MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As a rule, the various groups and quotients of the divisor group on a variety have coefficients in $\mathbb{Z}$. That is, you take $\mathbb{Z}$-linear combinations of Weil divisors or Cartier divisors, and then to construct other groups you take quotients.

However, in some cases, people tensor with $\mathbb{Q}$ and $\mathbb{R}$. So my question is:

Are these the only rings that people use as coefficients for divisors on a variety?

My vague intuition is that it probably is, because $\mathbb{Z}$ is initial in commutative rings with identity, $\mathbb{Q}$ is a field of characteristic zero, so we can use it to kill torsion, and $\mathbb{R}$ is complete, so we can guarantee that there is an $\mathbb{R}$-divisor, plus with orbifolds, rational coefficients seem to show up naturally. But is this it? More generally, what about for cycles and cocycles? There's an analogy with cohomology and the Chow ring, and we do sometimes take cohomology with coefficients either in an arbitrary ring or in some other rings (finite fields, for instance, when studying things like nonorientable manifolds), which is why I started wondering about this.

share|cite|improve this question
If $\mathbb{R}$ is complete, so are the $\mathbb{Q}_p$ for each prime $p$. – Anweshi Jan 3 '10 at 16:20
Well, then I guess I should say "complete ordered field" then, to capture what I really meant. – Charles Siegel Jan 3 '10 at 17:08
up vote 11 down vote accepted

The answer to the question is yes. For example, if $\omega$ is a meromorphic 1-form on a curve (smooth and projective, say) over a field $k$, then one can naturally form a degree zero divisor with coefficients in $k$, namely the residue divisor of $\omega$.

share|cite|improve this answer

If you view a $\mathbb{Z}$-valued Cartier divisor (on say, an integral separated scheme X) as a $\mathbb{G}\_m$-torsor on X with generic trivialization, then for any torus T with character group $X^\ast(T)$, an $X^\ast(T)$-valued divisor on X is a T-torsor with generic trivialization. The analogous construction will work for any group of multiplicative type, if you replace the character group with a suitable fpqc sheaf of abelian groups on the base. This shows up for X a curve in some treatments of geometric class field theory, since the space of these divisors is the affine Grassmannian (in the sense of Beilinson-Drinfeld) for T over the Ran space of X. This space is used in, e.g., Gaitsgory's Twisted Whittaker model paper, where it forms a home (together with some similar objects) for factorizable sheaves.

If R is a number ring, you can demand that $X^\ast(T)$ be a sheaf of R-modules, so in this case, you're looking for torsors under tori with CM, with generic trivializations. I don't know where or if this is used, but it seems interesting enough.

share|cite|improve this answer

I think there could be different definitions of what a divisor is, but if want to keep the property that divisors are connected to line bundles, you are forced to relate to combinations of subvarieties with coefficients in $\mathbb Z$.

There is nothing preventing you from considering this group tensored with $\mathbb Q$, $\mathbb R$ (as you mention) or $\mathbb F_q$ (as Emerton's example does), but you won't get anything essentially new by different coefficients here in contrast with the cohomology, where in general $H^i(X, k) \ne H^i(X, \mathbb Z) \otimes k$.

Thus cohomology with the coefficients in sheaves other than $\mathbb Z$ appears more often than group of divisors with coefficients other than $\mathbb Z$, where it doesn't present any new phenomenon.

share|cite|improve this answer

I think that line bundles over gerbes have a notion of associated divisor where the coefficients are not integers.

share|cite|improve this answer

Your Answer


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.