Inertia group $I_K$ is isomorphic to $\operatorname{Gal}(\bar K /K^{nr})$ as a group

Question

Let $K$ be a local field. Then, elements of $\operatorname{Gal}(\bar K /K)$ which induces identity map on residue field forms group, and the group is called inertia group of $\operatorname{Gal}(\bar K /K)$, and written like $I_K$.

I heard it is known that $I_K$ is isomorphic to $\operatorname{Gal}(\bar K /K^{nr})$ as a group. In other words, inertia field is maximal unramified extension. But I couldn't find another reference except for Silverman's 'the arithmetic of elliptic curves'.

　Could you give me self-contained strategy of this isomorphism or references ? (If possible, online accessible one is appreciated)

Thank you in advance.

Answer 1

You can do this by first proving/recalling the following:

If $L/K$ is a finite unramified extension then the map $\operatorname{Gal}(L/K)\to\operatorname{Gal}(\ell/k)$ is injective (in fact a bijection but we just need the former).

This can be proved by recalling one elementary fact about $L/K$ being finite unramified is that $L=K(a)$ for some element $a\in\mathcal O_L$ for which $\ell=k(\overline a)$ where $\overline a$ is the residue of $a$ in $\ell$. Then the hypothesis $\sigma\in\ker(\operatorname{Gal}(L/K)\to\operatorname{Gal}(\ell/k))$ means that $\sigma(a)\equiv a\mod\mathfrak m_L$, and using uniqueness from Hensel's lemma then you can deduce that $\sigma(a)=a$ so $\sigma=\operatorname{id}_L$.

With this claim proved it is easy to show that $\sigma\in I_K\iff \sigma|_{K^{nr}}=\operatorname{id}$ (you'll want to recall that finite extensions of $k$ correspond to finite unramified extensions of $K$).

Answer 2

You can find this (in slightly greater generality) for example in Neukirch, Algebraic Number Theory, Chapter II §9, Proposition 9.11, or in J.S.Milne's lectures notes on Algebraic Number Theory, Theorem 7.58.