The Stacks project

Proof. Observe that $\mathop{\mathrm{Spec}}(F) : \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A)$ is the identity map. Being regular is defined in terms of the local rings and being flat is something about local rings, see Algebra, Lemma 10.39.18. Thus we may and do assume $A$ is a Noetherian local ring with maximal ideal $\mathfrak m$.

Assume $A$ is regular. Let $x_1, \ldots , x_ d$ be a system of parameters for $A$. Applying $F$ we find $F(x_1), \ldots , F(x_ d) = x_1^ p, \ldots , x_ d^ p$, which is a system of parameters for $A$. Hence $F$ is flat, see Algebra, Lemmas 10.128.1 and 10.106.3.

Conversely, assume $F$ is flat. Write $\mathfrak m = (x_1, \ldots , x_ r)$ with $r$ minimal. Then $x_1, \ldots , x_ r$ are independent in the sense defined above. Since $F$ is flat, we see that $x_1^ p, \ldots , x_ r^ p$ are independent, see Lemma 51.17.5. Hence $\text{length}_ A(A/(x_1^ p, \ldots , x_ r^ p)) = p^ r$ by Lemma 51.17.4. Let $\chi (n) = \text{length}_ A(A/\mathfrak m^ n)$ and recall that this is a numerical polynomial of degree $\dim (A)$, see Algebra, Proposition 10.60.9. Choose $n \gg 0$. Observe that

as can be seen by looking at monomials in $x_1, \ldots , x_ r$. We have

By flatness of $F$ this has length $\chi (n) \text{length}_ A(A/F(\mathfrak m)A)$ (Algebra, Lemma 10.52.13) which is equal to $p^ r\chi (n)$ by the above. We conclude

Looking at the leading terms this implies $r = \dim (A)$, i.e., $A$ is regular. $\square$