The Stacks project

Lemma 69.20.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ separated and of finite type. The following are equivalent

  1. The morphism $f$ is proper.

  2. For any morphism $Y \to Z$ which is locally of finite presentation the map $|X \times _ Y Z| \to |Z|$ is closed, and

  3. there exists a scheme $V$ and a surjective ├ętale morphism $V \to Y$ such that $|\mathbf{A}^ n \times (X \times _ Y V)| \to |\mathbf{A}^ n \times V|$ is closed for all $n \geq 0$.

Proof. In view of the fact that a proper morphism is the same thing as a separated, finite type, and universally closed morphism, this lemma is a special case of Lemma 69.20.2. $\square$

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Your email address will not be published.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