Proof. This follows from [Reference Chen and Zhang13, 3.2]. ◻

Proof. Let $X{\dashrightarrow}Z$ be the rational map defined by the linear system $|mL|$ for some sufficiently divisible $m>0$ . Then $\dim Z=1$ . We can replace $X$ with the normalization of the graph of $X{\dashrightarrow}Z$ ; hence, assume $X{\dashrightarrow}Z$ is a morphism. We can in addition assume $L\geqslant H\geqslant 0$ where $H$ is the pullback of some ample $\mathbb{Q}$ -divisor on $Z$ . Since $L$ is not big, its support does not intersect the generic fiber of $X\rightarrow Z$ .

Let $F$ be a fiber of $X\rightarrow Z$ which has a common component with $L$ . Let $a$ be the smallest rational number such that $L-aF\leqslant 0$ near $F$ . Then $L-aF$ has no common component with $F$ ; otherwise there would be two components $C,D$ of $F$ such that $C$ intersects $D$ , $C$ is not a component of $L-aF$ but $D$ is a component of $L-aF$ which implies $(L-aF)\cdot C<0$ , a contradiction. These arguments show that $L$ is the pullback of some $\mathbb{Q}$ -divisor on $Z$ which is necessarily ample, hence $L$ is semi-ample. ◻

Proof. Since $D$ is effective, it is enough to prove the statement with $H$ replaced by any positive multiple and $D$ replaced by $D+lf^{\ast }H$ for some $l>0$ . If $V\rightarrow X$ is a morphism, we denote the pullback of $D$ to $V$ by $D_{V}$ (similar notation for other divisors). Let $m$ be a sufficiently divisible natural number and let $d=\dim _{K}H^{0}(mD_{F})-1$ where $K$ is the function field of $Z$ . Let $S$ be the normalization of the image of $\unicode[STIX]{x1D719}_{mD_{F}}:F{\dashrightarrow}\mathbb{P}_{K}^{d}$ whose dimension is equal to $\unicode[STIX]{x1D705}(D_{F})$ . Moreover, $\unicode[STIX]{x1D719}_{mD_{F}}$ induces a (not unique) map $\unicode[STIX]{x1D713}:X{\dashrightarrow}\mathbb{P}_{Z}^{d}$ over $Z$ which restricts to $\unicode[STIX]{x1D719}_{mD_{F}}$ . Let $T$ be the normalization of the image of $\unicode[STIX]{x1D713}$ . Let $Y$ be the normalization of the graph of $X{\dashrightarrow}\mathbb{P}_{Z}^{d}$ and $G$ the generic fiber of $Y\rightarrow Z$ . We have induced morphisms $Y\rightarrow T$ , $G\rightarrow S$ , and $G\rightarrow F$ .

Let $A$ on $\mathbb{P}_{Z}^{d}$ be the pullback of a hyperplane via the projection $\mathbb{P}_{Z}^{d}\rightarrow \mathbb{P}_{k}^{d}$ . Perhaps after changing $D$ up to $\mathbb{Q}$ -linear equivalence we can assume $mD_{G}\geqslant A_{G}$ . Thus replacing $D$ with $D+lf^{\ast }H$ for some $l$ we can assume $mD_{Y}\geqslant A_{Y}$ . Therefore, we may replace $X$ with $T$ and replace $D$ with $A_{T}$ . But then the statement is trivial in this case because we can assume $A+f^{\ast }H$ is ample. ◻

Proof. By assumption $\deg _{k}K_{X}<0$ , we get $h^{1}({\mathcal{O}}_{X})=h^{0}(K_{X})=0$ which implies $p_{a}(X)\leqslant 0$ . Then (i) and (ii) follow from [Reference Liu31, Chapter 9, Proposition 3.16], and (iii) is [Reference Cascini, Tanaka and Xu11, Lemma 6.5]. ◻

Proof. Let $S$ be a component of $\left\lfloor B\right\rfloor$ and let $x\in S$ be a closed point. As $(X,S)$ is plt, $S$ is regular at $x$ by [Reference Kollár27, 3.35]. ◻

