Assume that $R$ is a semiperfect ring in characteristic $p$, i.e the frobenius is surjective on $R$. I think one can prove that an etale cover of $R$ should again be semiperfect by considering the basic etale maps similar to the fact that an algebraic extension of a perfect field is again perfect. Is this true and has anyone a better proof or a reference for this fact?
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Yes, it is true.
Definition. We say that a map $A\to B$ of (commutative) $\mathbb F_p$-algebras is relatively perfect (resp. relatively semiperfect if the relative Frobenius map $\varphi_{B/A}\colon B\otimes_{A,\varphi_A}A\to B$ is an isomorphism (resp. surjective).
The result that you want is a consequence of the following:
Lemma. A composite of two composable relatively semiperfect maps of $\mathbb F_p$-algebras is still relatively semiperfect.
Proof. Let $A\to B\to C$ be two composable relatively semiperfect maps of $\mathbb F_p$-algebras. Then the relative Frobenius map $\varphi_{C/A}\colon C\otimes_{A,\varphi_A}A\to C$ coincides with the composite $$ C\otimes_{A,\varphi_A}A=C\otimes_BB\otimes_{B,\varphi_{B/A}}A\xrightarrow{1_C\otimes\varphi_{B/A}}C\otimes_{B,\varphi_{B/A}}B\xrightarrow{\varphi_{C/B}}C. $$ Q.E.D.
Proposition. (Stacks project, Tag 0F6W) Every weakly étale map of $\mathbb F_p$-algebras is relatively perfect.
