In their article https://www.sciencedirect.com/science/article/pii/0001870891900378 , Auslander and Reiten suggested a generalisation of Gorenstein algebras for Artin algbras which they call Cohen-Macaulay (=CM) Artin algebras. By definition an Artin algebra $A$ is CM iff there are adjoint functors $F$ and $G$ between the subcategories of mod-A consisting of modules of finite projective dimension and finite injective dimension. We can write $F=Hom_A(w,-)$ for an $A$-bimodule $w$ (which must be the strong cotilting module).

It seems that since then there were no other articles related to this subject (besides a survey article by Auslander and Reiten one year later). I think the reason might be that no non-trivial example of CM Artin algebras were known besides Gorenstein algebras and finitistic dimension 0 algebras (and a tensor product construction by tensoring a CM algebra with the algebra of upper triangular 2x2 matrices) which might be called trivial examples.

Despite searching with the computer for a long time, I never found a non-trivial example but it seems the computer found now such an example using QPA.

Namely let $A$ be the quiver algebra with quiver Quiver( ["v1","v2","v3"], [["v1","v2","a1"],["v2","v1","a2"],["v2","v3","a3"],["v3","v2","a4"]] )

and relations: [ a2*a1-a3*a4, a4*a3, a2*a1*a3,a4*a2*a1, a1*a2*a1*a2,a2*a1*a2*a1 ].

Then $A$ should be CM (with infinite Gorenstein dimension) with strong cotilting module $w=eA \oplus \Omega^2(D(A))$ when $eA$ is the minimal faithful projective-injective $A$-module (note that $A$ has dominant dimension equal to two).

Question 1: Is this algebra really CM? (the computer can not test it for sure, but I think it should be correct in case I understand the results of Auslander and Reiten correctly).

Note that the algebra is not of any "trivial form": It is not Gorenstein, has finitistic dimension 2 and its dimension is 13 which is not divisble by 3 (so it is not obtained by tensoring with the algebra of upper triangular 2x2 matrices). As a sidequestion it might be interesting to know whether this is really the first "nontrivial" examples or any example of CM Artin algebras outside the articles of Auslander and Reiten.

Question 2: It is an open question by Auslander and Reiten whether for a CM Artin algebra $A$ with strong cotilting module $w$ the trivial extension $B$ of $A$ by $w$ is Gorenstein. What is the algebra $B$ in this case?

(Note that one should have $End_A(w) \cong A$, giving $w$ a bimodule structure. But I can hardly imagine this in practise so I do not find a good way to calculate with the bimodule structure in order to obtain $B$)

Question 3: It is another open question by Auslander and Reiten whether for CM Artin algebras $A$ and $B$, one has that also $A \otimes B$ is CM. Is in this example the enveloping algebra $A \otimes A$ CM?

edit: Some additional information on A. It has 20 indecomposable modules (thus it is reprentation-finite), it has Loewy length 4 and it has 6 indecomposable modules of finite projective (injective) dimension.

  • $\begingroup$ I think I found a positive answer to 3 and it seems that question 2 was answered already by Beligiannis and Reiten in the positive. $\endgroup$ – Mare Dec 26 '19 at 22:47

Here is my solution for question 1 and a negative answer to a question of Auslander and Reiten. Maybe someone is interested to check it or maybe someone is even able to simplify stuff so one does not need a computer to verify it. I checked parts using QPA. We will use 2 references. [AR1]: https://www.sciencedirect.com/science/article/pii/0001870891900378 and [AR2]: https://link.springer.com/chapter/10.1007/978-3-0348-8658-1_8 .

After proposition 3.1. in [AR2], the authors asked whether $CM(A)=\Omega^d(mod-A)$ implies $A$ being Gorenstein (when $d$ is the finitistic dimension of $A$ and $A$ is CM. I guess they assume $d>0$ to avoid trivialities.). The present example gives a counterexample to this accordings to QPA. See at the end for details.

