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.
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 $H\subset \mathbb P^N_k$.