Every projective $R$-module $P$ is free

Question

I have come across a theorem which states that if the underlying ring $R$ is a principal ideal domain then every $R$-module $P$ which is projective is free also.

But the problem is I have encountered the example of $\mathbb{Z}_2$ over $\mathbb{Z}_6$ which is projective but not free. But $\mathbb{Z}_6$ is a PID.

Then how is the theorem true?

Answer 1

$\mathbb{Z}/6\mathbb{Z}$ is not a PID, because it's not a domain.