Proof. By perturbing the coefficients of $B$ (i.e., by replacing $B$ with $B-P$ and replacing $A$ with $A+P$ for some appropriate $P$ ) we can assume $S=\left\lfloor B\right\rfloor$ . Since $K_{X}+B+A$ is nef and big, we can write $K_{X}+B+A{\sim}_{\mathbb{Q}}H+D$ where $H$ is ample and $D\geqslant 0$ . Since $(H+D)\cdot S=0$ , $S$ is a component of $D$ and $S^{2}<0$ . Let $\unicode[STIX]{x1D716}>0$ be a small rational number such that $A^{\prime }:=A+\unicode[STIX]{x1D716}S$ is ample. Then $S$ is the only curve on $X$ such that $(K_{X}+B+A^{\prime })\cdot S<0$ . Let $t$ be the smallest real number such that $L:=K_{X}+B+A^{\prime }+tA$ is nef. We want to show $L$ is semi-ample and that $L\cdot S=0$ . If char $k=0$ , the last sentence and the other claims of the proposition can be reduced to the algebraically closed case by passing to the algebraic closure. So we assume char $k>0$ .

By definition $L$ is nef and big but not ample; then there is a curve $C$ with $L\cdot C=0$ by Nakai–Moishezon criterion [Reference Debarre14, Theorem 1.21], which implies $t$ is a rational number. Actually $C=S$ by construction.

By Lemma 3.4, $S$ is regular. Then by adjunction (2.8) we can write $K_{S}+B_{S}=(K_{X}+B)|_{S}$ where $B_{S}\geqslant 0$ . Since $\deg _{k}(K_{S}+B_{S})<0$ , we have $\deg _{k}K_{S}<0$ . This implies $\text{Pic}^{0}(S)=0$ by Lemma 3.2. Therefore, $L|_{S}{\sim}_{\mathbb{Q}}0$ which implies that $L$ is semi-ample [Reference Keel26], so it defines a birational contraction $\unicode[STIX]{x1D70E}:X\rightarrow Y$ contracting exactly $S$ so that $L_{Y}$ is ample where $L_{Y}$ is the pushdown of $L$ .

The dlt property of $(Y,B_{Y})$ is obvious once we show $Y$ is $\mathbb{Q}$ -factorial where $B_{Y}$ is the pushdown of $B$ . Let $R_{Y}$ be a prime divisor on $Y$ and $R$ its birational transform on $X$ . There is $s\geqslant 0$ such that $(R+sS)\cdot S=0$ . Since $L$ is the pullback of an ample divisor on $Y$ , the divisor $M:=mL+R+sS$ is nef and big on $X$ , and $\mathbb{E}(M)=S$ for any $m\gg 0$ . Moreover, $M|_{S}{\sim}_{\mathbb{Q}}0$ , so by [Reference Keel26, Theorem 0.2], $M$ is semi-ample, thus it is the pullback of some ample divisor $M_{Y}$ on $Y$ . But then $R_{Y}=M_{Y}-mL_{Y}$ is $\mathbb{Q}$ -Cartier. This shows $Y$ is $\mathbb{Q}$ -factorial. Finally

$$\begin{eqnarray}(K_{X}+B)\cdot S=\deg _{k}(K_{S}+B_{S})\geqslant \deg _{k}K_{S}=-2.\square\end{eqnarray}$$

Proof. If char $k=0$ , we can pass to the algebraic closure of $k$ in which case the theorem is well-known. So we assume char $k>0$ .

Since $L$ is nef and big, by [Reference Keel26, Theorem 1.9], there exist a birational morphism $X\rightarrow V$ to a proper algebraic space $V$ and a reduced divisor $D$ on $X$ such that, the exceptional locus is equal to $D$ , and that $L\equiv 0/V$ .

