TITLE: $R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=I\cap J$
QUESTION [12 upvotes]: 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

REPLY [8 votes]: There are $i\in I, j\in J$ such that $i+j=1$. Then, for all $a\in I\cap J$,
$$a=a1=a(i+j)=ai+aj\in JI+IJ=IJ.$$