\subseteq A$ and the generators of $J$ are contained in $B$. Let $(V,tV)$ be a discrete valuation ring attached to $(B,Q)$ for $Q \in {\operatorname{Spec}}B$, $J \subseteq Q$. Let $v$ be the extended valuation to $A$. But then we get $v(s^{1/q})=\frac{1}{q} \cdot v(s) \to 0$ $(q \to \infty)$ and $\inf\{v(a)~|~a \in J\} \ge v(t)=1$, which are not compatible with each other, as we know $r^{1/q} \cdot R^{\infty} \subseteq {\operatorname{Ker}}\phi$. Hence $R^{\infty}=T$, which proves that $R^{\infty}=(\overline{R})^{\infty}$.

\[coherent-5\] Let $R$ be a one-dimensional reduced excellent ring of characteristic $p>0$. Then $R$ is $F$-coherent if and only if the normalization of $R$ is purely inseparable over $R$.

As we will see later, this result gives a geometric interpretation of one-dimensional singularities of “$F$-coherent type” in characteristic zero. Roughly speaking, if the normalization separates a single singular point of a variety into at least two smooth points, then it is not of $F$-coherent type.

Let $R=k[x^{4},x^{2}y,xy^{2},y^{4}]$ for a field $k$ of characteristic $p > 2$. Then we see that $(xy^{2})^{2}/y^{4}=x^{2}$ is in the normalization of $R$. Now assume that $R$ is $F$-coherent. In view of Theorem \[coherent-4\], $x^{2}$ is purely inseparable over $R$. But then $(x^{4})^{n}=(x^{2})^{p^k}$ for some $k > 0$ and thus $2n=p^k$, 