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.
Best Answer
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.