Fact 2.2. Let A be a Cartier divisor on a projective variety Y such that the complete linear system $|A|$ has a non-empty fixed part B. Then $h^0(Y,{\mathcal O}_Y(B))=1$ .

Fact 2.3. Let X be a Fano manifold, and let D be a nef Cartier divisor on X. Then there exists a morphism with connected fibres $\varphi :X \rightarrow T$ and an ample Cartier divisor $H_T$ on T such that $D \simeq \varphi ^* H_T$ .

Proof. By the basepoint free theorem, we have $mD \simeq \varphi ^* H_m$ for some very ample divisor $H_m \rightarrow T$ for every $m \gg 0$ . Thus, $H_T := H_{m+1}-H_m$ is a Cartier divisor such that $D \simeq \varphi ^* H_T$ . Since D is numerically the pull-back of an ample class on T, the divisor $H_T$ is ample.

Fact 2.4. Let S be an irreducible projective surface with Gorenstein singularities, and let $C \subset S$ be an irreducible curve that is not contained in the singular locus of S. If $-K_S \cdot C \geq 2$ , the curve C deforms in S in a positive-dimensional family.

Proof. Let $\tau :S' \rightarrow S$ be the composition of normalisation and minimal resolution. Then we have $K_{S'} \simeq \tau ^* K_S - E$ , where E is an effective divisor that maps into the singular locus of S. Thus, if $C' \subset S'$ is the strict transform, it is not contained in the support of E. Thus, we have

$$ \begin{align*}-K_{S'} \cdot C' \geq -K_S \cdot C \geq 2. \end{align*} $$

The statement now follows from [Reference Kollár34, Theorem 1.15].

Fact 2.5 [Reference Beltrametti and Sommese8, Lemma 2.4.1].

Let X be a normal projective variety with rational singularities such that $q(X)>0$ . Then the Albanese map $\alpha :X \rightarrow A$ to the Albanese torus exists and is determined by the Albanese map of some resolution of singularities.

Fact 2.6. Let S be a normal projective surface with rational singularities, and let A be a Cartier divisor on S. Then the Riemann-Roch formula

$$ \begin{align*}\chi(S, {\mathcal O}_S(A)) = \frac{1}{2} A^2 + \frac{1}{2} (-K_S) \cdot A + \chi(S, {\mathcal O}_S) \end{align*} $$

holds.

Let Y be a normal projective threefold with terminal singularities, and let A be a Cartier divisor on Y. Then the Riemann-Roch formula

$$ \begin{align*}\chi(Y, {\mathcal O}_Y(A)) = \frac{1}{12} A \cdot (A-K_Y) \cdot (2A-K_Y) + \frac{1}{12} A \cdot c_2(Y) + \chi(Y, {\mathcal O}_Y) \end{align*} $$

holds.

Proof. It is sufficient to verify that all the terms are well-defined and can be calculated on a resolution of singularities. This is well-known; cf. [Reference Höring22, Section 2]. The statement then follows via the projection formula.

Proof. Consider the natural morphism

$$ \begin{align*}\alpha: H^0(X, {\mathcal O}_X(A)) \simeq H^0(X, {\mathcal I}_B \otimes {\mathcal O}_X(A)) \rightarrow H^0(B, {\mathcal I}_B/{\mathcal I}_B^2 \otimes {\mathcal O}_X(A)). \end{align*} $$

By [Reference Beltrametti and Sommese8, Lemma 1.7.4], the divisor Y is smooth along B if and only if the section $\alpha (s) \in H^0(B, {\mathcal I}_B/{\mathcal I}_B^2 \otimes {\mathcal O}_X(A))$ is nowhere zero, where $0 \neq s \in H^0(X, {\mathcal O}_X(A))$ is a section vanishing on Y. Since ${\mathcal I}_B/{\mathcal I}_B^2 \otimes {\mathcal O}_X(A) \simeq {\mathcal N}_{B/X}^* \otimes {\mathcal O}_X(A)$ is a tensor product of a nef vector bundle and an ample line bundle, it is ample [Reference Lazarsfeld40, Theorem 6.2.12.(iv)]. Set $d=\mathrm {codim}_X B$ . If $\alpha (s)$ does not vanish in any point of B, the top Chern class $c_d({\mathcal N}_{B/X}^* \otimes {\mathcal O}_X(A))$ is zero. Yet since $d \leq \dim B$ , this is a contradiction to the positivity of Chern classes of ample vector bundles [Reference Lazarsfeld40, Theorem 8.3.9].

Proof. Let

$$ \begin{align*}-K_X|_Y \simeq M + B \end{align*} $$

be the decomposition into fixed and mobile part. Since Y is smooth, the Weil divisors M and B are Cartier. Since $\dim X \geq 4$ and $\rho (X)=1$ , the Lefschetz hyperplane theorem implies $\operatorname *{\mathrm {Pic}}{Y} \simeq \operatorname *{\mathrm {Pic}}{X} \simeq \mathbb {Z} H$ , where H is the ample generator of the Picard group. Thus, we have

$$ \begin{align*}M \simeq a H|_Y, \qquad B \simeq b H|_Y \end{align*} $$

with $a,b \in \mathbb {N}^*$ . Since $h^0(Y, {\mathcal O}_Y(B))=1$ and $h^0(Y, {\mathcal O}_Y(M)) \geq 2$ , we have $a>b$ , and therefore, $a+b \geq 3$ . Since

$$ \begin{align*}-K_X|_Y \simeq M+B \simeq (a+b) H|_Y, \end{align*} $$

the statement follows from $\operatorname *{\mathrm {Pic}}{Y} \simeq \operatorname *{\mathrm {Pic}}{X}$ .

Proof. If D has numerical dimension one, there exists a fibration $\varphi : X \rightarrow \mathbb {P}^1$ such that $D \simeq \varphi ^* H$ with H an ample Cartier divisor on $\mathbb {P}^1$ (cf. Fact 2.3). Thus, $h^0(\mathbb {P}^1, {\mathcal O}_{\mathbb {P}^1}(H)) \geq 2$ yields the claim.

If D has numerical dimension at least two, we have $D^2 \cdot K_X^2= D^2 \cdot (-K_X)^2>0$ . Moreover, $h^0(X, {\mathcal O}_X(D))=\chi (X, {\mathcal O}_X(D))$ by Kodaira vanishing, and by Riemann-Roch,

$$ \begin{align*} \chi(X, {\mathcal O}_X(D)) &= \frac{D^4}{4!} - \frac{D^3 \cdot K_X}{2 \cdot 3!} + \frac{D^2 \cdot (K_X^2+c_2(X))}{12 \cdot 2 !} - \frac{D \cdot K_X \cdot c_2(X)}{24 \cdot 1!} + \chi(X, {\mathcal O}_X) \\ & \geq \frac{D^2 \cdot (-K_X)^2}{24} + \frac{D^2 \cdot c_2(X)}{24} - \frac{D \cdot K_X \cdot c_2(X)}{24} + 1. \end{align*} $$

Since $-K_X \cdot D \cdot c_2(X) \geq 0$ and $D^2 \cdot c_2(X) \geq 0$ by [Reference Peternell46, Theorem 1.3], [Reference Miyaoka43, Theorem 6.1] we obtain that the second line is at least two.

Proof. Since $\dim X \leq 4$ , the anticanonical system $|-K_X|$ has no fixed component; cf. Theorem 2.8. In particular, Y has no common component with D.

We argue by contradiction and assume that the non-lc locus Z is not empty.

1st step. Assume that Z has positive dimension. Since Y is an ample divisor, it intersects Z in at least one point, so the pair $(X,D)$ is not lc near Y. Thus, the pair $(X, D+Y)$ is not lc near Y. By Theorem 2.7, this is a contradiction to Corollary 2.12.

