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Though the answer is "yes" for Abelian varieties up to isogeny, the stronger statement is, perhaps, false. Does anyone know a counterexample?

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    $\begingroup$ No curve has a Jacobian that is a product as a ppav. I suppose a Jacobian might be a product not respecting polarization, but it seems unlikely to me. $\endgroup$ Mar 15, 2011 at 4:03
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    $\begingroup$ Vadik, could you explain or give a reference for "yes" in the case of Abelian varieties up to isogeny? $\endgroup$ Mar 15, 2011 at 9:09
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    $\begingroup$ The standard argument is to take a curve $C\subset A$ which is an intersection of ample divisors, then $Jac(C)$ surjects onto $A$. This map splits up to isogeny by Poincare reducibility. Milne has some articles on abelian varieties and Jacobians, it's probably in there. $\endgroup$ Mar 15, 2011 at 11:23
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    $\begingroup$ @Ben It may happen that a jacobian is isomorphic to a product of abelian varieties (notv respecting polarization). See for instance mathoverflow.net/questions/35060/…, in particular David Hansen comment and my answer. $\endgroup$ Mar 15, 2011 at 12:52
  • $\begingroup$ Thanks, Francesco, but your answer does not seem compatible with David's. Maybe he left out an isogeny? $\endgroup$ Mar 16, 2011 at 19:27

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