Rationality of $\mathbb{Q}$-Fano threefolds of large Fano index

TL;DR

This paper proves that all $Q$-Fano threefolds with Fano index at least 8 are rational, establishing a significant classification result in algebraic geometry.

Contribution

It demonstrates the rationality of $Q$-Fano threefolds with large Fano index, a previously unresolved classification problem.

Findings

All $Q$-Fano threefolds with Fano index ≥ 8 are rational.

Provides a classification criterion based on Fano index.

Advances understanding of the structure of Fano threefolds.

Abstract

We prove that $\mathbb{Q}$-Fano threefolds of Fano index $\ge 8$ are rational.

Equations

$$q_{\mathrm{W}}(X)$$

$${\mathrm{q}}_{\mathbb{Q}}(X)$$

$${\mathrm{q}}_{\mathbb{Q}}(X)\in \{1,\dots ,11,13,17,19\}$$

$$K_X+qA\sim (1+qt)\Xi .$$

$$f:(\tilde{Y}\supset E)\to (Y\ni P)$$

$$K_{\tilde{Y}}=f^{\ast }{K}_Y+\alpha E.$$

$$\mathrm{g}(X^{\prime \prime })\ge \mathrm{g}(X^{\prime }).$$

$$(q,n)=(3,3)\text{or}(4,2).$$

$$0<-K_X\cdot c_2(X)=24-\sum _{P\in \mathbf{B}}\frac{r_P-1}{r_P}.$$

$$A^{3}=\frac{12}{(q-1)(q-2)}\Bigl{(}1-\frac{A\cdot c_{2}(X)}{12}+\sum_{P\in B}c_{P}(-A)\Bigr{)}$$

$$\left(4q^2-3q\right)A^3\le -4K_X\cdot c_2(X).$$

$$\chi (tA+s\Xi )=h^0(tA+s\Xi )=0.$$

$$\alpha \le \{\begin{array}{ll}1& \text{if Cl(Y)t is of order 2},\\ 2/9& \text{if Cl(Y)t is of order 3}.\end{array}$$

$$\begin{array}[]{lll}K_{\tilde{X}}&\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}&f^{*}K_{X}+\alpha E,\\[2.0pt] \tilde{\mathscr{M}}&\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}&f^{*}{\mathscr{M}}-\beta E.\end{array}$$

$$\tilde{X}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$${\bar{\mathcal{M}}}_k{\sim }_{\mathbb{Q}}{f}^{\ast }{\mathcal{M}}_k-{\beta }_{k}E.$$

$$\hat{E}{\sim }_{\mathbb{Q}}e\Theta ,{\hat{\mathcal{M}}}_k{\sim }_{\mathbb{Q}}{s}_k\Theta ,$$

$$\dim |-K_{\hat{X}}|\ge \dim |-K_{\bar{X}}|=\dim |-K_{\tilde{X}}|\ge \dim |-K_X|.\text{qed}$$

$$\mathrm{Cl}(\hat{X}{)}_{\mathrm{t}}\simeq \mathbb{Z}/n\mathbb{Z},\text{where n=d/e.}$$

$$-K_{\bar{X}}{|}_{\bar{F}}=-K_{\bar{F}}\sim \hat{q}\Lambda ,\bar{E}{|}_{\bar{F}}\sim e\Lambda ,{\bar{\mathcal{M}}}_k{|}_{\bar{F}}\sim s_k\Lambda ,$$

$$k{K}_{\tilde{X}}+q{\tilde{\mathcal{M}}}_k\sim f^{\ast }(k{K}_X+q{\mathcal{M}}_k)+(k\alpha -q{\beta }_k)E\sim (k\alpha -q{\beta }_k)E$$

$$k\hat{q}=q{s}_k+(q{\beta }_k-k\alpha )e.$$

$$K_{\bar{X}}{\sim }_{\mathbb{Q}}{\bar{f}}^{\ast }{K}_{\hat{X}}+b\bar{F},{\bar{\mathcal{M}}}_k{\sim }_{\mathbb{Q}}{\bar{f}}^{\ast }{\hat{\mathcal{M}}}_k-{\gamma }_k\bar{F},\bar{E}{\sim }_{\mathbb{Q}}{\bar{f}}^{\ast }\hat{E}-\delta \bar{F}.$$

$$\begin{array}[]{lll}s_{k}K_{\bar{X}}+\hat{q}\bar{\mathscr{M}}_{k}&\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}&(bs_{k}-\hat{q}\gamma_{k})\bar{F},\\[2.0pt] eK_{\bar{X}}+\hat{q}\bar{E}&\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}&(be-\hat{q}\delta)\bar{F}.\end{array}$$

$$-q{s}_k+\hat{q}k$$

$$-q$$

$$\mathbf{B}(X)=(2,2,3,r),$$

$$r=5,A^3=1/15,\mathrm{g}(X)=11;$$

$$r=12,A^3=1/12,\mathrm{g}(X)=13.$$

$$\dim |kA|=k-1\text{for k=1, 2, 3.}$$

$$t=\{\begin{array}{ll}4& \text{if r=5, }\\ 9& \text{if r=12}.\end{array}$$

$${\beta }_3\ge t\alpha .$$

Full text

The author was partially supported by the HSE University Basic Research Program, Russian Academic Excellence Project ’5-100’.

1. Introduction

Recall that a projective algebraic variety $X$ called $\mathbb{Q}$-Fano if it has only terminal $\mathbb{Q}$-factorial singularities, $\operatorname{Pic}(X)\simeq\mathbb{Z}$, and the anticanonical divisor $-K_{X}$ is ample. $\mathbb{Q}$-Fano varieties plays a very important role in the higher dimensional geometry since they appears naturally in the minimal model program as building blocks in so-called Mori fiber spaces. It is known that $\mathbb{Q}$-Fano varieties of given dimension are bounded, i.e. they form an algebraic family [Ka92], [Bi16]. Moreover, the method of [Ka92] allows to produce a finite but very huge list of numerical candidates (Hilbert series) of $\mathbb{Q}$-Fanos [GRD]. In dimension three there are a lot of classificational results of $\mathbb{Q}$-Fanos of special types (see e.g. [Sa96], [Su04], [Ta06], [Pr10], [BKR], [PR16]) but the full classification is very far from being complete.

An important invariant of a $\mathbb{Q}$-Fano variety $X$ is its $\mathbb{Q}$-Fano index $\operatorname{q_{\mathbb{Q}}}(X)$ which is the maximal integer $q$ such that $-K_{X}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}qA$ for some integral Weil divisor $A$, where $\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}$ defines the $\mathbb{Q}$-linear equivalence. In this paper we prove the following.

1.1 Theorem**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)\geq 8$. Then $X$ is rational.

Note that in some sense our result is optimal: according to [Ok19] a very general weighted hypersurface $X_{14}\subset\mathbb{P}(2,3,4,5,7)$ is a non-rational (and even non-stably rational) $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)=7$. On the other hand, the result of Theorem 1.1 can be essentially improved. We hope that non-rational $\mathbb{Q}$-Fano threefolds of large indices admit a reasonable classification.

The structure of the paper is as follows. Section 2 is preliminary. In Section 3 we list certain kinds of $\mathbb{Q}$-Fano threefolds with torsions in the Weil divisor class group $\operatorname{Cl}(X)$. In Section 4 the main birational construction is introduced. The proof of the main theorem is given in Sections 5-9 by case by case analysis.