2nd step. Assume that Z has dimension zero. We endow $Z \subset X$ with its natural scheme structure; cf. [Reference Fujino17, \S 7]. Since

$$ \begin{align*}L - (K_X+D) \simeq -K_X \end{align*} $$

is ample, we can apply Fujino’s extension theorem (with W the empty set; cf. [Reference Fujino17, Theorem 8.1]) to see that the restriction map

$$ \begin{align*}H^0(X, L) \rightarrow H^0(Z, L) \end{align*} $$

is surjective. Since Z is a finite scheme, the restriction $L|_Z$ is globally generated. Thus, L is globally generated near Z; in particular, a general $D \in |L|$ is smooth near Z. Hence, the pair $(X,D)$ is lc, a contradiction.

Proof. Since canonical surface singularities are rational, we can replace S with a resolution of singularities. Thus, S is a smooth surface with $p_g(S)=0$ and $q(S) \geq 1$ . If the general $\alpha $ -fibre is not $\mathbb {P}^1$ , the surface S is not uniruled. Thus, we can apply [Reference Beauville5, Theorem VI.13] to the minimal model of S to obtain that $q(S)=1$ .

Proof of Proposition 4.2.

Since $A \cdot \Delta _S>0$ , we have $\Delta _S \neq 0$ . Thus,

$$ \begin{align*}H^2(S, {\mathcal O}_S(K_S+\Delta_S)) \simeq H^0(S, {\mathcal O}_S(-\Delta_S))=0, \end{align*} $$

and we have the inequalities

$$ \begin{align*}q(S) \geq h^0(S, {\mathcal O}_S(K_S+\Delta_S)) \geq \chi(S, {\mathcal O}_S(K_S+\Delta_S)). \end{align*} $$

1st step. Assume that S has canonical singularities. By Fact 2.6, we have the Riemann-Roch formula

$$ \begin{align*}\chi(S, {\mathcal O}_S(K_S+\Delta_S)) = \frac{1}{2} (K_S+\Delta_S) \cdot \Delta_S + \chi({\mathcal O}_S). \end{align*} $$

Hence, $A \cdot \Delta _S>0$ implies that $q(S)> \chi (S, {\mathcal O}_S) = 1 - q(S)$ . In particular, we have $q(S)>0$ , so there is a non-trivial Albanese morphism $\alpha $ . If $\dim \alpha (S)=2$ , the ramification formula shows that $p_g(S)>0$ which we excluded. Thus, the Albanese morphism gives a fibration [Reference Beltrametti and Sommese8, Lemma 2.4.5]

$$ \begin{align*}\alpha:S \rightarrow C \end{align*} $$

onto a curve C of genus $g:=q(S)>0$ . Since the pair $(S, \Delta _S)$ is lc, the direct image sheaf

$$ \begin{align*}\alpha_* {\mathcal O}_S(K_{S/C}+\Delta_S) \simeq \alpha_* {\mathcal O}_S(A) \otimes {\mathcal O}_C(-K_C) \end{align*} $$

is a nef vector bundle [Reference Fujino19, Theorem 1.1] of rank r. Thus,

$$ \begin{align*}V:= \alpha_* {\mathcal O}_S(A) \simeq \alpha_* {\mathcal O}_S(K_{S/C}+\Delta_S) \otimes {\mathcal O}_C(K_C) \end{align*} $$

has $c_1(V) \geq r (2g-2)$ . By the Riemann-Roch formula on curves, we have

(2) $$ \begin{align} h^0(S, {\mathcal O}_S(A)) = h^0(C, V) \geq \chi(C, V) = c_1(V) + r \chi(C, {\mathcal O}_C) \geq r (g-1). \end{align} $$

Now observe that $r(g-1)> g$ unless $(r-1) (g-1) \leq 1$ .

Finally, let us show that we have $h^0(S, {\mathcal O}_S(A)) = q(S)$ : if $h^0(S, {\mathcal O}_S(A)) \leq q(S)-1$ , the Riemann-Roch estimate (2) becomes $g-1 \geq r (g-1)$ .

Subcase a) Assume that $g>1$ . Then the unique possibility $r=1$ . Since A is nef and big, this implies that the general $\alpha $ -fibre is not $\mathbb {P}^1$ . Yet this contradicts Lemma 4.1.

Subcase b) Assume that $g=1$ . In this case, the Riemann-Roch inequality becomes

$$ \begin{align*}h^0(S, {\mathcal O}_S(K_S+\Delta_S)) \geq \frac{1}{2} (K_S+\Delta_S) \cdot \Delta_S>0, \end{align*} $$

so $h^0(S, {\mathcal O}_S(K_S+\Delta _S)) \geq 1 = g$ .

2nd step. We show that S has canonical singularities. Since S is Gorenstein, it has canonical singularities if and only if it has rational singularities. Arguing by contradiction, we assume that S has non-canonical singularities and denote by $\tau :S_c \rightarrow S$ the canonical modification. Then we have $K_{S_c} \simeq \tau ^* K_S - E$ with E an effective Weil divisor. Observe that

$$ \begin{align*}K_{S_c} + E + \tau^* \Delta_S \simeq \tau^* (K_S + \Delta_S), \end{align*} $$

so the pair $(S_c, \Delta _{S_c}) := (S_c, E + \tau ^* \Delta _S)$ is lc. Moreover,

$$ \begin{align*}A_c := \tau^* A \simeq K_{S_c} + \Delta_{S_c} \end{align*} $$

is a nef and big Cartier divisor with $A_c \cdot \Delta _{S_c}=A \cdot \Delta _S>0$ and $h^0(S, {\mathcal O}_S(A)) = h^0(S_c, {\mathcal O}_{S_c}(A_c))$ . Finally, by the remark before the proof, one has $p_g(S_c) \leq p_g(S)=0$ and $q(S_c)>q(S)$ . Therefore, by Step 1 of the proof,

$$ \begin{align*}h^0(S, {\mathcal O}_S(A)) = h^0(S_c, {\mathcal O}_{S_c}(A_c)) = q(S_c)> q(S), \end{align*} $$

a contradiction to our assumption.

Proof. We argue by contradiction, so let $l \subset \Delta _S$ be a smooth rational curve. In particular, $\Delta _S \neq 0$ , and therefore, $A\cdot \Delta _S>0$ . By Proposition 4.2, the surface has canonical singularities, and we have an Albanese fibration $\alpha :S \rightarrow C$ onto a smooth curve C of genus $g=q(S)$ . Moreover, we have

(3) $$ \begin{align} (r-1) \cdot (q(S)-1) \leq 1, \end{align} $$

where $r= \mathrm {rk} (\alpha _* {\mathcal O}_S(A))$ .

1st case. Assume that $g>1$ . By Lemma 4.1, the general $\alpha $ -fibre F is $\mathbb {P}^1$ . Since $A \cdot F \geq 1$ , we have $r \geq 2$ , and thus, (3) yields $g=2$ and $r=2$ . Moreover, $A \cdot F=1$ implies that $\alpha :S \rightarrow C$ is a $\mathbb {P}^1$ -bundle over the genus two curve C [Reference Kollár34, II,Theorem 2.8]. In particular, the rational curve l is a fibre of $\alpha $ , and thus a nef divisor on S. The pair $(S, \Delta _S-l)$ is lc, so the direct image

$$ \begin{align*}\alpha_* {\mathcal O}_S(K_{S/C}+\Delta_S-l) \simeq \alpha_* {\mathcal O}_S(A) \otimes {\mathcal O}_C(-K_C-l) \end{align*} $$

is a nef vector bundle [Reference Fujino19, Theorem 1.1]. Thus, $\alpha _* {\mathcal O}_S(A)$ has degree at least six. By the Riemann-Roch formula on the curve C, this implies $h^0(S, {\mathcal O}_S(A))=h^0(C, \alpha _* {\mathcal O}_S(A)) \geq 4$ , a contradiction.

