implies that $(S^{-1}R)^{\infty}$ is also coherent.

For $(3)$, we pick ${{\mathfrak p}}, {{\mathfrak q}}\in {\operatorname{Spec}}R$ such that ${{\mathfrak p}}\subseteq {{\mathfrak q}}$ and ${{\mathfrak q}}\in U$. Then since the localization of an $F$-coherent ring is $F$-coherent, it follows that ${{\mathfrak p}}\in U$, which implies that $U$ is stable under generization. Hence $U$ is open by assumption. That $U$ is non-empty follows from the fact that the total ring of fractions of $R_{\mathrm{red}}$ is the finite product of fields, which is obviously $F$-coherent.

For $(4)$, we prove the following fact: Let $R \subseteq S$ be a faithfully flat extension and let $M$ be an $R$-module. Then $M$ is finitely generated over $R$ if and only if $M \otimes_{R} S$ is finitely generated over $S$. The “only if” part is obvious. Conversely, assume that the finite set of elements $s_{1},\ldots,s_{n}$ of $M$ generates the $S$-module $M \otimes_{R} S$ under the inclusion $M \to M \otimes_{R} S$. Let $N$ be an $R$-submodule of $M$ that is generated by $s_{1},\ldots,s_{n}$. Then the exact sequence: $0 \to N \to M \to M/N \to 0$ gives that $(M/N) \otimes_{R} S=0$, hence $M/N=0$ by faithful flatness of $R \to S$ and this implies that $M$ is generated by 