Dn singularity, 730
Ep,q (entry in double complex), 58
E6 , 698, 704
E8 , 556, 730
EX Y (exceptional divisor), 599
En , 610, 706
En singularity, 730
Ep,q(entry (p, q) in rth page of spectralrsequence), 59
G(k, n) (Grassmannian), 199
GLn (group scheme), 198, 296, 445
H ((co)homology), 50
H0 (U, F ), 73
I(S), 127
K(A), 32, 156
K(X), 156
K(X) (Grothendieck group of coherent
sheaves on X), 554
K(X)≤d , 554
KX , 549
Li F (left derived functors), 626
M(n)• , 412
M• , 411
M• (I), 364
Mtors (torsion submodule), 386
NS(X) (Néron-Severi group), 484
N1 (X) = Pic X/ Picτ X, 484
NX Y (normal cone), 606
Ri F (right derived functor), 626

Ri π∗ (higher pushforward), 497, see sheaf,
higher pushforward
as derived functor, 635
quasicoherent definition, 498
Rk (regular in codimension k), 692
SLn (group scheme), 199, 296
S−1 A (localization), 32
S+ , 148
Sk , see property Sk
S• , 147
Sn• , 188
T • (M) tensor algebra, 384
T • (F ), 384
V(S) (vanishing set), 115
V(T ), 149
Xs , 209[p] (point of Spec A corresponding to p), 102
A2 , 105An
X , 449
Aut(A) (automorphisms of A), 26∆ (diagonal locally closed subscheme of
X ×Y X), 280
Fn (Hirzebruch surface), 452
Γadjoint to Spec, 183
Γ (U, F ), 73
Γ• , 417, 417
Mor, 26ΩiX/Y (sheaf of relative i-forms), 583ΩA/B (Kähler differentials), 573ΩX/Y (relative cotangent sheaf), 572
Pic0 X, 671
Spec 0, 104
Spec k, 104Sym• , 384%, 432, 437, 439, 475, 712
KX , see canonical, bundle
TX/Y (relative tangent sheaf), 572·∨ (dual), 372
χ(X, F ) (Euler characteristic), 399, 478!
(coproduct), 37
δ-functor, see delta-functor
775

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deg F (degree of coherent sheaf on curve),481
δ-invariant, 731
δπ (diagonal morphism), 279ℓ(M) (module length), 482^ (completion), 725
A!→, see [open, closed, l cl] embedding, see[open, closed, locally closed] embedding,202, 225, 228see[open, closed, lococally closed]
embedding, 228
κ(X) (Kodaira dimension), 585
κ(p) (residue field of point p), 138
λ-invariant, 526, 526
A1 , see also affine line, 102, 104, 105
A2 , see also affine plane
An , see also affine n-space, 106
functorial characterization, 196G(k − 1, n − 1) (Grassmanian), 200
Gm (multiplicative group scheme), 196, 296,536
PF (projectivization of finite type
quasicoherent sheaf), 452
P(d1 , . . . , dn ) (weighted projective space),235
PV (projectivization of vector space), 152
P1k (projective line), 142
Pn (projective space), 151
Pn -bundle (projective bundle), 452, 453Pn
A (projective space over A), 144
is quasicompact, 153
Q-Cartier, 406, 407
Q-Cartier divisor, 406
Q-line bundles, 484, 484, 485, 556
Z-graded ring, see graded ring, ZZ≥0 -graded ring, see graded ring, Z≥0Zp (p-adics), 41
An
X , 254
N (nilradical), 113
F (U), 73
F (n), 413
F |U (restriction of sheaf to open subset), 76
F |p , 139
F sh (sheafification), 84
F ∨ (dual of OX -module), 80
Flf , 391
Ftor , 391
G |X (pullback), 423
KX (canonical sheaf/bundle), 485
Mg (moduli space of curves), 523
O (structure sheaf), 129
O-connected morphisms+ proper implies surjective, 738
O-connected morphisms, 710, 710, 711,
735–738+ proper implies connected fibers, 733+ proper implies surjective, 710, 738
not preserved by base change, 710

preserved by flat base change, 710
O(1)on relative Proj, 451
OK (ring of integers in number field K), 276
OX (structure sheaf), 77
OX -modules, 78
Ext-functors for, see Ext functors, for
O-modules
not always quasicoherent, 375
tensor product, 92
OX,p , 78
OSpec A , 129
Hom , 389
of quasicoherents not necessarily
quasicoherent, 389
Proj , see relative Proj
Spec , see relative Spec
| L | (complete linear series), 416µn (group scheme), 199
ω (dualizing sheaf), 749, 751
Ass M (associated primes of M), 172
Cl X (class group), 404
Cl(Ank ) = 0, 404
Homneed not commute with localization, 53
Hom (“sheaf Hom”)
is left-exact, 53
LocPrin, 403∼
Pic(Pnk ) = Z, 405
Pic X (Picard group), 372
Pic(Ank ) = 0, 404
Pic0 (E), 525, 532, see also Picard group, 540∼ E, 531=
Picτ X (group of numerically trivial line
bundles), 484
Prin (group of principal Weil divisors), 403
Proj S• , 148
Proj, 145
Spec A, 102
Spec Z, 104Specadjoint to Γ , 183
Supp D (support of divisor), 400
Supp G (support of sheaf of sets), 94
Supp(s), 83
Sym• (F ), 384
Tor functors, 621
adj(M) (adjugate matrix), 204
codim (codimension), 305
coker (cokernel), 48
cokerpre (cokernel presheaf), 81
det F (determinant bundle), 385
div, 400
idA (identity morphism), 26
ker (kernel), 48
kerpre (presheaf kernel), 80
obj (object in a category), 26
tr. deg (transcendence degree), 308

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

November 18, 2017 draft⊕ (direct sum), 48⊗ (tensor product), 33⊗OX (tensor product of OX -modules), 92
φpq (cohomology and base change map), 707
π-ample, see relatively ample, see ampleness,in relative setting
π-very ample, see very ample, relatively, seevery ample, relativelyπ∗ G , 424
π∗ , 422, 425
π! , 759
right-adjoint to π∗ , 759
π!sh , 758
π−1 F (inverse image sheaf), 92
π−1 , 92
is right-exact, 94
π! (extension by zero), 95
ρ(X) (Picard number), 484
Ext functors
computable with locally free resolutions,756
σ-process &, see blow-up√√−5, 165, 166
I (radical of ideal), 116
Ab (category of abelian groups), 27
AbX (category of sheaves of abelian groups on
X), 79has enough injectives, 634
CohX (category of coherent sheaves on X), 388
ModA (category of A-modules), 27
has enough injectives, 18
ModOX , 79, 381
not enough projectives, 636
QCohX (category of quasicoherent sheaves onX), 381
Rings (category of rings), 27
Sets (category of sets), 26
SetsX (category of sheaves of sets on X), 79
Top (category of topological spaces), 27
Veck (category of vector spaces over k), 26
×A (sloppy notation), 247
S (constant sheaf), 76
lim (colimit), 41−→
lim (limit), 40←−∧• , 384∧• (F ), 384!
M• (quasicoherent sheaf corresponding tograded module M• ), 411
in terms of “compatible germs”, 411
is an exact functor, 412
need not be an isomorphism, 412" (quasicoherent sheaf corresponding to aM
module M), 132
d : OX → ΩX/Y , 572
d : A → ΩA/B , 562
d-uple embedding (Veronese embedding), 233
hA (functor of maps from A), 28

777

hA (functor of maps to A), see functor, ofpointshF (m) (Hilbert function), 488
ip,∗ , 76
k-smooth, see smoothness, over a fieldk-variety, see variety
k[[x1 , . . . , xn ]], 725
k[ϵ]/(ϵ2 ), see dual numbers
n-fold, 304
n-plane, 231
opp (opposite category), 29
p-adics, 39, 40, 55, 351, 362, 655, 725
pa (X) (arithmetic genus), 480
pg (X) (geometric genus), 584
t : Hn (X, ωX ) → k (trace map), 749
xi/j , 143, 149, 397
OX (D), 401
ModOX (category of O-modules)
has enough injectives, 634
Hom , 79is left-exact, 92
need not commute with stalkification, 80
Abel, N. H. &, 297
abelian category, 25, 27, 38, 47, 48, 48, 49,
51–55, 57, 80, 81, 90–92, 369, 374, 381, 382,
388, 395, 625–627, 632, 635, 716, 757
enough injectivese.g., modules, 626, 627
enough projectives, 716
has enough projectives, 625
of complexes, 51
abelian cone associated to coherent sheaf &,449
abelian group scheme &, 197
abelian semigroup &, 45
abelian varieties &, 297, 297, 298, 532
are abelian, 298
absolute Frobenius &, see Frobenius
abstract nonsense &, see category theory
abut &, 61
action of group variety/scheme &, see groupvariety/scheme, action
acyclic &, see also resolution (complexes),
acyclic
object, 629, 631
additive category &, see category, additive
additive functor &, see functor, additive
adjoints, 43
of forgetful functors, 46
rightpreserves injectives, 638
table of, 46
adjugate matrix &, 204
adjunction formula, 584, 699, 766
for canonical bundle, 584, 699
for curve on a surface, 549
for dualizing sheaf, 763, 766, 766

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The Rising Sea: Foundations of Algebraic Geometry

affine 3-space, 658
affine cover cohomology vanishing, 466, 473
affine line, 103affine n-spaceover a scheme X (AnX ), 254
affine 3-space, 115, 128, 332, 334, 336, 353, 568,
610, 658, 659, 690
Affine Communication Lemma &, 132, 157,
158, 161, 179, 201, 208, 210, 213, 375, 376,
388, 406
affine complex varieties, 12, 13, see varieties,
affine complex, 108, 110, 114
affine cone &, 146, 152, 235, 235, 236, 308, 316,
317, 341, 406, 605, 690
relative version, 453
affine cover cohomology vanishing, 467
affine cover cohomology vanishing, 466
affine cover cohomology vanishing, 468, 475,
497, 500
affine line, see also line, 102, 102, 103, 104, 104,
105, 105, 107, 116, 117, 120, 133, 140–142,
160, 179, 191–193, 195–199, 211, 213, 218,
219, 237, 247, 259, 290, 293, 295, 296, 303,
304, 319, 343, 345, 352, 375, 382, 384, 389,
392, 399, 401, 402, 422, 427, 436, 481, 488,
505, 506, 509, 515, 516, 525, 526, 536, 579,
590, 593, 645, 650, 658–660, 674, 681, 701
affine morphisms &are quasicompact and quasiseparated, 208
affine morphisms &, 208
affine-local on target, 208
are separated, 285
as Spec , 449
preserved by base change, 262
preserved by composition, 208
affine n-space, 375, 449
affine n-space, 106, 107, 108, 112, 121, 132, 143,
153, 155, 164, 190, 196, 215, 217, 232, 255,
259, 308–311, 317, 332, 338–340, 344–346,
355, 356, 404, 436, 449, 460, 500, 580, 598,
603, 612, 649, 652, 676, 689(over a field) is regular, 345(over any base) is flat, 647
minus origin, 120, 132, 139–142, 155, 213,
215, 237, 279, 281, 284, 291, 359, 360, 475
with doubled origin, 141, 142, 154, 279, 281,
284, 359
affine open subscheme/subset &, 137
affine plane, 105, 106, 108, 110, 115, 117, 119,122–124, 128, 129, 132, 139, 140, 142, 170,
196, 213, 215, 217, 227, 236, 247, 259, 260,
291, 292, 310, 343, 353, 373, 406, 436, 475,
515, 524, 540, 548, 549, 564, 568, 569, 597,
598, 605–607, 609–611, 615, 650, 657, 691,
705, 729
affine scheme &, see scheme, affine
affine variety &, see variety, affine
affine-local &, 157, 158

on the source, 202
on the target, 202
Alexandrov space &, 636
algebraic field extension &, see field extension,
algebraic
algebraic equivalence, 671
implies numerical equivalence, 671
algebraic group &, 296
Algebraic Hartogs’s Lemma &, 144, 145, 161,
163, 315, 320, 320, 353, 373, 400, 402, 407,
583, 693, 705
algebraic space &, 435
Alteration Theorem &, 611
ample cone, 483, 552, 556, 558
ampleness, 414, 437, see also Kodaira’s
criterion for ampleness, see also Kleiman’s
numerical criterion for ampleness, seealso Serre’s cohomological criterion for
ampleness, 494, see also Nakai-Moishezon
criterion for ampleness, 552, see also
ample cone, 558⊗ base-point-free is ample, 439= positive degree on smooth projective
integral curve, 510
better-behaved than very ampleness, 438
implies nef, 484
in relative setting, 456, 457
in the absolute setting, 441, 441
properties of, 441
Kodaira’s criterion, see Kodaira Embedding
Theoremof vector bundle, 458
open condition, 438
preserved by ⊗, 439
preserved by %, 439
preserved by finite pullback, 438, 457, 481,
494, 495
relative, see relatively ample, 442
Serre’s cohomological criterion, 496
Serre’s cohomological criterion, 438,
494–496vector bundle, 458
with respect to π, see ampleness, in relative
settinganalytic geometry, 725, 731
analytification &of a morphism, 186analytification &, 160, 160, 186, 284, 297, 373,
480, 484, 492, 539, 547, 576, 586
André-Quillen homology &, 566, 567
anticanonical bundle &, see also
canonical,bundle, 706
arithmetic genus &, see genus, arithmeticarrow &, see morphism
Artin-Rees Lemma &, 331
Artin-Rees Lemma &, 350, 364, 364, 365, 665,
666, 728
Artin-Schreier cover &, 593

