# Does $M\otimes_R N \cong N \otimes_R M$ hold for modules $M, N$ over noncommutative ring $R$?

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.

• 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. – Jackozee Hakkiuz Oct 11 '20 at 9:30
• @JackozeeHakkiuz I guess you mean $N$ needs to be a left $R$-module. – Berci Oct 11 '20 at 9:33
• @JackozeeHakkiuz I see it in this question math.stackexchange.com/questions/2091423/… – likemath Oct 11 '20 at 9:33
• @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. – Berci Oct 11 '20 at 9:39
• @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. – 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).