The Stacks project

Lemma 66.25.2. Let $S$ be a locally Noetherian and universally catenary scheme. Let $\delta : S \to \mathbf{Z}$ be a dimension function. Let $X$ be a decent algebraic space over $S$ such that the structure morphism $X \to S$ is locally of finite type. Let $\delta _ X : |X| \to \mathbf{Z}$ be the map sending $x$ to $\delta (f(x))$ plus the transcendence degree of $x/f(x)$. Then $\delta _ X$ is a dimension function on $|X|$.

Proof. Let $\varphi : U \to X$ be a surjective ├ętale morphism where $U$ is a scheme. Then the similarly defined function $\delta _ U$ is a dimension function on $U$ by Morphisms, Lemma 29.51.3. On the other hand, by the definition of relative transcendence degree in (Morphisms of Spaces, Definition 65.33.1) we see that $\delta _ U(u) = \delta _ X(\varphi (u))$.

Let $x \leadsto x'$ be a specialization of points in $|X|$. by Lemma 66.12.2 we can find a specialization $u \leadsto u'$ of points of $U$ with $\varphi (u) = x$ and $\varphi (u') = x'$. Moreover, we see that $x = x'$ if and only if $u = u'$, see Lemma 66.12.1. Thus the fact that $\delta _ U$ is a dimension function implies that $\delta _ X$ is a dimension function, see Topology, Definition 5.20.1. $\square$


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Your email address will not be published.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 0ED5. Beware of the difference between the letter 'O' and the digit '0'.