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Timeline for isomorphism of abelian varieties

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Apr 13, 2010 at 4:30 comment added Fei YE @Bjorn: In the remark (3) in your paper "The Grothendieck ring of varieties is not a domain", you mentioned elliptic curve over $\mathbb{C}$. Does that mean that there exist elliptic curves $A\not\cong B$ over $\mathbb{C}$ such that $A\times A \cong B\times B$?
Apr 11, 2010 at 18:39 comment added Torsten Ekedahl @Bjorn: You are right of course, I misremembered. (In my defence, an example of non-cancellation also gives examples of zero-divisors in the Grothendieck ring, the tricky thing is to get an example over $\mathbb Q$.)
Apr 10, 2010 at 1:02 comment added Bjorn Poonen @Torsten: I think the example in my paper was of something slightly different, namely A x A = B x B with A and B not isomorphic.
Apr 9, 2010 at 4:16 vote accept Tuan
Apr 9, 2010 at 4:04 comment added Torsten Ekedahl It is false also in characteristic $0$ though there it is true for elliptic curves (see Angelo's reply). It is a question of finding an example of non-cancellation for projective modules over a suitable ring and then mirror it for abelian varieties. See Poonen, "The Grothendieck ring of varieties is not a domain" for an example. (Bjorn's example is somewhat involved as he wants an example over $\mathbb Q$. An example over $\mathbb C$ is easier to construct.)
Apr 9, 2010 at 3:44 answer added Angelo timeline score: 20
Apr 9, 2010 at 3:23 comment added Tuan I meant the complete reducibility theorem. Thanks.
Apr 9, 2010 at 0:21 comment added Pete L. Clark I think that rather than "irreducibility" you mean "complete reducibility" or "Poincare's complete reducibility theorem". As for the question itself: I seem to recall that it is famously false for supersingular abelian varieties, i.e., in positive characteristic. Over C, I might guess that it's true, but that's just a guess.
Apr 8, 2010 at 21:56 history asked Tuan CC BY-SA 2.5