Left vs right ring module definitions

Question

In the book Algebra Vol. 2 by Cohn we have the following definitions

Let $R$ be any ring. A right $R$-module is an abelian group $M$ with a mapping $M\times R$ to $M$, $(x,r)\mapsto xr$ such that

• M.1 $\ (x+y)r=xr+yr,\quad x,y\in M,$
• M.2 $\ x(r+s)=xr+xs,\quad r,s\in R$,
• M.3 $\ x(rs)=(xr)s$,
• M.4 $\ x1=x$.

[...]

If in place of M.3 we have $x(rs)=(xs)r$, we shall usually write $r.x$ instead of $xr$, so that our module now satisfies

• M.3$^\circ$ $\ (rs).x=r.(s.x)$,

instead of M.3. This is called a left $R$-module, in contrast to the right $R$-modules satisfying M.3.

I have two questions for the above:

1. Why does he write the multiplication as $xr$ when it is a right module, but with a dot $r.x$ when it is a left module, rather than $rx$?
2. When he defines the left $R$-module, why does he only mention how M.3 changes? The other three conditions must change to a left multiplication of elements in $R$ as well.

Answer 1

I guess he uses a dot for left multiplication as a way of indicating that it is a left module. This can of course also be inferred by knowing that $r$ is a ring element and $x$ is a module element, but with the dot, it can be seen directly.

He only mentions how M3 changes because the others change in the obvious way, by replacing $xr$ by $rx$. But that is not how M3 changes. The original M3 was $x(rs) = (xr)s$. If we just change right multiplication to left multiplication, we would get $(rs)x = s(rx)$ which is not what we want. We must also exchange $r$ and $s$ on the right side to get the desired $(rs)x = r(sx)$.

This is the main difference between left and right modules. If you right multiply by $rs$, then it is equal to right multiplication by $r$, then right multiplication by $s$. But with left multiplication by $rs$, it is the other way around: left multiplication by $rs$ equal left multiplication by $s$, then left multiplication by $r$. So it is not just changing right multiplication to left multiplication.

The reason for this difference is purely to make the symbol manipulation natural, as if we have an "associative law" (but of course this is not really an associative law, since the elements $r$, $s$ come from a different set than $x$).