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Let $R$ be a unitary ring and $I_1,I_2$ be nontrivial ideals of $R$. I know that if $I_1+I_2=R$, then $I_1I_2=I_1 \cap I_2$. I wonder if this inverse holds.

When $R=\mathbb{Q}[X,Y]$ and $I_1=(X),I_2=(Y)$, then $I_1I_2=(XY)=I_1 \cap I_2$, but $I_1+I_2=(X,Y) \neq \mathbb{Q}[X,Y]=R$. According to this example, it is known that more conditions for $R$ are needed for the inverse to hold.

Then I heard that if $R$ is PID, the inverse holds, that is, $I_1I_2=I_1 \cap I_2 \Longrightarrow I_1+I_2=R$. However, I cannot prove it.

Would you have any ideas or hints for the proof? Thank you.

certain_integral
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    Consider $R = \mathbb{Z}/6$ with ideals $I = J = (2)$. Then $I \cdot J = (2) = I \cap J$, even though $I + J \ne R$. – Dietrich Burde Aug 11 '22 at 15:19
  • In a PID it's simply $,{\rm lcm}(a,b) = ab\Rightarrow \gcd(a,b)=1,,$, which has been proved here many times. More generally it holds in Prufer domans, see $(14)$ here. – Bill Dubuque Aug 11 '22 at 15:21
  • See the 2nd dupe for $(a)\cap (b) = ({\rm lcm}(a,b))$ and $,(a)+(b) = (\gcd(a,b)),,$ and see the first for $(\gcd(a,b){\rm lcm}(a,b)) = (ab),,$ which holds in any gcd domain). – Bill Dubuque Aug 11 '22 at 15:33

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If $R$ is a PID and hence a UFD, write $I_1=(a), I_2=(b)$ and $a=p_1^{a_1}...p_k^{a_k}$ and $b=p_1^{b_1}...p_k^{b_k}$ where $p_i$'s are primes and $a_i,b_i\geq 0$ integers. Now what does the equality tell you?
For simplicity think in $\mathbb{Z}$ first.

user26857
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Naba Kumar Bhattacharya
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