I want to show that there is a unique isomorphism $M \otimes N \to N \otimes M$ such that $x\otimes y\mapsto y\otimes x$. (Prop. 2.14, i), Atiyah-Macdonald)

My proof idea is to take a bilinear $f: M \times N \to N \otimes M$ and then use the universal property of the tensor product to get a unique linear map $l : M \otimes N \to N \otimes M$. Then show that $l$ is bijective.

Can you tell me if my proof is correct:

Let $M,N$ be two $R$-modules. Let $(M \otimes N, b)$ be their tensor product.

Then $$ \varphi: M \times N \to N \otimes M$$ defined as $$ (m,n) \mapsto n \otimes m$$ and $$ (rm , n) \mapsto r(n \otimes m)$$ $$ (m , rn) \mapsto r(m \otimes n)$$

is bilinear. Hence by the universal property of the tensor product there exists a unique $R$-module homomorphism ($\cong$ linear map) $l: M \otimes N \to N \otimes M$ such that $l \circ b = \varphi$.

$l$ is bijective:

$l$ is surjective: Let $n \otimes m \in N \otimes M$. Then $l(m \otimes n) = l(b(m,n)) = \varphi (m,n) = n \otimes m$.

$l$ is injective: Let $l(m\otimes n) = l(b(m,n)) = 0 = \varphi(m,n) = n \otimes m$. Then $n \otimes m = 0$ implies that either $n$ or $m$ are zero and hence $m \otimes n = 0$.

  • $\begingroup$ Are you sure that $n \otimes m = 0$ implies either $n = 0$ or $m = 0$? $\endgroup$ – Cocopuffs Jun 8 '12 at 12:41
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    $\begingroup$ Your heading and first line of posting is misleading. There are many such isomorphisms, in general. Each isomorphism of $M$ or $N$ produces another one, when composed with an isomorphism $M\otimes N \rightarrow N\otimes M $ $\endgroup$ – user20266 Jun 8 '12 at 12:41
  • $\begingroup$ @Thomas What title do you suggest? $\endgroup$ – Rudy the Reindeer Jun 8 '12 at 12:46
  • $\begingroup$ Maybe just omit the word unique. Or add some constraint which enforces uniqueness. $\endgroup$ – user20266 Jun 8 '12 at 13:55
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    $\begingroup$ If you're going to use the universal property of tensor products, you may as well prove directly that $M \otimes N$ and $N \otimes M$ have the same universal property. $\endgroup$ – Zhen Lin Jun 8 '12 at 14:27

It is not true that $n\otimes m = 0$ implies either $n$ or $m =0$ (see example below). To prove injectivity you should define a map going the other way and show that these maps are inverse.


$\bar1\otimes \bar2 \in \mathbb{Z}/2\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}/3\mathbb{Z}$ satisfies $\bar1\otimes \bar2=\bar1\otimes (2\cdot\bar1)=(\bar1\cdot 2)\otimes \bar1= \bar0\otimes \bar1=0$ but $\bar1\in\mathbb{Z}/2\mathbb{Z}$ and $\bar2\in\mathbb{Z}/3\mathbb{Z}$ are not zero.

  • $\begingroup$ So I don't show injectivity and surjectivity respectively but instead show that the map $$k : N \otimes M \to M \otimes N$$ defined as $$n \otimes m \mapsto m \otimes n$$ is a left and right inverse of $l$ and hence $l$ is an isomoprhism? $\endgroup$ – Rudy the Reindeer Jun 8 '12 at 12:52
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    $\begingroup$ Yes, exactly. In general it is hard to show injectivity on tensor products, so this is a standard trick. $\endgroup$ – Seth Jun 8 '12 at 12:57
  • $\begingroup$ Nice, thank you very much! $\endgroup$ – Rudy the Reindeer Jun 8 '12 at 13:06

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