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Mar 22, 2011 at 6:27 vote accept Marc Palm
Mar 21, 2011 at 18:39 comment added Dan Petersen See also mathoverflow.net/questions/12814/…
Mar 21, 2011 at 16:35 comment added Martin Brandenburg More generally, every abelian scheme is abelian in the honest sense of the word. The standard proof uses the rigidity lemma: Every morphism between abelian schemes fixing $e$ is a group morphism. In particular, $x \mapsto x^{-1}$ is a group morphism, i.e. the group is abelian.
Mar 21, 2011 at 13:23 answer added Emerton timeline score: 35
Mar 21, 2011 at 13:23 answer added Francesco Polizzi timeline score: 9
Mar 21, 2011 at 13:17 answer added Qfwfq timeline score: 3
Mar 21, 2011 at 13:04 history asked Marc Palm CC BY-SA 2.5