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 $K$).