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

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

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