Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a scheme and $D$ be a Cartier divisor on $X$. Then $D$ determines a line bundle $\mathcal{O}(D)$ on $X$. Under which condition, is the converse true? That is, when does a line bundle come from a Cartier divisor. This is equivalen to saying when does a line bundle have a meromorphic section?

I know that when $X$ is a non-projective manifold line bundles do not have sections in general.

share|improve this question
    
Look at Hartshorne chapter II section 6. It is all about this. If $X$ is an integral scheme, then the Cartier class divisor group is isomorphic to the group of line bundles (up to isomorphism), i.e. $CaCl(X)\to Pic(X)$ is an isomorphism which maps $D\mapsto \mathcal{O}(D)$. I don't think it's equivalent to having a section, because Cartier divisors are locally principal and hence don't have to be given by the zero locus of a global section. –  Matt Oct 11 '12 at 21:33
    
See arxiv.org/pdf/math/0001104 –  Andrew Oct 11 '12 at 21:35
    
@Matt: he/she meant rational sections (see the title). –  user18119 Oct 11 '12 at 21:37
    
Matt If your line bundle is given by a Cartier divisor, then take a trivial section on one local trivialization and extends to a rational section of the line bundle. On the other hand, if there is a rational, its zero and poles gives a divisor which gives back the original line bundle to you. –  M. K. Oct 12 '12 at 4:56
add comment

2 Answers

up vote 4 down vote accepted

The map you describe $Cacl(X)\to Pic(X)$ sending the linear equivalence class $[D]$ of a Cartier divisor $D$ to the line bundle $\mathcal O(D)$ is always injective.
It is very often surjective: it is the case if $X$ is integral or if $X$ is projective over a field.
However Kleiman has given a complicated example of a complete non-projective 3-dimensional irreducible scheme on which there is a line bundle not having any non-zero rational section and thus not coming from a Cartier divisor.
The scheme $X$ is obtained from Hironaka's complete, integral, non-singular, non projective variety of dimension 3 (which is already a strange beast!) by adding nilpotents to the local ring of just one point.
The details can be found in Hartshorne's Ample Subvarieties of Algebraic Varieties , Chapter I, Example 1.3, page 9.

Here is a picture (in blue) of Hironaka's strange beast . The description is on page 185 of Shafarevich's book.

share|improve this answer
add comment

The most general assertion I know is:

Any invertible sheaf on $X$ is isomorphic to a $O_X(D)$ when $X$ is locally noetherian and if the associated points of $X$ are contained in an affine open subset of $X$ (EGA IV.21.3.4).

The condition on the associated points is satisfied if for instance $X$ is quasi-projective over a noetherian ring (then any finite subset of $X$ is contained in an affine open subset), or if $X$ is noetherian and reduced.

share|improve this answer
add comment

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.