# Solutions to Hartshorne's Algebraic Geometry/Printable version
## Separated and Proper Morphisms

The reference for this section is EGA II.5, EGA II.6, EGA II.7. For the discrete vaulation ring questions at the end see Samula and Zariski's Commutative Algebra II.

## Exercise II.4.1

Let $f:X\to Y$ be the finite morphism. Finite implies finite type so we only need to show that $f$ is universally closed and separated.
$f$ is separated. We want to show that $X\to X\times _{Y}X$ is a closed immersion. To check that a morphism is a closed immersion it is enough to check for each element of an open cover of the target. Let $\{Spec\ B_{i}\}$ be an open affine cover of $Y$. The pull-back of $X\to X\times _{Y}X$ along each $Spec\ B_{i}\to Y$ is $Spec\ A_{i}\to Spec\ A_{i}\otimes _{B_{i}}A_{i}$ where $Spec\ A_{i}=f^{-1}Spec\ B_{i}$. The ring homomorphism corresponding to these morphisms of affine schemes is surjective, and so they are all closed immersions according to Exercise II.2.18(c).
$f$ is universally closed. The proof of Exercise II.3.13(d) goes through to show that finite morphisms are stable under base change (in fact, the proof becomes easier). Secondly, we know that finite morphisms are closed (Exercise II.3.5) and therefore finite morphisms are universally closed.

## Exercise II.4.2

Let $U$ be the dense open subset of $X$ on which $f$ and $g$ agree. Consider the pullback square(s):
Since $Y$ is separated, the lower horizontal morphism is a closed immersion. Closed immersions are stable under base extension (Exercise II.3.11) and so $Z\to X$ is also a closed immersion. Now since $f$ and $g$ agree on $U$, the image of $U$ in $Y\times _{S}Y$ is contained in the diagonal and so the pullback is, again $U$ (at least topologically. But this means that $U\to X$ factors through $Z$, whose image is a closed subset of $X$. Since $U$ is dense, this means that $sp\ Z=sp\ X$. Since $Z\to X$ is a closed immersion, the morphism of sheaves ${\mathcal {O}}_{X}\to {\mathcal {O}}_{Z}$ is surjective. Consider an open affine $V=Spec\ A$ of $X$. Restricted to $V$, the morphism $Z\cap V\to V$ continues to be a closed immersion and so $Z\cap V$ is an affine scheme, homeomorphic to $V$, determined by an ideal $I\subseteq A$. Since $Spec\ A/I\to Spec\ A$ is a homeomorphism, $I$ is contained in the nilradical. But $A$ is reduced and so $I=0$. Hence, $Z\cap V=V$ and therefore $Z=X$.
* Consider the case where $X=Y=Spec\ k[x,y]/(x^{2},xy)$, the affine line with nilpotents at the origin, and consider the two morphisms $f,g:X\to Y$, one the identity and the other defined by $x\mapsto 0$, i.e. killing the nilpotents at the origin. These agree on the complement of the origin which is a dense open subset but the sheaf morphism disagrees at the origin.
* Consider the affine line with two origins, and let $f$ and $g$ be the two open inclusions of the regular affine line. They agree on the complement of the origin but send the origin two different places.

## Exercise II.4.3

Consider the pullback square
Since $X$ is separated over $S$ the diagonal is a closed immersion. Closed immersions are stable under change of base (Exercise II.3.11(a)) and so $U\cap V\to U\times _{S}V$ is a closed immersion. But $U\times _{S}V$ is affine since all of $U,V,S$ are. So $U\cap V\to U\times _{S}V$ is a closed immersion into an affine scheme and so $U\cap V$ itself is affine (Exercise II.3.11(b)).
For an example when $X$ is not separated consider the affine plane with two origins $X$ and the two copies $U,V$ of the usually affine plane inside it as open affines. The intersection of $U$ and $V$ is $\mathbb {A} ^{2}-\{0\}$ which is not affine.

