Question: Let $\mathbb{Z} \subseteq R \subseteq \mathbb{Q}$ where $R$ is an integral domain. If $I$ is an ideal in $R$ such that $I \cap \mathbb{Z} = d\mathbb{Z}$, show that $I = dR$.
Here is my incomplete attempt:
Proof. Note that $\frac{a}{b} \in R \implies \frac{1}{b} \in R$; this follows from applying the Bezout's Lemma $$an+bm=1$$ and using the fact that $\mathbb{Z} \subseteq R$. Therefore, $R$ is an Euclidean domain with norm($\frac{a}{b}$) = $|a|$, thus $R$ is a Principal Ideal domain.
We know that $I = kR$, where $k \mid d$ in $R$ because $d \in I$. But if $k \notin \mathbb{Z}$ then clearly $kR = R$. $\square$
I am stuck here. We have got division in $R$ and in $\mathbb{Z}$. Also, I believe that there is a more fluid method to solve this question. Any help is greatly appreciated, thanks!
Suppose that $\frac{a}{b}\in I$, then $a\in \mathbb{Z}\cap I$, which implies $a\in d \mathbb{Z}$. Since $\frac{1}{b}\in R$, you have that $\frac{a}{b}\in dR$