= Solution
{c}
Every multiple of $dh$ is a multiple of both $d$ and $h$, so $dh\mathbb Z\subseteq d\mathbb Z\cap h\mathbb Z$.
For the reverse inclusion, let $z=du=hv$ lie in this <intersection of ideals>. Since $d,h$ are <coprime integers>, <Bézout's identity> supplies $r,s\in\mathbb Z$ with $rd+sh=1$. Multiplying by $u$ gives
$$
u=rdu+shu=rhv+shu=h(rv+su).
$$
Thus $h\mid u$, and $dh\mid du=z$. Consequently
$$
\boxed{d\mathbb Z\cap h\mathbb Z=dh\mathbb Z}.
$$
Here <Bézout's identity> follows from the allowed <principal ideal domain> property of $\mathbb Z$: the <ideal> $d\mathbb Z+h\mathbb Z$ is generated by their <greatest common divisor>, which is $1$.
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