Let $S= \operatorname{Spec} A$ be a local Dedekind scheme of dimension $1$, (eg spectrum of localization at a prime of the ring of integers of a number field). Let $s \in S$ it's unique closed point and $\eta \in S$ the generic one.
Let $f:\mathcal{X} \to S$ be a surjective proper morphism with generic fiber $X:= \mathcal{X}_{\eta}$, and $x \in X$ a closed point of $X$.
Question: Why the Zariski closure $\overline{x} \subset \mathcal{X} $ has dimension $1$?
(Motivation: Then we could easily conslude that the restriction of $f$ to $\overline{x}$ is quasi-finite and proper, so finite.)
But I'm wondering how to conclude that the Zariski closure $\overline{x}$ has dimension $1$ purely topologically without trying to proof directly that the restriction of $f$ to $\overline{x}$ is finite.
This problem appears in Qing Liu's "Algebraic Geometry and Arithmetic Curves" in 10.1.3.
Best Answer
Probably the easiest way to prove this is via flatness.
The closure $\bar{x}$ is integral and dominates $S$, thus is flat over $S$ (see Proposition III.9.7 in Hartshorne). The dimension of the fibres is thus constant, hence equal to $0$. Thus $\bar{x}$ is flat, proper, and quasi-finite over $S$, hence is finite over $S$.