I am trying to understand an argument of Corollary 7.3 from Silverman's The Arithmetic of Elliptic Curves. I am stuck and I would appreciate any explanations.
Let $E$ be an elliptic curve over $K$, a local field with valuation $v$. Let $K'$ be a finite Galois extension over $K$ with valuation $v'$. Suppose $I_{v'}$ (the inertia subgroup) acts trivially on $T_{\ell}(E)$, the Tate module. How do we know that
1) $I_{v'}$ is a normal subgroup of $I_v$?
2) $I_{v}/I_{v'}$ is finite?
It says in the proof below "Hence, the action of $I_v$ on $T_{\ell}(E)$ factors through the finite quotient $I_{v}/I_{v'}$. " so I think these two things must be true but I am not quite seeing how this is the case.
Any comments are appreciated. Thank you.
Corollary 7.3. Let $E/K$ be an elliptic curve. Then $E$ has potential good reduction if and only if the inertia group $I_v$ acts on the Tate module $T_\ell(E)$ through a finite quotient for some (all) prime(s) $\ell\ne\operatorname{char}(k)$.
Proof. Suppose that $E$ has potential good reduction, and let $K'/K$ be a finite extension such that $E$ has good reduction over $K'$. Extending $K'$, we may assume that $K'/K$ is a Galois extension. Let $v'$ be the valuation on $K'$ and let $I_{v'}$ be the inertia group of $G_{\overline K'/K'}$. We know from (VII.7.1) that $I_{v'}$ acts trivially on $T_\ell(E)$ for any prime $\ell\ne\operatorname{char}(k)$. Hence the action of $I_v$ on $T_\ell(E)$ factors through the finite quotient $I_v/I_{v'}$. This proves one implication
Assume now that for some prime $\ell\ne\operatorname{char}(k)$, the inertia group $I_v$ acts on $T_\ell(E)$ through a finite quotient, say $I_v/J$. Then the fixed field of $J$, which we denote by $\overline K^J$, is a finite extension of $K^{nr}=\overline K^{I_v}$. Hence we can find a finite extension $K'/K$ such that $\overline K^J$ is the compositum $$\overline K^J = K'K^{nr}.$$ Then the inertia group of $K'$ is equal to $J$, and by assumption $J$ acts trivially on $T_\ell(E)$. Now (VII.7.1) implies that $E$ has good reduction over $K'$. $\square$