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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 $K'$. $\square$

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    $\begingroup$ This is just Galois and ramification theory, nothing to do with elliptic curves. I suspect you want to read up on the definition and properties of the inertia groups in field extensions. $\endgroup$ Jan 28, 2018 at 14:50
  • $\begingroup$ I have tried it and I couldn't figure it out myself, so I decided to try asking it here... $\endgroup$
    – Johnny T.
    Jan 28, 2018 at 15:01
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    $\begingroup$ @JohnnyT. I have included the whole proof you refer to - in case it helps to clarify the context of the question - I hope it's ok. Of course, feel free to edit the question further as needed. $\endgroup$ Jan 28, 2018 at 15:06
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    $\begingroup$ @MartinSleziak I guess it's okay to post a verbatim copy of the proof, but as Chris says in his comment, the questions really have nothing to do with the proof, nor indeed with the statement of the theorem. What is needed is basic facts about local fields. If $K'/K$ is a finite Galois extension of finite fields with inertia groups $I$ and $I'$, then (1) $I$ is normal in $I'$ and (2) $I'/I$ is finite. For (1), just use the definition of inertia group as fixing things modulo the prime ideal, then for (2), since $I'/I$ injects into Gal($K'/K$), the finiteness is clear. $\endgroup$ Jan 28, 2018 at 17:33
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    $\begingroup$ @JoeSilverman I finally got this! Thank you!! $\endgroup$
    – Johnny T.
    Jan 28, 2018 at 20:10

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