Unramified cocycles and the Selmer group of an ellptic curve In Silverman's book on elliptic curves, he gives a procedure to compute the Selmer group of elliptic curve $E$ relative to an isogeny $\phi:E\to E'$. I am confused about one step in the discussion. The particular point in the book where I am confused is Remark X.4.4.5
Let $K$ be a number field, let $M_K$ be the set of places of $K$, and let $S$ be a finite set of places of $K$ which includes all archimedean places, all places where $E$ has bad reduction, and all places dividing the degree of $\phi$.
Let $E[\phi]$ denote the kernel of $\phi$.
First we have that $S^{(\phi)}(E/K)\subseteq H^1(G_{\bar{K}/K},E[\phi];S)$ (where the $S$ means that we consider only cocycles unramified outside of $S$).
To check if an element of $H^1(G_{\bar{K}/K},E[\phi];S)$ is in in $S^{(\phi)}(E/K)$ we must see if it its image in $\prod_{v\in M_K} WC(E/K_v)$ is trivial (where $WC$ is the Weil-Chatlet group). 
Silverman asserts that for any element in $H^1(G_{\bar{K}/K},E[\phi];S)$ it suffices to check only the that the image in $WC(E/K_v)$ is trivial for $v\in S$. 
This is the point at which I am confused. Why is the image automatically trivial for $v\notin S$. Does the unramifiedness somehow imply this? I cannot find anything in the book that address this, but maybe I am overlooking something.
Any help or direction would be much appreciated.
 A: I had the same question. I've solved the problem partially, so I share my idea.
I extend the set $S$ to $S' = S \cup \{v \in M_K : E' \mbox{ has bad reduction at } v\}$.
I use the notation same as Silverman's AEC.
Let $v \in S'$, and consider the localization at $v$.
A commutative diagram of exact sequences of $G_{\bar{k_v}/k_v} = G_v/I_v$-module
\begin{eqnarray}
0 \rightarrow &E[\phi]^{I_v}& \rightarrow &E^{I_v}& \rightarrow &E'^{I_v}& \rightarrow 0\\
&\downarrow&&\downarrow&&\downarrow&\\
0 \rightarrow &\tilde{E_v}[\phi]& \rightarrow &\tilde{E_v}& \rightarrow &\tilde{E'_v}& \rightarrow 0
\end{eqnarray}
induces a commutative diagram of exact sequences of cohomology
\begin{eqnarray}
0 \rightarrow &E'(K_v)/\phi(E(K_v))& \rightarrow &H^1(G_v/I_v,E[\phi]^{I_v})& \rightarrow &H^1(G_v/I_v,E^{I_v})[\phi]& \rightarrow 0\\
&\downarrow& &\downarrow& &\downarrow&\\
0 \rightarrow &\tilde{E_v}'(k_v)/\phi(\tilde{E_v}(k_v))& \rightarrow &H^1(G_{\bar{k_v}/k_v},\tilde{E_v}[\phi])& \rightarrow &H^1(G_{\bar{k_v}/k_v},\tilde{E_v})[\phi]& \rightarrow 0.
\end{eqnarray}
As proof of Theorem X.4.2. in AEC, $E[\phi]^{I_v} = E[\phi]$ and $E[\phi] \cong \tilde{E}[\phi]$ as $G_v/I_v.$-module.
Therefore the center vertical arrow in the above diagram of cohomology is isomorphism.
The left vertical arrow is surjective, so, from the five lemma, the right vertical arrow is an inclusion.
$H^1(G_{\bar{k_v}/k_v},\tilde{E_v})$ can be regarded as Weil-Chatelet group, so this is finite.
The map of cohomology
$$H^1(G_{\bar{k_v}/k_v},\tilde{E_v}) \xrightarrow{\phi} H^1(G_{\bar{k_v}/k_v},\tilde{E_v})$$
is surjective between same finite sets, so an inclusion.
Therefore $H^1(G_{\bar{k_v}/k_v},\tilde{E_v})[\phi] = 0$.
And from injectivity of the right vertical arrow of cohomology diagram,
$H^1(G_v/I_v,E^{I_v})[\phi] = 0$.
Now we obtain a commutative diagram of cohomology from a inflation map
\begin{eqnarray}
0 \rightarrow &E'(K_v)/\phi(E(K_v))& \rightarrow &H^1(G_v/I_v,E[\phi]^{I_v})& \rightarrow &H^1(G_v/I_v,E^{I_v})[\phi] (= 0)& \rightarrow 0\\
&\downarrow& &\downarrow& &\downarrow&\\
0 \rightarrow &E(K_v)/\phi(E(K_v))& \rightarrow &H^1(G_v,E[\phi])& \rightarrow &H^1(G_v,E)[\phi]& \rightarrow 0.
\end{eqnarray}
and an inflation-restriction sequence
$$0 \rightarrow H^1(G_v/I_v,E[\phi]^{I_v}) \rightarrow H^1(G_v,E[\phi]) \rightarrow H^1(I_v,E[\phi]) .$$
Let $\xi \in H^1(G_{\bar{K}/K}, E[\phi]; S')$ and $\bar{\xi}$ the image of $\xi$
in $H^1(G_v,E[\phi])$.
Then by definition $Res(\bar{\xi}) \in H^1(I_v,E[\phi])$ is $0$.
So the image of $\bar{\xi}$ in $H^1(G_v,E)[\phi]$ of the above diagram come from
$H^1(G_v/I_v,E[\phi]^{I_v})$, therefore $0$.
This means the image of $\xi$ in $WC(E/K_v)$ is trivial.
