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

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 $$K$$.

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
$$$$R:=\bigcap_{p: \ prime} R_p=\mathbb Z$$$$
where the field of fractions of $$\mathbb Z$$ is $$\mathbb Q$$.