Explicit isomorphism of free $R$-modules Let $R$ be a ring. Consider $R^n$ as a left $R$-module and prove that $$R^n \otimes_R R^m \cong R^{mn}.$$
I use Atiyah´s definition of tensor products (page 24), where the tensor product is regarded as a quotient of two modules. I am trying to define the isomorphism explicit but I am stuck, any hints?
 A: By definition, we have that 
$$R^{mn}\cong Re_{11}\oplus...\oplus Re_{1n} \oplus Re_{21}\oplus ... \oplus Re_{2n} \oplus ... \oplus Re_{m1} \oplus ... \oplus Re_{mn}$$ That is, $R^{mn}$ has a basis consisting on $mn$ elements which we have chosen to denote by $\{e_{ij}\}$, where $i=1,...,m$ and $j=1,...,n$. 
In the same fashion, pick $\{a_1,...,a_m\}$ a generating set for $R^m$ and $\{b_1,...,b_n\}$ a generating set for $R^n$. You can check (just by applying bilinearlity) that precisely $\{a_i\otimes b_j\}_{i,j}^{m,n}$ is a generating set for the $R$-module $R^m\otimes_R R^n$.
Now consider the $R$-linear extension of the map $f:R^{mn}\to R^m\otimes_R R^n$ given by $e_{ij}\mapsto a_i\otimes b_j$. This is an isomorphism, whose inverse is precisely the $R$-linear extension of $g:R^m\otimes_R R^n \to R^{mn}$ given by $a_i\otimes b_j \mapsto e_{ij}$.
A: The universal property of $R^J$ for a set $J$ is that there is a natural isomorphism $$ \hom_R(R^J,M)\simeq \hom_{\rm Set}(J,M)$$ Now, by the usual adjunctions in ${}_R\,\mathbb {mod}$ and $\mathrm{Set}$ we have natural isomorphisms
$$\hom_R(R^I\otimes_R R^J,M)\simeq \hom_R(R^J,\hom_R(R^I,M))\\\simeq \hom_{\rm Set}(J,\hom_{\rm Set}(I,M))\simeq \hom_{\rm Set}(I\times J,M)\simeq \hom_R(R^{I\times J},M)$$
The Yoneda lemma gives $R^I\otimes_R  R^J \simeq R^{I\times J}$
