2
$\begingroup$

Let $X$ be a normal, projective variety and $U$ be the regular locus of $X$. Let $\mathcal{F},\mathcal{G}$ be reflexive sheaves on $X$ and $f:\mathcal{F} \to \mathcal{G}$ be a morphism. Suppose that the restriction of $f$ to $U$ is surjective i.e., $f|_U:\mathcal{F}|_U \to \mathcal{G}|_U$ is surjective.

Is it true that $f:\mathcal{F} \to \mathcal{G}$ is surjective? The problem that I have is, $U$ need not be affine.

The second question is: Is there any criterion when $U$ is going to be affine? More generally, does there exist an open subcheme $V$ contained in $U$ which is affine and satisfies $U\backslash V$ is of codimension at least $2$?

$\endgroup$

1 Answer 1

4
$\begingroup$

No, that is not true. Let $X$ be a normal, quasi-projective variety over a field $k$. Let $U\subset X$ be the regular locus. Denote by $Z\subset X$ the closed complement of $U$ with its reduced, induced structure. Assume that $Z$ is not empty. Denote by $\mathcal{I}\subset \mathcal{O}_X$ the ideal sheaf of $Z$. Since $X$ is quasi-projective, there exists an integer $d\geq 0$, and integer $N>0$, and a surjection $\widetilde{f}:\mathcal{O}_X(-d)^{\oplus N} \to \mathcal{I}$.

Denote $\mathcal{O}_X(-d)^{\oplus N}$ by $\mathcal{F}$. Denote by $\mathcal{G}$ the structure sheaf $\mathcal{O}_X$. Denote by $f$ the composition of $\widetilde{f}$ and the inclusion $\mathcal{I}\subset \mathcal{O}_X$. Then $f$ is a morphism of reflexive sheaves that is surjective on $U$, yet it is not surjective on all of $X$.

$\endgroup$

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.