Let $S$ be a finite set of places of a number field $k$ and let $E$ be an elliptic curve over $k$. Define the ''$S$-Tate-Shafarevich group" of $E$ to be

$$Ш(E,S) = \ker\left(H^1(k,E) \to \prod_{v \not \in S}H^1(k_v,E_v)\right).$$

Note that the normal Tate-Shafarevich group is $Ш(E) = Ш(E,\emptyset)$. Recall that the Tate-Shafarevich conjecture states that $Ш(E)$ is finite. My question concerns the corresponding problem for $Ш(E,S)$.

Is $Ш(E,S)$ conjecturally finite?

If the answer to this is yes, then an obvious next question is whether this is equivalent to finiteness of $Ш(E)$, i.e.

Is it known that $Ш(E) \subset Ш(E,S)$ has finite index?

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    $\begingroup$ Did you just post the Shafarevich conjecture as a Math Overflow question? $\endgroup$ – Jason Starr Jan 22 '14 at 13:37
  • $\begingroup$ No, my question is about what happens when $S$ is non-empty. I have edited the question to make it clearer. $\endgroup$ – Daniel Loughran Jan 22 '14 at 14:38
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    $\begingroup$ No, once you allow non-empty $S$ it is often (maybe always?) provably infinite, in contrast with the case of linear algebraic groups, for which the "$S$-version" is provably always finite. $\endgroup$ – user76758 Jan 22 '14 at 16:54
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    $\begingroup$ Example 7.5.1 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio Math. 148 (2012), which uses just standard methods in global Galois cohomology (duality theorems, etc.) There must be earlier references on this issue as well. $\endgroup$ – user76758 Jan 22 '14 at 22:30
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    $\begingroup$ The problem is only for torsion at primes in S. More precisely, let m be an integer not divisible by any prime in S. Then the subgroup of the S Tate-Shafarevich group killed by some power of m coincides with the similar subgroup of the usual Tate-Shafarevich group. This is a standard result (see for example Milne's Arithmetic Duality Theorems I, 6.6., but note that his S is the complement of yours). $\endgroup$ – abz Jan 23 '14 at 7:58

Global duality (as for instance on page 29 of Rubin's "Euler systems") gives you a description of the cokernel of your inclusion. Let $p$ be a prime. Then the $p$-primary part of the quotient of $Ш(E,S)$ by $Ш(E)$ is dual to the cokernel of the map $$ \mathfrak S_p(E/k) = \varprojlim_n \,\mathrm{Sel}_{p^n}(E/k) \ \to \ \bigoplus_{v\in S} E(k_v)^{*}.$$ Here the source is the compact $p$-adic Selmer group of $E/k$, which is a finitely generated $\mathbb{Z}_p$-module of rank $r$, believed to be the rank of $E(k)$. The target is the sum of the $p$-adic completions of the local points $E(k_v)$. If $v$ is not above $p$, then this group is finite and otherwise it is a finitely generated $\mathbb{Z}_p$-module of rank $[k_v:\mathbb{Q}_p]$.

One can make quite precise conjectures as to what the corank of the $p$-primary part of $Ш(E,S)$ should be in terms of the rank of $E(k)$ and $k/\mathbb{Q}$ by the above duality. Roughly speaking some sort of independence of elliptic logarithms tells you that the above map should have image as large as possible. For instance over $k=\mathbb{Q}$, it is clear that the corank of the $p$-primary part of $Ш(E,S)$ will be $0$ is $p$ if is not in $S$ or if $r\geq 1$, and $1$ otherwise. For larger $k$ it is a bit harder to formulate as it will depend on the field of definition of $E$, but generally speaking it tends to be large as soon as a $p$-adic place of large degree is in $S$.

  • $\begingroup$ I know nothing about all that stuff, but isn't there a typo in your answer? Logically the quotient of $Sha(E,S)$ by itself should be trivial, shouldn't it? $\endgroup$ – Sylvain JULIEN Jan 23 '14 at 11:41
  • $\begingroup$ Sure, sorry. I corrected this. $\endgroup$ – Chris Wuthrich Jan 23 '14 at 11:55

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