which cannot be the case since $p > 2$. Therefore, $R$ is not $F$-coherent. However, if $k$ has characteristic 2, then $R$ is $F$-coherent. Finally, the normalization of $R$ is just $k[x,y]$ for any field $k$, so it is $F$-coherent.

\[coherent-6\] Let $(R,{{\mathfrak m}})$ be a reduced local ring which is a residue class ring of a Gorenstein local ring. Then if $R$ is an $F$-coherent ring, every system of parameters of $R$ is a regular sequence on $R^{\infty}$.

The claim is that $R^{\infty}$ is a big Cohen-Macaulay $R$-algebra. First of all, we need to show that every system of parameters of $R$ forms an almost regular sequence on $R^{\infty}$ ([@RSS], Proposition 1.4 for the case when $R$ is a domain). Let $d=\dim R$ and $x_{1},\ldots,x_{d}$ be a system of parameters. Then by ([@BH], Corollary 8.1.4), there is a non-nilpotent element $c \in R$ for which $$c \cdot \big((x_{1}^{p^e},\ldots,x_{i}^{p^e}):_{R} x_{i+1}^{p^e}\big) \subseteq (x_{1}^{p^e},\ldots,x_{i}^{p^e})$$ for $0 \le i \le d-1$ and any integer $e > 0$. Since $R^{1/p^e} \simeq R$ under the $e$-th Frobenius map, we have $c^{1/p^e} \cdot \big((x_{1},\ldots,x_{i}):_{R^{1/p^e}} x_{i+1}\big) \subseteq (x_{1},\ldots,x_{i})R^{1/p^e}$. Since $e > 0$ is arbitrary, we have $$c^{1/q} \cdot \big((x_{1},\ldots,x_{i}):_{R^{\infty}} x_{i+1}\big) \subseteq (x_{1},\ldots,x_{i})R^{\infty}$$ for all $q=p^e$. For a 