@>>>  R^{1/p^2} @>>> \cdots \\
@AAA @AAA @AAA \\
r \overline{R} @>>> r^{1/p}(\overline{R})^{1/p} @>>>  r^{1/p^2}(\overline{R})^{1/p^2} @>>> \cdots \\
\end{CD}$$ and taking the direct limit, we have $r^{1/q}(\overline{R})^{\infty} \subseteq R^{\infty}$ for any $q=p^{e}$. What we need to show is $(\overline{R})^{\infty}=R^{\infty}$. Let $R^{\infty} \subseteq T \subseteq (\overline{R})^{\infty}$ be any module-finite extension. Then $$\begin{CD}
0 @>>> R^{\infty} @>>> T @>>> T/R^{\infty} @>>> 0 \\
\end{CD}$$ is a short exact sequence of coherent $R^{\infty}$-modules ([@Glaz], Theorem 2.2.1). Next, let $N:=T/R^{\infty}=R^{\infty} \cdot u_{1}+\cdots+R^{\infty} \cdot u_{k}$ and define the $R^{\infty}$-module map $\phi:R^{\infty} \to N^{\oplus k}$ by $\phi(a)=(au_{1},\ldots,au_{k})$. Then we see that ${\operatorname{Ker}}\phi=(0:_{R^{\infty}}N)$ and $r^{1/q} \in {\operatorname{Ker}}\phi$ for all $q=p^{e}$. But since $\phi$ is a map of coherent $R^{\infty}$-modules, ${\operatorname{Ker}}\phi$ is finitely generated in $R^{\infty}$ by ([@Glaz], Corollary 2.2.2).

Assume that ${\operatorname{Ker}}\phi  \ne R^{\infty}$. Then choosing a minimal prime $P$ of $R^{\infty}$ that is contained in a maximal ideal containing ${\operatorname{Ker}}\phi$, we consider the natural map $\Psi:R^{\infty} \to R^{\infty}/P$. Then $J=\Psi({\operatorname{Ker}}\phi)$ is a finitely generated proper ideal and for $r \in R$ as above, we see that $s=\Psi(r)$ is non-zero because $r \notin P$. $A'=R/(R \cap P) \to A=R^{\infty}/P$ is an integral extension and $A'$ is Noetherian. Extend $A'$ to its module-finite extension $B$ such that $B 