Let $R$ be a (not necessarily commutative) ring, $M$ a left $R$-module, and $N$ a right $R$-module. We say that a pairing $$ \langle -,-\rangle:M \otimes_R N \to R $$ is non-degenerate if, for all $n \in N$ there exists an $m \in M$ such that $\langle m,n\rangle \neq 0$, \textbf{and} for all $m \in M$, there exists an $n \in N$ such that $\langle m,n\rangle \neq 0$.

Such a pairing will give embeddings $$ M \hookrightarrow N^*, ~~~~~ N \hookrightarrow M^*, $$ where $M^*$ and $N^*$ denote the dual modules of $M$ and $N$ respectively. In general (even for infinite-dimensional vector spaces) this will not give isomorphisms $$ N \simeq M^*, ~~~~ M \simeq N^*. $$ However, if we assume that $M$ and $N$ are finitely-generalted projective, then does non-degeneracy imply that we get isomorphisms?

If it fails in the general noncommutative setting, I would still be interested in a positive answer in the commutative setting.

anytwo rank one projective modules over a ring admit a non-degenerate pairing, as such a pairing is the same as a nonzero element of $(M\otimes N)^\vee$. $\endgroup$