## Exercise II.4.4

Since $Z\to S$ is proper and $Y\to S$ separated it follows from Corollary II.4.8e that $Z\to Y$ is proper. Proper morphisms are closed and so $f(Z)$ is closed.
$f(Z)\to S$ is finite type. This follows from it being a closed subscheme of a scheme $Y$ of finite type over $S$ (Exercise II.3.13(a) and (c)).
$f(Z)\to S$ is separated. This follows from the change of base square and the fact that closed immersions are preserved under change of base.
$f(Z)\to S$ is universally closed. Let $T\to S$ be some other morphism and consider the following diagram
Our first task will be to show that $T\times _{S}Z\to T\times _{S}f(Z)$ is surjective. Suppose $x\in T\times _{S}f(Z)$ is a point with residue field $k(x)$. Following it horizontally we obtain a point $x'\in f(Z)$ with residue field $k(x')\subset k(x)$ and this lifts to a point $x''\in Z$ with residue field $k(x'')\supset k(x')$. Let $k$ be a field containing both $k(x)$ and $k(x'')$. The inclusions $k(x''),k(x)\subset k$ give morphisms $Spec\ k\to T\times _{S}f(Z)$ and $Spec\ k\to Z$ which agree on $f(Z)$ and therefore lift to a morphism $Spec\ k\to T\times _{S},Z$ giving a point in the preimage of $x$. So $T\times _{S}Z\to T\times _{S}f(Z)$ is sujective.
Now suppose that $W\subseteq T\times _{S}f(Z)$ is a closed subset of $T\times _{S}f(Z)$. Its vertical preimage $(f')^{-1}W$ is a closed subset of $T\times _{S}Z$ and since $Z\to S$ is universally closed the image $s'\circ f'((f')^{-1}(W))$ in $T$ is closed. As $f'$ is surjective, $f'((f')^{-1}(W))=W$ and so $s'\circ f'((f')^{-1}(W))=s'(W)$. Hence, $T\times _{S}f(Z)$ is closed in $T$.

## Exercise II.4.5

* Let $R$ be the valuation ring of a valuation on $K$. Having center on some point $x\in X$ is equivalent to an inclusion ${\mathcal {O}}_{x,X}\subseteq R\subseteq K$ (such that ${\mathfrak {m}}_{R}\cap {\mathcal {O}}_{x,X}={\mathfrak {m}}_{x}$) which is equivalent to a diagonal morphism in the diagram
But by the valuative criterion for separability this diagonal morphism (if it exists) is unique. Therefore, the center, if it exists, is unique.
* Same argument as the previous part.
* The argument for the two cases is the same so we will prove: Suppose that every valuation ring $R$ of $K$ has a unique center in $X$, then $X$ is proper. This is clearly true for integral $k$-schemes of finite type of dimension zero. Suppose that it is true for integral $k$-schemes of dimension less than $n$ and that $X$ is an integral $k$-scheme of dimension $n$. We will use the valuative criteria. Suppose that we have a diagram
with $S$ a valuation ring of function field $L$. If the image of the unique point of $Spec\ L$ is not the generic point of $X$ then let $Z$ be the closure of its image with the reduced structure. We have a diagram
The scheme $Z$ is an integral $k$-scheme of dimension less than $n$ and so the square on the lest admits a lifting, which gives a lifting for the outside rectangle. Moreover, as closed immersions are proper, any lifting of the outside rectangle factors uniquely through $Z$ by the valuative criteria and so the lifting is unique.
Now suppose that the image of the point of $Spec\ L$ is the generic point of $X$. Then we have a tower of field extensions $L/K/k$ and the valuation on $L$ induces a valuation on $K$. We then have the following diagram.
By assumption the valuation ring $R$ has a unique center $x$ on $X$ and so there is a unique extension of the diagram above
Hence, there is a unique lifting of our original square. By the valuative criteria, the scheme $X$ is then proper.
* Suppose that there is some $a\in \Gamma (X,{\mathcal {O}}_{X})$ such that $a\not \in k$. Consider the image $a\in K$. Since $k$ is algebraically closed, $a$ is transcendental over $k$ and so $k[a^{-1}]$ is a polynomial ring. Consider the localization $k[a^{-1}]_{(a^{-1})}$. This is a local ring contained in $K$ and therefore there is a valuation ring $R\subset K$ that dominates it. Since ${\mathfrak {m}}_{R}\cap k[a^{-1}]_{(a^{-1})}=(a^{-1})$ we see that $a^{-1}\in {\mathfrak {m}}_{R}$.
Now since $X$ is proper, there exists a unique dashed morphism in the diagram on the left.
Taking global sections gives the diagram on the right which implies that $a\in R$ and so $v_{R}(a)\geq 0$. But $a^{-1}\in {\mathfrak {m}}_{R}$ and so $v_{R}(a^{-1})>0$ This gives a contradiction since $0=v_{R}(1)=v_{R}({\frac {a}{a}})=v_{R}(a)+v_{R}({\frac {1}{a}})>0$.

