Lemma 37.40.1. 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 29.44.11, 29.20.10, and 29.44.10. We have (2) implies (3) by Morphisms, Lemma 29.20.7. We have (3) implies (4) by the definition of proper morphisms and Morphisms, Lemmas 29.20.9 and 29.20.10.

Assume (4). Pick $s \in S$. By Morphisms, Lemma 29.20.7 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 37.37.6. 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 34.4.1. Hence it is an fpqc covering, see Topologies, Lemma 34.9.6. Hence we conclude $f$ is finite by Descent, Lemma 35.20.23. $\square$

Comment #4586 by fherzig on

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

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).