2nd case. Assume that $g=1$ . By Proposition 4.2, we have $h^0(S, {\mathcal O}_S(A))=1$ , and thus by Riemann-Roch,

$$ \begin{align*}1 \geq \chi(S, {\mathcal O}_S(A)) =\frac{1}{2} (K_S+\Delta_S) \cdot \Delta_S>0. \end{align*} $$

Therefore, $A \cdot \Delta _S = (K_S+\Delta _S) \cdot \Delta _S=2$ , and Lemma 4.5 applies: all the irreducible components of $\Delta _S$ are rational curves, so they are contracted by the Albanese map. Moreover, the support of $\Delta _S$ is connected, so it is contained in an $\alpha $ -fibre $F_0$ . Since $p_a(\Delta _S)=2$ , we deduce that $p_a(F_0) \geq 2$ . Since the arithmetic genus is constant in the flat family $\alpha :S \rightarrow C$ , we obtain that S is not uniruled. Let $\tau :S' \rightarrow S$ be the minimal resolution, and let $\nu :S' \rightarrow S_m$ the minimal model of $S'$ . Since S has canonical singularities, we have

$$ \begin{align*}p_g(S_m)=p_g(S')=p_g(S)=0, \qquad q(S_m)=q(S')=q(S)>0. \end{align*} $$

By [Reference Beauville5, Proposition VI.6], the minimal surface $S_m$ does not contain any rational curves. Since the exceptional divisors of the resolution $\tau $ are rational curves, the rigidity lemma [Reference Debarre11, Lemma 1.15] yields a birational map $\nu :S \rightarrow S_m$ that contracts all the rational curves on S. In particular, it contracts the divisor $\Delta _S$ . Yet $p_a(\Delta _S)=2>0$ , a contradiction to the fact that the smooth surface $S_m$ has rational singularities.

Proof. Recall first that canonical surface singularities are Gorenstein, so both $K_S$ and $\Delta _S = A-K_S$ are Cartier. We have $p_a(\Delta _S)=2$ , so the inclusion $l \subset \Delta _S$ must be strict. Since A is ample with $A \cdot \Delta _S=2$ , we obtain

$$ \begin{align*}\Delta_S = l + R \end{align*} $$

with R an irreducible curve such that $A \cdot R=1$ . In particular, we have by subadjunction

$$ \begin{align*}\deg K_R \leq (K_S+R) \cdot R \leq (K_S+l+R) \cdot R = A \cdot R = 1, \end{align*} $$

so $p_a(R)=\frac {1}{2} \deg K_R + 1 \leq 1$ . In particular, $\Delta _S$ is connected because otherwise

$$ \begin{align*}2= p_a(\Delta_S) = p_a(R) + p_a(l) \leq 1 \end{align*} $$

yields a contradiction.

Let $\tau :S' \rightarrow S$ be the minimal resolution and set

$$ \begin{align*}\Delta_{S'} := \tau^* \Delta_S. \end{align*} $$

Since $\Delta _S$ is connected, the cycle $\Delta _{S'}$ is connected. Denote by $R' \subset \Delta _{S'}$ the strict transform of R. We claim that $R'$ is a rational curve, and this concludes the proof.

Proof of the claim. Since S has canonical singularities, the minimal resolution $\tau $ is crepant, so we have $K_{S'} \simeq \tau ^* K_S$ and

$$ \begin{align*}K_{S'}+\Delta_{S'} = \tau^* (K_S+\Delta_S) = \tau^* A. \end{align*} $$

Thus, we have

$$ \begin{align*}1 = \tau^* A \cdot R' = (K_{S'}+\Delta_{S'}) \cdot R' = (K_{S'}+R') \cdot R' + (\Delta_{S'}-R') \cdot R'. \end{align*} $$

If we show that $(\Delta _{S'}-R') \cdot R'\geq 2$ , then the adjunction formula yields the claim. Since $\Delta _{S'}$ is connected, we have $(\Delta _{S'}-R') \cdot R'\geq 1$ . We will argue by contradiction to exclude the case where we have an equality.

In this case, $R'$ is a curve of arithmetic genus one that meets $\Delta _{S'}-R'$ transversally exactly in one point. In particular, we have

$$ \begin{align*}2 = p_a(\Delta_{S'}) = p_a(R') + p_a(\Delta_{S'}-R') = 1 + p_a(\Delta_{S'}-R'). \end{align*} $$

The curve $l \subset S$ being smooth, it meets the exceptional divisor over every point $p \in l \cap S_{sing}$ in exactly one point, and the intersection is transverse. Thus, we have

$$ \begin{align*}p_a(\Delta_{S'}-R') = p_a(l) + \sum_{p \in l \cap S_{sing}} \tau^{-1} (p) = 0 \end{align*} $$

since the exceptional divisors of the minimal resolution of an ADE-singularity have arithmetic genus zero. This gives the final contradiction.

Proof. Consider the exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(-M_X) \rightarrow {\mathcal O}_X(B_X) \rightarrow {\mathcal O}_Y(B) \rightarrow 0. \end{align*} $$

We have $h^0(X, {\mathcal O}_X(-M_X))=0$ since otherwise $-M_X$ , and thus its restriction $-M$ to the general anticanonical divisor Y, is effective. Thus, we have an injection

$$ \begin{align*}H^0(X, {\mathcal O}_X(B_X)) \hookrightarrow H^0(Y, {\mathcal O}_Y(B)). \end{align*} $$

Since $h^0(Y, {\mathcal O}_Y(B))=1$ by Fact 2.2, we have $h^0(X, {\mathcal O}_X(B_X)) \leq 1$ . By Lemma 3.7, this implies that $B_X$ is not nef.

If $B_X$ is effective, it is a normal prime divisor since its restriction to the ample divisor Y is the normal prime divisor B.

Proof. We choose a general element M in the mobile part, so by Theorem 2.11, the pair $(Y, M+B)$ is lc. If $A+M$ is not nef, there exists an irreducible curve $\gamma $ such that $(A+M) \cdot \gamma <0$ . Since A is ample, we have $M \cdot \gamma \leq -2$ . In particular, we have $\gamma \subset \mathrm {Bs}(|M|)$ . Since $|M|$ is the mobile part of $|A|$ , we have $\mathrm {Bs}(|M|) \subset \mathrm {Bs}(|A|)$ . By assumption, $\mathrm {Bs}(|A|)$ has pure dimension two, so it coincides with B. Thus, we have $\gamma \subset B,$ and hence,

$$ \begin{align*}\gamma \subset M \cap B. \end{align*} $$

The surface $M+B$ is slc by Theorem 2.7, so it has normal crossing singularities in codimension one. Therefore, B and M are smooth in the generic point of $\gamma $ ; in particular, $\gamma $ is not contained in the singular locus of M. Since

$$ \begin{align*}- K_M \cdot \gamma = - M \cdot \gamma \geq 2, \end{align*} $$

we know by Fact 2.4 that the curve $\gamma $ deforms in M. In particular, M contains a positive-dimensional family of curves $\gamma _t$ such that $M \cdot \gamma _t<0$ . Yet M is a mobile divisor and $\dim Y=3$ , so this is impossible.

