Rationality of weighted hypersurfaces of special degree

Michael Chitayat

Abstract

Let $X\subset\mathbb{P}(w_{0},w_{1},w_{2},w_{3})$ be a quasismooth well-formed weighted projective hypersurface and let $L=\operatorname{{\rm lcm}}(w_{0},w_{1},w_{2},w_{3})$ . We characterize when $X$ is rational under the assumption that $L$ divides $\deg(X)$ . Furthermore, we give a new family of normal rational weighted projective hypersurfaces with ample canonical divisor, valid in all dimensions, adding to the list of examples discovered by Koll $\rm{\acute{a}}$ r. Finally, we determine precisely which affine Pham-Brieskorn threefolds are rational, answering a question of Rajendra V. Gurjar.

Author information. Michael Chitayat, University of Ottawa, [email protected], ORCID: 0000-0002-0590-539X

Keywords. Commutative algebra, algebraic geometry, weighted complete intersections, rational varieties, Pham-Brieskorn varieties.

Statements and declarations. The author has no competing interests or funding to declare.

Acknowledgements. The author wishes to thank Rajendra V. Gurjar for proposing Question 1.1, Pedro Garcia Sanchez for sharing some useful comments as well as the reference [1], and Daniel Daigle for multiple comments and suggestions, the most important of which significantly shortened the proof of Theorem A.

1 Introduction

Let $X\subseteq\mathbb{P}^{3}$ be a smooth projective hypersurface over $\mathbb{C}$ of degree $d$ . It is well known that

\text{$X$ is rational}\iff d\leq 3\iff\text{the geometric genus of $X$ is zero.}

(1)

We consider the more general situation of weighted hypersurfaces $X$ in the weighted projective space $\mathbb{P}(w_{0},w_{1},w_{2},w_{3})$ . This additional generality makes understanding when $X$ is rational more difficult; we give new results in this direction with subsequent applications.

A Pham-Brieskorn ring is a $\mathbb{C}$ -algebra of the form

B_{a_{0},a_{1},\dots,a_{n}}=\mathbb{C}[X_{0},X_{1},\dots,X_{n}]/\langle X_{0}^{a_{0}}+X_{1}^{a_{1}}+\dots+X_{n}^{a_{n}}\rangle

where each $a_{i}\in\mathbb{N}^{+}$ . These rings, their corresponding affine varieties and their singular points have been studied extensively for decades from many perspectives (see [20], [11], [18], [15], etc.). This work was motivated by the following question of Rajendra V. Gurjar, to which we manage to provide a complete answer.

1.1 Question.

For which 4-tuples $(a_{0},a_{1},a_{2},a_{3})$ is $\operatorname{{\rm Spec}}B_{a_{0},a_{1},a_{2},a_{3}}$ rational?

We begin by defining an $\mathbb{N}$ -grading on $B_{a_{0},a_{1},a_{2},a_{3}}$ by declaring that $X_{i}$ is homogeneous of degree $w_{i}=L/a_{i}$ , where $L=\operatorname{{\rm lcm}}(a_{0},a_{1},a_{2},a_{3})$ . Thus, $\operatorname{{\rm Proj}}B_{a_{0},a_{1},a_{2},a_{3}}$ is a hypersurface of the weighted projective space $\mathbb{P}(w_{0},w_{1},w_{2},w_{3})$ . In view of Castelnuovo’s Theorem, which states that rationality and unirationality are equivalent in dimension 2 over $\mathbb{C}$ , it is clear that $\operatorname{{\rm Spec}}B_{a_{0},a_{1},a_{2},a_{3}}$ is rational if and only if $\operatorname{{\rm Proj}}B_{a_{0},a_{1},a_{2},a_{3}}$ is rational. As such, we instead answer the following question, which has the same answer as Question 1.1.

1.2 Question.

For which 4-tuples $(a_{0},a_{1},a_{2},a_{3})$ is $\operatorname{{\rm Proj}}B_{a_{0},a_{1},a_{2},a_{3}}$ rational?

To answer Question 1.2, we require a theorem on numerical semigroups. We say that a tuple $(d_{1},\dots,d_{n})\in(\mathbb{N}^{+})^{n}$ is well-formed if $\gcd(d_{1},\dots,d_{i-1},\hat{d_{i}},d_{i+1},\dots d_{n})=1$ for all $i\in\{1,\dots,n\}$ .

Theorem A.

