On the situation of intersections along a proper morphism

Question

The short question is: Say $p:\bar{X}\rightarrow S$ is a proper and normal morphism with the following properties:

1. S is integral and smooth over a certain base field $k$,
2. $\bar{X}$ has a smooth and dense open subscheme $X$, which is a smooth affine curve over S,
3. $\bar{X}_s$ is of dimension $1$ on every fibre $\kappa(s)\rightarrow S$.

Then according to the literature, every integral closed subscheme $Z$ of $\bar{X}$ of codimension $1$ which is supported on $X$ should be finite and surjective over $S$. But why?

This is an assertion without proof in Voevodsky and Suslin's paper Singular homology of abstract algebraic varieties, Theorem 3.1. I can see that this is true for the bundle case, that is, $X=\mathbb{A}_S^1\rightarrow\mathbb{P}_S^1=\bar{X}$. In general, the closed subscheme $\bar{X}\backslash X$ should be non-empty on every fibre of $S$ since $X\rightarrow S$ is affine, so the problem seems related to something quite intuitive that

• For a proper morphism $\bar{X}\rightarrow S$ smooth of relative dimension $1$, do two elementary Weil divisors of $\bar{X}$ meet when one is over the generic point of $S$ and the other is over a elementary Weil divisor of $S$?

Answer 1

There is a general fact that a proper affine morphism is finite, and a general trick that for $\overline{X} \to S$ proper containing an open subset $X\to S$ affine, a closed subscheme $Z$ of $\overline{X}$ which is in fact closed in $X$ is a closed subscheme of $\overline{X}$, hence proper, and also a closed subscheme of $X$, hence affine, and thus finite. Combined with the surjectivity you note this gives what you want.

Depending on the definition of affine curve, I think your argument with Weil divisors won't work. Consider $S$ a curve and $X$ a family of $\mathbb P^1$s that degenerates to two $\mathbb P^1$'s joined with a node over a single point. We can take one of the Weil divisors to be one of the $\mathbb P^1$s and the other to be a section passing through the other $\mathbb P^1$s. Then they don't intersect, the problem being that the complement of the section is not affine. (Concretely, work over the base $k[t]$ and take the family of curves in $\mathbb P^2$ with equation $xy-t z^2$ where one Weil divisor is the section $(0:1:0)$ and the other is the curve where $t=y=0$.)