For a subcategory $X$ we denote by $\hat{X}$ the modules $C$ such that there is an exact sequence $0 \rightarrow X_n \rightarrow ... \rightarrow X_0 \rightarrow C \rightarrow 0$ with $X_i \in X$. A cotilting module $T$ is called strong in case $\hat{addT}= I^{< \infty}$ (the category of modules of finite injective dimension). Such a module always exists for example for representation-finite algebras.

Now we use proposition 1.3. of [AR2], which tells us that an algebra $A$ is CM(=Cohen-Macaulay) iff there is an $A$-bimodule $W$ such that $W$ is on both sides a strong cotilting module and $End_A(W) \cong A$.

Let A=Quiver( ["v1","v2","v3"], [["v1","v3","a1"],["v2","v3","a2"],["v3","v1","a3"],["v3","v2","a4"]] ) [ (Z(3)^0)*a1*a3, (Z(3))*a3*a1+(Z(3)^0)*a4*a2, (Z(3)^0)*a2*a3*a1*a4 ]

be the quiver algebra in the thread ( the relations were simplified a bit). Then A is representation finite and the module $W=eA \oplus \Omega^2(D(A))$ is the unique cotilting module of injective dimension 2, and thus the strong cotilting module of $A$ as a right module.

By proposition 6.5. of [AR1] (see the correction of it in the answer by Alex Dugas in Question on strong cotilting modules) $W$ is a left strong cotilting module as well since the top of $W$ as a right module contains all simple $A$-modules. (I hope I applied this correct)

Now what is left to do is to show that $End_A(W) \cong A$, this was done with QPA.

Now we type in QPA: L:=ARQuiver([A,1111])[2];

[ <[ 1, 0, 0 ]>, <[ 1, 1, 1 ]>, <[ 0, 1, 1 ]>, <[ 1, 1, 1 ]>, <[ 0, 1, 1 ]>, <[ 1, 1, 0 ]>, <[ 1, 1, 2 ]>, <[ 1, 2, 2 ]>, <[ 0, 0, 1 ]>, <[ 1, 2, 1 ]>, <[ 1, 1, 1 ]>, <[ 2, 2, 2 ]>, <[ 1, 1, 0 ]>, <[ 1, 1, 2 ]>, <[ 1, 2, 2 ]>, <[ 1, 1, 1 ]>, <[ 0, 1, 2 ]>, <[ 0, 1, 0 ]>, <[ 1, 2, 2 ]>, <[ 1, 2, 2 ]> ]

(This gives all indecomposable $A$-modules, but it is not yet available in QPA. The output are the dimension vectors.)

U1:=Filtered(L,x->Size(ExtOverAlgebra(x,W)[2])=0 and Size(ExtOverAlgebra(NthSyzygy(x,1),W)[2])=0);

[ <[ 1, 1, 1 ]>, <[ 1, 1, 0 ]>, <[ 1, 1, 2 ]>, <[ 0, 0, 1 ]>, <[ 1, 2, 1 ]>, <[ 1, 1, 1 ]>, <[ 1, 2, 2 ]>, <[ 1, 2, 2 ]> ]

(so U1 filters all indecomposable modules $X$ with $Ext_A^i(X,W)=0$ for all $i>0$. Here we use that $W$ has injective dimension 2, so we only have to look at $Ext^1$ and $Ext^2$).


[ <[ 1, 1, 1 ]>, <[ 1, 1, 0 ]>, <[ 1, 1, 2 ]>, <[ 0, 0, 1 ]>, <[ 1, 2, 1 ]>, <[ 1, 1, 1 ]>, <[ 1, 2, 2 ]>, <[ 1, 2, 2 ]> ]

(U2 filters all 2. syzygy modules from L)


Thus we have $CM(A)=\Omega^2(mod-A)$ but $A$ is not Gorenstein, giving a negative answer to the question by Auslander and Reiten.


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