A New Criterion for Affineness

We show that an irreducible quasiprojective variety Y of dimension d ≥ 1 defined over an algebraically closed field with characteristic zero is an affine variety if and only if H(Y,O Y ) = 0 and H (Y,O 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.


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)  is an affine variety if and only if for all coherent sheaves  on  and all  > 0,   (, ) = 0. Goodman and Hartshorne proved that  is an affine variety if and only if  contains no complete curves and the dimension ℎ 1 (, ) of the linear space  1 (, ) is bounded for all coherent sheaves  on  [3].Let  be the completion of .In 1969, Goodman also proved that  is affine if and only if after suitable blowing up of the closed subvariety on the boundary  − , the new boundary   −  is a support of an ample divisor, where   →  is the blowing up with center in  −  ([4], [2, Chapter 2, Theorem 6.1]).For any quasiprojective variety , we may assume that the boundary  −  is the support of an effective divisor  with simple normal crossings by blowing up the closed subvariety in  − . is affine if  is ample.So, if we can show the ampleness of , then  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  contains no complete curves and the linear system || is base point free, then  is affine [2, Chapter 2, Page 64].Therefore, we can apply base point free theorem if we know the numerical condition of  [6, Chapter 3, Page 75, Theorem 3.3].Neeman proved that if  is a quasicompact Zariski open subset of an affine scheme Spec, then  is affine if and only if   (, O  ) = 0 for all  > 0 [7].The significance of Neeman's theorem is that it is not assumed that the ring  is Noetherian.
In [8], we show that if a quasiprojective variety  is Stein,   (, O  ) = 0 for all  > 0, and  has  = dim  algebraically independent nonconstant regular functions, then  is an affine variety.In this note, we give a new criterion for affineness.Notice that in Theorem 1, the two vanishing conditions imply that any hypersurface section of  is an affine variety.On the other hand, if every hypersurface section of  is affine,  may not be affine (Example 12).

Theorem 1. An irreducible quasiprojective variety 𝑌 of dimension 𝑑 ≥ 1 is an affine variety if and only if for all
If  is an affine variety, then the ring Γ(, O  ) 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  to be affine is different from the behavior of the boundary divisor , in particular, the numerical condition of  like nefness and finitely generated property of the graded ring The reason is that We will give two examples to demonstrate this difference in Section 3. One example (due to Zariski) is an affine surface  =  −  such that the the corresponding graded ring ⨁ ∞ =0  0 (, O  ()) is not finitely generated for an effective divisor .The other example [2, Page 232] is a surface  =  −  such that A necessary condition for the affineness of  with dimension  is that  has plenty of nonconstant regular functions.More precisely,  has  = dim  algebraically independent nonconstant regular functions.This means that the corresponding effective boundary divisor  must be big, that is, for some positive number  and  ≫ 0. So, this surface  is not affine but ⨁ ∞ =0  0 (, O  ()) is finitely generated.We will prove Theorem 1 in Section 2 and give examples in Section 3.

