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.



Computations of divisor class monoids
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.


3. 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 $x\_ix\_j = x\_{\max\{i,j\}}$.


