Take the 2-minute tour ×

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.

share|improve this question
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
add comment

1 Answer

up vote 6 down vote accepted

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.

share|improve this answer
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
@QiL: It is not true over fields. One has to require that the dimensions match. (or that it is non-degenerate in both directions) –  Martin Brandenburg Jul 5 '12 at 15:25
I thought non-degenerate means on both sides. –  user18119 Jul 5 '12 at 22:20
I've edited my answer. –  Martin Brandenburg Jul 6 '12 at 17:42
add comment

Your Answer


