Lemma 66.35.1. 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 66.33.1 and Properties of Spaces, Definition 65.10.2 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. $\square$

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