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Let $A/k$ be an abelian variety over a number field $k$ with a polarization of minimal degree $d>1$. (Assume all Tate-Shafarevich groups to be finite.)

What can one say about the order of $\mathrm{III}(A/k)$ in terms of being a multiple of a square?

Is it true that the order of $\mathrm{III}(A/k)$ equals $km^2$ for some $k$ dividing $2d$? If yes, why? Is this even true if $A/k$ is not isogenous to a principally polarized abelian variety?

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    $\begingroup$ I think this is true. See the introduction of William Stein's article "Shafarevich-Tate Groups of Nonsquare Order". He mentions a theorem by Tate and Flach which I think answers your question. The link is modular.math.washington.edu/papers/nonsquaresha/final2.pdf $\endgroup$
    – François Brunault
    Commented Feb 1, 2011 at 17:03
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    $\begingroup$ See also the previous MO question mathoverflow.net/questions/9924/… $\endgroup$
    – François Brunault
    Commented Feb 1, 2011 at 17:04
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    $\begingroup$ Thank you for your very fast answer. Indeed, the theorem of Tate and Flach (and maybe Cassels) answers the question. They state that if $\mathrm{III}(A/k)[p^\infty]$ is finite and p is an odd prime not dividing the degree of any polarization, then $\sharp \mathrm{III}(A/k)[p^\infty]$ is a square. On the other hand, assuming $\mathrm{III}(A/k)[p^\infty]$ finite, it's cardinality is $p^m$, for some $m \geq 0$, so primes in the prime factorization of $\sharp \mathrm{III}(A/k)$ can only possibly have odd exponent if $p=2$ or $p$ divides $d$, which proves the statement. $\endgroup$
    – Stefan Keil
    Commented Feb 9, 2011 at 10:34

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