(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 $Ш(E/\mathbb{Q})\rightarrow \mathrm{coker}~a~~$?

  • 2
    $\begingroup$ What is your definition of $Sel(E/\mathbb{Q})$ ? My first guess would be the inductive limit of $n$-Selmer groups. But then I can't see how you defined the map you want to be injective. Did you mean the target to be $E(\mathbb{Q}_p)\otimes \mathbb{Q}/\mathbb{Z}$ ? I think this question needs some improvement to be understandable. $\endgroup$ – Chris Wuthrich Apr 1 '15 at 15:25
  • 2
    $\begingroup$ And the title of the question als odoes not seem to have much relation to the question itself. $\endgroup$ – Chris Wuthrich Apr 1 '15 at 15:28
  • $\begingroup$ (After edit): Take $\ell$-primary parts everywhere. Then the local product of the first term is just $E(\mathbb{Q}_{\ell})\otimes\mathbb{Q}_{\ell}/\mathbb{Z}_{\ell}$ which is cofree of rank $1$. So $\operatorname{coker}(a)[\ell^{\infty}]$ is finite if the rank of $E(\mathbb{Q})$ is positive and is cofree of corank $1$ otherwise. That does not look analogous to your huge local product in $r$. $\endgroup$ – Chris Wuthrich Apr 3 '15 at 19:09
  • $\begingroup$ As to the kernel of your map: This is what I would call the "fine Tate-Shafarevich group". There are examples of when it is trivial, half or all of the Tate-Shafarevich group when the latter has 4 elements. In general I would think the kernel could just be anything. Proc. Camb. Soc. 142 (2007), no. 1, p. 1-12. $\endgroup$ – Chris Wuthrich Apr 4 '15 at 20:29
  • $\begingroup$ There is some evidence that the fine Sha is much smaller. For instance even over infinite extension considered in Iwasawa theory the fine Sha should be finite (all the time ?) while the full Sha can get very large. $\endgroup$ – Chris Wuthrich Apr 4 '15 at 20:32

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.