2. Preliminaries

We work over the complex number field $\mathbb{C}$ throughout.

2.1. Notation

• •

$\operatorname{Cl}(X)$ denotes the Weil divisors class group of a normal variety;

• •

$\operatorname{Cl}(X)_{\mathrm{t}}$ denotes the torsion part of $\operatorname{Cl}(X)$;

• •

${\mathbf{B}}(X)$ is the basket of a terminal threefold $X$ [Re87];

• •

$\mathrm{{r}}(X,P)$ is the singularity index of a terminal point $P\in X$;

• •

${\mathrm{g}}(X):=\dim|-K_{X}|-1$ is the genus of a $\mathbb{Q}$-Fano threefold $X$.

For a $\mathbb{Q}$-Fano threefold $X$ we define its Fano and $\mathbb{Q}$-Fano index by:

[TABLE]

where $\sim$ (resp. $\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}$) is the linear (resp. $\mathbb{Q}$-linear) equivalence. Clearly, $\operatorname{q}_{\mathrm{W}}(X)$ divides $\operatorname{q_{\mathbb{Q}}}(X)$, and $\operatorname{q}_{\mathrm{W}}(X)=\operatorname{q_{\mathbb{Q}}}(X)$ unless $K_{X}+qA\in\operatorname{Cl}X$ is a nontrivial torsion element. Throughout this paper, for a $\mathbb{Q}$-Fano threefold $X$, by $A$ we denote a Weil divisor such that $-K_{X}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}\operatorname{q_{\mathbb{Q}}}(X)A$. If $\operatorname{q_{\mathbb{Q}}}(X)=\operatorname{q}_{\mathrm{W}}(X)$ we take $A$ so that $-K_{X}\sim\operatorname{q}_{\mathrm{W}}(X)A$.

2.2 Theorem** ([Su04]).**

Let $X$ be a $\mathbb{Q}$-Fano threefold. Then

[TABLE]

and all the possibilities do occur.

The following easy observation will be used freely.

2.3 Lemma** ([Ka88, Lemma 5.1]).**

Let $(X\ni P)$ be a threefold terminal singularity and let $\operatorname{Cl}^{\mathrm{sc}}(X,P)$ be the subgroup of the (analytic) Weil divisor class group consisting of . Weil divisor classes which are $\mathbb{Q}$-Cartier. Then the group $\operatorname{Cl}^{\mathrm{sc}}(X,P)$ is cyclic of order $\mathrm{{r}}(X,P)$ and is generated by the canonical class $K_{X}$.

2.4 Lemma**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold and let $\mathrm{{r}}(X)$ be the global Gorenstein index of $X$. Then the equality $\operatorname{q_{\mathbb{Q}}}(X)=\operatorname{q}_{\mathrm{W}}(X)$ holds if and only if $\operatorname{q_{\mathbb{Q}}}(X)$ and $\mathrm{{r}}(X)$ are coprime.

Proof.

The “only if” part of the statement immediately follows from Lemma 2.3 (see [Su04, Lemma 1.2(3)]). Let us prove the “if” part. So, we assume that $\gcd(\operatorname{q_{\mathbb{Q}}}(X),\mathrm{{r}}(X))=1$. Put $q:=\operatorname{q_{\mathbb{Q}}}(X)$ and write $-K_{X}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}qA^{\prime}$, where $A^{\prime}$ is a Weil divisor. Then $\Xi:=K_{X}+qA^{\prime}$ is a torsion element in $\operatorname{Cl}(X)$. Take $A=A^{\prime}+t\Xi$, $t\in\mathbb{Z}$. Then

[TABLE]

Since the order of $\Xi$ in $\operatorname{Cl}(X)$ divides $\mathrm{{r}}(X)$, there exists $t\in\mathbb{Z}$ such that $(1+qt)\Xi\sim 0$. ∎

The following proposition a consequence of the classification of $\mathbb{Q}$-Fano threefolds of large degree (see [Pr07], [Pr10], [Pr13]).

2.5 Proposition**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)=\operatorname{q}_{\mathrm{W}}(X)\geq 3$. Assume that $X$ is not rational. Then $X$ belongs to one of the following classes below.

[TABLE]

Proof.

Given $q$, $\mathbb{Q}$-Fano threefolds $X$ with $\operatorname{q_{\mathbb{Q}}}(X)=q$ and genus ${\mathrm{g}}(X)\geq{\mathrm{g}}_{q}$ are completely described in [Pr10], [Pr13], [Pr16], where the number ${\mathrm{g}}_{q}$ is given by the third column in the table. It is easy to see that all these varieties are rational. The rest can be checked by a computer search as explained in [Su04], [Pr10, Lemma 3.5] or [PR16, 2.4] (see also [GRD]). ∎

2.6 Proposition** ([Ka96], [Kaw05]).**

Let $Y\ni P$ be a threefold terminal point of index $r>1$ and let

[TABLE]

be a divisorial Mori extraction, where $E$ is the exceptional divisor and $f(E)=P$. Write

[TABLE]

Then the following assertions hold.

1. (i)

If $Y\ni P$ is cyclic quotient singularity of type $\frac{1}{r}(1,a,r-a)$, then $\alpha=1/r$ and $f$ is a weighted blowup with weights $(1,a,r-a)$. 2. (ii)

If $Y\ni P$ is a point of type other than $\mathrm{cA}/r$ and $r>2$, then $\alpha=1/r$. 3. (iii)

If $Y\ni P$ is of type $\mathrm{cA}/r$ and its basket ${\mathbf{B}}(Y,P)$ consists of $m$ points of index $r$, then $\alpha=a/r$, where $m\equiv 0\mod a$.

3. $\mathbb{Q}$-Fano threefolds with torsion in the divisor class group

3.1.

Let $X$ be a $\mathbb{Q}$-Fano threefold and let $\Xi\in\operatorname{Cl}(X)_{\mathrm{t}}$ be a non-trivial torsion element of order $n$. Then $\Xi$ defines a finite étale in codimension two cover $\pi:X^{\prime}\to X$ such that $X^{\prime}$ has only terminal singularities, $K_{X^{\prime}}=\pi^{*}K_{X}$ and $\pi^{*}\Xi=0$ (see [Re87, 3.6]). Clearly, $X^{\prime}$ is a Fano variety. However, in general, we cannot say that $X^{\prime}$ is $\mathbb{Q}$-factorial neither $\operatorname{Pic}(X^{\prime})\simeq\mathbb{Z}$. Let $q:=\operatorname{q_{\mathbb{Q}}}(X)$. Take $A$ so that $-K_{X}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}qA$ and let $A^{\prime}:=\pi^{*}A$. Then $-K_{X^{\prime}}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}qA^{\prime}$. Hence, $\operatorname{q_{\mathbb{Q}}}(X^{\prime})$ is divisible by $q$.

3.1.1 Remark**.**

In the above notation, assume that $q\geq 5$. Run the MMP on $X^{\prime}$. On each step the relation $-K_{X^{\prime}}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}qA^{\prime}$ is preserved. Therefore, at the end we obtain a $\mathbb{Q}$-Fano threefold $X^{\prime\prime}$ such that $-K_{X^{\prime\prime}}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}qA^{\prime\prime}$, where $q\geq 5$. Then by (2.2.1) we have $\operatorname{q_{\mathbb{Q}}}(X^{\prime\prime})=q$ and so $\operatorname{q_{\mathbb{Q}}}(X^{\prime})=q$. Moreover,

[TABLE]

3.2 Proposition**.**

Notation as in * 3.1. Assume that $q\geq 3$ and $q\neq\operatorname{q}_{\mathrm{W}}(X)$. Take $\Xi:=K_{X}+qA$. Then*

[TABLE]

Proof.

