Let $R$ be a ring A $(left) \ R-module$ consists of an additive group $R$ equipped with a mapping $R\times M\to M$, which is an action, meaning
$ (rr')m=r(r'm)$ for $r,r'\in R$ and $m\in M$,
it is distributive over addition, and
it is unital: $1m=m$ for all $m$.
Dually there is the notion of a $right \ R-module$ with an action $M\times R \to R$. Apart from notation, it is the same thing as a left $R^{op}-module$. If R is commutative, the notion coincide.
If $R$ is commutative, a left $R-module$ $M$ has a natural $R-R$ bimodule structure, by simply defining the right scalar action as $mr:=rm$ for $r\in R, m \in M$.
What i don't get is why we need to distinguish left and right module since they are just a notation transposition? Why does the notation coincide when the ring is commutative.
What is the meaning of opposite ring in simple words?
And why do we need the ring to be commutative for the bimodule to make sense?
Thanks for your help. I am a bit lost on this definition.
For the commutativity part, we'd want $(rs)\cdot m=m\cdot (rs)$. Which if we have commutativity, we can get by $(rs)=(sr)$. So $m\cdot (rs)=m\cdot (sr)=(m\cdot s)\cdot r$.
If, however, the ring is not commutative we are stuck with $(m\cdot r)\cdot s=m\cdot(rs)$.
We distinguish left and right, since we could have bimodules where a ring acts differently from each side.
Finally, for uses of bimodules, we can construct a special kind of product of modules called tensor products.