Lemma 36.39.1. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. $f$ is finite,

2. $f$ is proper with finite fibres,

3. $f$ is proper and locally quasi-finite,

4. $f$ is universally closed, separated, locally of finite type and has finite fibres.

Proof. We have (1) implies (2) by Morphisms, Lemmas 28.42.11, 28.19.10, and 28.42.10. We have (2) implies (3) by Morphisms, Lemma 28.19.7. We have (3) implies (4) by the definition of proper morphisms and Morphisms, Lemmas 28.19.9 and 28.19.10.

Assume (3). Pick $s \in S$. By Morphisms, Lemma 28.19.7 we see that all the finitely many points of $X_ s$ are isolated in $X_ s$. Choose an elementary étale neighbourhood $(U, u) \to (S, s)$ and decomposition $X_ U = V \amalg W$ as in Lemma 36.36.6. Note that $W_ u = \emptyset$ because all points of $X_ s$ are isolated. Since $f$ is universally closed we see that the image of $W$ in $U$ is a closed set not containing $u$. After shrinking $U$ we may assume that $W = \emptyset$. In other words we see that $X_ U = V$ is finite over $U$. Since $s \in S$ was arbitrary this means there exists a family $\{ U_ i \to S\}$ of étale morphisms whose images cover $S$ such that the base changes $X_{U_ i} \to U_ i$ are finite. Note that $\{ U_ i \to S\}$ is an étale covering, see Topologies, Definition 33.4.1. Hence it is an fpqc covering, see Topologies, Lemma 33.9.6. Hence we conclude $f$ is finite by Descent, Lemma 34.20.23. $\square$

Comment #4586 by fherzig on

The second paragraph should say "Assume (4)", not (3).

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