As in Proposition 2.5 we use a computer search. But in this case the algorithm should be modified as follows (cf. [Ca08]). For short, we denote $r_{P}:=\mathrm{{r}}(X,P)$. Let $r:=\operatorname{lcm}(\{r_{P}\})$ be the global Gorenstein index of $X$.

Step 1

By [Ka92] we have the inequality

[TABLE]

This produces a finite (but huge) number of possibilities for the basket ${\mathbf{B}}(X)$ and the number $-K_{X}\cdot c_{2}(X)$.

Step 2.

(2.2.1) implies that $q\in\{3,\dots,11,13,17,19\}$. In each case we compute $A^{3}$ by the formula

[TABLE]

(see [Su04]), where $c_{P}$ is the correction term in the orbifold Riemann-Roch formula [Re87]. The number $rA^{3}$ must be a positive integer [Su04, Lemma 1.2].

Step 3.

Next, by [Su04, Prop. 2.2] the Bogomolov–Miyaoka inequality (see [Ka92]) implies that

[TABLE]

Step 4.

In a neighborhood of each point $P\in X$ we can write $A\sim l_{P}K_{X}$ by Lemma 2.3, where $0\leq l_{P}<r_{P}$. There is a finite number of possibilities for the collection $\{(l_{P})\}$.

Step 5.

The number $n$ is determined as minimal positive such that $\chi(n\Xi)=1$ (by the Kawamata–Viehweg vanishing). Hence, $n$ can be computed by using orbifold Riemann-Roch.

Step 6.

Finally, applying Kawamata–Viehweg vanishing we obtain

[TABLE]

for $-q<t<0$ and $0\leq s<n$. Again, we check this condition using orbifold Riemann-Roch.

To run this algorithm the author used the computer algebra system PARI/GP [PARI]. As the result, we get a short list from which one can see that (3.2.1) holds. ∎

3.3 Proposition**.**

Notation as in * 3.1. Assume that $q\geq 5$ and $\operatorname{Cl}(X)_{\mathrm{t}}$ contains an element $\Xi$ of order $n\geq 2$. Then $n\leq 3$, $\operatorname{q_{\mathbb{Q}}}(X)=q$, and one of the following holds:*

[TABLE]

Moreover, the group $\operatorname{Cl}(X)_{\mathrm{t}}$ is cyclic and generated by $\Xi$.

Proof.

Similar to Proposition 3.2. But in this case, $\operatorname{q_{\mathbb{Q}}}(X)=\operatorname{q}_{\mathrm{W}}(X)$ and we have to modify one step:

Step 4′.** In this case $\gcd(q,r)=1$ by Lemma 2.4. Since $K_{X}+qA\sim 0$, the numbers $l_{P}$ are uniquely determined by $1+ql_{P}\equiv 0\mod r_{P}$. But for $\Xi$ there are several choices. Again, near each point $P\in X$ we can write $\Xi\sim k_{P}K_{X}$ by Lemma 2.3, where for the collection $\mathbf{k}=(\{k_{P}\})$ there are only a finite number of possibilities.

We obtain a list $\{(n,q,{\mathbf{B}}(X),{\mathrm{g}}(X),A^{3},\mathbf{k})\}$. In each case we compute the basket ${\mathbf{B}}(X^{\prime})$ of a (terminal) Fano threefold $X^{\prime}$ with $A^{\prime 3}=nA^{3}$. By Remark 3.1.1 we have $\operatorname{q_{\mathbb{Q}}}(X^{\prime})=q$. Then we can compute ${\mathrm{g}}(X^{\prime})$ by orbifold Riemann-Roch. At the end we get the list in the table and several extra possibilities which do not occur because ${\mathrm{g}}(X^{\prime})\leq 32$ in the case $\operatorname{q_{\mathbb{Q}}}(X^{\prime})=5$ by [Pr13, Th. 1.2(v)] and Remark 3.1.1. ∎

We do not assert that all the possibilities in Proposition 3.3 occur. We are able only to provide several examples for 3.3, 3.3-3.3.

3.4 Examples**.**

The following quotient of weighted hypersurfaces are $\mathbb{Q}$-Fano threefolds as in 3.3, 3.3-3.3.

• 3.3

$\{x_{1}^{6}+x_{2}^{3}+x_{2}^{\prime 3}+x_{3}x_{3}^{\prime}=0\}\subset\mathbb{P}(1,2^{2},3^{2})/{\boldsymbol{\mu}}_{3}(0,1,-1,1,-1)$;

• 3.3

$\{x_{1}^{6}+x_{1}^{\prime 6}+x_{2}x_{4}+x_{3}^{2}=0\}\subset\mathbb{P}(1^{2},2,3,4)/{\boldsymbol{\mu}}_{2}(0,1,1,1,1)$;

• 3.3

$\{x_{1}^{4}+x_{1}^{\prime 4}+x_{1}x_{3}+x_{2}^{2}+x_{2}^{\prime 2}=0\}\subset\mathbb{P}(1^{2},2^{2},3)/{\boldsymbol{\mu}}_{2}(0,1,1,1,0)$;

• 3.3

$\{x_{1}^{8}+x_{2}^{4}+x_{3}x_{5}+x_{4}^{2}=0\}\subset\mathbb{P}(1,2,3,4,5)/{\boldsymbol{\mu}}_{2}(0,1,1,1,1)$;

• 3.3

$\{x_{1}^{6}+x_{2}x_{4}+x_{3}^{2}+x_{3}^{\prime 2}=0\}\subset\mathbb{P}(1,2,3^{2},4)/{\boldsymbol{\mu}}_{2}(0,1,0,1,1)$.

One can expect also that the variety 3.3 is a quotient of a codimension four $\mathbb{Q}$-Fano (see [GRD, No. 41418] and [CD18, § 5.4]).

Using the orbifold Riemann-Roch one can compute dimensions of linear systems on $X$:

3.5 Corollary**.**

In the cases * 3.3 and * 3.3* of Proposition * 3.3* the dimension of the linear systems $|kA+s\Xi|$ are as follows*

[TABLE]

Combining 3.3 and 2.6 we obtain.

3.6 Corollary**.**

Let $Y$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)\geq 5$. Assume that $\operatorname{Cl}(Y)_{\mathrm{t}}\neq 0$. Let $P\in Y$ be a non-Gorenstein point and let $f$ be a divisorial Mori extraction of $P$. Then for the discrepancy $\alpha$ of the exceptional divisor $E\subset\tilde{Y}$ we have

[TABLE]

4. Main construction

4.1.

Let $X$ be a $\mathbb{Q}$-Fano threefold. For simplicity, we assume that the group $\operatorname{Cl}(X)$ is torsion free (this is the only case that we need in this paper). Denote $q=\operatorname{q_{\mathbb{Q}}}(X)=\operatorname{q}_{\mathrm{W}}(X)$. Thus $-K_{X}\sim qA$ and $A$ is the ample generator of the group $\operatorname{Cl}(X)\simeq\mathbb{Z}$.

