Finite étale covers admit connected Galois trivializations and subgroup quotients

Given $u:Y\to Z$, its graph in $Y{\times }_{X}Z$ is the pullback of the diagonal of $Z\mathrm{/}X$, hence is an open and closed immersion by [F1]. The projection $Y{\times }_{X}Z\to Z$ is finite étale, so the graph factorization shows that $u$ is finite étale. The equalizer of two maps is likewise the pullback of an open and closed diagonal. If two maps agree on $F(Y)$, their equalizer contains the entire fibre; its complementary open and closed cover has rank zero at $\bar{x}$ and hence everywhere on connected $X$ by [F2], so it is empty. This proves faithfulness.

A nonempty open and closed piece of a cover is itself finite étale. Its image is open and closed by [F1] and nonempty, hence all of $X$. Each such piece therefore contributes at least one point to every geometric fibre. A cover of rank $n$ cannot have more than $n$ disjoint nonempty open and closed pieces. Repeatedly split a disconnected piece: every split increases their number, so after at most $n-1$ splits all pieces are connected. This proves the finite connected decomposition without a Noetherian assumption. A map to a connected cover is open and closed by step 1.1 and [F1], and thus is surjective if its source is nonempty.

If $F(u)$ is a bijection, its source-to-target map has degree one on each component of $Z$, by [F2], step 2.1 and connectedness. A finite locally free map of degree one is an isomorphism: locally its algebra is free of rank one, and the unit generates its residue fibre at every prime; Nakayama as used in [F2] makes the unit a local basis, so the ring map is an isomorphism. Hence $F$ reflects isomorphisms. For maps from a connected cover, agreement at one fibre point makes their open and closed equalizer nonempty and therefore the whole source.

Let $Y$ have rank $n>0$. In $Y^n=Y{\times }_X\cdots {\times }_{X}Y$, remove the finitely many open and closed loci on which two coordinates agree. The resulting open and closed cover $P$ has as its fibre the ordered lists of all $n$ distinct points of $F(Y)$. Choose one such list $p$, and let $T$ be the connected component of $P$ containing it. By step 2.1, $T\to X$ is nonempty finite étale and surjective. The coordinate sections of $T{\times }_{X}Y\to T$ have disjoint open and closed graphs by [F1]. On every geometric fibre they exhaust the $n$ points, since the list in $P$ has no repeated coordinate. Their disjoint union is thus the whole $T{\times }_{X}Y$, and $T$ trivializes $Y$. For $n=0$ take $T=X$; the empty cover is already trivialized.

Permuting the $n$ coordinates gives an action of $S_n$ on $P$. Any other point $p^{\mathrm{\prime }}\in F(T)$ is obtained from $p$ by a unique permutation $\sigma$. The image $\sigma (T)$ is a connected component of $P$ meeting $T$ at $p^{\mathrm{\prime }}$, and therefore equals $T$. Thus the subgroup preserving $T$ acts transitively on $F(T)$. An automorphism of $T$ fixing one geometric point is the identity by step 3.1, so $H={\mathrm{Aut}}_X(T)$ acts freely as well. This proves simple transitivity. Any map $T\to Y$ is a section of the trivial cover $T{\times }_{X}Y$; connectedness of $T$ makes it one of the coordinate sections. These are in bijection with $F(Y)$ by evaluation at a fixed point of $T$, proving the uniqueness and existence assertion.

The map ${\coprod }_{h\in H}T\to T{\times }_{X}T$ given on the $h$-summand by the graph of $h$ is finite étale by step 1.1. It is bijective on the geometric fibre because $H$ acts simply transitively; step 3.1 makes it an isomorphism. Consequently $T\mathrm{/}X$ is an $H$-torsor, and $T\to X$ is an fpqc cover. Any finite $H^{\mathrm{o}\mathrm{p}}$-set $E$ determines a descent datum on ${\coprod }_{E}T$ over that fpqc cover: on the overlap identified with ${\coprod }_{H}T$, use the permutation of $E$ associated to $h$, with the opposite-group convention matching composition of changes of trivialization. The group-action law is precisely the cocycle identity. By [F3] this datum descends to a finite étale $X$-cover with fibre $E$.

For $K\le H$, use the coset action in step 5.1. The quotient map of the finite trivial fibre is compatible with the descent datum, so [F3] descends it to $q:T\to T\mathrm{/}K$. Its fibre is $F(T)\mathrm{/}K$, and $q$ is finite étale and surjective by step 1.1 and fibrewise surjectivity; thus it is fpqc. The relation $T{\times }_{T\mathrm{/}K}T$ is the disjoint union of the graphs of $k\in K$, as seen after the faithfully flat cover $T\to X$ and then descended by [F3]. A $K$-invariant map $a:T\to W$ to any $X$-scheme has equal pullbacks on that relation. For each affine open of $W$, its preimage in $T$ is saturated, hence descends to an open of $T\mathrm{/}K$ by submersiveness in [F3]. Cover that open by affine opens $U$: since $q$ is finite, $q^{-1}(U)$ is affine, and the ring equalizer argument of [F3] descends $a$ on $U$. The unique descended morphisms agree on overlaps and glue. Thus $T\mathrm{/}K$ satisfies the universal property of the scheme quotient. This completes the assertions with the AC use recorded in [F4].