# Ample divisor on abelian variety

just a short question:

if one has an abelian variety $X$ over a field $k$ and an ample irreducible divisor $D$ on $X$, then why is $H^1(X-D,\mathcal O_X)$ zero?

Should it be that $X-D$ is affine? But I don't see an argument.

Thanks!

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## 1 Answer

Yes, your conjecture is correct: $X\setminus D$ is affine. Actually, this has nothing to do with abelian varieties, since we have quite generally:

Theorem Let $X$ be a complete integral scheme over the algebraically closed field $k$ (of any characteristic). Then for any ample divisor $D$ on $X$, the open subscheme $X\setminus |D|$ is affine.

Proof For some large integer $n\gt \gt 0$, $nD$ is very ample and embeds $X$ in $\mathbb P^N_k$, so that you can assume $X \subset \mathbb P^N_k$ and $|D|=|nD|=X\cap H$ for some hyperplane $H\subset \mathbb P^N_k$.
But then $X\setminus |D|=(\mathbb P^N_k \setminus H) \cap X=\mathbb A^N_k \cap X$, which is clearly affine.

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Hi Georges, this is a very satisfying answer. Just one question: why is $\mid D \mid =X \cap H$ for a hyperplane? – Veen Sep 27 '11 at 12:09
Dear @Veen,this is because in the embedding $i:X \to \mathbb P^N_k$, the bundle $\mathcal O_{\mathbb P^N_k}(1)$ gets pulled back to $L=\mathcal O_X(nD)$, so a section on $X$ of with zeros along $nD$ corresponds to a section of $\mathcal O_{\mathbb P^N_k}(1)$, with zeros along a hyperplane of $\mathbb P^N_k$. – Georges Elencwajg Sep 27 '11 at 14:27