Lemma 65.22.5. Let $S$ be a scheme. Let $f : X \to Y$ be an étale morphism of algebraic spaces over $S$. Let $x \in X$. Then (1) $\dim _ x(X) = \dim _{f(x)}(Y)$ and (2) the dimension of the local ring of $X$ at $x$ equals the dimension of the local ring of $Y$ at $f(x)$. If $f$ is surjective, then (3) $\dim (X) = \dim (Y)$.

**Proof.**
Choose a scheme $U$ and a point $u \in U$ and an étale morphism $U \to X$ which maps $u$ to $x$. Then the composition $U \to Y$ is also étale and maps $u$ to $f(x)$. Thus the statements (1) and (2) follow as the relevant integers are defined in terms of the behaviour of the scheme $U$ at $u$. See Definition 65.9.1 for (1). Part (3) is an immediate consequence of (1), see Definition 65.9.2.
$\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: