# Cohomology of elliptic curves

Assume $K$ is an imaginary quadratic extension of $\mathbb{Q}$, and $E$ an elliptic curve defined over $\mathbb{Q}$. Let $p\neq l$ be primes in $\mathbb{Q}$ where $E$ has good reduction. Assume $p$ splits in $K/\mathbb{Q}$ and $l$ stays inert. Denote by $K_{l}$ the localization of $K$ at the place $l$. Is there always an isomorphism

$$H^1(K_{l},E)_{p^n}\rightarrow E(K_l)/p^nE(K_l),$$ at least for large enough $n$?

• I think that it is customary for downvotes to be accompanied with a little word of explanation. I encourage those who have cast downvotes to write a comment explaining why they think that this question is inappropriate: this might help the OP. – André Henriques Mar 14 '16 at 23:27
• Indeed, André. It could be said that the question lacks the description of the background and the reason why the question is interesting. With further detail it will become clear if the question was intended as written currently or if there is a confusion as suggested by user84144. But I would think that the question is ok and should not be closed. – Chris Wuthrich Mar 15 '16 at 10:55

More generally let k be a local field of residual characteristic $\ell$ and $E/k$ an elliptic curve with good reduction. Then for any prime $p\neq\ell$, there is a perfect duality between $E(k)/p^n$ and $H^1(k,E)[p^n]$ discovered by Tate. As both are finite abelian groups, they are indeed isomorphic. Knowing the structure of elliptic curves over local fields, it is easy to see that $E(k)/p^n$ is isomorphic to the same group in the reduction $\tilde E(\tilde k)/p^n$. So it is clear when these groups are non-trivial.
Yet, there is, in general, no canonical choice of an isomorphism. In some places like when using Euler systems, and it may be that you are referring to this, it is useful to fix an isomorphism between them so that one can compare classes. Typically one wants even a complement in $H^1(k,E[p^n])$ to the image of the Kummer map.