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Jul 2, 2015 at 9:15 comment added ACL My mistake! Sorry for the confusion and thank you for your answer and your reply to my comment.
Jul 1, 2015 at 12:39 comment added Yuri Zarhin Actually, $A^4$ does not have to be principally polarized. In fact, it does not have to be isomorphic to its dual. (As an example, you may take an abelian surface $A$ over an algebraically closed field $K$ with $\End(A)=\Z$ and such that $\Hom(A,A^t)$ is generated by the polarization $\lambda: A \to A^t$ with $\ker(\lambda)$ being a product of two cyclic groups of prime order $\ell \ne char(K)$.) It is $(A \times A^t)^4$, which is always principally polarized.
Jul 1, 2015 at 8:42 comment added ACL It seems that one can also combine Joe Silverman's answer with your observation that $A^4$ has a principal polarization.
Jun 4, 2015 at 8:32 vote accept David84
Jun 4, 2015 at 7:35 history edited Yuri Zarhin CC BY-SA 3.0
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Jun 4, 2015 at 7:27 history answered Yuri Zarhin CC BY-SA 3.0