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I thought that the order of the Tate-Shafarevich group should always be a square (it's also supposed to be finite, but for the purposes of this question let's assume we know this) but I don't seem to find a good explanation; Wikipedia is silent on the matter.

While I know it may be an open problem, is there a good argument pro or contra this?

(I made this a community wiki as I mistakenly thought this was an open problem)

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This should not be community wiki. As to the answer, there's more to be said... – Pete L. Clark Dec 28 '09 at 1:53
I thought is was an open problem -- I even put the tag [open-problem] originally. Once I realized it's solved, I undid the tag, but unfortunately CW cannot be undone :) You're welcome to post an answer, and if you want we can even do a new question/answer pair :) – Ilya Nikokoshev Dec 28 '09 at 2:41
It might be nice to start afresh with a new question. As to the answer, I have to give first dibs to Bjorn Poonen (and/or Michael Stoll). – Pete L. Clark Dec 28 '09 at 3:48
I'll supplement Pete's comment with this link: – David Zureick-Brown Dec 28 '09 at 7:47
up vote 32 down vote accepted

The first example of an abelian variety with nonsquare Sha was discovered in a computation by Michael Stoll in 1996. He emailed it to me and Ed Schaefer, because his calculation depended on a paper that Ed and I had written. At first none of us believed that it was what it was: instead we thought it must be due to either an error in Stoll's calculations or an error in the Poonen-Schaefer paper. Stoll and I worked together over the next few weeks to develop a theory that explained the phenomenon, and this led to the paper - that paper contains a detailed answer to your question.

To summarize a few of the key points: If the abelian variety over a global field $k$ has a principal polarization coming from a $k$-rational divisor (as is the case for every elliptic curve), then the order of Sha is a square (if finite), because it carries an alternating pairing - this is what Tate proved, generalizing Cassels' result for elliptic curves. For principally polarized abelian varieties in general, the pairing satisfies the skew-symmetry condition $\langle x,y \rangle = - \langle y,x \rangle$ but not necessarily the stronger, alternating condition $\langle x,x \rangle=0$, so all one can say is that the order of Sha is either a square or twice a square (if finite). Stoll and I gave an explicit example of a genus 2 curve over $\mathbf{Q}$ whose Jacobian had Sha isomorphic to $\mathbf{Z}/2\mathbf{Z}$ unconditionally (in particular, finiteness could be proved in this example).

If the polarization on the abelian variety is not a principal polarization, then the corresponding pairing need not be even skew-symmetric, so there is no reason to expect Sha to be even within a factor of $2$ of a square. And indeed, William Stein eventually found explicit examples and published them in the 2004 paper cited by Simon.

A final remark: Ironically, my result with Stoll quantifying the failure of Sha to be a square is used by Liu-Lorenzini-Raynaud to prove that the Brauer group $\operatorname{Br}(X)$ of a surface over a finite field is a square (if finite)!

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Brian Conrad told me that this is not always the case; Tate's paper where he claimed this was misunderstood until some counterexamples were found. William Stein has a paper "Shafarevich-Tate groups of nonsquare order'' with counterexamples; it's available online at

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Okay, that's not an open-problem them, I was wrong :) – Ilya Nikokoshev Dec 28 '09 at 1:46

Incidentally, it was thought for some time that the Brauer group $B(X)$ of a surface $X$ over a finite field could have order which was not a square. This turned out to be false : if $B(X)$ is finite, then its order is a square (Liu--Lorenzini--Raynaud, Inventiones, 2005). Conjecturally, the group $B(X)$ is always finite.

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