Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
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 – François Brunault Feb 1 '11 at 17:03
See also the previous MO question… – François Brunault Feb 1 '11 at 17:04
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. – Stefan Keil Feb 9 '11 at 10:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.