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November 18, 2017 draft

ascending chain condition &, 125, 126, 127,219
associated primes &, 173
associated graded &, 334, 555
associated points &, 170
associated points &, 135, 166–168, 169,
169–171, 173, 174, 239, 241, 269, 292, 388,
405, 409, 462, 483, 489, 491, 544–546, 548,
554, 577, 601, 650, 657, 663, 685
and flatness, 650
correspond to irreducible components ofsupport of sections, 167of coherent sheaf, 388
of integral schemes, 173of locally Noetherian scheme, 169
associated primes &, 167–169, 172, 172, 173,
650, 685, 687, 693, 695
properties of, 172
weakly, 168
Associated-to-Separated Theorem &, 293
Asymptotic Riemann-Roch &, see
Riemann-Roch, Asymptotic
Atiyah flop &, 614
Atiyah, M. &, 17, 203, 614
Auslander, M. &, 362
Auslander-Buchsbaum Theorem &, 363, 363,
410, 547, 685, 693
automorphism group &, 26, 27, 518, 519, 521,
523, 535, 536, 587, 594
bounds for curves, 594
infinitesimal, 587
of a scheme, 181
of an object in a category, 26, 26
Axiom of Choice &, 18, 113, 123, 263, 623, 625,
see also Zorn’s Lemma, 627
base, 256base gluability axiom, see also gluability axiom
base change, 256
base change diagram, see Cartesian
diagram/squarebase field extension, 255
base gluability axiom, 130base gluability axiom &, 130base gluability axiom &, 87, 131base identity axiom &, 87, see also identity
axiom, 130, 131base locus, 415, 416, 612–614
scheme-theoretic, see scheme-theoretic, base
locus
base of a topology, 87
presheaf, see presheaf, on a basesheaf, see sheaf, on a base
base point &, 415, 416, 509, 519, 557, 741
base ring, 148base scheme, 181, 256, 416
base-point-free locus, 415779

base-point-freeness, see also global generation,
413, 414, 415, 415, 416⊗ ample is ample, 439⊗ very ample is very ample, 437
and maps to projective space, 415
implied by very ampleness, 437
in relative setting, 456
preserved by ⊗, see global generation (ofquasicoherent sheaf), preserved by ⊗relative, 442
with respect to π, see base-point-freeness, in
relative setting
Bertini’s Theorem &, 348
Bertini’s Theorem &, 331, 343, 346, 346, 492,
494, 520, 613, 684, see also Kleiman-Bertini
Theorem, 699
improved version, 683
weaker hypotheses, 348
Betti numbers &, 586, 708
Bézout’s Theorem &, 531
Bézout’s Theorem &, 231, 234, 274, 344, 431,
432, 491, 492, 512, 514, 522, 523, 543, 544,
549, 704
for plane curves, 491
bilinear &, 34bilinear form &, 552
symmetric (index of), 553
birational &, 587(rational) map, 190, 190, 192, 193, 273, 462,
583, 611, 612, 614, 705, 743
factorization of, 614
invariant, 583–586, 743
model, 459
morphism, 190, 274, 277, 460, 502, 598, 704,
705, 739, 740, 742, 743
Zariski’s Main Theorem version, 735, 736
varieties/schemes, 190, 190, 460, 462, 550,
584, 597, 607, 614, 743
need not be isomorphic, 406
birational, right?, 736
bitangents &, 697
blow down &, 740
blow-up &, 259, 334, see also proper transform(of blow-up), 597, see also universal
property, of blow-up, 599, see also totaltransform (of blow-up)
and line bundles, 614
and the canonical bundle, 615
commutes with flat base change, 652, 730
computable locally, 600
does not commute with general basechange, 652
exists, 603
is projective, 603
locally principal closed subscheme, 600
not flat, 650
of affine space, 450, 599
of nonreduced subscheme, 611

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780

The Rising Sea: Foundations of Algebraic Geometryof regular embeddings, 606
of smooth in smooth is smooth, 607
of the plane, 259, 406, 549, 550, 597, 609, 697,
698, 703–706
as graph of rational map, 292
preserves irreducibility and reducedness,
600
Proj description, 604
to resolve base locus, 612
Blow-Up Closure Lemma &, 600, 602, 602,
603, 605, 609, 613
Borel-Moore homology and Chow groups,
555
boundary (homological algebra) &, 50
box-times, see %branch &divisor, 590, 592
locus, 588
points, 514, 515, 516, 526, 591, 591, 594
branched cover &, see also double cover, 211,
515, 517, 593
Buchsbaum, D. &, 362
Calabi-Yau variety &, 322
Calabi-Yau variety &, see also trichotomy(classification of varieties), 585, 585, 586
Cancellation Theorem &, 288, 456
for closed embeddings, 437
for finite type morphisms, 327
for flat morphisms, 677
for projective morphisms, 456, 735, 742, 760
for proper morphisms, 294, 296, 739
for quasicompact separated morphisms,466
for quasicompact morphisms, 208, 459
in Sets, 288
canonical &bundle KX , 485, 584, see also anticanonicalbundleis Serre-dualizing, 485, 524, 579, 584,
586–588, 593, 749, 763
curve, see curve, canonicalembedding, 521
map, 517
sheaf, see canonical, bundle KX
Caroll, L. &, 115
Cartan-Eilenberg resolution &, 630, 631, 633,
633, 634
Cartesian diagram/square, 36, 256
Cartier divisors &, see also effective Cartierdivisors, 17Castelnuovo’s Criterion &, 605, 697, 705, 725,
740, 740, 742
category &, 26, see also subcategory
abelian, see abelian category
additive, 47, 48, 79–81, 90
concrete, 27

equivalence, 30, 30, 88, 89, 182, 191, 378,
412, 462
index, 40, 41, 42, 74
locally small, 26
of A-schemes, 183
of k-schemes, 183
of S-schemes, 183of complexes, 51
of open subsets, 28
of schemes, 181of subsets of a set, 28
opposite, 29, 191, 461
rings and affine schemes, 182
small, 30, 39, 43, 51
category theory &, 13, 19, 23, 78
catenary rings &, 311
Cayley, A. &, 697
Cayley-Bacharach Theorem &, 512Čech cohomology, 470
Čech cohomology, 17, 279, 465, 470, 471, 472,
628, 637–642= derived functor cohomology, 621, 635, 637
agrees with derived functor, 637Čech complex &, 470, 470–473, 475–477, 498,
516, 639–641, 669, 713, 715, 752Čech cover &, 638, 640, 754Čech resolution, 670
chase of base, see base change
Chasles’ Theorem &, 512
Chern class &, 480, 546, 554, 555
Chevalley’s Theorem &, 216, 217, 218,
218–221, 236, 260, 262, 321, 546, 662
for finitely presented morphisms, 218, 261
for locally finitely presented morphisms,
261
Chinese Remainder Theorem &, 133, 144, 254
Chow group &, 555
Chow groups, 555
Chow’s Lemma &, 501, 501–503, 544
other versions, 504
Churchill, W. &, 645
class (set theory) &, 26class group &, 166, 404, 404–408
and unique factorization, 407
excision exact sequence, 404
in number theory, 374, 405, 407
Clifford, W. &, 526
closed embeddings &preserved by base change, 254
closed embeddings &, 211, 225, 225
affine-local on target, 226
are finite, 225
are monomorphisms, 255
are proper, 293, 294
are separated, 281, 286
base change by, 254
closed under composition, 225
in projective space, 230

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November 18, 2017 draft

intersections of, 254
preserved by base change, 262, 285
closed immersions &, see closed embeddings
closed map &, 211, 221, see also universallyclosed morphisms, 293
closed point &, 505, 691, 744
of Spec A correspond to maximal ideals,
121
closed point &, 107, 108, 121, 121, 123, 142,
144, 152, 153, 155, see also degree, ofclosed point, 160, 179, 212, 219, 226, 232,
248, 260, 267, 268, 290, 296, 298, 306, 314,
327, 335, 336, 339, 340, 342, 343, 345, 347,
348, 357, 359, 361, 363, 391, 415, 422, 432,
435, 440, 460, 464, 481, 491, 495, 496, 505,
507–509, 514, 524, 537–540, 550, 564, 568,
571, 576–578, 606, 607, 616, 636, 657, 665,
667, 677, 678, 682, 701, 705, 710, 727, 729,
730
checking Cohen-Macaulayness (forquasicompact schemes), 691
checking for singularities of varieties, 338
checking normality (for quasicompact
schemes), 162
checking reducedness (for quasicompact
schemes), 155
dense for finitely generated k-algebras, 121
dense for varieties, 154
of Pn , 143, 144, 151, 232
kof Spec Q(t) ⊗Q C, 256of affine n-space, 121
of locally finite type k-schemes, 160
quasicompact schemes have one, 153
schemes need not have one, 153closed subscheme, 202closed subscheme exact sequence &, 479
closed subscheme exact sequence &, 384, 386,
408, 412, 427, 428, 490, 496, 651, 723, 741,
765
closed subschemes &, 133, see also closed
embeddings, 138, 147, 151, 225, 225
arbitrary intersections, 227
correspond to quasicoherent sheaf of ideals,386
cut out by s, 227
finite union of, 227
locally principal
have codimension 0 or 1, 316
locally principal, 241, 241, 262, 602, 603, 612,
see also divisor, locally principal, 668
pull back, 262
closurescheme-theoretic, see scheme-theoretic,
closurecocycle condition &, 89, 141, 251, 370, 371, 376,
399, 411
codimension &, 244, 305
at most difference of dimensions, 306781

odd behavior, 306
one, 315
pathologies, 318
sometimes difference of dimensions, 311
cofinal &, 726, 728, 738, 741, 745
Cohen Structure Theorem &, 726
Cohen-Macaulayness &, 363, 364, 486, 548,
685, 688, 689–693, 701, 750, 753, 754,
760–762, 766
and adding a variable, 691
fancy properties, 691
implies no embedded points, 689
implies no embedded points, 689, 690
of all dimension 0 Noetherian schemes, 689
of all dimension 1 Noetherian schemes
without embedded points, 689
of all dimension 2 normal varieties, 692
of regular local rings, 685
open condition, 691
preserved by localization, 691
slicing criterion, see slicing criterion, for
Cohen-Macaulayness
Coherence Theorem &, see Grothendieck’s
Coherence Theorem
coherent sheaves &, 152, 369, 374, 388, see also
modules, coherent
CohX is abelian category, 395
abelian cone of, 449
affine-local nature, 388
dual, 389
Euler characteristic, 478
flat = locally free, see finitely presented
sheaves, flat = locally freecohomological criterion of affineness &, seeSerre’s cohomological criterion for
affinenesscohomology, see also Čech cohomology, seealso derived functor cohomology, see also
homology, 17, 50, 478, 499
Čech = derived functor, 637Čech = derived functor, 621, 635, 637
and change of base field, 474
behaves well in flat families, 669
commutes with flat base change, 651, 708,
730
cup product, see cup productdimensional vanishing, 660, 764
more generally, 474
dimensional cohomology vanishing, 466
dimensional vanishing, 474, 615, 616
relative, 500
of double complex, 58
of line bundles on projective space, 475
pullback, 466
relative dimensional vanishing, 500
with supports, 17
Cohomology and Base Change Theorem &,720

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Cohomology and Base Change Theorem &,364, 497, 499, 709, 709–711, 713, 715, 716,
718, 719, 722
cokernel &, 48, see also universal property, of
cokernelpresheaf, 81
colimits
do not necessarily commute withright-exact functors, 56
colimits &, 41
are right-exact, 54
commute with colimits, 54
commute with left adjoints, 54
e.g., cokernels, 54
filtered, 42
are exact for A-modules, 94
commute with cohomology, 56, 655
commute with homology in ModA , 56
in ModA are exact, 54–56
localization as, 43
not always exact, 55
non-filtered, 43
collapse (of spectral sequence) &, 61
comathematician &, 41
compatible germs &, 49, 83, 83, 84, 87, 88, 94,
151, 227, 378, 411, 434
complete intersections &, 584
complete k-scheme &, 293
complete intersections &, 245, 494, 521–523,
531, 532, 585, 766
in affine space
are Cohen-Macaulay, 689
implies no embedded points, 689
in projective space, 494
positive-dimensional ones are connected,
494
complete linear series &, see linear series,
completecomplete with respect to an ideal I &, 726
completion &, 14, 41, 334, 393, 658, 666, 680,
725, 725, 726, 726–730, 732–734, 744
and Cohen-Macaulayness, 691
and exactness, 727
commutes with finite direct sums, 729
is flat, 667, 727, 730
not always exact, 728
of A along I, 725
preserves exact sequences of finitely
generated modules, 727, 729
complex (abelian category) &, 49, 49
double, 57, 58, 58, 63–65, 69, 243, 244,
629–631, 638, 641(hyper)cohomology of, 58
first quadrant, 59
total complex of, 58
factorization into short exact sequences, 50
homotopic maps give same map on
homology, 623

complex (affine) n-space, see n-spacecomplex (affine) plane, see affine planecomplex manifold, see manifold, complex
complex varieties, see also affine complex
varieties, see varieties, complex, 145, 160,
161, 186, 373, 541, 586
component &connected, 153
irreducible, 153
component &connected, 123, 123–125, 159, 266, 267, 371
irreducible, 123, 123, 125, 269= minimal primes for Spec A, 128
Noetherian spaces have finitely many,
124
composition series &, see modules,
composition series
composition-of-functors spectral sequence &,see Grothendieckcomposition-of-functors spectral
sequencecompositum &, 308
comultiplication &, 199cone, 333
affine, see affine cone
projective, see projective cone
cone over smooth quadric surface, see quadricsurface, 145, 163, 165, 272, 309, 334, 341,
408, 613, 614
cone point, 235, 333, 538, 610, 612, 613
conic, 147, 149, 151, 191–193, 231, 233, 235,
236, 321, 348, 349, 433, 482, 492, 511, 513,
514, 519, 520, 537, 538, 671, 706, 712
universal, 481
connected &, 119, 153
component, see component, connected
fibers, see fibers, connectedconnecting homomorphism &, 52, 60, 466,
471, 473, 498, 622, 624, 637
conormalbundle, 567
cone/bundle/sheaf, 597
exact sequence, 567, 574, 676
affine version, 567
for smooth varieties, 577
left-exact in good circumstances, 567, 575,
577
left-exactness, 680
module, 567, 567
sheaf, 567, 568, 569
of regular embedding is locally free, 569
Conrad, B. &, 25constant (pre)sheaf &, 76
constant rank implies locally free for finite
type quasicoherent sheaves &, see finite
type, quasicoherent sheaves, constantrank implies locally free

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constructible subsets &, see also locallyconstructible subsets, 217, 218
are finite disjoint unions of locally closed
subsets, 217
in more general situations, 262
locally, 262
contravariant functor &, see functor,
contravariant
coordinates &, 102
projective, 151, 533, 581
coproduct &, see also fibered coproduct, 37, 42
of schemes, 247
of sets, 37
cotangentbundle, 485
complex, 566, 680
map, 335
sheaf, 561, see also tangent sheaf
relative, 572
space, 73, 331, 331–333, 335, 341, 561, 568,
570, 577, 678, 682
Jacobian criterion, 335
vector, 332, 561, 561= differential, 332
relative, 561
vertical, 561, 565
cotangent exact sequence
left-exact in good circumstances, 566
counterexamples (list of) &, 132
covariant functor &, see functor, covariant
cover, 72
by open subfunctors, 252
covering space &, 355, 674
Cremona transformation &, 193, 292, 437, 613,
706
cross-ratio &, 526
cross-ratio &, 526
classifies 4 points on P1 , 526
cubic, see also twisted cubic curve, see also
nodal cubic, 274, 513, 514, 520, 522,
527–535, 540, 541, 613, 658, 660
curves, see also cuspidal cubic, 613
hypersurface, 230
nodal
normalization of, 658
plane curves, 343
surface, 14, 16, 315, 521, 522, 613, 697–706
Fermat, 699, 703
is blown-up plane, 704
singular, 699
cup product &, 475, 750, 754
Yoneda, 757, 762
curve singularity, 730
Curve-to-Projective Extension Theorem &,435Curve-to-Projective Extension Theorem &,191–193, 436, 460–462, 526, 530
necessity of hypotheses, 436

783

curves &genus 1classified by j-invariant, 528
genus 3, 521
genus 4, 521
nonhyperelliptic, 521
plane, 519
curves &, 230, 304
canonical, 521
elliptic, see elliptic curves, 528
exists regular projective birational model of
each, 459
genus, 505
genus 0, 493, 510–512, 514, 517, 518, 523,
588, 593, 711, 731
are conics, 511
genus 1, 493, 514, 517, 518, 521, 523, 524,
528, 535–537, 540, 660
not classified by j-invariant, 528
genus 2, 517–519, 592
hyperelliptic in precisely one way, 517
genus 3, 480, 493, 517, 519–523, 592, 731
not all hyperelliptic, 517
genus 4, 494, 521, 522, 531
genus 5, 521–523
genus 6, 493, 731
hyperelliptic, see hyperelliptic curves
moduli space, 646
non-rational, 193
nonhyperelliptic, 517, 519–522, 531
of every genus, 516
plane, 343, 491, 512, 520, 537, 549, 564, 602,
609, 660, 697, 729, 731
proper implies projective, 495
trigonal, 520, 522
various categories are equivalent, 461
cusp, 274, 343, see also normalization, of cusp,
505, 588, 609, 612, 730, 730
cuspidal cubic, 274, 343, 588, 609
normalization, 663
cycle (homological algebra) &, 50de Jong, A.J. &, 611
de Rham cohomology &, 578, 579
Dedekind domain &, 165, 166, 276, 311, 354,
374
unique factorization of ideals, 354
deformation theory &, 336, 587, 646, 655, 697,
701
deformation to the normal cone &, 616, 616,
657
deformations (infinitesimal) &, 587
degenerate linear series &, see linear series,(non)degenerate
degreed map, 431
algebraic = analytic for maps of curves, 463
in projective space, 491, 645

