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If $I$ and $J$ are two coprime ideals in a unitary commutative ring $R$, i.e. $I+J=R$, then $IJ=I\cap J$.

The above fact has been stated without proof in almost every textbook I have referred. No one seems to be giving any direction for proof.

I know that one direction of the proof ( $IJ\subseteq I\cap J$ ) is trivial.

How do I go about the other direction?

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  • $\begingroup$ If $I+J = R$ then $I \cap J = (I \cap J)(I + J) = (I \cap J)I + (I \cap J) J \subseteq JI + IJ = IJ$ $\endgroup$ – Badam Baplan Mar 24 '19 at 23:08
  • $\begingroup$ user26857, I've edited the question to reflect that. $\endgroup$ – Insignificant Mar 26 '19 at 0:39
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To see $IJ\supseteq I\cap J$ you should use the assumption on $I$ and $J$. Take $x\in I$ and $y\in J$ so that $x+y=1$. Now, suppose $a\in I\cap J$, then $ax+ay=a$. $ax\in IJ$ because $a\in J$ and $x\in I$ and similarly $ay\in IJ$ because $a\in I$ and $x\in J$. So, $ax+ay\in IJ$ so, $a\in IJ$.

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Because $I+J=R$, you can write $1=a+b$ for $a\in I$ and $b\in J$.

Now if $x\in I\cap J$ you have $x=1x=ax+bx$; use now the fact that $x\in J$ to get $ax\in IJ$, and use that $x\in I$ to get $bx\in IJ$.

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You have a Bézout's relation: $\;ui+vj=1$ for some $u,v\in R,i\in I, j\in J$. Now take a $k\in I\cap J$, and write $k=k(ui+vj)$.

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