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Let $R$ be a principal ideal domain and $I, J$ ideals in $R$. Prove that $IJ = I \cap J$ if and only if $I+J =R$.

I can't prove that $IJ = I \cap J$ implies $I+J = R$. The rest I can prove. Can someone help me with this?

user10024395
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    Can you check the problem statement, it seems that if $I$ or $J$ is the $0$-ideal, then $IJ=I\cap J$, but $I+J\not=R$. – Michael Burr Apr 30 '15 at 13:44
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    Hint: Think about the integers: suppose $I$ and $J$ are nonzero ideals in $\mathbb{Z}$. Then $I=\langle p\rangle$ and $J=\langle q\rangle$. Then $IJ=\langle pq\rangle$ and $I\cap J=\langle\text{lcm}(p,q)\rangle$. Then $IJ=I\cap J$ iff $pq=\text{lcm}(p,q)$. $pq=\text{lcm}(p,q)$ iff $\gcd(p,q)=1$, so $1\in\langle p\rangle+\langle q\rangle$. – Michael Burr Apr 30 '15 at 13:45
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    Most proofs for $,\Bbb Z,$ generalize to PIDs, e.g. see here for a quick one. – Bill Dubuque Apr 30 '15 at 14:31
  • Hi, I don't quite understand how to generalise it to PID. Isn't this the right proof? – user10024395 May 01 '15 at 01:51
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    @Aha $ $ Translate via: $\ (i)\supseteq (j)\iff i\mid j,,$ and $,(i)+(j) = (\gcd(i,j)),$ and $,(i)\cap(j) = ({\rm lcm}(i,j))\ \ $ Note you need to ping a username or no one will be notified by your comment. – Bill Dubuque May 01 '15 at 03:40
  • @MichaelBurr , and Bill_Dubuque , would any one of you please post an answer so I can at least upvote it? Thanks – evaristegd Jun 13 '19 at 02:16

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