Suppose $(d_{1},d_{2},d_{3},d_{4})\in(\mathbb{N}^{+})^{4}$ is well-formed, let $\Gamma=\langle d_{1},d_{2},d_{3},d_{4}\rangle\subseteq\mathbb{N}$ be the numerical semigroup generated by $d_{1},d_{2},d_{3},d_{4}$ and let $L=\operatorname{{\rm lcm}}(d_{1},d_{2},d_{3},d_{4})$ .

(i)

If $N\geq\max\{2L-\sum_{m=1}^{4}d_{m},0\}$ then $N\in\Gamma$ .
(ii)

If $n\in\mathbb{N}^{+}$ and $n\cdot L-\sum_{i=1}^{4}d_{i}\in\mathbb{N}\setminus\Gamma$ , then $n=1$ and $|\{d_{1},d_{2},d_{3},d_{4}\}|=2$ .

In addition to Theorem A, we prove the rationality of a special family of projective varieties in Proposition 3.6.

Proposition.

Let $a,c\in\mathbb{N}^{+}$ and consider the graded polynomial ring $R={\rm\bf k}_{c,\dots,c,a,\dots,a}[X_{1},\dots,X_{k},Y_{1},\dots,Y_{\ell}]$ where ${\rm\bf k}$ is a field, $k\geq 1$ , $\ell\geq 1$ , $\deg(X_{i})=c$ and $\deg(Y_{j})=a$ for all $i,j$ . Suppose $f=g(X_{1},\dots,X_{k})+h(Y_{1},\dots,Y_{\ell})$ is irreducible and homogeneous of degree $L=\operatorname{{\rm lcm}}(a,c)$ and both $g(X_{1},\dots,X_{k})$ and $h(Y_{1},\dots,Y_{\ell})$ are non-zero. Then $\operatorname{{\rm Proj}}(R/\langle f\rangle)$ is rational over ${\rm\bf k}$ .

Using Theorem A together with Proposition 3.6, we eventually obtain the following (Theorem 3.14) which generalizes (1):

Theorem.

Let $w_{0},w_{1},w_{2},w_{3}\in\mathbb{N}^{+}$ where $w_{0}\leq w_{1}\leq w_{2}\leq w_{3}$ and let $L=\operatorname{{\rm lcm}}(w_{0},w_{1},w_{2},w_{3})$ . Suppose $X=X_{f}\subset\mathbb{P}(w_{0},w_{1},w_{2},w_{3})$ is a well-formed quasismooth hypersurface of degree $nL$ for some $n\in\mathbb{N}^{+}$ . Then the following are equivalent:

(a)

$X$ is rational,
(b)

$h^{2}(X,\operatorname{\mathcal{O}}_{X})=0$ ,
(c)
one of the following holds
1. (i)
  
  $n=1$ , $w_{0}=w_{1}$ , $w_{2}=w_{3}$ and $\gcd(w_{0},w_{2})=1$ ;
2. (ii)
  
  $nL-\sum_{i=0}^{3}w_{i}<0$ .

With Theorem 3.14 available to us, we are able to address Gurjar’s Question 1.2. We will see that Question 1.2 reduces to the special case where $\operatorname{{\rm cotype}}(a_{0},a_{1},a_{2},a_{3})=0$ (see Definition 4.4) and hence is completely answered by the following, which is Theorem 4.7.

Theorem.

Suppose $a_{0}\leq a_{1}\leq a_{2}\leq a_{3}$ and $\operatorname{{\rm cotype}}(a_{0},a_{1},a_{2},a_{3})=0$ . Then $\operatorname{{\rm Proj}}B_{a_{0},a_{1},a_{2},a_{3}}$ is rational if and only if one of the following holds:

(a)

$a_{0}=a_{1}$ , $a_{2}=a_{3}$ and $\gcd(a_{0},a_{2})=1$ ;
(b)

$\frac{1}{a_{0}}+\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}>1$ .

Another interesting application of our results is Example 3.18, which gives a new family of rational quasismooth weighted hypersurfaces with ample canonical divisor. We believe these are the first such examples since Koll $\rm{\acute{a}}$ r introduced in [14] what are now known as Koll $\acute{a}$ r hypersurfaces. Our examples are simple and exist in any dimension. As a very special case, if $a,c\in\mathbb{N}^{+}$ are such that $\gcd(a,c)=1$ and $ac-2a-2c>0$ then $\operatorname{{\rm Proj}}B_{a,a,c,c}$ is a quasismooth normal rational projective surface with quotient singularities and ample canonical divisor.

