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Let $R$ be a commutative ring and $A$ and $B$ two $R$-module. Suppose that $A$ is free of rank $n$ with basis $a_1,\dots,a_n$. Then there is an isomorphism $\Phi: Hom_R(A,B) \to Hom_R(A,R)\otimes_R B$ defined by $\Phi(\sigma)=\sum_{i=1}^n \psi_i\otimes \sigma_i$, where $\sigma_i=\sigma(a_i)$ and $\psi_i$ is the map such that $\psi_i(a_j)=\delta_{ij}$. Is there another way to describe $\Phi$?

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3 Answers

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Such an isomorphism is worthless if it is written down with a choice of a basis, because then naturality is unclear (which is, of course, very important if you need this isomorphism not just as an isolated relation). There is a homomorphism of $R$-modules

$\alpha : A^* \otimes B \to \hom(A,B)$

defined by $\phi \otimes b \mapsto \phi(-)b$, which is natural in $A$ and in $B$, which are arbitrary $R$-modules. It is a natural question when this is an isomorphism for all $B$, when we fix $A$. Note that the inverse will be, restricted to these $A,B$, also natural due to general reasons, although perhaps we need to make choices to write down the inverse (without making reference to $\alpha$)! Also remember the slogan "You can work locally, if you are given something globally".

Now both $(-)^* \otimes B$ and $\hom(-,B)$ are functors which transform finite direct sums into finite products, and transform split cokernels into split kernels. In particular, the set of $A$s for $\alpha$ is an isomorphism is closed under these operations and since $R$ is an example, we see that every finitely generatd projective $R$-module is an example. Now, the converse is also true: If $A^* \otimes - \cong \hom(A,-)$, then the right hand side is preserving all colimits (since this is true for the left hand side). Restricting to coequalizers shows that $A$ is projective, and restricting to filtered colimits shows that $A$ is finitely presented (in the categorical sense, thus also in the algebraic sense).

You can view this also as a special case of the Theorem of Eilenberg-Watts: If $A$ is finitely generated projective, then $\hom(A,-) : \text{Mod}(R) \to \text{Mod}(R)$ is a cocontinuous functor, thus is given by tensoring with a $R$-module, namely $\hom(A,R) = A^*$.

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I don't think that "naturality is unclear", if the iso "is written down with a choice of a basis". Given $A,A'$ with bases $\lbrace a_i \rbrace, \lbrace a'_j\rbrace$ and a hom $\alpha: A \to A'$ of $R$-modules, then by writing $\alpha(a_i) = \sum_j r_{ij}a'_j$ it's a direct computation to show that $\phi \circ \alpha^{\ast} = (\alpha^{\ast} \otimes \operatorname{id}_B) \circ \phi'$ (where $\phi$ is taken from the OP). Similarly nat. in $B$ can be shown. In the general case of $A$ being f.g. projective just replace the bases through projective bases. – Ralph Jul 11 2011 at 16:11
I know that you can do a computation, but you cannot see naturality without this computation when you write the inverse map as above. Thus it is somewhat unclear. – Martin Brandenburg Jul 11 2011 at 18:09
I agree in this point of view. Thanks for your explication. – Ralph Jul 11 2011 at 19:16
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By definition of freeness, $Hom_R(A,X)$ is naturally (with respect to $X$) isomorphic to the direct product of $n$ copies of $X$. Apply this in both the domain and codomain of $\Phi$, and use the fact that $-\otimes B$ distributes over finite products (because they're the same as finite sums in abelian categories). Is that "another" description? The content is the same but the viewpoint seems a bit different.

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 When I said another way, what I had in mind was something like the inverse map, i.e. not involving explicitely $a_1,\dots,a_n$. In fact $\Phi^{-1}(\nu\otimes b)=(a\mapsto \nu(a)b)$. Instead, if I'm not wrong, identifying $Hom(A,X)$ with $X^n$ require the basis of $A$. – Michele Torielli Jul 10 2011 at 14:20
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 What about "the inverse of the map $\nu\otimes b\mapsto \left(a\mapsto \nu\left(a\right)b\right)$ ? Because I don't think there is anything more explicit. If there was, it would work for any module, not necessarily free... – darij grinberg Jul 10 2011 at 14:24
I agree with darij. If the inverse of a naturally defined map doesn't always exist, there's no reason to expect that it has a more natural definition than "the inverse, when it exists." – Qiaochu Yuan Jul 10 2011 at 14:30
I agree too. That what I was thinking but I was hoping to be wrong.Thank you. – Michele Torielli Jul 10 2011 at 14:34
At least the result also holds if $A$ is a finitely generated projective $R$-module. – Ralph Jul 10 2011 at 14:42
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As Benjamin Steinberg says in his comment, the inverse map always exists and moreover it is natural in $A$ and $B$,

$\Psi\colon Hom_R(A,R)\otimes_RB\longrightarrow Hom_R(A,B)$

but it's only an isomorphism for f.g. projective $A$. A definition of $\Psi$ in terms of functors and adjunctions is as follows. By the adjointness between $-\otimes_RB$ and $Hom_R(B,-)$, the natural homomorphism $\Psi$ is the same as a natural morphism

$\Psi'\colon Hom_R(A,R)\longrightarrow Hom_R(B,Hom_R(A,B))\cong Hom_R(B\otimes_RA,B)$

This $\Psi'$ is the same as

$Hom_R(A,R)\mathop{\longrightarrow}\limits^{{B\otimes_R-}} Hom_R(B\otimes_RA,B\otimes_RR)\cong Hom_R(B\otimes_RA,B)$

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