Proof of the Theorem
Recall our notation:  is an open subset of a projective variety  with dimension  ≥ 1 and  is the effective boundary divisor with support  − .We may assume that  has simple normal crossings by further blowing up suitable closed subvariety of  − .
In the following lemmas,  is irreducible and satisfies   (, O  ) = 0 and   (, O  (−)) = 0 for all  > 0 and  =  ∩ , where  is any hypersurface with sufficiently large degree in the ambient projective space containing .Lemma 3. Let  be a hypersurface in Theorem 1,  =  ∩ , then  satisfies the same vanishing conditions:   (, O  ) = 0 and   (, O  (−  )) = 0 for all  > 0, where   is any hypersurface section on  with sufficiently large degree.
Proof.There is a short exact sequence, where  is considered as a Cartier divisor on .By the assumption, we have   (, O  ) = 0 and   (, O  (−)) = 0 for all  > 0. The corresponding long exact sequence gives   (, O  ) = 0 for all  > 0. Similarly, for any hypersurfaces  and   ,   =   ∩ , from the short exact sequence we have   (, O  (−  )) = 0 for all  > 0.
Lemma 4. If  is a curve with  1 (, O  (−  )) = 0 for all hypersurface sections   =   ∩ with sufficiently large degree, then  is an affine curve.
Proof.First, we assume that  is irreducible.
If  is complete, then by Riemann-Roch for singular curves [9, Page 298], we have where   is the arithmetic genus of  and  is the divisor with support in the set  reg of smooth points of  given by −  .Choose the hypersurface   with sufficiently large degree such that then ℎ 1 (, O  (−  )) > 0. This is a contradiction.Therefore,  is not a complete but an affine curve [2, Page 62].
If  is not irreducible, then  is still an affine curve.Assume that  has two irreducible components  1 and  2 ,  1 ̸ =  2 .Then, the dimension of  1 ∩  2 is at most 0. So, By Mayer-Vietoris sequence, for the ideal sheaf O  (−  ), we have where the first and last terms vanish.This gives Now  1 and  2 are affine curves.If  has more than 2 irreducible components, then by using mathematical induction, every irreducible component of  is an affine curve.Thus  is an affine curve.
By Lemmas 3 and 4 and mathematical induction, we may assume that every hypersurface section of  is an affine variety.
ISRN Algebra 3 Lemma 5.If  is an irreducible surface with   (, O  ) = 0 and for all  > 0   (, O  (−  )) = 0 for all hypersurface sections   with sufficiently large degree, then  is an affine surface.
Proof.By Lemma 3, any hypersurface section  with sufficiently large degree on  satisfies the same vanishing condition.So,  is an affine curve by Lemma 4. Since  is closed in ,  is not complete.
Let  be an irreducible curve on , then we may choose  such that  is irreducible [10] and  ∩  contains more than two points [10].Let  1 and  2 be two distinct points on  ∩ , then there is a regular function  on  such that ( 1 ) ̸ = ( 2 ) since  is an affine curve.From the exact sequence and  1 (, O  (−)) = 0, we have a surjective map from  0 (, O  ) to  0 (, O  ).Lift  from  to , we have a regular function on  such that it is not a constant on .By Goodman and Hartshorne's theorem [3],  is a quasiaffine variety.By Neeman's theorem [7],  is an affine surface.
Let  be an irreducible normal complete variety and let  be a Cartier divisor on .If  0 (, O  ()) = 0 for all  > 0, then we define the -dimension (, ) to be −∞.Otherwise, we define If  is not normal, we define (, ) = ( * ,   ), where  :   →  is the normalization.From the definition, we see that if  is an effective divisor, then 0 ≤ (, ) ≤ , where  is the dimension of .An effective divisor  is defined to be big if (, ) = .Lemma 6.Let  = dim  > 2 and  be a hypersurface such that  =  ∩  is irreducible, then  is a big divisor on .
Proof.Let  =  ∩  be the open irreducible hypersurface section, then it satisfies the same condition in Theorem 1 by the above lemmas.We may assume that  is an affine variety by Lemmas 4 and 5 and inductive assumption.So, the closure  in  has  − 1 algebraically independent nonconstant rational functions which are regular on .This implies that  is a big divisor on  =  ∩ , that is, (|  , ) =  − 1.

Lemma 7. 𝑌 has no complete curves.
Proof.If  has an irreducible complete curve , choose a hypersurface  such that  =  ∩  is irreducible [10] and  intersects  at more than 2 distinct points.Let  1 ,  2 ∈  ∩ ,  1 ̸ =  2 .By Lemmas 4 and 5, we may assume that any irreducible hypersurface section of dimension  − 1 with sufficiently large degree is an affine variety.By inductive assumption,  is an affine variety.So, there is a regular function  ∈  0 (, O  ) such that ( 1 ) ̸ = ( 2 ).Since we can lift  to .So there is a regular function  on  such that |  is not a constant.This is not possible since  is complete.The contradiction shows that  has no complete curves.
By Lemma 4 and inductive assumption, if  is not irreducible, the proof still works since  is affine.Lemma 8.For any irreducible curve  on , there is a regular function  on  such that  is not a constant on .
Proof.Let  be an irreducible complete curve in  containing .Then,  −  is a finite set and a general hypersurface does not contain any point in  − .Let  be a hypersurface away from  −  such that  intersects  at more than two distinct points and  =  ∩  is irreducible [10].Then,  is an affine variety and  ∩  =  ∩ .Let  1 and  2 be two distinct points in  ∩ , then there is a regular function  on  such that ( 1 ) ̸ = ( 2 ).Lift  to , we find a regular function  on  such that |  =  and |  is not a constant.Theorem 9.An irreducible quasiprojective variety  is an affine variety if and only if   (, O  ) = 0 and   (, O  (−)) = 0 for all  > 0,  =  ∩ , where  is any hypersurface with sufficiently large degree.
By Lemma 8 and Goodman and Hartshorne's theorem [3],  is a quasiaffine variety since for every irreducible curve  on , there is a global regular function  on  such that |  is not a constant.By Neeman's result [7], a quasiaffine variety  is affine if and only if   (, O  ) = 0 for all  > 0. So,  is an affine variety.
If  is not irreducible, then Theorem 9 still holds since the proof works by Lemma 4 and mathematical induction.Corollary 10.An irreducible quasiprojective variety  of dimension  ≥ 1 over C is a Stein variety if for all  > 0,  =  ∩ ,   (, O  ) = 0, and   (, O  (−)) = 0, where  is any hypersurface with sufficiently large degree,  =  ∩  and all sheaves are algebraic.

