Here $M$ and $N$ are both left $R$-module. I have seen that $M\otimes_R N \cong N \otimes_R M$ is meaningful only when $R$ is commutative, but I can't see the reason.

In the noncommutative case, tensor product of two left $R$-module $M,N$ could be defined as an left $R$-module$M\otimes_R N$, right(although it seems that it's useless)? And then we could ask if there always holds $M\otimes_R N \cong N \otimes_R M$ as left $R$-module. I think it's true but I can't see why this is meaningless. Could you give some hints? Thanks in advance.

  • 2
    $\begingroup$ Where did you read about tensor product of two left $R$-modules? In the non commutative case, $M$ needs to be a right $R$-module and $N$ needs to be a right $R$-module for $M\otimes_RN$ to be just an abelian group. $\endgroup$ – Jackozee Hakkiuz Oct 11 '20 at 9:30
  • 2
    $\begingroup$ @JackozeeHakkiuz I guess you mean $N$ needs to be a left $R$-module. $\endgroup$ – Berci Oct 11 '20 at 9:33
  • $\begingroup$ @JackozeeHakkiuz I see it in this question math.stackexchange.com/questions/2091423/… $\endgroup$ – likemath Oct 11 '20 at 9:33
  • 1
    $\begingroup$ @likemath If you use the definition provided in the post you linked, everything can be led through the commutative $R/[R,R]$. Otherwise, it's usually just not defined. $\endgroup$ – Berci Oct 11 '20 at 9:39
  • $\begingroup$ @Beci aghh yeah I meant that. I screw up as well hahah :(. The easy thing to remember is that the $R$-module structure needs to "touch the tensor", so $R$ acts on th right of $M$ and on the left of $N$. Thanks for the correction. $\endgroup$ – Jackozee Hakkiuz Oct 11 '20 at 10:19

First of all, note that $\Bbb Z$ acts naturally on the other side on every left or right module, and that if $R$ is a commutative ring, then we can regard any $R$-module as an $R$-$R$-bimodule.
Said that, every module can be regarded a bimodule.

The tensor product in the noncommutative setting rather serves as a composition-like operation of bimodules:

If $M$ is an $A$-$B$-bimodule and $N$ is a $B$-$C$-bimodule, then the thing we can naturally obtain is the tensor product $M\otimes_BN$ as an $A$-$C$-bimodule.
Its construction is similar, we just need to take care on the left and right actions, so that the free Abelian group on $M\times N$ can be quotiented out by $(mb,\,n)\sim (m,\,bn)$ (among the other rules to ensure distributivity).

Note that in this setting, the actions of $B$ are 'swallowed' by the tensor product, but the actions of $A$ (from left on $M$) and of $C$ (from right on $N$) are naturally preserved.


Formally the tensor product is defined between right $M$ and left $N$ module. That's in order to make this true: for $a\in M$, $b\in N$ and $r,s\in R$

$$ars\otimes b=ar\otimes sb=a\otimes rsb$$

Note that otherwise (i.e. both are left modules) we would have

$$rsa\otimes b=sa\otimes rb=a\otimes srb$$

for which you need commutativity of $R$. Now $M\otimes N$ is itself an abelian group, not an $R$ module. In order for $M\otimes N$ to be an $R$ module some additional structure on $M$ or $N$ is required, e.g. bimodule structure. Note that if $R$ is commutative (or more generally $R$ is equiped with an antihomomorphism $R\to R$) then every module is naturally a bimodule.

You could of course reverse sideness (i.e. $M$ is left, $N$ is right) and do

$$rsa\otimes b=sa\otimes br=a\otimes brs$$

and this is fine. In that setup $M\otimes N$ will be (group) isomorphic to $N\otimes M$. But given those additional bimodule structures I don't think the isomorphism has to preserve the $R$ action (in noncommutative case).


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.