The Stacks project

Lemma 69.23.2. (For a more general version see More on Morphisms of Spaces, Lemma 76.35.2). Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\overline{y}$ be a geometric point of $Y$. Assume

  1. $Y$ is locally Noetherian,

  2. $f$ is proper, and

  3. $|X_{\overline{y}}|$ is finite.

Then there exists an open neighbourhood $V \subset Y$ of $\overline{y}$ such that $f|_{f^{-1}(V)} : f^{-1}(V) \to V$ is finite.

Proof. The morphism $f$ is quasi-finite at all the geometric points of $X$ lying over $\overline{y}$ by Morphisms of Spaces, Lemma 67.34.8. By Morphisms of Spaces, Lemma 67.34.7 the set of points at which $f$ is quasi-finite is an open subspace $U \subset X$. Let $Z = X \setminus U$. Then $\overline{y} \not\in f(Z)$. Since $f$ is proper the set $f(Z) \subset Y$ is closed. Choose any open neighbourhood $V \subset Y$ of $\overline{y}$ with $Z \cap V = \emptyset $. Then $f^{-1}(V) \to V$ is locally quasi-finite and proper. Hence $f^{-1}(V) \to V$ has discrete fibres $X_ k$ (Morphisms of Spaces, Lemma 67.27.5) which are quasi-compact hence finite. Thus $f^{-1}(V) \to V$ is finite by Lemma 69.23.1. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A4W. Beware of the difference between the letter 'O' and the digit '0'.