5
$\begingroup$

I am working with the the tensor product $-\otimes_R -$ over some noncommutative ring $R$. Is the tensor product always commutative if $R$ is commutative? If so, can the tensor product be commutative if $R$ is noncommutative?

$\endgroup$
  • $\begingroup$ Do you mean whether $M\otimes_RN\simeq N\otimes_RM$? $\endgroup$ – user26857 Jul 6 '15 at 16:59
  • 1
    $\begingroup$ The thing is, if $R$ is not commutative, $M\otimes_R N$ makes sense only if $M$ is a right $R$-module and $N$ a left $R$-module. Hence the tensor product is not commutative, but worse, it is defined for only one way. $\endgroup$ – Roland Jul 6 '15 at 17:48
  • $\begingroup$ Well, I suppose we could consider $M,\,N$ to be $R$-$R$- bimodules? $\endgroup$ – Sam Williams Jul 7 '15 at 14:05
7
$\begingroup$

The tensor product's commutativity depends on the commutativity of the elements. If the ring is commutative, the tensor product is as well. If the ring R is non-commutative, the tensor product will only be commutative over the commutative sub-ring of R. There will always be tensors over the ring that will not commute if R is non-commutative.

$\endgroup$
  • $\begingroup$ Many thanks! I was thinking along those lines but was unsure. $\endgroup$ – Sam Williams Jul 7 '15 at 14:06
  • 2
    $\begingroup$ Actually, even if $R$ is commutative, then the tensor product is not always commutative! Indeed, we are talking about $R$-$R$-bimodules here, not about $R$-modules (= $R$-$R$-bimodules satisfying the identity $rm = mr$). In $M \otimes_R N$, you "use up" the right $R$-module structure on $M$ and the left $R$-module structure on $N$ for tensoring, whereas in $N \otimes_R M$ it is the "other two" structures that you are using. The results can easily be non-isomorphic. $\endgroup$ – darij grinberg Sep 13 '15 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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