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?

  • $\begingroup$ Do you mean whether $M\otimes_RN\simeq N\otimes_RM$? $\endgroup$ – user26857 Jul 6 '15 at 16:59
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    $\begingroup$ 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

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.

  • $\begingroup$ Many thanks! I was thinking along those lines but was unsure. $\endgroup$ – Sam Williams Jul 7 '15 at 14:06
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    $\begingroup$ 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

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