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:



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> 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 $n \in \mathbb{Z}$?
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> 
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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)



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


