# Product of two coprime ideals in a commutative ring.

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?

• If $I+J = R$ then $I \cap J = (I \cap J)(I + J) = (I \cap J)I + (I \cap J) J \subseteq JI + IJ = IJ$ – Badam Baplan Mar 24 '19 at 23:08
• user26857, I've edited the question to reflect that. – Insignificant Mar 26 '19 at 0:39

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, $$a\in IJ$$.
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 $$bx\in IJ$$.
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 $$k=k(ui+vj)$$.