# Prove that $R \otimes_R M \cong M$

Let $R$ be a commutative unital ring and $M$ an $R$-module.

I'm trying to prove $R \otimes_R M \cong M$ but I'm stuck. If $(R \otimes M, b)$ is the tensor product then I thought I could construct an isomorphism as follows:

Let $\pi: R \times M \to M$ be the map $rm$. Then there exists a unique linear map $l: R \otimes M \to M$ such that $l \circ b (r,m)= l(r \otimes m) =r l(1 \otimes m) = \pi(r,m) = rm$.

Now I need to show that $l$ is bijective. Surjectivity is clear. But I can't seem to show injectivity. In fact, by now I think it might not be injective. But I can't think of a different suitable map $\pi$. Then I thought perhaps I should show that $l$ has a two sided inverse but for an $m$ in $M$ I can't write down its inverse. How do I finish the proof?

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In those kind of questions it would be nice if you precised that $M$ was an $R$-module, what was $b$, etc. Some people might be able to help you even though they might not guess what $R$ and $M$ are precisely ; tensor products can consist of rings, vector spaces, $R$-module with $R$ a ring, a unital ring, a field (i.e. vector spaces), etc... – Patrick Da Silva Jun 11 '12 at 21:22
@PatrickDaSilva You're right. I was typing it up too quickly and forgot. – Rudy the Reindeer Jun 11 '12 at 21:26
FWIW, in my experience the "elementary" isomorphisms concerning tensor products (e.g. this one, the distributivity with respect to direct sum, the commutativity if $R$ is a commutative ring, etc.) are most easily established by first defining a map out of your tensor product with the universal property, and then defining a map the other way around and checking that it is its inverse. – Bruno Stonek Jun 11 '12 at 23:03

For an inverse, define $M \to R \otimes M$ by $m \mapsto 1 \otimes m$. I suppose you could also show directly that the map is injective, but point remains the same: $R$ contains $1$. If $\sum r_i \otimes m_i$ is in the kernel then $\sum r_im_i = 0$, and we can write the tensor as $\sum 1 \otimes r_im_i = 1 \otimes \sum r_im_i = 1 \otimes 0 = 0$.

Another way is to show that your map $R \times M \to M$ satisfies the universal property of the tensor product of $R$ and $M$.

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Oh noes. Of course. I'd actually thought of that but then discarded it because I thought it wasn't actually an inverse. But of course it is! Thank you very much! – Rudy the Reindeer Jun 11 '12 at 21:22
Apparently, I can only accept your answer in 9 minutes... – Rudy the Reindeer Jun 11 '12 at 21:22
@MattN I just mean that a tensor product of $R$ and $M$ over $R$ is an $R$-module $T$ together with an $R$-bilinear map $t\colon R \times M \to T$ such that any $R$-bilinear $f\colon R \times M \to N$ factors through $t$ uniquely. If you show that the map $R \times M \to M$ is such a $t$, then $M$ is (canonically) isomorphic to anything else that you're calling the tensor product. – Dylan Moreland Jun 12 '12 at 15:07
@MattN. Hey Matt. I would define $l$ by $l(m) = \psi(1, m)$. Is it easier to see uniqueness now? – Dylan Moreland Jul 24 '12 at 15:31
@MattN. I guess I would say that no matter what $l$ is, for the appropriate diagram to commute we must have $l(m) = l(p(1, m)) = \psi(1, m)$. – Dylan Moreland Jul 24 '12 at 16:18

I am following the nomenclature I learn in Dummit & Foote's Abstract Algebra.

To show that $R \otimes_R M \cong M$, let $\varphi : R \times M \to M$ defined by $\varphi(r,m) = rm$. To show that this can be extended to an $R$-module homomorphism, one can show that this map is $R$-balanced, i.e. that \begin{align} \varphi( (r_1,m) + (r_2,m) ) & = \varphi((r_1,m)) + \varphi((r_2,m)), \\\ \varphi( (r,m_1) + (r,m_2) ) & = \varphi((r,m_1)) + \varphi((r,m_2)), \\\ \varphi( (r_1 r_2,m) ) & = \varphi((r_1,r_2m)). \end{align} Once that this is done (trivially), then one can show that $\varphi : R \otimes_R M \to M$ defined by $\varphi( r \otimes m) = rm$ and extending by linearity, is well defined and is an $R$-module homomorphism.

