$R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=I\cap J$

Question

If $R$ is a commutative ring and $I$ and $J$ are ideals s.t. $I+J=R$ then show that $IJ=I\cap J$.

I've already shown that $IJ \subset I\cap J$, now I need to show the reverse inclusion.

I'm a bit lost, so far i'm just figuring out what pieces I have to work with.

Such as:

$\forall r\in R$ $\exists i\in I ,j\in J$ s.t. $i+j=r$

$\forall ij\in IJ$, $ij=i_1$ and $ij=j_1$ for some $i_1\in I$, $j_1\in J$.

Also, if I let $x\in I\cap J$, then $x=i_2=j_2=i+j$ for some $i_2\in I$, $j_2\in J$

Anyone, having problem getting to the conclusion here, thanks in advance

Answer 1

$I \cap J = (I \cap J) \cdot R \\ = (I \cap J) \cdot (I + J) \\ = (I \cap J) \cdot I + (I \cap J) \cdot J \\ \subseteq IJ + IJ \\ = IJ.$

Answer 2

There are $i\in I, j\in J$ such that $i+j=1$. Then, for all $a\in I\cap J$, $$a=a1=a(i+j)=ai+aj\in JI+IJ=IJ.$$