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additive for unions, 493
constant in flat families, 670
of curve, 431
topological definition, 492
of closed point, 160of coherent sheaf on curve, 481
additive in exact sequences, 481
of discriminant of degree d polynomials,594
of divisor on regular curve over k, 479
of finite flat morphism is locally constant,656
of finite flat morphism is locally constant,259, 462, 546, 593, 615, 657, 664, 701
of finite map from curve to regular curve isconstant, 463, 479, 514, 527, 670
of finite morphism at point, 393
of generically finite morphism, 664
of generically finite rational map, 309
multiplicative under composition, 309
of homogeneous element, 148of hypersurface, 230
of line bundle on curve
additive for ⊗, 479
of line bundle on curve, 479, 483, 543, 645
additive in exact sequences, 592
of line bundle on projective space, 405
of projective morphism from curve toregular curve, 657
of projective morphism of curves, 462, 462
delta-functors &, 762
universal, 753
delta-functors &, 627, 763
e.g., derived functors, 628
morphism of, 628
universal, 627, 628, 753, 762
depth &, 685, 685, 691–694
bounded by dimension of support, 685
derivation &, 332, 570, 570–574, 580
universal, 570
derived functors
left, 626
derived category &, 471, 760
derived functor cohomology, 17, 628, 637,
639–641= Čech cohomology, 621, 635, 637
derived functors, 17, 57, 465, 474, 497, 627,
754, 756
agrees with Čech cohomology, 637
and spectral sequences, 629
computed with acyclic resolutions, 629
left, 625
right, 626
derived pushforward, see higher pushforward
descending chain condition &, 124, 125, 327,
328, 483
descent &, 255, 534, 646

desingularization &, see resolution(singularities)
determinant (line) bundles &, 481
determinant (line) bundles &, 385, 485, 749
behave well in exact sequences, 386
determinant map GLn → Gm &, 198dévissage &, 17, 544
diagonal ∆ (locally closed subscheme of
X ×Y X) &, 280
diagonal morphism δ &, 279
is a locally closed embedding, 280
diagonalizing quadratic forms &, 164, 165,
193, 233, 234, 272, 333, 343, 552
diagram indexed by &, see category, index
Dieudonné, J. &, 243
completely functorised, 47
different ideal &, see also relative different anddiscriminant ideals, 589
differentiable functions &germ, 72
sheaf, 71differentiable manifolds &, see manifolds,
differentiable
differentials, 280, 332, see also relative
cotangent sequence, 592= cotangent vectors, 332
behave well with respect to base change,
571, 575
explicit description, 563
fiber at a rational point, 570
module of (for ring maps), 562
on hyperelliptic curves, 579
on projective space, see Euler exact
sequencepulling back, 571, 575
universal property, see universal property,of differentials
dilation &, see blow-up
dimensionof variety at a point, 313
dimension &pure, see equidimensional
dimension &, 106, 303= transcendence degree, 308
additive for products of varieties, 311, 663
constant in flat families, 670
of fibers, 322
never lower than expected, 322, 700
of scheme, 14, 155, 162
dimensional cohomology vanishing &for quasiprojective schemes, 616
dimensional cohomology vanishing &, 466,
474, 616, 660, 764
for quasiprojective schemes, 615
more generally, 474
relative, 500
Diophantine equations &, 185, 191–193, 518,
697

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November 18, 2017 draft

direct image sheaf, see sheaf, pushforward
direct limits &, see colimits
discrete valuation &, 361
discrete topology &, 76, 76, 84, 144, 212, 640,
726
discrete valuation &, see also discrete
valuation rings, see also valuations
discrete valuation &, 351, 359
discrete valuation rings &finitely generated modules over, 375
flatness criterion, 668
discrete valuation rings &, 110, 275, 331, 339,
345, 351, 358–362, 389, 391, 400, 437, 460,
463, 464, 538, 580, 591, 592, 647, 648, 655,
657, 658
classification of finite generated modules
over, 580
differentials over a field, 580, 592
finitely generated modules over, 354, 391,
399, 464, 525, 580, 648
flatness criterion, 655
discriminant ideal &, see also relative different
and discriminant ideals, 589
disjoint union of schemes &, see also infinite
disjoint union of schemes, 137, 137, 141,
146, 156, 157, 162, 218, 240, 247, 267, 305,
575, 588, 674, 676, 720
distinguished affine base &, 377, 377, 378
distinguished open set, see open set,
distinguisheddivisible abelian group &, 627
divisor &Q-Cartier, see Q-Cartier divisor
branch, see branch, divisor
Cartier, 17effective Cartier, see effective Cartierdivisors
exceptional, see exceptional divisor
Weil, see Weil divisor
domain of definition &, 171, 171, 188, 192, 193,
289, 291, 291, 583
dominant (or dominating) &, 189
dominant (or dominating) &, 273
dominant (or dominating) &, 189, 189–191,
193, 218, 219, 275, 308, 309, 323, 326, 461,
462, 539, 546, 590, 593, 681, 682
Donagi, R. &, 697
double complex &, see complex (abeliancategory), double
double cover &, 525, 529
double cover &, see also branched cover, 111,
211, 514, 514–516, see also hyperelliptic
map, 517–519, 521, 525, 526, 529, 540, 579,
589, 645
dual &of OX -module, 80of coherent sheaf, 389
of locally free sheaf, 372

785

projective space, 346
variety, 348
dual numbers &, 104, 112, 113, 133, 134, 138,
156, 186, 225, 237, 290, 304, 336, 393, 418,
655, 674
flatness criterion, 655
dualizing sheaf &, 467, 485, 522, 749, 751, 761,
765
= canonical sheaf, 763
DVR &, see discrete valuation rings
Dwork, B. &, 362
Dynkin diagrams &, 610, 706
effaceable functor &, see functor, effaceable
effective Cartier divisors &, 17, 240, 241, 241,
242, 245, 260, 292, 337, 397, 405–410, 435,
474, 544–547, 549, 569, 598–607, 612–616,
651, 667, 668, 689, 704, 736, 742, 763, 765,
766
have pure codimension 1, 316
normal bundles to, 569
relative, 668
restriction to open sets, 405
effective cone, 552, 552
Eilenberg-Zilber Theorem &, 475
Eisenbud, D. &, 276
Eisenstein’s Criterion &, 343
elimination of quantifiers &, 221elimination theory &, 216, see also
Fundamental Theorem of Elimination
Theory, 222
ellipse &, 513, 536, 537
elliptic curves &, 525
elliptic curves &, 373
group law, 531
elliptic curves &, 163, 186, 193, 297, 516, 518,
523, 524, 524–527, 529, 532, 535–541, 550,
708
are group varieties, 532
counterexamples and pathologies, 538
degenerate, 536
degree 1 points = degree 0 line bundles, 524
group law, 524, 525, 530–532, 540, 705
level n structure, 525
non-torsion point, 539, 540
not same as genus 1 curves, 524
elliptic fibration &, 613
embedded points &, 168, 168, 170, 170, 171,388, 400, 462, 464, 491, 548, 549, 577, 687,
689, 690, 692, 694, 765
embedding (unwise terminology) &, 229
enough injectives, 630
enough injectives, 18, 626–630, 634, 642, 754,
758
enough projectives, 625, 626, 629, 636, 754
epimorphisms &, 37, 38, 47–49, 54, 57, 86, 89,
91, 226
equalizer exact sequence &, 75, 130, 251

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equidimensional &, 323
equidimensional &, 304, 304, 305, 314, 317,
326, 338, 341, 342, 356, 357, 435, 462, 555,
575, 613, 662, 663, 689–691, 751, 760–764,766
equivalence of categories &, see category,equivalence
equivariant morphism &, 683
espace étalé &, see sheaf, space of sections
essentially finitely presented &, 668
essentially surjective &, 30étale morphisms &, 703
étale cohomology &, 381, 621
étale cover &, 356, 676
étale morphisms &, 331, 340, 354, 355, 356,
589, 673, 675, 676, 677, see also formallyétale, 680, 701, 703, 727
closed under composition, 356
differential geometric motivation, 354
form “reasonable” class, 356
not same as “local isomorphism”, 356
open condition, 355étale topology &, 381, 674, 727
étale-locally &, 732
Euclid &, 105
Euclidean algorithm / Euclidean domain &,105, 106, 163
Euclidean domain, 104
Euler exact sequence &, 582
Euler characteristic &, 399, 492, 493, 548, 554,
645, 646, 661, 669
additive in exact sequences, 478, 479, 490,
554
constant in projective flat families, 647, 661,
669, 708, 715
constant in proper flat families, 497, 715
without Noetherian conditions, 720of coherent sheaf, 478
polynomiality, 488, 523
Euler exact sequence &, 580–583, 586, 587
generalizations, 582
Euler test for projective hypersurfaces &, 343,
343
exact, 49, 49
exact functor &, see functor, exact
exact modules, 645
exact morphisms, 645
exact sequence, see also exponential exact
sequence, see also Grassmannian,
universal exact sequence, see also longexact sequence, see also Euler exact
sequence
closed subscheme, see closed subscheme
exact sequenceequalizer, 75, 130, 251
excision for class groups, 404
grafting, 63
split, 634

excellent rings &, 327
exceptional divisor &, 260, 549, 598, 599,
604–607, 609–612, 614–616, 705, 706, 730,
736, 740, 742
normal bundle to, 605
exponential exact sequence &, 86, 89, 91, 480
Ext functors, 621, 749, 753
composition of, 757
first version, 626
for O-modules, 754
computable by acyclics, 754
long exact sequence, 754
second version, 626
two definitions are same, 629
Ext functors, 754
behaves like derived functor in firstargument, 756
needn’t preserve quasicoherence, 755
extending the base field &, 255
extension by zero &, 95, 637, 642, 758
extension of an ideal &, 254
extension of fields &, see field extension
extension of scalars &, 46
exterior algebra &, 384
factorial &, 363
factorial &, 163, 163–165, 334, 363, 403, 405,
406, 410, 538, 547, 558, 579, 685
but no affine open is UFD (example), 538
implies normal, 164
open condition, 163
weaker than regular (example), 363
faithful flatness &, 661
preserved by base change, 661
faithful functor &, see functor, faithful
faithful module &, 207
Faltings’ Theorem (Mordell’s Conjecture) &,518
Fano variety (classification theory) &, 322,
585, 585, 586, 704
Fano variety (of lines) &, 704
Fermat, 193
Fermat cubic surface, see cubic, surface,
FermatFermat curve &, 193, 343
Fermat’s Last Theorem &, 193, 518
Fermat, P. de &, 193
FHHF Theorem &, 54, 56, 474, 497–500, 652,
716, 717
fibered productin category of open sets, 36
fibered product &, 229
fibered coproduct &, see also coproduct, seealso universal property, of fiberedcoproduct, 37, 248
fibered diagram/square, 36
fibered productof schemes, 247

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November 18, 2017 draftwith open embedding, 202
fibered product &, see also product, 35, 35–38,
40, 127, 195, 201, 247–250, see also
universal property, of fibered product,251–256, 259, 262, 264, 273, 284, 285, 289,
290, 292, 294, 423, 425, 434, 447, 448, 450,
563, 601, 653“commutes with localization”, 255
of functors, 251
of schemes, 247
fibered productsof sets, 36
fibers, 201, 256, 257, see also generic fiberconnected, 267, 734, 735
from Zariski’s Connectedness Lemma,733
dimension of, 322
well-behaved for flat morphisms, 650,
661, 662
geometric, 264, 264, 265, 646, 676–678, 680,
711, 712
properties preserved by base change, 264
geometrically connected (etc.), 264
of OX -module, 139
scheme-theoretic, see scheme-theoretic,
fiber, 702
field &
perfect, 268, 269
field &
perfect, 578
field &
imperfect, 296
perfect, 268, 269, 342, 363, 576–578, 679, 693
separably closed, 264, 267, 268, 270
field extension
finite, 275
field extension &, 166, 191, 210, 255, 256, 265,
267–269, 273, 276, 277, 314, 339, 342, 345,
357, 415, 463, 474, 479, 494, 508, 510,
532–535, 539, 544, 576, 593, 594, 611, 679
algebraic, 203, 204, 275, 305, 307, 314, 565
finite, 107, 121, 160, 165, 212, 218, 254, 255,
265, 269, 276, 277, 305, 308, 309, 313, 342,
459, 515, 538, 565, 588, 674, 676
finitely generated, 191, 267, 268, see alsotranscendence degree, 308, 493, 681
Galois, 254, 255, 313, 593
inseparable, 264
normal, 313
purely inseparable, 270
purely inseparable, 266, 270, 277, 468, 681
purely transcendental, 191, 314
separable, 269, 277, 565, 570, 576, 588, 590,
674, 676
separably generated, 268, 572, 576, 681
filtered, see also colimits, filtered, 42, 42, 43,
54–57, 74, 94, 378, 466, 474, 655
final object &, see object, final

787

finite B-algebra &, 210
finite fibers not preserved by base change &,262
finite flat morphisms &, 259, 462, 546, 656,
664, 699, 701, 720, 760, 762
have locally constant degree, 656
have locally constant degree, 546, 657, 664,
701
finite global generation &, see global
generation (of quasicoherent sheaf), finite
finite length &, see length, finite
finite morphisms &, 210, see also finite flat
morphisms= integral + finite type, 214= projective + affine, 469= projective + finite fibers, 456, 462, 469,
469, 481, 507, 701, 742= projective + quasifinite, 469= proper + affine, 296, 469, 501, 739, 739= proper + finite fibers, 469, 739= proper + quasifinite, 296, 739= proper + quasifinite + locally finitely
presented, 739
affine-local on target, 211
are affine, 210
are closed, 213, 232
are integral, 204, 213
are projective, 213, 455
are proper, 294
are quasifinite, 213, 215
are separated, 285
closed under composition, 213
have finite fibers, 213, 257, 259
not same as finite fibers, 213
preserved by base change, 262
separable, 590
to Spec k, 212
finite type implied by quasiprojective, 159
finite type &A-scheme, 159
morphisms, 214, 293
are affine-local, 214
preserved by base change, 262
over a ring, 159
quasicoherent sheaves, see also modules,
finitely generated, 388
affine-local, 388
constant rank implies locally free, 392,
464, 656, 670, 681, 718
support is closed, 390
finitely generated &domain, 308
field extension, 191
modules, see modules, finitely generated
finitely globally generated &, see global
generation (of quasicoherent sheaf), finite