## Exercise II.4.6

Since $X$ and $Y$ are affine varieties, by definition they are integral and so $f$ comes from a ring homomorphism $B\to A$ where $A$ and $B$ are integral. Let $K=k(A)$. Then for valuation ring $R$ of $K$ that contains $\phi (B)$ we have a commutative diagram
Since $f$ is proper, the dashed arrow exists (uniquely, but we don't need this). From Theorem II.4.11A the integral closure of $\Phi (B)$ in $K$ is the intersection of all valuation rings of $K$ which contain $\phi (B)$. As the dashed morphism exists for any valuation ring $K$ containing $\phi (B)$ so it follows that $A$ is contained in the integral closure of $\phi (B)$ in $K$. Hence every element of $A$ is integral over $B$, and this together with the hypothesis that $f$ is of finite type implies that $f$ is finite.

## Exercise II.4.8

* Let $X{\stackrel {f}{\to }}Y$ and $X'{\stackrel {f'}{\to }}Y'$ be the morphisms. The morphism $f\times f'$ is a composition of base changes of $f$ and $f'$ as follows:
\tbd{\mathfrak{m}arginpar{Should really check that the all the claims made about pullbacks in here are true.}}
Therefore $f\times f'$ has property $P$.
* Same argument as above but we should also note that since $g$ is separated the diagonal morphism $Y\to Y\times _{Z}Y$ is a closed embedding and therefore satisfies $P$.
* Consider the factorization
The morphism $X_{red}\to X\to Y$ is a composition of a closed immersion and a morphism with property $scP$ and therefore it has property $P$. Therefore the vertical morphism out of the fibre product is a base change of a morphism with property $P$ and therefore, itself has property $P$. To se that $f_{red}$ has property $P$ it therefore remains only to see that the graph $\Gamma _{f_{red}}$ has property $P$ for then $f_{red}$ will be a composition of morphisms with property $P$. To see this, recall that the graph is following base change
But $Y_{red}\times _{Y}Y_{red}=Y_{red}$ and $\Delta =id_{Y_{red}}$ and so $\Delta$ is a closed immersion. Hence, $\Gamma$ is a base change of a morphism with property $P$.

## Exercise II.4.9

Let $X{\stackrel {f}{\to }}Y{\stackrel {g}{\to }}Z$ be two projective morphisms. This gives rise to a commutative diagram
where $f'$ and $g'$ (and therefore $id\times g'$) are closed immersions. Now using the Segre embedding the projection $\mathbb {P} ^{r}\times \mathbb {P} ^{s}\times Z\to Z$ factors as $\mathbb {P} ^{r}\times \mathbb {P} ^{s}\times Z\to \mathbb {P} ^{rs+r+s}\times Z\to Z$ So since the Segre embedding is a closed immersion then we are done since we have found a closed immersion $X\to \mathbb {P} _{Z}^{rs+r+s}$ which factors $g\circ f$.

