Is Ш a good parameter for the failure of Global-Local principle for abelian varieties?

Question

(Comparing to class group cases: we have an isomorphism $Cl(K)\rightarrow \prod \left(K^\times \backslash K_p^\times /O_p^\times \right)$ for a number field $K$.

Similarly, for an elliptic curve $E/\mathbb{Q}$, I expect the injectivity of $Sel(E/\mathbb{Q})\rightarrow \prod \left( (E/E_{tor})(\mathbb{Q}_p)\right) \left(\hookrightarrow \prod H^1(\mathbb{Q}_p,E_{tor})\right)$, but I cannot prove or disprove it.)

I'm sorry for unclear question. I use Sel$(E/\mathbb{Q})$ as the direct limit of $n$-Selmer groups.

My question comes from comparing two sequences:

$0\rightarrow E(\mathbb{Q})\otimes \mathbb{Q}/\mathbb{Z} \rightarrow Sel(E/\mathbb{Q})\rightarrow Ш(E/\mathbb{Q}) \rightarrow 0 $ and $0\rightarrow \underset{p}\prod E(\mathbb{Q}_p)\otimes \mathbb{Q}/\mathbb{Z} \rightarrow \underset{p}\prod H^1(\mathbb{Q}_p,E_{tor})\rightarrow \underset{p}\prod H^1(\mathbb{Q}_p,E) \rightarrow ... $

Let $a,b$ be the restriction maps from $E(\mathbb{Q})\otimes \mathbb{Q}/\mathbb{Z},Sel(E/\mathbb{Q})$ respectively. There is an exact sequence $0\rightarrow \ker a\rightarrow \ker b\rightarrow Ш(E/\mathbb{Q})\rightarrow \mathrm{coker}~ a\rightarrow \mathrm{coker}~b$.

$Ш(E/\mathbb{Q})\rightarrow \mathrm{coker}~a~~$ is an analogue of $r~:~Cl(K)\rightarrow K^\times \backslash \prod \left(K_p^\times /O_p^\times \right)$. The reciprocity map $r$ is known to be bijective and the proof does not assume the finiteness of class groups. Is there similar argument for $Ш(E/\mathbb{Q})\rightarrow \mathrm{coker}~a~~$?

Answer 1

All your groups are torsion, so we may split it into primary parts. Let $\ell$ be a prime. First the map $a\colon E(\mathbb{Q})\otimes \mathbb{Q}_{\ell}/\mathbb{Z}_{\ell} \to \prod_p E(\mathbb{Q}_p)\otimes \mathbb{Q}_{\ell}/\mathbb{Z}_{\ell}$. The target is in fact simply $E(\mathbb{Q}_{\ell})\otimes \mathbb{Q}_{\ell}/\mathbb{Z}_{\ell}$ since all other terms in the product are zero. Hence if the rank of $E(\mathbb{Q})$ is positive, then the cokernel of $a$ is finite. Otherwise the Pontryagin dual of the cokernel of $a$ is a free $\mathbb{Z}_{\ell}$-module of rank $1$. The kernel of $a$ is of corank $\max\{\operatorname{rank}(E(\mathbb{Q}))-1,0\}$.

The elements of $\operatorname{Sel}_{\ell^{\infty}}(E/\mathbb{Q})$ satisfy all local condition except that they may be non-trivial at $\ell$. The kernel of $b$ is called the fine or restricted Selmer group in the literature. One could call the quotient $\operatorname{coker}(b)/\operatorname{coker}(a)$ the fine Tate-Shafarevich group. This is the kernel of your map $r\colon Ш \to \operatorname{coker}(a)$.

I know of examples of curves with $ Ш \ \ $ having $4$ elements and where $\ker(r)$ is trivial, others where it has $2$ elements and yet others where it has $4$ elements. Somehow I would believe that if the rank of $E$ is zero, then $\ker(r)$ could be any subgroup of $Ш\ $. (See Proc. Camb. Soc. 142 (2007), no. 1, p. 1-12)

Of course, proving that $\ker(r)$ is finite would be very, very good. In fact, one may believe that this fine version is a finite subgroup even over infinite extensions of $\mathbb{Q}$. For instance, I believe that it is always finite on top of a $\mathbb{Z}_{\ell}$-extension.

Finally, the cokernel of $b$ is really big. It would be more natural to consider the restricted product over primes relative to the unramified subgroup in $H^1$. Then $b$ belongs to a long exact sequence in global duality (See Cohomology of number fields).