Proof. By Theorem 2.10, we know that $-K_X+M_X$ is nef if and only if the restriction to Y is nef. Thus, nefness follows from Proposition 5.2. The numerical dimension is at least three since the restriction to Y is big, so of numerical dimension three. Finally, the vanishing follows from

$$ \begin{align*}H^q(X, {\mathcal O}_X(M_X)) = H^q(X, {\mathcal O}_X(K_X + (-K_X+M_X))) \end{align*} $$

and the numerical Kawamata-Viehweg vanishing theorem [Reference Lazarsfeld39, Example 4.3.7].

Proof. Since the restriction $D|_{D'}$ is pseudoeffective and $\dim D'=2$ , there are at most finitely many curves $\gamma \subset Y$ such that $D \cdot \gamma <0$ . In particular, for a general very ample divisor $A \subset Y$ , the restriction $D|_A$ is nef. Moreover,

$$ \begin{align*}(D|_A)^2 = D^2 \cdot A = D|_D \cdot A|_D \geq 0, \end{align*} $$

and equality holds if and only if the pseudoeffective class $D|_D$ is zero. Hence, if $(D|_A)^2 =0$ , the restriction $D|_D$ is numerically trivial and D is nef with numerical dimension at most one. If $(D|_A)^2>0$ , the restriction $D|_A$ is nef and big and we obtain the vanishing by [Reference Küronya38, Theorem C].Footnote ³

Proof of Proposition 5.4.

We twist the exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(-Y) \simeq {\mathcal O}_X(K_X) \rightarrow {\mathcal O}_X \rightarrow {\mathcal O}_Y \rightarrow 0 \end{align*} $$

with $M_X$ to get

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(K_X+M_X) \rightarrow {\mathcal O}_X(M_X) \rightarrow {\mathcal O}_Y(M) \rightarrow 0. \end{align*} $$

Since M is mobile, it satisfies the conditions of Lemma 5.5; by assumption, $M_X$ (and thus M) is not nef, so we have $H^q(Y, {\mathcal O}_Y(M))=0$ for $q \geq 2$ . By the exact sequence in cohomology, this implies

$$ \begin{align*}H^q(X, {\mathcal O}_X(K_X+M_X)) = H^q(X, {\mathcal O}_X(M_X)) \qquad \forall \ q \geq 3. \end{align*} $$

Yet the right-hand side vanishes by Corollary 5.3.

Proof. We twist the restriction sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(-Y) \simeq {\mathcal O}_X(-(M_X+B_X)) \rightarrow {\mathcal O}_X \rightarrow {\mathcal O}_Y \rightarrow 0 \end{align*} $$

with $B_X$ to get

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(-M_X) \rightarrow {\mathcal O}_X(B_X) \rightarrow {\mathcal O}_Y(B) \rightarrow 0. \end{align*} $$

By Serre duality and Proposition 5.4, one has

$$ \begin{align*}H^1(X, {\mathcal O}_X(-M_X)) = H^3(X, {\mathcal O}_X(K_X+M_X)) = 0. \end{align*} $$

Therefore, $h^0(Y, {\mathcal O}_Y(B))=1$ implies that $B_X$ is effective.

Proof. If $M_X$ is nef, this is immediate from Lemma 3.7. If $M_X$ is not nef, we know by Corollary 5.6 that $B_X$ is effective. Thus, $B_X$ is normal prime divisor by Proposition 5.1, and we can twist the restriction sequence for $B_X$ by $K_X+B_X$ to get an exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(K_X) \rightarrow {\mathcal O}_X(K_X+B_X) \rightarrow {\mathcal O}_{B_X}(K_X+B_X) \simeq {\mathcal O}_{B_X}(K_{B_X}) \rightarrow 0. \end{align*} $$

Taking cohomology, we get a sequence

$$ \begin{align*} &\ldots \rightarrow H^3(X, {\mathcal O}_X(K_X)) \rightarrow H^3(X, {\mathcal O}_X(K_X+B_X)) \rightarrow H^3(B_X, {\mathcal O}_{B_X}(K_{B_X})) \\ &\rightarrow H^4(X, {\mathcal O}_X(K_X)) \rightarrow H^4(X, {\mathcal O}_X(K_X+B_X)) \rightarrow 0. \end{align*} $$

Using Serre duality on X and $B_X$ , this transforms into

$$ \begin{align*} &\ldots \rightarrow H^1(X, {\mathcal O}_X)=0 \rightarrow H^3(X, {\mathcal O}_X(K_X+B_X)) \rightarrow H^0(B_X, {\mathcal O}_{B_X})=\mathbb{C} \\ &\rightarrow H^0(X, {\mathcal O}_X)=\mathbb{C} \rightarrow H^0(X, {\mathcal O}_X(-B_X))=0 \rightarrow 0. \end{align*} $$

In conclusion, we get

$$ \begin{align*}H^q(X, {\mathcal O}_X(K_X+B_X))=0 \qquad \forall \ q \geq 3. \end{align*} $$

Now we twist the restriction sequence to Y by $M_X$ to get

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(-B_X) \rightarrow {\mathcal O}_X(M_X) \rightarrow {\mathcal O}_Y(M) \rightarrow 0. \end{align*} $$

By Serre duality and what precedes,

$$ \begin{align*}H^1(X, {\mathcal O}_X(-B_X)) = H^3(X, {\mathcal O}_X(K_X+B_X)) = 0. \end{align*} $$

Thus, the restriction map

$$ \begin{align*}H^0(X, {\mathcal O}_X(M_X)) \rightarrow H^0(Y, {\mathcal O}_Y(M)) \end{align*} $$

is surjective. In particular,

$$ \begin{align*}\mathrm{Bs}(|M|) = \mathrm{Bs}(|M_X|) \cap Y. \end{align*} $$

Since Y is ample and M is mobile, this shows that $\dim \mathrm {Bs}(|M_X|) \leq 2$ . Thus, $|M_X|$ is mobile.

Proof. We have $h^0(X, {\mathcal O}_X(B_X)) = 1$ by Proposition 5.1. Thus, $B_X$ is the unique effective divisor in its linear system, and hence general. Since $(Y, B) =(Y, Y \cap B_X)$ is log-canonical by Corollary 2.12, we know by Lemma 3.8 that $(X, B_X)$ is lc.

Let us now describe its lc centres: since $B_X-(K_X+B_X)=-K_X$ is ample, we know by [Reference Fujino17, Theorem 8.1] that for every lc centre $Z \subset X$ , the restriction morphism

$$ \begin{align*}H^0(X, {\mathcal O}_X(B_X)) \rightarrow H^0(Z, {\mathcal O}_Z(B_X)) \end{align*} $$

is surjective. Since $Z \subset B_X$ and $h^0(X, {\mathcal O}_X(B_X))=1$ , we obtain that $H^0(Z, {\mathcal O}_Z(B_X))=0$ .

This immediately excludes the possibility that $\dim Z=0$ . Since $B_X$ is normal by Proposition 5.1, there are no two-dimensional lc centres.

The unique remaining possibility is that $\dim Z=1$ and the centre is minimal. Thus, by Kawamata subadjunction [Reference Kawamata27], the curve Z is normal and there exists an effective divisor $\Delta _Z$ on Z such that

$$ \begin{align*}K_Z + \Delta_Z \sim_{\mathbb{Q}} (K_X+B_X)|_{Z}. \end{align*} $$

In particular,

$$ \begin{align*}B_X|_Z - (K_Z + \Delta_Z) \sim_{\mathbb{Q}} -K_X|_Z \end{align*} $$

is ample. Thus, if $B_X \cdot Z \geq 0$ , a Riemann-Roch computation shows that $h^0(Z, {\mathcal O}_Z(B_X))>0$ , a contradiction.

