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$Hom(V,W)\cong V^*\otimes W$ works for finite dimensional vector space, when will it also work for modules over ring $R$?

The motivation (i.e. the question is more general than this) is to know whether and why this works for the $C^\infty(M)$-module of sections of a vector bundle which is not vector space since $C^\infty(M)$ is not a field.

It seems that $R$ has to be commutative, and the module also need to be finitely generated or even projective (since Serre–Swan theorem seems to be related). Is that enough and what is the minimum requirement?

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  • $\begingroup$ You don't need $R$ commutative, you just want $V$ to be a finitely generated projective $R$-module i.e. a locally free finitely generated bundle. $\endgroup$
    – Pedro Tamaroff
    Sep 27 at 10:13
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Let $R$ be a ring (need not be commutative) and let $M$ and $N$ be left $R$-modules. As usual set $M^* = \hom_R(M,R)$ with its canonical right $R$-module structure. There is a natural map $$\Psi_{M,N} : M^* \otimes_R N \longrightarrow \hom_R(M,N)$$ such that $\Psi(f\otimes x)(y) = f(y)x$.

Lemma. Fixing $M$, this map is an isomorphism for all $N$ if and only if $M$ is finitely generated and projective.

You can find this, for example, proved in Bourbaki's algebra book. Indeed, it is standard the following are equivalent:

  1. $P$ is finitely generated and projective
  2. $P$ admits a finite dual basis, that is, there exist $(x_i,y_i) \in P\times P$ for $i\in [n]$ such that for each $x\in P$ we have that $x = \sum_{i=1}^n y_i(x) x_i$ or, what is the same,
  3. We have that $\Psi_{P,P}(u) = 1_P$ for some $u= \sum_{i=1}^n y_i\otimes x_i\in P^*\otimes_R P$.

Proof of the Lemma: if $\Psi_{P,Q}$ is an isomorphism for all $Q$, picking $Q=P$ and covering the identity gives you a dual basis for $P$, by 3. so that $M=P$ is projective.

Conversely, if you have a dual basis, given a map $f: M\to N$, consider the element $\Phi(f) = \sum_{i=1}^n y_i \otimes f(x_i)$. It follows that

$$\Psi(\Phi(f))(x) = \sum_{i=1}^n y_i(x)f(x_i) = f\left( \sum_{i=1}^n y_i(x)x_i\right) = f(x)$$

and that for $g = \Psi(f\otimes x)$

$$\Phi(g) = \sum_{i=1}^n y_i(-)\otimes f(x_i)x = \sum_{i=1}^n y_i(-)f(x_i)\otimes x = f \otimes x$$

so that $\Psi$ is an isomorphism.


Note that dual bases are a bit less canonical than actual bases of free $R$-modules, so the map $\Phi$ above is not unique.

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