Htykuut

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.

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.

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 {$O_K={xin K, v(x) ge 0}$}, which is a local ring with principal maximal ideal {$P_K={xin K , v(x) ge 1}$}, 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 $xin O_L$, induces a homomorphism $r:G to bar G$ whose kernel is the inertia subgroup {$I = {sin G, v_L(s(x)-x)ge 1}$ 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 $bar Gcong G/I$ and $mid bar Gmid$= 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 $G_0=I$.

The inertia group is usually defined as the kernel of the homomorphism
$$varepsilon:Dtwoheadrightarrow G(overline L/overline K),$$
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 $L$.