If $aR$ is free, how can I conclude $a$ is not a zero divisor?

Question

Let $M=R_R$. I need to show that every submodule of $M$ is free if and only if $R$ is a PID.
For the first implication, assume that every submodule of $M$ is free.
If I take an element $a \in R$, $a$ non-zero, then I know that $aR$ is free, as a submodule of $M$. How do I show that $a$ is not a zero divisor? Then I could conclude that $R$ is an integral domain.

Answer 1

First you need to note that a basis for a free ideal in a commutative ring can only have one generator. (There are actually some exotic cases in noncommutative rings that permit ideals of higher rank!)

For if $b_1, b_2$ were two different basis elements, then $b_1b_2\in b_1R\cap b_2R=\{0\}$. By uniqueness of representation this would imply $b_2=0$, a contradiction because $0$ cannot be a basis element.

So let $x\in aR$ be a basis element. In particular $x=ar$ for some $r\in R$. But if $ab=0$ for some nonzero $b\in R$, you'd have that $xr=0$. Since $x$ is the basis for $aR$, that would imply $r=0$ and in turn $x=0$, a contradiction.