shows that even if $R^{\infty}$ is Cohen-Macaulay, $R$ is not so in general. We will explore this issue later again. The following proposition is taken from ([@AH], Proposition 2.3) with a slight modification.

\[coherent-7\] Let $R$ be a reduced Noetherian ring of characteristic $p>0$. Suppose that $R$ is $F$-coherent and $I \subseteq R$ is any ideal. Then we have $I^{F}=I^{*}$. In particular, if $R$ is $F$-coherent, $F$-purity is equivalent to $F$-regularity.

We first recall that $I^{F} \subseteq I^{*}$. For a contradiction, let $u \in R$ be such that $u \in I^{*}$, but $u \notin I^{F}$. Then coherence implies that $J:=IR^{\infty}:_{R^{\infty}} u$ is a finitely generated non-unit ideal, and we have $c u^{q} \in I^{[q]}$ for $q=p^{e} \gg 0$ and $c \in R^{0}$ by our assumption. Hence $c^{1/q}u \in IR^{1/q} \subseteq IR^{\infty}$, or equivalently $c^{1/q} \cdot R^{\infty} \subseteq J$ for $q=p^{e} \gg 0$. Since $c \in R^{0}$, we can find a minimal prime $P$ of $R^{\infty}$ such that $c \notin P$ and $(J+P)$ is a proper ideal. Then we may argue as previously to conclude that $I^{F}=I^{*}$.

Suppose next that $R$ is $F$-coherent and $F$-pure. Then the above discussion gives that $I=I^{F}=I^{*}$ for any ideal $I$, which implies that every ideal 