0

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 $\pi|_{D} :D \to S$ 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?

Plantation
  • 2,417
  • 1
    Flatness of $\pi$ should tell you that all the fibers are 1-dimensional right? Also isn't it necessarily the case that a prime Weil divisor on a surface is 1-dimensional by definition? – Daniel Nov 22 '23 at 15:43
  • Prime weil divisor is of codimension 1, not gauranteed to have dimension 1. Do you think that we can prove that a prime weil divisor is 1-dimensional from definition? How? And why you mentioned the 1-dimensionality of fibers for flat morphism? Uhm I searched everywhere(?) but couldn't find any related information :) – Plantation Nov 22 '23 at 19:31
  • 1
    The Gortz Prop 5.30 you mentioned should imply that $D$ has dimension 1 if and only it has codimension 1. The fact about the dimension of the fibers of a flat map can be found as Corollary 4.3.14 in Liu's book. (pg. 138 in my pdf) – Daniel Nov 23 '23 at 05:15
  • If we try to apply the Gortz Prop. 5.30 directly, we may stuck since $X$ is not necessarily finite type over a field $k$. How can we breaktrhough this difficulty? Perhaps, can you give me a proof of your own? Or can you give me a hint? I want to understand this issue rigorously and desperately. ( P.s. And.. to apply Corollary 4.3.14 in Liu's book to show that the fiber of $\pi : X \to S$ is $1$-dimensional, we face at difficulty since each $X$ , $S$ are not always algebraic varieties. But by the Liu's Lemma 8.3.3., each fiber $X_s$ is a projective curve over $k(s)$ so that has dimension 1 !) – Plantation Nov 23 '23 at 07:53
  • 1
    Ah, forgive my ignorance. I was thinking that $S$ was finite type but I see now that is not an assumption. I think the idea of 4.3.14 should still work since the preceding theorem 4.3.13 has no finite-type or variety hypotheses. – Daniel Nov 23 '23 at 18:03
  • Yes. So how can we show that the dimension of vertical divisor D is 1 (My main question )? Do you mean that we can borrow idea of 4.3.14? I don't understand your intention. How can we apply Gortz prop.5.30 ? I am struggling with it until now~ – Plantation Nov 23 '23 at 18:46

1 Answers1

1

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.

Daniel
  • 5,274
  • 1
  • 10
  • 25
  • O.K. I got it. Thank you very much! C.f. Let $Z$ be the irreducible component of $X_s$ containing $D$. Then $Z \neq X$ since if $Z=X$, then $1=\dim X_s = \dim X =2 $, which is contradiction ~. – Plantation Nov 24 '23 at 01:25