Consider a non-empty linear system ${\mathscr{M}}$ on $X$ without fixed components. Let $c=\operatorname{ct}(X,{\mathscr{M}})$ be the canonical threshold of the pair $(X,{\mathscr{M}})$. Consider a log crepant blowup $f:\tilde{X}\to X$ with respect to $K_{X}+c{\mathscr{M}}$. One can choose $f$ so that $\tilde{X}$ has only terminal $\mathbb{Q}$-factorial singularities, i.e. $f$ is a divisorial extraction in the Mori category (see [Co95], [Al94]). Let $E$ be the exceptional divisor. Write

[TABLE]

where $\alpha,\,\beta\in\mathbb{Q}_{\geq 0}$, and $\tilde{\mathscr{M}}$ is the birational transform of ${\mathscr{M}}$. Then $c=\alpha/\beta$.

4.1.2 Lemma** (see [Pr10, Lemma 4.2]).**

Let $P\in X$ be a point of index $r>1$. In a neighborhood of $P$ we can write $\mathscr{M}\sim-tK_{X}$, where $0<t<r$. Then $c\leq 1/t$ and so $\beta\geq t\alpha$.

Assume that the log divisor $-(K_{X}+c{\mathscr{M}})$ is ample. Run the log minimal model program with respect to $K_{\tilde{X}}+c\tilde{\mathscr{M}}$. We obtain the following diagram (Sarkisov link, see [Al94], [Pr10], [Pr16])

[TABLE]

Here $\chi$ is a composition of $K_{\tilde{X}}+c\tilde{\mathscr{M}}$-log flips, the variety $\bar{X}$ has only terminal $\mathbb{Q}$-factorial singularities, $\uprho(\bar{X})=2$, $\uprho(\hat{X})=1$, and $\bar{f}:\bar{X}\to\hat{X}$ is an extremal $K_{\bar{X}}$-negative Mori contraction. In what follows, for the divisor (or linear system) $D$ on $X$ by $\tilde{D}$ and $\bar{D}$ we denote proper transforms of $D$ on $\tilde{X}$ and $\bar{X}$ respectively.

If $|kA|\neq\emptyset$, we put ${\mathscr{M}}_{k}:=|kA|$ (is it possible that ${\mathscr{M}}_{k}$ has fixed components in general). If $\dim{\mathscr{M}}_{k}=0$, then by $M_{k}$ we denote a unique effective divisor $M_{k}\in{\mathscr{M}}_{k}$. As in (4.1.1), we write

[TABLE]

4.2.

Assume that the contraction $\bar{f}$ is birational. Then $\hat{X}$ is a $\mathbb{Q}$-Fano threefold. In this case, we denote by $\bar{F}$ the $\bar{f}$-exceptional divisor, by $\tilde{F}\subset\tilde{X}$ its proper transform, $F:=f(\tilde{F})$, and $\hat{q}:=\operatorname{q_{\mathbb{Q}}}(\hat{X})$. Again we denote by $\hat{D}$ the proper (birational) transform of an object $D$ (resp. $\tilde{D}$, $\bar{D}$) on $X$ (resp. $\tilde{X}$, $\bar{X}$). Let $\Theta$ be an ample Weil divisor on $\hat{X}$ generating $\operatorname{Cl}(\hat{X})/\operatorname{Cl}(\hat{X})_{\mathrm{t}}$. Write

[TABLE]

where $e\in\mathbb{Z}_{>0}$, $s_{k}\in\mathbb{Z}_{\geq 0}$. If $\dim{\mathscr{M}}_{k}=0$ and $\bar{M}_{k}=\bar{F}$ (i.e. a unique element $M_{k}$ of the linear system $\bar{\mathscr{M}}_{k}$ is the $\bar{f}$-exceptional divisor), we put $s_{k}=0$.

4.2.1 Lemma**.**

If in the above notation $\alpha<1$, then ${\mathrm{g}}(\hat{X})\geq{\mathrm{g}}(X)$.

Proof.

We have $a(E,|{-}K_{X}|)<1$. On the other hand, $0=K_{X}+|{-}K_{X}|$ is Cartier. Hence, $a(E,|{-}K_{X}|)\leq 0$ and $K_{\tilde{X}}+f^{-1}_{*}|{-}K_{X}|$ is linearly equivalent to a non-positive multiple of $E$. Therefore, $f^{-1}_{*}|{-}K_{X}|\subset|{-}K_{\tilde{X}}|$ and so

[TABLE]

Note that in general, the group $\operatorname{Cl}(\hat{X})$ can have torsions:

4.2.2 Lemma** (see [Pr10, Lemma 4.12]).**

Write $F\sim dA$. Then

[TABLE]

4.3.

Assume that the contraction $\bar{f}$ is not birational. In this case, $\operatorname{Cl}(\hat{X})$ has no torsion. Therefore, $\operatorname{Cl}(\hat{X})\simeq\mathbb{Z}$. Denote by $\Theta$ the ample generator of $\operatorname{Cl}(\hat{X})$ and by $\bar{F}$ a general geometric fiber. Then $\bar{F}$ is either a smooth rational curve or a del Pezzo surface. The image of the restriction map $\operatorname{Cl}(\bar{X})\to\operatorname{Pic}(\bar{F})$ is isomorphic to $\mathbb{Z}$. Let $\Lambda$ be its ample generator. As above, we can write

[TABLE]

where $\hat{q},\,e\in\mathbb{Z}_{>0}$, $s_{k}\in\mathbb{Z}_{\geq 0}$.

If $\hat{X}$ is a curve, then $\hat{q}\leq 3$ and $\hat{X}\simeq\mathbb{P}^{1}$. If $\hat{X}$ is a surface, then $\hat{q}\leq 2$. In this case, $\hat{X}$ can have only Du Val singularities of type $\mathrm{A_{n}}$ [MP08, Theorem 1.2.7].

4.3.1 Lemma**.**

If the contraction $\bar{f}$ is not birational and $\hat{q}>1$, then $X$ is rational.

Proof.

Indeed, if $\hat{X}$ is a curve and $\hat{q}\geq 2$, then a general fiber $\bar{F}$ is a del Pezzo surface with divisible canonical class. Then $\bar{F}$ is either a projective plane or a quadric. Clearly, $\bar{X}$ is rational in this case. Similarly, if $\hat{X}$ is a surface and $\hat{q}=2$, then there is a divisor which is a generically section of $\bar{f}$ and $\bar{X}$ is again rational. ∎

4.4.

Since the group $\operatorname{Cl}(\bar{X})$ has no torsion, the numerical equivalence of Weil divisors on $\bar{X}$ coincides with linear one. Hence the relations (4.1.1) and (4.1.4) give us

[TABLE]

where $k\alpha-q\beta_{k}\in\mathbb{Z}$. From this we obtain the following important equality which will be used throughout this paper:

[TABLE]

4.5.

Suppose that the morphism $\bar{f}$ is birational. Similar to (4.1.1) and (4.1.4) we can write

[TABLE]

This gives us

[TABLE]

Taking proper transforms of these relations to $X$, we obtain

[TABLE]

4.5.3 Corollary**.**

If, in the above notation, $\gcd(n,q)=1$, then $\bar{f}(\bar{F})$ is a point on $\hat{X}$ whose index is divisible by $n$.

Proof.

Indeed, either the discrepancy $b$ of $\bar{F}$ or the multiplicity $\delta$ is fractional and its denominator is divisible by $n$ according to (4.5.2). ∎

5. $\mathbb{Q}$-Fano threefolds of Fano index $7$ and large genus

Now we apply the techniques outlined in the previous section to $\mathbb{Q}$-Fano threefolds of indices $\geq 7$. The following result will be used in subsequent sections.

