$R$ is a residue class ring of a Cohen-Macaulay ring.

2.  $R$ is excellent.

Then if $R$ is $F$-pure, it is Cohen-Macaulay.

By the previous proposition, we see that if $R$ is $F$-pure, then it is $F$-regular, so that our assumptions imply that it is Cohen-Macaulay due to ([@HH1], Theorem 4.2 and Proposition 6.27).

The following corollary is very similar to the above one, but there are minor differences in their assumptions.

Let $(R,{{\mathfrak m}})$ be a reduced local ring which is a residue class ring of a Gorenstein local ring. If $R$ is $F$-coherent and $F$-injective, then $R$ is Cohen-Macaulay.

We need only show that $H^{k}_{{{\mathfrak m}}}(R)=0$ for all $k < \dim R$. As was already shown, every system of parameters of $R$ is a regular sequence on $R^{\infty}$, so $H^{k}(R^{\infty})=0$ and the natural map $H^{k}_{{{\mathfrak m}}}(R) \to H^{k}_{{{\mathfrak m}}}(R^{\infty})$ is injective for all $k < \dim R$ since $R$ is $F$-injective, which clearly gives $H^{k}_{{{\mathfrak m}}}(R)=0$ for $k < \dim R$, as desired.

Let $R \to S$ be a ring extension. Then we say that $R$ is an $\textit{algebra retract}$ if there is a ring homomorphism $\phi:S \to R$ such that the restriction of $\phi$ to $R$ is 