# The Stacks Project

## Tag 09EH

Lemma 15.93.15. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite Galois extension. Let $\mathfrak m \subset B$ be a maximal ideal of the integral closure of $A$ in $L$. Let $I \subset G$ be the inertia group of $\mathfrak m$. Then $B^I$ is the integral closure of $A$ in $L^I$ and $A \to (B^I)_{B^I \cap \mathfrak m}$ is étale.

Proof. Write $B' = B^I$. It follows from the definitions that $B' = B^I$ is the integral closure of $A$ in $L^I$. Write $\mathfrak m' = B^I \cap \mathfrak m = B' \cap \mathfrak m \subset B'$. By Lemma 15.92.5 the maximal ideal $\mathfrak m$ is the unique prime ideal of $B$ lying over $\mathfrak m'$. As $I$ acts trivially on $\kappa(\mathfrak m)$ we see from Lemma 15.92.2 that the extension $\kappa(\mathfrak m)/\kappa(\mathfrak m')$ is purely inseparable (perhaps an easier alternative is to apply the result of Lemma 15.92.6). Since $D/I$ acts faithfully on $\kappa(\mathfrak m')$, we conclude that $D/I$ acts faithfully on $\kappa(\mathfrak m)$. Of course the elements of the residue field $\kappa$ of $A$ are fixed by this action. By Galois theory we see that $[\kappa(\mathfrak m') : \kappa] \geq |D/I|$, see Fields, Lemma 9.21.5.

Let $\pi$ be the uniformizer of $A$. Since $\text{Norm}_{L/K}(\pi) = \pi^{[L : K]}$ we see from Algebra, Lemma 10.120.8 that $$|G| = [L : K] = [L : K]~\text{ord}_A(\pi) = |G/D|~[\kappa(\mathfrak m) : \kappa]~\text{ord}_{B_\mathfrak m}(\pi)$$ as there are $n = |G/D|$ maximal ideals of $B$ which are all conjugate under $G$, see Remark 15.93.6 and Lemma 15.93.8. Applying the same reasoning to the finite extension the finite extension $L/L^I$ of degree $|I|$ we find $$|I|~\text{ord}_{B'_{\mathfrak m'}}(\pi) = [\kappa(\mathfrak m) : \kappa(\mathfrak m')]~\text{ord}_{B_\mathfrak m}(\pi)$$ We conclude that $$\text{ord}_{B'_{\mathfrak m'}}(\pi) = \frac{[\kappa(\mathfrak m') : \kappa]}{|D/I|}$$ Since the left hand side is a positive integer and since the right hand side is $\leq 1$ by the above, we conclude that we have equality, $\text{ord}_{B'_{\mathfrak m'}}(\pi) = 1$ and $\kappa(\mathfrak m')/\kappa$ has degree $|D/I|$. Thus $\pi B'_{\mathfrak m'} = \mathfrak m' B_\mathfrak m'$ and $\kappa(\mathfrak m')$ is Galois over $\kappa$ with Galois group $D/I$, in particular separable, see Fields, Lemma 9.21.2. By Algebra, Lemma 10.141.7 we find that $A \to B'_{\mathfrak m'}$ is étale as desired. $\square$

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\begin{lemma}
\label{lemma-inertial-invariants-unramified}
Let $A$ be a discrete valuation ring with fraction field $K$.
Let $L/K$ be a finite Galois extension. Let $\mathfrak m \subset B$
be a maximal ideal of the integral closure of $A$ in $L$.
Let $I \subset G$ be the inertia group of $\mathfrak m$.
Then $B^I$ is the integral closure of $A$ in $L^I$ and
$A \to (B^I)_{B^I \cap \mathfrak m}$ is \'etale.
\end{lemma}

\begin{proof}
Write $B' = B^I$. It follows from the definitions that $B' = B^I$
is the integral closure of $A$ in $L^I$. Write
$\mathfrak m' = B^I \cap \mathfrak m = B' \cap \mathfrak m \subset B'$.
By Lemma \ref{lemma-one-orbit} the maximal ideal $\mathfrak m$ is the
unique prime ideal of $B$ lying over $\mathfrak m'$.
As $I$ acts trivially on $\kappa(\mathfrak m)$ we see from
Lemma \ref{lemma-invariants-modulo} that the extension
$\kappa(\mathfrak m)/\kappa(\mathfrak m')$ is purely inseparable
(perhaps an easier alternative is to apply the result of
Lemma \ref{lemma-one-orbit-geometric}).
Since $D/I$ acts faithfully on $\kappa(\mathfrak m')$,
we conclude that $D/I$ acts faithfully on $\kappa(\mathfrak m)$.
Of course the elements of the residue field $\kappa$ of $A$
are fixed by this action.
By Galois theory we see that $[\kappa(\mathfrak m') : \kappa] \geq |D/I|$,
see Fields, Lemma \ref{fields-lemma-galois-over-fixed-field}.

\medskip\noindent
Let $\pi$ be the uniformizer of $A$. Since
$\text{Norm}_{L/K}(\pi) = \pi^{[L : K]}$ we see from
Algebra, Lemma \ref{algebra-lemma-finite-extension-dim-1}
that
$$|G| = [L : K] = [L : K]\ \text{ord}_A(\pi) = |G/D|\ [\kappa(\mathfrak m) : \kappa]\ \text{ord}_{B_\mathfrak m}(\pi)$$
as there are $n = |G/D|$ maximal ideals of $B$ which are all
conjugate under $G$, see
Remark \ref{remark-finite-separable-extension} and
Lemma \ref{lemma-galois}.
Applying the same reasoning to the finite extension
the finite extension $L/L^I$ of degree $|I|$ we find
$$|I|\ \text{ord}_{B'_{\mathfrak m'}}(\pi) = [\kappa(\mathfrak m) : \kappa(\mathfrak m')]\ \text{ord}_{B_\mathfrak m}(\pi)$$
We conclude that
$$\text{ord}_{B'_{\mathfrak m'}}(\pi) = \frac{[\kappa(\mathfrak m') : \kappa]}{|D/I|}$$
Since the left hand side is a positive integer and since the right hand
side is $\leq 1$ by the above, we conclude that we have equality,
$\text{ord}_{B'_{\mathfrak m'}}(\pi) = 1$ and
$\kappa(\mathfrak m')/\kappa$ has degree $|D/I|$.
Thus $\pi B'_{\mathfrak m'} = \mathfrak m' B_\mathfrak m'$ and
$\kappa(\mathfrak m')$ is Galois over $\kappa$ with
Galois group $D/I$, in particular separable, see
Fields, Lemma \ref{fields-lemma-finite-Galois}.
By Algebra, Lemma \ref{algebra-lemma-characterize-etale}
we find that $A \to B'_{\mathfrak m'}$ is \'etale
as desired.
\end{proof}

Comment #3200 by Qixiao on February 13, 2018 a 4:25 am UTC

Hi Johan, there's a small typo in the last formula on ord_{B'_{m'}}(\pi), the numerator and denominator should be switched.

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