The Stacks project

Proof. Let $f : X \to Y$ be as in the lemma. Let $\xi _0 \leadsto \xi _1 \leadsto \ldots \leadsto \xi _ e$ be a sequence of specializations in $X$. Set $x = \xi _ e$ and $y = f(x)$. Observe that $e \leq \dim (\mathcal{O}_{X, x})$ as the given specializations occur in the spectrum of $\mathcal{O}_{X, x}$, see Schemes, Lemma 26.13.2. By the dimension formula, Lemma 29.52.1, we see that

Hence we conclude that $e \leq \dim (Y) + \text{trdeg}_{R(Y)} R(X)$ as desired.

Next, assume $f$ is also closed. Say $\overline{\xi }_0 \leadsto \overline{\xi }_1 \leadsto \ldots \leadsto \overline{\xi }_ d$ is a sequence of specializations in $Y$. We want to show that $\dim (X) \geq d + r$. We may assume that $\overline{\xi }_0 = \eta $ is the generic point of $Y$. The generic fibre $X_\eta $ is a scheme locally of finite type over $\kappa (\eta ) = R(Y)$. It is nonempty as $f$ is dominant. Hence by Lemma 29.16.10 it is a Jacobson scheme. Thus by Lemma 29.16.8 we can find a closed point $\xi _0 \in X_\eta $ and the extension $\kappa (\eta ) \subset \kappa (\xi _0)$ is a finite extension. Note that $\mathcal{O}_{X, \xi _0} = \mathcal{O}_{X_\eta , \xi _0}$ because $\eta $ is the generic point of $Y$. Hence we see that $\dim (\mathcal{O}_{X, \xi _0}) = r$ by Lemma 29.52.1 applied to the scheme $X_\eta $ over the universally catenary scheme $\mathop{\mathrm{Spec}}(\kappa (\eta ))$ (see Lemma 29.17.5) and the point $\xi _0$. This means that we can find $\xi _{-r} \leadsto \ldots \leadsto \xi _{-1} \leadsto \xi _0$ in $X$. On the other hand, as $f$ is closed specializations lift along $f$, see Topology, Lemma 5.19.7. Thus, as $\xi _0$ lies over $\eta = \overline{\xi }_0$ we can find specializations $\xi _0 \leadsto \xi _1 \leadsto \ldots \leadsto \xi _ d$ lying over $\overline{\xi }_0 \leadsto \overline{\xi }_1 \leadsto \ldots \leadsto \overline{\xi }_ d$. In other words we have

which means that $\dim (X) \geq d + r$ as desired. $\square$