# Computations of divisor class monoids

Let me first recall some definitions from the very first pages of Bourbaki, Commutative Algebra, Chapter 7, "Divisors".

Let $$A$$ be a (commutative) domain, $$K$$ its field of fractions. A fractional ideal of $$A$$ is a finitely generated $$A$$-submodule of $$K$$. The set of all non-zero fractional ideals of $$A$$ is called $$I(A)$$. On $$I(A)$$ there is a natural equivalence $$\sim$$ relation: for two fractional ideals $$\mathfrak a$$ and $$\mathfrak b$$ we write $${\mathfrak a} \sim {\mathfrak b}$$ if every principal fractional ideal containing $$\mathfrak a$$ also contains $$\mathfrak b$$ and vice-versa. The set of equivalence classes in $$I(A)$$ for this equivalence relation $$\sim$$ is called $$D(A)$$; its elements are called divisors of $$A$$. The multiplication of fractional ideals induces a multiplication on $$D(A)$$, which makes it a monoid. So $$D(A)$$ isthe divisor monoid of $$A$$.

Now on $$D(A)$$ we define a second equivalence relation, where two elements $$d$$ and $$d'$$ of $$D(A)$$ are equivalent if for some (or equivalently any) representative $$\mathfrak a$$ of $$d$$ and for some (or equivalently any) representative $$\mathfrak a$$ of $$d$$, one has $$\mathfrak a =\mathfrak a' x \text{ for some }x \in K^\ast.$$ The quotient of $$D(A)$$ by this equivalence relation clearly inherits the monoid structure of $$D(A)$$ and is called the divisor class monoid of $$A$$. Bourbaki doesn't introduce a special notation for it but let us denote it by $$DC(A)$$.

Bourbaki proves that $$D(A)$$ is a group (hence also $$DC(A)$$) if and only if $$A$$ is totally integrally closed (Theorem 1 of chapter 7). But I am interested in the cases where $$A$$ is not integrally closed, especially to the cases where $$A$$ is a noetherian complete domain of Krull dimension 1, or even more especially to the case where $$A$$ is the completed local ring at a singular point of an algebraic curve over $$\mathbb C$$. My question is:

Has there been any systematic attempt to compute the divisor class monoid $$DC(A)$$ for $$A$$ the completed local ring at a singular point of an algebraic curve? Or at least some example of non trivial computations of such $$DC(A)$$?

It seems to me that $$DC(A)$$ is a very natural invariant of a singularity of an algebraic curve. People working in the theory of singularities of algebraic or analytic curves (a vast subject) have certainly met this invariant, but I can't find any reference in the literature. Any pointers, or any suggestion to attack the problem is very welcome.

Remark: I know how to compute $$DC(A)$$ in simple special cases, for example the case where $$A$$ is the complete local ring of a cusp, i.e $$A=\{f \in \mathbb C[[T]], f'(0)=0\}$$. This is Exercise 1 in the exercises of chapter 7, \S1 of Bourbaki. In this case $$DC(A)$$ is the monoid $$\{1,x\}$$, where $$x$$ satisfies $$x^2=x$$. (Here $$x$$ can be the class of the ideal $$(T^2,T^3)$$ of $$A$$, for instance). But I'd like to know the answer for more general situations.

• Nice question! "I'd like to know the answer for more general situations". How general? Feb 24 '21 at 22:41

Here are a few remarks about $$DC(A)$$ (assuming $$A$$ is a complete Noetherian local domain of dimension $$1$$).

1. The equivalence relation in $$D(A)$$ is just isomorphism as $$A$$-modules. So you can view $$DC(A)$$ as the monoid of isomorphism classes of nonzero ideals $$I$$ in $$A$$ under multiplication.

2. For any $$x\in DC(A)$$, $$x^{n+1}=x^n$$ for $$n$$ large enough. That is because $$aI^n = I^{n+1}$$ if $$n$$ is large enough for any minimal reduction $$a$$ of $$I$$ (here one must first enlarge the residue field, but this is safe).

From above, it follows immediately that if $$DC(A)$$ is cancellative if and only if it is trivial if and only if $$A$$ is regular. This generalizes the fact in Bourbaki about being a group.

1. Over the complex numbers, $$DC(A)$$ is finite if and only if $$A$$ has finitely many Cohen-Macaulay modules up to isomorphisms (for reference see the books on this topic by Yoshino or Leuschke-Wiegand). For instance, if $$A$$ is a simple singularity (ADE singularity) then this holds. In such case, you can work out the monoid explicitly with a bit of effort.

Here is an example that contains what you mentioned in the last paragrach. Consider $$A_n= k[[t^2,t^{2n+1}]]$$. Up to isomorphisms, the only ideals are $$I_i=(x^{2i}, x^{2n+1})$$ with $$i=0,1,...,n$$. One can check that $$I_iI_j \cong I_{\max\{i,j\}}$$. So the monoid is $$\{x_0=1,...,x_n\}$$ with $$x_ix_j = x_{\max\{i,j\}}$$.

• Thanks a lot, this is very helpful. I am not sure I understand your proof of 2. What is a "minimal reduction" of $I$?
– Joël
Feb 25 '21 at 21:52
• @Joël: a reduction of $I$ is an ideal inside $I$ that has the same integral closure. A minimal reduction is a reduction minimal w.r.t inclusion. For ideals of finite colength in a local ring $A$ with infinite residue field, they exist and can be generated by $\dim A$ elements. You can find discussion of them in the book "Integral Closure..." by Huneke-Swanson, I think it is available freely and legally online. Feb 25 '21 at 22:00
• @Joël: by the way, it might be worth adding the tag "semigrous-and-monoids" as people in that community may have thought about this question. Feb 26 '21 at 1:39