Proof. As k is perfect, we obtain that $A/\mathfrak {m}$ is also perfect. Moreover, by Cohen’s structure theorem [1, 032A], A contains a field of representatives $k_0$ . So we can write $A=k_0\oplus \mathfrak {m}$ and $a_i=\lambda _i+x_i$ with $\lambda _i\in k_0$ and $x_i\in \mathfrak {m}$ . By assumption the $d_{A/A^p}(a_i)$ ’s form a basis for $\Omega _{A/A^p}^1$ over A. Since $k_0=k_0^p$ and since $\Omega ^1_{A/A^p}=\Omega ^1_{A/k}$ , we obtain that the $d_{A/k}(x_i)$ ’s form an A-basis of $\Omega ^1_{A/k}$ . In particular $n=\dim A$ . Applying Nakayama’s lemma to the isomorphism $\mathfrak {m}/\mathfrak {m}^2\cong \Omega ^1_{A/k}\otimes A/\mathfrak {m}$ , we obtain that $\mathfrak {m}=(x_1,\dots ,x_n)$ .

Proof. Since $\widehat {A}$ is a flat A-module and $\Omega _{A/R}^1$ is a finitely presented A-module, the canonical morphism

$$ \begin{align*}\operatorname{\mathrm{Der}}_R(A)\otimes_A\widehat{A}\longrightarrow \operatorname{\mathrm{Hom}}_{\widehat{A}}(\Omega_{A/R}^1\otimes_A \widehat{A},\widehat{A})\end{align*} $$

is an isomorphism [Reference Bourbaki9, Chapter I, §2, n.10, Proposition 11]. In the rest of the proof, we describe the target of this isomorphism. By the universal property of the inverse limit, it can be written as

$$ \begin{align*}\operatorname{\mathrm{Hom}}_{\widehat{A}}(\Omega_{A/R}^1\otimes \widehat{A},\widehat{A})=\varprojlim \operatorname{\mathrm{Hom}}_{\widehat{A}}(\Omega_{A/R}^1\otimes \widehat{A},A/\mathfrak{m}^n).\end{align*} $$

Let us describe the Hom-module into $A/\mathfrak {m}^n$ . On the one hand, an $\widehat {A}$ -linear morphism $\Omega _{A/R}^1\otimes \widehat {A}\to A/\mathfrak {m}^n$ is always continuous for the natural topologies, since it is uniquely specified by an A-linear map $\Omega ^1_{A/R}\to A/\mathfrak {m}^n$ . On the other hand by [Reference Liu38, 6.Exercise 1.3], we have a canonical isomorphism

$$ \begin{align*}\Omega^1_{A/R}\otimes \widehat{A}\cong \varprojlim \left(\Omega^1_{\widehat{A}/R}/\mathfrak{m}^n\Omega^1_{\widehat{A}/R}\right).\end{align*} $$

Combining these two facts with [Reference Grothendieck14, 20.7.14.4], which we can apply as $A/\mathfrak {m}^n$ is discrete and is annihilated by $\mathfrak {m}^n$ , we obtain a canonical identification

$$ \begin{align*}\operatorname{\mathrm{Hom}}_{\widehat{A}}(\Omega^1_{A/R}\otimes \widehat{A},A/\mathfrak{m}^n)\cong \operatorname{\mathrm{Hom}}^{\text{cont}}_{\widehat{A}}(\Omega^1_{\widehat{A}/R}, A/\mathfrak{m}^n).\end{align*} $$

Let us apply the inverse limit along n: by the token already used above, it amounts to the same to apply the inverse limit on the second arguments of the Hom modules, and so we get an isomorphism

$$ \begin{align*}\operatorname{\mathrm{Hom}}_{\widehat{A}}(\Omega^1_{A/R}\otimes \widehat{A},\widehat{A})\cong \operatorname{\mathrm{Hom}}^{\text{cont}}_{\widehat{A}}(\Omega^1_{\widehat{A}/R}, \widehat{A}).\end{align*} $$

The right-hand side is a sub-module of $\operatorname {\mathrm {Hom}}_{\widehat {A}}(\Omega _{\widehat {A}/R}^1,\widehat {A})=\operatorname {\mathrm {Der}}_R(\widehat {A})$ which, by [Reference Grothendieck14, 20.4.8.2], corresponds to the set of continuous R-derivations of $\widehat {A}$ into itself. This completes the proof.

Proof. See [Reference Seshadri56, Proposition 6] or [Reference Yuan64, Proof of Theorem 12].