Notation

We use the following terminology and notation throughout the article.

•

The set of natural numbers includes $0$ and is denoted by $\mathbb{N}$ . The set of positive integers is denoted by $\mathbb{N}^{+}$ .
•

Let $\Phi$ denote the set of prime numbers and let $p\in\Phi$ . Given $n\in\mathbb{N}^{+}$ , define

$v_{p}(n)=\max\big{\{}\,u\in\mathbb{N}\,\mid\,p^{u}\text{ divides }n\,\big{\}}$

and observe that $n=\prod_{p\in\Phi}p^{v_{p}(n)}$ is the prime factorization of $n$ . Given $f,g\in\mathbb{N}^{+}$ , we define $v_{p}(\frac{f}{g})=v_{p}(f)-v_{p}(g)$ .
•

If $d_{1},\dots,d_{n}\in\mathbb{N}^{+}$ , we define $\langle d_{1},\dots,d_{n}\rangle$ to be the submonoid of $(\mathbb{N},+)$ generated by $d_{1},\dots,d_{n}$ .
•
A numerical semigroup $\Gamma$ is a subset of $(\mathbb{N},+)$ satisfying the following three conditions:
1. (i)
  
  $0\in\Gamma$
2. (ii)
  
  $\mathbb{N}^{+}\setminus\Gamma$ is finite
3. (iii)
  
  $x,y\in\Gamma\Rightarrow x+y\in\Gamma$
The Frobenius number of $\Gamma$ is defined as $F(\Gamma)=\max(\mathbb{Z}\setminus\Gamma)$ . If $\gcd(d_{1},\dots,d_{n})=1$ , we may write $F(d_{1},\dots,d_{n})$ instead of $F(\langle d_{1},\dots,d_{n}\rangle)$ .
•

Let ${\rm\bf k}$ be a field. If $K/{\rm\bf k}$ is a field extension, the notation $K={\rm\bf k}^{(n)}$ means that $K/{\rm\bf k}$ is purely transcendental of transcendence degree $n$ .
•

A variety is an integral separated scheme of finite type over an algebraically closed field ${\rm\bf k}$ . A surface is a two-dimensional variety.
•

Let $X$ be a normal variety over ${\rm\bf k}$ . Then $\operatorname{{\rm Div}}(X)$ is its group of Weil divisors and $K_{X}$ denotes a canonical divisor of $X$ . The group of $\mathbb{Q}$ -divisors is denoted by $\operatorname{{\rm Div}}(X,\mathbb{Q})$ . A divisor $D\in\operatorname{{\rm Div}}(X)$ is ample if there exists some $n\in\mathbb{N}^{+}$ such that $\operatorname{\mathcal{O}}_{X}(nD)$ is a very ample invertible sheaf (over ${\rm\bf k}$ ).
•

A del Pezzo surface is a normal projective surface with at most quotient singularities such that $-K_{X}$ is ample. It is well known that a del Pezzo surface over $\mathbb{C}$ is rational.

2 A Semigroup Theorem

We state a theorem used in Section 3. The proof of Theorem A is given in Section 5.

2.1 Definition.

A tuple $(d_{1},\dots,d_{n})\in(\mathbb{N}^{+})^{n}$ is well-formed if $\gcd(d_{1},\dots,d_{i-1},\hat{d_{i}},d_{i+1},\dots d_{n})=1$ for all $i\in\{1,\dots,n\}$ .

Theorem A.

Suppose $(d_{1},d_{2},d_{3},d_{4})\in(\mathbb{N}^{+})^{4}$ is well-formed, let $\Gamma=\langle d_{1},d_{2},d_{3},d_{4}\rangle$ and let $L=\operatorname{{\rm lcm}}(d_{1},d_{2},d_{3},d_{4})$ .

(i)

If $N\geq\max\{2L-\sum_{m=1}^{4}d_{m},0\}$ then $N\in\Gamma$ .
(ii)

If $n\in\mathbb{N}^{+}$ and $n\cdot L-\sum_{i=1}^{4}d_{i}\in\mathbb{N}\setminus\Gamma$ , then $n=1$ and $|\{d_{1},d_{2},d_{3},d_{4}\}|=2$ .

2.2 Remark.

With the assumptions of Theorem A(ii), if we assume in addition that $d_{1}\leq d_{2}\leq d_{3}\leq d_{4}$ , then $d_{1}=d_{2},d_{3}=d_{4}$ and $\gcd(d_{1},d_{3})=1$ .