5.1 Proposition**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)=7$ and ${\mathrm{g}}(X)\geq 11$. Then $X$ is rational.

Proof.

By Proposition 3.3 the group $\operatorname{Cl}(X)$ is torsion free. Assume that $X$ is not rational. According to [Pr16, Theorem 1.2, Proposition 2.1] we have

[TABLE]

where for $r$ there are only two possibilities:

[TABLE]

In particular, $X$ has only cyclic quotient singularities. By the orbifold Riemann-Roch in both cases we have

[TABLE]

Hence the linear system $|A|$ contains a unique irreducible surface $M_{1}$ and $|kA|$ has no fixed components for $k=2$ and $3$.

5.2.

Apply the construction (4.1.3) with ${\mathscr{M}}=|3A|$. In a neighborhood of the point of index $r$ ($r=5$ or $12$) we have ${\mathscr{M}}\sim-tK_{X}$, where

[TABLE]

Then by Lemma 4.1.2

[TABLE]

The relation (4.4.1) for $k=3$ has the form

[TABLE]

where $\hat{q}\leq 13$ by Proposition 2.5. If the contraction $\bar{f}$ is not birational, then $\hat{q}=1$ by Lemma 4.3.1. Hence, $\alpha\leq 3/(7t-3)$. On the other hand,

[TABLE]

The contradiction shows that the contraction $\bar{f}$ must be birational. In particular, the movable linear system ${\mathscr{M}}$ is not contracted, i.e.

[TABLE]

5.3.

If $\alpha\geq 1$, then the inequality (5.2.3) and Proposition 2.5 give us successively

[TABLE]

a contradiction. Taking (5.1.1) into account we see that $P:=f(E)$ is a non-Gorenstein point of $X$ and $f$ is the weighted blowup as in Proposition 2.6(i) (so-called Kawamata blowup). In particular, $\alpha=1/\mathrm{{r}}(X,P)$. In this case by Lemma 4.2.1 we have

[TABLE]

Since $\hat{X}$ is not rational, according to Proposition 2.5 we have

[TABLE]

Note that $(7t-3)\alpha e\geq 5$. Then (5.2.3) implies

[TABLE]

5.4. Case: $\mathrm{{r}}(X,P)=2$

Then $\alpha=1/2$ and $\beta_{3}=1/2+m_{3}$, where $m_{3}\geq 2$ by (5.2.2). We can rewrite (5.2.3) in the following form

[TABLE]

Since $\hat{q}\leq 7$, this equation has no solutions.

5.5. Case: $\mathrm{{r}}(X,P)=3$

Then, as above, $\alpha=1/3$, $\beta_{3}$ is an integer $\geq 2$, and (5.2.3) has the form

[TABLE]

Again, there are no solutions.

5.6. Case: $\mathrm{{r}}(X,P)=r$, $r=5$ or $12$

Then $\beta_{1}=t^{\prime}/r+m_{1}$, where $m_{1}\geq 0$, and $t^{\prime}=3$ if $r=5$ and $t^{\prime}=7$ if $r=12$. The relation (4.4.1) for $k=1$ has the form

[TABLE]

From this we obtain $s_{1}=0$ and $\hat{q}=4$. Then from (5.2.3) we obtain $s_{3}=1$. Since $s_{1}=0$, the group $\operatorname{Cl}(\hat{X})$ is torsion free by Lemma 4.2.2. Thus $\hat{\mathscr{M}}\sim 0$ and so $\dim|\Theta|\geq 2$. This contradicts Proposition 2.5. ∎

5.7 Corollary**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)=7$ and let $A$ be a Weil divisor such that $-K_{X}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}}7A$ (here we do not claim that $-K_{X}\sim 7A$). Assume that $\dim|2A|\geq 1$. Then $X$ is rational.

Proof.

By Corollary  3.5 the group $\operatorname{Cl}(X)$ is torsion free. Then a computer search gives us ${\mathrm{g}}(X)\geq 11$. ∎

6. $\mathbb{Q}$-Fano threefolds of Fano index $13$

6.1 Proposition**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)=13$. Then $X$ is rational.

Proof.

By Proposition 3.3 the group $\operatorname{Cl}(X)$ is torsion free. Assume that $X$ is not rational. According to [Pr10] we have to consider only one case:

[TABLE]

One can expect that all the varieties of this type are hypersurfaces $X_{12}\subset\mathbb{P}(3,4,5,6,7)$ (cf. [BS07]), but this is not known.

By the orbifold Riemann-Roch, (6.1.1) implies that $|A|=|2A|=\emptyset$, the linear system $|kA|$ for $k=3$, $4$, $5$ contains a unique irreducible surface $M_{k}$ and for $k=6$, $7$, $8$ the linear system $|kA|$ is a pencil ${\mathscr{M}}_{k}$ without fixed components [Pr10, Proposition 3.6].

6.2.

Apply the construction (4.1.3) with ${\mathscr{M}}=|8A|$. Then near the point of index 7 we have ${\mathscr{M}}\sim-6K_{X}$. By Lemma 4.1.2

[TABLE]

The relation (4.4.1) for $k=8$ has the form

[TABLE]

where $\hat{q}\leq 13$ by Proposition 2.5. Since $\alpha\geq 1/7$, we see that $\hat{q}>1$. By Lemma 4.3.1 this implies that the contraction $\bar{f}$ is birational and so $s_{8}>0$. We also have

[TABLE]

where $\beta_{3}+\beta_{5}\geq\beta_{8}$ and $2\beta_{4}\geq\beta_{8}$. Pushing forward this relation to $\hat{X}$ we obtain

[TABLE]

Since the $\bar{f}$-exceptional divisor is irreducible, only one of the numbers $s_{3}$, $s_{4}$, $s_{5}$ can be equal to [math]. Therefore,

[TABLE]

6.3.

If $\alpha\geq 2/3$, then the relation (6.2.2) gives us $\hat{q}\geq 10$. Then $\operatorname{Cl}(\hat{X})$ is torsion free by Proposition 3.3 and $\dim|k\Theta|\leq 0$ for $k=1$, $2$, $3$ by Proposition 2.5. Hence, $s_{8}\geq 4$. Then $\hat{q}\geq 13$ and so $s_{8}\geq 6$, $\hat{q}>13$, a contradiction. Therefore, $P:=f(E)$ is a non-Gorenstein point of $X$ and $f$ is the Kawamata blowup of $P$ by Proposition 2.6(i). In particular, $\alpha=1/\mathrm{{r}}(X,P)$, where $\mathrm{{r}}(X,P)=2$, $3$, $5$ or $7$.

6.4. Case: $\mathrm{{r}}(X,P)=2$

Then $\beta_{8}$ is an integer $\geq 3$ by (6.2.1). The relation (6.2.2) has the form

[TABLE]

It has no solutions satisfying the inequalities $s_{8}\geq 2$, $\beta_{8}\geq 3$, $\hat{q}\leq 13$.

6.5. Case: $\mathrm{{r}}(X,P)=3$

Assume that $\mathrm{{r}}(X,P)=3$. Then as above $\beta_{8}=2/3+m_{8}$, $m_{8}\geq 2$, and

[TABLE]

Again the equation has no suitable solutions.