Let $\unicode[STIX]{x1D719}:W\rightarrow X$ be a log resolution of $(X,B+D)$ . Let $\unicode[STIX]{x1D6E5}_{W}$ be the sum of the birational transform of $B_{V}$ plus the reduced exceptional divisor of $W\rightarrow V$ where $B_{V}$ is the pushdown of $B$ on $V$ . Let $R_{W}$ be an ample divisor on $W$ and let $L_{W}$ be the pullback of $L$ . Also let $G=L-(K_{X}+B)$ and $G_{W}$ be its pullback. Fix $m\gg 0$ and let $t$ be the smallest number such that

$$\begin{eqnarray}N_{W}:=K_{W}+\unicode[STIX]{x1D6E5}_{W}+G_{W}+tR_{W}\end{eqnarray}$$

is nef. Note that by construction, $K_{W}+\unicode[STIX]{x1D6E5}_{W}+G_{W}=L_{W}+E_{W}$ where $E_{W}\geqslant 0$ and its support is equal to the exceptional locus of $W\rightarrow V$ . Moreover, $N_{W}$ is nef and big but not ample, so by [Reference Cascini, McKernan and Mustaţă10], there is a curve $S$ with $N_{W}\cdot S=0$ . Since $(K_{W}+\unicode[STIX]{x1D6E5}_{W}+G_{W})\cdot S<0$ , $E_{W}\cdot S<0$ , hence $S$ is a component of $E_{W}$ which is contracted over $V$ , so it is a component of $\unicode[STIX]{x1D6E5}_{W}$ . In addition, $t$ is a rational number and $(K_{W}+\unicode[STIX]{x1D6E5}_{W})\cdot S<0$ . Therefore, by Proposition 3.5, $S$ can be contracted by a birational morphism $W\rightarrow W^{\prime }$ with an induced morphism $W^{\prime }\rightarrow V$ . Continuing this process gives an LMMP on $K_{W}+\unicode[STIX]{x1D6E5}_{W}$ over $V$ . It terminates with some model $Y$ on which $K_{Y}+\unicode[STIX]{x1D6E5}_{Y}$ is nef $/V$ .

Since $K_{Y}+\unicode[STIX]{x1D6E5}_{Y}\equiv E_{Y}/V$ , $E_{Y}$ is nef over $V$ and since $E_{Y}$ is exceptional over $V$ , we deduce $E_{Y}=0$ by the negativity lemma (which holds over arbitrary fields). Therefore, $Y=V$ because $E_{Y}$ contains all the exceptional curves of $Y\rightarrow V$ . Thus $V$ is projective and $\mathbb{Q}$ -factorial. Now $L_{V}$ is ample and it pulls back to $L$ , hence $L$ is semi-ample. ◻

Proof. By Proposition 3.7, we can assume $L$ is not big. Moreover, replacing $X$ with its minimal resolution we can assume $X$ is regular. Let $G:=L-(K_{X}+B)$ . By the Riemann–Roch theorem for regular surfaces (see 2.3), for any sufficiently divisible natural number $m$ we have

$$\begin{eqnarray}{\mathcal{X}}(mL)={\textstyle \frac{1}{2}}mL\cdot (mL-K_{X})+{\mathcal{X}}({\mathcal{O}}_{X}).\end{eqnarray}$$

Since $G$ is big and $L$ is not numerically trivial, $L\cdot G>0$ , and since

$$\begin{eqnarray}mL-K_{X}{\sim}_{\mathbb{Q}}(m-1)L+B+G,\end{eqnarray}$$

$L\cdot (mL-K_{X})>0$ , hence ${\mathcal{X}}(mL)$ is large when $m$ is large. This implies $h^{0}(mL)\geqslant 2$ for such $m$ because $h^{2}(mL)=h^{0}(K_{X}-mL)=0$ . Therefore, $L$ is semi-ample by Lemma 2.16. ◻

Proof of Theorem 1.4.

The proof is broken into several steps.

Step 1. Assume $K_{X}+B$ is pseudo-effective but not nef. First suppose $X$ is $\mathbb{Q}$ -factorial (we see in the next step that this is automatically satisfied). Let $H$ be an ample divisor on $X$ and let $t$ be the smallest number such that $L=K_{X}+B+tH$ is nef. Obviously $L$ is nef and big. Moreover, $t$ is rational which can be seen as in the proof of Proposition 3.5. Although we can apply Proposition 3.7 to deduce that $L$ is semi-ample and defines a contraction but we want to modify the situation so that the contraction contracts only one curve. Pick a curve $C$ such that $L\cdot C=0$ . Let $\unicode[STIX]{x1D6E5}=(1-\unicode[STIX]{x1D6FF})B+\unicode[STIX]{x1D716}C$ for certain small rational numbers $\unicode[STIX]{x1D716},\unicode[STIX]{x1D6FF}>0$ so that $(X,\unicode[STIX]{x1D6E5})$ is klt, $(K_{X}+\unicode[STIX]{x1D6E5})\cdot C<0$ , and $\unicode[STIX]{x1D6FF}B+tH$ is ample. Now let $t^{\prime }$ be the smallest number such that $L^{\prime }:=K_{X}+\unicode[STIX]{x1D6E5}+\unicode[STIX]{x1D6FF}B+t^{\prime }H$ is nef. Then $t<t^{\prime }$ because $C^{2}<0$ , so $L^{\prime }$ is nef and big, and $\unicode[STIX]{x1D6FF}B+t^{\prime }H$ is ample. Note that $C$ is the only curve satisfying $L^{\prime }\cdot C=0$ .

Now by Proposition 3.7, $L^{\prime }$ is semi-ample and it defines a nontrivial birational contraction $X\rightarrow Y$ contracting $C$ with $(K_{X}+B)\cdot C<0$ . Let $R_{Y}$ be a prime divisor on $Y$ and $R$ its birational transform on $X$ . Let $r$ be the number such that $(R+rC)\cdot C=0$ . If $m>0$ is sufficiently large, then $L^{\prime \prime }:=mL^{\prime }+R+rC$ is nef and big. Moreover, applying 3.7 to $L^{\prime }+L^{\prime \prime }$ shows that $L^{\prime \prime }$ is the pullback of an ample divisor on $Y$ , hence $R_{Y}$ is $\mathbb{Q}$ -Cartier. Therefore, $Y$ is $\mathbb{Q}$ -factorial and $(Y,B_{Y})$ is dlt. Now replace $(X,B)$ with $(Y,B_{Y})$ and repeat the argument.

Step 2. In this step we show that the dlt property of $(X,B)$ implies $X$ is $\mathbb{Q}$ -factorial. Since the pair is dlt, there is a log resolution $\unicode[STIX]{x1D719}:W\rightarrow X$ such that the log discrepancy $a(D,X,B)>0$ for every curve $D$ contracted by $\unicode[STIX]{x1D719}$ . Fix $m\gg 0$ . By Bertini Theorem (similar argument as in [Reference Xu47, Proposition 2.3]), there exists a divisor $H^{m}{\sim}_{\mathbb{Q}}mH$ on $X$ such that $(X,B+H^{m})$ is dlt. Let $\unicode[STIX]{x1D6E4}_{W}$ on $W$ be the sum of the birational transform of $B+H^{m}$ and the reduced exceptional divisor of $W\rightarrow X$ . Then $K_{W}+\unicode[STIX]{x1D6E4}_{W}=\unicode[STIX]{x1D719}^{\ast }(K_{X}+B+H^{m})+E_{W}$ where $E_{W}\geqslant 0$ is contracted over $X$ . Then applying a similar procedure as above, we can run an LMMP for $K_{W}+\unicode[STIX]{x1D6E4}_{W}$ . Since $m\gg 0$ , by projection formula the curves $C$ contracted by the LMMP intersect $\unicode[STIX]{x1D719}^{\ast }H$ trivially, and by Proposition 3.5, such curves are contracted by $\unicode[STIX]{x1D719}$ . In other words, the LMMP is over $X$ . The LMMP contracts $E_{W}$ so it ends with $X$ which means $X$ is $\mathbb{Q}$ -factorial. This and the previous step together prove the theorem when $K_{X}+B$ is pseudo-effective.

Step 3. From now on we assume $K_{X}+B$ is not pseudo-effective. If there is a curve $C$ such that $(K_{X}+B)\cdot C<0$ and such that there is a birational morphism $X\rightarrow Y$ with exceptional divisor equal to $C$ , then we replace $(X,B)$ with $(Y,B_{Y})$ . So we can assume there is no such $C$ .

Pick an ample divisor $A$ and let $t$ be the smallest number such that $K_{X}+B+tA$ is pseudo-effective. By the last paragraph $K_{X}+B+tA$ is nef: otherwise, we can run an LMMP for $K_{X}+B+tA$ which is also an LMMP for $K_{X}+B$ contracting some curve $C$ , a contradiction.

If $\unicode[STIX]{x1D70C}(X)=1$ , then we already have a Mori fiber space. So assume $\unicode[STIX]{x1D70C}(X)>1$ . Then there is another ample divisor $H$ such that $A$ is not numerically equivalent to $hH$ for any number $h$ . Let $s$ be the smallest number such that $K_{X}+B+sH$ is pseudo-effective. Arguing as above, $K_{X}+B+sH$ is nef. By our choice of $A$ and $H$ both $K_{X}+B+tA$ and $K_{X}+B+sH$ cannot be numerically trivial at the same time. We may assume $K_{X}+B+tA$ is not numerically trivial.

Step 4. In this step we assume $t$ is a rational number. By Proposition 3.8, $K_{X}+B+tA$ is semi-ample defining a contraction $f:X\rightarrow Z$ onto $Z$ of dimension one. Assume there is a fiber $F$ of $f$ which is not irreducible. Let $C$ be a component of $F$ . Then $C^{2}<0$ . We can find a $\mathbb{Q}$ -boundary $\unicode[STIX]{x1D6E5}$ such that $(X,\unicode[STIX]{x1D6E5})$ is klt, $K_{X}+\unicode[STIX]{x1D6E5}$ is pseudo-effective, and $(K_{X}+\unicode[STIX]{x1D6E5})\cdot C<0$ . So we can contract $C$ . But since $(K_{X}+B)\cdot C<0$ , this contradicts the first paragraph of Step 3. Therefore, we can assume all the fibers of $f$ are irreducible. But this means $f$ is extremal and so $f$ is a Mori fiber space.

Step 5. Finally we show $t$ is indeed a rational number. Assume not. We derive a contradiction. Let $L=K_{X}+B+tA$ . We can assume $A$ is an effective $Q$ -divisor. For each sufficiently divisible natural number $m$ , let $a_{m}$ be the number so that $\left\lfloor mL\right\rfloor =mL-a_{m}A$ . Since $t$ is not rational, there is an infinite set $\unicode[STIX]{x1D6F1}$ of such $m$ so that the $a_{m}$ form a strictly decreasing sequence with $\lim _{m\in \unicode[STIX]{x1D6F1}}a_{m}=0$ . On the other hand, for each $m\in \unicode[STIX]{x1D6F1}$ , let $a_{m}^{\prime }$ be the number so that

$$\begin{eqnarray}\displaystyle & & \displaystyle (mL-a_{m}^{\prime }A)\cdot ((m-1)L+B+(t-a_{m})A)\nonumber\\ \displaystyle & & \displaystyle \quad =(mL-a_{m}^{\prime }A)\cdot (\left\lfloor mL\right\rfloor -K_{X})=0.\nonumber\end{eqnarray}$$

Since $\lim _{m\in \unicode[STIX]{x1D6F1}}a_{m}=0$ and $L^{2}=0$ , we can see

$$\begin{eqnarray}\lim _{m\in \unicode[STIX]{x1D6F1}}a_{m}^{\prime }=\lim _{m\in \unicode[STIX]{x1D6F1}}\frac{mL\cdot ((m-1)L+B+(t-a_{m})A)}{A\cdot ((m-1)L+B+(t-a_{m})A)}=\frac{L\cdot (B+tA)}{A\cdot L}>0.\end{eqnarray}$$