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finitely presented, see also local finite
presentation, 214, see finitely presented
morphisms, see finitely presented= finite type under Noetherian hypotheses,
215, 216, 261
algebra (ring map), 215, 215, 216, 216, 563,
668
essentially, 668
modules, see modules, finitely presented
morphisms, 207, 215, 216, 438, 646, 657, 664,
677, 680, 710, 719–722= “finite in all ways”, 216
affine-local on source, 216
Chevalley’s Theorem, 218, 261
closed under composition, 216
locally pullbacks of nice morphisms, 261
preserved by base change, 262
finitely presented modules &, see modules,
finitely presented
finitely presented quasicoherent sheaves &,see also modules, finitely presented, 388,
669, 719
affine-local, 388
finitely presented sheaves
flat = locally free, 656
local freeness is stalk-local property, 391
finiteness of zeros and poles &, 400
finiteness of zeros and poles &, 353
first algebraic de Rham cohomology group &,578, see also genus, geometricof hyperelliptic curve, 579
first quadrant double complex &, 59
Five Lemma &, 49, 50, 63, 64
subtler, 64, 752
flabby sheaf &, see sheaf, flasque/flabby
flag variety &, 200, 446, 583
bundles, 583
flasque sheaf &, see sheaf, flasque/flabby
flatness, 217, 295, 428, 428, 462, 474, 484, 497,
499, 543, see also finite flat morphisms,
555, 616, 621, 622, 624, 625, see also
resolution (complexes), flat, 649, 672, seealso generic flatness, see also Miracle
Flatness Theorem= free = projective for finitely presented
modules over local rings, 647, 655= locally free for finitely presented sheaves,656= projective for finitely presented modules,656= torsion-free for PID, 655
and blowing up, 652
at a point, 649
commutes with base change, 667
constancy of Euler characteristics, see Euler
characteristic, constant ...
constancy of fiber arithmetic genus, 670
constancy of fiber degree, 670

constancy of fiber dimension, 670
equational criterion, 654
faithful, see faithful flatness
fiber dimension well-behaved, 650, 670
fibral, 668, 669
flat limit, 657, 658, 659–661
explicit example, 658
Going-Down Theorem, see Going-Down
Theorem, for flatness
good behavior of fiber dimension, 662
ideal-theoretic criteria, 647, 652, 654, 654,
655, 665, 729
finitely generated version, 654
stronger form, 729
implies constant Euler characteristic, see
Euler characteristic, constant inprojective flat families
implies torsion-free, 648, 652, 655, 665, 667,
691
infinitesimal criterion, 666
local criterion, 647, 665, 665–668, 709, 718,
722
more general version, 665
no higher Tor, 665
no higher Tor, 624, 625, 646, 652–654
of affine space, 647
of completion, 667, 727
of free modules, 647
of localization, 647
of modules, 54, 647
of open embeddings, 649
of projective modules, 647
of quasicoherent sheaf over a base, 649
of relative dimension n, 663, 663, 664, 675,
676, 691
preserved by base change, 663
of ring morphism, 647
open condition, 664, 664
open in good situations, 647, 662
over a field, 647
over Discrete Valuation Rings (criterion),
647, 655, 668
over dual numbers, 647, 649, 655
over integral domains, 647, 648
over principal ideal domains, 647, 648, 655
over regular curves, 657
preserved by base change, 648, 649
sends associated points to associatedpoints, 650
slicing criterion, see slicing criterion, forflatness
stalk/prime-local property, 648
topological aspects, 647, 661
transitivity of, 648, 649
valuative criterion, 657
variation of cohomology, 658
flex &, 528, 530
forgetful functor &, see functor, forgetful

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November 18, 2017 draft

formal neighborhood &, 727, 732, 732, 734,
744
formal power series &, see power series,
formalformal scheme &, 17, 732
formally étale &, 680, 680, 727
formally smooth &, 680, 680
formally unramified &, 588, 680, 680
Four Squares Theorem &, 193
fraction field &, see also total fraction ring
(total quotient ring), 32, 138, 139, 156, 162,
191, 273, 274, 351–354, 407, 576
fractional ideal &, 369, 373, 407
fractional linear transformations &, 430, 519
free resolution &, see resolution (complexes),
free
free sheaf &, 369
Freyd-Mitchell Embedding Theorem &, 49, 49
Frobenius &, 505
Frobenius &, 215, 505
absolute, 215
full functor &, see functor, full
full rank quadratic form &, see maximal rank
(quadratic form)
full subcategory &, see subcategory, full
fully faithful functor &, see functor, fully
faithful
Fulton, W. &, 616
function field &, 189
function field &, 156, 171, 174, 189–191, 193,
269, 273, 275, 276, 308–310, 325, 353, 361,
375, 459, 460, 463, 511, 515, 554, 590, 593,
611
determines irreducible regular projective
curve, 461
functions &, see also differentiable functions,171= map to A1 , 195
multiplicity at regular point, 612
on a scheme, 102
on a ringed space, 77
on a scheme, 136
on projective space, 144value at a point, 138
value at a point, 102, 138
functor, see also derived functors
additive, 48, 48, 52, 55, 627–630, 638
category, 39, 51, 251
composition of, 28
contravariant, 29
covariant, 28
effaceable, 628
exact, 53
faithful, 29, 30, 39, 46, 49, 290
forgetful, 28
full, 29, 30, 39, 46, 49, 290
fully faithful, 29
identity, 28789

left-exact, see left exact
natural isomorphism, 30
natural transformation, 30, 30, 39, 44, 55, 79,
251, 252, 444, 628, 750
of invertible functions, 196of maps, 29of points, 29, 184, 185
representable, 38, 39, 195, 195, 251–253
right-exact, see right-exact
functors
right-exact
do not always commute with colimits, 56
fundamental groupoid &, 26
fundamental point (of rational map) &, 291
Fundamental Theorem of Elimination Theory&, 217, 221, 222, 223, 232, 295, 322, 347,
436, 469, 672
GAGA &, 586, 645
Galois cohomology &, 381
Galois theory &, 255
Galois theory &, 105, 107, 108, 114, 254, 255,
277, 312, 313, 342, 515, 527, 593, 646, 704
Gauss’s Lemma &, 163, 164
Gaussian integers Z[i] &, 163, 259, 275, 345,
353
general point &, 309
general fiber, see also generic fiber, 260
general flatness &, 664
general point &, 349
general point &, 123, 348, 613, 704, 705
general type (variety) &, 322, see also
trichotomy (classification of varieties),
585, 585, 586
generated by a finite number of global
sections, see global generation (ofquasicoherent sheaf) ,finite
generated by global sections, see global
generation
generated in degree 1, 148, 417, 451
generic smoothness &on target, 683
generic fiber, 258, see also general fiber, 260,
327
generic flatness &, 681
generic flatness &, 123, 664
improved version, 664
Generic Freeness Lemma &, 219, 219, 220, 664
generic point &, 103, 153
generic point &, 105, 110, 114, 115, 120, 122,
122, 123, 128, 133, 139, 142, 156, 166–168,
170, 171, 173, 189, 191, 213, 218, 219, 226,
255, 258, 260, 261, 303, 306, 308, 311, 316,
325, 353, 373, 375, 386, 391, 392, 400, 422,
423, 437
generic smoothness &, 123, 576, 681–683
of varieties, 576
on source, 682

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on source, 681, 681, 682
on target, 681
generically finite &, 260, 611
implies generally finite, 326, 664
generically separable morphisms &, see also
separable finite morphisms, 590
generization &, 122, 122, 153, 162, 218, 662
genus &, 488
arithmetic, 480, 491, 493, 522, 584, 593, 645,
646, 670, 720, 730, 731, 736= geometric, 584= topological, 480
constant in projective flat families, 670
geometric, 578, 584, 584, 731= topological, 593
is birational invariant, 583, 584
topological, 480
geometric fiber &, see fiber, geometricgeometric genus &, see genus, geometric
Geometric Nakayama’s Lemma &, see
Nakayama’s Lemma, geometric version
geometric Noether normalization &, see
Noether normalization, geometricgeometric point &, 264, 264, 297, 298, 519, 676,
677
geometrically connected &, 264, 264, 265, 267,
468, 585, 593, 710
geometrically integral &, 264, 264, 265, 268,
297, 485, 505, 508, 511, 512, 533
geometrically irreducible &, 264, 264–266, 298,
461, 485, 508, 554, 585, 588
geometrically reduced &, 264, 264, 265, 268,
269, 296, 298, 469, 710
Germain, S. &, 102
germs &, 74
compatible, see compatible germs
determine section of sheaf, 82
of differentiable functions, 72
global generation (of quasicoherent sheaf) &finiteevery finite type quasicoherent sheaf on
every affine scheme, 414
global generation (of quasicoherent sheaf) &,see also base-point-freeness, 413, 414, 414,
494
at a point, 414
every quasicoherent sheaf on every affine
scheme, 414
finite, 414
open condition, 415
preserved by ⊗, 415, 432
relative, 442, 456
for locally ringed spaces, 456
with respect to π, see globally generation,
relative
global section, 74
gluability axiom &, 75, see also base gluability
axiom, 131

gluing ringed spaces along open sets &, 141
gluing schemes &along closed subschemes, 433
along open subsets, 140, 433
Gödel-Bernays set theory &, 30
Godement, R. &, 635
Going-Down Theorem &, 206
for finite extensions of integrally closed
domains, 206, 311, 312, 312, 662
for flat morphisms, 312, 661, 662, 662, 663
for integral extensions of integrally closed
domains, 312
Going-Up Theorem &, 205, 304, 312, 662
graded ideal, see homogeneous ideal
graded modules
saturation, 412
graded rings &, 46, 146, 147, 186
Z-, 147= Z≥0 -graded ring, 147
assumptions on, 148finitely generated over A, 148
generated in degree 1, 148maps of, 186
over A, 148graph (of a) morphism &, 286, 288, 503, 550
is a locally closed embedding, 286, 504
to a separated scheme is a closed
embedding, 286, 504
graph of a morphism, 504
graph of a rational map &, 286, 292, 292, 613
Grassmannian &, 152, 199, 200, 200, 432, 442,
443, 445, 582, 583, 700, 701, 720–723
G(2, 4) = G(1, 3), 314, 445
G(2, 5), 532
as moduli space, 442
bundle, 445, 445, 583
functor, 442, 443
universal exact sequence, 444
universal vector bundle, 583
Grauert’s Theorem &, 709, 709, 711, 713, 715,
716, 718, 720
Grothendieck composition-of-functors
spectral sequence &, 630
Grothendieck composition-of-functors
spectral sequence &, 630, 630, 631,
633–635, 757
Grothendieck functor &, 196
Grothendieck group of coherent sheaves on X
(K(X)) &, 554
Grothendieck topology &, 380, 381, 646, 674
Grothendieck’s “six operations” &, 760
Grothendieck’s Coherence Theorem &, 295,
414, 422, 456, 467, 468, 498, 501, 501, 502,
554, 715, 732, 735, 744–746
for projective morphisms, 469
Grothendieck’s form of Zariski’s Main
Theorem &, 737

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November 18, 2017 draft

Grothendieck’s Generic Freeness Lemma &,see Generic Freeness Lemma
Grothendieck’s Theorem &, 486
Grothendieck, A. &, 11, 12, 14, 17, 23, 25, 47,
57, 65, 98, 99, 112, 122, 152, 186, 201, 208,
219, 251, 280, 362, 363, 454, 475, 486, 621,
635, 645, 720, 736, 753
calls Serre a bastard, 749
Grothendieck-Riemann-Roch Theorem &, 546
group (perverse definition) &, 26
group law on elliptic curve &, see elliptic
curves, group lawgroup object (in a category) &, 196, 196, 283
group variety/scheme &, 195, 196, 197,
197–199, 283, 296, 296, 298, 432, 445, 532,
536, 682, 683, 712
abelian, 197
action, 199, 432, 445, 682
transitive, 682
morphism of, 198
structure on elliptic curves, 535
groupification &, 45, 46
groupoid &, 26
Hairy Ball Theorem &, 579
Harris, J. &, 646
Hartogs’s Lemma &, see also Algebraic
Hartogs’s Lemma, 140, 319
Hausdorff &, 100, 120, 141, 145, 160, 279, 279,
280, 293
height &, see codimension
Hensel’s Lemma &, 727
Hermitian metric &, 438
higher pushforward, 635
higher direct image sheaf, see sheaf, higher
pushforward
Hilbert Basis Theorem &, 125, 126
Hilbert function &, 126, 488, 488, 493, 523, 670
of coherent sheaf, 488
of projective scheme, 488
Hilbert polynomial &, 467, 488, 489, 489–491,
493, 522, 544, 546, 645, 720
locally constant in flat families, 670
Hilbert scheme &, 646, 704, 720, 723
Hilbert Syzygy Theorem &, 126, 414, 489
Hilbert’s (Weak) Nullstellensatz &, see
Nullstellensatz
Hilbert’s fourteenth problem &, 541
Hilbert, D. &, 107, 126, 414
Hironaka’s Theorem &, 584, 611, 614, 743
Hironaka, H. &, 295, 671
Hirzebruch surface &, see also ruled surface
Hirzebruch surface &, 452, 453, 551, 609, 731,
742
Hirzebruch-Riemann-Roch Theorem &, 546
Hodge bundle &, 712, 713
Hodge diamond &, 586, 587, 743
Hodge Index Theorem &, 484, 552, 553

791

Hodge numbers &, 385, 586, 586, 587, 708
invariance of some, 586
not all are birational invariants, 587
of projective space, 586
Hodge theory &, 583, 586, 587, 708
homogeneous element &, 148
homogeneous ideal, 148, 163, 334, 412, 417,
418
homogeneous ideal &, 148
homogeneous space &, 682
homology &, 50
commutes with exact functors, see also
FHHF Theorem, 54
commutes with filtered colimits in ModA ,56
homomorphism (in additive category) &, 48
homotopic maps of complexes &, 623
give same map on homology, 623
Hopf algebra &, 199
Horrocks, G. &, 486
horseshoe construction &, 624, 633
Hurwitz’s Automorphisms Theorem &, 594
hypercohomology of double complex &, 58
hyperelliptic
curves, 505, 514, 514, 516, see also curves,
nonhyperelliptic, 517–521, 529, 531, 532,
579, see also Riemann-Hurwitz formula,
hyperelliptic, 580
differentials on, 579
first algebraic de Rham cohomology, 579
moduli spaces, 517
most not if g > 2, 517
involution, 532
map, 514, 517
unique for g ≥ 2, 517
hyperplane, 147, 230, 231, 310, 315, 316,
346–348, 404, 405, 429, 431, 432, 489, 491,
492, 509, 517, 522, 532, 754
universal, 683
hyperplane class, 403, 492
hypersurface, 107, 145, 147, 171, 230, 230, 231,
234, 241, 306, 310, 312, 314–316, 321, 322,
338–340, 343, 344, 346, 348, 363, 405, 432,
445, 469, 490–492, 494, 532, 543, 585, 689,
692, 699, 700, 720, 722, 723, 760, 766
degree, 230
discriminant, 699
has no embedded points, 171
quadric, see also quadric surface, 230
I-adic completion &, 725
I-adic filtration &, 364
I-adically separated &, 726
I-depth &, see depth
I-filtration of a module &, 364
I-stable filtration &, 364
ideal
graded, see homogeneous ideal

