Lemma 68.25.6. Let $S$ be a scheme. Let $f : Y \to X$ be a surjective finite morphism of decent and locally Noetherian algebraic spaces. Let $\delta : |X| \to \mathbf{Z}$ be a function. If $\delta \circ |f|$ is a dimension function, then $\delta $ is a dimension function.

**Proof.**
Let $x \mapsto x'$, $x \not= x'$ be a specialization in $|X|$. Choose $y \in |Y|$ with $|f|(y) = x$. Since $|f|$ is closed (Morphisms of Spaces, Lemma 67.45.9) we find a specialization $y \leadsto y'$ with $|f|(y') = x'$. Thus we conclude that $\delta (x) = \delta (|f|(y)) > \delta (|f|(y')) = \delta (x')$ (see Topology, Definition 5.20.1). If $x \leadsto x'$ is an immediate specialization, then $y \leadsto y'$ is an immediate specialization too: namely if $y \leadsto y'' \leadsto y'$, then $|f|(y'')$ must be either $x$ or $x'$ and there are no nontrivial specializations between points of fibres of $|f|$ by Lemma 68.18.10.
$\square$

## 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Your email address will not be published.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)