Thus we can assume

$$\begin{eqnarray}(mL-a_{m}A)\cdot (\left\lfloor mL\right\rfloor -K_{X})>\unicode[STIX]{x1D70F}(m-1)A\cdot L\end{eqnarray}$$

for some $\unicode[STIX]{x1D70F}>0$ independent of $m$ . Therefore,

$$\begin{eqnarray}{\mathcal{X}}(\left\lfloor mL\right\rfloor )={\textstyle \frac{1}{2}}(mL-a_{m}A)\cdot (\left\lfloor mL\right\rfloor -K_{X})+{\mathcal{X}}({\mathcal{O}}_{X})\end{eqnarray}$$

is large when $m\in \unicode[STIX]{x1D6F1}$ is large. This in turn implies $h^{0}(\left\lfloor mL\right\rfloor )$ is large for such $m$ . Take a divisor $M_{1}\in |\left\lfloor mL\right\rfloor |$ . Then $mL{\sim}_{\mathbb{R}}M:=M_{1}+a_{m}A$ . If $C$ is a component of $\text{Supp}M$ , then $L\cdot C=0$ which means $t$ is a rational number, a contradiction. ◻

Proof. Let $F$ be the generic fiber of $f$ which is a regular curve since $X$ is regular in codimension one. Since $-(K_{X}+B)$ is ample over $Z$ , $-K_{F}$ is ample. On the other hand, since $f$ is a contraction, $H^{0}({\mathcal{O}}_{F})=K$ where $K$ is the function field of $Z$ . The assertions follow from Lemma 3.2 straightforwardly. ◻

Proof. Assume $\unicode[STIX]{x1D6FC}\in H^{0}(L)$ is nonzero. Then $\unicode[STIX]{x1D6FC}|_{D_{i}}$ is either nowhere vanishing or everywhere vanishing because $L|_{D_{i}}\equiv 0$ . Since $D$ is connected, either $\unicode[STIX]{x1D6FC}$ is nowhere vanishing on $D$ or $\unicode[STIX]{x1D6FC}|_{\text{Supp}\,D}=0$ . The former implies $L$ is generated by global sections which in turn implies $L\sim 0$ . So it is enough to treat the latter. Let $n_{i}$ be the order of vanishing of $\unicode[STIX]{x1D6FC}$ along $D_{i}$ . Let $N=\sum n_{i}D_{i}$ . Then by assumption $0<N<D$ .

We claim that for every component $D_{i}$ such that $n_{i}<d_{i}$ , $-N|_{D_{i}}$ is nef. Granted this claim, since $D$ is connected we have that $n_{i}>0$ for every $i\in \{1,2,\ldots ,r\}$ . Let $a$ be the smallest rational number such that $aN-D\geqslant 0$ . Then $aN-D=0$ because otherwise, since $D$ is connected, there will exist a component $D_{i}$ of $D$ such that $D_{i}$ is not contained in $aN-D$ , $n_{i}<d_{i}$ and $(aN-D)\cdot D_{i}>0$ , which contradicts that $-N|_{D_{i}}$ is nef. Since the $d_{i}$ have no common factor, $N=D$ . So $\unicode[STIX]{x1D6FC}=0$ , a contradiction.

The claim follows from local analysis. Assume $n_{i}<d_{i}$ , say for $i=1$ . Consider the exact sequence

$$\begin{eqnarray}0\rightarrow {\mathcal{O}}_{D_{1}}(L-n_{1}D_{1})\rightarrow {\mathcal{O}}_{(n_{1}+1)D_{1}}(L)\rightarrow {\mathcal{O}}_{n_{1}D_{1}}(L)\rightarrow 0.\end{eqnarray}$$

Since $\unicode[STIX]{x1D6FC}|_{n_{1}D_{1}}=0$ by definition of $n_{1}$ , the section $\unicode[STIX]{x1D6FC}|_{(n_{1}+1)D_{1}}$ is the image of a section $\unicode[STIX]{x1D6FD}$ of $(L-n_{1}D_{1})|_{D_{1}}$ . If $P$ is the zero divisor of $\unicode[STIX]{x1D6FD}$ , then a local computation of intersection numbers shows that $P\geqslant (N-n_{1}D_{1})|_{D_{1}}$ . More precisely, let $v\in D_{1}$ be a closed point, let $R={\mathcal{O}}_{X,v}$ , and let $f_{i}$ be a local equation of $D_{i}$ near $v$ . Then locally considering $\unicode[STIX]{x1D6FC}$ as an element of $\frac{R}{\langle f_{1}^{d_{1}}\cdots f_{r}^{d_{r}}\rangle }$ , it is easy to see that $\unicode[STIX]{x1D6FC}$ is represented by $\unicode[STIX]{x1D706}f_{1}^{n_{1}}\cdots f_{r}^{n_{r}}$ for some $\unicode[STIX]{x1D706}\in R$ , and that $\unicode[STIX]{x1D6FD}$ is represented by $\unicode[STIX]{x1D706}f_{2}^{n_{2}}\cdots f_{r}^{n_{r}}$ which gives the equation of $P$ near $v$ . Therefore, from

