$\Phi: Hom_R(A,B) \to Hom_R(A,R)\otimes_R B$ - MathOverflow

$\Phi: Hom_R(A,B) \to Hom_R(A,R)\otimes_R B$

2011-07-10T13:56:38Z

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 $\Phi$?

Answer by Andreas Blass for $\Phi: Hom_R(A,B) \to Hom_R(A,R)\otimes_R B$

2011-07-10T14:10:10Z

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.

Answer by Fernando Muro for $\Phi: Hom_R(A,B) \to Hom_R(A,R)\otimes_R B$

2011-07-11T07:04:56Z

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

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

Answer by Martin Brandenburg for $\Phi: Hom_R(A,B) \to Hom_R(A,R)\otimes_R B$

2011-07-11T07:18:39Z

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 $\hom(A,R) = A^*$.