1
$\begingroup$

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 $S$?
$\endgroup$

1 Answer 1

3
$\begingroup$

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 $t=y=0$.)

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.