The Stacks project

Lemma 66.10.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following quantities are equal:

  1. The dimension of $X$.

  2. The supremum of the dimensions of the local rings of $X$.

  3. The supremum of $\dim _ x(X)$ for $x \in |X|$.

Proof. The numbers in (1) and (3) are equal by Definition 66.9.2. Let $U \to X$ be a surjective ├ętale morphism from a scheme $U$. The supremum of $\dim _ x(X)$ for $x \in |X|$ is the same as the supremum of $\dim _ u(U)$ for points $u$ of $U$ by definition. This is the same as the supremum of $\dim (\mathcal{O}_{U, u})$ by Properties, Lemma 28.10.2. This in turn is the same as (2) by definition. $\square$


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Your email address will not be published.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