Express the inertia group in terms of the maximal unramified extension Suppose $L/K$ is a Galois extension of local fields with Galois group $G = \operatorname{Gal}(L/K)$. Let $K'$ be the maximal unramified Extension of $K$ in $L$.
The Definition of the inertia group of $L/K$ is given by $I = I_{L/K} = \operatorname{Gal}(L/K')$ which I understand.
In some notes I found that this is equivalent to
$$ I = \{ \sigma \in G : \sigma \text{ maps to } \iota \text{ in } \operatorname{Gal}(k_L/k_K) \}$$
or $$I = \{ \sigma \in G : \sigma x \equiv x \text{ (mod }M) \: \forall x \in \mathcal{O}_L \}.
$$
I know that $k_L$ and $k_K$ are the residue fields of $L$ and $K$.
Could you please explain me what $\iota$ and $M$ are supposed to be? If possible, could you also explain why all the conditions above are indeed equivalent then?
Thank you.
 A: Assume $L/K$ are local fields.  Let $\mathfrak p$ (resp. $\mathfrak P$) be the unique maximal ideal of $\mathcal O_K$ (resp. $\mathcal O_L$).  Let $\kappa(\mathfrak p) = \mathcal O_K/\mathfrak p$ and $\kappa(\mathfrak P) = \mathcal O_L/\mathfrak P$.  The inclusion of $\mathcal O_K$ into $\mathcal O_L$ induces an inclusion of $\kappa(\mathfrak p)$ into $\kappa(\mathfrak P)$.
Then for all $\sigma \in G = \operatorname{Gal}(L/K)$, we have $\sigma \mathfrak P = \mathfrak P$.  Then the ring isomorphism $\sigma: \mathcal O_L \rightarrow \mathcal O_L$, fixing $\mathcal O_K$ pointwise, induces a field isomorphism $\bar{\sigma}: \kappa(\mathfrak P) \rightarrow \kappa(\mathfrak P)$ fixing $\kappa(\mathfrak p)$ pointwise.  Thus we have a homomorphism
$$G \rightarrow \operatorname{Gal}(\kappa(\mathfrak P)/\kappa(\mathfrak p)), \sigma \mapsto \bar{\sigma}$$
It can be shown to be surjective.  Let $I$ be the kernel of this homomorphism.  By definition,
$$I = \{ \sigma \in G : \bar{\sigma} = 1_{\kappa(\mathfrak P)}\}$$
If $x = a + \mathfrak P$ is an element of $\kappa(\mathfrak P) = \mathcal O_L/\mathfrak P$, then by definition, $\bar{\sigma}(x) = \sigma(a) + \mathfrak P$.  Hence $\bar{\sigma} = 1_{\kappa(\mathfrak P)}$ if and only if for all $a \in \mathcal O_L$, we have $\sigma(a) + \mathfrak P = a + \mathfrak P$.  This shows that your two definitions of $I$ are equivalent.
Now $L/K$ is unramified, if and only if $[L : K] = [\kappa(\mathfrak P) : \kappa(\mathfrak p)]$, if and only if $I$ is trivial.  If $K'$ is the fixed field of $L/K$, then $G/I \cong \operatorname{Gal}(K'/K)$, and the corresponding "$I$" for $L/K'$ is trivial.  Hence $K'/K$ is unramified.  I forgot how to show it is maximal unramified, but it should not be too hard.  
A: Your question makes no sense if you don't specify what your fields $K, L$ are. Since you introduce the maximal unramified subextension $K'/K$ of $L/K$, there is only one possible interpretation: $K$ is a local field, i.e. it is complete w.r.t. a non archimedean discrete valuation $v$. The ring of integers of $K$ is the valuation ring {$O_K={x\in K, v(x) \ge 0}$}, which is a local ring with principal maximal ideal {$P_K={x\in K , v(x) \ge 1}$}, with residual field $k_K=O_K/P_K$. The same notations carry over to $L$. The maximal unramified subextension $K'/K$ of $L/K$ is the compositum of all the unramified subextensions of $L/K$, and because of the maximality property, $K'/K$ is Galois if $L/K$ is. Recall that the definition of "unramified" then requires that the residual extension $k_K'/k_K$ is also Galois (see Cassels-Fröhlich, p.21).
In the latter case, denote $G=Gal(L/K), G_0=Gal(K'/K), \bar G=Gal(k_K'/k_K)$. Since the action of $G$ respects the valuation, it is obvious that the passage to residual fields $s \in G \to \bar s \in \bar G$ s.t. $\bar s (\bar x)=\bar {s(x)}$ for $x\in O_L$, induces a homomorphism $r:G \to \bar G$ whose kernel is the inertia subgroup {$I = {s\in G, v_L(s(x)-x)\ge 1}$ for all $x \in O_L$}, as you noticed. It remains to show that $G_0=I$. Oviously $I$ contains $G_0$ because $P_L=P_{K'}$ (total ramification). The main point now is the surjectivity of $r$, which is a consequence of Hensel's lemma (op. cit., lemma 1, p.27), so that $\bar G\cong G/I$ and $\mid \bar G\mid$= the residual degree $f$. But $\mid G/G_0 \mid$= $f$ because of the classical relation $n=ef$ (op. cit., prop. 3, p.9), hence finally $G_0=I$.
A: The inertia group is usually defined as the kernel of the homomorphism
$$\varepsilon:D\twoheadrightarrow G(\overline L/\overline K),$$
where here $D$ denotes the decomposition group (in the case of a Galois extension of local fields as in your question, $D=G(L/K)$), and $\overline L=k_L$ in your notation, etc.
Then it is a fact that the extension of local fields $L^I/K$ is unramified, and moreover $\overline {L^I}$ coincides with the separable closure of $\overline K$ in $\overline L$ (Proposition 21 in I$\S$7 of Serre, Local Fields). Therefore, the fixed field of the inertia subgroup corresponds to the maximal unramified extension of the local field $K$ in $L$.
