Proving $I+J=R$ in a PID

Question

Let $R$ be a PID. Let $I$ and $J$ be two nonzero ideals such that $I\cap J=IJ$. How to prove that $I+J=R$?

So far, I have only been able to show that $I$ and $J$ will be contained inside different prime ideals (hence maximal ideals, since in a PID prime ideals are maximal):

Let $\langle p\rangle$ be a prime ideal which contains both $I$ and $J$. Then $I=\langle ap\rangle$ and $J=\langle bp\rangle$ for some $a,b\in R$. Then $abp\in I\cap J=IJ$. As $abp\in IJ$ we have $abp=cabp^2$ for some $c\in R$. Then $cp=1$, a contradiction.

Answer 1

Hint: If $I + J\neq R$, then there exists a maximal ideal $\mathfrak{m}\subseteq R$ with $I + J\subseteq\mathfrak{m}$.