Proof. We apply [Reference Hartshorne18, II, Lemma 8.9] to the cokernel $\mathcal {N}$ of $\mathcal {F}\hookrightarrow T_{X/k}$ : it is free at x if and only if $\operatorname {\mathrm {rk}}(\mathcal {N}\otimes k(x))=\dim \mathcal {O}_{X,x}-1$ . This is equivalent to left-exactness of the right-exact sequence

$$ \begin{align*}0\to \mathcal{F}\otimes k(x)\to T_{X/k}\otimes k(x)\to \mathcal{N}\otimes k(x)\to 0.\end{align*} $$

If $\mathcal {F}$ is generated by $\partial $ at x, then left-exactness holds if and only if $\partial \notin \mathfrak {m}_{X,x}T_{X/k}$ .

Construction 2.24 (Birational pullback)

Let $\pi \colon Y\to X$ be a birational morphism of normal connected k-schemes. The generic stalk $\mathcal {F}_{k(X)}$ determines a foliation on Y, which we will usually denote by $\pi ^{*}\mathcal {F}$ . If $\mathcal {F}$ is a $1$ -foliation, then so is $f^{*}\mathcal {F}$ by Remark 2.15.

Proof. This is a direct application of Remark 2.15 and Galois descent, except for the fact that if $\mathcal {F}\subset T_{X/k}$ is a $1$ -foliation on Y then $\mathcal {F}_{K(Y)}\otimes K(X)$ is closed under Lie brackets and pth powers. This is an immediate calculation (using Hochschild’s formula (2.1) for pth powers).

Proof. The quotient Y is $\mathbb {Q}$ -factorial, so $\mathcal {F}$ is automatically $\mathbb {Q}$ -Gorenstein. So consider a birational proper morphism $f\colon Y'\to Y$ , and let $X'$ be the integral closure of $Y'$ in $K(X)$ . Then G acts on $X'$ with quotient $Y'$ , and we have a commutative diagram

where $q,q'$ are the quotient morphisms. If $\mathcal {F}'$ and $\mathcal {H}'$ denote the birational pullback of $\mathcal {F}$ and $\mathcal {H}$ then by Lemma 3.8 we have $q^{*}\mathcal {F}=\mathcal {H}$ and $(q')^{*}\mathcal {F}'=\mathcal {H}'$ . By the foliated version of Riemann–Hurwitz [Reference Spicer57, Proposition 3.7] (Footnote ⁶ ) it holds that

$$ \begin{align*}K_{\mathcal{H}}=q^{*}K_{\mathcal{F}}+\sum_{D\subset X}\epsilon_{\mathcal{H}}(D)(r_D-1)D, \quad K_{\mathcal{H}'}=(q')^{*}K_{\mathcal{F}'}+\sum_{D'\subset X'}\epsilon_{\mathcal{H}}(D')(r_{D'}-1)D',\end{align*} $$

where $r_{\bullet }$ denotes the ramification index of divisors with respect to the action of G.

Let E be a prime f-exceptional divisor. We want to estimate $a(E;\mathcal {F})$ . If $q'$ is étale over the generic point of E then $a(E;\mathcal {F})=a(E';\mathcal {H})\geq 0$ where $E'\subset X'$ is any divisor lying over E. So we may assume that $r_{E'}>1$ . Notice that $(q')^{*}E=\sum _{E'}r_{E'}E'$ .

We compute $(q\circ f')^{*}K_{\mathcal {F}}=(f\circ q')^{*}K_{\mathcal {F}}$ at the generic points of $(q')^{-1}(E)$ . On the one hand

$$ \begin{align*}(q')^{*}f^{*}K_{\mathcal{F}}=K_{\mathcal{H}'} -\sum_{D'\subset X'}\epsilon_{\mathcal{H}}(D')(r_{D'}-1)D' -\sum_{E'}a(E;\mathcal{F})r_{E'}E'\end{align*} $$

and on the other

$$ \begin{align*}(f')^{*}q^{*}K_{\mathcal{F}}=K_{\mathcal{H}'} -\sum_{E'} a(E';\mathcal{H})E' -\sum_{D\subset Y}\epsilon_{\mathcal{G}}(D)(r_D-1)(f')^{*}D.\end{align*} $$

Equating the two, we find that

$$ \begin{align*}r_{E'}a(E;\mathcal{F})+\left(r_{E'}-1\right)\epsilon_{\mathcal{H}}(E') = a(E';\mathcal{H})+\delta,\end{align*} $$

where $\delta \geq 0$ . As $\epsilon _{\mathcal {H}}(E')=\epsilon _{\mathcal {F}}(E)$ and $a(E';\mathcal {H})\geq -\epsilon _{\mathcal {H}}(E')$ , we find that

$$ \begin{align*}a(E;\mathcal{F})\geq -\epsilon_{\mathcal{F}}(E)+\frac{\delta}{r_{E'}}\end{align*} $$

which shows that $\mathcal {F}$ is lc.

