Is $\operatorname{Frac}(\bigcap_{i \in I}R_i)=K$ when each $R_i$ is integrally closed and $\operatorname{Frac}(R_i)=K$?

Question

Let $K$ be a field. Let $\left \{ R_i\right \}_{i \in I} $ be a set of integrally closed domains in $K$ whose field of fractions equals $K$.

It is easy to show that $R:=\bigcap_{i \in I}R_i$ is an integrally closed ring in its field of fractions. I was wondering if the field of fractions of $R$ is always $K$.

Answer 1

It is not true in general. Here is a counter-example.

Consider the domains $R_p:=\mathbb Z[px]$, where $p$ is a prime number. Observe that the field of fractions of each $R_p$ is equal to $\mathbb Q(x)$. Since each $R_p$ is isomorphic to $\mathbb Z[x]$ then it is a UFD and hence integrally closed in its field of fractions. Then, observe that

\begin{equation} R:=\bigcap_{p: \ prime} R_p=\mathbb Z \end{equation}

where the field of fractions of $\mathbb Z$ is $\mathbb Q$.