# A New Criterion for Affineness

Authors: Jing Zhang
Journal: ISRN Algebra
 (2013)

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## Abstract

We show that an irreducible quasiprojective varietyY of dimension d≥1 defined over an algebraically closed field with characteristic zero is an affine variety if and only if Hi(Y,Y) = 0 and Hi (Y,Y(−Z)) = 0 for all i>0, Z=H∩Y, where H is any hypersurface with sufficiently large degree. A direct application is that an irreducible quasiprojective variety Y over ℂ is a Stein variety if it satisfies the two vanishing conditions. Here, all sheaves are algebraic.


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## Body

## 1. Introduction

We work over an algebraically closed field with characteristic zero.Affine varieties are important in algebraic geometry. J.-P. Serre introduced sheaf and cohomology techniques to algebraic geometry and discovered his well-known cohomology criterion ([1], [2, Chapter 2, Theorem 1.1]): a variety (or a Noetherian scheme) Y is an affine variety if and only if for all coherent sheaves F on Y and all i>0, Hi(Y,F)=0. Goodman and Hartshorne proved that Y is an affine variety if and only if Y contains no complete curves and the dimension h1(Y,F) of the linear space H1(Y,F) is bounded for all coherent sheaves F on Y [3]. Let X be the completion of Y. In 1969, Goodman also proved that Y is affine if and only if after suitable blowing up of the closed subvariety on the boundary X-Y, the new boundary X′-Y is a support of an ample divisor, where X′→X is the blowing up with center in X-Y ([4], [2, Chapter 2, Theorem 6.1]). For any quasiprojective variety Y, we may assume that the boundary X-Y is the support of an effective divisor D with simple normal crossings by blowing up the closed subvariety in X-Y. Y is affine if D is ample. So, if we can show the ampleness of D, then Y is affine. There are two important criteria for ampleness according to Nakai-Moishezon and Kleiman ([5], [6, Chapter 1, Section 1.5]). Another sufficient condition is that if  Y contains no complete curves and the linear system |nD| is base point free, then Y is affine [2, Chapter 2, Page 64]. Therefore, we can apply base point free theorem if we know the numerical condition of D [6, Chapter 3, Page 75, Theorem 3.3]. Neeman proved that if Y is a quasicompact Zariski open subset of an affine scheme SpecA, then Y is affine if and only if Hi(Y,𝒪Y)=0 for all i>0 [7]. The significance of Neeman’s theorem is that it is not assumed that the ring A is Noetherian.In [8], we show that if a quasiprojective variety Y is Stein, Hi(Y,𝒪Y)=0 for all i>0, and Y has d=dimX algebraically independent nonconstant regular functions, then Y is an affine variety. In this note, we give a new criterion for affineness.Theorem 1.
An irreducible quasiprojective varietyY of dimension d≥1 is an affine variety if and only if for all i>0, Hi(Y,𝒪Y)=0, and Hi(Y,𝒪Y(-Z))=0, where H is any hypersurface with sufficiently large degree and Z=H∩Y.By Cartan’s Theorem B, an analytic varietyY is a Stein variety if and only if for all analytic coherent sheaf F on Y, Hi(Y,F)=0 for all i>0. Since an algebraic affine variety over ℂ is a Stein variety [2, Page 232], we have the following application.Corollary 2.
An irreducible quasiprojective varietyY of dimension d≥1 over ℂ is a Stein variety if for all i>0, Hi(Y,𝒪Y)=0, and Hi(Y,𝒪Y(-Z))=0, where H is any hypersurface with sufficiently large degree, Z=H∩Y, and all sheaves are algebraic.Notice that in Theorem1, the two vanishing conditions imply that any hypersurface section of Y is an affine variety. On the other hand, if every hypersurface section of Y is affine, Y may not be affine (Example 12).IfY is an affine variety, then the ring Γ(Y,𝒪Y) is finitely generated [9, Page 20]. However, in our proof of Theorem 1, we do not directly check the finitely generated property of this ring, which is very hard in general. And the question of a quasiprojective variety Y to be affine is different from the behavior of the boundary divisor D, in particular, the numerical condition of D like nefness and finitely generated property of the graded ring
(1)⨁n=0∞H0(X,𝒪X(nD)).The reason is that(2)Γ(Y,𝒪Y)≠⨁n=0∞H0(X,𝒪X(nD)).We will give two examples to demonstrate this difference in Section3. One example (due to Zariski) is an affine surface Y=X-D such that the the corresponding graded ring ⨁n=0∞H0(X,𝒪X(nD)) is not finitely generated for an effective divisor D. The other example [2, Page 232] is a surface Y=X-D such that
(3)H0(Y,𝒪Y)=H0(X,𝒪X(nD))=ℂ.A necessary condition for the affineness ofY with dimension d is that Y has plenty of nonconstant regular functions. More precisely, Y has d=dimY algebraically independent nonconstant regular functions. This means that the corresponding effective boundary divisor D must be big, that is,
(4)h0(X,𝒪X(nD))≥and
for some positive number a and n≫0. So, this surface Y is not affine but ⨁n=0∞H0(X,𝒪X(nD)) is finitely generated.We will prove Theorem1 in Section 2 and give examples in Section 3.


