The Stacks project

Proof. In the statement $\dim _ u(U)$ is the dimension of $U$ at $u$ as defined in Topology, Definition 5.10.1 as the minimum of the Krull dimensions of open neighbourhoods of $u$ in $U$. Similarly for $\dim _ v(V)$.

Let us show that $\dim _ v(V) \geq \dim _ u(U)$. Let $V'$ be an open neighbourhood of $v$ in $V$. Then there exists an open neighbourhood $U'$ of $u$ in $U$ contained in $f^{-1}(V')$ such that $\dim _ u(U) = \dim (U')$. Suppose that $Z_0 \subset Z_1 \subset \ldots \subset Z_ n$ is a chain of irreducible closed subschemes of $U'$. If $\xi _ i \in Z_ i$ is the generic point then we have specializations $\xi _ n \leadsto \xi _{n - 1} \leadsto \ldots \leadsto \xi _0$. This gives specializations $f(\xi _ n) \leadsto f(\xi _{n - 1}) \leadsto \ldots \leadsto f(\xi _0)$ in $V'$. Note that $f(\xi _ j) \not= f(\xi _ i)$ if $i \not= j$ as the fibres of $f$ are discrete (see Morphisms, Lemma 29.36.7). Hence we see that $\dim (V') \geq n$. The inequality $\dim _ v(V) \geq \dim _ u(U)$ follows formally.

Let us show that $\dim _ u(U) \geq \dim _ v(V)$. Let $U'$ be an open neighbourhood of $u$ in $U$. Note that $V' = f(U')$ is an open neighbourhood of $v$ by Morphisms, Lemma 29.25.10. Hence $\dim (V') \geq \dim _ v(V)$. Pick a chain $Z_0 \subset Z_1 \subset \ldots \subset Z_ n$ of irreducible closed subschemes of $V'$. Let $\xi _ i \in Z_ i$ be the generic point, so we have specializations $\xi _ n \leadsto \xi _{n - 1} \leadsto \ldots \leadsto \xi _0$. Since $\xi _0 \in f(U')$ we can find a point $\eta _0 \in U'$ with $f(\eta _0) = \xi _0$. Consider the map of local rings

which is a flat local ring map by Morphisms, Lemma 29.36.12. Note that the points $\xi _ i$ correspond to primes of the ring on the left by Schemes, Lemma 26.13.2. Hence by going down (see Algebra, Section 10.41) for the displayed ring map we can find a sequence of specializations $\eta _ n \leadsto \eta _{n - 1} \leadsto \ldots \leadsto \eta _0$ in $U'$ mapping to the sequence $\xi _ n \leadsto \xi _{n - 1} \leadsto \ldots \leadsto \xi _0$ under $f$. This implies that $\dim _ u(U) \geq \dim _ v(V)$. $\square$