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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 $\mathbb P(V)$ 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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up vote 15 down vote accepted

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

I have a silly answer to your second question. Take a disjoint union of an elliptic curve with any other curve, and set L to be a nontrivial degree 0 bundle on the elliptic curve and trivial on the other curve.

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I see, so maybe I mean to ask a much harder question. For nonsingular curves over C I think its true that degree(L) = 0 plus h^0(L) = 1 implies that L = O_X. So I guess I'm wondering how many assumptions you can relax (if any) before this stops being true. – solbap Oct 12 '09 at 6:09
Yeah, see Lemma 1.2 and its proof in section 4.1 in Hartshorne. Your divisor has to be effective to get h^0 positive, and if it has degree 0, then it is the zero divisor. I believe this extends to all proper varieties. – S. Carnahan Oct 12 '09 at 22:04

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