Orthogonal idempotents from a prime-order endomorphism

Let $F$ be a field and $A$ a commutative ring which is an $F$-algebra, finite as an $F$-module and reduced (its nilradical is zero). Let $s \colon A \to A$ be a ring homomorphism, and let $n$ be a prime number such that the $n$-fold iterate of the underlying map of $s$ is the identity on $A$. Assume that every element fixed by $s$, that is every $a \in A$ with $s(a) = a$, lies in the image of the structure map $F \to A$. Assume finally that $A$ is not a field. Then there exists $e \in A$ with $e^2 = e$ such that $e \cdot s^i(e) = 0$ for every index $i$ with $0 < i < n$, and such that $\sum_{i=0}^{n-1} s^i(e) = 1$, the iterates being those of the underlying function of $s$. Thus $e, s(e), \dots, s^{n-1}(e)$ form a complete family of pairwise orthogonal (by the stated relations together with the action of the iterates of $s$) idempotents summing to $1$, so that $A$ decomposes as a product of $n$ copies of $eA$ permuted cyclically by $s$.

This is the structural input for a descent argument concerning a finite reduced algebra over a field carrying an endomorphism of prime order whose fixed ring is as small as possible: either the algebra is a field, or it splits into $n$ factors cyclically permuted by the endomorphism. It is cited by IsDedekindDomain.HeightOneSpectrum.exists_units_prod_tensor_map_iterate_eq_tmul_one_of_finrank_dvd_valuation_norm.