Proof. By Proposition 5.9, the pair $(X,B_X)$ is lc, and $B_X$ is an lc centre for this pair. By Fujino’s extension theorem [Reference Fujino18, Theorem 1.11], we have to show that

$$ \begin{align*}-K_X - (K_X+B_X) = -K_X + M_X \end{align*} $$

is nef and log big (i.e., the restriction of $-K_X+M_X$ to every lc centre of $(X, B_X)$ is big).

By Corollary 5.8, the nef divisor $-K_X+M_X$ is big on X. Since $M_X$ is mobile by Proposition 5.7, the restriction $M_X|_{B_X}$ is pseudoeffective, so $(-K_X+M_X)|_{B_X}$ is big. By Proposition 5.9, an lc centre Z that is distinct from $B_X$ is a curve such that $B_X \cdot Z<0$ . Therefore, $M_X \cdot Z>0$ , and hence $(-K_X+M_X)|_{Z}$ , is ample.

Proof. We have $B=Y \cap B_X$ , so $B \subset B_X$ is a Cartier divisor that is linearly equivalent to $-K_X|_{B_X}$ . Thus, we have an exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_{B_X} \rightarrow {\mathcal O}_{B_X}(-K_X) \rightarrow {\mathcal O}_B(-K_X) \rightarrow 0. \end{align*} $$

Let us first show that the restriction map

$$ \begin{align*}H^0(B_X, {\mathcal O}_{B_X}(-K_X)) \rightarrow H^0(B, {\mathcal O}_B(-K_X)) \end{align*} $$

is zero. Since $B \subset \mathrm {Bs}(|-K_X|)$ , the restriction map

$$ \begin{align*}H^0(X, {\mathcal O}_X(-K_X)) \rightarrow H^0(B, {\mathcal O}_B(-K_X)) \end{align*} $$

is zero. Since $H^0(X, {\mathcal O}_X(-K_X)) \rightarrow H^0(B_X, {\mathcal O}_{B_X}(-K_X))$ is surjective by Lemma 5.10, we get the claim.

This already implies the second statement since the kernel of the restriction map is $H^0(B_X, {\mathcal O}_{B_X})=\mathbb {C}$ . Since the restriction map is zero, we have an injection

$$ \begin{align*}H^0(B, {\mathcal O}_B(-K_X)) \rightarrow H^1(B_X, {\mathcal O}_{B_X}). \end{align*} $$

Now consider the exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_{B_X}(K_X) \rightarrow {\mathcal O}_{B_X} \rightarrow {\mathcal O}_B \rightarrow 0. \end{align*} $$

The pair $(X, B_X)$ is lc by Proposition 5.9, and $B_X$ is normal by Proposition 5.1, so $B_X$ is a threefold with lc singularities. Since $K_X|_{B_X}$ is an antiample Cartier divisor, we can apply Kodaira vanishing [Reference Kovács, Schwede and Smith36, Corollary 6.6] to get

$$ \begin{align*}H^1(B_X, {\mathcal O}_{B_X}(K_X)) = 0 = H^2(B_X, {\mathcal O}_{B_X}(K_X)). \end{align*} $$

Thus, we have an isomorphism $H^1(B_X, {\mathcal O}_{B_X}) \simeq H^1(B, {\mathcal O}_B),$ and the first statement follows.

Proof. By Proposition 5.11, we have $h^0(B, {\mathcal O}_B(-K_X)) \leq q(B)$ . By Corollary 2.12, the conditions of Lemma 4.2 are satisfied. Thus, we know that B has canonical singularities. By inversion of adjunction (Theorem 2.7), the pair $(B_X, B)=(B_X, Y \cap B_X)$ is plt near B. Thus, $(B_X, 0)$ is plt (i.e., klt) near B. Since $B_X$ is Gorenstein, this implies that $B_X$ has canonical singularities near B. Since $B = B_X \cap Y$ is an ample divisor, the non-canonical locus of $B_X$ is at most a finite set.

By Proposition 5.9, the pair $(X, B_X)$ is lc, and the lc centres of dimension at most two are irreducible curves C. Again by inversion of adjunction, the non-canonical locus of $B_X$ coincides with the union of lower-dimensional lc centres which has pure dimension one. By the first paragraph, the non-canonical locus has dimension at most zero, so it is empty.

Proof. We argue by contradiction and assume that $M_X$ is not nef. By the cone theorem, there exists a $K_X$ -negative extremal ray $\mathbb {R}^+ \gamma $ such that $M_X \cdot \gamma <0$ . Since $M_X$ is mobile by Proposition 5.7, the extremal ray is small. Thus, by Kawamata’s classification [Reference Kawamata26, Theorem 1.1], a connected component of the exceptional locus is a $\mathbb {P}^2 \subset X$ such that ${\mathcal O}_{\mathbb {P}^2}(-K_X) \simeq {\mathcal O}_{\mathbb {P}^2}(1)$ . Thus, the intersection $Y \cap \mathbb {P}^2$ is either a line or the whole surface $\mathbb {P}^2$ . In the latter case, we would have $\mathbb {P}^2 \subset M \subset Y$ , in contradiction to the mobility of M. Therefore, $Y \cap \mathbb {P}^2$ is a smooth rational curve l such that $M \cdot l<0$ . Thus,

$$ \begin{align*}l \subset \mathrm{Bs}(|M|) \subset \mathrm{Bs}(|-K_X|_Y|)=B \end{align*} $$

shows that the effective divisor $M|_B$ contains a smooth rational curve.

Since $M_X$ is not nef, the divisor $B_X$ is effective by Corollary 5.6. Thus, Proposition 5.11 implies that we have an injection

$$ \begin{align*}H^0(B, {\mathcal O}_B(-K_X)) \hookrightarrow H^1(B, {\mathcal O}_B). \end{align*} $$

By Corollary 2.12, the surface B satisfies the conditions of Proposition 4.4. Thus, the support of $M|_B$ does not contain a smooth rational curve, in contradiction to the first paragraph.

Proof. It is clearly sufficient to show that

$$ \begin{align*}H^1(Y, {\mathcal O}_Y(A - F)) = 0. \end{align*} $$

Since Y has klt singularities and $(Y, B)$ is lc, the pair $(Y, \epsilon B)$ is klt for every $\epsilon < 1$ [Reference Kollár and Mori31, Corollary 2.35(5)]. In particular, $(Y, \frac {1}{m} B)$ is klt. Now we write

$$ \begin{align*}A - F \equiv K_Y + (m-1) F + B = (K_Y + \frac{1}{m} B) + (m-1) (F+\frac{1}{m}B). \end{align*} $$

Since $A \equiv m (F+\frac {1}{m}B)$ and $m>1$ , the $\mathbb {Q}$ -divisor class $(m-1) (F+\frac {1}{m}B)$ is nef and big. Now we conclude with Kawamata-Viehweg vanishing [Reference Debarre11, Theorem 7.26].

Proof. Since Y is a terminal, Gorenstein, $\mathbb {Q}$ -factorial threefold, it is factorial [Reference Kawamata25, Lemma 5.1]. Thus, all the Weil divisors on Y are Cartier.