Proof. Since the vertical arrows in (4.1) are étale and induce isomorphisms on residue fields, they induce isomorphisms of completions. So we may assume that A is complete and that D is given by a continuous $\mu _p$ -action. The action of $\mu _p$ on can be linearized, see the proof of [Reference Satriano54, Corollary 1.8]. Then it is given by the action of a single matrix $M\in \operatorname {\mathrm {GL}}_n(k(A))$ . We have $M^p=M$ and thus the minimal polynomial of M divides $T^p-T$ . So it cannot have multiple roots, and we deduce that M is semi-simple. As $M^p=M$ we see that its eigenvalues are elements of $\mathbb {F}_p$ . So after a $k(A)$ -linear change of coordinates we may assume that M is diagonal. Therefore the derivation D can be given a normal form

$$ \begin{align*} D=\sum_i\lambda_ix_i\frac{\partial}{\partial x_i},\quad \lambda_i\in\mathbb{F}_p. \end{align*} $$

From this it is easily seen that $A^D$ is generated by monomials, and thus it is toric.

Proof. All these properties can be checked on localizations, and passing to the ring of constants commutes with localization at prime ideals. So we may assume that A is local with maximal ideal $\mathfrak {m}$ . Then $A^D$ is also local with maximal ideal $\mathfrak {n}=\mathfrak {m}^D$ . By F-finiteness the map $A^D\hookrightarrow A$ is finite, so $\dim A=\dim A^D$ .

Our main observation is the following one: since $\sqrt {\mathfrak {n} A}=\mathfrak {m}$ , we have

$$ \begin{align*}H^{\bullet}_{\mathfrak{m}}(A)=H^{\bullet}_{\mathfrak{n} A}(A)=H^{\bullet}_{\mathfrak{n}}(A), \end{align*} $$

where on the right-hand side we consider A as an $A^D$ -module [Reference Iyengar, Leuschke, Leykin, Miller, Miller, Singh and Walther23, Propositions 7.3 and 7.15(2)]. As mentioned above, by (4.1) we can write $A=A^D\oplus B$ for some finite $A^D$ -module B, and so the local cohomology splits accordingly, that is

(4.2) $$ \begin{align} H^{\bullet}_{\mathfrak{m}}(A)=H^{\bullet}_{\mathfrak{n}}(A^D\oplus B)=H^{\bullet}_{\mathfrak{n}}(A^D)\oplus H^{\bullet}_{\mathfrak{n}}(B). \end{align} $$

Let us show the first statement. Serre’s property $S_r$ states that, after localizing at any prime, the depth along the maximal ideal is at least $\min \{\dim , r\}$ . Since $\dim A=\dim A^D$ , it suffices to show that the depth does not decrease upon passing to the sub-ring of constants. By [Reference Iyengar, Leuschke, Leykin, Miller, Miller, Singh and Walther23, Theorem 9.1], we have

$$ \begin{align*}\operatorname{\mathrm{depth}}_{\mathfrak{m}}(A)=\inf_i\{ H^i_{\mathfrak{m}}(A)\neq 0\},\quad \operatorname{\mathrm{depth}}_{\mathfrak{n}}(A^D)=\inf_i\{H^i_{\mathfrak{n}}(A^D)\neq 0\}.\end{align*} $$

So it follows from (4.2) that $\operatorname {\mathrm {depth}}_{\mathfrak {n}} A^D\geq \operatorname {\mathrm {depth}}_{\mathfrak {m}} A$ as desired.

Next we discuss descent of F-singularities. F-purity and F-regularity descend to split subrings [Reference Ma and Polstra39, Theorem 3.9 and Exercise 9, p. 13]. F-injectivity and F-rationality do not in general (see [Reference Ma and Polstra39, Section 8] and [Reference Watanabe63]). Hopefully, the key fact is our situation is that $\dim A=\dim A^D$ . We prove that F-rationality descends, the F-injective case is similar.

