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Jul 2, 2015 at 9:17 comment added ACL @JoeSilverman: Unfortunately, I made a confusion and (see Zarhin's comment above), it is $(A\times A^{\vee })^4$ which is principally polarized.
Jul 1, 2015 at 10:58 comment added Joe Silverman @ACL Very nice!
Jul 1, 2015 at 8:41 comment added ACL But one knows (“Zarhin's trick”) that $A^4$ has a principal polarization. Since $K(A[n])=K(A^4[n])$, this implies that the result holds in general!
Jun 4, 2015 at 7:05 vote accept David84
Jun 4, 2015 at 8:32
Jun 4, 2015 at 2:52 history answered Joe Silverman CC BY-SA 3.0