Exercise 1, § 8

Let A be a semisimple ring, and let M be a finitely generated A-module. For any submodule N of M, we denote the orthogonal of N in $M^*$ by $N'$. Prove that the orthogonal of $N'$ in M is equal to N and that we have long$_A(N) +$ long$_A(N') =$ long$_A(M)$. The mapping $N\mapsto N'$ is a bijection from the set of submodules of M to the set of submodules of $M^*$. If L is another submodule of M, then the orthogonal of $L\cap N$ is $L'+ N'$.