Irreducible vertical divisor on a (normal) fibered surface has dimension $1$ ? ( Liu’s Algebraic Geometry )

algebraic-geometrydivisors-algebraic-geometry

I am reading the book Liu's Algebraic Geometry and arithmetic curves and some question arises.

Let $S$ be a Dedekind scheme. We call an integral, projective, flat $S$-scheme $\pi : X \to S$ of dimension $2$ a fibered surface over $S$. We will say that $X$ is normal fibered surface if $X$ is normal.

Let $D$ be an irreducible Weil divisor ( Definition 7.2.4 ) ; i.e., a cycle of codimension $1$ on $X$. We say that $D$ is horizontal if $\dim S=1$ and if $\pi|_{D} 😀 \to S$ is surjective (hence finite). If $\pi(D)$ is reduced to a point, we say that $D$ is vertical. More generally, an arbitrary Weil divisor will be called horizontal ( resp. vertical ) if its components are horizontal ( resp. vertical).

Q.1 My question is, if $D$ is irreduiclbe vertical divisor on a (normal) fibered surface $\pi : X \to S$ over a Dedekind scheme of dimension $1$, then $\dim D =1$?

This question has origination and I will upload that after considerting it more. I think that this question seems easy but I can't make rigorous proof until now. Is there a counter example? An issue that makes me stuck is, for closed subset $Z$ of $X$, the inequality $ \operatorname{codim}(Z,X) + \dim Z \le \dim X$ ( Liu's book p.69 ) is not always equality.

Q.2. Perhaps, if we add next condition, then $\dim D =1$ is true?

: there are fibered surface $Y \to S$ and a birational projective morphism $f:X\to Y$ such that $f(D)$ is a point.

I know that for an irreducible scheme $X$ of finite type over a field $k$ and a closed subset $Y$ of $X$, $\dim Y + \operatorname{codim}_XY =\dim X$ ( Gotz's Algebraic Geometry book, Proposition 5.30 ).

Can anyone help?

Best Answer

We claim that if $D$ is an irreducible vertical divisor, then $\dim D = 1$.

By assumption, $f(D) = \{s\}$ for some closed point $s \in S$, so we have a sequence of embeddings $D \subset X_s \subset X.$ Since the codimension of $D$ is $1$, it follows that $D$ is an irreducible component of $X_s$.

Hence, it suffices to show $X_s$ is purely $1$-dimensional, but this is established as part of Lemma 8.3.3 in Liu's book.

Related Question