# Proving $I+J=R$ in a PID

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.

• If $I = (0)$, then $I\cap J = I = IJ$ for any ideal $J$, and $I + J = J$. Are you sure this is the statement you want? Do you want instead if $I + J = R$, then $IJ = I\cap J$? – Stahl Jul 19 '17 at 4:48
• See question number 11, Part 2. math.indiana.edu/graduate/tier1/Algebra/2009-01-Algebra-T1.pdf – learning_math Jul 19 '17 at 5:31
• It would be preferable for you to include what you are able to show by simply applying the definition of PID to the nonzero ideals $I$ and $J$ (and $IJ$, for the sake of completeness). – hardmath Jul 19 '17 at 5:36

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