3
$\begingroup$

Let $X$ be a normal projective rational surface over $\mathbb{C}$ with finitely generated divisor class group $\text{Cl}(X)$. Consider the exact sequence $$0 \rightarrow \text{Pic}(X) \rightarrow \text{Cl}(X) \rightarrow \bigoplus_{x \in X^{\text{sing}}} \text{Cl}(\mathcal{O}_{X,x})$$ and the injections $\text{Cl}(\mathcal{O}_{X,x}) \hookrightarrow \text{Cl}(\hat{\mathcal{O}}_{X,x})$ where $\hat{\mathcal{O}}_{X,x}$ is the completion of the local ring of $x$.

What criteria are there for the right arrow to be surjective? Or generally, can you say anything specific about the image of $\text{Cl}(X)$?

My situation is pretty specific so I am willing to assume further that $\text{Pic}(X)$ is free and $[\text{Cl}(X) : \text{Pic}(X)] < \infty$ if that helps.

$\endgroup$

1 Answer 1

1
$\begingroup$

The cokernel of the right arrow is in general $H^2(X, \, \mathcal{O}_X^*)$. In many cases, for instance when the strict henselization of every $\mathcal{O}_{X, \, x}$ is a factorial ring, one can conclude that this group is a torsion group.

See this MathOverflow question and the related comments and answers.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer! That's interesting. Do you also know what happens when the right group is torsion in the first place? For example when $X$ has at worst rational double points as singularities. $\endgroup$
    – user269218
    Sep 23, 2015 at 14:41
  • $\begingroup$ This clearly answers my original question so I've accepted it. Thanks again. However this has led me to a follow-up question. $\endgroup$
    – user269218
    Sep 24, 2015 at 12:02

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.