Let $A$ be a finite dimensional $k$ algebra, $k$ is a field. It is evident that the duality functor ($(-)^\star=hom_k(-,k)$) preserves the simplicity in the case of finitely generated $A$- modules (left $A$-modules to right $A$- modules ) here is a proof in a related question
My question is, could you possibly provide me with a counterexample of the infinite dimensional case. That is $_AM$ is simple but $M^\star_A$ is not, where $A$ is still finite dimensional?
If $A$ is a finite dimensional algebra, then every simple $A$-module is finite dimensional, since it is a quotient of the regular module (if $S$ is a simple left module, then for any $0\neq s\in S$, $a\mapsto as$ is a non-zero homomorphism $A\to S$, which is surjective since $S$ is simple).