Lemma 29.56.2 (01TI)—The Stacks project

Lemma 29.56.2. Let $f : X \to S$ be a morphism of schemes. The set of points of $X$ where $f$ is quasi-finite is an open $U \subset X$. The induced morphism $U \to S$ is locally quasi-finite.

Proof. Suppose $f$ is quasi-finite at $x$. Let $x \in U = \mathop{\mathrm{Spec}}(A) \subset X$, $V = \mathop{\mathrm{Spec}}(R) \subset S$ be affine opens as in Definition 29.20.1. By either Theorem 29.56.1 above or Algebra, Lemma 10.123.13, the set of primes $\mathfrak q$ at which $R \to A$ is quasi-finite is open in $\mathop{\mathrm{Spec}}(A)$. Since these all correspond to points of $X$ where $f$ is quasi-finite we get the first statement. The second statement is obvious. $\square$

Comments (4)

Comment #3963 by Manuel Hoff on January 26, 2019 at 10:54

Typo: It should say $x \in U = \mathop{\mathrm{Spec}}(A) \subset X$ , $V = \mathop{\mathrm{Spec}}(R) \subset S$ , not the other way around.

Comment #4099 by Johan on April 04, 2019 at 19:12

Thanks and fixed here.

Comment #5429 by slogan_bot on July 30, 2020 at 07:39

Suggested slogan: "The locally quasi-finite locus of a morphism is open"

P.S. This result is surprisingly difficult to locate!

Comment #5656 by Johan on November 14, 2020 at 17:22

Yes and it is a surprisingly hard thing to prove! OK, I've added the slogan and hopefully this will help people find it more easily. Google: are you listening?

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2 comment(s) on Section 29.56: Zariski's Main Theorem (algebraic version)

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