Suppose that A is F-rational: this means that A is Cohen–Macaulay and that given any $c\in A$ not contained in any minimal prime, there exists an $e>0$ such that the composition

(4.3) $$ \begin{align} H^d_{\mathfrak{m}}(A)\overset{f_A^e}{\to} H^d_{\mathfrak{m}}(F^e_*A)\overset{F^e_*c}{\longrightarrow} H^d_{\mathfrak{m}}(F^e_*A) \end{align} $$

is injective, where $d=\dim A$ and $f_A^e=H^d_{\mathfrak {m}}(A\to F_*^eA)$ . As seen above, $A^D$ is also Cohen–Macaulay. If we assume that $c\in A^D$ is not contained in any minimal prime, then by going-up c is not contained in any minimal prime of A either, and the above sequence is thus injective. Each local cohomology module in the sequence splits accordingly to (4.2). Since $c\in A^D$ , the action of $F_e^{*}c$ preserves each one of the summands. In other words, (4.3) splits as

Since it is injective, the sequence given by the first summands is also injective. Thus $A^D$ is F-rational, as claimed.

Proof. Recall that two-dimensional linearly reductive quotient singularities are the same as F-regular ones [Reference Liedtke, Martin and Matsumoto36, Theorem 5.11]. So $\mathcal {O}$ is F-regular, and then $\mathcal {O}^D$ is F-regular by Theorem 4.3.

Proof. We can work on an étale neighborhood of the generic point of E, where X and E are regular and $\mathcal {F}$ a sub-bundle of the tangent sheaf. Then by Example 2.18 we may assume that $X=\mathbb {A}^n_{x_1,\dots ,x_n}$ , that $\mathcal {F}$ is generated by $\frac {\partial }{\partial x_1},\dots , \frac {\partial }{\partial x_r}$ with $r<n$ , and that E is cut out by a linear polynomial. The invariant subring, whose spectrum gives Y, is given by $k[x_1^p,\dots ,x_r^p,x_{r+1},\dots ,x_r]$ .

• • If the polynomial is not a linear combination of $x_1,\dots ,x_r$ , we make a change of coordinate and assume that E is given by $(x_{r+1}=0)$ . Then E is $\mathcal {F}$ -invariant and $E_Y=(x_{r+1}=0)$ so $q^{*}E^Y=E$ .

• • If the polynomial is cut out by the $x_1,\dots ,x_r$ , we may similarly assume that $E=(x_r=0)$ . Then E is not $\mathcal {F}$ -invariant, as $\frac {\partial }{\partial x_r}|_E\notin T_E$ , and $E^Y=(x_r^p=0)$ so $q^{*}E^Y=pE$ .

This completes the proof.

Proof. Let E be a prime divisor on $\operatorname {\mathrm {Supp}}(\Delta )$ . Then by Lemma 4.7, we have in any case

$$ \begin{align*}q^{*}\left(1-\epsilon(E)\frac{p-1}{p}\right)q(E)=E\end{align*} $$

and so $q^{*}\Delta _Y=\Delta $ . Combining this equality with the adjunction formula Proposition 2.41 yields the result.

Proof. Since Y is normal, it has a well-defined canonical divisor $K_Y$ which is invertible on a big open subset U. Over that locus, the log adjunction reads $q^{*}(K_Y+\Delta _Y)|_U=(K_X+\Delta +(p-1)K_{\mathcal {F}})|_{q^{-1}U}$ . Now $K_X+\Delta $ and $K_{\mathcal {F}}$ are by assumption $\mathbb {Q}$ -Cartier, so for $n>0$ big enough the Weil divisor $n(K_X+\Delta +(p-1)K_{\mathcal {F}})$ is Cartier. Thus its pullback on $Y^{(-1)}$ is Cartier as well. Since it is represented on the big open subset $U^{(-1)}$ by the divisor $np(K_{Y^{(-1)}}+\Delta _{Y^{(-1)}})$ , it follows that $K_{Y^{(-1)}}+\Delta _{Y^{(-1)}}$ is $\mathbb {Q}$ -Cartier, and therefore $K_Y+\Delta _Y$ is $\mathbb {Q}$ -Cartier as claimed.

Proof. By Lemma 4.9 the $\mathbb {Q}$ -Weil log canonical divisor $K_Y+\Delta _Y$ is $\mathbb {Q}$ -Cartier, and thus we may investigate its birational singularities. Consider a proper birational morphism $\mu \colon Y'\to Y$ , and write $K_{Y'}+\mu _*^{-1}\Delta _{Y}=\mu ^{*}(K_Y+\Delta _Y)+\sum a_{E}E$ where E runs through the exceptional prime divisors of $\mu $ . We are interested in the numbers $a_E$ . To compute them we look at the following commutative diagram

where $X'$ is the normalization of $Y'$ in $K(X)$ . The morphism $\nu \colon X'\to X$ indeed exists and is uniquely determined by Zariski’s Main theorem. Notice that $q'$ is an homeomorphism; in particular $q'$ induces a bijection between the $\mu $ -exceptional prime divisors and the $\nu $ -exceptional ones. So if E is $\mu $ -exceptional, let us write $E'$ the corresponding $\nu $ -exceptional divisor. Let also $\mathcal {F}'=\nu ^{*}\mathcal {F}$ be the foliation induced by $\mathcal {F}$ on $X'$ , so that $q'$ is the quotient morphism.

