The Stacks project

Lemma 29.28.5. Let $f : X \to Y$ be a morphism of finite type with $Y$ quasi-compact. Then the dimension of the fibres of $f$ is bounded.

Proof. By Lemma 29.28.4 the set $U_ n \subset X$ of points where the dimension of the fibre is $\leq n$ is open. Since $f$ is of finite type, every point is contained in some $U_ n$ (because the dimension of a finite type algebra over a field is finite). Since $Y$ is quasi-compact and $f$ is of finite type, we see that $X$ is quasi-compact. Hence $X = U_ n$ for some $n$. $\square$

Comments (1)

Comment #722 by Kestutis Cesnavicius on

Shouldn't the inequality in the proof be reversed?

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Your email address will not be published.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 0A3V. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 29.28.5 as pdf