Examples
Again  is an irreducible open variety contained in a projective variety  such that  =  − , where  is an effective boundary divisor with support −.In this section, we assume that the ground field is C.

ISRN Algebra
Example 11.There is an affine surface  such that the graded ring is not finitely generated.This example is according to Zariski [11,.
Let  be a smooth curve of degree 3 in P 2 .Let Λ be a divisor class cut out on  by a curve of degree 4 in P 2 .There exist 12 is not finitely generated.
Example 12.A nonaffine surface  such that the graded ring is finitely generated.
Let  be an elliptic curve and  the unique nonsplit extension of O  by itself.Let  = P  () and  be the canonical section, then  =  −  is not affine and is finitely generated.Since the surface  has no complete curves, any hypersurface section of  is an affine curve.But  is not affine since  0 (, O  ) = C [2, Page 232].It shows that if every hypersurface section of  is affine,  may not be affine.
The above two examples demonstrate that the affineness of  and the finitely generated property of the graded ring are different in nature.The reason is that In fact, we have the following.
Lemma 13 (see [3]).Let  be a scheme and  be an effective Cartier divisor on .Let  = −Supp and  be any coherent sheaf on , then, for every  ≥ 0, So, we have 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  is affine, the boundary divisor can be very bad.For example,  may not be nef.It is easy to see this by blowing up P 2 at a point.Let  be a line in P 2 , let  be a point on .Let  :  → P 2 be the blow up of P 2 at .Let  be the exceptional divisor and  =  −1 () + , where  −1 () is the strict transform of  and  is a large positive integer.Then,  ⋅  = 1 −  < 0 [2, Chapter V, Corollary 3.7].Therefore,  is not nef.
Example 14.If a smooth threefold  such that  contains no complete curves, then   (, Ω   ) = 0 for all  > 0 and  ≥ 0 but is not affine.
Let   be a smooth projective elliptic curve defined by  2 = ( − 1)( − ),  ̸ = 0, 1.Let  be the elliptic surface defined by the same equation, then we have surjective morphism from  to  = C − {0, 1} such that for every  ∈ , the fiber  −1 () =   .In [13], we proved that there is a rank 2 vector bundle  on  such that when restricted to   , |   =   is the unique nonsplit extension of O   by O   , where  is the morphism from  to .We also proved that there is a divisor  on  = P  () such that when restricted to   = P   (  ), |   =   is the canonical section of   .Let  =  − , we have   (, Ω   ) = 0 for all  > 0 and  ≥ 0. We know that this threefold  contains no complete curves [13] and (, ) = 1.So,  is not affine.
Example 15.A surface  without complete curves such that (, ) = 2 but is not affine.
Remove a line  from P 2 , then we have C 2 = P 2 − .Remove the origin  from C 2 , let  = C 2 − {}.Then,  is

Corollary 2 .
> 0,   (, O  ) = 0, and   (, O  (−)) = 0, where  is any hypersurface with sufficiently large degree and  =  ∩ .By Cartan's Theorem B, an analytic variety  is a Stein variety if and only if for all analytic coherent sheaf  on ,   (, ) = 0 for all  > 0. Since an algebraic affine variety over C is a Stein variety [2, Page 232], we have the following application.An irreducible quasiprojective variety  of dimension  ≥ 1 over C is a Stein variety if for all  > 0,   (, O  ) = 0, and   (, O  (−)) = 0, where  is any hypersurface with sufficiently large degree,  =  ∩ , and all sheaves are algebraic.