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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.

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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
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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.

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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).

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