Lemma 67.35.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $Y$ is locally Noetherian and $f$ locally of finite type. Let $x \in |X|$ with image $y \in |Y|$. Then we have
\begin{align*} & \text{the dimension of the local ring of }X\text{ at }x \leq \\ & \text{the dimension of the local ring of }Y\text{ at }y + E - \\ & \text{ the transcendence degree of }x/y \end{align*}
Here $E$ is the maximum of the transcendence degrees of $\xi /f(\xi )$ where $\xi \in |X|$ runs over the points specializing to $x$ at which the local ring of $X$ has dimension $0$.
Proof.
Choose an affine scheme $V$, an étale morphism $V \to Y$, and a point $v \in V$ mapping to $y$. Choose an affine scheme $U$ , an étale morphism $U \to X \times _ Y V$ and a point $u \in U$ mapping to $v$ in $V$ and $x$ in $X$. Unwinding Definition 67.33.1 and Properties of Spaces, Definition 66.10.2 we have to show that
\[ \dim (\mathcal{O}_{U, u}) \leq \dim (\mathcal{O}_{V, v}) + E - \text{trdeg}_{\kappa (v)}(\kappa (u)) \]
Let $\xi _ U \in U$ be a generic point of an irreducible component of $U$ which contains $u$. Then $\xi _ U$ maps to a point $\xi \in |X|$ which is in the list used to define the quantity $E$ and in fact every $\xi $ used in the definition of $E$ occurs in this manner (small detail omitted). In particular, there are only a finite number of these $\xi $ and we can take the maximum (i.e., it really is a maximum and not a supremum). The transcendence degree of $\xi $ over $f(\xi )$ is $\text{trdeg}_{\kappa (\xi _ V)}(\kappa (\xi _ U))$ where $\xi _ V \in V$ is the image of $\xi _ U$. Thus the lemma follows from Morphisms, Lemma 29.52.2.
$\square$
Comments (2)
Comment #1535 by Pieter Belmans on
Comment #1565 by Johan on