2.3 Remark.

Not only is the well-formedness assumption in Theorem A natural given the geometric context of weighted projective varieties discussed in the following section, it is also the case that various general questions about numerical semigroups can be reduced to the special case where $(d_{1},\dots,d_{n})$ is well-formed. (See for instance Section 3 of [13], Proposition 8 in [10], as well as Lemma 2.16 and Proposition 2.17 in [17].)

3 Rationality of Some Weighted Hypersurfaces

3.1.

Let $S=\mathbb{C}_{w_{0},\dots,w_{n}}[X_{0},\dots,X_{n}]$ denote the graded polynomial ring where $n\geq 1$ and $\deg(X_{i})=w_{i}$ for each $i=0,\dots,n$ . The weighted projective space $\mathbb{P}=\mathbb{P}(w_{0},\dots,w_{n})=\operatorname{{\rm Proj}}(S)$ is well-formed if $(w_{0},\dots,w_{n})$ is well-formed. Every weighted projective space is a projective variety and is isomorphic to a well-formed weighted projective space. Note also that every weighted projective space is rational.

3.2.

Let $I$ be a homogeneous prime ideal of the graded ring $S=\mathbb{C}_{w_{0},\dots,w_{n}}[X_{0},\dots,X_{n}]$ and define $X_{I}=\operatorname{{\rm Proj}}(S/I)\subseteq\mathbb{P}(w_{0},\dots,w_{n})$ . If $f_{1},\dots,f_{k}$ are homogeneous elements of $S$ , we abbreviate $X_{\langle f_{1},\dots,f_{k}\rangle}$ by $X_{f_{1},\dots,f_{k}}$ . If $I$ is generated by a regular sequence $(f_{1},\dots,f_{k})$ of homogeneous elements of $S$ of respective degrees $d_{i}=\deg(f_{i})$ where $i=1,\dots,k$ , then $X_{I}$ is called a weighted complete intersection of multidegree $(d_{1},\dots,d_{k})$ . In particular, if $k=1$ , $f_{1}=f$ and $d_{1}=d$ , we will say that $X_{I}=X_{f}$ is a weighted hypersurface of degree $d$ . The closed subset $C_{X_{I}}=V(I)\subseteq\mathbb{A}^{n+1}$ is called the affine cone over $X_{I}$ ; note that $C_{X_{I}}$ passes through the origin of $\mathbb{A}^{n+1}$ and that $C_{X_{I}}\cong\operatorname{{\rm Spec}}(S/I)$ is an integral affine scheme. The variety $X_{I}$ is quasismooth if $C_{X_{I}}$ is nonsingular away from the origin. A weighted complete intersection $X_{I}\subseteq\mathbb{P}(w_{0},\dots,w_{n})$ is well-formed if both

(i)

$\mathbb{P}(w_{0},\dots,w_{n})$ is well-formed,
(ii)

$\operatorname{{\rm\bf codim}}_{X_{I}}(X_{I}\cap\operatorname{{\rm Sing}}(\mathbb{P}))\geq 2$ .

The amplitude of a well-formed weighted complete intersection $X_{f_{1},\dots,f_{k}}\subseteq\mathbb{P}(w_{0},\dots,w_{n})$ is the quantity $\alpha=\sum_{i=1}^{k}d_{i}-\sum_{j=0}^{n}w_{j}$ where $d_{i}=\deg(f_{i})$ . (We will preserve this notation for the amplitude $\alpha$ through to the end of Section 3.) It is well known that if $X_{I}$ is a well-formed quasismooth weighted complete intersection, then $X_{I}$ is a normal, Cohen-Macaulay, $\mathbb{Q}$ -factorial variety with at most cyclic quotient (hence rational) singularities.

We make use of 3.3 to 3.5 in the proof of Corollary 3.17. Paragraph 3.3 appears in Section 5.4 of [5], Lemma 3.4 (a) is due to Demazure. Lemma 3.4 (b) and Proposition 3.5 are unpublished results of Daigle.

3.3.

If $B=\bigoplus_{i\in\mathbb{N}}B_{i}$ is an $\mathbb{N}$ -graded domain, the number $e(B)=\gcd\big{\{}\,i\in\mathbb{N}\,\mid\,B_{i}\neq 0\,\big{\}}$ is called the saturation index of $B$ . One says that $B$ is saturated in codimension $1$ if $e(B/\mathfrak{p})=e(B)$ for every height $1$ homogeneous prime ideal $\mathfrak{p}$ of $B$ . Let $X=\operatorname{{\rm Spec}}B$ and $Y=\operatorname{{\rm Proj}}B$ . Let $Y^{(1)}$ be the set of height $1$ homogeneous prime ideals of $B$ . If $\mathfrak{p}\in Y^{(1)}$ then $B_{(\mathfrak{p})}\subset B_{\mathfrak{p}}$ is an extension of discrete valuation rings whose ramification index we denote by $e_{\mathfrak{p}}$ . For each $\mathfrak{p}\in Y^{(1)}$ , let $C_{\mathfrak{p}}^{Y}$ (resp. $C_{\mathfrak{p}}^{X}$ ) be the closure of $\{\mathfrak{p}\}$ in $Y$ (resp. in $X$ ); note that $C_{\mathfrak{p}}^{Y}$ (resp. $C_{\mathfrak{p}}^{X}$ ) is a prime divisor of $Y$ (resp. of $X$ ), and that each prime divisor on $Y$ is a $C_{\mathfrak{p}}^{Y}$ for some $\mathfrak{p}\in Y^{(1)}$ . For each prime divisor $C_{\mathfrak{p}}^{Y}\in Y^{(1)}$ , define $\varphi(C_{\mathfrak{p}}^{Y})=e_{\mathfrak{p}}C_{\mathfrak{p}}^{X}$ , and extend linearly to a $\mathbb{Q}$ -linear map $\varphi:\operatorname{{\rm Div}}(Y,\mathbb{Q})\to\operatorname{{\rm Div}}(X,\mathbb{Q})$ , where $D\mapsto\varphi(D)$ .

3.4 Lemma.

Let $B$ be an $\mathbb{N}$ -graded normal domain that is finitely generated over $\mathbb{C}$ such that $e(B)=1$ and $0pt(B_{+})>1$ .

(a)

There exists an ample $\mathbb{Q}$ -divisor $D$ of $Y=\operatorname{{\rm Proj}}B$ such that $\operatorname{\mathcal{O}}_{Y}(n)=\operatorname{\mathcal{O}}_{Y}(nD)$ for all $n\in\mathbb{Z}$ .
(b)

If $B$ is saturated in codimension $1$ , then $D\in\operatorname{{\rm Div}}(Y)$ (i.e. $D$ has integral coefficients).

Proof.

We use the notation of 3.3. Let $f,g$ be nonzero homogeneous elements of $B$ such that $\deg(f)-\deg(g)=1$ and let $W=\operatorname{{\rm div}}_{X}(f/g)\in\operatorname{{\rm Div}}(X)$ . By the Theorem below Section 3.5 in [7], there exists a unique $D\in\operatorname{{\rm Div}}(Y,\mathbb{Q})$ such that $\varphi(D)=W$ and this $D$ is ample and satisfies $\operatorname{\mathcal{O}}_{X}(nD)\cong\operatorname{\mathcal{O}}_{X}(n)$ for all $n\in\mathbb{Z}$ . This shows (a).

If $B$ is saturated in codimension $1$ , then Corollary 9.4 of [6] implies that $e_{\mathfrak{p}}=1$ for all $\mathfrak{p}\in Y^{(1)}$ ; since $W$ has integral coefficients, it follows that $D$ has integral coefficients, proving (b). ∎

3.5 Proposition.

Let $n\geq 2$ , let $\mathbb{P}=\mathbb{P}(w_{0},\dots,w_{n})$ be a well-formed weighted projective space and let $I$ be a homogeneous prime ideal of $R=\mathbb{C}_{w_{0},\dots,w_{n}}[X_{0},\dots,X_{n}]$ with $0ptI<n$ . Consider the graded ring $B=R/I$ and the closed subvariety $X=V_{+}(I)$ of $\mathbb{P}$ . Then $X$ is well-formed if and only if $B$ is saturated in codimension $1$ .

Proof.

