# 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?

• 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 $M$.

• 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
• No harm in waiting to see what others come up with, although I can't see much more to say, other than fleshing out the universal property argument (which is a line or two). – Dylan Moreland Jun 11 '12 at 21:28
• @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,

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 $M$.

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

• 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