# Duality of a finitely generated projective modules

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.

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You need (at least) the bilinear pairing to be surjective. Otherwise you get examples like: $R$ the ring of algebraic integers of a number field $K$, $[K:\mathbb{Q}]<\infty$. $M,N$ any non-zero ideals of $R$ (they are all projective), $\langle\ ,\ \rangle =$ the product. – Jyrki Lahtonen Jul 6 '12 at 18:10

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.
Thank you. Is the condition that $<x,y>=0$ for all $y$ implies $x=0$ (and similar for $y$) weaker than the definition of non degeneracy of yours? – Primo Jul 4 '12 at 18:31
I think the equivalence is not correct. The original question has clearly a positive answer when $R$ is field, while the statement you give (injectivity implies isomorphism) is false. So the two properties are not equivalent. – user18119 Jul 4 '12 at 21:17