6.6. Case: $\mathrm{{r}}(X,P)=5$

Then near the point of index $5$ we have $-K_{X}\sim{\mathscr{M}}_{8}$. Hence $\beta_{8}=1/5+m_{8}$, where $m_{8}\geq 1$. The relation (6.2.2) has the form

[TABLE]

We get only one solution: $\hat{q}=5$, $e=1$, $s_{8}=2$. Since $e=1$, we have $d=n$ by Lemma 4.2.2. Since $|A|=|2A|=\emptyset$, we have $d\geq 3$ and so $n=d=3$ by Proposition 3.3. Thus $\operatorname{Cl}(\hat{X})_{\mathrm{t}}\simeq\mathbb{Z}/3\mathbb{Z}$. Then the image $\bar{f}(\bar{F})$ is a non-Gorenstein point according to Corollary 4.5.3. For $k=8$ the relation (4.5.1) yields $b\geq 7/3$. This contradicts Corollary 3.6.

6.7. Case: $\mathrm{{r}}(X,P)=7$

Finally we assume that $\mathrm{{r}}(X,P)=7$. Then $\beta_{8}=6/7+m_{8}$, where $m_{8}\geq 0$. Hence,

[TABLE]

If $e\leq 2$, then the torsion part of $\operatorname{Cl}(\hat{X})$ is non-trivial Lemma 4.2.2 because $|2A|=\emptyset$. By Proposition 3.3 we have $\hat{q}\leq 7$ and then (6.7.1) has no solutions. Thus $e\geq 3$ and then there is only one possibility: $\hat{q}=7$, $s_{8}=2$. Then $\hat{X}$ is rational by Corollary 5.7. This concludes the proof of Proposition 6.1. ∎

7. $\mathbb{Q}$-Fano threefolds of Fano index $11$

7.1 Proposition**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)=9$. Then $X$ is rational.

Proof.

By Proposition 3.3 the group $\operatorname{Cl}(X)$ is torsion free. According to Proposition 2.5 and [Pr10] we have to consider only two cases:

[TABLE]

There are examples of varieties of these types: they are hypersurfaces $X_{12}\subset\mathbb{P}(1,4,5,6,7)$ and $X_{10}\subset\mathbb{P}(2,3,4,5,7)$ in cases 7 and 7, respectively [BS07].

7.1.1**.**

From the table above one can see that in both cases the linear systems $|kA|$ have no fixed components for $k=4$, $5$, $6$. Apply the construction (4.1.3) with ${\mathscr{M}}=|5A|$. Then near the point of index 7 we have $A\sim-2K_{X}$, ${\mathscr{M}}\sim-3K_{X}$. By Lemma 4.1.2

[TABLE]

The relation (4.4.1) for $k=5$ has the form

[TABLE]

Assume that $X$ is not rational. Then $\hat{q}\leq 11$ by Propositions 2.5 and 6.1.

7.2.

Assume that $\alpha\geq 1$. Then $\hat{q}\geq 6$ and $\alpha$ is an integer by Proposition 2.6. Moreover, $\alpha=e=1$ and $s_{5}+\beta_{5}e$ is also an integer. Hence, $\hat{q}\equiv-1\mod 11$. This contradicts (2.2.1). Therefore, $\alpha<1$ by Proposition 2.6(i). In particular,

[TABLE]

7.3.

Assume that $\bar{f}$ is not birational. Since $\bar{X}$ is not rational by our assumptions, $\hat{q}=1$ (see Lemma 4.3.1). Then $s_{5}=0$ and $5=(11\beta_{5}-5\alpha)e$, where $11\beta_{5}-5\alpha\in\mathbb{Z}$. Then $\beta_{5}=l/r$, $l\in\mathbb{Z}$ and $l\geq 3$ by (7.1.1). Thus we can write $5r=(11l-5)e$. But this equation has no solutions satisfying (7.2.1). Therefore, the contraction $\bar{f}$ is birational. In particular, $s_{5}>0$.

7.4. Cases 7 and 7 with $\mathrm{{r}}(X,P)=2$

Then $\beta_{5}=1/2+m_{5}$, $m_{5}\geq 1$. Thus (4.4.1) for $k=5$ has the form

[TABLE]

We get one possibility: $\hat{q}=5$, $e=1$, $s_{5}=1$.

In the case 7 the linear system $|A|$ contains a unique member $M_{1}$. Then (4.4.1) for $k=1$ has a similar form

[TABLE]

We obtain $s_{1}=0$. So, $\operatorname{Cl}(\hat{X})$ is torsion free by Lemma 4.2.2. Since $\dim|\Theta|=2$, the variety $\hat{X}$ is rational by Proposition 2.5.

In the case 7 the map $\bar{f}\mathbin{\scriptstyle{\circ}}\chi\mathbin{\scriptstyle{\circ}}f^{-1}$ contracts a divisor $F\sim dA$ with $d>1$ (because $|A|=\emptyset$). Since $e=1$, by Lemma 4.2.2 we have $\operatorname{Cl}(\hat{X})_{\mathrm{t}}\simeq\mathbb{Z}/n\mathbb{Z}$ with $n=d>1$. Apply (4.5.1)-(4.5.2). Recall that $n\leq 3$ (see Proposition 3.3). In particular, $\gcd(n,11)=1$. Then the image $\bar{f}(\bar{F})$ is a non-Gorenstein point according to Corollary 4.5.3. For $k=5$ the relation (4.5.1) yields $14=n(b-5\gamma_{5})$. According to Corollary 3.6 this is impossible.

7.5. Cases 7 and 7 with $\mathrm{{r}}(X,P)=7$.

Then $\beta_{5}=3/7+m_{5}$, $\beta_{6}=5/7+m_{6}$, where $m_{5},m_{6}\geq 0$. The relation (4.4.1) for $k=5$ and $6$ has the form

[TABLE]

Here $s_{5}\leq 3$ because $\hat{q}\leq 11$. By Proposition 2.5 we have $\hat{q}\neq 9$ because ${\mathrm{g}}(\hat{X})\geq{\mathrm{g}}(X)\geq 7$. Then the system of equations (7.5.1) one has $\hat{q}=3e$, $s_{5}=s_{6}=e=1$ or $2$.

Assume that $\hat{q}=6$ (and $e=s_{5}=s_{6}=2$). In the case 7 we have

[TABLE]

Hence the divisor $\bar{M}_{1}$ is contracted (otherwise the class of $\Theta$ in the group $\operatorname{Cl}(\hat{X})/\operatorname{Cl}(\hat{X})_{\mathrm{t}}$ would be divisible). Since $e=2$, this contradicts Lemma 4.2.2. In the case 7 from the relation

[TABLE]

we see that the divisor $\bar{M}_{2}$ must be contracted. Since $e=2$, the group $\operatorname{Cl}(\hat{X})$ is torsion free by Lemma 4.2.2. Since $s_{6}=2$ and $\dim{\mathscr{M}}_{6}=2$, we have $\dim|2\Theta|\geq 2$. This contradicts Proposition 2.5.

Finally, assume that $\hat{q}=3$ (and $e=s_{5}=s_{6}=1$). In the case 7 we have

[TABLE]

As above, the divisor $\bar{M}_{1}$ must be contracted and the group $\operatorname{Cl}(\hat{X})$ is torsion free. Since $s_{6}=1$ and $\dim{\mathscr{M}}_{6}=3$, we have $\dim|\Theta|\geq 3$. This contradicts Proposition 2.5.

In the case 7 we have

[TABLE]

where $3\beta_{2}\geq\beta_{6}$, $2\beta_{3}\geq\beta_{6}$. Since both $\bar{M}_{2}$ and $\bar{M}_{3}$ cannot be contracted simultaneously, this gives a contradiction.

