# 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?

• The map $F$ sending the pair $\left(m,n\right)$ to $mr \otimes n$ does not in general satisfy $R$-bilinearity (it is the $F\left(ms,n\right) = F\left(m,sn\right)$ condition that can fail); thus you cannot usually define a map from $M \otimes_R N$ to $M \otimes_R N$ sending $m\otimes n$ to $mr\otimes n$. Nov 20, 2014 at 3:41
• I don't understand how that condition can fail. $F(ms,n)=ms \otimes n=m \otimes sn=F(m,sn)$ is one of the basic properties of tensors I thought? Nov 20, 2014 at 3:51
• You forgot the $r$ in the definition of $F$. Nov 20, 2014 at 4:20
• Ah ok, it makes sense now. Nov 20, 2014 at 5:00

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.