## Exercise II.4.10

Chow's Lemma is in EGA II.5.6.

## Exercise II.4.11

See Samula and Zariski's Commutative Algebra II.
Suppose that $L=K(t)$. Then define: ${\mathfrak {m}}_{R}=\{a_{0}+a_{1}t+\dots +a_{n}t^{n}\in {\mathcal {O}}[t]:a_{0}\in {\mathfrak {m}}\}$ The ring ${\mathcal {O}}[t]$ is a discrete noetherian local domain with maximal ideal ${\mathfrak {m}}_{R}$ and quotient field $L$. By induction then, we can reduce to the case when $L$ is a finite field extension of $K$. Now consider a set of generators $\{x_{1},\dots ,x_{n}\}$ of ${\mathfrak {m}}$ such that $x_{1}\not \in {\sqrt {(x_{2},\dots ,x_{n})}}$\mathfrak{m}arginpar{does such a set always exist?} (if ${\mathfrak {m}}$ is principal wait for the next step). We claim that the ideal $(x_{1})$ is not the unit ideal in ${\mathcal {O}}'={\mathcal {O}}[{\frac {x_{2}}{x_{1}}},\dots ,{\frac {x_{n}}{x_{1}}}]$. If it were then there would be some polynomial $f$ of degree, say $d$, in the ${\frac {x_{i}}{x_{1}}}$ such that $1=x_{1}f$. Let $f_{0}$ be the degree 0 part of $f$ and $f_{1}$ be the higher degree part. Since $x_{1}\in {\mathfrak {m}}$ the element $1-x_{1}f_{0}$ has an inverse, say $a$. Now with this in mind, our equality $1=x_{1}f_{0}+x_{1}f_{1}$ implies that $1=ax_{1}f_{1}$ which then implies that $x_{1}^{d}=ax_{1}^{d+1}f_{1}$. Since $f_{1}$ is made up of terms of degree higher than zero, the element $ax_{1}^{d+1}f_{1}\in (x_{2},\dots ,x_{n})$ which implies that $x_{1}\in {\sqrt {(x_{2},\dots ,x_{n})}}$ contradicting our assumption. So $(x_{1})$ is not the unit ideal in ${\mathcal {O}}'$. Now let ${\mathfrak {p}}$ be a minimal prime ideal of $(x_{1})$, and consider the localization $({\mathcal {O}}')_{\mathfrak {p}}$.

## Exercise II.4.12

See Samuel and Zariski's Commutative Algebra II.

