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We all know that Tate-Shafarevich group is defined as $$Ш(E/K)=\mathrm{Ker}(H^1(K,E)\mapsto \prod_{v}H^1(K_v,E))$$ for an abelian variety $A$ defined over a number field $K$, the non-trivial elements of the Tate-Shafarevich group can be thought of as the homogeneous spaces of A that have $K_v$-rational points for every place $v$ of $K$, but no $K$-rational point. And we also have the $p$-part or $n$-part of the group which occurs in the method of Descent, suppose if $\phi\colon E/K\to E'/K$ is an isogeny of elliptic curves, then you have an exact sequence, $$0\to E'(K)/\phi(E(K)) \to S^{(\phi)}(E/K) \to \text{Ш}(E/K)[\phi]\to 0$$ So one can define the $\phi$-part of Sha-group as $Ш_2(E/\mathbb{Q})=[\rm{Sel_{2}}(E/\mathbb{Q})/(E(\mathbb{Q})/\rm{2}E(\mathbb{Q})]$. We know already that $Ш_2(E/\mathbb{Q})$ can be easily proven to be finite as both $\rm{Sel_2}(E/\mathbb{Q})$ and $E(\mathbb{Q})/\rm{2}E(\mathbb{Q})$ are finite.

So my main query is :

"Is there any way to profitably write $Ш(E/\mathbb{Q})$ in terms of $Ш_m(E/\mathbb{Q})$ when we perform an infinite descent. If so that will account for decomposition on Sha."

To add some flavor to it, suppose think that $2$-descent works on elliptic curve $E$, then we obtain the $Ш_2(E/\mathbb{Q})$ and if we proceed for the "Infinite-Descent" we obtain for each $E(\mathbb{Q})/\rm{m} E(\mathbb{Q})$ ($m\ge2$) a corresponding Sha-part i.e $Ш_m(E/\mathbb{Q})$ . So can we write the group as $$Ш(E/\mathbb{Q})=\bigcup_m Ш_m(E/\mathbb{Q})$$ (that might not represent the union but some way of unifying the local-sha(not to be confused with Hasse-Weil word "local", here local seems to the one at 'm'), that may be either a Direct sum or something like that.

To be sharp

"Are there any references or previously done work for decomposing Tate-Shafarevich group ?". (Please give me the references)

Please correct me if I am in a wrong perspective of these groups and Descent method.

Thank you.

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I see a lot of improvement in both formatting which was only possible by the suggestions gave by group members, thank you one and all, now you can answer sir as I followed your suggestions in formatting and editing @Alex B. –  Iyengar Nov 10 '11 at 16:38
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1 Answer

up vote 5 down vote accepted

The Tate--Shafarevich group of $A$ is a torsion group, and so is the union of its $m$-torsion subgroups, as $m$ ranges over integers $\geq 1$. In fact the same is true of the Weil--Chatelet group of $A$ (the group of $A$-torsors), of which Sha is a subgroup (the group of locally trivial torsors).

One way to see this is to observe that any $A$-torsor becomes trivial over some finite extension of $K$ (since it has points defined over some finite extension of $A$). If the $A$-torsor $X$ has a rational point over $L$, and $L$ is a degree $n$ extension of $K$, then $n\cdot X = 0$ in the Weil--Chatelet group of $A$. This gives an upper bound on the order of $X$ as an $A$-torsor, in terms of the fields of definition of its points. (The precise relationship between these two invariants of $X$ --- its order as an $A$-torsor and the degrees of the field extensions of $K$ over which its points are defined --- is the subject of the period-index problem; see e.g.this paper of Pete Clark's for a more detailed discussion of this.)

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Neat answer sir, Thanks a lot! . But let me wait and see whether someone adds something to it, before selecting it as answer.@Matt E –  Iyengar Nov 11 '11 at 5:54
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