Weil divisors on Noetherian local ring of dimension $1$

Question

Ler $A$ be a Noetherian local ring of dimension $1$, with maximal ideal $\mathfrak m$ and minimal prime ideals $\mathfrak p_1,\dots, \mathfrak p_r$. In exercise 11.18 of the book "Algebraic Geometry 1" by Görtz and Wedhorn, they claim that in this setup we have $Z^1(\operatorname{Spec} A)=\mathbb Z^r$ where $Z^1(-)$ denotes the group of Weil divisors. I don't see how this is possible. Doesn't $A$ being 1-dimensional immediately imply that the only prime Weil divisor is $V(\mathfrak m)$, therefore implying $Z^1(\operatorname{Spec} A)=\mathbb Z$?

Answer 1

You are absolutely right: for a local noetherian ring $A$ of dimension $1$ with associated scheme $X=\operatorname {Spec}(A)$ we have indeed $$Z^1(X)=\mathbb Z$$ The authors of the book probably had in mind the correct equality $$Z_1(X)=\mathbb Z^r$$ where $Z_1(X)$ is the free group on the 1-dimensional (not 1-codimensional) integral subschemes of $X$.
This group $Z_1(X)$ is not defined in their book but they certainly know it and this might have induced their false statement.
(In Chow group theory, a vast generalization of class group theory, subscripts denote dimension and upper indices codimension, mimicking the convention in topology of homology versus cohomology).

Beware also that the definition on page 300 of the sheaf of rational functions $\mathcal K_X$ is false : the thing defined there is not even a presheaf!
For an analysis of that error and a correct definition consult Kleiman's article Misconceptions about $K_X$.