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.
Best Answer
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).