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Suppose $C$ is an algebraic curve (which has singular points) over an algebraically closed field $k$, and that $f$ is a rational function on $C$. How does one defines the Weil divisor of $f$?

The problem is that the local rings of $C$ at singular points are not DVR's, so I do not have an obvious candidate for an order at a point.

Thanks!

Edit: Let me give an example, inspired by an answer from below. Suppose $C$ is curve $y^2=x^3$. What would be the order of the rational function $x/y$ at the origin?

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You can define the "order" of a regular function at a point $x$ to be the length of the quotient: $\mathcal O_{X,x}/(f)$ (which will be Artinian, so has finite length). You can probably find the details in Hartshorne's book or Fulton's "Intersection theory".

Details added: here is a sketch that this is well-defined: Suppose your rational function can be represented by $f/g $ and f'/g' at the stalk, which I will call $R$. $R$ is one-dimensional domain (you can assume less, but let's make it simple). We know f/g =f'/g' and want to prove l(R/fR) - l(R/gR) = l(R/f'R) - l(R/g'R) or l(R/fR) + l(R/g'R) = l(R/f'R) + l(R/gR)

Since fg' = f'g, we are done by the following general fact:

$l(R/abR) = l(R/aR) + l(R/bR) $

Hint: look at the sequence $0 \to aR/abR \to R/abR \to R/aR \to 0$

PS: One can also give a definition by using the normalization of $X$, but I think the above is more down-to-earth and computable, if less sexy.

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    @anonymous: please see the edit.2011-03-03