Let $B$ be a finite dimensional selfinjective algebra over a field $k$ with a finite dimensional non-projective $B$-module $M$ and $$A=\pmatrix{k&M\\0&B}.$$

A module $N$ over an algebra $C$ is called Gorenstein projective in case $Ext_C^i(N,C)=0=Ext_C^i(D(C),\tau(N))$ for all $i >0$.


  1. Is there a general description of Gorenstein projective $A$-modules depending on $B$ and $M$?

  2. Can $A$ have only finitely many indecomposable Gorenstein projectives when $M$ satisfies $Ext_B^j(M,M)=0$ for all $j >t$ for some $t$?

Together with (the answer in) Question on Ext for finite dimensional algebras a positive answer to 2. would give a negative answer to the first question in chapter 8 of https://arxiv.org/pdf/1808.01809.pdf.

There is some literature on triangular matrix algebras and their Gorenstein projective modules, but it seems that the assumptions are always too strong to apply here.


A (right) $A$-module consists of a pair $\pmatrix{V&X}$, where $V$ is a vector space and $X$ a $B$-module, together with a $B$-module map $V\otimes_kM\to X$.

Another criterion for $\pmatrix{V&X}$ to be Gorenstein projective is that it has a complete projective resolution: i.e., an acyclic complex $$P^\bullet:=\dots\longrightarrow P^{-1}\longrightarrow P^0\stackrel{d^0}{\longrightarrow}P^1\longrightarrow\dots$$ of projective $A$ modules with $\text{im}(d^0)\cong\pmatrix{V&X}$, such that $\text{Hom}_A(P^\bullet,A)$ is also acyclic.

Up to removing projective summands of $\pmatrix{V&X}$ we can assume that $P^\bullet$ is minimal (i.e., contains no contractible summands), which means that it cannot involve the projective module $\pmatrix{k&M}$, and so $V=0$ and $P^\bullet$ is just a complete projective $B$-module resolution of some $B$-module $X$.

In order that $\text{Hom}_A(P^\bullet,A)$ is acyclic, it is necessary and sufficient that $\widehat{\text{Ext}}^i_B(X,M)=0$ for all $i\in\mathbb{Z}$.

So the non-projective indecomposable Gorenstein projective $A$-modules are those of the form $\pmatrix{0&X}$, where $X$ is an indecomposable non-projective $B$-module such that $\widehat{\text{Ext}}^i_B(X,M)=0$ for all $i\in\mathbb{Z}$.

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  • $\begingroup$ Thanks. One question: Is $\hat{Ext_B}^0(X,M))=\underline{Hom_B}(X,M)$? Ill try my luck a bit to look for question 2 with $B=K<x,y>/(x^2,y^2,xy-qyx)$ since there such modules $M$ exists when q is not a root of unity and maybe in some cases there can be only finitely many such X, at least it looks like a very strong condition that this ext vanishes for all $i \in \mathbb{Z}$. $\endgroup$ – Mare Nov 10 '18 at 12:17
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    $\begingroup$ @Mare Yes, $\widehat{\text{Ext}}^0$ is stable $\text{Hom}$. $\endgroup$ – Jeremy Rickard Nov 10 '18 at 12:22

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