Surjectivity is clear since $\varphi(1 \otimes m) = m$ for every $m \in M$. For injectivity, you will want to show that $\varphi$ is invertible, and $\varphi^{-1} : M \to R \otimes_R M$ would be such that $\varphi^{-1}(m) = 1 \otimes m$. Clearly $\varphi$ and $\varphi^{-1}$ are inverses of each other, so that all you need to do is to verify that $\varphi^{-1}$ is an $R$-module homomorphism. Really not that hard, since it follows from the properties of the tensor product.

Hope that helps,

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As Dylan suggested, we can use the universal property of the tensor product to show that $R \otimes_R M \cong M$. Now this means we would like to show that the $R$ - module $M$ is equipped with a bilinear map $\pi : R \times M \rightarrow M$ such that for any bilinear map $B : M \times N \rightarrow P$, where $P$ is some other $R$ - module, there exists a unique linear map $L : M \rightarrow P$ such that

$$B = L \circ \pi.$$

This shouldn't be too hard. Say we are given an $R$ - bilinear map $B : M \times N \rightarrow P$. We can define the map $\pi : R \times M \rightarrow M$ by mapping the pair $(r,m)$ to $rm$. One easily checks that the map $\pi$ is well-defined and bilinear. Now define the map $L : M \rightarrow P$ on "elementary elements" by

$$L(m) = B(1,m)$$

and extend it additively. I said "elementary elements" because usually we define maps on elementary tensors - but there are none here so I just coined this term. Now your map $L$ is easily seen to be well-defined. For linearity, the fact that it is additive comes from definition of how we defined $L$. Let's check that it is compatible with scalar multiplication: For any $r \in R$, we have that

$$\begin{eqnarray*} L(rm) &=& B(1,rm) \\ &=& rB(1,m) \hspace{5mm} \text{(By definition of bilinearity)} \\ &=& r L(m) \end{eqnarray*}$$

completing the claim that $L$ was $R$ - linear. Now it remains to check that our map $B : R \times M \rightarrow P$ factors uniquely through the tensor product. The question whether $B$ factors through $M$ is equivalent to asking if first sending $(r,m)$ to $B(r,m)$ in $P$ is equal to first sending $(r,m) \mapsto rm$ in $M$, and then sending $rm$ in $M$ to $P$. But this is clear because

$$\begin{eqnarray*} L(rm) &=& B(1,rm) \\ &=& rB(1,m) \\ &=& B(r,m). \end{eqnarray*}$$

It is possible to do manipulations like that because $M$ is an $R$ module and the other guy in the direct product is $R$ itself.

It now remains to see why given our maps $B$ and $\pi$, there is a unique $R$ - linear map $L : M \longrightarrow P$. If you have any linear map $L$ out of $M$, in order for an appropriate diagram in question to commute we must have that $L(m) = L(\pi(1,m)) = B(1,m)$. There really is no choice for what $L$ is because it is defined by $B$. Hence there is only one $L$ in question for a given bilinear map $B: R \times M \longrightarrow P$ and uniqueness is proven.

We have shown that the $R$ - module $M$ satisfies the universal property of the tensor product $R \otimes_R M$, and hence must be isomorphic to $M$.

$$\hspace{6in} \square$$

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I think you need to say one more thing about the uniqueness. To me the point is that $m = \pi(1, m)$, so for commutativity of the right diagram we need $L(m) = L(\pi(1, m)) = B(1, m)$. So there's no choice in the definition of $L$. – Dylan Moreland Jun 12 '12 at 5:29
@DylanMoreland I have edited my answer, is that paragraph on uniqueness a little better now? – user38268 Jun 12 '12 at 6:01
Looks good!${}$ – Dylan Moreland Jun 14 '12 at 2:17
@DylanMoreland Thanks :-) – user38268 Jun 14 '12 at 4:46