Question on $\operatorname{Ext}$ in a local Frobenius algebra

Let $$A$$ be a finite dimensional local Frobenius algebra with simple module $$k$$ and an indecomposable non-projective module $$M$$ (that is also finite dimensional).

Question:

Is there an example of such an $$M$$ such that every map $$f: \Omega^n(M) \rightarrow k \rightarrow \Omega^{n-1}(M)$$ is zero in the stable module category of $$A$$ for all $$n \in \mathbb{Z}$$?

Note that $$\operatorname{Ext}_A^1(M,M) \cong \underline{\operatorname{Hom}_A}(\Omega^{n}(M),\Omega^{n-1}(M))$$ and it seems to be unknown whether this can be zero.

$$\operatorname{Ext}_A^1(M,M)$$ is for example always non-zero for radical cube zero algebras or Hopf algebras.

(I reformulated the question as suggested by a comment of Jeremy Rickard)

• You want this for every $f$? In other words, you want an $M$ so that all compositions of the forms $M\to k\to\Omega^{-1}M$ and $\Omega M\to k\to M$ are zero in the stable module category? Also, you ask for an easy example: do you know a difficult example? – Jeremy Rickard Nov 29 '18 at 9:34
• @JeremyRickard yes every $f$, Ill add it. I do not know an example. Every example is welcome, Im just a little afraid of verifing very complicated examples. – Mare Nov 29 '18 at 9:38
• @JeremyRickard Your formulation of the question seems actually to be better/more compact. Maybe even asking the question for all $n \in \mathbb{Z}$ and $\Omega^{n}(M) \rightarrow k \rightarrow \Omega^{n-1}(M)$ might be interesting. – Mare Nov 29 '18 at 9:54

Let $$A$$ be the six-dimensional algebra $$\langle x,y\mid x^2=y^2=0, xyx=yxy\rangle$$, and $$M$$ the three dimensional uniserial (right) module $$A/yA$$.
Then $$\Omega M\cong M$$, and so $$\Omega^n M\cong M$$ for all $$n\in\mathbb{Z}$$.
Up to scalar multiplication, there is a unique map $$M\to M$$ that factors through $$k$$, and this also factors through the free module, and so is zero in the stable module category.
• Thanks. It doesnt look so easy to come up with such an example as the algebra is quite exotic. Do you also know an example of a module that is not 1-periodic? Being 1-periodic immediately implies $Ext_A^1(M,M) \neq 0$ so the question is not so interesting for 1-periodic modules. – Mare Nov 29 '18 at 21:38