# Showing that $R^n \otimes_R R^m$ is isomorphic to $R^{nm}$ when $R$ is commutative

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?

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

• 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? – Sigurd Mar 18 at 18:15
• @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. – Javi Mar 18 at 18:20
• Oh indeed I see. – Sigurd Mar 18 at 18:37
• 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 $\vec{r} \otimes \vec{s} = \sum_{k=1}^n \sum_{j=1}^m (r_k e_k) \otimes (s_j h_j) = \sum_{k,j} (r_k s_j) (e_k \otimes h_j) = 0$. – jawheele Mar 18 at 19:00
• 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 $e_k \in R^n$. – jawheele Mar 18 at 20:31

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.

• Yes nice! That's the inverse I was looking for in my proof. – Sigurd Mar 18 at 18:43
• Just wondering where exactly you used commutativity of $R$ here? – Sigurd Mar 18 at 19:23
• 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. – jawheele Mar 18 at 19:55