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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.

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    $\begingroup$ 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 $(M\otimes N)^\vee$. $\endgroup$
    – Daniel Litt
    Commented Mar 8, 2022 at 14:06
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    $\begingroup$ Just take $R = M = N = \mathbb{Z}$ and $\langle m, n \rangle = 2mn$. $\endgroup$
    – Najib Idrissi
    Commented Mar 8, 2022 at 14:48
  • $\begingroup$ @Najib: Excellent example! This illustrates exactly what goes wrong. $\endgroup$
    – Adam Bondal
    Commented Mar 9, 2022 at 11:57

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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.)

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    $\begingroup$ By an odd coincidence, I've spent the last few weeks in exactly this situation, and mildly grumpy that this isn't true as it would save a pile of work... $\endgroup$
    – Jan Grabowski
    Commented Mar 8, 2022 at 14:03
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    $\begingroup$ Even for $R = \mathbb{Z}$ . . . I didn't know life was so cruel. $\endgroup$
    – Adam Bondal
    Commented Mar 8, 2022 at 14:14
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    $\begingroup$ Could you please provide a reference for this? $\endgroup$
    – Adam Bondal
    Commented Mar 8, 2022 at 14:14
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    $\begingroup$ No I can't sorry. I struggled to find this set out anywhere - it took a bit of digging even to turn up that non-degenerate and perfect were the right terminology. $\endgroup$
    – Jan Grabowski
    Commented Mar 8, 2022 at 14:17

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