\[coherent-3\] Let $R \to S$ be a purely inseparable extension of Noetherian rings. Then $R$ is $F$-coherent if and only if $S$ is so. In particular, if $R$ is a purely inseparable extension, or subextension of a polynomial algebra over a field of characteristic $p>0$, it is $F$-coherent.

The proof of this corollary is immediate from the definition.

\[coherent-2\] Affine semigroup rings provide non-trivial (non-regular) examples of $F$-coherent rings. They even include rings which are neither Cohen-Macaulay, nor normal. Let $k$ be any field of characteristic $2$ and let $R=k[x^{4},x^{3}y,xy^{3},y^{4}]$. Then $(x^{3}y)^{2}/x^{4}=x^{2}y^{2}$, which is integral over $R$, but is not in $R$. The failure of Cohen-Macaulay property is easy to see. On the other hand, we have $k[x^{4},y^{4}] \subseteq R \subseteq k[x,y]$, which is a tower of purely inseparable extensions, hence $R$ is $F$-coherent. Later on, we prove an easy-to-use criterion in terms of Cohen-Macaulay property to see that $R$ is not $F$-pure. Is the normalization $k[x^{4},x^{3}y,x^{2}y^{2},xy^{3},y^{4}]$ of $R$ an $F$-coherent ring?

We shall use a valuative method for deriving various properties for $F$-coherent rings. Let $R$ be any Noetherian domain and let $P$ be its prime ideal. Then there exists a discrete valuation domain $(V,tV)$ such that $R \subseteq 