### Projective Morphisms

[Hartshorne 5.14]
Proof:
Proof: \textbf{ Claim I}: $S$ is a domain.\\ It is easy!\\
Let $S_{x_{i}}, 0\leq i\leq r,$ be a localization of $S$. It is a graded ring. For any element $x\in S_{x_{i}}$ we define the degree of x, $deg(x)$, by the degree of the lowest homogeneous part of $x.$ Construct a ring $\Gamma=\{x|x\in \cap_{0\leq i\leq r}S_{x_{i}}, deg(x)\geq 0\}$,
\textbf{ Claim II}: $\Gamma$ is integral over $S$, and $\Gamma_{\geq n}=S_{\geq n}$ for sufficiently large enough $n$.\\
Let $y\in \Gamma.$ Then by the definition of $\Gamma,$ we have for any $x_{i}, 0\leq i\leq r,$ there exists an integer $n_{i}$, such that $x_{i}^{n_{i}}y\in S.$ Thus there is an integer $N,$ such that $ yS_{\geq N}\subseteq S.$ Especially, we have $x_{0}^{N}y\in S.$ On the other hand, since the degree of $y,$ $deg(y)\geq 0,$ we have for any integer $r$, $x^{N}_{0}y^{r}\in S.$ Thus we have $S[y]\subseteq S\frac{1}{x_{0}^{N}},$ where $S\frac{1}{x_{0}^{N}}$ is a finite generated $S$ module. So $y$, and furthermore the whole $\Gamma$ is integral over $S.$\\
Further, since $S$ is a finite generated domain over a filed $k$, we have that $\Gamma$ as a $S$ module, must be finite. And from the method we have used above, we could prove that for any $y\in \Gamma,$ there is an integer $N$, such that $yS_{\geq N}\subseteq S.$ Thus, we get for sufficiently large enough $n,$ $\Gamma_{\geq n}=S_{\geq n}.$ \\
Construct rings $^{i}\Gamma=\{x|x\in S_{x_{i}}, deg(x)\geq 0\}$, $0\leq i\leq r.$
On the other hand, it is easy to see that $^{i}\Gamma$ equals to the ring $S_{(x_{i})}[x_{i}]$, which are polynomial rings over $S_{(x_{i})}$, respectively. Since $S_{(x_{i})}$ are integral closed domains, so $^{i}\Gamma$, and further more $\Gamma$ are also integral closed domains. \\
Since $k$ is algebraic closed, we have $\Gamma^{(d)}=S^{(d)}$ for sufficiently large enough $d$. \\
Now let $y\in Q(\Gamma^{(d)}),$ the quotient filed of $\Gamma^{(d)}$, which is integral over $\Gamma^{(d)}.$ Then since $\Gamma$ is integral closed, we have $y\Gamma.$ On the other hand, since $y$ is almost integral over $\Gamma^{(d)},$ it is easy to see that $y\in \Gamma^{(d)}.$ So we proved that for sufficiently large $d$, the -uple embedding $S^{(d)}$ is integral closed.--

### Cech Cohomology
## Exercise 4.7
### Problem Statement:

Let $f$ be an equation cutting out a degree d curve $X$ in $\mathbb {P} _{k}^{2}$. Suppose that $X$ doesn't contain the point $(1,0,0)$. Use \v{C}ech cohomology to calculate the dimensions of $H^{0},H^{1}$ of $X$.