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ideal sheaves &, 380
ideal &, see also fractional ideal, see also
different ideal, see also discriminant ideal
extension of, 254
graded, see homogeneous ideal
homogeneous, see homogeneous ideal
irrelevant, 148, 149, 150, 232, 235
minimal prime, see minimal prime idealof denominators, 162, 163, 320, 320
of nilpotents, 680
principal, see principal ideal
product of two, 116
symbolic power, 328
ideal sheaves &, 226
product of, 652
product of, 409, 569, 604
idempotent &, 119, 267, 695, 734
identity axiom &, 75, see also base identity
axiom, 131
identity functor &, 28
identity morphism idA &, 26
image (of morphism in a category), 49
image (of morphism of schemes)of proper is proper, 295
scheme-theoretic, see scheme-theoretic,
image
set-theoretic, 217
image presheaf &, 81
immersion(differential geometry), 673
of manifolds, 101
of schemes (unwise terminology), 229
incidence correspondence/variety &, 314, 315,
346, 699, 700
inclusion-exclusion via cohomology ofsheaves &, 640
index category &, see category, index
index of symmetric bilinear form &, 553
indicator sheaf &, 85
induced reduced subscheme &, see reduced
subscheme structure
inductive limits &, see colimits
infinite disjoint union of schemes &, see also
disjoint union of schemes, 132, 137, 140,
156, 237, 240, 380, 728
infinitesimal neighborhood &, see formalneighborhood
initial object &, see object, initial
injective &, see also resolution (complexes),
injective
objects
closed under direct product, 635
in abelian category, 626
preserved by right adjoints, 638
injective limits &, see colimits
injective morphisms of schemes notpreserved by base change &, see also

universally injective (radicial)morphisms, 263
integral closure &, 207, 273, 274, 276, 312, 320,
735
finiteness of, 276, 277, 460, 462
of Noetherian need not be Noetherian, 276
integral element of B-algebra &, 203
integral extensions of rings &, 203, 312
does not imply finite, 204preserved by localization of source, 203
integral homomorphisms of rings &, 203closed under composition, 204preserved by localization of source,
quotient of source and target, 203
integral morphisms &, 213affine-local, 213
are affine, 213
are closed, 213closed under composition, 213
fibers of, 214, 304
need not be finite, 213
preserved by base change, 214, 262
integral scheme &, 156= irreducible and reduced, 156
almost stalk-local property, 157, 159
integrally closed &, see also integral closure,162
intersection, see scheme-theoretic, intersection
intersection number/product of L1 , . . . , Lnwith F &, 546
intersection number/product of L1 , . . . , Lnwith F &, 543, 543–547, 553–555
deformation invariance, 543
is symmetric multilinear, 544
locally constant in flat families, 671
on a surface, 548
intersection theory &, 431, 616, 645, 697
on a surface, 547
inverse image
scheme, see scheme-theoretic, pullback
scheme-theoretic, see scheme-theoretic,
inverse image
sheaf, 92, 92, 93, 422, 425
construction, 93
exact, 94
left adjoint to pushforward, 92
right adjoint to extension by zero, 95
inverse limit &, see limit
invertible sheaf &, see line bundles
irreducible &, 119, 119, 153
component, see component, irreducible
criterion to be, 325
irregularity (surface) &, 587
irrelevant ideal &, see ideal, irrelevant
Ischebek, F. &, 688
isomorphism &in a category, 26
of ringed spaces, 136

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November 18, 2017 draftof schemes, 136, 181
j-invariant &, 528
j-invariant &, 516, 523, 526, 527, 528, 530
and cross-ratio, 526
Jacobiancriterion, 335, 338–340, 342–344, 347, 357,
515, 564, 571, 576, 578
for projective hypersurfaces, 343
for smoothness, 678
description of Ω, 565, 574
map, 335, 344
matrix, 336, 336, 338, 339, 347, 348, 355, 357,
564, 565, 590, 678
Jacobson radical &, 206
joke (bad) &, 12
joke (bad) &, 39, 41, 583
K2 &, 585
K3 surface &, 584, 585, 585
Kähler differentials, see differentials, 573
universal property, 563, 571, 573
Kähler, E. &, 585
Kedlaya, K. &, 11
kernels &, 48, see also universal property, ofkernelare limits, 54
commute with limits, 55
presheaf, 80
Kim, M. &, 539
Kisin, M. &, 16
Kleiman’s numerical criterion for ampleness&, 485, 552, see also Nakai-Moishezon
criterion for ampleness, 556, 558, 558
Kleiman’s Theorem &, 558, 558
Kleiman, S. &, 682
Kleiman-Bertini Theorem &, 346, 682, 683,
683, see also Bertini’s Theorem
poor man’s, 683
Klein quartic curve &, 521
Klein, F. &, 303
Knutson, A. &, 13
Kodaira dimension κ(X) &, 585
is birational invariant, 585
Kodaira Embedding Theorem &, 438
Kodaira vanishing &, 467, 586, 586
Kodaira’s criterion for ampleness &, 438
Kodaira, K. &, 438, 585
Koszul complex &, 243, 581, 685
Krull dimension &, see dimension
Krull Intersection Theorem &, 365, 665, 666,
726
Krull’s Height Theorem &, 244, 307, 318, 318,
329, 337, 345
for irreducible varieties, 318
Krull’s Principal Ideal Theorem &for tangent spaces, 333

793

Krull’s Principal Ideal Theorem &, 20, 244,
309, 315, 316, 316–318, 321, 328, 329, 332,
337, 338, 345, 349, 357, 482, 490, 685, 690
for tangent spaces, 333
Krull, W. &, 315, 337
Kummer, E. &, 585
Künneth formula for quasicoherent sheaves&, 475, 477
Lasker, E. &, 174
Lefschetz principle &, 586, 708
left derived functors &, see derived functors,
left
left-adjoints, 43
are right-exact, 54
left-exact, 52, 53
T 0 , 628
Hom, 53
Hom , 92
and derived functors, 465
and FHHF Theorem, 54
and right derived functors, 626
and universal δ-functors, 628
e.g., right adjoint, 55
limits, 54
not ⊗, 34
not pullback, 427
pushforward, 91
right adjoints, 54
sections over U, 91
Leibniz rule &, 332, 562, 567, 570, 573, 574
length &, 482, 482
finite, 482, 483, 665, 666
of dimension 0 scheme, 548
Leray spectral sequence &, 466, 499, 501, 502,
621, 630, 635, 636
Leray, J. &, 57, 71, 642
Lie groups &, 297, 299, 697, 698
limits &, 40, see also universal property, oflimit
are left-exact, 54, 744
commute with kernels, 55
commute with limits, 54, 55
commute with right adjoints, 54, 55
e.g., kernels, 54
projective, 40
line, 14–16, see also affine line, 102–105, 111,
114, 135, 141–143, 145–147, 160, 179, 191,
226, 230, 231, 231–234, 247, 248, 259, 272,
274, 279, 284, 291, 310, 314, 315, 321,
332–334, 341, 344, 348, 349, 359, 360, 403,
406, 408, 433, 445, 449, 482, 491, 492,
512–514, 522, 523, 526–531, 533–538, 540,
549, see also projective line, 597, 598, 607,
609, 611–615, 650, 658, 690, 699–701, 703,
see also Fano variety of lines, 704–706
on cone, 333
tangent, see tangent, line

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with doubled origin, 132, 141, see also affine
space, with doubled origin, 142, 160, 279,
281, 284, 291, 359, 360
line bundles, see also ampleness, 510, 711
semiample, 441
line bundles, 17, 151, 185, 320, 369, 370, 371,
371–374, 380, 384, 386, 397–406, 408–410,
413–417, 426, 429–433, 436, see also very
ampleness, 437–439, 441, 442, 444,
449–455, 457, 458, 463, 474, 475, 479–481,
see also Picard group, 482, 483, see also
Q-line bundle, 484–486, 490–492, 494, 495,
501, 508–513, 517–519, 521–525, 528,
530–532, 535, 538, 539, 541, 542, 544–547,
549–556, 558, 559, 564, 569, 579, 582, 584,
586, 590, 592, 594, 604, 605, 607, 612, 613,
615, 645, 670–672, 683, 708, 711, 712, 723,
740, 749, 750, 752, 765, 766
additive notation, 547
algebraically equivalent, see algebraic
equivalence
and maps to projective schemes, 429
base-point-free, see base-point-freenessdegree, 543
additive in exact sequences, 592
degree 0, 508
degree 2g − 2, 508
globally generated, see global generation (ofquasicoherent sheaf)
nef, 483, see also nef cone, 484, 558
cone, 485, 550
negative degree, 508
numerical equivalence, 484
numerical property, 484
numerically effective, see line bundles, nef
numerically trivial, 483, 483, 484, 552, 553,
671
on projective schemes, 412
on projective spacecohomology, 475
total space, see total space, of line bundle
twisting by, 413
very ample, see very ample
line sheaf &, 371
linear form &, 165, 522, 532, 534, 568
linear independence of characters &, 254, 277
linear series &, 416, 550, 552, 612, 683, 684, 699(non)degenerate, 431, 431
anticanonical, 706
canonical, 517
complete, 416, 430, 432, 507, 517, 531, 535,
740, 741
always nondegenerate, 431
linear space &, 231, 231, 234, 235, 316, 532, 602,
760
Lipman, J. &, 611
Littlewood, J.E. &, 113
local &, see also affine-local, see also stalk-local

on the source, 202
on the target, 201
local finite presentation, see also finite
presentation, 214morphisms, 207, 215, 216, 355, 588, 589, 662,
665, 669, 674–677, 739”limit-preserving”, 216
affine-local on target, 216
and semicontinuity, 327
Chevalley’s Theorem, 261
closed under composition, 216
preserved by base change, 262
local freeness of finitely presented sheaf isstalk-local, 391
local isomorphism, 673
local rings &, 18
morphisms of, 180, 661
localization &, see also universal property, of
localization, 31, see also universal
property, of localization, 362
commutes with direct sums, 33
is exact, 53–55, 209, 376, 395, 590, 647
is flat, 647
need not commute with Hom, 33, 53
need not commute with infinite products,
33
of schemes are monomorphisms, 255
preserved by base change, 255
locally closed embeddings &finite intersections of, 254
locally closed embeddings &, 348
locally closed embeddings &, 202, 225, 228,
228–230, 239, 240, 244, 254, 255, 270, 279,
280, 287, 288, 290, 360, 572, 588, 683
are locally of finite type, 229
are monomorphisms, 255
are separated, 281
closed under composition, 230
finite intersections of, 229, 254
preserved by base change, 254
locally closed immersions &, see locally closed
embeddings
locally closed subschemes &, see also locallyclosed embeddings, 228
locally closed subset &, 218
locally constructible subsets &, see also
constructible subsets, 262
locally factorial &, 163
locally finite presentation
morphism, 675
locally finite typeA-scheme, 159
morphisms, 214
are affine-local, 214closed under composition, 214
preserved by base change, 262
to locally Noetherian schemes, 215
over a ring, 159

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November 18, 2017 draft

locally free resolution &, see resolution(complexes), locally free
locally free sheaf &, see also vector bundle,
369–371, 374= finitely presented sheaf with free stalks,391
associated to vector bundle, 449
rational/regular section, 373
locally Noetherian scheme, see scheme, locally
Noetherian
locally of finite presentation, see local finite
presentation
locally of finite type &, see locally finite type
locally principal closed subscheme, see closed
subscheme, locally principal
locally ringed spaces &, 102, 138, 138, 179–183,
196, 335, 371, 434
affine schemes as examples, 180morphisms of, 180, 456
locus of indeterminacy &, 291
locus where two morphisms agree &, 290, 291,
298
long exact sequence, 52, 57, 59, 60, 64, 465,
466, 468, 471–473, 477, 478, 480, 486, 487,
497, 498, 509, 555, 557, 566, 623–627, 636,
637, 642, 654, 660, 687, 744, 747, 756, 757
for Ext, 626, 686–688, 754
for Ext-functors of O-modules, 754
for Tor, 622, 623, 653
for higher pushforward sheaves, 498, 722,
723
in cohomology, 752
lower semicontinuity &, see semicontinuity
Lüroth’s Theorem &, 527, 593, 593
Lutz-Nagell Theorem &, 540
Lying Over Theorem &, 205, 205, 206, 213, 274,
309, 312, 325, 360
MacPherson, R. &, 616
magic diagram &, 37, 38, 283, 284, 286, 287
manifolds, 25, 75, 99–102, 138, 180, 186, 195,
199, 200, 279, 303, 369, 373, 404, 541, 562,
568
complex, 101, 111, 140, 186, 187, 193, 293,
388, 539, 576, 598
differentiable, 13, 25, 71, 78, 99–101, 177,
178, 190, 332, 335, 354, 374, 572, 673
immersion, see immersion, of manifolds
morphisms of, 100smooth, 73, 99, 190smooth morphisms of, 201
submersion, see submersion
vector bundle on, 369
map (in a category) &, 26
mapping cone &, 59, 64, 64, 473, 714, 715
Max Noether’s AF + BG Theorem &, 690
maximal rank (quadratic form) &, 165795

maximal rank (quadratic form) &, 234, 272,
343
maximum principle &, 295
McLarty, C. &, 9, 98, 99
minimal models of surfaces &, 743
minimal prime ideal &, 128, 128, 174, 268, 305,
316, 320, 328, 329, 341, 694= irreducible component of Spec A, 128
Noetherian ring has finite number, 128
Miracle Flatness Theorem &, 685, 690, 691,
701, 760
algebraic version, 690
Möbius strip &, 369
modular curve &, 526
modular form &, 526
modules &finitely generated
over Discrete Valuation Rings, 375
graded, 417
modules &Artinian, 327, 328
category ModA has enough injectives, 626,
627
classification of finitely generated over
DVRs, 580
Cohen-Macaulay, see also
Cohen-Macaulayness, 688
coherent, see also coherent sheaves, 387,
393–395
Homs are coherent, 395
form an abelian category, 395
over non-Noetherian rings, 393
composition series, 172, see module,
composition series, 482, 483
depth, see depth
finitely generated, see also finite type
quasicoherent sheaves, 387
over a Principal Ideal Domain, 18, 391
over Discrete Valuation Rings, 354, 391,
399, 464, 525, 580, 648
finitely presented, 53, 387, 387, 388, 393–395,
563, 655, 656= coherent for Noetherian rings, 395“finitely presented implies always
finitely presented”, 387, 388, 656
and commutation of localization and
Hom, 33
graded, 412
flatness, see flatness
free implies projective implies flat, 647
graded, 46, 380, 411, 414, 417, 418, 745, 746
quasicoherent sheaf corresponding to,411, 417
saturated, see saturated S• -module
Noetherian, see Noetherian, moduleof Kähler differentials, see Kähler
differentials
projective, 625