We now write down the pullback formulas for every canonical divisor in sight:

$$ \begin{align*} K_{Y'}+\mu_*^{-1}\Delta_Y &= \mu^{*}(K_Y+\Delta_Y)+\sum a_EE, \\K_X+\Delta &= q^{*}(K_Y+\Delta_Y)+(1-p)K_{\mathcal{F}}, \\K_{X'}+\nu^{-1}_*\Delta &=(q')^{*}(K_{Y'}+\mu_*^{-1}\Delta_Y)+(1-p)K_{\mathcal{F}'},\\K_{\mathcal{F}'}&=\nu^{*}K_{\mathcal{F}}+\sum b_{E'}E', \\K_{X'}+\nu_*^{-1}\Delta &= \nu^{*}(K_X+\Delta)+\sum c_{E'}E'. \end{align*} $$

Here we use the notations of Definition 4.6, and the third equality follows from Proposition 4.8 and the observation that $(\nu ^{-1}_*\Delta )_{Y'}=\mu ^{-1}_*\Delta _Y$ (which can be verified over the generic points of $\mu ^{-1}\Delta _Y$ , where $\mu $ is an isomorphism). To simplify the notations, we may localize at the generic point of some E, and assume that there is a unique exceptional divisor. If we apply $(q')^{*}$ to the very first equation in the above list, we find

$$ \begin{align*}a_E\delta(E') E'=(q')^{*}(K_{Y'}+\mu^{-1}_*\Delta_Y)-(q')^{*}\mu^{*}(K_Y+\Delta_Y),\end{align*} $$

where

$$ \begin{align*}\delta(E')=\begin{cases} 1 & \text{if }E'\text{ is }\mathcal{F}'\text{-invariant}, \\ p & \text{otherwise}. \end{cases}\end{align*} $$

Using the canonical isomorphism $(q')^{*}\mu ^{*}\cong \nu ^{*}q^{*}$ and substituting in the other formulae on the right-hand side, we find the equation

$$ \begin{align*}a_E\delta(E') = c_{E'}+(p-1)b_{E'}.\end{align*} $$

In other words,

(4.4) $$ \begin{align} \begin{cases} \text{if }E'\text{ is }\mathcal{F}'\text{-invariant}:& a_E=c_{E'}+(p-1)b_{E'}\\ \text{if }E'\text{ is not }\mathcal{F}'\text{-invariant}:& a_E=\frac{1}{p}\left(c_{E'}+(p-1)b_{E'}\right). \end{cases} \end{align} $$

The number $c_{E'}$ has a lower bound according to the birational singularities of $(X,\Delta )$ . The number $b_{E'}$ has a lower bound according to the birational singularities of the foliation $\mathcal {F}$ (and in the log canonical and klt cases, it might depend on the $\mathcal {F}'$ -invariance of $E'$ ; in particular, if $b_{E'}$ is negative then $E'$ is not $\mathcal {F}'$ -invariant). A simple case-by-case analysis based on (4.4) concludes the proof.

Proof. By [Reference Kollár30, Corollary 2.30] a terminal variety is regular in codimension $2$ . So if $\mathcal {F}$ is canonical, then by Theorem 4.10 the quotient $X/\mathcal {F}$ is terminal, hence regular in codimension $2$ . Thus $\mathcal {F}$ is regular in codimension $2$ by Lemma 2.37.

Now if $\mathcal {F}\subsetneq T_{X/k}$ was terminal then it would be generically regular by the previous paragraph, but by Example 3.4 regular foliations on regular varieties are not terminal: contradiction.

Proof. Let $T=S/\mathcal {F}$ , let $\pi \colon S'\to S$ be the minimal resolution of S, and let $T'$ be quotient of $S'$ by $\pi ^{*}\mathcal {F}$ . Then we have a commutative diagram

Let $E\subset S'$ be a prime $\pi $ -exceptional divisor, with image $E'=q(E)\subset T'$ . Then as in the proof of Theorem 4.10, we find