Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a regular, irreducible projective scheme, of finite type over an arbitrary field $k$. Both Weil divisors and Cartier divisors are defined on $X$ and naturally correspond. If $k$ is algebraically closed (but of arbitrary characteristic), then we know (Hartshorne, Prop II.7.7) that global sections of $\mathcal{O}_X(D)$ correspond to effective divisors linearly equivalent to $D$. Is the same thing true for schemes such as $X$ above?

In particular, if $k=\bar{k}$ we can tell that $D$ is not effective if $\dim_k \Gamma (\mathcal{O}_X(D) = 0$. In general, do we at least know that $D$ is linearly equivalent to an effective divisor iff $\dim_k \Gamma (\mathcal{O}_X(D) \ne 0$.

It doesn't seem to be very hard to just repeat Hartshorne's proof to a more general setting, but I would also like to know what the most general versions of this correspondence are. For example, what if $X$ is only proper over $k$, or maybe over something like $\textrm{Spec } \mathbb{Z[1/2]}$. Is essentially the same thing true - or is it actually false? Any references will be very appreciated!

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Dear LMN, I think the answer to all your questions is yes. Everything is true.

See for example, Proposition 2.9 in Hartshorne's Generalized divisors on Gorenstein schemes and for some corrections, see Remark 2.9 Hartshorne's Generalized divisors and biliaison.

Here's a statement (more general versions can be found in those references). This version is basically 2.9 in the first reference.

Theorem: if $X$ is a normal Noetherian separated scheme, then for any Weil divisor $D$ on $X$, there is a 1-to-1 correspondence between non-degenerate global sections $s \in \Gamma(X, O_X(D))$, modulo sections of $\Gamma(X, O_X^*)$, and the set of effective divisors $D' \sim D$.

Here non-degenerate means not being contained in any prime associated to a generic point of any irreducible component of $X$ if I recall correctly.

In particular, I think the reason why Hartshorne generally assumed that $k = \overline{k}$ is that in these cases, these global sections can be used to induce maps to various projective spaces easily.

share|improve this answer
    
Thanks a lot! I've been wondering this sort of thing for a while. –  LMN Sep 12 '12 at 6:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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