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796

The Rising Sea: Foundations of Algebraic Geometrysimple, 482
topological, 726, 744
moduli space &, 200, see also parameter space,
360, 361, 429, 517, 520, 522, 523, 526, 532,
646, 699, 701, 704, 712, 720of curves, 523
monoidal transformation &, see blow-up
monomorphisms &, 48
monomorphisms &, 37, 38, 47–50, 85, 86, 91,
255, 258, 270, 281, 628, 629
are separated, 281
closed under composition, 37, 255diagonal characterization, 38
Mordell’s conjecture (Faltings’ Theorem) &,518
Mordell-Weil Theorem, group, and rank &,525, 539
morphisms, 26, 177
and tangent spaces, 335
criterion to be a closed embedding, 505, 506
diagonal, see diagonal morphism
dominant, 189
from Spec of local ring, 184
manifolds, 100of (pre)sheaves, 78, 79
of graded rings, 186of group schemes, 198of local rings, 180of locally ringed spaces, 180
glue, 180of ringed spaces, 178
glue, 178of schemes, 14, 179, 181
glue, 180, 182, 249, 251
of sheaves
determined by stalks, 83of topological spaces
glue, 76
quasiaffine, see quasiaffine morphisms
multiplicative group, 196
multiplicative subset &, 31
multiplicity of function at regular point &, 612
multiplicity of zero or pole &, see order ofpole/zero
Mumford, D. &, 11, 112, 558, 645, 713
Nagata’s Compactification Theorem &, 738
Nagata’s Lemma &, 163, 407, 407, 408
Nagata, M. &, 307, 310
Nagell-Lutz Theorem &, 540
Nakai-Moishezon criterion for ampleness &,510, 543, 547, see also Kleiman’s numericalcriterion for ampleness, 556, 556, 558, 559
Nakayama’s Lemma &, 203, 206, 207, 207, 244,
329, 337, 341, 350, 351, 358, 365, 390–392,
415, 507, 580, 589, 615, 655, 656, 679, 688,
710, 711
geometric version, 390, 392

natural isomorphism of functors &, 30
natural transformation of functors &, seefunctor, natural transformationnef, see line bundles, nef, 558
nef cone, 483, 484, 485, 550, 552, 558
Néron-Severi group NS(X) &, 483, 484, 671
Néron-Severi Theorem &, see Theorem of theBase (Néron-Severi Theorem)
Nietzsche, F. &, 23
nilpotent &, 13, 112, 113, 113, 117, 119, 133,
144, 155, 166, 174, 186, 240, 304, 352, 380,
390, 514, 680, 692
nilradical &, 113, 116, 121, 155, 190, 206, 240,
380
is intersection of all primes, 113
nodal cubic, 274, 606, 609, 646, 656, 658–660
normalization of, 274, 658
node, 274, 343, 612, 725, 729, 729, 730
normalization, 506
Noether normalization &, 459
geometric, 309
Noether normalization &, 277, 307, 309,
309–312, 324, 325
Noether, E. &, 174, 276
Noether, M. &, see also Max Noether’s
AF + BG Theorem, 690
Noether-Lefschetz Theorem &, 315
Noetherian, see also scheme, locally
Noetherian
conditions, 497
hypotheses, 14, 17, 127, 152, 168, 173, 174,
220, 221, 243–245, 315, 350, 422, 439, 458,
467, 469, 494, 496, 589, 657, 658, 662, 665,
668, 669, 675, 709–711, 715, 718, 719, 722,
735, 738–740
induction, 124, 220
local ring, 243, 307, 311, 316, 318, 337, 349,
350, see also discrete valuation rings, 351,
352, 363, 365, 483, 589, 650, 665–667,
685–689, 691, 693, 726, 734
has finite dimension, 307, 318
module, 126, 126
rings, 120, 124, 125, 125–128, 148, 158, 159,
163, 167, 171–174, 219, see also Noetherian,
local ring, 243, 244, 261, 275–277, 304,
316–318, 320, 328, 337, 353, 364, 365, 387,
388, 393, 483, 494, 685, 692, 716, 726, 727
finite number of minimal primes, 128
finitely generated module over, 172
important facts about, 125, 126
infinite-dimensional, 304, 307, 318
scheme, 158, 159, 162, 163, 170, see scheme,
Noetherian, 288
fibered product need not be Noetherian,248
finite number of irreducible components,
159
finitely many associated points, 169

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November 18, 2017 draftwith ring of sections not Noetherian, 542
topological space, 124, 124–126, 153, 159,
217, 218, 474
finitely many irreducible components,
124
Noetherian topological space, 153
non-affine scheme (example) &, 137, 140,
142–144
non-archimedean &, 351, 381, 388, 393
non-zerodivisor &, 32, 241–244, 409, 601, 605,
607, 608, 612, 647, 650, 667, 686, 687,
689–691, 694
nondegenerate linear series &, see linear
series, (non)degenerate
nonprojective proper variety &, see proper
nonprojective variety (examples)
nonquasiseparated scheme (example) &, 154
nonreduced scheme (example) & , 155
nonsingular &, see regularnormal form for quadrics, see diagonalizing
quadratic forms
normal (scheme) &, 140, see also
normalization, 161–165, 273, 275–277,
320, 334, 352, 353, 363, 373, 400–406, 408,
515, see also Serre’s criterion for
normality, 685, 692, 694, 735, 739
implied by factorial, 164
may not be UFD, 165
need not be integral, 162
normal bundle, 550, 551, 569, 573, 582,
605–607, 610, 740
projectivized, 604, 607
to effective Cartier divisor, 569
to exceptional divisors, 605
to regular embedding, 569
normal cone, 606, 616
projective completion, 616
normal exact sequence for smooth varieties &,577
normal sheaf to subscheme &, see also normal
bundlenormal sheaf to subscheme &, 569, 766
to regular embedding is locally free, 569
normalization &, 211, see also universal
property, of normalization, 273, 273–275,
277, 305, 353, 360, 447, 448, 450, 451, 461,
462, 495, 515, 520, 584, 588, 656, 658, 659,
731, 735, 741, 754
in field extension, 276
in field extension, 275, 276, 277, 345, 459
is birational morphism, 274, 277
is integral and surjective, 274
not flat, 656
of cusp, 505, 588, 663
of node, 462, 506
Nullstellensatz &, 107, 107, 108, 114, 115, 121,
128, 154, 160, 217, 218, 221, 248, 263, 308,
342, 524

797

from dimension theory, 308
number field &, see also ring, of integers in
number field, 276
numerical criterion &for ampleness, see Kleiman’s numericalcriterion for amplenessfor line bundle on curve to bebase-point-free, 509, 510
for line bundle on curve to be very ample,
510
numerical equivalence, 483, 484, 545, 556
implied by algebraic equivalence, 671
numerically effective, see line bundles, nef
numerically trivial line bundles Picτ X, seeline bundle, numerically trivial, 484
object &initial, 40
object &acyclic, 754, 757
final, 31, 31, 36, 75, 76, 94, 182, 196, 208, 255,
280, 416, 442, 600
in category of schemes, 182
in a category, 26
initial, 31, 31, 47
zero, 31, 31, 47, 48, 198
Oka’s Theorem &, 388, 393
open set
distinguished, 118, 377
open cover &, 72
open embeddings &, 202, 739
are flat, 649
are locally finitely presented, 216
are locally of finite type, 214
are monomorphisms, 255
are separated, 281, 286
base change by, 253
of ringed spaces, 178
preserved by base change, 262
open immersion &, see open embeddings
open set
distinguished, 109, 114, 118, 119, 130, 131,
137, 139, 140, 143, 151, 157, 158, 166, 169,
181, 188, 210, 237, 255, 266, 276, 291, 323,
375, 377, 379, 380, 383, 405, 411, 424, 473,
497, 540
form base for Zariski topology, 118
projective distinguished, 150, 150
open subfunctors, 252
open subscheme, 137, 202
opposite category &, see category, opposite
orbifolds &, 14, see also stacks, 523
order of pole/zero &, 353
order of pole/zero &, 352
ordinary double point &, 730
ordinary multiple (m-fold) point &, 730
orientation of spectral sequence &rightward, 58

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The Rising Sea: Foundations of Algebraic Geometry

upward, 61
page of spectral sequence &, 58
paleolithic era &, 349
Pappus of Alexandria &, 512
Pappus’s Theorem &, 512, 513, 514
parabola &, 106, 109–111, 123, 135, 258, 259,
264, 464, 492, 673
parameter space &, see also moduli space, 314,
315, 321, 482, 699, 712, 721, 723
partial flag variety &, see flag variety
partially ordered set &, 27, 27, 28, 40, 42, 123,
319
partition of unity &, 130, 161, 203, 472
Pascal’s “Mystical Hexagon” Theorem &,512–514, 697
Pascal, B. &, 512
perfect pairing of vector bundles &, 385
Pfaffian &, 532
Picard group &, 372, 373, 374, 397, 398, 401,
403–406, 409, 480, see also line bundles,
482, 484, 493, 525, 535, 538, 539, 541,
547–552, 555, 614, 671, 712
of Spec of UFD is 0, 404
of cone, 406
torsion example, 405
Picard number ρ(X) &, 484, 587, 743
finiteness of, see also Theorem of the Base(Néron-Severi Theorem), 484
complex case, 484
PID &, see principal ideal domain
pinched plane &, 353, 354, 413, 692
normalization of, 275, 354
not Cohen-Macaulay, 693
plane, 231
pinched, see pinched planewith doubled origin, see affine space, withdoubled origin
plane curves, see curves, planeBézout’s Theorem, 491
Plücker &embedding, 200, 444, 445, 532
equations, 445
plurigenus &, 585
is birational invariant, 585
Poincaré, H. &, 136
point &
closed, see closed point
field-valued/ring-valued/scheme-valued,
184
general, see general point
generic, see generic point
geometric, see geometric point
pole &of order nfor Noetherian scheme, 353
pole &, 162, 320, 352, 352, 353, 353, 373, 398,
400–403, 405, 463, 479, 508, 511, 540, 579

of order nfor DVR, 352
Poncelet’s Porism &, 536, 536, 538
Poonen, B. &, 11, 520
poset &, see partially ordered set
power series &, 610, 725, 726, 740
formal, 725
preimage, see also scheme-theoretic preimage,257
presheaf &section of, 73
presheaf &, 73
cokernel, 81
espace étalé, see sheaf, space of sections
germ, see germ, 74
image, 81
kernel, 80
not necessarily sheaf, 75
on a base, 87
stalk, 87
quotient, 81
separated, 75
stalk, 74
prevariety (complex analytic) &, 160
primary &decomposition, 167, 169, 174
ideal, 174
prime avoidance &, 312, 313, 317, 318, 650
principal divisor, see Weil divisor, principal
principal fractional ideal &, 374
principal ideal &, see also Krull’s Principal
Ideal Theorem, 105, 140, 241, 333, 334, 374
principal ideal domain &, 18, 105, 345, 351,
354, 375, 399, 484, 525, 647, 655
finitely generated modules over, 18, 391
Principal Ideal Theorem &, see Krull’sPrincipal Ideal Theorem
principality not affine-local, 241, 540
product &, see also fibered product, 23, 24, seealso universal property, of product, 40
of irreducible varieties over k is irreducible,263
of quasicoherent ideal sheaves, see ideal
sheaves, product ofof schemes, 247
isn’t products of sets, 247
projection formula &, 428, 490, 499, 546, 555
more general, 547
projective
module= free for finitely generated over local
rings, 655
is flat, 625
moduleu= direct summand of free module, 625
object in an abelian category, 625
projective module &, see modules, projective
projective S-scheme, 435

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November 18, 2017 draft

projective Y-scheme, 454
projective bundle, 347
projective bundle &, 186, 314, 347, 452, 482,
551, 582, 605–607, 672, 700, 711, 712, 721,
742, 743
projective completion, 236, 616, 660
of normal cone, 616
projective cone, 146, 152, 235, 236, 236
relative version, 453
projective coordinates &, see coordinates,
projective
projective limit &, see limit
projective line, 142, see also line, 578, 598, 613
projective morphisms &, 435, 447, 453, 454,
454, 455, 460, 462, 469, 498, 500, 502, 503,
506, 507, 657, 669, 670, 672, 701, 735, 742,
760
“sort of” local on the base, 453
are proper, 455
not a “reasonable” class, 456
notion not local on target, 672
preserved by base change, 453, 455
usually closed under composition, 453, 455,
456, 460, 503, 603, 672
usually preserved by product, 456
projective over Y, 454
projective resolution &, see resolution(complexes), projective
projective schemes, see scheme, projective
are proper, 295
line bundles on, 412
projective space &, 143, 151, 397
automorphisms, 430, 519
cohomology of line bundles on, 475
coordinates, see coordinates, projectivefunctions on, 144
functorial definition, 429
is finite type, 214
is separated, 281
Serre duality for, 751, 753
tautological bundle, 450
projective variety &, see variety, projective
projectivization &of finite type sheaf, 452
of finite type quasicoherent sheaf, see also
normal bundle, projectivized, see also
tangent cone, projectivizedof vector space, 152proper nonprojective variety (examples), 295,
433, 550, 671
proper transform (of blow-up) &, 549, 598,
600, 600, 602, 609, 610, 615, 671, 706, 742,
743
of nodal curve, 609
properness &, 120, 146, 279, 293, 293, 455
closed under composition, 294
closed under product, 294
local on target, 294

799

of projective schemes, 295
preserved by base change, 294, 298
property Rk , see regular, in codimension kproperty R0 means generically reduced, 692
property Sk , 692, 692
S1 means no embedded points, 692
pseudo-morphisms &, 189
Puiseux series &, 351
pullback, see also base change
construction via affines, 423
in cohomology, 466
is not left-exact, 427
is right-exact, 427
of ideal sheaves, 428= pullback as sheaf for flat morphisms,428
of quasicoherent sheaves, 421, 422
explicit construction, 425
universal property, 424
various properties, 426–427
of schemes, 256
scheme-theoretic, see scheme-theoretic,pullback
universal property, 424–426
pullback diagram/square, 36, see Cartesian
diagram/squarepure dimension, see equidimensional
purely inseparable &, see field extension,
purely inseparable
push-pull formula &, 427
push-pull formula &, 499
pushforward
occasional right-adjoint π! , 764
occasional right-adjoint π! , 759, 760
occasional right-adjoint of, 759
of O-modules, 178of coherents
need not be coherent, 422
of quasicoherent sheaves, 421
of quasicoherents
is exact for affine morphisms, 422
is usually quasicoherent, 422
right adjoint to inverse image, 92
sheaf, see sheaf, pushforward
Pythagorean triples &, 191, 192, 194, 511
qcqs (quasicompact and quasiseparated) &,17, 154, 207, 207–210, 216, 362, 379, 380,
422, 426–428, 439, 456, 458, 459, 466, 738,
740
Qcqs Lemma &, 207, 209, 209, 210, 379
quadratic trasformation &, see blow-up
quadric hypersurface, 147, 234, 445, 522
quadric surface, 132, 145, 163, 164, 193, 234,
272, 309, 334, 341, 354, 408, 521, 522, 537,
613, 614
cone over, see cone over quadric surface
quartic, 266, 519–522, 530, 584, 607, 697

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The Rising Sea: Foundations of Algebraic Geometry

hypersurface, 230
Klein, 521
surface, 315, 584, 607
most contain no lines, 315, 607
quasiaffine &morphisms, 439, 458, 459
properties of, 459
schemes, 458= OX ample, 459
quasicoherent sheaves &, 14, see also finite
type quasicoherent sheaf, 131, 369, 374,
375, 375, 377, see also finitely presented
quasicoherent sheaves
characterization by distinguishedinclusions, 379
corresponding to graded modules, 411, 417
corresponding to modules, 375
form an abelian category, 381
globally generated, see global generationKünneth formula, 475, 477
local nature of, 411
morphisms of, 381
not all locally free, 375
not necessarily preserved by Hom , 389
on affines have no derived functor
cohomology, 640
on projective schemes, 411
on ringed spaces, 382
pullback of, 425
pushforwards usually quasicoherent, 380
restriction of, 423
quasicompact &schemes, 153
quasicompact &morphisms, 207affine-local on the target, 208, 285
closed under composition, 207
preserved by base change, 262
schemeshave closed points, 153
topological space, 120
quasifinite morphisms &, 212, 215, 739
from proper to separated is finite, 296
how to picture, 215, 739
most are open subsets of finite morphisms,739
need not be finite, 215
preserved by base change, 262
to fields are finite, 215, 219, 260, 263, 739
quasiinverse &, 31
quasiisomorphism &, 471, 713, 714
quasiprojective
morphisms, 454
scheme, see scheme, quasiprojective
variety, see variety, quasiprojective
quasiprojective implies finite type, 159
quasiseparatedness, 17, 154, 155, 159, 201,
207–210, 216, 284, 284–286, 288, 289, 361,

