# Is $H^i(X,F)$ finitely generated over $\Gamma(O_X)$ if $F$ is coherent?

Suppose $\mathcal{X}$ is a smooth quasi-projective variety over $\mathbb{C}$ (I apologize if these hypotheses have little to do with the question at hand). Let $\mathcal{F}$ be a coherent sheaf on $\mathcal{X}$. Is $H^i(X,\mathcal{F})$ finitely generated over $\Gamma(O_X)$ if $\mathcal{F}$ is coherent ? This statement is simple enough that I probably would have heard it if it were true.

If it makes a difference, the statement that I really would like to understand is whether $\Gamma(O_X) \to Ext^i(\mathcal{G},\mathcal{G})$ is module finite for any coherent sheaf.

Most of the schemes that I am "friendly with" are either projective or affine. The statements are correct in those particular cases, so probably I just need to learn more examples... Thanks!

• Probably could have posted this on stackexchange since it's probably not a "research question". My apologies if it's not appropriate. Feb 25 '12 at 10:06
• Is the following true: If $X$ is a quasi-projective variety and all cohomology groups of coherent sheaves on $X$ are finitely generated, and $\Gamma(X)=k$, is then $X$ projective? Feb 25 '12 at 13:18
• @Martin: the case $\dim X=1$ is well-known, otherwise choose an embedding $X\subset\mathbb P^N$, suppose there is $x\in\bar X\setminus X$, choose a hyperplane $H$ containing $x$ such that $x\in\overline{H\cap X}$. By induction on the dimension, $H\cap X$ is projective, contradiction. Feb 25 '12 at 15:00
• Daniel: see tea.mathoverflow.net/discussion/1300/1/… I don't know how others will feel about the appropriateness of your question, but I personally am glad you asked it here. Feb 25 '12 at 15:46

This is false even for $\mathrm H^0$: take $X$ to be $\mathbb A^2 \smallsetminus \{0\}$, and as $F$ the structure sheaf of $L \smallsetminus \{0\}$, where $L$ is a line through $0$.
Counterexample: For $X=\mathbb A^2\setminus\{0\}$, one has $\Gamma(X,O_X)=\mathbb C[u,v]$, but $H^1(X,O_X)\cong\mathbb C[u^{\pm1},v^{\pm1}]/(\mathbb C[u,v^{\pm1}]+\mathbb C[u^{\pm1},v])$ is not finitely generated over $\mathbb C[u,v]$.