# Nondegenerate pairings versus perfect pairings for finitely generated projective modules

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.

• In general $M, N$ need not even be abstractly isomorphic. For example, any two rank one projective modules over a ring admit a non-degenerate pairing, as such a pairing is the same as a nonzero element of $(M\otimes N)^\vee$. Commented Mar 8, 2022 at 14:06
• Just take $R = M = N = \mathbb{Z}$ and $\langle m, n \rangle = 2mn$. Commented Mar 8, 2022 at 14:48
• @Najib: Excellent example! This illustrates exactly what goes wrong. Commented Mar 9, 2022 at 11:57

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.)
• Even for $R = \mathbb{Z}$ . . . I didn't know life was so cruel. Commented Mar 8, 2022 at 14:14