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?



does a line bundle always have a degree
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 $H^2(X;\mathbb{Z})$.


