On Frobenius extensions of the centralizer matrix algebras (Q6093313)

Given a ring \(R\) and \(\emptyset\ne C\subseteq R\), the \textit{centralizer of \(C\) in \(R\)} is \[ S(C,R) := \{r \in R \mid rc = cr,\ \text{ for all } c \in C\}. \] Notice that \(S(C,R)\) is a subring of \(R\). An \(R\)-\(S\)-bimodule \({}_SP_R\) is a \textit{Frobenius bimodule} if both \({}_SP\) and \(P_R\) are finitely generated projective modules, and there is an \(R\)-\(S\)-bimodule isomorphism \[ \mathrm{Hom}_S (P,S) \cong \mathrm{Hom}_{R^{op}}(P,R). \] A ring extension \(S \subseteq R\) is called a \textit{Frobenius extension} if \({}_SR_R\) is a Frobenius bimodule. The main result of this short note is the following. Theorem. Let \(R\) be a commutative ring, \(P\) a progenerator over \(R\) and \(A = \mathrm{End}_R(P)\). Let \(C\) be an \(R\)-subalgebra of \(A\). Then \(A\) is a Frobenius extension of \(S(C,A)\) if and only if \begin{itemize} \item[1.] \(P\) is a generator as a left \(C\)-module, and \item[2.] \(\mathrm{Hom}_R({}_CC_C,R) \cong {}_CC_C\). \end{itemize} As a consequence, the following result is obtained. Corollary. Let \(\Bbbk\) be a field and \(C\) a commutative subalgebra of \(M_n(\Bbbk)\). Then \(M_n(\Bbbk)\) is a Frobenius extension of \(S(C, M_n(\Bbbk))\) if and only if \(C\) is a Frobenius algebra. The author gives a counterexample showing that the previous result is false over commutative rings.