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?

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    $\begingroup$ An easier way than doing so would be to show that $R^{mn}$ satisfies the universal mapping property (Proposition $2.12$). Atiyah and Macdonald prove uniqueness of the tensor product, so any $A$-module satisfying the UMP is isomorphic to the tensor product as $A$-modules. $\endgroup$ – Marcus M Sep 19 '15 at 22:56
  • $\begingroup$ I thought about this, but I decided it seemed harder than trying to construct the isomorphism. I will try this, thanks for the hint. $\endgroup$ – Gonzales Sep 19 '15 at 23:02
  • $\begingroup$ @MarcusM, I have tried to work out your hint, but I dont manage to. In accordance with atiyah, let $T=R^{mn}$ and $g: R^n \times R^m \to R^{mn}$. Now let $P$ be a $A$-module and let $f:R^n \times R^m \to P$ be an $A$-bilinear map. I manage to prove the existence of a map $f´ : T \to P$ such that $f=f´ \circ g$, but I cant prove the uniqueness. I have defined $g$ as $(a,b) \mapsto (a,b,0...0)$. Any hint? Thanks for your answer. $\endgroup$ – Gonzales Sep 20 '15 at 3:36

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 $a_i\otimes b_j \mapsto e_{ij}$.

  • $\begingroup$ The key step in this argument is verifying that there really does exist a bilinear map $R^m\times R^n\to R^{mn}$ sending $(a_i,b_j)$ to $e_{ij}$; that is, that the map $(\sum_i r_i a_i, \sum_j s_j b_j)\mapsto \sum_{i,j} r_is_j e_{ij}$ really is bilinear. $\endgroup$ – Eric Wofsey Sep 20 '15 at 21:27

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 $R^I\otimes_R R^J \simeq R^{I\times J}$

  • $\begingroup$ (I hope this element free proof pleases Martin Brandenburg.) $\endgroup$ – Pedro Tamaroff Sep 20 '15 at 21:42
  • $\begingroup$ It is a nice solution, although your milage will vary depending on readers' definition of 'explicit isomorphism' :) $\endgroup$ – rschwieb Sep 25 '15 at 16:00

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