362, 379, 380, 422, 426–428, 439–442, 456,
458, 459, 466, 642, 738, 740
affine-local on target, 208
all locally Noetherian schemes, 159
closed under composition, 207, 216, 285
closed under product, 286
e.g., affine schemes, 154e.g., locally Noetherian schemes, 159, 208
e.g., morphisms from locally Noetherian
schemes, 208
local on the target, 285
of topological space, 154preserved by base change, 262, 284
Quillen, D. &, 404
Quillen-Suslin Theorem &, 404
quotient &, 49
object, 48
presheaf, 81
sheaf, see sheaf, quotient
radical ideal &, 116, 128, 133, 160, 239
radical of ideal &, 116= intersection of primes containing ideal,
116
commutes with finite intersections, 116
radicial morphisms &, see universally
injective morphisms
ramification
points, 525
ramification &divisor, 590, 593, 594
and number of preimages, 592
locus, 588
order, 592
point, 514, 591, 591
tame, 592
wild, 592
rank &of coherent sheaf, 392
additive in exact sequences, 480
of finite type quasicoherent sheaf at point,392
of quadratic form, 165, see also maximal
rank (quadratic form)of vector bundle on manifold, 370
rational function &, 463
rational function &, 77, 102–104, 140, 156, 162,
171, 171, 173, 188, 189, 191, 194, 195, 291,
320, 353, 400–402, 479, 482, 533, 540
rational map &, 188
rational map &, 188, see also dominant, 189,
see also birational, (rational) map,
190–193, 195, 286, 289, 291, 292, see also
degree, of rational map, see also graph ofa rational map, 309, 430, 437, 461, 463,
530, 539, 583, 597, 612–614, 664, 704
rational normal curve &, 120, 233, 431, 494,
517, see also twisted cubic curve

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November 18, 2017 draft

not complete intersection for d > 2, 494
rational section of locally free sheaf &, 373
rational variety &, 190, 607
Raynaud, M. &, 586“reasonable” class of morphisms &, 201
reduced subscheme structure &, 240, 240, 616
Reduced-to-Separated Theorem &, 188, 289,
291, 291, 292, 358, 359, 435, 705
reducedness &, 155
is stalk-local, 155
not open condition in general, 155of ring, 113
often open condition, 240, 390
reducible &, 153
reducible &, see irreducible
reduction (of scheme) &, 240
Rees algebra &, 364, 604, 745
reflexive sheaf &, 389, 569
regular &does not imply smooth, 340
e.g., Ank , 345in codimension 1 (property R1 ), 353, 400
in codimension k (property Rk ), 692
local rings, 337
are Cohen-Macaulay, 689
are Cohen-Macaulay, 364, 685, 693
are integral domains, 340
are integrally closed, 363, 685, 693
are unique factorization domains, see
Auslander-Buchsbaum Theorem
preserved by localization, 338, 362, 363,
564, 576, 578, 692, 693
of An
k , 346
ring, 337
scheme, 337
regular embeddings &, 240, 244, 244, 245, 317,
335, 341, 569, 570, 577, 604, 606, 607, 651,
685, 689, 692, 750, 766
open condition, 244
regular function &, see function, 171
regular section of locally free sheaf &, 373
regular sequence &, 569
regular sequence &, 240, 241, 242, 242–245,
569, 606, 608, 685–687, 689
cohomological criterion for existence, 686
maximal, 686
order matters, 242
relative Proj &universal property, 453
relative Proj &, 447, 450, 451
O(1), 451
and affine base change, 450
commutes with base change, 452
finite generation in degree 1, 451
universal property, 450
relative Spec &, 447, 447
and base change, 449
universal property, 447, 448

801

relative i-forms, see sheaf of (relative) i-forms
relative conormal exact sequence, 566
relative cotangent exact sequence, 566, 574,
677
affine version, 565, 566
left-exact in good circumstances, 575, 590
left-exactness, 680
relative cotangent sheaf, 572
relative different and discriminant ideals &,589
relative dimension, 326, 354, 355
relative tangent sheaf, 572
relatively ample, 437, see ampleness, in
relative setting
relatively base-point-free, seebase-point-freeness, in relative setting
relatively globally generated &, see global
generation (of quasicoherent sheaf),
relative
relatively very ample &, see very ample,
relatively
Remainder Theorem &, 608
representable functor &, 195, see functor,
representable
residue field &, 138
Residue Theorem &, 463, 753
resolution (complexes) &, see also
Cartan-Eilenberg resolution
acyclic, 629
flat, 629, 668
free, 622, 629, 652, 667, 756
injective, 629, 631–634, 638, 640, 686, 755,
756, 758, 764
locally free, 755, 756, 765
projective, 625, 626, 630
resolution (singularities) &, see also
Hironaka’s Theorem, 597, 598, 609, 610,
see also Alteration Theorem, 611, see also
Weak Factorization Theorem, 730, 743
for surfaces, 743
for curves, 735
for surfaces, 736
of indeterminacy of rational maps, 597
small, 591, 614, 614
non-isomorphic, 614
restriction map, 74
restriction of a quasicoherent sheaf &, 423
seealsoquasicoherent sheaves, restriction of,423
restriction of scalars, 46
restriction of sheaf to open subset G |U , 76
resultant &, 222
Riemann surface &, 101, 462, 518, 593, 753
genus, 480
Riemann-Hurwitz formula &hyperelliptic, 525
Riemann-Hurwitz formula &, 505, 514, 528,
579, 580, 584, 589, 592, 592–594

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applications, 592
hyperelliptic, 514, 516, 518, 580
Riemann-Roch &for surfaces, 553
Riemann-Roch &, 477, 479, 479, 485, 486, 491,
508, 519, 521, 524, 528, 546, 549
Asymptotic, 546, 546, 548, 557
common formulation, 480
for coherent sheaves, 481
for nonreduced curves, 482, 483
for surfaces, 546, 549
in higher dimension, seeHirzebruch-Riemann-Roch Theorem
restatement using Serre duality, 485
vast generalization, see
Grothendieck-Riemann-Roch Theorem
right adjoints &, 43, 758
are left-exact, 54
commute with limits, 55
right derived functors &, see derived functors,
rightright-exact, 34, 52, 53
π−1 , 94and derived functors, 465
and FHHF Theorem, 54
and left derived functors, 626
e.g., left adjoint, 55
left adjoints, 54
need not commute with colimit, 56
pullback, 427
tensor product, 34, 53
rightward orientation (of spectral sequence)&, 58
rigid &, 588
Rigidity Lemma &, 296, 297, 298, 469
ringlocal, 655
ring &, see also local rings, see also graded ring,see also Noetherian ring
Artinian, 327, 328, 483
catenary, 311, 327, 691
examples, 311
excellent, 327
of integers in number field, 276, 276, 369,
373, 407, 589
topological, 726
ring scheme &, 197, 199
ring-theoretic
branch locus, 589
ramification locus, see different ideal
ringed spaces, 13, 77, 78, 79, 92, 99, 100, 136,
see also locally ringed spaces, 138, 141,
149, 177, 178, 180, 214, 227, 369, 371, 374,
385, 398, 421, 451, 635–637, 754–758(locally) free sheaf on, 369, 371(quasi)coherent sheaves on, 374, 382
functions on, 77, 102, 138
higher pushforwards for, 635

isomorphism of, 136, 136, 137, 178, 738
Leray spectral sequence, 635
morphisms of, 177, 178, 178–181, 424, 425,
448, 635, 636
glue, 178
open embedding of, 78, 178, 181, 202, 425
Picard functor for, 372
pullback for, 423
pushforward of O-modules, 178
structure sheaf, see structure sheaf
Rosenbluth, A. &, 133
ruled surface &, see also Hirzebruch surface,452, 550, 551, 743
elementary transformation, 742, 743
self-intersection of section of, 551
ruling &, 145, 233, 234, 234, 272, 315, 341, 408,
445, 537
of higher-dimensional quadrics, 235
of quadrics in P5 , 445
Salmon, G. &, 697
Samuel, P. &, 363
saturated S• -module &, 419
saturation functor &, 417, 417, 418
adjoint to forgetful functor, 417, 418
need not be injective or surjective, 418
scheme-theoretic
base locus, 415, 416
closure, 236, 239, 239, 292, 460, 503, 601, 602,
657, 658
fiber, 201, 257, 258, 259, 266, 677, 702
image, 236, 236–240, 292, 295, 296, 431
and set-theoretic image, 238
computable affine-locally, 238
intersection, 136, 227, 245, 333, 348, 415, 435,
492, 548, 600, 602
inverse image, see scheme-theoretic,pullbackpreimage, 463, 549
preimage [right name?], 393
pullback, 256, 256, 599, 600, 603, 604, 741
support, 240, 483, 502
union, 227, 333, 435, 575, 616, 676
scheme-theoretically dense, 168, 168
schemes, 136affine, 129, 136
cohomological criterion, see Serre’s
cohomological criterion for affinenessis quasiseparated, 154disjoint union, 380
finite type, see finite type, 159
functions on, 136
gluing, 249
isomorphism, see isomorphism, scheme
locally Noetherian, 154, 158, 159
are quasiseparated, 159
associated points, 169
implies quasiseparated, 159

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November 18, 2017 draft

locally of finite type, see locally of finite
type
morphisms, see morphisms, of schemes
Noetherian, 17, 153, 202, 208, 209, 218, 220,
229, 239, 260, 261, 304, 320, 340, 352, 353,
358, 363, 404, 405, 422, 435, 501, 506, 589,
658
finiteness of zeros and poles, 353
non-affine, 137
nonquasiseparated (example), 154
normal, see normal scheme
over a ring, 159
projective, 46, 145–147, 151, 185, 186, 191,
231, 235, 279, 291, 293, 429, 501, 546, 583,
605, 612, 766
are qcqs, 154maps to, 429
quasiaffine, see quasiaffine, schemes
quasiprojective, 145, 151, 159
regular, 337
visualizing, 106, 110, 113, 133, 170, 186, 215,
236, 393
Schubert cell &, 199, 200
Second Riemann Extension Theorem &, 320
section &of a morphism, 289
section &of a sheaf, see sheaf, section ofsections of F
sometimes means sections of F over X, 74
Segre embedding &, 271, 272, 282, 432, 437,
507
as complete linear series, 432
coordinate-free description, 272
Segre variety &, 272
semiample line bundle, 441
semicontinuity &, 322(lower) rank of matrix, 327(lower) rank of matrix, 713(upper) degree of finite morphism, 327, 393,
507
(upper) dimension of tangent space, 571(upper) dimension of tangent space atclosed points, 327(upper) fiber dimension, 326(upper) fiber dimension, 325, 327, 469, 507,
701
(upper) of fiber dimension, 322(upper) rank of cohomology groups, 327,
see Semicontinuity Theorem(upper) rank of sheaf, 327, 392, 393, 576,
681, 708
list of examples, 326
Semicontinuity Theorem &, 658, 661, 669, 708,
708, 709, 713, 715, 720
separable extension &, see field extension,
separable
separable finite morphisms &, 590

803

separable generation &, see field extension,
separably generated
separably closed &, see field, separably closed
separated presheaf &, 75
separatedness &, 284, 498
separatedness &, 142, 145, 154, 160, see also
Associated-to-Separated Theorem, seealso Reduced-to-Separated Theorem, 189,
207, 270, 279, 280, 280, 281, 283–286,
288–296, 298, 331, 358–361, 435, 439, 441,
442, 455–459, 461, 465, 466, 470–475,
496–499, 501, 503, 504, 572, 598, 603, 621,
628, 635, 637, 642, 643, 651, 680, 705,
738–740
closed under composition, 285, 286
closed under product, 286
implies quasiseparated, 284
local on target, 285
preserved by base change, 284
topological characterization, 280
Serre duality &for Homfunctorial, 761
for Ext
functorial, 761
for Homtrace version, 761
Serre duality &, 467, 477, 485, 487, 505, 508,
519, 521, 524, 549, 553, 579, 584, 586, 587,
626, 627, 685, 690, 749, 753, 754
for Ext, 749, 763
functorial, 753, 753, 757, 760, 762–766
functorial (for projective space), 753
trace version, 753, 754
for Hom, 749, 751, 753
for projective space, 751
functorial, 750, 751, 761, 763
trace version, 750, 751, 761
for Cohen-Macaulay projective schemes,486
for projective schemes, 758
for smooth projective varieties, 485
for vector bundles, 749, 753
functorial, 750, 753, 757, 760–762
functorial (for projective space), 757
trace version, 750, 753, 762
weak, 750
trace version
for Ext, 758
for vector bundles, 750
Serre vanishing &, 467, 478, 487, 489, 495, 557,
669, 740, 754, 762, 764
relative, 501
Serre’s cohomological criterion for ampleness,
494
Serre’s cohomological criterion for affineness,
466, 475, 496

