TITLE: Duality of a finitely generated projective modules
QUESTION [9 upvotes]: Let $M$ and $N$ be a finitely generated projective module over a ring $R$. Suppose we have a non degenerate bilinear pairing $\langle \ \cdot \ ,\ \cdot\ \rangle: M \times N \to R$.
I want to show $M$ is isomorphic to the dual $N^*$ of $N$.
The injectivity of $M$ into $N^*$ follows from the non degeneracy of the pairing by definiting a map $x \mapsto \langle x, \cdot\rangle$. What I cannot prove is surjectivity.
If I impose a condition that $R$ is an injective module then I think surjectivity also follows. But I want to prove it without any condition on $R$
Any help is appriciated.

REPLY [8 votes]: If by non-degenerate you mean that $\langle x , - \rangle = 0 \Leftrightarrow x = 0$ (or $\langle - , y \rangle \Leftrightarrow y = 0$), your question is equivalent to: Is every injective homomorphism between finitely generated projective $R$-modules an isomorphism? The answer is no, even for maps $R \to R$.
If by non-degenerate you mean that $\langle x , - \rangle = 0 \Leftrightarrow x = 0$ and $\langle - , y \rangle \Leftrightarrow y = 0$, then your question is equivalent to: If $M \to N$ is an injective homomorphism of f.g. proj. modules such that $N^* \to M^*$ is also injective, then $M \to N$ is an isomorphism. But again this is false for maps $R \to R$.
The "correct" definition of a perfect pairing $M \times N \to R$ is the following: Both maps $M \to N^*$ and $N \to M^*$ are isomorphisms.