# Show that $M\otimes N$ is isomorphic to $N\otimes M$

I want to prove the following:

Let $$A$$ be a ring and $$M,N$$ be $$A$$-modules. Show that the tensor products $$M\otimes N$$ and $$N\otimes M$$ are isomorphic.

I have consulted this question but it did not answer my quetsion: Prove that $M \otimes N$ is isomorphic to $N \otimes M$.

The definition I have is:

Let $$A$$ be a ring. Let $$M,N$$ be $$A$$-modules. The tensor product $$M\otimes_A N$$ is another A-module together with an A-bilinear map $$\phi: M\times N\to M\otimes_A N$$ such that:

if $$P$$ is an A-module and $$f: M\times N \to P$$ is A-bilinear, then there exists a unique homomorphism $$\tilde{f}$$ such that $$f = \tilde{f}\circ\phi$$

My confusions are :

1. Can I just say that "any two tensor products of $$M$$ and $$N$$ are isomorphic, so these two have to be the same", given that I've proved the claim already?
2. What is the difference between $$M\otimes N$$ and $$N\otimes M$$? According to the definition I have, the only difference is that one starts with $$M\times N$$ and the other starts with $$N\times M$$. Is my understanding correct?
• The point here is that $M\times N$ and $N\times M$ are not the same things. Jul 2 '16 at 21:50
• Which definition of tensor product are you using? Jul 2 '16 at 21:50
• Re: 1: What is your definition of a "tensor product of $M$ and $N$"? Is it an object satisfying the universal property? Re: 2: It's analogous to the difference between $M \times N$ and $N \times M$. I.e., the order the elements appear in is swapped. Jul 2 '16 at 21:50
• @Misakov Well, $(a,b)\in M\times N$ iff $a\in M$ and $b\in N$, and similar for $N\times M$. So regardless the order you use the same objects (and this is the key point in the proof of the theorem you've stated in your question). Jul 2 '16 at 21:58
• It should be noted that you have to assume $A$ is a commutative ring. If the ring is not commutative, one of the tensor products doesn't make sense. Jul 2 '16 at 22:31

If you have already proved that the tensor product $M\otimes N$ is unique up to isomorphism, the heavy lifting is already done.

What you need to prove now is that $M\otimes N$ satisfies the condition for being $N\otimes M$. That is, you need to show that a $\phi':N\times M\to M\otimes N$ such that (etc etc etc). This ought to be easy enough because you already have $\phi:M\times N\to M\otimes N$ -- just show that the obvious way to construct $\phi'$ from $\phi$ gives you the required universal property.

If you want to go with the universal property route, assuming you've defined $N \otimes M$ as a quotient of the free module with basis $N \times M$ by the bilinearity relations:

Consider the $A$-bilinear map $\phi:M \times N \to N \otimes M$ given by $\phi(m,n) = n \otimes m$. Given $P$ and a bilinear map $f:M\times N \to P$, the only way to define a suitable $\overline{f}:N \otimes M \to P$ is to impose $\overline{f}(n \otimes m):=f(m,n)$ and to extend by linearity. (We have just inserted a generic element $(m,n) \in M \times N$ in the relation $f = \overline{f} \phi$.) Now you just have to check that this map is well-defined and a module homomorphism, and then you are done.

If instead you accept the existence of $N \otimes M$ without constructing it explicitly (as some module together with a map $\phi':N \times M \to N \otimes M$ satisfying the universal property), you may proceed as follows: the reversal isomorphism $f:M \times N \to N \times M$, $f(m,n) = (n,m)$, is $A$-bilinear; therefore the universal property of $M \otimes N$ gives us a map $\overline{f}:M \otimes N \to N \times M$ such that $f = \overline{f}\phi$. (Here $\phi:M \times N \to M \otimes N$ is the data that comes with $M \otimes N$.) We can compose with $\phi'$ to obtain $\phi' \overline{f}:M \otimes N \to N \otimes M$.

The same procedure with $M$ and $N$ reversed gives us a map $\phi\overline{f^{-1}}: N\otimes M \to M \otimes N$. Now we check that these are inverses of each other. For instance, let's verify that $g := (\phi'\overline{f})(\phi\overline{f^{-1}}) = 1_{N \otimes M}$ by showing that $g\phi' = \phi'$ and invoking the uniqueness part of the universal property of $N \otimes M$:

$$g\phi' = (\phi'\overline{f})(\phi\overline{f^{-1}})\phi' = \phi'(\overline{f}\phi)(\overline{f^{-1}}\phi') = \phi' f f^{-1} = \phi'.$$

The analogous argument for $h := (\phi\overline{f^{-1}})(\phi'\overline{f})$ shows that $h = 1_{M \otimes N}$, and we are done.

• I am actually confused by the notation $x\otimes y$. Is it defined as $x\otimes y = \phi (x,y)$ according to the definition I have? Jul 2 '16 at 22:37
• @Misakov I (mistakenly) assumed you defined $N \otimes M$ explicitly using generators and relations. See my edit. Jul 2 '16 at 23:23