Lemma 66.12.6. Let $S$ be a scheme. Let $X \to Y$ be a locally quasi-finite morphism of algebraic spaces over $S$. Let $x \in |X|$ with image $y \in |Y|$. Then the dimension of the local ring of $Y$ at $y$ is $\geq$ to the dimension of the local ring of $X$ at $x$.

Proof. The definition of the dimension of the local ring of a point on an algebraic space is given in Properties of Spaces, Definition 64.10.2. Choose an étale morphism $(V, v) \to (Y, y)$ where $V$ is a scheme. Choose an étale morphism $U \to V \times _ Y X$ and a point $u \in U$ mapping to $x \in |X|$ and $v \in V$. Then $U \to V$ is locally quasi-finite and we have to prove that

$\dim (\mathcal{O}_{V, v}) \geq \dim (\mathcal{O}_{U, u})$

This is Algebra, Lemma 10.124.4. $\square$

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).