A first application is the characterization of finite morphisms as proper morphisms with finite fibres.
Lemma 37.44.1. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:
$f$ is finite,
$f$ is proper with finite fibres,
$f$ is proper and locally quasi-finite,
$f$ 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 37.41.6. 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 34.9.7. Hence we conclude $f$ is finite by Descent, Lemma 35.23.23. $\square$
As a consequence we have the following useful results.
Lemma 37.44.2. Let $f : X \to S$ be a morphism of schemes. Let $s \in S$. Assume that $f$ is proper and $f^{-1}(\{ s\} )$ is a finite set. Then there exists an open neighbourhood $V \subset S$ of $s$ such that $f|_{f^{-1}(V)} : f^{-1}(V) \to V$ is finite.
Proof. The morphism $f$ is quasi-finite at all the points of $f^{-1}(\{ s\} )$ by Morphisms, Lemma 29.20.7. By Morphisms, Lemma 29.56.2 the set of points at which $f$ is quasi-finite is an open $U \subset X$. Let $Z = X \setminus U$. Then $s \not\in f(Z)$. Since $f$ is proper the set $f(Z) \subset S$ is closed. Choose any open neighbourhood $V \subset S$ of $s$ with $f(Z) \cap V = \emptyset $. Then $f^{-1}(V) \to V$ is locally quasi-finite and proper. Hence it is quasi-finite (Morphisms, Lemma 29.20.9), hence has finite fibres (Morphisms, Lemma 29.20.10), hence is finite by Lemma 37.44.1. $\square$
Lemma 37.44.3. Consider a commutative diagram of schemes Let $s \in S$. Assume
$X \to S$ is a proper morphism,
$Y \to S$ is separated and locally of finite type, and
the image of $X_ s \to Y_ s$ is finite.
\[ \xymatrix{ X \ar[rr]_ h \ar[rd]_ f & & Y \ar[ld]^ g \\ & S } \]
Then there is an open subspace $U \subset S$ containing $s$ such that $X_ U \to Y_ U$ factors through a closed subscheme $Z \subset Y_ U$ finite over $U$.
Proof. Let $Z \subset Y$ be the scheme theoretic image of $h$, see Morphisms, Section 29.6. By Morphisms, Lemma 29.41.10 the morphism $X \to Z$ is surjective and $Z \to S$ is proper. Thus $X_ s \to Z_ s$ is surjective. We see that either (3) implies $Z_ s$ is finite. Hence $Z \to S$ is finite in an open neighbourhood of $s$ by Lemma 37.44.2. $\square$
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