I'm flipping through Silverman's Arithmetic of Elliptic Curves and I'm on Proposition VII.4.1.
Let $E/K$ be an elliptic curve over a local field $K$ with valuation $v$ and residue field $k$. If $E$ has good reduction at $v$ and $m\geq 1$ is an integer such that $v(m)=0$, then $E[m]$ is unramified at $v$.
In the proof, we extends K to K' such that $E[m]\subset E(K')$. Let $v'$ be the valuation of $K'$ and $k'$ its residue field.
If $E$ has good reduction, then a minimal Weierstrass equation has $v(\Delta)=0$. Now $v'(\Delta) = e(v'|v)v(\Delta) =0$. So $E$ has good reduction at $v'$. This implies that the reduction map
$$E[m] \to \tilde{E}(k')$$
is injective.
Now let $\sigma \in I_v$ and $P\in E[m]$. Silverman claims that $\sigma$ acts trivially on $\tilde{E}(k')$ by definition. However, by definition $\sigma$ acts trivially on $k$ and hence trivially on $\tilde{E}(k)$. I don't see why the action is also trivial on the extension $k'$.
Best Answer
We identify $Gal(\bar{k}/k)$ with $Gal(K^{ur}/K)$ since $k' \subset \bar{k}$ we're done because $I_v$ is the kernel of $G_K \to Gal(K^{ur}/K)$ i.e., $\sigma$ acts tivially on the algebraic closure of the residue field, not just the residue field itself.