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Let $M$, $N$ be finitely generated projective modules over a ring $R$. Suppose that we have a non degenerate form $\langle\cdot\;,\,\cdot\rangle: N \times N \to R$. ($N$ is isomorphic to its dual $N^*$.)

I want to show that this form induces an isomorphism $N \otimes M \cong \rm{Hom}(N, M)$.

Since $N$ is isomorphic to its dual we have $N \cong \rm{Hom}(N, R)$. Tensoring $M$, we have $N \otimes M \cong \rm{Hom}(N, R) \otimes M$.

So it suffices to show that the last module is isomorphic to $ \rm{Hom}(N, M)$. But I don't know how to prove this.

I haven't used the conditions that $M$, $N$ are finitely generated projective modules.

Could you give me some advice? Thank you in advance.

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    We generally have $\operatorname{Hom}(N,M) \cong M\otimes N^*,$ so if $N$ is self-dual then we're done. Unless if you're asking how to prove this more general fact?2012-07-04
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    Yes I want to know how to prove that general fact.2012-07-04
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    @ Andrew: Certainly it doesn't hold not in general, see http://math.stackexchange.com/questions/69184/a-counterexample-to-the-isomorphism-m-otimes-m-rightarrow-hom-r-m-m2012-07-04
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    Ah, thanks, yes I knew there were conditions, but was just curious about why the self-duality hypothesis is stated.2012-07-04
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    @MartinWanvik Yes. I just fixed. Thanks.2012-07-04
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    See http://math.stackexchange.com/questions/81429/hom-and-tensor-with-a-flat-module2012-07-04

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