Selfextensions for modules of commutative Frobenius algebras

Question

Let $A$ be a commutative finite dimensional Frobenius algebra and $M$ a non-projective $A$-module.

Can we have $Ext_A^i(M,M)=0$ for some $i>0$?

Can we have $Ext_A^i(M,M)=0$ for some $i>0$ in case $A=kG$ is a group algebra?

Answer 1

For (commutative) group algebras $A=kG$ the answer is no.

Suppose $k$ has characteristic $p$, and $G=P\times H$, where $P$ is a Sylow $p$-subgroup. Then $kG\cong kP\otimes_kkH$, which, since $kH$ is a commutative semisimple algebra, is a product of algebras of the form $KP$, where $K$ is a field extension of $k$. Therefore we can reduce to the case $A=KP$.

$\text{Ext}^i_{KP}(M,M)\cong \text{Ext}^i(K,M\otimes_K M^*)$, and the theory of varieties implies that $M\otimes M^*$ is not projective (= injective), since $V_P(M\otimes M^*)=V_P(M)\neq\{0\}$. Since $KP$ is local and $M\otimes_KM^*$ is not injective, $\text{Ext}^i(K,M\otimes_KM^*)\neq0$.