Module finite extension of one-dimensional local domain which is PID

Question

Let $R$ be a Noetherian local domain of dimension $1$. Let $R \subseteq B$ be a module finite ring extension such that $B\subseteq Q(R)$. Then clearly, $B \subseteq \overline R$, where $\overline R$ is the integral closure of $R$ in $Q(R)$. My question is: If $B$ is a principal ideal domain, then is $B=\overline R$ ?

Answer 1

Yes, principal ideal domains are normal domains (see Every PID is integrally closed), so integrally closed in their field of fractions. Since $R\subset B\subset Q(R)$, we have $Q(B)=Q(R)$, and since $R\subset B \subset \overline{R}$, we have $\overline{R} \subset \overline{B} \subset \overline{\overline{R}} = \overline{R}$, so $\overline{B}=\overline{R}$. Finally, since $B$ is integrally closed in $Q(B)=Q(R)$, we must have $B=\overline{B}=\overline{R}$.