Let $R$ be a commutative ring. I want to show that $R^n \otimes_R R^m$ is isomorphic to $R^{nm}$ as $R$-module.
So far, I've tried to define the only natural map that I can think of:
$$
f : R^n \times R^m \to R^{nm} : (\vec{r},\vec{s}) \mapsto (r_1\vec{s},\ldots,r_n\vec{s}),
$$
I think it is $R$-bilinear and hence we use the universal property of the tensor product to obtain a group homomorphism
$$
f' : R^n \otimes_R R^m \to R^{nm} : \vec{r} \otimes \vec{s} \mapsto (r_1\vec{s},\ldots,r_n\vec{s}).
$$
However, I'm stuck in trying to define an inverse, I'm not even sure if this map is surjective in the first place. Am I on the right track, how to proceed from here?
Best Answer
First, I'll provide a less direct proof using the distributivity of the tensor product with respect to the direct sum.
We have, by the mentioned property that
$R^n\otimes R^m=(\underbrace{R\oplus\cdots \oplus R}_{n})\otimes_R R^m=(R\otimes_R R^m)\oplus\cdots\oplus (R\otimes_R R^m).$
Now, since $R^m$ is a $R$-module, $R\otimes_R R^m\cong R^m$, so the latter is just
$\underbrace{R^m\oplus\cdots\oplus R^m}_{n}=R^{nm}.$
Now, I'll make a few comments on your map. It is surjective if you define it as the linear extension of what you explicitly described. Expanding $(r_1\vec{s},\dots, r_n\vec{s})$ you can vew the image as $nm$-tuples whose coordinates are of the form $s_ir_j$ for some $1\leq i\leq n$, $1\leq j\leq m$. Then, if you want to produce an element $(0,\dots, 0,r,0,\dots, 0)$ for $r\in R$ in the $k$-th coordinate, then take the element $r_is_j$ corresponding to the $k$-th coordinate, and declare $r_i=r$, $s_j=1$, and every other $r_l,s_h=0$. Any other element of $R^{mn}$ is hit if you extend linearly.
Instead of trying to find an inverse, I would recommend you to try to show that it is injective, though I don't have a proof for that at the moment. Edit See the comment of jawheele below