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.



Nondegenerate pairings versus perfect pairings for finitely generated projective modules
No: for $R=\mathbb{Z}$ and even for $M,N$ both free (i.e. free Abelian groups), non-degenerate doesn't imply perfect. (You get finite index sublattices, so torsion quotients, unlike the case of vector spaces.)


