 Solution

Take $R=\mathbb{Z}$, $I=a\mathbb{Z}$ and $J=b\mathbb{Z}$. Since $a,b$ are coprime integers, Bézout's identity gives $ra+sb=1$, so $I+J=\mathbb{Z}$. Their intersection is $ab\mathbb{Z}$: it consists of integers divisible by both $a$ and $b$, and coprimality makes this equivalent to divisibility by $ab$. Thus the Chinese remainder theorem givesas a ring isomorphism. This includes $a=1$ or $b=1$, where the corresponding factor is the zero ring.