Lemma 29.28.5 (0A3V)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 29: Morphisms of Schemes
Section 29.28: Morphisms and dimensions of fibres
Lemma 29.28.5 (cite)

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Lemma 29.28.5. Let $f : X \to Y$ be a morphism of finite type with $Y$ quasi-compact. Then the dimension of the fibres of $f$ is bounded.

Proof. By Lemma 29.28.4 the set $U_ n \subset X$ of points where the dimension of the fibre is $\leq n$ is open. Since $f$ is of finite type, every point is contained in some $U_ n$ (because the dimension of a finite type algebra over a field is finite). Since $Y$ is quasi-compact and $f$ is of finite type, we see that $X$ is quasi-compact. Hence $X = U_ n$ for some $n$ . $\square$

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Comment #722 by Kestutis Cesnavicius on June 21, 2014 at 20:07

Shouldn't the inequality in the proof be reversed?

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