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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?

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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

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    $\begingroup$ Nice! I hadn't thought about that proof but it's much easier. Btw where exactly do you use commutativity of $R$ in that proof? $\endgroup$ – Sigurd Mar 18 at 18:15
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    $\begingroup$ @Sigurd When $R$ is non commutative, the tensor product $M\otimes_R N$ a right $R$-module $M$ and a left $R$-module $N$ is in general just an abelian group (see here ), so I used commutativity to get a $R$-module isomorphism. $\endgroup$ – Javi Mar 18 at 18:20
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    $\begingroup$ Oh indeed I see. $\endgroup$ – Sigurd Mar 18 at 18:37
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    $\begingroup$ To directly show injectivity, consider that $f'(\vec{r} \otimes \vec{s})=0$ implies $r_k s_j = 0$ for each $1 \leq k \leq n$, $1 \leq j \leq m$, and so if $e_k$ is as in my answer and we similarly define $h_j \in R^m$, then $\vec{r} \otimes \vec{s} = \sum_{k=1}^n \sum_{j=1}^m (r_k e_k) \otimes (s_j h_j) = \sum_{k,j} (r_k s_j) (e_k \otimes h_j) = 0$. $\endgroup$ – jawheele Mar 18 at 19:00
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    $\begingroup$ It used $R$-bilinearity of the tensor map $\otimes : R^n \times R^m \to R^n \otimes_R R^m$. Your latter statement about some kind of commutativity was not explicitly required in that computation. Remember $r_k, s_j \in R$ while $e_k \in R^n$. $\endgroup$ – jawheele Mar 18 at 20:31
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Write $e_k \in R^n$ for each $1 \leq k \leq n$ for $e_k=(0,...,1,...,0)$, with the $1$ in the $k$th slot, and simply define $g:R^{nm} \to R^n \otimes_R R^m$ by, for each $\vec{r} = (r_i)_{i=1}^{nm} \in R^{nm}$

$$g(\vec{r})=\sum_{k=1}^n e_k \otimes (r_{(k-1)m+i})_{i=1}^m $$ I.e., tensor $e_k$ with the $k$th disjoint set of $m$ consecutive coordinates in $\vec{r}$, and sum over $k$. It is straightforward to check that $f' \circ g$ and $g \circ f'$ are the identity maps.

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  • $\begingroup$ Yes nice! That's the inverse I was looking for in my proof. $\endgroup$ – Sigurd Mar 18 at 18:43
  • $\begingroup$ Just wondering where exactly you used commutativity of $R$ here? $\endgroup$ – Sigurd Mar 18 at 19:23
  • $\begingroup$ Your map $f$ was not $R$-bilinear if $R$ is not commutative, as then $f(\vec{r},\alpha \vec{s}) =(r_1 \alpha \vec{s},...r_n \alpha \vec{s}) \neq \alpha f(\vec{r},\vec{s})$, so $f$ does not induce an $R$-module homomorphism on the tensor product $R^n \otimes_R R^m$ in the noncommutative case. $\endgroup$ – jawheele Mar 18 at 19:55

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