Lemma 29.8.6 (01RM)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 29: Morphisms of Schemes
Section 29.8: Dominant morphisms
Lemma 29.8.6 (cite)

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Lemma 29.8.6. Let $f : X \to S$ be a morphism of schemes. Suppose that $X$ has finitely many irreducible components. Then $f$ is dominant (if and) only if for every irreducible component $Z \subset S$ the generic point of $Z$ is in the image of $f$ . If so, then $S$ has finitely many irreducible components as well.

Proof. Assume $f$ is dominant. Say $X = Z_1 \cup Z_2 \cup \ldots \cup Z_ n$ is the decomposition of $X$ into irreducible components. Let $\xi _ i \in Z_ i$ be its generic point, so $Z_ i = \overline{\{ \xi _ i\} }$ . Note that $f(Z_ i)$ is an irreducible subset of $S$ . Hence

$S = \overline{f(X)} = \bigcup \overline{f(Z_ i)} = \bigcup \overline{\{ f(\xi _ i)\} }$

is a finite union of irreducible subsets whose generic points are in the image of $f$ . The lemma follows. $\square$

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