In a PID, every principal ideal is free

Question

Suppose $R$ is a PID and let $a \in R$ be non-zero. I want to prove that $aR$ is free as an $R$-module.
If I consider the map $\phi:R \rightarrow aR$, given by $\phi(r)=ar$, then this is an isomorphism of $R$-modules. Does this mean that $aR$ is free?

Answer 1

For this result to hold, would it not be sufficient for $R$ to be an integral domain? You're not really using the PID condition here.