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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.

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    $\begingroup$ Have you tried drawing a picture of the problem? $\endgroup$
    – Daniel Loughran
    Commented Oct 9, 2023 at 16:02
  • $\begingroup$ @DanielLoughran: Yes, for $\mathcal{X}$ of dimension two this is clear (otherwise \overline{x}$ would be the complete space as one immediatelly see "drawing a picture"), but I don't see how to "extend" this "idea" to higher dimension. $\endgroup$
    – user267839
    Commented Oct 9, 2023 at 16:19
  • $\begingroup$ @DanielLoughran: Maybe I see now what you mean: since $\overline{x} \cap X={x}$, then $x$ must be open in $\overline{x}$ as preimage of $\eta$ (which is open in $S$). How can we conclude from this that $\overline{x}$ has dimension at most $1$. Note that if $\overline{x}$ would be Jacobson scheme we are done (cp with 33.20.3 (3) in stacks.math.columbia.edu/tag/0A21), but I not see any reason why $\overline{x}$ should have this property here. $\endgroup$
    – user267839
    Commented Oct 9, 2023 at 19:18

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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 $S$.

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