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Serre’s cohomological criterion for affineness&, 466, 475, 496
Serre’s cohomological criterion for ampleness,see ampleness, Serre’s cohomologicalcriterion
Serre’s criterion for normality &, 320, 685, 691,
692, 693, 695
Serre’s Theorem A &, 416, 416, 438, 764
Serre, J.-P. &, 11, 17, 47, 57, 71, 99, 130, 186,
362, 404, 416, 438, 497, 621, 645, 749, 753
sextic, 230, 521, 527, 528
sheaf, see also subsheaf, 13, 75
coherent, see coherent sheaf
determined by sheaf on affine base, 87, 378
direct image, see sheaf, pushforward
dualizing, see dualizing sheaf
espace étalé, see sheaf, space of sections
extension by zero, 95
left adjoint to inverse image, 95
fiber, 422
finite type, see finite type, quasicoherent
sheaf
finitely presented, see finite presentation,
quasicoherent sheaf
flasque (flabby), 635, 757
have no Čech cohomology, 640
implies acyclic, 636
preserved by pushforward, 636
germ, see germs
gluability axiom, 75
higher pushforward, 669
higher pushforward, 413, 497–502, 555, 637,
642, 651, 669, 707, 709, 713, 723
and base change, 499
commutes with affine flat base change,
499
commutes with flat base change, 651
ideal, see also pullback, of ideal sheaves
identity axiom, 75
indicator, 85
injective
have no Čech cohomology, 639
implies flasque, 636
inverse image, see inverse image sheaf
invertible, see line bundle
locally free, see locally free sheafof (relative) i-forms ΩiX/Y , 583
of differentiable functions, 71, 100of nilpotents, 380
has closed support for locally NoetherianX, 390is coherent for locally Noetherian X, 390
of relative differentials, 561, 677
of sections, 370
on a base, 87, 87
determines sheaf, 87
on the big Zariski site, see Zariski sheaf

pushforward, 77, 77, 79, 91, 92, 100, 178,
380, 421, 422, 424, 425, 427, 469, 499, 555,
635, 709, 711, 744, 746, 758
preserves coherence for projective
morphisms, 469
preserves coherence for propermorphisms, see Grothendieck’s
Coherence Theorem
quasicoherent, see quasicoherent sheaves
quotient, 86
restriction to open subset G |U , 422
section of, see section of sheaf, 73
determined by germs, 82
support of, see support, of section
skyscraper, 76, 76, 77, 86, 94, 375, 635, 660
space of sections, 71, 73, 77, 85, 94
stalk, 75
structure, see structure sheaf
sheaf Hom, 79
sheaf Proj, see relative Proj
universal property, 450
sheaf Spec, see relative Spec
sheafification &, 46, 81, see also universal
property, of sheafification, 84
construction, 84
induces isomorphism of stalks, 85
is a functor, 84
universal property, 84
short exact sequence &, 49
should be (x + y)/(x − y), someone said —maybe Aaron L? 2015, 536
should we have π♯ B/J?, 213
signature of symmetric bilinear form &, 553
simple group of order 168 &, 521
singular (point), see regular
singularities, see also resolution (singularities),
337, 725, 729
site &, see Grothendieck topology
skyscraper sheaf &, see sheaf, skyscraperslicing criterion &for Cohen-Macaulayness, 689, 689
geometric interpretation, 689
for flatness, 667, 667, 668, 675, 689, 690
in source, 668, 676
in source (more general version), 668
for regularity, 338, 533, 607, 676, 689
small category &, see category, small
small resolution &, see resolution(singularities), small
Smith, R. &, 697
smooth quadric surface, 165, 334
smoothness &, 14, 101, 110, 295, 296, 331, 354,
355, 355, 663, 673, 676, 677, see also
formally smooth, see also genericsmoothness, 682
closed under product, 356
differential geometric motivation, 354
form “reasonable” class, 356

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

November 18, 2017 draftof relative dimension n, 355, 355–357, 574,
589, 675, 675, 677, 683
open condition, 355
of varieties preserved by base field
extension, 339
open condition, 520
over a field= geometrically regular, 576
over a field, 338, 338, 575, 598
of dimension d, 338
over a ring, 356
Smoothness-Regularity Comparison Theorem&, 691
Smoothness-Regularity Comparison Theorem&, 340, 342, 607, 676
Snake Lemma &, 50, 52, 60, 62, 62, 63, 387,
394, 717
source (of morphism in category) &, 26
space of sections of sheaf &, see sheaf, space ofsections
specialization &, 122, 122, 123, 153, 218, 319
spectral functor &, 65, 642
spectral sequence &, 52, 57, 58, 58–65, 69, 243,
244, 472, 501, 629–631, 634, 635, 637–639,
641, 642, 756, 757
and derived functors, 629
collapse of, 61
convergence, 61Grothendieck composition-of-functors, see
Grothendieck composition-of-functors
spectral sequence
Leray, see Leray spectral sequence
local-to-global for Ext, 757, 757, 764
orientation, see orientation of spectral
sequence
page, 58
spectrum of ring (Spec) &, 101
split exact sequence &, 634
stacks &, 14, 89, 381, see also orbifolds, 523, 526
stalk-local &, 155
implies affine-local, 158
stalkification &doesn’t commute with Hom , 80
is exact, 55, 91
stalks &determine isomorphisms, 83
determine morphisms, 83of (pre)sheaf on base, 87of a sheaf, 75
of presheaf, 74
Stein factorization &, 733, 734, 735, 735, 737
Stein Factorization Theorem &, 725, 734, 735
Stein space &, 497, 539
Stillwell, J. &, 136
strict transform (of blow-up) &, see propertransform (of blow-up)
structure morphism &, 183
structure sheaf &, 129805

of ringed space, 77
subcategory &, 28
full, 29, 29, 45, 49, 79, 381
fully faithful, 29, 46, 251
submersion &, 25, 101, 562, 673, 674
subobject &, 48
subscheme
closed, see closed subschemes
cut out by section of locally free sheaf, 372
locally closed, see locally closed subschemes
open, 137
subsheaf &, 85
subtler Five Lemma &, see Five Lemma,
subtlersubvariety (incl. open and closed), 283
support &of a section s
is closed, 83
of a sheaf, 390
support &of a section s, 82, 83, 390
of a sheaf, 94
of finite type quasicoherent sheaf is closed,390
of Weil divisor, 400
scheme-theoretic, see scheme-theoretic,
supportsurfaces, 231, 304, 584, see also cubic, surface
cubic
Fermat, 701, 703
Enriques-Kodaira classification, 743
intersection theory, 547
minimal (model), 743
proper but no nontrivial line bundles, 540
proper nonprojective, 550
quadric, 494
smooth projective, 406
surjective morphisms &, 205
checking on closed points, 219
preserved by base change, 263
Sylvester’s law of intertia, 553
Sylvester, J. &, 553
symbolic power of an ideal &, 328
symmetric algebra &, 148, 384
system of parameters &, 318, 323
tacnode &, 343, 611, 730, 730
tame ramification &, 592
tangent
bundle, 369
cone, 606, 606= tangent space for regular point, 606,
607
projectivized, 606
line, 332, 333, 344, 348, 528, 530, 533, 536,
538
map, 505
plane, 344, 344

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806

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sheaf, see also cotangent, sheaf
relative, 566, 572
space, 101, 327, 331–333, 335–337, 341, 344,
347, 353, 357, 403, 506, 561, 562, 566, 567,
571, 578, 606, 607, 673, 674, 678
and morphisms, 335
vector, 332, 561, 561
relative, 561
vertical, 561, 564, 565
tangent cone
projectivized, 606
Taniyama-Shimura Conjecture &, 518
Tao, T. &, 13
target (of morphism in category) &, 26
Tarski, A. &, 221
Tate curve &, 527
Tate, J. &, 362, 381
tautological bundle &Grassmannian, 583
on projective space, 450
Taylor, R. &, 518
tensor algebra &, 384
tensor product &, 33, see also universal
property, of tensor product, see also
universal property, of tensor productas fibered coproduct of rings, 248
commutes with direct sum, 35, 56, 622, 729
is not left-exact, 34
is right-exact, 34, 53, 648, 729
of OX -modules, 92
tensor-finiteness trick &, 266, 267, 269
Theorem of the Base (Néron-Severi Theorem)&, see also Picard number ρ(X), 484, 484
Theorem on Formal Functions &, 725, 732,
733, 733, 740–744
thickening &, see formal neighborhood
topological ring/module &, 726
topology &base of, 87
discrete, see discrete topologyétale, see étale topology
Grothendieck, see Grothendieck topology
profinite, 255
Zariski, see Zariski topology
topos &, 23, 381
Tor functors, 646, 647, 652
long exact sequence, 623
symmetry of, 629, 653
torsion &(sub)module, 386
quasicoherent sheaf on reduced scheme,386
sheaf, 391
torsion-freeness &= flat for PID, 655
implied by flatness, 648, 652, 655, 665, 667,
691
module, 386

sheaf, 386, 391
torsor &, 255
total complex (of a double complex) &, 58
total fraction ring (total quotient ring) &, seealso fraction field, 171, 607, 694, 695
total space &of line bundle, 539, 541, 551, 741
of vector bundle, 371, 449, 542, 606
total transform (of blow-up) &, 600, 602, 609
trace map &, 389, 582, 749, 750, 751, 751, 752,
754, 761, 763
transcendence basis &, 681
transcendence basis &, 308, 314, 572
transcendence degree &, 310
transcendence degree &, 307, 308, 308–312,
324, 461, 493, 576, 681= dimension, 308
transcendence theory &, 204, 268, 307, 324,
576, 681
transition functions/matrices &, 370, 370–372,
383, 385, 397–400, 413, 423, 426, 427, 429,
449, 450, 480, 488
trichotomy (classification of varieties) &, 518,
585, 586
triple point &, 730, 730
trivial bundle/sheaf of rank n &, 370
trivialization of vector bundle&, 370
tropical geometry &, 127
Tsen’s Theorem &, 712
twisted cubic curve &, 490
twisted cubic curve &, see also rational normal
curve, 120, 231, 231, 233, 308, 343, 412,
490, 491, 658, 661
two planes meeting at a point &, 691
two planes meeting at a point &, 132, 486, 494,
657, 668, 687–689, 692, 754
not S2 or Cohen-Macaulay, 692
ultrafilter &, 121
uniformizer &, 350, 351, 354, 389, 436, 463,
464, 580, 591, 592, 655
union (scheme-theoretic), see
scheme-theoretic, union
unique factorization domain &, 406
unique factorization and class groups &, 407
unique factorization domain &, 102, 104,
163–166, 171, 194, 275, 315, 317, 334, 349,
352, 354, 363, 403–408, 538, 693= all codimension 1 primes principal, 163,
307, 317, 334, 403, 404, 407
codimension 1 primes are principal, 306,
307
how to check, 163
not affine-local condition, 166
uniruled &, 321
universal delta-functor &, see delta-functors,
universal

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November 18, 2017 draft

universal hyperplane, see hyperplane,
universal
universal property &, 24, 31, 32, 35, 40, 45,
195, 196, 250, 251, 253, 273, 290, 384, 424,
447, 448, 451, 570, 604, 613, 628, 646, 751
and Yoneda’s Lemma, 38
of Sym and ∧, 384
of blow-up, 597, 598, 599, 600–604, 652, 742
of cokernel, 48, 81, 90of colimit, 90
of differentials, 570
of fibered coproduct, 37
of fibered product, 35, 36, 248–251, 262
of inverse image sheaf, 93
of Kähler differentials, 563, 571, 573
of kernel, 48, 81, 82
of limit, 40, 55
of localization, 31, 32, 33, 209, 379
of normalization, 273, 273, 274, 460, 741
of product, 24, 31
of projective space, 723
of pullback, 423, 424, 424–426
of relative Proj, 453
of relative Proj, 450
of relative Spec, 447, 447, 448
of sheafification, 84, 84, 85, 90of stalks, 77of tensor product, 33, 34, 35, 44, 573
universally closed morphisms &, 293, 293,
361, 362, 435, 455, 503
universally injective (radicial) morphisms &,270, 270, 271, 289= “injective on geometric points”, 271
and the diagonal morphism, 271, 289
are separated, 289
preserved by composition, base change,
products, 271
universally open morphisms &, 266“universally P” &, 293
unramifiedness &, 101, 225, 229, 506, 583, 588,
588, 592, 673, 674, 676, 737
characterizations by fibers, 588
closed under composition, 588
diagonal characterization, 589
examples, 588
formal, see formally unramified
in number theory, 588
open condition, 589
upper semicontinuity &, see semicontinuity
upward orientation (of spectral sequence) &,61
Valéry, P. &, 767
valuation rings &, 127, 320, see also discrete
valuation rings, 351, 358, 361, 362, 393
importance of, 362
valuations &, see also discrete valuations, 351,
see also valuation rings, 400, 460, 579

807

valuative criteria &, 17, 331, 358
for flatness, 657
for properness, 361, 361, 362, 436, 680
for separatedness, 359, 360, 361, 362, 680
for universal closedness, 361, 361, 362
value of a function at a point &, see functions,value at a point
vanishing theorems &, 500
vanishing scheme &, 151, 151, 227, 230, 590
vanishing set &, 115, 149
vanishing theorems &, see also dimensional
cohomology vanishingvanishing theorems &, see also Kodairavanishing, see also Serre vanishing, 478
varieties, 160, 283
affine, 160
affine complex, 108classically, 121
complex, 99
projective, 160
quasiprojective, 160
with non-finitely-generated ring of global
sections, 541
Vaserstein, L. &, 404
vector bundles &, 14, 77, 78, 152, 369–371, 374,
375, 382, 388, 389, 393, 413, 449, 452, 481,
488, 550, 568, 569, 582, 604, 646, 653, 709,
712, 713, see also Serre duality, for vector
bundles, 749, 750, 753, 757, 760–762
ample, 458
associated to locally free sheaves, 449
classification on P1 , 486
not always direct sum of line bundles, 582
on manifold, 369
perfect pairing, 385
pulling back, 423, 426
total space, see total space, of vector
bundles
trivialization, 370
Veronese embedding &, see also rational
normal curve, see also twisted cubic
curve, 120, 232, 233, 272, 282, 346, 430,
431, 490–492, 507, 517, 760
as complete linear series, 430, 507
Veronese subring Sn• &, 188, 188, 232
Veronese surface &, 233, 233
very ample &, 437, 489, 545–547, 553, 554, 556,
557, 559, 672, 684, 740⊗ base-point-free is very ample, 437
behaves worse than ample, 438
implies base-point-free, 437
in relative setting, 456
preserved by %, 437
preserved by ⊗, 437
relatively, 437, 452
with respect to π, see very ample, relatively
very general &, 315
Viète’s formula &, 533

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.

808

The Rising Sea: Foundations of Algebraic Geometry

Voevodsky, V. &, 381
Weak Factorization Theorem &, 614
Weak Nullstellensatz &, see Nullstellensatz
weakly associated prime &, 168
Weierstrass coordinates &, 530
Weierstrass normal form &, 529, 531, 532, 535
weighted projective space &, 235
Weil divisor &locally principal, 403
Weil divisor &, 397, 400, 400, 401, 403–408,
549, 579, 614
effective, 400
irreducible, 400
locally principal, 403, 405, see also closed
subscheme, locally principal, 614
not necessarily locally principal, 403
not necessarily locally principal, 407
of zeros and poles, 400
principal, 403
restriction to open set, 400
support of, 400
Weil, A. &, 362, 532, 585
Weyl group &, 698, 704, 706
Weyl, H. &, 144
Wiener, N. &, 133
wild ramification &, 592
Wiles, A. &, 518
Włodarczyk, W. &, 614

topological and power series forms, 740
Zariski, O. &, 295, 337, 736, 743
zero of order n &for Noetherian scheme, 353
zero object &, see object, zero
zero of order n &for discrete valuation rings, 352
zerodivisor (for ring or module) &, 32
Zorn’s Lemma &, 113, 123, 627

Yoda embedding &, 39
Yoneda cup product &, see cup product,
YonedaYoneda embedding &, 39
Yoneda’s Lemma &, 13, 38, 39, 185, 195, 248,
251, 358, 429, 444, 599
Zariski cotangent space &, see cotangent,
space
Zariski sheaf (sheaf on the big Zariski site) &,251, 252, 252, 253, 443
Zariski site &, 251
Zariski tangent space &, see tangent space
Zariski topology &, 114–119, 133, 136, 137,
142, 150, 221, 222, 255, 304, 377, 378, 402,
418, 571
on Proj S• , 150
on Spec A, 115n
on k , 221, 313
on scheme, 136
Zariski’s Lemma, see also Nullstellensatz
Zariski’s Connectedness Lemma &, 710, 725,
733, 733–737
Zariski’s Lemma &, 107
Zariski’s Main Theorem &, 215, 296, 705, 736,
739, 740
for birational morphisms, 735, 736
Grothendieck’s form, 725, 737, 738
misnamed, 736

Early (out-of-date) version of The Rising Sea: Foundations of Algebraic Geometry (c) 2024 Ravi Vakil. Published by Princeton University Press.