\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

The Stacks project

Lemma 15.96.15. 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 ├ętale.

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 15.95.5 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 15.95.2 that the extension $\kappa (\mathfrak m)/\kappa (\mathfrak m')$ is purely inseparable (perhaps an easier alternative is to apply the result of Lemma 15.95.6). 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 15.96.6 and Lemma 15.96.8. 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

\[ \text{ord}_{B'_{\mathfrak m'}}(\pi ) = \frac{|D/I|}{[\kappa (\mathfrak m') : \kappa ]} \]

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 ├ętale as desired. $\square$


Comments (2)

Comment #3197 by Qixiao on

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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