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.

  • $\begingroup$ Nice question! "I'd like to know the answer for more general situations". How general? $\endgroup$ 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 $x_ix_j = x_{\max\{i,j\}}$.

  • $\begingroup$ Thanks a lot, this is very helpful. I am not sure I understand your proof of 2. What is a "minimal reduction" of $I$? $\endgroup$
    – Joël
    Feb 25 '21 at 21:52
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    $\begingroup$ @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. $\endgroup$ Feb 25 '21 at 22:00
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    $\begingroup$ @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. $\endgroup$ Feb 26 '21 at 1:39

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