The Stacks project

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 15.110.8 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 15.110.2 that the extension $\kappa (\mathfrak m)/\kappa (\mathfrak m')$ is purely inseparable (perhaps an easier alternative is to apply the result of Lemma 15.110.9). 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 9.21.6.

Let $\pi $ be the uniformizer of $A$. Since $\text{Norm}_{L/K}(\pi ) = \pi ^{[L : K]}$ we see from Algebra, Lemma 10.121.8 that

as there are $n = |G/D|$ maximal ideals of $B$ which are all conjugate under $G$, see Remark 15.111.6 and Lemma 15.112.1. Applying the same reasoning to the finite extension the finite extension $L/L^ I$ of degree $|I|$ we find

We conclude that

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 10.143.7 we find that $A \to B'_{\mathfrak m'}$ is étale as desired. $\square$