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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?
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3x89g2
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    The point here is that $M\times N$ and $N\times M$ are not the same things. – Nicolas Jul 02 '16 at 21:50
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    Which definition of tensor product are you using? – hmakholm left over Monica Jul 02 '16 at 21:50
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    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. – Alex Provost Jul 02 '16 at 21:50
  • @HenningMakholm Updated. Please see the definition I have. – 3x89g2 Jul 02 '16 at 21:52
  • @Nicolas Is the order the only difference? – 3x89g2 Jul 02 '16 at 21:55
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    @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). – Nicolas Jul 02 '16 at 21:58
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    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. – egreg Jul 02 '16 at 22:31

2 Answers2

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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.

Alex Provost
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  • 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? – 3x89g2 Jul 02 '16 at 22:37
  • @Misakov I (mistakenly) assumed you defined $N \otimes M$ explicitly using generators and relations. See my edit. – Alex Provost Jul 02 '16 at 23:23
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    (the universal property method) How do we show that $n_1 \otimes m_1 = n_2 \otimes m_2 \Rightarrow \overline{f}(n_1 \otimes m_1):=f(m_1,n_1) = \overline{f}(n_2 \otimes m_2)=f(m_2,n_2)$ ? (which proves that that $\overline{f}$ is well-defined). – MUH Aug 02 '22 at 11:04
  • @MUH You can use universal property of $N \otimes M$: the map $N \times M \to P$ that maps $(n,m)$ to $f(m,n)$ is $A$-bilinear, so it induces the map $\overline f(n \otimes m) = f(m,n)$. Or via generators and relations, you can map $(n,m)$ in the free module on $N \times M$ to $f(m,n)$, and check that the relators are mapped to zero. For example, $(n+n',m) - (n,m) - (n',m) \mapsto f(m,n+n') - f(m,n) - f(m,n') = 0$ since $f$ is $A$-bilinear. – Alex Provost Aug 10 '22 at 00:00
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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.

hmakholm left over Monica
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