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Let $Y$ be a connected, nonsingular, positive dimensional subvariety of $\mathbb{P}^n_k$ over an algebraically closed field $k$, and let $\mathcal{I}$ be the ideal sheaf of $Y$.

Why are the followings true?

1) $\Gamma(Y,\mathcal{I}^r/\mathcal{I}^{r+1})=0$ for any $r\ge 1$. 2) $\Gamma(Y,\mathcal{O}_X/\mathcal{I}^r)=k$ for any $r\ge 1$.

Could someone explains this for me, thanks.

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2 Answers 2

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Put $\mathcal N=(\mathcal I/\mathcal I^2)^\vee$, the normal bundle to $Y$ in $X=\mathbb P^n$. The tangent bundle sequence $0\to T_Y\to (T_X)\vert_Y\to \mathcal N\to 0$ and the ampleness of $T_X$ shows that $\mathcal N$ is ample. Now $\mathcal I^r/\mathcal I^{r+1}$ is the dual of the $r$th symmetric power of $\mathcal N$, so cannot have sections.

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  • $\begingroup$ Note that Kodaira's vanishing theorem fails in positive characteristics (counterexamples are given by Mumford with normal surfaces and Raynaud with smooth surfaces). So this nice proof (and also that of J.C. Ottem) works only in characteristic 0. $\endgroup$
    – Qing Liu
    Commented Mar 2, 2011 at 13:35
  • $\begingroup$ Liu: Unless there is something I missed in the proof, no H^1 is used here; only the fact that the dual of an ample line bundle has no sections. Am I wrong? $\endgroup$
    – ACL
    Commented Mar 2, 2011 at 17:49
  • $\begingroup$ You are right. I am sorry ! $\endgroup$
    – Qing Liu
    Commented Mar 2, 2011 at 18:53
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Write for simplicity $X=\mathbb{P}^n$.

The easiest way of showing 1) is probably by noting that $\mathcal I / \mathcal I^2$ injects as a subbundle of $\mathcal O_Y(-1)^{n+1}$ (this follows from combining the conormal sequence with the Euler sequence) and so none of it's symmetric powers $S^r(\mathcal I / \mathcal I^2)=\mathcal I^r / \mathcal I^{r+1}$ can have any global sections. Now taking the cohomology sequence of $$ 0 \to \mathcal{I}^r/\mathcal{I}^{r+1}\to\mathcal{O}_X/\mathcal{I}^{r+1} \to\mathcal{O}_X/\mathcal{I}^{r} \to 0, $$shows that $\Gamma(Y,\mathcal{O}_X/\mathcal{I}^{r+1})$ injects into $\Gamma(Y,\mathcal{O}_X/\mathcal{I}^{r})$, and so by induction on $r$, we get 2).

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    $\begingroup$ Sorry, but how do you know $rY$ has no global functions? Connectivity does not imply this, because the scheme is non-reduced, so it may have global nilpotent functions. $\endgroup$
    – t3suji
    Commented Mar 1, 2011 at 14:13
  • $\begingroup$ @t3suji: JC is saying that $\Gamma(Y,\mathcal{O}_X/\mathcal{I}^{r+1})$ injects into $\Gamma(Y,\mathcal{O}_X/\mathcal{I}^{r})$ for all $r$, so by iterating this you get that $\Gamma(Y,\mathcal{O}_X/\mathcal{I}^{r})$ injects into $\Gamma(Y,\mathcal{O}_X/\mathcal{I})$. There are no nilpotents there. $\endgroup$
    – Sándor Kovács
    Commented Mar 1, 2011 at 17:54
  • $\begingroup$ ps: In other words, the nilpotents would have to show up in $\Gamma(Y,\mathcal{I}^r/\mathcal{I}^{r+1})$ for some $r$. $\endgroup$
    – Sándor Kovács
    Commented Mar 1, 2011 at 17:55
  • $\begingroup$ @Sandor Kovacs. Of course it is true that 1) and 2) are equivalent. I was objecting to the original answer (which has been edited away): the idea was to get 2) from connectivity and then derive 1). $\endgroup$
    – t3suji
    Commented Mar 1, 2011 at 18:55
  • $\begingroup$ I edited the answer above to make it more precise. Thanks to t3suji and Sandor for their comments. $\endgroup$
    – J.C. Ottem
    Commented Mar 1, 2011 at 19:46

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