The Stacks project

Lemma 70.3.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is proper and $Y$ is locally Noetherian. Let $y \in Y$ be a point of codimension $\leq 1$ in $Y$. Let $X^0 \subset |X|$ be the set of points of codimension $0$ on $X$. Assume in addition one of the following conditions is satisfied

  1. for every $x \in X^0$ the transcendence degree of $x/f(x)$ is $0$,

  2. for every $x \in X^0$ with $f(x) \leadsto y$ the transcendence degree of $x/f(x)$ is $0$,

  3. $f$ is quasi-finite at every $x \in X^0$,

  4. $f$ is quasi-finite at a dense set of points of $|X|$,

  5. add more here.

Then there exists an open subspace $Y' \subset Y$ containing $y$ such that $Y' \times _ Y X \to Y'$ is finite.

Proof. By Lemma 70.3.1 the morphism $f$ is quasi-finite at every point lying over $y$. Let $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ be a geometric point lying over $y$. Then $|X_{\overline{y}}|$ is a discrete space (Decent Spaces, Lemma 66.18.10). Since $X_{\overline{y}}$ is quasi-compact as $f$ is proper we conclude that $|X_{\overline{y}}|$ is finite. Thus we can apply Cohomology of Spaces, Lemma 67.22.2 to conclude. $\square$


Comments (0)

There are also:

  • 1 comment(s) on Section 70.3: Generically finite morphisms

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Your email address will not be published.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 0AD2. Beware of the difference between the letter 'O' and the digit '0'.