### Solution:

The degree d curve $X$ is the vanishing locus of $f$, so we have a short exact sequence:
    $0\rightarrow {\mathcal {O}}(-d)\rightarrow {\mathcal {O}}\rightarrow {\mathcal {O}}_{X}\rightarrow 0$
where ${\mathcal {O}}$ without further decoration denotes the structure sheaf of $\mathbb {P} _{k}^{2}$. Specifically, the map on the left is multiplication by our polynomial f, which is a degree d map ${\mathcal {O}}\rightarrow {\mathcal {O}}$, but a degree 0 map ${\mathcal {O}}(-d)\rightarrow {\mathcal {O}}$. This is an injective map, and we are quotienting out precisely by its image, so it's equivalent to the usual short exact sequence associated to a closed subscheme.
Then apply the H functor to get a long exact sequence:
    $0\rightarrow \Gamma ({\mathcal {O}}(-d))\rightarrow \Gamma ({\mathcal {O}})\rightarrow \Gamma ({\mathcal {O}}_{X})\rightarrow H^{1}({\mathcal {O}}(-d))\rightarrow H^{1}({\mathcal {O}})\rightarrow H^{1}({\mathcal {O}}_{X})\rightarrow H^{2}({\mathcal {O}}(-d))\rightarrow H^{2}({\mathcal {O}})\rightarrow H^{2}({\mathcal {O}}_{X})\rightarrow 0$
Which vanishes in higher degrees by dimensional vanishing.
Now to figure out what these things are:
$H^{i}({\mathcal {O}}(e))=0$ for $0<i<r$ in projective space $\mathbb {P} _{A}^{r}$.
This gives us that $H^{1}({\mathcal {O}}(-d))=H^{1}({\mathcal {O}})=0$. Furthermore, assuming degrees must be positive $\Gamma ({\mathcal {O}}(-d))=0$.
$H^{2}({\mathcal {O}}_{X})$ actually vanishes again by dimensional vanishing. $\Gamma ({\mathcal {O}})=k$, either by general knowledge (constants are the only globally defined homogeneous polynomials with degree zero on any of the standard open affines) or by the fact that $h^{0}({\mathcal {O}}(e))={\dbinom {r+e}{r}}$ in general; when e = 0, this gives dimension 1 over k. ($h^{i}:={\text{dim}}H^{i}$).
Our last trick we shall use is Serre duality (here just for projective space):
$\Gamma ({\mathcal {O}}(-e))\cong (H^{r}(e-r-1))^{\vee }$, where $\cdot ^{\vee }$ represents the dual.
Since the dimension of a vector space (these H's are vector spaces in this context because of III 5.2 in Hartshorne, pg 228) is the same as its dual, ${\text{dim}}H^{2}({\mathcal {O}})={\text{dim}}(H^{2}({\mathcal {O}}))^{\vee }$. Moreover, $(H^{2}({\mathcal {O}}))^{\vee }\cong \Gamma ({\mathcal {O}}(-r-1))$, which has dimension ${\dbinom {-1}{r}}=0$, so it's 0. Hence $H^{2}({\mathcal {O}})=0$.
Moreover, ${\text{dim}}H^{2}({\mathcal {O}}(-d))={\text{dim}}(H^{2}({\mathcal {O}}(-d)))^{\vee }$, and by the same trick (Serre duality), $(H^{2}({\mathcal {O}}(-d)))^{\vee }\cong \Gamma ({\mathcal {O}}(d-r-1))$, which has well-known dimension (e.g., Vakil 14.1.c) of ${\dbinom {r+d-r-1}{r}}={\dbinom {d-1}{2}}={\dbinom {d-1}{d-3}}={\dfrac {1}{2}}(d-1)(d-2)$.
Combining all of the above results, we get two short exact sequences:
    $0\rightarrow k\rightarrow \Gamma ({\mathcal {O}}_{X})\rightarrow 0$
    $0\rightarrow H^{1}({\mathcal {O}}_{X})\rightarrow \Gamma ({\mathcal {O}}(d-r-1))\rightarrow 0$
So we have $h^{0}({\mathcal {O}}_{X})=1$ and $h^{1}({\mathcal {O}}_{X})={\dfrac {1}{2}}(d-1)(d-2)$.

#### Riemann-Roch Theorem
## Exercise IV.1.1

Let $X$ have genus $g$. Since $X$ is dimension 1, there exists a point $Q\in X$, $Q\neq P$. Pick an $n>\max(g,2g-2,1)$. Then for the divisor $D=n(2P-Q)$ of degree $n$, $l(K-D)=0$(Example 1.3.4), so Riemann-Roch gives $l(D)=n+1-g>1$. Thus there is an effective divisor $D'$ such that $D-D'=(f)$. Since $(f)$ is degree 0 (II 6.10), $D'$ has degree $n$, so $D'$ cannot have a zero of order large enough to kill the pole of $D$ of order $2n$. Therefore, $f$ is regular everywhere except at $P$.

### Glossary of notation

$\mathbb {A} ^{n}$ - affine $n$-space $\mathbb {C}$ - the complex numbers ${\mathfrak {m}}$ - a/the maximal ideal $\mathbb {N}$ - the natural numbers ${\mathcal {O}}_{X}$ - the sheaf of rings on a ringed space $X$. $\mathbb {P} ^{n}$ - projective $n$-space ${\mathfrak {p}}$ - a prime ideal ${\mathfrak {q}}$ - another prime ideal $\mathbb {Q}$ - the rational numbers $\mathbb {R}$ - the real numbers $\mathbb {Z}$ - the integers