9
$\begingroup$

In algebraic geometry, there are two standard "kinds" of divisors: Weil divisors and Cartier divisors. Weil divisors provide better geometric intuition, while Cartier divisors are more general (if not precisely a generalization). In both of these kinds of divisors, the "key" results seem to be their relation to the Picard group (of isomorphism classes of line bundles).

However, one could also define a "divisor" to be an equivalence class of pairs $(s, \mathcal{L})$, where $s$ is an invertible rational section of $\mathcal{L}$. (By "invertible" I mean that there exists a rational section $s'$ of $\mathcal{L}^{\vee}$ such that $s' s = 1$.) We say $(s, \mathcal{L}) \sim (s', \mathcal{L}')$ if there is an isomorphism $\mathcal{L} \to \mathcal{L}'$ taking $s \mapsto s'$.

This definition works well at least for all Noetherian schemes (on which associated points behave nicely), and possibly more generally. It also seems less confusing than the definition of a Cartier divisor, and the relationship between divisors and line bundles is already embodied in the definition.

So why have I never seen this definition given as a kind of divisor?

[Note: I am aware of what "data" defines a Cartier divisor (the $(U_i, f_i)$, etc.) and how this provides a reasonably natural way to think of Cartier divisors geometrically (the "subscheme" defined locally by the $f_i$). So while I appreciate the thought, please don't waste your time writing a note for the sole purpose of explaining this.]

Edit: Since this issue has come up in an answer, I thought I would explain that by "rational section," I mean a (maximally extended) section over an open subset containing all the associated points; or equivalently, a section of $\mathcal{K} \otimes \mathcal{L}$, where $\mathcal{K}$ is the sheaf of total quotient rings of $\mathcal{O}_X$. (Actually, $\mathcal{K} \otimes \mathcal{L}$ is isomorphic to $\mathcal{K}$, but not naturally so.) A rational section can fail to be invertible if it vanishes on one or more associated points.

$\endgroup$
8
  • 2
    $\begingroup$ You should look in Fulton's book "Intersection Theory". I believe your definition is (almost) what he calls a pseudo-divisor (Chapter 2). $\endgroup$
    – mdeland
    Commented Jan 28, 2011 at 5:37
  • 5
    $\begingroup$ EGA IV-4, (21.2.11)? $\endgroup$ Commented Jan 28, 2011 at 8:27
  • 3
    $\begingroup$ Since every Cartier divisor is a Weil divisor but not conversely, I would rather say that Weil divisors are a generalization of Cartier divisors... $\endgroup$ Commented Jan 28, 2011 at 9:58
  • 3
    $\begingroup$ I thought the point of pseudo-divisors is that the section $s$ can be zero (which allows you to pull-back pseudo-divisors without any problems). So at least the intention of their definition is rather orthogonal to Charles' intention. $\endgroup$ Commented Jan 28, 2011 at 10:10
  • 3
    $\begingroup$ If $K\otimes L$ is isomorphic to $K$, then $L$ injects into $K$ and then $L$ is associated to a Cartier divisor. This however is not true in general. There is an old counterexample by Kleiman (published later in Comm. in Algebra, 2000) and by Schröer (Arkiv der Math. 2000). $\endgroup$
    – Qing Liu
    Commented Jan 28, 2011 at 10:32

1 Answer 1

1
$\begingroup$

Sorry to revive an old question, but I think Ravi Vakil uses exactly this notation on page 356 in his online notes:

http://math.stanford.edu/~vakil/216blog/FOAGmay1612public.pdf

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .