Proof. Since a klt surface is automatically quasi-projective (see e.g. [Reference FujinoFuj12, Lemma 2.2]), by taking suitable compactifications, we may assume that $X$ and $Y$ are projective. Let $h:\tilde{X}\rightarrow X$ be the minimal resolution and let $g:\tilde{X}\rightarrow Y$ be the induced morphism. It is enough to show that there exists a Cartier divisor $L_{Y}$ on $Y$ such that ${\mathcal{O}}_{\tilde{X}}(h^{\ast }L)\simeq {\mathcal{O}}_{\tilde{X}}(g^{\ast }L_{Y})$ . Let $E$ be the sum of the exceptional prime divisors of $g$ . By running a $(K_{\tilde{X}}+E)$ -MMP over $Y$ , we may assume that $f$ is a birational morphism between projective klt surfaces such that $\unicode[STIX]{x1D70C}(X/Y)=1$ .

By using the base point free theorem over surfaces [Reference TanakaTan15, Theorem 3.2 and Corollary 3.6], we may apply the same argument as [Reference Kollár and MoriKM98, Theorem 3.7(4)] to conclude. ◻

Proof. The first part of the lemma follows from [Reference KollárKol13, Theorem 3.36]. We are left to show that if $D$ is a Weil divisor on $X$ , then $m_{q}D$ is Cartier at $q$ . Let $f:Y\rightarrow X$ be the minimal resolution of $X$ at $q$ . Then the exceptional divisor is a chain of rational curves $C_{1},\ldots ,C_{r}$ . We may write

$$\begin{eqnarray}f^{\ast }D=D_{Y}+\mathop{\sum }_{i=1}^{r}b_{i}C_{i},\end{eqnarray}$$

where $D_{Y}$ is the strict transform of $D$ in $Y$ and $b_{1},\ldots ,b_{r}\in \mathbb{Q}$ . Thus, for any $j=1,\ldots ,r$ , we have

$$\begin{eqnarray}\mathop{\sum }_{i=1}^{r}b_{i}C_{i}\cdot C_{j}=-D_{Y}\cdot C_{j}.\end{eqnarray}$$

Since $D_{Y}\cdot C_{j}\in \mathbb{Z}$ , it follows that $m_{q}b_{i}\in \mathbb{Z}$ for each $i=1,\ldots ,r$ . Since $X$ is klt at $q$ , Lemma 2.1 implies the claim.◻

Proof. See [Reference AlexeevAle93, Theorem 5.3] and [Reference BirkarBir16, Proposition 11.7]. ◻

Proof. We have

$$\begin{eqnarray}K_{Y}+f_{\ast }^{-1}\unicode[STIX]{x1D6E5}+\mathop{\sum }_{i=1}^{n}(1-a_{i})E_{i}=f^{\ast }(K_{X}+\unicode[STIX]{x1D6E5}).\end{eqnarray}$$

Since $X$ is klt and $\ell f_{\ast }^{-1}\unicode[STIX]{x1D6E5}$ is integral, by Lemma 2.1, it is enough to show that $2\ell a_{j}\in \mathbb{Z}$ for any $j=1,\ldots ,n$ .

We may assume that $x\in X$ is a cyclic or a dihedral singularity, as otherwise, by the classification of two-dimensional klt singularities, we have that $n\leqslant 8$ (see e.g. [Reference Kollár and MoriKM98, Theorem 4.16]), and the claim follows immediately.

We first assume that $x\in X$ is a cyclic singularity and $E_{1},\ldots ,E_{n}$ is a chain of rational curves. For each $i\in \{2,\ldots ,n-1\}$ , by taking the intersection with $E_{i}$ on both sides of the equality above, we obtain

$$\begin{eqnarray}(-E_{i}^{2})a_{i}+f_{\ast }^{-1}\unicode[STIX]{x1D6E5}\cdot E_{i}=a_{i-1}+a_{i+1}.\end{eqnarray}$$

In particular, the numbers $a_{1},\ldots ,a_{n}$ satisfy the following convexity inequality:

$$\begin{eqnarray}a_{i}\leqslant \frac{a_{i-1}+a_{i+1}}{-E_{i}^{2}}\leqslant \frac{a_{i-1}+a_{i+1}}{2}.\end{eqnarray}$$

Thus, after possibly replacing $\unicode[STIX]{x1D70E}(1)$ by another index $j\in \{1,\ldots ,n\}$ such that $a_{\unicode[STIX]{x1D70E}(1)}=a_{j}$ , we may assume that $\unicode[STIX]{x1D70E}(2)$ is equal to $\unicode[STIX]{x1D70E}(1)-1$ or $\unicode[STIX]{x1D70E}(1)+1$ , say $\unicode[STIX]{x1D70E}(1)+1$ . In particular, $\ell a_{\unicode[STIX]{x1D70E}(1)},\ell a_{\unicode[STIX]{x1D70E}(1)+1}\in \mathbb{Z}$ .

Thus, for each $i=2,\ldots ,n-1$ , we have

$$\begin{eqnarray}(-E_{i}^{2})\ell a_{i}+(\ell f_{\ast }^{-1}\unicode[STIX]{x1D6E5})\cdot E_{i}=\ell a_{i-1}+\ell a_{i+1},\end{eqnarray}$$

and it follows inductively that $\ell a_{j}\in \mathbb{Z}$ for any $j=1,\ldots ,n$ . Thus, the claim follows.

Let us assume now that $x\in X$ is a dihedral singularity. Let $E_{n-1}$ and $E_{n}$ be two tails of self-intersection $(-2)$ and let $E_{1},E_{2},\ldots ,E_{n-2}$ be the remaining chain of rational curves [Reference KollárKol13, 3.35(3)], so that $E_{n-2}$ intersects $E_{n-1}$ and $E_{n}$ . For $m\in \{n-1,n\}$ , by taking the intersection with $E_{m}$ on both sides of the equality

$$\begin{eqnarray}K_{Y}+f_{\ast }^{-1}\unicode[STIX]{x1D6E5}+\mathop{\sum }_{i=1}^{n}(1-a_{i})E_{i}=f^{\ast }(K_{X}+\unicode[STIX]{x1D6E5}),\end{eqnarray}$$

we obtain

$$\begin{eqnarray}2a_{m}+f_{\ast }^{-1}\unicode[STIX]{x1D6E5}\cdot E_{m}=a_{n-2}+1.\end{eqnarray}$$

Thus, it is enough to show that $\ell a_{i}\in \mathbb{Z}$ for any $i=1,\ldots ,n-2$ . By assumption, it follows that

$$\begin{eqnarray}\ell \min _{1\leqslant i\leqslant n-2}a_{i},\quad \ell \min _{\substack{ 1\leqslant i\leqslant n-2 \\ i\neq j_{1}}}a_{i}\in \mathbb{Z}\end{eqnarray}$$

for some $j_{1}\in \{1,\ldots ,n-2\}$ such that $a_{j_{1}}=\min _{1\leqslant i\leqslant n-2}a_{i}$ . By applying the same argument as above to the chain $E_{1},E_{2},\ldots ,E_{n-2},$ we obtain that $\ell a_{j}\in \mathbb{Z}$ for every $j\in \{1,\ldots ,n-2\}$ . Thus, the claim follows.◻

Proof. By Lemma 2.3, we may assume that $I$ is a finite set. Further, we may assume that $1\in I$ . Let $(X,\unicode[STIX]{x1D6E5})$ be a log pair as in the proposition.

First, we show that there exists $m_{1}\in \mathbb{Z}_{{>}0}$ , depending only on $I$ , such that $m_{1}(K_{X}+\unicode[STIX]{x1D6E5})$ is Cartier around any closed point $q\in X$ such that $(X,\unicode[STIX]{x1D6E5})$ is not klt at $q$ . By Lemma 2.1, after taking a dlt modification, we may assume that $(X,\unicode[STIX]{x1D6E5})$ is dlt at $q$ . If the support of $\llcorner \unicode[STIX]{x1D6E5}\lrcorner$ is singular at $q$ , then $X$ is smooth at $q$ (see [Reference Kollár and MoriKM98, Theorem 4.15(1)]), and there is nothing to show. Thus, we may assume that $(X,\unicode[STIX]{x1D6E5})$ is plt around $q$ . Let $C$ be the component of $\lfloor \unicode[STIX]{x1D6E5}\rfloor$ containing $q$ . Since $(X,\unicode[STIX]{x1D6E5})$ is plt around $q$ , we know that $C$ is smooth, and so we may write

$$\begin{eqnarray}K_{C}+\unicode[STIX]{x1D6E5}_{C}=(K_{X}+\unicode[STIX]{x1D6E5})|_{C}\equiv 0,\end{eqnarray}$$

where, by inversion of adjunction, $\unicode[STIX]{x1D6E5}_{C}$ is a $\mathbb{Q}$ -divisor on $C$ such that $(C,\unicode[STIX]{x1D6E5}_{C})$ is klt. By [Reference McKernan and ProkhorovMP04, Lemma 4.3], $\unicode[STIX]{x1D6E5}_{C}$ has coefficients in $D(I)$ . Since $D(I)$ is a DCC set [Reference McKernan and ProkhorovMP04, Lemma 4.4] and $\operatorname{Diff}_{C}\leqslant \unicode[STIX]{x1D6E5}_{C}$ , Lemma 2.3 implies that there are only finitely many possibilities for the coefficients of $\operatorname{Diff}_{C}$ at $q$ , and the existence of $m_{1}$ follows from Lemma 2.2.

Now we show that there exists $m_{2}\in \mathbb{Z}_{{>}0}$ , depending only on $I$ , such that $m_{2}(K_{X}+\unicode[STIX]{x1D6E5})$ is Cartier around any closed point $q\in X$ such that $(X,\unicode[STIX]{x1D6E5})$ is klt at $q$ . Let $f:Y\rightarrow X$ be the minimal resolution at $q$ . We have

$$\begin{eqnarray}K_{Y}+f_{\ast }^{-1}\unicode[STIX]{x1D6E5}+\mathop{\sum }_{i=1}^{n}(1-a_{i})E_{i}=f^{\ast }(K_{X}+\unicode[STIX]{x1D6E5})\end{eqnarray}$$

where $E_{1},\ldots ,E_{n}$ are the $f$ -exceptional prime divisors, ordered so that

$$\begin{eqnarray}a_{1}\leqslant a_{2}\leqslant \cdots \leqslant a_{n}.\end{eqnarray}$$

By Lemma 2.4, if $\ell$ is a positive integer such that

$$\begin{eqnarray}\ell a_{1},\ell a_{2},\ldots ,\ell a_{8}\in \mathbb{Z}\end{eqnarray}$$

and such that $\ell \unicode[STIX]{x1D6E5}$ is a $\mathbb{Z}$ -divisor, then $2\ell (K_{X}+\unicode[STIX]{x1D6E5})$ is Cartier around $q$ . Thus, it is enough to find $n_{0}\in \mathbb{Z}_{{>}0}$ , depending only on $I$ , such that $n_{0}a_{1},\ldots ,n_{0}a_{8}\in \mathbb{Z}$ .

To this end, we first extract a prime divisor $E_{1}$ such that $a(E_{1},X,\unicode[STIX]{x1D6E5})=a_{1}$ and we obtain a birational morphism $g:Z\rightarrow X$ such that

$$\begin{eqnarray}K_{Z}+g_{\ast }^{-1}\unicode[STIX]{x1D6E5}+(1-a_{1})E_{1}=g^{\ast }(K_{X}+\unicode[STIX]{x1D6E5})\equiv 0.\end{eqnarray}$$

By ACC for the minimal log discrepancy [Reference AlexeevAle93, Theorem 3.2], the set $\{a_{1}\}_{(X,\unicode[STIX]{x1D6E5}),x\in X}$ is an ACC set, hence $\{1-a_{1}\}$ is a DCC set. By Lemma 2.3, there are finitely many possibilities for $a_{1}$ . Then, we extract $E_{2}$ with $a(E_{2},X,\unicode[STIX]{x1D6E5})=a_{2}$ , and apply the same argument, to show that there are only finitely many possibilities for $a_{2}$ . Repeating the same argument eight times, we see that there are finitely many possibilities for $a_{1},\ldots ,a_{8}$ . Thus, we may find a positive integer $n_{0}$ as above and the claim follows.◻

Proof. Let $F:=\unicode[STIX]{x1D70B}^{-1}(t)_{\text{red}}$ . Note that $F$ is irreducible, because $Z$ is $\mathbb{Q}$ -factorial and $\unicode[STIX]{x1D70C}(Z/T)=1$ . Since $C$ is $\unicode[STIX]{x1D70B}$ -horizontal, we obtain

$$\begin{eqnarray}(K_{Z}+F)\cdot F=K_{Z}\cdot F<(K_{Z}+C+B)\cdot F=0.\end{eqnarray}$$

Thus, [Reference TanakaTan14, Theorem 3.19(1)] implies part (1).

Let $G$ be a general fibre of $\unicode[STIX]{x1D70B}$ . Then $G$ is integral [Reference BădescuBăd01, Corollary 7.3]. Thus, $G\simeq \mathbb{P}^{1}$ by part (1). In particular, it follows that

$$\begin{eqnarray}(K_{Z}+B)\cdot G\geqslant K_{Z}\cdot G=(K_{Z}+G)\cdot G=\operatorname{deg}(K_{G}+\operatorname{Diff}_{G})\geqslant -2,\end{eqnarray}$$

which implies $C\cdot G=-(K_{Z}+B)\cdot G\leqslant 2.$ Thus, part (2) holds.

We now prove part (3). Note that $B$ is $\unicode[STIX]{x1D70B}$ -vertical, since

$$\begin{eqnarray}0=(K_{Z}+C+B)\cdot G\geqslant -2+2+B\cdot G=B\cdot G,\end{eqnarray}$$

where $G$ is a general fibre of $\unicode[STIX]{x1D70B}$ . Since $\unicode[STIX]{x1D70C}(Z/T)=1$ , we obtain $K_{Z}+C\equiv _{\unicode[STIX]{x1D70B}}0$ . Since the divisor $\unicode[STIX]{x1D70B}^{\ast }(t)|_{C}$ is $\text{Gal}(K(C)/K(T))$ -invariant for every closed point $t\in T$ , so is $B|_{C}$ . In particular, we may assume $B=0$ and $B_{C}=\operatorname{Diff}_{C}$ .

Let $Q\in T$ be such that $\unicode[STIX]{x1D70B}^{-1}(Q)\cap C$ consists of two distinct points $Q_{1},Q_{2}$ . It is enough to show that the coefficients of $\operatorname{Diff}_{C}$ at the points $Q_{1}$ and $Q_{2}$ coincide. We may write $\unicode[STIX]{x1D70B}^{\ast }Q=mF$ where $F$ is a prime divisor and $m$ is a positive integer. By part (1), $F\simeq \mathbb{P}^{1}$ .

We show that the pair $(Z,C+F)$ is log canonical. We have

$$\begin{eqnarray}0{\sim}_{\mathbb{Q}}(K_{Z}+C+F)|_{F}=K_{F}+\operatorname{Diff}_{F}+C|_{F}.\end{eqnarray}$$

Since $(Z,C+F)$ is not plt at the points $Q_{1}$ and $Q_{2}$ , inversion of adjunction implies that the pair $(F,\operatorname{Diff}_{F}+C|_{F})$ is not klt at the points $Q_{1}$ and $Q_{2}$ . Thus,

$$\begin{eqnarray}\operatorname{Diff}_{F}+C|_{F}\geqslant Q_{1}+Q_{2},\end{eqnarray}$$

and since $\operatorname{deg}(\operatorname{Diff}_{F}+C|_{F})=\operatorname{deg}(-K_{F})=2$ , equality holds. Thus, inversion of adjunction implies that $(Z,C+F)$ is log canonical.

Since $\unicode[STIX]{x1D70B}|_{C}:C\rightarrow T$ is étale over $Q$ , we have $mF|_{C}=\unicode[STIX]{x1D70B}^{\ast }Q|_{C}=Q_{1}+Q_{2}$ and

$$\begin{eqnarray}(K_{Z}+C+F)|_{C}=K_{C}+\operatorname{Diff}_{C}+\frac{1}{m}(Q_{1}+Q_{2}).\end{eqnarray}$$

In particular, by inversion of adjunction again, the coefficients of $\operatorname{Diff}_{C}$ at $Q_{1}$ and $Q_{2}$ are equal to $1-1/m$ , and the claim follows.◻

Proof. Let ${\mathcal{L}}={\mathcal{O}}_{X}(-(p^{e}-1)(K_{X}+S+B))$ . We have the following diagram.

Assumption (2) implies that the right vertical arrow $\unicode[STIX]{x1D719}_{B_{S}}^{S}$ is surjective. By Serre vanishing, after possibly replacing $e$ by a larger multiple, assumption (1) implies that $H^{1}(X,F_{\ast }^{e}({\mathcal{L}}(-S)))=0$ . By a diagram chase, it follows that $\unicode[STIX]{x1D719}_{S+B}^{X}$ is surjective. Thus, $(X,S+B)$ is globally $F$ -split.◻

Proof. Let ${\mathcal{L}}:={\mathcal{O}}_{Z}(-(p^{e}-1)(K_{Z}+C+B))$ . We have the following commutative diagram.

By assumption (3), the right vertical arrow $\unicode[STIX]{x1D719}_{B_{C}}^{C}$ is surjective.

By assumptions (2) and (4), we can find an ample Cartier divisor $A^{\prime }$ on $T$ such that ${\mathcal{L}}\simeq {\mathcal{O}}_{Z}(\unicode[STIX]{x1D70B}^{\ast }A^{\prime })$ (see [Reference TanakaTan15, Theorem 0.4]). Since $\unicode[STIX]{x1D70B}_{\ast }{\mathcal{O}}_{Z}={\mathcal{O}}_{T}$ , we have that

$$\begin{eqnarray}H^{0}(Z,{\mathcal{L}})\simeq H^{0}(T,{\mathcal{O}}_{T}(A^{\prime })),\end{eqnarray}$$

and so the image of the upper horizontal arrow is exactly the $G$ -invariant part: $H^{0}(C,F_{\ast }^{e}{\mathcal{L}}|_{C})^{G}$ , where $G$ is the Galois group of $K(C)/K(T)$ .

Let $f\in H^{0}(C,F_{\ast }^{e}{\mathcal{L}}|_{C})$ be such that $\unicode[STIX]{x1D719}_{B_{C}}^{C}(f)=1$ . Since $B_{C}$ is $G$ -invariant by Lemma 2.6, we have that $\unicode[STIX]{x1D719}_{B_{C}}^{C}$ is $G$ -equivariant. In particular,

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{B_{C}}^{C}\biggl(\frac{1}{|G|}\mathop{\sum }_{g\in G}g(f)\biggr)=1.\end{eqnarray}$$

We can divide by $|G|$ , because $\text{char}\;k>2$ and $|G|\leqslant 2$ by Lemma 2.6. Thus, $\unicode[STIX]{x1D719}_{C+B}^{Z}$ is surjective and $(Z,C+B)$ is globally $F$ -split.◻

Proof. We only prove the case $n=2$ , as the case $n=1$ is easier.

Let $\unicode[STIX]{x1D708}:C^{\prime }\rightarrow C$ be the normalisation. We may write

$$\begin{eqnarray}\unicode[STIX]{x1D708}^{\ast }((K_{Z}+C+B)|_{C})=K_{C^{\prime }}+B_{C^{\prime }}\end{eqnarray}$$

for some effective $\mathbb{Q}$ -divisor $B_{C^{\prime }}$ on $C^{\prime }$ (cf. § 2.1). By assumption (3) and by inversion of adjunction, it follows that $\llcorner B_{C^{\prime }}\lrcorner \neq 0$ . Thus, by assumption (1), it follows that $C^{\prime }$ is isomorphic to $\mathbb{P}^{1}$ . Therefore, by the same lemma in dimension one, $(C^{\prime },B_{C^{\prime }})$ is globally $F$ -split.

Let $e$ be a positive integer satisfying (1) and let

$$\begin{eqnarray}{\mathcal{L}}:={\mathcal{O}}_{Z}(-(p^{e}-1)(K_{Z}+C+B)).\end{eqnarray}$$

By assumption (1), we have that ${\mathcal{L}}\simeq {\mathcal{O}}_{Z}$ . We consider the following commutative diagram.

We now show that $\unicode[STIX]{x1D713}$ is surjective. We have the following commutative diagram.

Since $\unicode[STIX]{x1D6FC}$ is induced from the natural homomorphism ${\mathcal{O}}_{C}\rightarrow \unicode[STIX]{x1D708}_{\ast }{\mathcal{O}}_{C^{\prime }}$ and ${\mathcal{L}}\simeq {\mathcal{O}}_{Z}$ , it follows that $\unicode[STIX]{x1D6FC}$ is an isomorphism. Since $(C^{\prime },B_{C^{\prime }})$ is globally $F$ -split, $\unicode[STIX]{x1D719}_{B_{C^{\prime }}}^{C^{\prime }}$ is surjective. Thus, $\unicode[STIX]{x1D713}$ is surjective as well.

Since ${\mathcal{L}}\simeq {\mathcal{O}}_{Z}$ , assumption (2) implies

$$\begin{eqnarray}\displaystyle H^{1}(Z,F_{\ast }^{e}({\mathcal{L}}(-C))) & \simeq & \displaystyle H^{1}(Z,{\mathcal{L}}(-C))\nonumber\\ \displaystyle & \simeq & \displaystyle H^{1}(Z,{\mathcal{O}}_{Z}(-C))\{H^{1}(Z,{\mathcal{O}}_{Z})=0.\nonumber\end{eqnarray}$$

Since $\unicode[STIX]{x1D713}$ is surjective, by a diagram chase it follows that $\unicode[STIX]{x1D719}_{C+B}^{Z}$ is surjective as well, and, in particular, $(Z,C+B)$ is globally $F$ -split. Thus, the claim follows.◻

Proof. This follows from the same proof as [Reference Schwede and SmithSS10, Theorem 4.3(ii)]. ◻

Proof. Let $U\subseteq X$ be an affine open subset such that none of the irreducible components of $\unicode[STIX]{x1D6E5}$ are contained in $X\setminus U$ and such that, around each point of $X\setminus U$ , we have that $X$ is smooth, the support of $\unicode[STIX]{x1D6E5}$ is simple normal crossing and $(X,\unicode[STIX]{x1D6E5})$ is plt. Since $(X,\unicode[STIX]{x1D6E5})$ is sharply $F$ -pure and $U$ is affine, it follows that $(U,\unicode[STIX]{x1D6E5}|_{U})$ is globally sharply $F$ -split.

By Lemma 2.10, there exists an effective $\mathbb{Q}$ -divisor $H^{\prime }$ on $X$ such that $(U,(\unicode[STIX]{x1D6E5}+H^{\prime })|_{U})$ is globally sharply $F$ -split and

$$\begin{eqnarray}(p^{e}-1)(K_{U}+(\unicode[STIX]{x1D6E5}+H^{\prime })|_{U})\sim 0,\end{eqnarray}$$

for some positive integer $e$ . Since $X$ is smooth around each point of $X\setminus U$ , it follows that $(p^{e}-1)(K_{X}+\unicode[STIX]{x1D6E5}+H^{\prime })$ is Cartier on $X$ . For every positive integer $d$ , we have

$$\begin{eqnarray}(K_{X}+\unicode[STIX]{x1D6E5}+H^{\prime })-\biggl(K_{X}+\unicode[STIX]{x1D6E5}+\frac{1}{p^{d}+1}H^{\prime }\biggr)=\frac{p^{d}}{p^{d}+1}H^{\prime }.\end{eqnarray}$$

Thus, for every sufficiently large positive integer $d$ , there exists $e(d)\in \mathbb{Z}_{{>}0}$ such that

$$\begin{eqnarray}(p^{e(d)}-1)\biggl(K_{X}+\unicode[STIX]{x1D6E5}+\frac{1}{p^{d}+1}H^{\prime }\biggr)\end{eqnarray}$$

is Cartier.

Let $H:=(1/(p^{d}+1))H^{\prime }$ . It is enough to show that, after possibly replacing $d$ by a larger value, the pair $(X,\unicode[STIX]{x1D6E5}+H)$ is sharply $F$ -pure around any point $q\in X\setminus U$ . By assumption, around $q$ we have that $X$ is smooth, the support of $\unicode[STIX]{x1D6E5}$ is simple normal crossing and $(X,\unicode[STIX]{x1D6E5})$ is plt. Thus, if $q$ is not contained in the support of $\llcorner \unicode[STIX]{x1D6E5}\lrcorner$ and $d$ is sufficiently large, then $(X,\unicode[STIX]{x1D6E5}+H)$ is strongly $F$ -regular. On the other hand, if $q$ is contained in the support of $\llcorner \unicode[STIX]{x1D6E5}\lrcorner$ then the claim follows from inversion of adjunction (see e.g. [Reference SchwedeSch09, Main Theorem] and [Reference DasDas15, Theorem A]).◻

Proof. By Lemma 2.6, there exists a non-empty open subset $T^{0}\subseteq T$ such that the induced morphism $\unicode[STIX]{x1D70B}^{-1}(T^{0})\rightarrow T^{0}$ is a $\mathbb{P}^{1}$ -bundle. By assumptions (2) and (5), it follows that, after possibly shrinking $T^{0}$ , there exists a positive integer $d$ such that

$$\begin{eqnarray}(p^{d}-1)(K_{Z}+C+B)|_{\unicode[STIX]{x1D70B}^{-1}(T^{0})}\sim 0.\end{eqnarray}$$

By Lemma 2.6(1), every fibre of $\unicode[STIX]{x1D70B}$ is irreducible. Thus, we may write

$$\begin{eqnarray}(p^{d}-1)(K_{Z}+C+B)\sim \mathop{\sum }_{i=1}^{r}a_{i}F_{i}+\mathop{\sum }_{j=1}^{s}b_{j}F_{j}^{\prime },\end{eqnarray}$$

where $a_{i},b_{j}\in \mathbb{Z}$ , $F_{i}=\unicode[STIX]{x1D70B}^{-1}(t_{i})_{\text{red}}$ , and $F_{j}^{\prime }=\unicode[STIX]{x1D70B}^{-1}(t_{j}^{\prime })_{\text{red}}$ , with

$$\begin{eqnarray}t_{1},\ldots ,t_{r},t_{1}^{\prime },\ldots ,t_{s}^{\prime }\in T,\end{eqnarray}$$

so that $(Z,C+B)$ is plt (respectively not plt) along $F_{i}$ (respectively at some point of $F_{j}^{\prime }$ ). Since $\llcorner C+B\lrcorner =C$ is $\unicode[STIX]{x1D70B}$ -horizontal and $(Z,B)$ is klt, by inversion of adjunction we can find $\unicode[STIX]{x1D6FC}_{i}\in \mathbb{Q}_{{>}0}$ for $i=1,\ldots ,r$ such that $(Z,C+B+\sum _{i=1}^{r}\unicode[STIX]{x1D6FC}_{i}F_{i})$ is log canonical and $(p^{e_{1}}-1)(K_{Z}+C+B+\sum _{i=1}^{r}\unicode[STIX]{x1D6FC}_{i}F_{i})$ is Cartier around $\bigcup _{i=1}^{r}F_{i}$ for some $e_{1}\in \mathbb{Z}_{{>}0}$ .

Fix an index $1\leqslant j\leqslant s$ . It is enough to find $e_{2}\in \mathbb{Z}_{{>}0}$ such that $(p^{e_{2}}-1)b_{j}F_{j}^{\prime }$ is Cartier. By construction, there exists a zero-dimensional log canonical centre $z\in F_{j}^{\prime }$ of $(Z,C+B)$ . Since $(Z,B)$ is klt, it follows that $z\in C$ . Let $m_{z}$ be the Cartier index of $F_{j}^{\prime }$ at $z$ . Since $z\in C$ , it follows that $m_{z}F_{j}^{\prime }\cdot C\geqslant 1$ . We can write $\unicode[STIX]{x1D70B}^{\ast }(\unicode[STIX]{x1D70B}(z))=mF_{j}^{\prime }$ for some positive integer $m$ . Lemma 2.6 implies that $mF_{j}^{\prime }\cdot C=\unicode[STIX]{x1D70B}^{\ast }(\unicode[STIX]{x1D70B}(z))\cdot C\leqslant 2$ . Thus,

$$\begin{eqnarray}\frac{1}{m_{z}}\leqslant F_{j}^{\prime }\cdot C\leqslant \frac{2}{m},\end{eqnarray}$$

and, in particular, $2m_{z}\geqslant m$ . Since $m_{z}$ divides $m$ , it follows that either $m=m_{z}$ or $m=2m_{z}$ .

By applying Lemma 2.10 to an affine open neighbourhood $U$ of $z$ , we can find $e_{3}\in \mathbb{Z}_{{>}0}$ and an effective $\mathbb{Q}$ -divisor $H$ on $U$ such that $(U,(C+B)|_{U}+H)$ is sharply $F$ -pure and

$$\begin{eqnarray}(p^{e_{3}}-1)(K_{U}+(C+B)|_{U}+H)\sim 0.\end{eqnarray}$$

Since $z\in Z$ is a zero-dimensional log canonical centre of $(Z,C+B)$ , it follows that $z$ is not contained in the support of $H$ . Therefore, $(p^{e_{3}}-1)(K_{Z}+C+B)$ is Cartier around $z$ . Since

$$\begin{eqnarray}(p^{d}-1)(K_{Z}+C+B)\sim \mathop{\sum }_{i=1}^{r}a_{i}F_{i}+\mathop{\sum }_{j=1}^{s}b_{j}F_{j}^{\prime },\end{eqnarray}$$

it follows that $(p^{e_{4}}-1)b_{j}F_{j}^{\prime }$ is Cartier around $z$ for some $e_{4}\in \mathbb{Z}_{{>}0}$ . Since $m\in \{m_{z},2m_{z}\}$ , we have that $2(p^{e_{4}}-1)b_{j}F_{j}^{\prime }$ is Cartier. Since $\text{char}\;k\neq 2$ , we can find $e_{2}\in \mathbb{Z}_{{>}0}$ such that $(p^{e_{2}}-1)b_{j}F_{j}^{\prime }$ is Cartier. Thus, the claim follows.◻