## 2. Proof of the Theorem

Recall our notation:Y is an open subset of a projective variety X with dimension d≥1 and D is the effective boundary divisor with support X-Y. We may assume that D has simple normal crossings by further blowing up suitable closed subvariety of X-Y.In the following lemmas,Y is irreducible and satisfies Hi(Y,𝒪Y)=0 and Hi(Y,𝒪Y(-Z))=0 for all i>0 and Z=H∩Y, where H is any hypersurface with sufficiently large degree in the ambient projective space containing X.Lemma 3.
LetH be a hypersurface in Theorem 1, Z=H∩Y, then Z satisfies the same vanishing conditions: Hi(Z,𝒪Z)=0 and Hi(Z,𝒪Z(-Z′))=0 for all i>0, where Z′ is any hypersurface section on Y with sufficiently large degree.Proof.
There is a short exact sequence,(5)0⟶𝒪Y(-Z)⟶𝒪Y⟶𝒪Z⟶0,
where Z is considered as a Cartier divisor on Y. By the assumption, we have Hi(Y,𝒪Y)=0 and Hi(Y,𝒪Y(-Z))=0 for all i>0. The corresponding long exact sequence gives Hi(Z,𝒪Z)=0 for all i>0. Similarly, for any hypersurfaces H and H′, Z′=H′∩Y, from the short exact sequence
(6)0⟶𝒪Y(-Z-Z′)⟶𝒪Y(-Z′)⟶𝒪Z(-Z′)⟶0,
we have Hi(Z,𝒪Z(-Z′))=0 for all i>0.Lemma 4.
IfZ is a curve with H1(Z,𝒪Z(-Z′))=0 for all hypersurface sections Z′=H′∩Z with sufficiently large degree, then Z is an affine curve.Proof.
First, we assume thatZ is irreducible.
IfZ is complete, then by Riemann-Roch for singular curves [9, Page 298], we have
(7)h0(Z,𝒪Z(-Z′))-h1(Z,𝒪Z(-Z′))=degD+1-pa,
where pa is the arithmetic genus of Z and D is the divisor with support in the set Zreg of smooth points of Z given by -Z′. Choose the hypersurface H′ with sufficiently large degree such that
(8)degD+1-g<0,
then h1(Z,𝒪Z(-Z′))>0. This is a contradiction. Therefore, Z is not a complete but an affine curve [2, Page 62].
IfZ is not irreducible, then Z is still an affine curve. Assume that Z has two irreducible components Z1 and Z2, Z1≠Z2. Then, the dimension of Z1∩Z2 is at most 0.  So,
(9)H1(Z1∩Z2,𝒪Z(-Z′))=0.
By Mayer-Vietoris sequence, for the ideal sheaf 𝒪Z(-Z′), we have
(10)H1(Z,𝒪Z(-Z′))⟶H1(Z1,𝒪Z(-Z′))⨁H1(Z2,𝒪Z(-Z′))⟶H1(Z1∩Z2,𝒪Z(-Z′)),
where the first and last terms vanish. This gives
(11)H1(Z1,𝒪Z(-Z′))=H1(Z2,𝒪Z(-Z′))=0.
NowZ1 and Z2 are affine curves. If Z has more than 2 irreducible components, then by using mathematical induction, every irreducible component of Z is an affine curve. Thus Z is an affine curve.By Lemmas3 and 4 and mathematical induction, we may assume that every hypersurface section of Y is an affine variety.Lemma 5.
IfZ is an irreducible surface with Hi(Z,𝒪Z)=0 and for all i>0  Hi(Z,𝒪Z(-Z′))=0 for all hypersurface sections Z′ with sufficiently large degree, then Z is an affine surface.Proof.
By Lemma3, any hypersurface section A with sufficiently large degree on Z satisfies the same vanishing condition. So, A is an affine curve by Lemma 4. Since A is closed in Z, Z is not complete.
LetC be an irreducible curve on Z, then we may choose A such that A is irreducible [10] and C∩A contains more than two points [10]. Let P1 and P2 be two distinct points on C∩A, then there is a regular function ϕ on A such that ϕ(P1)≠ϕ(P2) since A is an affine curve. From the exact sequence
(12)0⟶𝒪Z(-A)⟶𝒪Z⟶𝒪A⟶0,
and H1(Z,𝒪Z(-A))=0, we have a surjective map from H0(Z,𝒪Z) to H0(A,𝒪A). Lift ϕ from A to Z, we have a regular function on Z such that it is not a constant on C. By Goodman and Hartshorne’s theorem [3], Z is a quasiaffine variety. By Neeman's theorem [7], Z is an affine surface.LetX be an irreducible normal complete variety and let D be a Cartier divisor on X. If H0(X,𝒪X(nD))=0 for all n>0, then we define the D-dimension κ(D,X) to be -∞. Otherwise, we define