$$\begin{eqnarray}\text{length}_{R/\langle f_{1}\rangle }\frac{R}{\langle f_{1},\unicode[STIX]{x1D706}f_{2}^{n_{2}}\cdots f_{r}^{n_{r}}\rangle }\geqslant \text{length}_{R}\frac{R}{\langle f_{1},f_{2}^{n_{2}}\cdots f_{r}^{n_{r}}\rangle }\end{eqnarray}$$

we deduce that locally near $v$ we have $P\geqslant (N-n_{1}D_{1})|_{D_{1}}$ because the left hand side of the displayed formula is the coefficient of $v$ in $P$ and the right hand side is nothing but the local intersection number $(N-n_{1}D_{1})\cdot D_{1}$ at $v$ which is in turn equal to the coefficient of $v$ in $(N-n_{1}D_{1})|_{D_{1}}$ . As $P\sim (L-n_{1}D_{1})|_{D_{1}}$ , we deduce that $\deg N|_{D_{1}}\leqslant 0$ . ◻

Proof. By definition of curves of canonical type $K_{D}=(K_{X}+D)|_{D}\equiv 0$ . By [Reference Liu31, Chapter 7, Corollary 3.31], $0=\deg _{k}K_{D}=2(p_{a}(D)-1)$ . Thus $p_{a}(D)=1$ which means ${\mathcal{X}}({\mathcal{O}}_{D})=0$ , hence $h^{1}({\mathcal{O}}_{D})=h^{0}({\mathcal{O}}_{D})>0$ . So by duality $h^{0}(K_{D})=h^{1}({\mathcal{O}}_{D})>0$ which implies $K_{D}\sim 0$ by Proposition 3.13. ◻

Proof. Let $r$ be the order of $D|_{D}$ in $\text{Pic}(D)$ . First we want to show $rD|_{rD}\sim 0$ . This is trivially true if $r=1$ , so assume $r>1$ . Assume we already know $rD|_{lD}\sim 0$ for some $0<l<r$ . Consider the exact sequence

$$\begin{eqnarray}0\rightarrow {\mathcal{O}}_{D}(rD-lD)\rightarrow {\mathcal{O}}_{(l+1)D}(rD)\rightarrow {\mathcal{O}}_{lD}(rD)\rightarrow 0.\end{eqnarray}$$

Now $h^{0}({\mathcal{O}}_{D}(rD-lD))=0$ by Proposition 3.13, and since ${\mathcal{X}}({\mathcal{O}}_{D})=0$ , by Riemann–Roch we get

$$\begin{eqnarray}{\mathcal{X}}({\mathcal{O}}_{D}(rD-lD))=\deg _{k}(rD-lD)|_{D}+{\mathcal{X}}({\mathcal{O}}_{D})=0\end{eqnarray}$$

which implies $h^{1}({\mathcal{O}}_{D}(rD-lD))=0$ . So any nowhere vanishing section of ${\mathcal{O}}_{lD}(rD)$ lifts to ${\mathcal{O}}_{(l+1)D}(rD)$ which shows $rD|_{(l+1)D}\sim 0$ . Inductively one shows $rD|_{rD}\sim 0$ . Finally applying [Reference Totaro42, Lemma 4.1], we deduce $D$ is semi-ample. ◻

Proof of Theorem 1.5.

