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Let $E$ be an elliptic curve defined over a number field $K$. Let $v$ be a finite place of $K$ not dividing $n$ such that $E$ has good reduction at $v$. This only excludes a finite set of places $\Sigma$ of $K$ (containing the infinite places). Let $k_v$ be the residue field of $v$ and let $\ell$ be the characteristic of $k_v$ with $\ell\nmid n$. Let $\widetilde{E}$ be the reduction of $E$ modulo $v$. I want to show that the kernel $$\ker \left( H^1(K,E[n])\overset{Res}{\longrightarrow}\underset{v\notin\Sigma}{\prod} H^1(I_v,E[n])\right)$$(where $I_v$ is the inertia group) is isomorphic to $H^1(G_{{K_\Sigma}/K},E[n]).$ I started with the inflation-restriction sequence for any $v$ $$0\rightarrow H^1(G_K/I_v,E[n])\rightarrow H^1(K,E[n])\rightarrow H^1(I_v,E[n])$$so my idea is to show that $\underset{v\notin\Sigma}{\cap}H^1(G_K/I_v,E[n])\simeq H^1(G_{{K_\Sigma}/K},E[n]).$ Do someone have an hint or a different method for that?

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Let $H^1(K,E[n],\Sigma)=\{\xi\in H^1(K,E[n])\ |\ \xi \text{ is unramified for } v\notin \Sigma\}$. A global cocycle $\xi$ is unramified at a place $v$, if it becomes trivial when restricted to $I_v$. The group $H^1(K,E[n],\Sigma)$ is the kernel you are interested in.

Denote by $\delta_v\colon E(K_v)/nE(K_v) \longrightarrow H^1(K_v,E[n])$ the connecting homomorphism. For $v\notin \Sigma$, an element $\xi_v\in H^1(K_v,E[n])$ lies in the image of the connecting homomorphism if and only if $\xi_v$ becomes trivial when restricted to $I_v$ (not trivial). In this way we can also write

$H^1(K,E[n],\Sigma)=\{\xi\in H^1(K,E[n])\ |\ \xi_v \in image(\delta_v) \text{ for }v\notin \Sigma\}$

Maybe that helps! I don't know what you mean by $K_\Sigma$ by the way.

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  • $\begingroup$ Thanks, and sorry I forgot! $K_\Sigma$ is the maximal extension of $K$ unramified outside $\Sigma$. $\endgroup$ – Ashley Courtney Mar 14 '18 at 18:17
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Well, ok. For every $v\notin \Sigma$ you have $K_\Sigma\subset \overline{K}^{I_v}$ and hence $I_v \subset Gal(\overline{K}/K_\Sigma)$. Moreover, we have an isomorphism $(G_K/I_v)/(G_{\overline{K}/K_\Sigma}/I_v)\simeq G_{K_\Sigma/K}$. The inflation map then gives us an injection

$H^1(G_{K_\Sigma/K}, E[n]) \hookrightarrow H^1(G_K/I_v, E[n])$ which implies $H^1(G_{K_\Sigma/K}, E[n]) \subset \cap_{v \notin \Sigma} H^1(G_K/I_v, E[n])$. Note that the intersection is taken in $H^1 (K,E[n])$.

This gives at least half of the proof. If you could prove that $\cup_{v\notin \Sigma} I_v = Gal(\overline{K}/K_\Sigma)$, you should get equality. But I don't know whether the last equality is valid, although it does not seem to hard to prove it since we already have $K_\Sigma = \cap_{v\notin \Sigma} \overline{K}^{I_v}$.

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