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Let $R = k[[x, y]]/(f)$, where $k$ is algebraically closed of characteristic zero. I'm particularly interested in studying the endomorphism ring of indecomposable MCM (maximal Cohen-Macaulay) modules when $R$ has finite type (ie. finitely many indecomposable MCM modules). For example, if $I$ is an ideal of $R$, then $End_R(I)$ can be identified with a subring of the integral closure of $R$ in its total quotient ring. In this case, it's not particularly difficult to compute $End_R(I)$ using well-known methods.

A particular example I would be interested in would be when $f = x^3 + y^4$. Here $R$ has two indecomposable MCM $R$-modules that are not isomorphic to ideals. I would like to know if there are any explicit computations of the endomorphism ring for such modules $M$.

My motivation for doing so is to study the Quillen $K$-groups $K_1(mod \hspace{.125 cm} R)$ ($mod\hspace{.125 cm} R =$ finitely generated modules). In particular, I would like to study $Aut(M)$ and $Aut(M)_{ab}$. See this paper for details.

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  • $\begingroup$ If the answer given is helpful, why not vote it up? If it isn't, probably you should leave a brief comment explaining why $\endgroup$ – Yemon Choi Sep 7 '14 at 0:13
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When $R$ is a simple curve singularity, the category $MCM(R)$ has finite type and the structure of its Auslander algebra is known. (When $\mathcal{C}$ is a Krull-Schmidt category with only finitely many indecomposables up to isomorphism, one can form a basic representation generator $M$ as the direct sum of one copy of each indecomposable object. The Auslander algebra of $\mathcal{C}$ is then defined as the endomorphism ring of $M$.) A good reference for information about the Auslander algebras of simple curve singularities is Chapter 9 of Yoshino's book Cohen-Macaulay Modules over Cohen-Macaulay Rings.

In particular, Yoshino computes the Auslander-Reiten quivers of $MCM(R)$ for the A-D-E singularities, which can also be viewed as the Ext-quiver (or bound quiver) of the associated Auslander algebra, call it $\Gamma$. Thus $\Gamma$ is a quotient of the (completed) path algebra of this quiver, and I believe the relations of $\Gamma$ can be chosen to be the "mesh relations" (i.e., there is one relation coming from each almost split sequence in $MCM(R)$. Unfortunately, I don't have a good reference for this fact, but it should not be too hard to verify by choosing nice irreducible maps corresponding to the arrows of the quiver and computing the relations they satisfy.)

Now the endomorphism ring of a single indecomposable MCM $R$-module can be computed as a corner ring of $\Gamma$. For instance, I would start by identifying the minimal nonzero cycles at the vertex corresponding to the indecomposable module of interest. I haven't tried to do this myself for the $f=x^3+y^4$ example, so perhaps it is harder than it sounds. In fact, in this case it seems nontrivial to compute even the endomorphism ring of $R$ itself in this way.

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  • $\begingroup$ Most of my education in the way of MCM modules has been through Yoshino's book. Another great (and slightly more modern) resource is Wiegand and Lueschke's book, "Cohen-Macaulay Representations". On a somewhat related note, it appears one might be able to compute the abelianization of the automorphism group without obtaining an explicit computation of the automorphism group or endomorphism ring (see Theorem 2.2 of this paper math.nju.edu.cn/~guoxj/articles/gp.pdf). $\endgroup$ – Zach Flores Oct 18 '14 at 20:55
  • $\begingroup$ It's not immediately clear Theorem 2.2 is applicable to the abelianization of the automorphism group, but a theorem of Leuschke shows that the Auslander Algebra has finite global dimension in the case of finite. Moreover, we are working in characteristic zero, so a theorem of Vaserstein applies (see the paper referenced in my original question). So, to summarize, understanding the endomorphism ring of these individual MCM modules allows us to understand something quite a bit bigger. Do you have a reference for these combinatorial methods? I would like to take a look. $\endgroup$ – Zach Flores Oct 18 '14 at 21:01

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