Grothendieck in SGA 1 introduces a proposition in expose 5 (proposition 3.1) which states:


Let $X$ be etale, separated of finite type over $Y$, locally noetherian, and let $G$ be a finite group which operates on $X$ by $Y$-automorphisms. Then $G$ operates admissibly and the quotient scheme $X/G$ is etale over $Y$.


The hint he gives is that we may show this for $X$ quasi-projective, and to use proposition 1.8, which states that $G$ operates admissibly on $X$ iff $X$ is the union of open affines that are invariant under the action of $G$.


I am unsure how to show this. Help please?


EDIT: so I understand how to make the reduction to the quasi-projective case (since every etale morphism is quasi-finite, and then just apply Zariski's main theorem, and we get thus that this is quasi-projective), but I am unsure how to show that if a finite group operates on a quasi-projective scheme then it operates admissibly. I have a rough idea for how it could be done for a projective scheme, but I do not see how I could alter it for quasi-projective :( I could post this proof if it is of any help to readers?? 



admissible group operation on etale, separated, finite type scheme
We can reduce to the case where $Y$ is affine, and in this case (as you observe), $X$ is quasi-projective over $Y$ by Zariski's main theorem. Consequently, the key step is to show that a finite group operating on a quasi-projective scheme does so admissibly (meaning that we can form the quotient nicely). 


Admissibility is equivalent to the condition that every $G$-orbit is contained in an open affine, but any finite subset of projective space is contained in an open affine (e.g. find a hypersurface not containing any element of the subset and take the complement). 


A proof of étaleness in the case of a finite morphism is on p. 56 of Murre's "Lectures on an introduction to Grothendieck's theory of the fundamental group." It is based on a sequence of reductions, which I'll try to outline (without assuming finiteness of $X \to Y$, but I will assume that $Y$ has finite Krull dimension). First, étaleness descends under faithfully flat base change, and it can be checked on the stalks. Moreover, the process of taking the quotient by a finite group (which, on the level of rings, amounts to taking fixed points) is preserved under flat base change. (This is essentially the following observation: if $G$ acts on an $A$-algebra $B$, then if $A'$ is a flat extension of $A$, then $(A' \otimes\_A B)^G \simeq B^G \otimes\_A A'$; one proves this by fitting both into exact sequences.)
What all this means is that if you want to check that a map $X/G \to Y$ is étale, then you may as well base-change the whole problem to $T \to Y$ where $T$ is a faithfully flat extension of the local scheme $\mathrm{Spec} \mathcal{O}\_y$ (for any $y \in Y$). 


Anyway, a lot of things simplify when you're allowed to make these kinds of base changes. So let's assume for the sake of argument that $Y$ is the $\mathrm{Spec}$ of a complete local ring, noetherian, whose residue field is algebraically closed. Then $X$ splits into two pieces: the first is a piece *finite* étale over $Y$ (and thus a disjoint union of copies of $Y$, since etale covers of $T$ split) and a second piece whose image does not contain the closed point of $Y$. 


Let's consider what happens to the piece which is finite étale over $Y$, and so looks like $Y \sqcup Y \sqcup \dots \sqcup Y$. Here the group $G$ just acts by permuting the factors, and the quotient is what you get by identifying a bunch of pieces of $Y$, so in particular looks like $Y \sqcup Y \sqcup \dots \sqcup Y$ with a different number of pieces. 


Now what can we say about the piece $X'$ over $Y - \{ \ast\}$ for $\ast \in Y$ the closed point? Not that much, but we do know that it is of smaller Krull dimension, and using induction on the dimension, we can assume that $X'/G$ is étale over $Y - \{ \ast\}$. It follows that $X/G$ is etale over $Y$, completing the (sketched) proof. 


The key lemmas that make possible this replacement of $Y$ by such a nice ring are the following: 


**Lemma:** Let $(A, \mathfrak{m})$ be a local noetherian ring and $L/k$ an algebraic extension. Then there exists a flat extension $(\widetilde{A}, \widetilde{m})$ of $(A, \mathfrak{m})$ such that the $\widetilde{A}/\widetilde{m} \simeq L$ of the same Krull dimension. 


This is essentially the result of EGA III.10.3. 


**Corollary:** If $(A, \mathfrak{m})$ is any local noetherian ring of finite Krull dimension, then there exists a flat extension $(B, \mathfrak{n})$ of the same Krull dimension and such that $B$ is complete local and such that $B/\mathfrak{n}$ is algebraically closed. 


This follows from the lemma applied to the completion of $A$. The whole point is that the etale maps to $\mathrm{Spec} B$ are a lot simpler than the maps to $\mathrm{Spec} A$ (for instance, the etale *covers* of $\mathrm{Spec} B$ are all trivial). In the present case, the strict henselianization would also do. 


To recap, the point of the argument is to induct on the dimension of $Y$. Then, one reduces to checking the claim of etaleness by making a very strong base change, such that $X$ splits into an easily understood finite piece and a less easily understood but inductively controlled non-finite piece. 


These types of reductions are quite useful. For instance, they're integral to the proof of Zariski's main theorem given in EGA. Also, I'm pretty sure that the hypothesis of finite Krull dimension on $Y$ can be removed by a "noetherian descent" type argument (i.e., regarding $X, Y$ as a limit of schemes of finite type over $\mathbb{Z}$), but I'm not completely sure how noetherian descent is supposed to work for étaleness. 


