In Eisenbud's commutative algebra book, he defines a Weil divisor to be an element of the free abelian group generated by codimension 1 prime ideals of a ring $R$. He then goes on to define a Cartier divisor as an element of the group of invertible ideals. (p. 261)

The relationship between his definition of Weil divisor here and the usual definition seems fairly straightforward. However, I'm having trouble seeing any sort of relationship between what he calls a Cartier divisor and the usual definition. This is the only place I've seen this group called the group of Cartier divisors. Any help would be appreciated.


You are right: Eisenbud's definition is not the standard one .

Standard definition : a Cartier divisor on the scheme $X$ is a global section $D \in \Gamma(X,\mathcal K^*_X/ O^*_X)$, where $\mathcal K$ is the sheaf of rational functions on $X$.
In the affine case $X=Spec(R)$, this translates into $D\in (TotR)^*/R^*$, where $Tot R$ is the ring of fractions $S^{-1}R$ with $S=$ the set of non zero-divisors.

In practice, $D$ is given by an open covering$(U_i)$ of $X$ and rational functions $s_i\in \mathcal K^* (U_i)$, such that $s_i=g_{ij}\cdot s_j$ on $U_i\cap U_j$ and $g_{ij}\in \mathcal O^*(U_{ij})$.
To these data you associate Eisenbud's invertible ideal sheaf $\mathcal O(D)\subset \mathcal K_X$, characterized by: $\mathcal O(D)|U_i=\frac{1}{s_i}\mathcal K_X|U_i$.
This correspondence between Cartier divisors and invertible subsheaves of the rational functions goes over to isomorphism classes and yields an isomorphism between $Cacl(X)$, the additive group of classes of Cartier divisors, and $Inv(X)$, the multiplicative group of classes of invertible fractional ideal sheaves, in other words between the standard point of view and Eisenbud's.

I think Eisenbud adopted his point of view because in the affine case $X=Spec(R) $, fractional ideals of $R$ are a fairly elementary concept and are already known to students having followed a first course on algebraic number theory.

  • $\begingroup$ Thanks this helped clear it up. $\endgroup$ – user23214 Jan 20 '12 at 19:09

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