The Stacks project

Proof. Let $X$, $Z$, $\xi$ be as in the lemma. Since $X$ is locally of finite type over $B$ we see that $X$ is Jacobson, see Decent Spaces, Lemma 68.23.1. Hence $X_{\text{ft-pts}} \subset |X|$ is the set of closed points by Decent Spaces, Lemma 68.23.3. Given a chain $T_0 \supset \ldots \supset T_ e$ of irreducible closed subsets of $|Z|$ we have $T_ e \cap X_{\text{ft-pts}}$ nonempty by Morphisms of Spaces, Lemma 67.25.6. Thus we can always assume such a chain ends with $T_ e = \{ z\}$ for some $z \in |Z|$ closed. It follows that $\dim (Z) = \sup _ z \text{codim}(\{ z\} , |Z|)$ where $z$ runs over the closed points of $|Z|$. We have $\text{codim}(\{ z\} , Z) = \delta (\xi ) - \delta (z)$ by Topology, Lemma 5.20.2. By Morphisms of Spaces, Lemma 67.25.4 the image of $z$ is a finite type point of $B$, i.e., a closed point of $|B|$. By Morphisms of Spaces, Lemma 67.33.4 the transcendence degree of $z/b$ is $0$. We conclude that $\delta (z) = \delta (b) = 0$ by assumption. Thus we obtain equality

for all $z \in |Z|$ closed. Finally, we have that $\text{codim}(\{ z\} , Z)$ is equal to the dimension of the local ring of $Z$ at $z$ by Decent Spaces, Lemma 68.20.2 which in turn is equal to $\dim (\mathcal{O}_{Z, \overline{z}})$ by Properties of Spaces, Lemma 66.22.4. $\square$