7.6. Case 7 with $\mathrm{{r}}(X,P)=3$

Then $\beta_{5}=1/3+m_{5}$, $m_{5}\geq 1$. Thus

[TABLE]

and we obtain $\hat{q}=7$ and $s_{5}\leq 2$. Then $\hat{X}$ is rational by Corollary 5.7.

7.7. Case 7 with $\mathrm{{r}}(X,P)=4$

Then $\beta_{5}=3/4+m_{5}$, $m_{5}\geq 0$. If $m_{5}=0$, then $\mathrm{ct}(X,{\mathscr{M}})=1/3$. In this case $(X,\frac{1}{3}{\mathscr{M}})$ is canonical and points of indices $4$ and $7$ are canonical centers. Then we can apply our construction (4.1.3) starting with the point of index $7$, as in 7.5. This gives a rationality construction.

Thus we assume that $m_{5}\geq 1$. The relation (4.4.1) for has the form

[TABLE]

and then $\hat{q}=8$, $s_{5}\leq 2$. By Proposition 2.5 the variety $\hat{X}$ is rational.

7.8. Case 7 with $\mathrm{{r}}(X,P)=5$

Then ${\mathscr{M}}$ is a Cartier at $P$ and so $\beta_{5}$ must be a positive integer. The relations (4.4.1) has the form

[TABLE]

Since $\hat{q}\leq 11$, this equation has no solutions. This concludes the proof of Proposition 7.1. ∎

8. $\mathbb{Q}$-Fano threefolds of Fano index $9$

8.1 Proposition**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)=9$. Then $X$ is rational.

Proof.

By Proposition 3.3 the group $\operatorname{Cl}(X)$ is torsion free. Assume that $X$ is not rational. According to [Pr10, Proposition 3.6] we have to consider only one case:

[TABLE]

By the orbifold Riemann-Roch (8.1.1) implies that

[TABLE]

Thus the linear system $|kA|$ contains a unique irreducible surface $M_{k}$ for $k=2$ and $3$ and $|kA|$ for $k=4$ and $5$ is a pencil without fixed components.

8.2.

Apply the construction (4.1.3) with ${\mathscr{M}}=|5A|$. Then near the point of index ${7}$ we have ${\mathscr{M}}\sim-6K_{X}$. By Lemma 4.1.2

[TABLE]

The relation (4.4.1) for $k=5$ has the form

[TABLE]

8.3.

By Propositions 2.5, 6.1, and 7.1 we have $\hat{q}\leq 9$. Then, obviously, $\alpha<1$. Therefore, $P:=f(E)$ is a non-Gorenstein point of $X$ by Proposition 2.6(i) and $\alpha=1/\mathrm{{r}}(X,P)$, where $\mathrm{{r}}(X,P)=2$, $5$ or $7$.

8.4.

If $\bar{f}$ is not birational, then $\hat{q}=1$ by Lemma 4.3.1 and so $s_{5}=0$, i.e. $\bar{\mathscr{M}}$ is $\bar{f}$-vertical. Note that $9\beta_{5}-5\alpha$ is an integer (because $9{\mathscr{M}}_{5}+5K_{X}$ is Cartier). Hence, $9\beta_{5}-5\alpha=1$ or $5$. Let $r:=\mathrm{{r}}(X,P)$. Then $\beta_{5}=l/r$ for some $l$ and $9l=r+5$ or $5(r+1)$. For $r=2$, $5$, $7$ this equation has no solutions. The contradiction shows that $\bar{f}$ is birational. In particular, $s_{5}>0$.

8.5. Case: $\mathrm{{r}}(X,P)=2$

Then $\beta_{5}=1/2+m_{5}$, $m_{5}\geq 3$ and the relation (4.4.1) for $k=5$ has the form

[TABLE]

Since $\hat{q}\leq 9$, this is impossible.

8.6. Case: $\mathrm{{r}}(X,P)=5$

Then $\beta_{5}$ is an integer $\geq 2$ and, as above,

[TABLE]

We get one possibility: $\hat{q}=7$, $e=1$, $s_{5}+\beta_{5}=4$. Since $|A|=\emptyset$, the group $\operatorname{Cl}(\hat{X})_{\mathrm{t}}$ is non-trivial by by Lemma 4.2.2. By Proposition 3.3 we have $\operatorname{Cl}(\hat{X})_{\mathrm{t}}\simeq\mathbb{Z}/2\mathbb{Z}$. By Corollary 4.5.3 the image $\bar{f}(\bar{F})$ is a point of even index. The relation (4.5.1) for $k=5$ has the form

[TABLE]

Then we obtain a contradiction by Corollary 3.6.

8.7. Case: $\mathrm{{r}}(X,P)=7$

Then $\beta_{5}=6/7+m_{5}$, $m_{5}\geq 0$,

[TABLE]

We get the following possibilities:

[TABLE]

If $\hat{q}=6$, then the group $\operatorname{Cl}(\hat{X})$ is torsion free by Proposition 3.3. Since $s_{5}+3m_{5}=1$, we have $s_{5}=1$. Hence, $\dim|\Theta|\geq 1$. This contradicts Proposition 2.5.

Consider the case $\hat{q}=5$. Then $s_{5}\leq 2$ Since $e=1$ and $|A|=\emptyset$, by Lemma 4.2.2 we have $\operatorname{Cl}(\hat{X})_{\mathrm{t}}\simeq\mathbb{Z}/n\mathbb{Z}$ with $n=d>1$. Apply (4.5.1) with $k=5$. We obtain $25-9s_{5}\leq ns_{5}b$ and so $b\geq 7/2n$. Since $n\leq 3$, we get a contradiction by Corollary 3.6. This concludes the proof of Proposition 8.1. ∎

9. $\mathbb{Q}$-Fano threefolds of Fano index $8$

9.1 Proposition**.**

Let $X$ be a $\mathbb{Q}$-Fano threefold with $\operatorname{q_{\mathbb{Q}}}(X)=8$. Then $X$ is rational.

Proof.

By Proposition 3.3 the group $\operatorname{Cl}(X)$ is torsion free. Assume that $X$ is not rational. Using a computer search and taking Proposition 2.5 into account we obtain the following possibilities:

[TABLE]

Note that existence of varieties with ${\mathbf{B}}(X)=(7,13)$ and $(5,7)$ is not known. Varieties with ${\mathbf{B}}(X)=(3,5,11)$ can be realized as hypersurfaces $X_{12}\subset\mathbb{P}(1,3,4,5,7)$ which are rational. But again we do not know if this is the only family with corresponding invariants.

Apply the construction (4.1.3) with ${\mathscr{M}}=|4A|$. Since $X$ is not rational by our assumption, we have $\hat{q}\leq 8$ (see Propositions 2.5, 6.1, 7.1, and 8.1).

