TITLE: Definition of $R$-modules
QUESTION [5 upvotes]: I have a dumb question on the definition of $R$-modules. I borrow the definitions from Wikipedia.
A left $R$-module $M$ over the ring $R$ consists of an Abelian group $(M,+)$ and an operation $R\times M \rightarrow M$ such that for all $r,s\in R$ and $x,y\in M$, we have

$r(x+y) = rx + ry$
$(r+s)x  = rx+sx$.
$(rs)x  = r(sx)$.
$1_{R} x = x$ if $R$ has a multiplicative identity $1_R$.

The article later says, a right $R$-module is defined similarly, and then "If $R$ is commutative, then left $R$-modules are the same as right $R$-modules and are simply called $R$-modules."
I have a slight issue with the third item above. How does the above definitions define $rs$? (As we have only defined multiplication of an element of $R$ by an element of $M$). I am assuming that this multiplication operation is the multiplication operation of the ring $R$ itself, say $\circ$, and actually the third line should perhaps read $(r\circ s)x = r(sx)$  instead. Is this true?
And similarly, for the second item we have the addition of two ring elements. This addition operation is the addition operation of the ring $R$, am I right?
In this context, my biggest problem is with this sentence "If $R$ is commutative, then left $R$-modules are the same as right $R$-modules and are simply called $R$-modules." Since we apparently have two different multiplications (one for the definition of the commutative ring $R$, and another for the definition of the module itself), how can the commutativeness of $R$ imply that the left $R$-module above is also a right $R$-module?

REPLY [4 votes]: If we have a left $R$-module $M$, denoted $r\cdot m$, then we can define a right multiplication by $m\star r:=r\cdot m$.  Now we must check that this indeed defines a right $R$-module structure, in particular, it must satisfy:
$$m\star(r\circ s)=(m\star r)\star s$$
If you expand the left hand side using the definition, you'll obtain $(r\circ s)\cdot m$, while on the right hand side, you'll obatin $s\cdot(r\cdot m)=(s\circ r)\cdot m$.  For these two to be equal for all $r,s\in R$, we require that the ring is commutative.