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.
Best Answer
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.