Lemma 29.29.5 (0397)—The Stacks project

Table of contents
Part 2: Schemes
Chapter 29: Morphisms of Schemes
Section 29.29: Morphisms of given relative dimension
Lemma 29.29.5 (cite)

Lemma 29.29.5. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is locally of finite type. Let $x \in X$ with $s = f(x)$. Then $f$ is quasi-finite at $x$ if and only if $\dim _ x(X_ s) = 0$. In particular, $f$ is locally quasi-finite if and only if $f$ has relative dimension $0$.

Proof. If $f$ is quasi-finite at $x$ then $\kappa (x)$ is a finite extension of $\kappa (s)$ (by Lemma 29.20.5) and $x$ is isolated in $X_ s$ (by Lemma 29.20.6), hence $\dim _ x(X_ s) = 0$ by Lemma 29.28.1. Conversely, if $\dim _ x(X_ s) = 0$ then by Lemma 29.28.1 we see $\kappa (s) \subset \kappa (x)$ is algebraic and there are no other points of $X_ s$ specializing to $x$. Hence $x$ is closed in its fibre by Lemma 29.20.2 and by Lemma 29.20.6 (3) we conclude that $f$ is quasi-finite at $x$. $\square$

Comments (1)

Comment #8421 by Ryo Suzuki on May 23, 2023 at 22:12

The proof can be simplified.

By virtue of Lemma 01TH, it is sufficient to prove that: Let $Y$ be a locally Noetherian scheme, and $x \in Y$ . Then $x$ is isolated in $Y$ if and only if ${\rm dim}_x Y=0$ .

Proof. If $x$ is an isolated point, $\{x\}$ is open (by Definition 06RM). Hence {\rm dim}_x Y = 0 (by Definition 0055). Conversely, if ${\rm dim}_x(Y) =0$ , then there exists an open subset $x\in U\subset Y$ such that ${\rm dim} U = 0$ . We can take $U$ Noetherian because $Y$ is locally Noetherian. Since $U$ is 0-dimensional Noetherian scheme, it is discrete. Hence $x$ is isolated point of $Y$ . q.e.d.

Comments (1)

Post a comment