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May 2, 2012 at 2:19 history edited Felipe Voloch CC BY-SA 3.0
corrected mistake; edited body
Apr 27, 2012 at 15:25 comment added aglearner Felipe, thank you for giving more details! Once we know that $D$ is ample we can indeed use Kodaira vanishing to deduce $H^i(A,O(D))=0$ for $i>0$.
Apr 27, 2012 at 15:15 history edited Felipe Voloch CC BY-SA 3.0
grammar
Apr 27, 2012 at 6:56 comment added Alexander Chervov Is there some simple way to explain that theta divisor is principal ? and ample ? I cannot remember now, it seems to me there was such an argument... Riemann-Roch for abelian varieties - we can use just Hircebruch's version and we are lucky that canonical class is equal to zero for any abelian variety so we get just ch( D) which is exp(D) = \sum_i D^i /i! - so the g-dimensional component is $D^g/g! $.
Apr 27, 2012 at 2:14 history answered Felipe Voloch CC BY-SA 3.0