# Does vanishing of cohomology of locally free sheaves imply affiness of scheme

We have Serre criterion of affiness of a scheme which states that if a quasi compact scheme has higher cohomology vanishing for all the quasi coherent sheaves,then the scheme is affine. I wonder whether we have similar statement for locally free sheaves as following:

Let $X$ be a noetherian scheme,let $F$ be arbitrary locally free sheaf on $X$,if higher cohomology of $F$ vanishing(for $i\geq 1$),then $X$ is affine scheme.Is this statement true?

For $X$ be quasi compact scheme,I think it is not true,but for noetherian scheme,I do not know Maybe it is a stupid question.

• IF $X$ is regular and has enough locally frees one can take syzygies, and use long exact sequence of cohomology to show that higher cohomology vanish for all coherent sheaves, which would be enough. – Hailong Dao Jan 4 '14 at 4:04

There are many versions of Serre's criterion for affineness. One version states that for every quasi-compact, quasi-separated scheme $X$, $X$ is affine if and only if $H^1(X,\mathcal{F})$ vanishes for every quasi-coherent sheaf $\mathcal{F}$ that is locally finitely generated. In particular, if $X$ is Noetherian, then $X$ is affine if and only if $H^1(X,\mathcal{F})$ vanishes for every coherent sheaf.
For this criterion, it does not suffice to consider only locally free sheaves (of finite rank). Let $n>1$ be any integer. Let $X$ be the quasi-compact, quasi-separated, yet non-separated scheme obtained by glueing two copies, $X_1$ and $X_2$, of $\mathbb{A}^n$ along the common open $X_{1,2} = \mathbb{A}^n\setminus\{0\}$. There is a unique morphism of schemes, $f:X\to \mathbb{A}^n$ that restricts to the identity on $X_1$ and $X_2$. Using the S2 property, every locally free sheaf on $X$ is of the form $f^*E$. Thus, by the Quillen-Suslin theorem, every locally free sheaf on $X$ is a direct sum of copies of the structure sheaf. By straightforward computation, $H^1(X,\mathcal{O}_X)$ vanishes. Yet $X$ is not affine, since $X$ is not separated.
This is an incomplete answer, which shows that finding a counter-example would be quite difficult: the result holds if your scheme is divisorial and of finite dimension. Indeed the first hypothesis guarantees that for any coherent $\mathcal{F}$, there is an exact sequence $0\rightarrow \mathcal{K}\rightarrow E\rightarrow \mathcal{F}$ with $E$ locally free. Then $H^{i}(X,\mathcal{F})\cong H^{i+1}(X,\mathcal{K})$; applying this again to $\mathcal{K}$ and going on, you'll arrive eventually to $H^{i}(X,\mathcal{F})\cong H^{N}(X,\mathcal{G})$ with $\mathcal{G}$ coherent and $N>\dim(X)$, hence $H^{i}(X,\mathcal{F})=0$.