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Let $X$ be a closed subscheme of $\mathbb{P}^n_k$, where $k$ is a field. Let $\mathcal{O}(1)$ be the standard line bundle on projective space, and $\mathcal{O}(m)$ its tensor powers. We know that for $m \gg 0$, the natural map $H^0(\mathbb{P}^n_k, \mathcal{O}(m)) \to H^0(X, \mathcal{O}(m))$ is surjective (this is "Theorem B"). Are there general techniques for showing that this is a surjection for $m$ in a certain interval $[n_0, \infty)$?

This boils down to showing that $H^1(\mathbb{P}^n_k, \mathcal{I}(m)) = 0$ where $\mathcal{I}$ is the sheaf of ideals of $X$. Consequently, when $X$ is a divisor, I know what to do: $\mathcal{I}$ is a line bundle and the cohomology of line bundles is known. But when $X$ is not a divisor, I don't really know any techniques for showing this vanishing fact. I suspect Castelnuovo-Mumford regularity might be relevant, but I don't know how to apply it here. For instance, one can figure out how regular $\mathcal{O}_{\mathbb{P}^n}$ is, and with luck how regular $\mathcal{O}/\mathcal{I}$ is (e.g. if you know what the subvariety $X$ looks like), but I'm not sure why this implies (if it does imply) that $\mathcal{I}$ will satisfy similar regularity conditions.

(Note: this is motivated by certain homework problems I have. I'm just curious about general techniques though, and possibly references.)

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Dear Akhil, What comes out of just looking at the proof? Regards, – Matt E Sep 21 '11 at 18:51
@Matt: Could you perhaps give another hint? I thought about your suggestion and am not really sure how to make it effective (e.g. if $X$ is known). Am I missing something? – Akhil Mathew Sep 23 '11 at 3:36
Dear Akhil, My question wasn't meant to be rhetorical; rather, I just assumed that the proof itself would give something effective (if unwieldy in practice). It's been a while since I've thought about the argument in any careful way; if I do so and something comes to me, I'll report back. Regards, – Matt E Sep 24 '11 at 6:08
@Matt: Oh, OK. Actually, in class yesterday I learned that the problem is apparently very hard when $X$ is a finite subscheme, so maybe in general it shouldn't be easy either... – Akhil Mathew Sep 24 '11 at 16:36

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