3
$\begingroup$

Note: There are already several related questions, without any definite answer.

I want to find an example of a Noetherian integral scheme $X$ which contains a Cartier divisor that is not linearly equivalent to a difference of two effective Cartier divisors. It will be great if $X$ is a variety.

This stack project exercise suggests a proper variety that is not projective. However, I cannot prove why this is a counterexample. It is not hard to show that $X$ does not have any nontrivial effective Cartier divisor. Hence it is enough to find an example of a nontrivial Cartier divisor, or a nontrivial line bundle, on $X$, but it is hard to construct such an example.

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ OK,I think your notion of a Cartier divisor is different from Hartshorne's definition: a Cartier divisor on a variety $X$ is an element of $\Gamma(X, \mathcal{K}^*/\mathcal{O}^*)$ where $\mathcal{K}$ is the sheaf of rational functions. For a variety of positive dimension, this is always very big. $\endgroup$
    – Johan
    Commented Sep 15, 2022 at 0:33
  • $\begingroup$ Sorry. I mean "Cartier divisor up to linear equivalence". Fixed! $\endgroup$
    – Daebeom Choi
    Commented Sep 15, 2022 at 1:27
  • 1
    $\begingroup$ The Picard group of the example is nontrivial as it equals $k^*$. The Picard group of a variety is equal to the group of Cartier divisors up to rational equivalence. $\endgroup$
    – Johan
    Commented Sep 15, 2022 at 1:40
  • $\begingroup$ How can I calculate that Picard group of the example? $\endgroup$
    – Daebeom Choi
    Commented Sep 15, 2022 at 1:44
  • 1
    $\begingroup$ Take a few days to think about it. Fitst do some easier examples of glueings. Etc, etc. $\endgroup$
    – Johan
    Commented Sep 15, 2022 at 1:47

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .