When is the Tensor product of Modules itself a Module? If $M$ is a right $R$ module and $N$ is a left $R$ module then $M \otimes_R N$ is an abelian group. Wikipedia says that if $M$ is an $R$ bimodule then $M \otimes_R N$ can take on the structure of a left $R$ module under the operation $r(m \otimes n)=rm\otimes n$. I tried to confirm that this actually does make $M \otimes_R N$ an $R$ module but the properties didn't seem obvious to verify as they required working with some non-pure tensors. I'm wiling to take Wikipedia's word for it, but the question this definition gave me was: why doesn't $M \otimes_R N$ automatically inheret the structure of a left $R$ module under the operation $r(m \otimes n)=mr \otimes n$? What goes wrong with this definition which does not go wrong with the $r(m \otimes n)=rm \otimes n$ definition?

First of all, setting $r(m\otimes n)=mr\otimes n$ would have no chance in general of giving us a left $R$-module structure, since $M$ is a right $R$-module, so let's try defining $r(m\otimes n) = m\otimes rn$. In order for this to give a left $R$-module structure to $M\otimes N$, we need (at least) that the maps $\theta_r \colon M\times N \to M\otimes N$ given by $\theta_r(m,n)=m\otimes rn$ ($r\in R$) be $R$-middle-linear. This fails in general:

Let $r, s\in R$. Then for all $m\in M$, $n\in N$, $$ \theta_r(m,sn) = m\otimes rsn,$$ while $$ \theta_r(ms,n) = ms \otimes rn = m\otimes srn.$$ If $R$ isn't commutative, these two quantities are not necessarily going to be the same.