I am reading the Liu's Algebraic Geometry and Arithmetic Curves, p.348, Lemma 8.3.3. and stuck at some point :
Lemma 8.3.3. Let $S$ be a Dedekind scheme of dimension $1$, with generic point $\eta$. Let $X \to S$ be a fibered ( resp. normal fibered ) surface. Then $X_{\eta}$ is an integral (resp. normal) curve over $K(S)$. For any $s\in S$, $X_s$ is a projective curve over $k(s)$.
Here, we call an integral, projective, flat $S$-scheme $\pi :X \to S$ of dimension $2$ a fibered surface over $S$. Note that the flatness of $\pi$ is equivalent to the surjectivity of $\pi$ ( C.f. Fibered surface is flat if and only if it surjects onto the base ). And 'algebraic curve over $k$' means an alebraic variety over $k$ whose irreduible components are of dimension $1$. ( C.f. his book p.75 )
Proof. Let $\xi$ be the generic point of $X$. As $X\to S$ is dominant, $\xi \in X_{\eta}$. Hence $\xi$ is the generic point of $X_{\eta}$, which shows that $X_{\eta}$ is irreducible. In addition, for ant $x\in X_{\eta}$, we have $\mathcal{O}_{X,x} \cong \mathcal{O}_{X_{\eta},x} $ (Exercise 3.7); hence $X_{\eta}$ is normal if $X$ is normal. We know that for any closed point $x\in X_s$, we have $\dim\mathcal{O}_{X_s,x} =\dim X_{\eta}$ (Proposition 4.4.16). It therefore remains to show that $\dim X_{\eta}=1$. Let $x\in X$ be a point such that $\dim \mathcal{O}_{X,x}=2$. Then $x$ is closed and therefore belongs to a closed fiber $X_s$. It follows that $\dim\mathcal{O}_{X_s,x} = \dim \mathcal{O}_{X,x} -\dim \mathcal{O}_{S,s}=1$ (Theorem 4.3.12 ). This proves that $\dim X_{\eta} =1$. QED.
My question is,
Q. Why the bold statement is true? There, the Proposition 4.4.16 is,
Proposition 4.4.16. Let $S$ be a Dedekind scheme with generic point $\eta$, and let $X\to S$ be a dominant morphism of finite type with $X$ irreducible. Then for any $s\in S$ such that $X_s \neq \varnothing$, we have $X_s$ equidimensional of dimension $\dim X_{\eta}$.
Let me expand my own argument to show the bold statement.
Since $X_s \to \operatorname{Spec}k(s)$ is of finite type, $X_s$ is noetherian. Let $X_s = X_1 \cup \cdots \cup X_n$ be its irreducible components, each of which is endowed with the closed subscheme structure. By the Proposition 4.4.16, we have $$\dim X_1 = \cdots = \dim X_n = \dim X_{\eta}.$$ Let $x\in X_s$ be a closed point. Then there exists $i$ such that $x \in X_i$. Note that the composition $X_i \hookrightarrow X_s \to \operatorname{Spec}k(s)$ is also finite type. Then by the Gortz's Algebraic Geometry book, Theorem 5.22. :
We have that $$\dim \mathcal{O}_{X_i, x} = \dim X_i = \dim X_{\eta}.$$
From this, can we show that $\dim \mathcal{O}_{X_s,x} = \dim X_{\eta}$ ? If so, how?
Or is there any other route to show this ?
Can anyone help ?
