The Stacks project

Proof. Choose a presentation $A^{\oplus m} \to A^{\oplus n} \to M$ which induces an isomorphism $\kappa ^{\oplus n} \to M/\mathfrak m M$. Let $T = (a_{ij})$ be the matrix of the map $A^{\oplus m} \to A^{\oplus n}$. Observe that $a_{ij} \in \mathfrak m$. Applying base change by $F$, using right exactness of base change, we get a presentation $A^{\oplus m} \to A^{\oplus n} \to M$ where the matrix is $T = (a_{ij}^ p)$. Thus we have a presentation with $a_{ij} \in \mathfrak m^ p$. Repeating this construction we find that for each $e \geq 1$ there exists a presentation with $a_{ij} \in \mathfrak m^ e$. This implies the fitting ideals (More on Algebra, Definition 15.8.3) $\text{Fit}_ k(M)$ for $k < n$ are contained in $\bigcap _{e \geq 1} \mathfrak m^ e$. Since this is zero by Krull's intersection theorem (Algebra, Lemma 10.51.4) we conclude that $M$ is free of rank $n$ by More on Algebra, Lemma 15.8.7. $\square$

Proof. Say $\sum a_ if_ i = 0$. Then $\sum a_ ig_ rf_ i = 0$. Hence $a_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$. Write $a_ r = \sum _{i < r} b_ i f_ i + b f_ rg_ r$. Then $0 = \sum _{i < r} (a_ i + b_ if_ r)f_ i + bf_ r^2g_ r$. Thus $a_ i + b_ i f_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$ which implies $a_ i \in (f_1, \ldots , f_ r)$ as desired. $\square$

Proof. We claim there is an exact sequence

Namely, if $a f_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$, then $\sum _{i < r} a_ i f_ i + (a + bg_ r)f_ r = 0$ for some $b, a_ i \in A$. Hence $\sum _{i < r} a_ i g_ r f_ i + (a + bg_ r)g_ rf_ r = 0$ which implies $a + bg_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$ which means that $a$ maps to zero in $A/(f_1, \ldots , f_{r - 1}, g_ r)$. This proves the claim. To finish use additivity of lengths (Algebra, Lemma 10.52.3). $\square$

Proof. Use Lemmas 51.17.2 and 51.17.3 and induction. $\square$

Proof. Let $I = (f_1, \ldots , f_ r)$ and $J = \varphi (I)B$. By flatness we have $I/I^2 \otimes _ A B = J/J^2$. Hence freeness of $I/I^2$ over $A/I$ implies freeness of $J/J^2$ over $B/J$. $\square$

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. $\square$