Timeline for Tate-Shafarevich group of non-principally polarized abelian variety
Current License: CC BY-SA 2.5
4 events
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| Feb 9, 2011 at 10:34 | comment | added | Stefan Keil | 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. | |
| Feb 1, 2011 at 17:04 | comment | added | François Brunault | See also the previous MO question mathoverflow.net/questions/9924/… | |
| Feb 1, 2011 at 17:03 | comment | added | François Brunault | 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 | |
| Feb 1, 2011 at 16:46 | history | asked | Stefan Keil | CC BY-SA 2.5 |