We can assume $K_{X}+B$ is not big by Proposition 3.7. Replacing $X$ with its minimal resolution we can assume $X$ is regular. By assumption $m(K_{X}+B)\sim M$ for some integer $m>0$ and $M$ is an effective Cartier divisor. Let $n$ be a sufficiently large natural number. We can run an LMMP on $K_{X}+nM$ because

$$\begin{eqnarray}K_{X}+nM\sim (1+nm)\left(K_{X}+\frac{nm}{1+nm}B\right)\end{eqnarray}$$

and because $(X,\frac{nm}{1+nm}B)$ is klt. Moreover, we claim that $M$ is numerically trivial on each step of the LMMP and the nefness of $M$ is preserved in the process. Indeed assume the first step of the LMMP is a birational contraction $X\rightarrow Z$ contracting a curve $E$ . Then $\deg _{k}K_{E}=(K_{X}+E)\cdot E<0$ , hence by Lemma 3.2, if setting $l=H^{0}({\mathcal{O}}_{E})$ then

$$\begin{eqnarray}-2=\deg _{l}K_{E}=\deg _{l}(K_{X}+E)|_{E}\end{eqnarray}$$

which implies $\deg _{l}K_{X}|_{E}=-1$ . Thus from $\deg _{l}(K_{X}+nM)|_{E}<0$ we deduce $M\cdot E=0$ , because otherwise, $\deg _{l}M|_{E}\geqslant 1$ since $M$ is Cartier, this is impossible since we have assumed $n\gg 0$ . On the other hand, if $X\rightarrow Z$ is a Mori fiber space, then we stop the LMMP and in this case $M\equiv 0/Z$ by Proposition 3.11 and calculations similar to those above. Applying this argument to every step of the LMMP proves the claim. Note that the regularity of $X$ is also preserved by the LMMP because the LMMP is a $K_{X}$ -MMP; hence $X$ remains with terminal singularities which implies regularity.

Replacing $X$ with the end product of the LMMP we can assume either $K_{X}+nM$ is nef or that there is a Mori fiber structure $X\rightarrow Z$ so that $M\equiv 0/Z$ . First assume $K_{X}+nM$ is nef. There is a divisor $0\leqslant D=\sum d_{i}D_{i}\leqslant M$ such that $D$ is connected, the $d_{i}$ have no common prime factor, and $M=aD$ in a neighborhood of $D$ for some number $a$ . In particular, $D$ is nef. We show $D$ is an indecomposable curve of canonical type. It is enough to show $K_{X}|_{D}\equiv 0$ because $M$ not being big implies $D|_{D}\equiv 0$ . Since $M$ is not big and since $m(K_{X}+B)\sim M$ , we deduce $K_{X}+n^{\prime }M$ is not big for any $n^{\prime }$ . Therefore, $(K_{X}+nM+M)^{2}=0$ from which we deduce $(K_{X}+nM)\cdot M=0$ , hence $(K_{X}+nM)|_{D}\equiv 0$ , so $K_{X}|_{D}\equiv 0$ .

In order to apply Proposition 3.15 we need to show $D|_{D}$ is torsion. By construction, $B|_{D}\equiv 0$ which implies $B=bD$ in some neighborhood of $D$ because $D$ is connected, where $b<1$ is a rational number. Taking positive integer $m^{\prime }$ so that $m^{\prime }b\in \mathbb{Z}$ we get

$$\begin{eqnarray}\displaystyle 0 & {\sim} & \displaystyle m^{\prime }K_{D}=m^{\prime }(K_{X}+D)|_{D}=m^{\prime }(K_{X}+B+D)|_{D}-m^{\prime }B|_{D}\nonumber\\ \displaystyle & {\sim} & \displaystyle (a+m^{\prime })D|_{D}-m^{\prime }bD|_{D}\nonumber\end{eqnarray}$$

which implies $D|_{D}$ is torsion because $a+m^{\prime }-m^{\prime }b>0$ . Therefore, $D$ is semi-ample, hence $\unicode[STIX]{x1D705}(M)=1$ which implies $M$ is semi-ample by Lemma 2.16.

Now assume we have a Mori fiber structure $X\rightarrow Z$ with $M\equiv 0/Z$ . If $F$ is the generic fiber, then $M|_{F}\sim 0$ . This implies $M$ is the pullback of some effective divisor $N$ on $Z$ . Either $N$ is ample or $N=0$ , hence in any case $M$ is semi-ample. ◻