contradiction, consider the multiplication map: $$\begin{CD}
R^{\infty}/(x_{1},\ldots,x_{i})R^{\infty} @>x_{i+1}>> R^{\infty}/(x_{1},\ldots,x_{i})R^{\infty}
\end{CD}$$ and assume that for some $i>0$, the kernel of the above map contains a non-zero cyclic $R^{\infty}$-submodule $N$. Since $R^{\infty}/(x_{1},\ldots,x_{i})R^{\infty}$ is finitely presented over $R^{\infty}$, it is coherent by ([@Glaz], Theorem 2.3.2), and thus we find that $N$ is a finitely presented cyclic $R^{\infty}$-module. So we may write it as $R^{\infty}/J$ for some finitely generated ideal $J$. Note that $J$ is not a unit ideal.

Knowing that $c^{1/q} \cdot R^{\infty} \subseteq J$, $R^{\infty}$ is local, and $c$ is not nilpotent, we can find a minimal prime $P$ of $R^{\infty}$ such that $c \notin P$ and $(J+P)$ is a proper ideal. After reducing $R^{\infty}$ modulo $P$, we may argue as previously to get a contradiction by finding some optimal valuation on $R^{\infty}/P$. Hence $x_{i+1}$ must be a non-zero divisor of $R^{\infty}/(x_{1},\ldots,x_{i})R^{\infty}$. As our initial choice of a system of parameters was arbitrary, we proved the theorem.

For example, if $R$ is a complete local domain and $F$-pure, but not Cohen-Macaulay, then in view of Theorem \[coherent-6\], a system of parameters of $R$ is not a regular sequence on $R^{\infty}$. Hence $R$ is not $F$-coherent. Rings of this type abound. Example \[coherent-2\] 