(13)κ(D,X)=tr·degk⨁n≥0H0(X,𝒪X(nD))-1.IfX is not normal, we define κ(D,X)=κ(π*D,X′), where π:X′→X is the normalization. From the definition, we see that if D is an effective divisor, then 0≤κ(D,X)≤d, where d is the dimension of X. An effective divisor D is defined to be big if κ(D,X)=d.Lemma 6.
Letd=dimX>2 and H be a hypersurface such that Z-=H∩X is irreducible, then D is a big divisor on Z-.Proof.
LetZ=H∩Y be the open irreducible hypersurface section, then it satisfies the same condition in Theorem 1 by the above lemmas. We may assume that Z is an affine variety by Lemmas 4 and 5 and inductive assumption. So, the closure Z- in X has d-1 algebraically independent nonconstant rational functions which are regular on Z. This implies that D is a big divisor on Z-=X∩H, that is, κ(D|Z-,Z-)=d-1.Lemma 7.
Y has no complete curves.Proof.
IfY has an irreducible complete curve C, choose a hypersurface H such that Z=H∩Y is irreducible [10] and H intersects C at more than 2 distinct points. Let P1,P2∈C∩H, P1≠P2. By Lemmas 4 and 5, we may assume that any irreducible hypersurface section of dimension d-1 with sufficiently large degree is an affine variety. By inductive assumption, Z is an affine variety. So, there is a regular function ϕ∈H0(Z,𝒪Z) such that ϕ(P1)≠ϕ(P2). Since
(14)0⟶H0(Y,𝒪Y(-Z))⟶H0(Y,𝒪Y)⟶H0(Z,𝒪Z)⟶0,
we can lift ϕ to Y. So there is a regular function f on Y such that f|C is not a constant. This is not possible since C is complete. The contradiction shows that Y has no complete curves.By Lemma4 and inductive assumption, if Z is not irreducible, the proof still works since Z is affine.Lemma 8.
For any irreducible curveC on Y, there is a regular function f on Y such that f is not a constant on C.Proof.
LetC- be an irreducible complete curve in X containing C. Then, C--C is a finite set and a general hypersurface does not contain any point in C--C. Let H be a hypersurface away from C--C such that H intersects C at more than two distinct points and Z=H∩Y is irreducible [10]. Then, Z is an affine variety and H∩C-=Z∩C. Let P1 and P2 be two distinct points in Z∩C, then there is a regular function ϕ on Z such that ϕ(P1)≠ϕ(P2). Lift ϕ to Y, we find a regular function f on Y such that f|Z=ϕ and f|C is not a constant.Theorem 9.
An irreducible quasiprojective varietyY is an affine variety if and only if Hi(Y,𝒪Y)=0 and Hi(Y,𝒪Y(-Z))=0 for all i>0, Z=H∩Y, where H is any hypersurface with sufficiently large degree.Proof.
By Serre’s criterion, one direction is trivial: ifY is affine, then it satisfies Hi(Y,𝒪Y)=0 and Hi(Y,𝒪Y(-Z))=0 for all i>0 and Z=H∩Y, where H is any hypersurface.
We assume thatY satisfies Hi(Y,𝒪Y)=0 and Hi(Y,𝒪Y(-Z))=0 for all i>0.
By Lemma8 and Goodman and Hartshorne’s theorem [3], Y is a quasiaffine variety since for every irreducible curve C on Y, there is a global regular function f on Y such that f|Y is not a constant. By Neeman’s result [7], a quasiaffine variety Y is affine if and only if Hi(Y,𝒪Y)=0 for all i>0. So, Y is an affine variety.IfY is not irreducible, then Theorem 9 still holds since the proof works by Lemma 4 and mathematical induction.Corollary 10.
An irreducible quasiprojective varietyY of dimension d≥1 over ℂ is a Stein variety if for all i>0, Z=H∩Y, Hi(Y,𝒪Y)=0, and Hi(Y,𝒪Y(-Z))=0, where H is any hypersurface with sufficiently large degree, Z=H∩Y and all sheaves are algebraic.


