I've read in several places a statement like "a principle divisor on a projective curve has degree zero" (roughly).
A closer examination of some proofs (Fulton, Hartshorne) shows that this is based on the assumption of algebraic closure.
What happens if the coefficient field is not algebraically closed?
sage: F.<x> = FunctionField(QQ)
sage: R.<Y> = F[]
sage: L.<y> = F.extension(Y^2 - x^2 - 1)
sage: y.divisor()
- Place (1/x, 1/x*y - 1)
- Place (1/x, 1/x*y + 1)
+ Place (x^2 + 1, y)
(note: this calculation was done with a development version of Sage; it won't work on a stock Sage)
The function $y$ on the curve $y^2 = x^2 + 1$ seems to have two poles and only one zero, at least over the rational field (not its algebraic closure). The zero, however, is over a place of degree 2 (due to the $x^2+1$ generator).
So, it seems that we have to multiply our divisor multiplicities by the degree of the place in order to compute the degree of the divisor.
Does anybody have a reference that explains this in detail?
One reference for degrees of zero cycles (and hence in particular for divisors for curves) can be found in Chapter 1 of Fulton's intersection theory.
The story is this: Let $X,Y$ be algebraic schemes over $k$ and suppose we have a proper map $f:X\rightarrow Y$. This induces a pushforward map on cycles $f_*: A_k(X) \rightarrow A_k(Y)$ which maps a $k$-dimensional subvariety $[V]$ to $\deg(V:f(V))[f(V)]$, where $\deg(V:f(V))$ is the degree of the field extension of function fields $k(f(V)) \hookrightarrow k(V)$, if $\dim(V) = \dim(f(V))$, and $0$ otherwise.
Specifically in your context, you have a proper map $X\rightarrow \mathrm{Spec} k$ where $X$ is your projective curve, and your divisors correspond to $A_0(X)$. So you see that you find that you need to account for for the degree of the field extension.
The claim that a principal divisor on a projective curve has degree $0$ is still true, since the above map respects linear equivalence.