Let $A$ be an abelian variety over the rational numbers $\mathbf{Q}$. Let $V=T_p A \otimes \mathbf{Q}_p$ be the $\mathbf{Q}_p$-Tate module of $A$. Let $G$ be the absolute Galois group of $\mathbf{Q}$. (added in edit)

I keep seeing a natural map $A\to H^1(G,V)$. How is this map constructed? What does it have to do with "Kummer theory"?

What is the image of this map? That is, how can one describe it? Does it have to do with Selmer groups?

Sorry for the vagueness.

  • $\begingroup$ What is G? The absolute galois group of $\mathbb{Q}$? $\endgroup$ Mar 18 '12 at 15:26
  • $\begingroup$ Yes. I was going to put that. $\endgroup$
    – Harry
    Mar 18 '12 at 15:39

Does Silverman, The Arithmetic of Elliptic Curves, X.1 or Cornell-Silverman-Stevens, p. 33 help? Form the long exact sequence of $0 \to A[\ell^n] \to A \to A \to 0$ (analogue of the Kummer sequence if you replace $A$ by $\mathbf{G}_m$) and take the inverse limit.

Relation to the Selmer group: $\mathrm{Sel}(A/K)_m \subseteq H^1(K,A[m])$ and I suspect $\mathrm{Sel}(A/K)_m = H^1(\mathrm{Spec}\,\mathcal{O}_K,\mathcal{A}[m])$ (disregarding the archimedean places).

  • $\begingroup$ I must be missing something very elementary. We have inclusions $A[\ell^n]\to A$ for all $n>0$. Taking the inverse limit is a covariant functor, right? So this gives me a map from $T_\ell A $ to $A$? And this functor is not left exact? $\endgroup$
    – Harry
    Mar 18 '12 at 17:37
  • $\begingroup$ You get maps $A(K) \stackrel{\ell^n}{\to} A(K) \to H^1(K,A[\ell^n]) \to H^1(K,A)$. $\endgroup$
    – user19475
    Mar 18 '12 at 17:40
  • $\begingroup$ @Harry I think your question in the comment above asks if the inclusions of $A[\ell^n]$ into $A(\overline{K})$ induce a homomorphism from $T_\ell A$ to $A(\overline{K})$. They don't, the issue being that your inclusions aren't compatible with the transition maps in the Tate module (which are multiplication by $\ell$). $\endgroup$ Mar 18 '12 at 21:35

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.