9.2. Case ${\mathbf{B}}(X)=(5,7)$

In a neighborhood of the point of index $7$ we have ${\mathscr{M}}\sim-4K_{X}$. Thus by Lemma 4.1.2

[TABLE]

The relation (4.4.1) for $k=4$ has the form

[TABLE]

We claim that the contraction $\bar{f}$ is birational. Indeed, otherwise $\hat{q}=1$ by Lemma 4.3.1 and so $s_{4}=0$, i.e. $\bar{\mathscr{M}}$ is the pull-back of some linear system on $\hat{X}$. Since $\dim\bar{\mathscr{M}}=2$, $\dim\hat{X}\neq 1$ (otherwise $\bar{\mathscr{M}}=\bar{f}^{*}|2p|$, where $p$ is a point on $\hat{X}\simeq\mathbb{P}^{1}$, and then $\bar{M}_{2}=\bar{f}^{*}p$ must be movable). Further, $4\bar{M}_{1}\sim\bar{\mathscr{M}}$ and so $\bar{M}_{1}$ is also the pull-back of some divisor, say $\Lambda$, on the surface $\hat{X}$. Thus $\bar{M}_{1}=\bar{f}^{*}\Lambda$ and $\bar{\mathscr{M}}=\bar{f}^{*}|4\Lambda|$. Clearly, $\Lambda$ is a generator of the group $\operatorname{Cl}(\hat{X})$. Recall that $\hat{X}$ is a del Pezzo surface with at worst Du Val singularities of type $\mathrm{A_{n}}$ [MP08, Theorem 1.2.7]. According to the classification (see e.g. [MZ88, Lemmas 3 & 7]) for $\hat{X}$ there are only four possibilities:

[TABLE]

where $\operatorname{DP}_{5}^{\mathrm{A}_{4}}$ is a del Pezzo surface of degree $5$ whose singular locus consists of one point of type $\mathrm{A_{4}}$. Since $\dim|\bar{M}_{1}|=\dim|\bar{M}_{2}|=0$, the divisors $\Lambda$ and $2\Lambda$ are not movable. But one can easily check that $\dim|2\Lambda|>0$ in all cases. The contradiction shows that the contraction $\bar{f}$ is birational. In particular,

[TABLE]

Then from (9.2.1) we immediately see that $\alpha<1$. Therefore, $P:=f(E)$ is a non-Gorenstein point of $X$ and $\alpha=1/\mathrm{{r}}(X,P)$, where $\mathrm{{r}}(X,P)=5$ or $7$ (see Proposition 2.6(i)).

9.2.2**.**

Subcase $\mathrm{{r}}(X,P)=7$. Then we can write $\beta_{1}=1/7+m_{1}$ and $\beta_{4}=4/7+m_{4}$, where $m_{1}$ and $m_{4}$ are non-negative integers. We can rewrite the relation (4.4.1) for $k=1$ and $4$ as follows

[TABLE]

This yields $\hat{q}=e$ and $s_{4}=0$, a contradiction.

9.2.3**.**

Subcase $\mathrm{{r}}(X,P)=5$. As above, $\beta_{4}=3/5+m_{4}$, $\beta_{1}=2/5+m_{1}$, where $m_{1}\geq 0$ and $m_{4}\geq 1$. Therefore,

[TABLE]

This yields $s_{1}=0$ and $\hat{q}=3e$. Since $s_{1}=0$, by Lemma 4.2.2 we have $e=1$, $\hat{q}=3$, and $s_{4}=0$, a contradiction.

9.3. Case ${\mathbf{B}}(X)=(7,13)$

Near the point of index $13$ we have ${\mathscr{M}}\sim-7K_{X}$. Thus by Lemma 4.1.2

[TABLE]

The relation (4.4.1) for $k=4$ has the form

[TABLE]

From this, one immediately sees that $\alpha<1$. Therefore, $P:=f(E)$ is a non-Gorenstein point of $X$ and $f$ is the Kawamata blowup of $P$ by Proposition 2.6(i). In particular, $\alpha=1/\mathrm{{r}}(X,P)$, where $\mathrm{{r}}(X,P)=7$ or $13$.

9.3.3**.**

Subcase $\mathrm{{r}}(X,P)=13$. Then we can write $\beta_{1}=5/13+m_{1}$, where $m_{1}$ is a non-negative integer. Therefore,

[TABLE]

This gives us $s_{1}=0$ and $\hat{q}=3e$. Since $s_{1}=0$, Lemma 4.2.2 we have $e=1$, $\hat{q}=3$, and $\operatorname{Cl}(\hat{X})$ is torsion free. Similarly, we can compute

[TABLE]

Therefore, $\dim|\Theta|\geq\dim{\mathscr{M}}_{5}=3$. This contradicts Proposition 2.5.

9.3.4**.**

Subcase $\mathrm{{r}}(X,P)=7$. Then we can write $\beta_{1}=1/7+m_{1}$, where $m_{1}$ is a positive integer. Then

[TABLE]

which is a contradiction.

9.4. Case ${\mathbf{B}}(X)=(3,5,11)$

Near the point of index $11$ we have ${\mathscr{M}}\sim-6K_{X}$. Thus by Lemma 4.1.2

[TABLE]

The relation (4.4.1) for $k=4$ has the form

[TABLE]

From this we immediately see that $\alpha<1$. Therefore, $\alpha=1/\mathrm{{r}}(X,P)$, where $\mathrm{{r}}(X,P)=3$, $5$ or $11$ (see Proposition 2.6(i)).

9.4.3**.**

Subcase $\mathrm{{r}}(X,P)=3$. Then we can write $\beta_{4}=2/3+m_{4}$, where $m_{4}\geq 2$. Therefore,

[TABLE]

In particular, $\bar{f}$ is birational and $s_{4}>0$. We get only one solution: $\hat{q}=7$, $e=s_{4}=1$. By Corollary 5.7 the variety $\hat{X}$ is rational.

9.4.4**.**

Subcase $\mathrm{{r}}(X,P)=5$. Then we can write $\beta_{2}=4/5+m_{2}$ and $\beta_{4}=3/5+m_{4}$, where $m_{2}\geq 0$ and $m_{4}>0$. Therefore,

[TABLE]

In particular, $\bar{f}$ is birational and $s_{4}>0$. We obtain $\hat{q}=7$ and $s_{4}\leq 2$. Then $\hat{X}$ is rational again by Corollary 5.7.

9.4.5**.**

Subcase $\mathrm{{r}}(X,P)=11$. Then we can write $\beta_{4}=6/11+m_{4}$, where $m_{4}$ is a non-negative integer. Therefore,

[TABLE]

Similarly, the relation (4.4.1) for $k=3$ has the form

[TABLE]

One can see that there are only two solutions:

[TABLE]

If $\hat{q}=7$, then by (9.4.6) we have $s_{4}=2$. This contradicts Corollary 5.7. Hence, $\hat{q}=5$ and $e=1$. Since $|A|=\emptyset$ and $e=1$, we have $\operatorname{Cl}(\hat{X})_{\mathrm{t}}\simeq\mathbb{Z}/n\mathbb{Z}$ with $1<n\leq 3$ by Proposition 3.3. If $n=3$, then $s_{3}=0$ by Lemma 4.2.2. Then $\bar{f}(\bar{F})$ is a non-Gorenstein point by Corollary 4.5.3. The relation (4.5.1) gives us

[TABLE]

where $s_{4}\leq 2$ by (9.4.6). Hence, $b\geq 2/3$. This contradicts Corollary 3.6.

Assume that $n=2$. Then $s_{2}=0$ by Lemma 4.2.2. The relation (4.5.1) for $k=3$ has the form

[TABLE]

where $s_{3}=1$ by (9.4.7). We see that $\bar{f}(\bar{F})$ is a non-Gorenstein point and $b\geq 7/2$. Again, this contradicts Corollary 3.6. Proposition 9.1 is proved.∎

Now Theorem 1.1 follows from Propositions 6.1, 7.1, 8.1, and 9.1.