Let $k$ be a field. If $A$ is an abelian variety over $k$, does $A$ have a unique decomposition as a direct product of directly indecomposable abelian varieties, up to rearrangement and isomorphism of the factors?
I know that the corresponding statement in the isogeny category is true (as the isogeny category is semisimple), but that the actual category of abelian varieties is not semisimple (or even abelian).
I believe the answer is no, based on an example my friend pointed out in Orlov's paper on derived equivalences of abelian varieties. (Link, Example 4.16.) In that example, it is assumed that $A \times \hat{A} \cong B \times \hat{B}$, and the hypotheses on $A$ imply that each of the four summands is directly indecomposable. I'm assuming from the discussion there that such varieties exist (other than $B = A$ and $B = \hat{A}$ of course), but it isn't clear to me how to construct them.
Motivation: a positive answer would simplify the last step of Zarhin's proof that there are only finitely many abelian varieties of a given dimension over a given finite field--a key step in the proof of the Tate conjecture for abelian varieties over finite fields. (Edit: Tate himself used a weaker version of the finiteness statement; Zarhin's trick simplifies his proof.) A negative answer would explain the importance of passing to the isogeny category.