Consider the exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_Y(M) \rightarrow {\mathcal O}_Y(A) \rightarrow {\mathcal O}_B(A) \rightarrow 0 \end{align*} $$

and the long exact sequence in cohomology

$$ \begin{align*}H^0(Y, {\mathcal O}_Y(A)) \rightarrow H^0(B, {\mathcal O}_B(A)) \rightarrow H^1(Y, {\mathcal O}_Y(M)) \rightarrow H^1(Y, {\mathcal O}_Y(A)). \end{align*} $$

By Kawamata-Viehweg vanishing, we have $H^1(Y, {\mathcal O}_Y(A))=0$ . Since $B \subset \mathrm {Bs}(|A|)$ , the restriction map $H^0(Y, {\mathcal O}_Y(A)) \rightarrow H^0(B, {\mathcal O}_B(A))$ is zero, so we have an isomorphism

(5) $$ \begin{align} H^0(B, {\mathcal O}_B(K_B+M)) \simeq H^0(B, {\mathcal O}_B(A)) \simeq H^1(Y, {\mathcal O}_Y(M)). \end{align} $$

By the Leray spectral sequence, we have an exact sequence

$$ \begin{align*}0 \rightarrow H^1(T, \varphi_* {\mathcal O}_Y(M)) \rightarrow H^1(Y, {\mathcal O}_Y(M)) \rightarrow H^0(T, R^1 \varphi_* {\mathcal O}_Y(M)). \end{align*} $$

By the canonical bundle formula [Reference Ambro2], there exists a boundary divisor $\Delta _T$ on T such that $(T, \Delta _T)$ is klt and $K_T+\Delta _T \sim _{\mathbb {Q}} 0$ . Since $\varphi _* {\mathcal O}_Y(M) \simeq {\mathcal O}_T(M_T)$ is nef and big, we can apply the Kawamata-Viehweg vanishing theorem [Reference Debarre11, Theorem 7.26] to see that

$$ \begin{align*}H^1(T, \varphi_* {\mathcal O}_Y(M)) = 0. \end{align*} $$

By the projection formula, we have

$$ \begin{align*}R^1 \varphi_* {\mathcal O}_Y(M) \simeq R^1 \varphi_* {\mathcal O}_Y \otimes {\mathcal O}_T(M_T), \end{align*} $$

so $R^1 \varphi _* {\mathcal O}_Y(M)$ is torsion-free by [Reference Kollár32, Theorem 2.1].

Since $K_Y \simeq {\mathcal O}_Y$ and $\varphi $ is flat over a big open susbset of T, we have by relative duality [Reference Kleiman30]

$$ \begin{align*}(R^1 \varphi_* {\mathcal O}_Y) \simeq (\varphi_* {\mathcal O}_Y(K_{Y/T}))^* \simeq {\mathcal O}_T(K_T) \end{align*} $$

in the complement of finitely many points. Since $R^1 \varphi _* {\mathcal O}_Y(M)$ is torsion-free, the morphism

$$ \begin{align*}R^1 \varphi_* {\mathcal O}_Y(M) \simeq {\mathcal O}_T(K_T) \otimes {\mathcal O}_T(M_T) \rightarrow {\mathcal O}_T(K_T+M_T) \end{align*} $$

is thus injective and yields an inclusion

$$ \begin{align*}H^0(T, R^1 \varphi_* {\mathcal O}_Y(M)) \rightarrow H^0(T, {\mathcal O}_T(K_T+M_T)). \end{align*} $$

The statement now follows from (5).

Proof. Since $K_S \cdot A=-1$ , the canonical bundle is not pseudoeffective, so S is uniruled. Since $H^1(S, {\mathcal O}_S) \neq 0$ , the surface is not rationally connected, and the MRC fibration coincides with the Albanese map $f: S \rightarrow \text {Alb}(S)$ . In particular, f contracts all the rational curves on S and $\pi _1(S) \simeq \pi _1(C)$ , where C is the image of f. Our goal is to show that $q(S)=1$ and f is a $\mathbb {P}^1$ -bundle.

Since the Cartier divisor A is effective and $A^2=1$ , the divisor A is an integral curve. By the adjunction formula, the arithmetic genus of A is one, so either A is smooth or its normalisation is $\mathbb {P}^1$ . Since the ample divisor A is not contracted by the fibration f, we see that A is a smooth elliptic curve. Now recall that canonical surface singularities are hypersurface singularities, so by Goresky-MacPherson’s Lefschetz theorem for homotopy groups [Reference Lazarsfeld39, Theorem 3.1.21, Remark 3.1.41], the morphism

$$ \begin{align*}\mathbb{Z}^2 \simeq \pi_1(A) \rightarrow \pi_1(S) \end{align*} $$

is surjective; in particular, $q(S)=1$ and C is also an elliptic curve. Since

$$ \begin{align*}\pi_1(A) \rightarrow \pi_1(S) \simeq \pi_1(C) \end{align*} $$

is surjective, the étale map map $A \rightarrow C$ is an isomorphism. Thus, A is a section of f. Since A is ample and has degree one on the fibres, all the f-fibres are integral curves. Thus, f is a $\mathbb {P}^1$ -bundle by [Reference Kollár34, II,Theorem 2.8]. Since $A \cdot f=1$ , we have $S \simeq \mathbb {P}(f_* {\mathcal O}_S(A))$ .

Proof. Let $\tau :Z' \rightarrow Z$ be a terminalisation of Z. Then $-K_{Z'} \simeq \tau ^* (-K_Z)$ is nef and $h^i(Z', {\mathcal O}_{Z'})=h^i(Z, {\mathcal O}_Z)$ for all $i \in \mathbb {N}$ . Since $\tau ^* S \cdot (-K_{Z'})^2 = S \cdot (-K_Z)^2>0$ , the nef divisor $-K_{Z'}$ is not trivial. Thus, $Z'$ is uniruled, and hence, $h^3(Z', {\mathcal O}_{Z'})=0$ . Since $q(Z)=1$ by assumption, we obtain $\chi (Z', {\mathcal O}_{Z'}) \geq 0$ . Since $\tau ^* S$ is nef and big and $-K_{Z'}$ is nef, we have

$$ \begin{align*}H^j(Z', {\mathcal O}_{Z'}(\tau^* S)) = H^j(Z', {\mathcal O}_{Z'}(K_{Z'} + (-K_{Z'}+\tau^* S))) = 0 \qquad \forall \ j \geq 1 \end{align*} $$

by Kawamata-Viehweg vanishing. Thus, the Riemann-Roch formula (Fact 2.6) yields

(6) $$ \begin{align} 1 = h^0(Z', {\mathcal O}_{Z'}(\tau^* S)) &= \chi(Z', {\mathcal O}_{Z'}(\tau^* S)) \\ &\geq \frac{1}{12} \tau^* S \cdot \tau^* (S-K_Z) \cdot \tau^* (2S-K_Z) + \frac{1}{12} \tau^* S \cdot c_2(Z').\nonumber \end{align} $$

Since $-K_{Z'}$ is nef and $\tau ^* S$ is nef, we have $\tau ^* S \cdot c_2(Z') \geq 0$ by [Reference Ou45, Corollary 1.5]. Applying the projection formula, we obtain

$$ \begin{align*}S \cdot (S-K_Z) \cdot (2S-K_Z) \leq 12. \end{align*} $$

Set $A:=S|_S$ and $H=-K_Z|_S$ . Then A is an ample Cartier divisor on S, and H is a nef and big Cartier divisor on S such that $H^2 \geq 2$ . By the Hodge index inequality $(A \cdot H)^2 \geq A^2 \cdot H^2$ [Reference Beltrametti and Sommese8, Proposition 2.5.1], this implies $A \cdot H \geq 2$ . Since

$$ \begin{align*}12 \geq S \cdot (S-K_Z) \cdot (2S-K_Z) = 2 A^2 + 3 A \cdot H + H^2 \geq 4 + 3 A \cdot H, \end{align*} $$

we actually have $A \cdot H=2$ . Moreover, we have $A^2 \leq 2$ .

1st case. Suppose that $A^2=2$ . Then we have equality in the Hodge index inequality, and therefore, $A \equiv H$ by [Reference Beltrametti and Sommese8, Corollary 2.5.4]. Since

$$ \begin{align*}K_S \simeq (K_Z+S)|_S \simeq -H + A, \end{align*} $$

we obtain $K_S \equiv 0$ .

