# does a line bundle always have a degree

For curves there is a very simple notion of degree of a line bundle or equivalently of a Weil or Cartier divisor. Even in any projective space P(V) divisors are cut out by hypersurfaces which are homogeneous polynomials of a certain degree.

Is there a more general notion of degree that applies to schemes with less structure?

Also, say you have a nice enough scheme X so line bundles correspond to Cartier divisors under linear equivalence. In whatever the most general setting is so that the degree of a line bundle makes sense, is there an example of a line bundle L \ne O_X that is degree 0 and has h^0(L) = 1?

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One generalization of degree is first Chern class: A Cartier divisor corresponds to a class in $H^1(X;\mathcal{O}_X^{\times})$, and you take its image under the boundary map of the long exact sequence corresponding to the exponential exact sequence $\mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}_X^{\times}$ where the second map is taking exponential (if you want to work in the algebraic category, there is a fix for this, using the exact sequence $\mathbb{Z}/n\mathbb{Z} \to \mathcal{O}_X^{\times} \to \mathcal{O}_X^{\times}$, where the second map is nth power).

Geometrically, on a smooth thing, this means you take the sum of all the Weil divisors as a homology class, and then take the Poincare dual class in $H^2(X;\mathbb{Z})$.

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So in the algebraic version of the exponential sequence what do you take as n? Is it the dimension of the space? –  solbap Oct 10 '09 at 18:39
You have to look at arbitrarily large n. Each one lets you see the reduction of the Chern class mod n. –  Ben Webster Oct 10 '09 at 21:30