= Solution
{c}
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> gives
$$
\boxed{\mathbb Z_{ab}\longrightarrow\mathbb Z_a\times\mathbb Z_b,\qquad
[n]_{ab}\longmapsto([n]_a,[n]_b)}
$$
as a <ring isomorphism>. This includes $a=1$ or $b=1$, where the corresponding factor is the zero ring.
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