Algebraic Geometry – When Does a Line Bundle Have a Meromorphic Section?

algebraic-geometrycomplex-geometryvector-bundles

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.

Best Answer

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.

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