1
$\begingroup$

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 $i >0$.

Questions:

  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.

$\endgroup$

1 Answer 1

3
$\begingroup$

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 $i\in\mathbb{Z}$.

$\endgroup$
2
  • $\begingroup$ 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, 2018 at 12:17
  • 1
    $\begingroup$ @Mare Yes, $\widehat{\text{Ext}}^0$ is stable $\text{Hom}$. $\endgroup$ Nov 10, 2018 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.