## isomorphism of abelian varieties

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Let $A, B, C$ and $D$ be abelian varieties (over $\mathbb{C}$) such that $A \times B \cong C \times D$, and $A \cong C$. From the irreducibility of abelian varieties, we can say that $B$ and $D$ are isogeneous. But do we actually have $B \cong D$?

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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. – Pete L. Clark Apr 9 2010 at 0:21
I meant the complete reducibility theorem.  Thanks. – Tuan Apr 9 2010 at 3:23
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.) – Torsten Ekedahl Apr 9 2010 at 4:04
@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. – Bjorn Poonen Apr 10 2010 at 1:02
@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 $\mathbb Q$.) – Torsten Ekedahl Apr 11 2010 at 18:39
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## 1 Answer

This is false even for elliptic curves over $\mathbb{C}$. This was proved by T. Shioda in "Some remarks on abelian varieties" J. Fac. Sci. Univ. Tokyo Sect. IA Math. 24 (1977), no. 1, 11-21, http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6164/1/jfs240102.pdf.

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Thank you.  That was very helpful! – Tuan Apr 9 2010 at 4:31
This is a very interesting paper.  I look forward to reading it carefully.   – Pete L. Clark Apr 9 2010 at 5:04
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