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I am wondering because Milne here in Proposition 10.1, page 42, takes any abelian subvariety $B$ of an abelian variety $A$, with $0\neq B\neq A$, then he takes an ample line bundle $\mathcal{L}$ on $A$ and he says that $\mathcal{L}|B$ is ample because of his result 6.6b (page 31), which says that the restriction of an ample line bundle to a closed subvariety is again ample.

Hence I suppose that $B$ is closed, but we didn't make any assumption about $B$. So every abelian subvariety of an abelian variety is closed, right? Or am I missing some hypotheses in this proof?

Thanks!

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  • $\begingroup$ Welcome Math.SE! Take the tour to get familiar with this site. If you receive useful answers, consider accepting one. $\endgroup$ – Shailesh Jun 14 '16 at 8:57
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    $\begingroup$ Abelian varieties are proper over the base field. The image of a morphism from a proper variety to any other variety is always closed. $\endgroup$ – Mohan Jun 14 '16 at 12:13

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