Call a finite dimensional selfinjective algebra $A$ nice in case $Ext^{1}(X,Y) \neq 0$ (or equivalently $\underline{Hom}(X,Y) \neq 0$) for arbitrary indecomposable modules $X,Y$. Is there a classificaiton of nice selfinjective algebras? I think the only examples might be $K[x]/(x^n)$, but Jeremey Rickard found another example in his answer.
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I don't know a classification, but another example is the group algebra (over a field $K$ of characteristic $2$) of the quaternion group $Q_8$ (or more generally any generalized quaternion group).
answered Jan 11, 2017 at 15:17
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$\begingroup$ Do you have a reference or quick argument for that? $\endgroup$– MareCommented Jan 11, 2017 at 15:18
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$\begingroup$ For a local (nonsemisimple) Hopf algebra with nonprojective indecomposable modules X,Y one has $Ext^{1}(X,Y)=Ext^{1}(k \otimes_k X,Y) =Ext^{1}(k, Hom_k(X,Y))$ which is always nonzero if $Hom_k(X,Y)$ is not projective. So can $Hom_k(X,Y)$ ever be projective in case the hopf algebra is local? $\endgroup$– MareCommented Jan 11, 2017 at 15:28
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1$\begingroup$ @Mare Your second comment makes the answer to your first comment easier! It follows from Chouinard's Theorem (a module for a finite group algebra $KG$ over a field of characteristic $p$ is projective iff its restriction to every elementary abelian $p$-subgroup of $G$ is projective). The only non-trivial elementary abelian $2$-subgroup of $Q_8$ is its cyclic centre $Z\cong C_2$. So if $M$, $N$ are non-projective, then their restrictions to $Z$ are non-projective, and so $\text{Hom}_K(M,N)$ is projective on restriction to $Z$, and hence projective. $\endgroup$ Commented Jan 11, 2017 at 15:31
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1$\begingroup$ @Mare And in answer to your second comment, $\text{Hom}_K(M,N)$ can be projective for the group algebra of any $p$-group of $p$-rank greater than one (i.e., containing a subgroup $C_p\times C_p$). This follows from the theory of varieties for modules. $\endgroup$ Commented Jan 11, 2017 at 15:33
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1$\begingroup$ We need $Hom_k(X,Y)$ nonprojective but you said at the end of your 2. comment that it is projective. I think you mean "non-projective" instead. The argument should be taht $M$ and $N$ contain $k$ as a direct summand as representations over $Z$ and thus $Hom_K(M,N)$ has also $K$ as a direct summand and is not projective. $\endgroup$– MareCommented Jan 11, 2017 at 15:36