The Stacks project

Proof. The equivalence of (1) and (2) follows from Lemma 68.18.10 (and the fact that a quasi-separated morphism is decent by Lemma 68.17.2).

Assume the equivalent conditions of (1) and (2). Choose an affine scheme $V$ and an étale morphism $V \to Y$ mapping a point $v \in V$ to $y$. Then $v$ is a generic point of an irreducible component of $V$ by Properties of Spaces, Lemma 66.11.1. Choose an affine scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. Then $U \to V$ is of finite type. The morphism $U \to V$ is quasi-finite at every point lying over $v$ by (2). It follows that the fibre of $U \to V$ over $v$ is finite (Morphisms, Lemma 29.20.14). By Morphisms, Lemma 29.51.1 after shrinking $V$ we may assume that $U \to V$ is finite. Let

Since $f$ is quasi-separated, we see that $V \times _ Y X$ is quasi-separated and hence $R$ is a quasi-compact scheme. Moreover the morphisms $R \to V$ is quasi-finite as the composition of an étale morphism $R \to U$ and a finite morphism $U \to V$. Hence we may apply Morphisms, Lemma 29.51.1 once more and after shrinking $V$ we may assume that $R \to V$ is finite as well. This of course implies that the two projections $R \to V$ are finite étale. It follows that $V/R = V \times _ Y X$ is an affine scheme, see Groupoids, Proposition 39.23.9. By Morphisms, Lemma 29.41.9 we conclude that $V \times _ Y X \to V$ is proper and by Morphisms, Lemma 29.44.11 we conclude that $V \times _ Y X \to V$ is finite. Finally, we let $Y' \subset Y$ be the open subspace of $Y$ corresponding to the image of $|V| \to |Y|$. By Morphisms of Spaces, Lemma 67.45.3 we conclude that $Y' \times _ Y X \to Y'$ is finite as the base change to $V$ is finite and as $V \to Y'$ is a surjective étale morphism.

If $Y$ is decent and $f$ is quasi-separated, then we see that $X$ is decent too; use Lemmas 68.17.2 and 68.17.5. Hence Lemma 68.18.10 applies to show that (4) implies (1) and (2). On the other hand, we see that (2) implies (4) by Morphisms of Spaces, Lemma 67.27.9. $\square$

Proof. Since quasi-separated algebraic spaces are decent, the equivalence of (1) and (2) follows from Lemma 68.18.10. To prove that (1) and (2) imply (3) we may and do replace $Y$ by a quasi-compact open containing $y$. Since $f^{-1}(\{ y\} )$ is finite, we can find a quasi-compact open subspace of $X' \subset X$ containing the fibre. The restriction $f|_{X'} : X' \to Y$ is quasi-compact and quasi-separated by Morphisms of Spaces, Lemma 67.8.10 (this is where we use that $Y$ is quasi-separated). Applying Lemma 68.21.1 to $f|_{X'} : X' \to Y$ we see that (3) holds. We omit the proof that (3) implies (2). $\square$

Proof. Let $V$ be a scheme and let $V \to Y$ be a surjective étale morphism. Let $U$ be a scheme and let $U \to V \times _ Y X$ be a surjective étale morphism. Then $f$ is quasi-finite at the image $x$ of a point $u \in U$ if and only if $U \to V$ is quasi-finite at $u$. Moreover, $x \in X^0$ if and only if $u$ is the generic point of an irreducible component of $U$ (Properties of Spaces, Lemma 66.11.1). Thus the lemma reduces to the case of the morphism $U \to V$, i.e., to Morphisms, Lemma 29.51.4. $\square$

Proof. By Lemmas 68.20.4 and 68.20.1 we may assume $Y$ is a scheme with finitely many irreducible components. Shrinking further we may assume $Y$ is an irreducible affine scheme with generic point $y$. Then the fibre of $f$ over $y$ is finite.

Assume $f$ is quasi-compact and $Y$ affine irreducible. Then $X$ is quasi-compact and we may choose an affine scheme $U$ and a surjective étale morphism $U \to X$. Then $U \to Y$ is of finite type and the fibre of $U \to Y$ over $y$ is the set $U^0$ of generic points of irreducible components of $U$ (Properties of Spaces, Lemma 66.11.1). Hence $U^0$ is finite (Morphisms, Lemma 29.20.14) and after shrinking $Y$ we may assume that $U \to Y$ is finite (Morphisms, Lemma 29.51.1). Next, consider $R = U \times _ X U$. Since the projection $s : R \to U$ is étale we see that $R^0 = s^{-1}(U^0)$ lies over $y$. Since $R \to U \times _ Y U$ is a monomorphism, we conclude that $R^0$ is finite as $U \times _ Y U \to Y$ is finite. And $R$ is separated (Properties of Spaces, Lemma 66.6.4). Thus we may shrink $Y$ once more to reach the situation where $R$ is finite over $Y$ (Morphisms, Lemma 29.51.5). In this case it follows that $X = U/R$ is finite over $Y$ by exactly the same arguments as given in the proof of Lemma 68.21.1 (or we can simply apply that lemma because it follows immediately that $X$ is quasi-separated as well).

Assume $f$ is separated and $Y$ affine irreducible. Choose $V \subset Y$ and $U \subset X$ as in Lemma 68.21.2. Since $f|_ U : U \to V$ is finite, we see that $U \subset f^{-1}(V)$ is closed as well as open (Morphisms of Spaces, Lemmas 67.40.6 and 67.45.9). Thus $f^{-1}(V) = U \amalg W$ for some open subspace $W$ of $X$. However, since $U$ contains all the codimension $0$ points of $X$ we conclude that $W = \emptyset $ (Properties of Spaces, Lemma 66.11.2) as desired. $\square$