The Stacks project

Lemma 66.33.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is locally of finite type, $Y$ is Jacobson (Properties of Spaces, Remark 65.7.3), and $x \in |X|$ is a finite type point of $X$, then the transcendence degree of $x/f(x)$ is $0$.

Proof. Choose a scheme $V$ and a surjective ├ętale morphism $V \to Y$. Choose a scheme $U$ and a surjective ├ętale morphism $U \to X \times _ Y V$. By Lemma 66.25.5 we can find a finite type point $u \in U$ mapping to $x$. After shrinking $U$ we may assume $u \in U$ is closed (Morphisms, Lemma 29.16.4). Let $v \in V$ be the image of $u$. By Morphisms, Lemma 29.16.8 the extension $\kappa (u)/\kappa (v)$ is finite. This finishes the proof. $\square$


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Your email address will not be published.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