Lemma 72.15.1 (0EDB)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 72: Algebraic Spaces over Fields
Section 72.15: Dimension
Lemma 72.15.1 (cite)

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Lemma 72.15.1. Let $S$ be a scheme. Let $f : X \to Y$ be an integral morphism of algebraic spaces. Then $\dim (X) \leq \dim (Y)$ . If $f$ is surjective then $\dim (X) = \dim (Y)$ .

Proof. Choose $V \to Y$ surjective étale with $V$ a scheme. Then $U = X \times _ Y V$ is a scheme and $U \to V$ is integral (and surjective if $f$ is surjective). By Properties of Spaces, Lemma 66.22.5 we have $\dim (X) = \dim (U)$ and $\dim (Y) = \dim (V)$ . Thus the result follows from the case of schemes which is Morphisms, Lemma 29.44.9. $\square$

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