of characteristic $p>0$. Suppose that every finitely generated ideal of $A$ satisfies the Krull’s intersection property; that is $\bigcap_{n>0} I^{n}=0$. Then every finitely generated ideal of $A$ is tightly closed.

Let $I$ be any finitely generated ideal and let $z \in I^{*}$. Then there exists $c \in A^{0}$ such that $cz^{[q]} \in I^{[q]}$ for $q=p^{e} \gg 0$. If we assume $z \notin I$, then $c \in (I^{[q]}:_{A}z^{[q]})$ is a non-zero proper ideal of $A$. By the flatness of $\mathbf{F}_{A}$, we have $$(I:_{A}z)^{[q]}=(I^{[q]}:_{A}z^{[q]}),$$ which is finitely generated, due to ([@Glaz], Theorem 2.3.2). Hence $$c \in \bigcap_{q>0} (I:_{A}z)^{[q]} \subseteq \bigcap_{q>0} (I:_{A}z)^{q}=0.$$ But this is a contradiction.

\[Krull\] Let $R$ be any Noetherian domain and let $I$ be its proper ideal. Then $R^{+}$ is $I$-adically separated.

If we assume the contrary, there is a non-zero element $c \in \bigcap_{n>0} I^{n}R^{+}$. Extend $R$ to its module-finite extension domain $S$ so that $c \in IS$, and we may choose a prime ideal $P$ of $S$ such that $IS \subseteq P$ and a discrete valuation ring $(V,tV)$ attached to $(S,P)$. Then we must have $c \in \bigcap_{n>0} I^{n}R^{+} \cap V \subseteq \bigcap_{n>0} t^{n}V^{+} \cap V=\bigcap_{n>0} t^{n}V=0$, which is a contradiction.

The following proposition looks similar to 