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Proposition 2.2 in Exposé V of SGA I states the following (schemes assumed locally noetherian).

Let $X$ be a scheme with an admissible action of a finite group $G$, let $Y$ be the quotient scheme; suppose $X$ is finite over $Y$. Let $H$ be a subgroup of $G$, consider $X'=X/H$, let $x\in X$, $x'$ its image in $X'$ and $y$ its image in $Y$. Then (i) If $H\supset G_d(x)$, then the homomorphism $\mathscr O_y\rightarrow\mathscr O_{x'}$ induces an isomorphism on the completions. (ii) If $H\supset G_i(X)$, then the homomorphism $\mathscr O_y\rightarrow\mathscr O_{x'}$ is étale.

Here $G_d(x)$ is the decomposition group of $x$ and $G_i(x)$ is the inertia group of $x$.

The proof reduces to the case where $Y$ is the spectrum of a complete local ring $B$ and $X$ is the spectrum of a finite $B$-algebra $A$. It then claims that after passing to a finite flat extension of $A$, and using Prop. 2.1, one reduces to the case where the residual extension $k(x)/k(y)$ is trivial, and since $G_i(x)=G_d(x)$, one is reduced to (i).

How to fill in the details of this proof (of (ii))?

Prop. 2.1 says, if $Y$ is a $Z$-scheme,

Let $Z'\rightarrow Z$ be a base extension, and $X'=X\times_Z Z'$. Let $X'$ a point of $X'$, $x$ its image in $X$, then $G_i(x)=G_i(x')$.

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I don't quite follow the sketch of the proof. Instead, I want to make a suitable étale cover $Y'\rightarrow Y$ such that in the fiber over $y$ there is exactly one point, $y'$, and $k(x)/k(y')$ is purely inseparable. To do this, I let $k(y)^{\text{sep}}$ denote the maximal separable subextension of $k(x)/k(y)$ and I first consider the case where it is generated by a root of a single separable polynomial with coefficients in $k(y)$. I may assume this polynomial is monic and I lift it to a monic irreducible polynomial $p\in\mathscr O_y[x]$ in such a way that there is still a map $\mathscr O_y[x]/(p)\rightarrow A$. Then $\mathscr O_y[x]/(p)$ is free of rank $\deg p$ over $\mathscr O_y$, hence finite and flat, and the residual extension is a separable extension of fields, hence the $y$-fiber has one point and the morphism is net/unramified on $y$. By repeating this process I produce $Y'$. Now, I still have $G_i(x)=G_d(x)$ since $\operatorname{Aut}(k(x)/k(y)^{\text{sep}})=1$ as it is a purely inseparable extension.

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