Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $m \otimes n = 0$.

share|improve this question
    
Are you sure that $n \otimes m = 0$ implies either $n = 0$ or $m = 0$? –  Cocopuffs Jun 8 '12 at 12:41
8  
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 $ –  user20266 Jun 8 '12 at 12:41
    
@Thomas What title do you suggest? –  Rudy the Reindeer Jun 8 '12 at 12:46
    
Maybe just omit the word unique. Or add some constraint which enforces uniqueness. –  user20266 Jun 8 '12 at 13:55
3  
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. –  Zhen Lin Jun 8 '12 at 14:27

1 Answer 1

up vote 9 down vote accepted

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.

example:

$\bar1\otimes \bar2 \in \mathbb{Z}/2\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}/3\mathbb{Z}$ 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.

share|improve this answer
    
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? –  Rudy the Reindeer Jun 8 '12 at 12:52
5  
Yes, exactly. In general it is hard to show injectivity on tensor products, so this is a standard trick. –  Seth Jun 8 '12 at 12:57
    
Nice, thank you very much! –  Rudy the Reindeer Jun 8 '12 at 13:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.