# Weil and Cartier divisors on a curve

I'm trying to understand the relationship between Weil divisors and Cartier divisors, and I would like to see why these are the same in the simple case where $X$ is a nonsingular projective curve over an algebraically closed field. What is the most concrete way of explaining the equivalence between the two sorts of divisors in this situation? In particular, if $P \in X$ is a closed point thought of as a prime Weil divisor, what is the Cartier divisor corresponding to $P$? What about the invertible sheaf corresponding to $P$?

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I can only suggest Hartshorne's Algebraic Geometry, Chapter 2 Paragraph 6. But I suppose you read it before! Waiting with you for a different answer... – Giovanni De Gaetano Dec 20 '11 at 18:38
This is explained in Hartshorne: see Chapter II, Proposition 6.11. The idea is that a Cartier divisor is essentially a meromorphic function modulo local regular functions, and so can be thought of as its zero/singular locus with multiplicities attached. – Zhen Lin Dec 20 '11 at 18:42
Yes, this question occurred to me while reading Hartshorne. But his explanation is not so clear, and I got confused when trying to specialize to the case of a smooth curve. – Justin Campbell Dec 20 '11 at 20:55

1) The Cartier divisor corresponding to the Weil divisor $1.P$ is given by the pair $s=\lbrace (U,z),(V, 1) \rbrace$ described as follows:
$\bullet$ $U$ is an open neigbourhood of $P$ and $z$ is a regular function on $U$ whose sole zero is $P$ with multiplicity one.
$\bullet \bullet$ $V=X\setminus \lbrace P\rbrace$ and of course $1$ is the constant function equal to $1$ on $V$.
(This pair determines a section $s\in \Gamma(X,\mathcal K^* _X/\mathcal O^*_X)$ if you unravel what it means to be a section of a quotient sheaf.)

2) The invertible sheaf corresponding to $P$ is the sheaf denoted by $\mathcal O(P)$.
Its $k$-vector space of sections $\Gamma(W,\mathcal O(P))$ over an open subset $W\subset X$ consists of those rational functions $f\in Rat(W)=Rat(X)$ regular on $W$ except perhaps at $P$, where $f$ is allowed a pole of order at most $1$.

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Thanks, very illuminating! To be clear: such a regular function $z$ in the first bullet point is obtained by lifting a uniformizer in the local ring at $P$ to some open neighborhood, then discarding any other points where it may happen to have a zero? – Justin Campbell Dec 20 '11 at 20:56
Dear Justin: that's it, exactly! – Georges Elencwajg Dec 20 '11 at 21:32