2nd case. Suppose that $A^2=1$ . Since

$$ \begin{align*}K_S \simeq (K_Z+S)|_S \simeq -H + A, \end{align*} $$

we have $K_S \cdot A=-1$ . Finally, we have the Riemann-Roch inequality

$$ \begin{align*}h^0(S, {\mathcal O}_S(A)) \geq \frac{1}{2} A^2 + \frac{1}{2} (-K_S \cdot A) + \chi({\mathcal O}_S). \end{align*} $$

Since S is an ample divisor in a threefold and Z has canonical singularities, we can use Kodaira vanishing to show that $q(S)=q(Z)=1$ . In particular, $\chi ({\mathcal O}_S) \geq 0$ , and by the Riemann-Roch inequality, $h^0(S, {\mathcal O}_S(A))>0$ , so A is an effective divisor. Thus, the polarised surface $(S, A)$ satisfies the conditions of Proposition 6.3.

Setup 7.2. For the proof of Theorem 7.1, we will argue by contradiction and assume that $M_X$ is nef. By the Fact 2.3, there exists a morphism with connected fibres

(7) $$ \begin{align} \varphi:X \rightarrow T \end{align} $$

such that $M_ X \simeq \varphi ^* M_T$ for some ample Cartier divisor $M_T \rightarrow T$ . Note that $B_X$ is $\varphi $ -ample since $B_X \sim _\varphi -K_X$ .

Since X is a Fano, there exists a boundary divisor such that $(X, \Delta )$ is klt and $\Delta \sim _{\mathbb {Q}} -K_X$ . Thus, we apply Ambro’s theorem [Reference Ambro2] to see that T is klt. Note that it is not clear whether T is $\mathbb {Q}$ -factorial.

Proof. Since $-K_X$ is $\varphi $ -ample, we have $R^j \varphi _* {\mathcal O}_X=0$ for every $j \geq 1$ by relative Kodaira vanishing. Since $M_X \simeq \varphi ^* M_T$ , we deduce that $R^j \varphi _* {\mathcal O}_X(-M_X)=0$ for $j \geq 1$ by the projection formula. Thus, we have

$$ \begin{align*}H^1(X, {\mathcal O}_X(-M_X)) \simeq H^1(T, {\mathcal O}_T(-M_T)), \end{align*} $$

and the latter is zero by Kodaira vanishing on the klt space T unless T is a curve. The rest is now straightforward.

Proof. Immediate from Lemma 7.3 and the exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(-M_X) \rightarrow {\mathcal O}_X(B_X) \rightarrow {\mathcal O}_Y(B) \rightarrow 0. \end{align*} $$

Proof. If $\dim T=1$ , the fibration $\varphi $ has as general fibre a smooth Fano threefold F (the divisor $-K_F \simeq -K_X|_F$ is ample) and $T \simeq \mathbb {P}^1$ .

Choose $Y \in |-K_X|$ a general element, and denote by $\psi :Y \rightarrow \mathbb {P}^1$ the restriction of $\varphi $ to Y. Denote by $F_Y$ the general $\psi $ -fibre; that is, $F_Y= F \cap Y$ . We have a commutative diagram

and the horizontal maps are surjective since $q(X)=0=q(F)$ .

1st case. Suppose that ${\mathcal O}_T(M_T) \simeq {\mathcal O}_{\mathbb {P}^1}(m)$ with $m \geq 2$ . We have $-K_X|_Y \simeq \psi ^* M_T + B$ ; moreover, the pair $(Y,B)$ is lc by Corollary 2.12. Thus, by Proposition 6.1, the restriction map $r_Y$ is surjective. Since $-K_X|_Y \simeq \psi ^* M_T + B$ and B is a fixed component of the linear system $|-K_X|_Y|$ , the image of $r_Y$ is generated by a section vanishing on $B \cap F_Y$ . Thus, $r_Y$ is a surjective map of rank one, and therefore, $h^0(F_Y, {\mathcal O}_{F_Y}(-K_X))=1$ . The restriction map $H^0(F, {\mathcal O}_F(-K_X)) \rightarrow H^0(F_Y, {\mathcal O}_{F_Y}(-K_X))$ has kernel $H^0(F, {\mathcal O}_F) \simeq \mathbb {C}$ , so we deduce $h^0(F, {\mathcal O}_F(-K_X))=2$ . Yet $-K_F \simeq -K_X|_F$ , and for a smooth Fano threefold, we always have $h^0(F, {\mathcal O}_F(-K_F)) \geq 3$ by [Reference Iskovskikh and Prokhorov24, Corollary 2.1.14].

2nd case. Suppose that ${\mathcal O}_T(M_T) \simeq {\mathcal O}_{\mathbb {P}^1}(1)$ . By Corollary 7.4, this implies that $B_X$ is an effective divisor. Since $-K_X \simeq \varphi ^* M_T + B_X\simeq F + B_X$ , the divisor $B_X$ is relatively ample. We claim that

$$ \begin{align*}H^1(Y, {\mathcal O}_Y(-K_X|_Y-F_Y)) = H^1(Y, {\mathcal O}_Y(B)) = 0. \end{align*} $$

As in the first case, this yields the surjectivity of $H^0(Y, {\mathcal O}_Y(-K_X)) \rightarrow H^0(F_Y, {\mathcal O}_{F_Y}(-K_X))$ and the desired contradiction.

Proof of the claim. By Proposition 5.1, we know that $B_X$ is not nef, so there exists a $K_X$ -negative extremal ray $\mathbb {R}^+ \gamma $ such that $B_X \cdot \gamma <0$ . Since $B_X$ is effective, the corresponding contraction

$$ \begin{align*}\mu:X \rightarrow X' \end{align*} $$

must be birational with exceptional locus contained in $B_X$ . Since $B_X$ is $\varphi $ -ample and $\mu $ -antiample, the intersection of any $\mu $ -fibre with a $\varphi $ -fibre must be finite. Thus, the fibres of $\mu $ have dimension at most one. By Ando’s theorem [Reference Ando3, Theorem 2.3], this implies that $\mu $ is a smooth blowup along a surface. In particular, we have $(-K_X+B_X) \cdot \gamma =0$ , so $-K_X+B_X$ is non-negative on the extremal ray $\mathbb {R}^+ \gamma $ . Since $\gamma $ was an arbitrary $B_X$ -negative extremal ray, the cone theorem implies that $-K_X+B_X$ is nef. Since $B_X$ is effective, the divisor $-K_X+B_X$ is nef and big. Therefore,

$$ \begin{align*}H^j(X, {\mathcal O}_X(B_X)) = H^j(X, K_X + {\mathcal O}_X(-K_X+B_X)) = 0 \end{align*} $$

for $j \geq 1$ by Kawamata-Viehweg vanishing. Now consider the exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_X(-M_X) \rightarrow {\mathcal O}_X(B_X) \rightarrow {\mathcal O}_Y(B) \rightarrow 0. \end{align*} $$

Since $M_X \simeq \varphi ^* M_T$ and $\dim T=1$ , we have $H^2(X, {\mathcal O}_X(-M_X)) \simeq H^2(T, {\mathcal O}_T(-M_T)) = 0.$ By the long exact sequence in cohomology, the map $0=H^1(X, {\mathcal O}_X(B_X)) \rightarrow H^1(Y,{\mathcal O}_Y(B))$ is surjective, and we are finally done.