## 3. Examples

AgainY is an irreducible open variety contained in a projective variety X such that Y=X-D, where D is an effective boundary divisor with support X-Y. In this section, we assume that the ground field is ℂ.Example 11.
There is an affine surfaceY such that the graded ring
(15)⨁n=0∞H0(X,𝒪X(nD))
is not finitely generated. This example is according to Zariski [11, Pages 562–564].
LetC be a smooth curve of degree 3 in ℙ2. Let Λ be a divisor class cut out on C by a curve of degree 4 in ℙ2. There exist 12 distinct points p1,p2,…,p12 on C such that
(16)m(p1+p2+⋯+p12)∉mΛ
for all positive integers m. Let X be a surface obtained by blowing up ℙ2 at these 12 points p1,p2,…,p12. Let C- be the strict transformation of C (i.e., the closure of the inverse image of C-{p1,p2,…,p12} in X). Let L be a line not passing through any point pi in these 12 points. Let L- be the strict transform of L. Then, the complete linear system
(17)|m(C-+L-)|
has a fixed locus C- for all m≥1 and
(18)|mC-+(m-1)L-|
has no fixed components and is base point free. By Nakai-Moishenzon’s ampleness criterion [9, Chapter V, Section 1], the divisor
(19)mC-+(m-1)L-
is ample. Hence, the complement Y=X-(mC-+(m-1)L-) is affine but the graded ring
(20)R=⨁m=0∞H0(X,𝒪X(m(C-+L-)))
is not finitely generated.Example 12.
A nonaffine surfaceY such that the graded ring
(21)⨁n=0∞H0(X,𝒪X(nD))
is finitely generated.LetC be an elliptic curve and E the unique nonsplit extension of 𝒪C by itself. Let X=ℙC(E) and D be the canonical section, then Y=X-D is not affine and H0(X,𝒪X(nD))=ℂ [2, Page 232]. So,
(22)⨁n=0∞H0(X,𝒪X(nD))
is finitely generated.Since the surfaceY has no complete curves, any hypersurface section of Y is an affine curve. But Y is not affine since H0(Y,𝒪Y)=ℂ [2, Page 232]. It shows that if every hypersurface section of Y is affine, Y may not be affine.The above two examples demonstrate that the affineness ofY and the finitely generated property of the graded ring
(23)⨁n=0∞H0(X,𝒪X(nD))
are different in nature. The reason is that
(24)Γ(Y,𝒪Y)≠⨁n=0∞H0(X,𝒪X(nD)).In fact, we have the following.Lemma 13 (see [3]).
LetV be a scheme and D be an effective Cartier divisor on V. Let U=V-SuppD and F be any coherent sheaf on V, then, for every i≥0,
(25)limn→Hi(V,F⊗𝒪(nD))≅Hi(U,F|U).So, we have(26)Γ(Y,𝒪Y)≅limn→H0(X,𝒪X(nD)).The direct limit is the quotient of the direct sum and its subring, so it is much “smaller” than direct sum [12, Chapter II, Section 10]. And even though Y is affine, the boundary divisor can be very bad. For example, D may not be nef. It is easy to see this by blowing up ℙ2 at a point. Let L be a line in ℙ2, let O be a point on L. Let π:X→ℙ2 be the blow up of ℙ2 at O. Let E be the exceptional divisor and D=π-1(L)+mE, where π-1(L) is the strict transform of L and m is a large positive integer. Then, D·E=1-m<0 [2, Chapter V, Corollary 3.7]. Therefore, D is not nef.Example 14.
