The Stacks project

Lemma 66.21.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type. Let $X^0 \subset |X|$, resp. $Y^0 \subset |Y|$ denote the set of codimension $0$ points of $X$, resp. $Y$. Assume

  1. $Y$ is decent,

  2. $X^0$ and $Y^0$ are finite and $f^{-1}(Y^0) = X^0$,

  3. either $f$ is quasi-compact or $f$ is separated.

Then there exists a dense open $V \subset Y$ such that $f^{-1}(V) \to V$ is finite.

Proof. By Lemmas 66.20.4 and 66.20.1 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 ├ętale 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 64.11.1). 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 29.50.1). Next, consider $R = U \times _ X U$. Since the projection $s : R \to U$ is ├ętale 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 64.6.4). Thus we may shrink $Y$ once more to reach the situation where $R$ is finite over $Y$ (Morphisms, Lemma 29.50.5). 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 66.21.1 (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 66.21.2. 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 65.40.6 and 65.45.9). 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 64.11.2) as desired. $\square$


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