Proof. Assume that $\dim T \geq 3$ . Since Y is an ample divisor, the morphism $\varphi |_Y: Y \rightarrow T$ is generically finite onto its image, so $M \simeq \varphi ^* M_T|_Y$ is nef and big. Consider the exact sequence

$$ \begin{align*}0 \rightarrow {\mathcal O}_Y(M) \rightarrow {\mathcal O}_Y(-K_X) \rightarrow {\mathcal O}_B(-K_X) \rightarrow 0. \end{align*} $$

Since $H^1(Y, {\mathcal O}_Y(M))=0$ , by Kawamata-Viehweg vanishing, the restriction morphism

$$ \begin{align*}H^0(Y, {\mathcal O}_Y(-K_X)) \rightarrow H^0(B, {\mathcal O}_B(-K_X)) \end{align*} $$

is surjective. Since B is in the base locus of $|-K_X|_Y|$ , we obtain $H^0(B, {\mathcal O}_B(-K_X))=0$ .

Since $\dim\ T \geq 3$ , we have $h^0(X, {\mathcal O}_X(B_X)) \neq 0$ by Corollary 7.4. Thus, Proposition 5.11 shows that we have an injection

$$ \begin{align*}H^0(B, {\mathcal O}_B(-K_X)) \hookrightarrow H^1(B, {\mathcal O}_B). \end{align*} $$

By Corollary 2.12, the surface B satisfies the conditions of Proposition 4.2. Thus, we have $q(B)>0$ , and the inclusion $H^0(B, {\mathcal O}_B(-K_X)) \hookrightarrow H^1(B, {\mathcal O}_B)$ is an equality. Since the first space has dimension zero, this is a contradiction.

Proof. Assume that $\dim T=2$ . By Corollary 7.4, the divisor $B_X$ is effective. By Corollary 5.12, the divisor $B_X$ has canonical Gorenstein singularities, and by Proposition 5.11, we have

$$ \begin{align*}h^0(B_X, {\mathcal O}_{B_X}(-K_X))=1 \quad \text{and} \quad q(B_X)=q(B). \end{align*} $$

Since $Y \subset X$ is an ample divisor, the fibration $\varphi $ induces an elliptic fibration

$$ \begin{align*}\psi:Y \rightarrow T \end{align*} $$

such that $-K_X|_Y \simeq \psi ^* M_T + B$ . By Theorem 2.8 and Corollary 2.9, the threefold Y satisfies the conditions of Lemma 6.2. Hence, we have an injection

(8) $$ \begin{align} H^0(B, {\mathcal O}_B(K_B+M)) \hookrightarrow H^0(T, {\mathcal O}_T(K_T+M_T)). \end{align} $$

Observe that the induced morphism $\varphi |_{B_X}$ is surjective onto T since $B_X$ is $\varphi $ -ample.

Step 1. We show that $h^0(T, {\mathcal O}_T(K_T+M_T))=0$ unless $K_T \simeq -M_T$ . By the adjunction formula,

$$ \begin{align*}-K_{B_X} \simeq M_X|_{B_X} \simeq (\varphi^* M_T)|_{B_X} \end{align*} $$

is nef with numerical dimension two. Since $B_X$ has canonical singularities, we can apply [Reference Fujino and Gongyo13, Theorem 3.1] to see that there exists a boundary divisor $\Delta _T$ on T such that $(T, \Delta _T)$ is klt and

$$ \begin{align*}K_{B_X} \sim_{\mathbb{Q}} (\varphi^* (K_T+\Delta_T))|_{B_X}. \end{align*} $$

Thus, we have $K_T+\Delta _T \equiv -M_T$ , and hence, $K_T+M_T=-\Delta _T$ is not pseudoeffective unless $\Delta _T=0$ . Therefore, $h^0(T, {\mathcal O}_T(K_T+M_T)) =0$ unless $\Delta _T=0$ . In the latter case, $h^0(T, {\mathcal O}_T(K_T+M_T)) \neq 0$ implies that $K_T \simeq -M_T$ ; in particular, $K_T$ is Cartier.

Step 2. We reach a contradiction. By Proposition 5.11, we have an injection

$$ \begin{align*}H^0(B, {\mathcal O}_B(-K_X)) \hookrightarrow H^1(B, {\mathcal O}_B). \end{align*} $$

By Corollary 2.12, the surface B satisfies the conditions of Proposition 4.2. In particular, B has canonical singularities, positive irregularity $q(B)>0$ , and the inclusion above is an equality. Since

$$ \begin{align*}H^0(B, {\mathcal O}_B(-K_X)) = H^0(B, {\mathcal O}_B(K_B+M)), \end{align*} $$

the injection (8) and the first step show that

$$ \begin{align*}H^0(B, {\mathcal O}_B(-K_X)) = H^1(B, {\mathcal O}_B) = H^0(T, {\mathcal O}_T(K_T+M_T) = \mathbb{C}. \end{align*} $$

In particular, $K_T \simeq -M_T$ by the first step (i.e., T is a del Pezzo surface with at most canonical singularities).

Since B is the complete intersection of $B_X$ and Y, the surface $B \subset B_X$ is an ample Cartier divisor, and the restriction of $\varphi |_{B_X}$ to B is still surjective onto the surface T. Since $q(T)=0$ and $q(B)=1$ , and the irregularity is a birational invariant of varieties with rational singularities, the map

$$ \begin{align*}\tau:= \varphi|_T: B \rightarrow T \end{align*} $$

is not birational. Thus, $\tau $ is generically finite of degree at least two and

$$ \begin{align*}(M_X|_B)^2 = (\tau^* M_T)^2 = \deg \tau \cdot M_T^2 \geq 2. \end{align*} $$

Thus, $q(B_X)=q(B)=1$ , $-K_{B_X} \simeq M_X|_{B_X}$ is nef, $B\subset B_X$ is an ample Cartier divisor with $h^0(B_X, {\mathcal O}_{B_X}(B))=h^0(B_X, {\mathcal O}_{B_X}(-K_X))=1$ and $B\cdot (-K_{B_X})^2\geq 2$ . Thus, the threefold $B_X$ and the surface $B \subset B_X$ satisfy the conditions of Proposition 6.5, and we have two cases:

Case a) We have $K_B \equiv 0$ . Since $K_B \simeq B|_B$ , this implies that B is nef. Thus, $B_X$ is nef by Theorem 2.10, a contradiction to Proposition 5.1.

Case b) B is a ruled surface over an elliptic curve. Since $\rho (\mathbb {P}(V))=2$ and the map $\tau $ is surjective, the del Pezzo surface T has Picard number at most two (canonical singularities are $\mathbb {Q}$ -factorial, [Reference Kollár and Mori31, Proposition 4.11]). By Remark 6.4, the nef and big class $M_X|_B$ is ample and $(M_X|_B)^2=3$ . Since $M_X|_B \simeq \tau ^* M_T \simeq \tau ^* (-K_T)$ is ample, the generically finite map $\tau $ is actually finite. Moreover, $\tau $ having degree at least two, we deduce that $\tau $ has degree three and $(-K_T)^2=1$ . Thus, T is a del Pezzo surface of degree one and Picard number at most two; in particular, it is not smooth. By [Reference Andreatta and Wiśniewski4, Proposition 1.3], this implies that the fibration $\varphi :X \rightarrow T$ is not equidimensional, so there exists a prime divisor $D \subset X$ that is contracted onto a point in T.

Since the effective divisors $B_X$ and Y are $\varphi $ -ample, the intersection

$$ \begin{align*}D \cap B_X \cap Y = D \cap B \end{align*} $$

is non-empty. Hence, there exists a curve $E \subset B$ that is contracted by $\tau $ . Yet we showed above that $\tau $ is finite, a contradiction.

Proof of Theorem 1.3.

Assume that a general anticanonical divisor $Y \in |-K_X|$ is $\mathbb {Q}$ -factorial, so we satisfy the Assumption 1.4 from the Setup 1.2. By Theorem 5.13, the divisor $M_X$ is nef, yet this contradicts Theorem 7.1.