If a smooth threefoldY such that Y contains no complete curves, then Hi(Y,ΩYj)=0 for all i>0 and j≥0 but is not affine.LetCt be a smooth projective elliptic curve defined by y2=x(x-1)(x-t), t≠0,1. Let Z be the elliptic surface defined by the same equation, then we have surjective morphism from Z to C=ℂ-{0,1} such that for every t∈C, the fiber f-1(t)=Ct. In [13], we proved that there is a rank 2 vector bundle E on Z such that when restricted to Ct, E|Ct=Et is the unique nonsplit extension of 𝒪Ct by 𝒪Ct, where f is the morphism from Z to C. We also proved that there is a divisor D on X=ℙZ(E) such that when restricted to Xt=ℙCt(Et), D|Xt=Dt is the canonical section of Xt. Let Y=X-D, we have Hi(Y,ΩYj)=0 for all i>0 and j≥0. We know that this threefold Y contains no complete curves [13] and κ(D,X)=1. So, Y is not affine.Example 15.
A surfaceY without complete curves such that κ(D,X)=2 but is not affine.Remove a lineL from ℙ2, then we have ℂ2=ℙ2-L. Remove the origin O from ℂ2, let Y=ℂ2-{O}. Then, Y is not affine since the boundary is not connected [2, Chapter II, Section 3 and Section 6]. Blow up ℙ2 with center O, let E be the exceptional divisor and π:X→ℙ2 be the blowup. Let D=π-1(L)+E, where π-1(L) is the strict transformation of L. Then by Iitaka’s result, on X, κ(D,X)=2 and X-D≅Y has no complete curves, but Y is not affine.Lemma 16 (see [6]).
Letπ:X→Z be a surjective map between projective varieties, X smooth, Z normal. Let F be the geometric generic fiber of π and assume that F is connected. The following two statements are equivalent:(i)
Riπ*𝒪X=0 for all i>0;(ii)
Z has rational singularities and Hi(F,𝒪F)=0 for all i>0.Example 17.
A smooth varietyY of dimension d≥1 with Hi(Y,𝒪Y)=0 for all i>0 and κ(D,X)=d but is not affine.LetX be the smooth projective variety obtained by blowing up a point O in ℙd. Let π:X→ℙd be the blowup. Let H be a hyperplane not passing through O. Let D=π*(H) and Y=X-D. Then, κ(D,X)=d [14, Chapter 2, Theorem 5.13]. Let E be the exceptional divisor on X, then E≅ℙd-1. So, Hi(E,𝒪E)=0 for all i>0. By Lemma 16, we have Riπ*𝒪X=0 for all i>0.LetU=ℙd-H, then U≅ℂd. For the global sections on affine space U, we have [9, Chapter III, Proposition 8.1, 8.5 and Chapter II, Proposition 5.1(d)]
(27)0=Riπ*𝒪X(U)=Hi(Y,𝒪Y),
for all i>0.It is obvious thatY is not affine since it contains a projective space ℙd-1.Example 18.
A threefoldY satisfies the following three conditions but is not affine:(1)
Y contains no complete curves;(2)
the boundaryX-Y is connected;(3)
κ(D,X)=3.LetH be a hyperplane in ℙ3. Let L be a line not contained in H. Blow up ℙ3 along L, let π:X→ℙ3 be the blowup. Define a divisor D on X such that D=π-1H+E, where E is the exceptional divisor on X. Let Y=X-D, then Y≅ℙ3-H-L.It is easy to see that the above three conditions are satisfied.Y is an open subset of ℂ3 and ℂ3-Y is a line, which is not of codimension 1. So, Y is not affine [2, 7].Example 19.
A quasiprojective varietyY with a surjective morphism f:Y→U such that U is affine, a general fiber is affine and Hi(Y,𝒪Y)=0 for all i>0, but Y is not affine.LetY, U be the varieties defined in Example 17. Then the fiber space π:Y→U satisfied the above requirements. Y is not affine because it has a projective space ℙd-1.



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