Ideals of subrings of the rational numbers ($\mathbb{Z}\subseteq R\subseteq \mathbb{Q}⟹I=(I\cap \mathbb{Z})R$)

Every subring $R$ of the rational numbers containing $\mathbb{Z}$ is a principal ideal domain. If $a/b\in R$ is reduced, Bézout's identity gives integers $u,v$ with $ua+vb=1$, hence $1/b=u(a/b)+v\in R$. Thus every reduced denominator occurring in $R$ is a unit.

For an ideal $I$, write $I\cap \mathbb{Z}=d\mathbb{Z}$. Then $dR\subseteq I$. If $a/b\in I$ is reduced, $a=b(a/b)\in I\cap \mathbb{Z}$, so $a=dc$ for some integer $c$. Since $1/b\in R$, $a/b=d(c/b)\in dR$. Therefore $I=dR$, including the zero ideal. This is a concrete instance of extending and contracting ideals under localization of a ring.