The Stacks project

Proof. Note that the assertion makes sense by the result of Lemma 79.12.7 on the diagonal of $(X/Y)_{fin}$, see Spaces, Lemma 65.5.10. Let $V$ be a scheme and let $V \to Y$ be a surjective étale morphism. By Lemma 79.12.9 the morphism $(V \times _ Y X/V)_{fin} \to (X/Y)_{fin}$ is a base change of the map $V \to Y$ and hence is surjective and étale, see Spaces, Lemma 65.5.5. Hence it suffices to prove the lemma for $(V \times _ Y X/V)_{fin}$. (Here we implicitly use that the composition of representable, surjective, and étale transformations of functors is again representable, surjective, and étale, see Spaces, Lemmas 65.3.2 and 65.5.4, and Morphisms, Lemmas 29.9.2 and 29.36.3.) Note that the properties of being separated and locally quasi-finite are preserved under base change, see Morphisms of Spaces, Lemmas 67.4.4 and 67.27.4. Hence $V \times _ Y X \to V$ is separated and locally quasi-finite as well, and by Morphisms of Spaces, Proposition 67.50.2 we see that $V \times _ Y X$ is a scheme as well. Thus we may assume that $f : X \to Y$ is a separated and locally quasi-finite morphism of schemes.

Pick a point $y \in Y$. Pick $x_1, \ldots , x_ n \in X$ points lying over $y$. Pick an étale neighbourhood $a : (U, u) \to (Y, y)$ and a decomposition

as in More on Morphisms, Lemma 37.41.5. Pick any subset

Given these choices we obtain a pair $(a, Z)$ with $Z = \bigcup _{(i, j) \in I} V_{i, j}$ which satisfies conditions 79.12.0.1. In other words we obtain a morphism $U \to (X/Y)_{fin}$. The construction of this morphism depends on all the things we picked above, so we should really write

This morphism is étale by Lemma 79.12.8.

Claim: The disjoint union of all of these is surjective onto $(X/Y)_{fin}$. It is clear that if the claim holds, then the lemma is true.

To show surjectivity we have to show the following (see Spaces, Remark 65.5.2): Given a scheme $T$ over $S$, a point $t \in T$, and a map $T \to (X/Y)_{fin}$ we can find a datum $(y, n, x_1, \ldots , x_ n, a, I)$ as above such that $t$ is in the image of the projection map

To prove this we may clearly replace $T$ by $\mathop{\mathrm{Spec}}(\overline{\kappa (t)})$ and $T \to (X/Y)_{fin}$ by the composition $\mathop{\mathrm{Spec}}(\overline{\kappa (t)}) \to T \to (X/Y)_{fin}$. In other words, we may assume that $T$ is the spectrum of an algebraically closed field.

Let $T = \mathop{\mathrm{Spec}}(k)$ be the spectrum of an algebraically closed field $k$. The morphism $T \to (X/Y)_{fin}$ is given by a pair $(T \to Y, Z)$ satisfying conditions 79.12.0.1. Here is a picture:

Let $y \in Y$ be the image point of $T \to Y$. Since $Z$ is finite over $k$ it has finitely many points. Thus there exist finitely many points $x_1, \ldots , x_ n \in X$ such that the image of $Z$ in $X$ is contained in $\{ x_1, \ldots , x_ n\} $. Choose $a : (U, u) \to (Y, y)$ adapted to $y$ and $x_1, \ldots , x_ n$ as above, which gives the diagram

Since $k$ is algebraically closed and $\kappa (y) \subset \kappa (u)$ is finite separable we may factor the morphism $T = \mathop{\mathrm{Spec}}(k) \to Y$ through the morphism $u = \mathop{\mathrm{Spec}}(\kappa (u)) \to \mathop{\mathrm{Spec}}(\kappa (y)) = y \subset Y$. With this choice we obtain the commutative diagram:

We know that the image of the left upper arrow ends up in $\coprod V_{i, j}$. Recall also that $Z$ is an open subscheme of $\mathop{\mathrm{Spec}}(k) \times _ Y X$ by definition of $(X/Y)_{fin}$ and that the right hand square is a fibre product square. Thus we see that

is an open subscheme. By construction (see More on Morphisms, Lemma 37.41.5) each $V_{i, j}$ has a unique point $v_{i, j}$ lying over $u$ with purely inseparable residue field extension $\kappa (v_{i, j})/\kappa (u)$. Hence each scheme $\mathop{\mathrm{Spec}}(k) \times _ U V_{i, j}$ has exactly one point. Thus we see that

for a unique subset $I \subset \{ (i, j) \mid 1 \leq i \leq n, \ 1 \leq j \leq m_ i\} $. Unwinding the definitions this shows that

with $I$ as found above is nonempty as desired. $\square$