Let $\pi:R\to B$ be the canonical epimorphism. Let $w=\prod_{i=0}^{n}w_{i}$ . For each prime factor $p$ of $w$ , let $J_{p}=\big{\{}\,j\,\mid\,0\leq j\leq n\text{ and }p\nmid w_{j}\,\big{\}}$ and $\mathfrak{q}_{p}=\sum_{j\in J_{p}}RX_{j}$ . It is well known that $\operatorname{{\rm Sing}}\mathbb{P}=\bigcup_{p\mid w}V_{+}(\mathfrak{q}_{p})$ . Also note that

given any homogeneous prime ideal

\mathfrak{h}

R

p\mid e(R/\mathfrak{h})

\Leftrightarrow

\mathfrak{q}_{p}\subseteq\mathfrak{h}

(2)

Suppose that $B$ is not saturated in codimension $1$ . Then there exists a height $1$ homogeneous prime ideal $\mathfrak{p}$ of $B$ such that $e(B/\mathfrak{p})\neq 1$ . Let $\mathfrak{h}=\pi^{-1}(\mathfrak{p})$ ; then $R/\mathfrak{h}\cong B/\mathfrak{p}$ , so $e(R/\mathfrak{h})=e(B/\mathfrak{p})\neq 1$ . Choose a prime factor $p$ of $e(R/\mathfrak{h})$ ; then $p\mid w$ and (by (2)) $\mathfrak{q}_{p}\subseteq\mathfrak{h}$ . We also have $I\subseteq\mathfrak{h}$ , so $V_{+}(\mathfrak{h})\subseteq V_{+}(I)\cap V_{+}(\mathfrak{q}_{p})\subseteq X\cap\operatorname{{\rm Sing}}\mathbb{P}$ . Since $0pt\mathfrak{h}=1+0ptI$ , we find $\dim V_{+}(\mathfrak{h})=\dim X-1$ , so $\operatorname{{\rm\bf codim}}_{X}(X\cap\operatorname{{\rm Sing}}\mathbb{P})\leq 1$ and hence $X$ is not well-formed.

Conversely, suppose that $X$ is not well-formed. Then $\operatorname{{\rm\bf codim}}_{X}(X\cap\operatorname{{\rm Sing}}\mathbb{P})\leq 1$ , so there exists a homogeneous prime ideal $\mathfrak{h}$ of $R$ such that $V_{+}(\mathfrak{h})\subseteq X\cap\operatorname{{\rm Sing}}\mathbb{P}$ and $\dim V_{+}(\mathfrak{h})=\dim X-1$ . Note that $I\subseteq\mathfrak{h}$ and that $0pt\mathfrak{h}=0ptI+1$ ; so $\mathfrak{p}=\pi(\mathfrak{h})$ is a homogeneous prime ideal of $B$ of height $1$ . Since $V_{+}(\mathfrak{h})\subseteq\operatorname{{\rm Sing}}\mathbb{P}=\bigcup_{p\mid w}V_{+}(\mathfrak{q}_{p})$ , there exists a prime divisor $p$ of $w$ such that $V_{+}(\mathfrak{h})\subseteq V_{+}(\mathfrak{q}_{p})$ and hence $\mathfrak{q}_{p}\subseteq\mathfrak{h}$ . So (2) implies that $p$ divides $e(R/\mathfrak{h})=e(B/\mathfrak{p})$ , which implies that $B$ is not saturated in codimension $1$ . ∎

We generalize an unpublished result of Michela Artebani. See Proposition 4.3.4 in [3].

3.6 Proposition.

Proof.

Let $B=R/\langle f\rangle$ and write $B={\rm\bf k}[x_{1},\dots x_{k},y_{1},\dots y_{\ell}]$ where $x_{i}$ and $y_{j}$ are the canonical images of $X_{i}$ and $Y_{j}$ in $B$ . Let $g=\gcd(a,c)$ . Since $\operatorname{{\rm Proj}}B\cong\operatorname{{\rm Proj}}B^{(g)}$ , we may assume without loss of generality that $\gcd(a,c)=1$ and that $L=ac$ . We define the following elements of the function field $K$ of $\operatorname{{\rm Proj}}B$ :

t_{i}=\frac{x_{i}}{x_{k}}\text{ for each }i=1,\dots,k-1,\quad u=\frac{x_{k}^{a}}{y_{\ell}^{c}},\quad v_{j}=\frac{y_{j}}{y_{\ell}}\text{ for each }j=1,\dots,\ell-1.

We claim that

B_{(y_{\ell})}\subseteq{\rm\bf k}(t_{1},\dots,t_{k-1},u,v_{1},\dots,v_{\ell-1})\subseteq K

(3)

where the second inclusion is obvious. To prove the first inclusion, observe that $B_{(y_{\ell})}$ is generated by elements of the form

\frac{x_{1}^{\alpha_{1}}\cdots x_{k}^{\alpha_{k}}y_{1}^{\beta_{1}}\cdots y_{\ell-1}^{\beta_{\ell-1}}}{y_{\ell}^{\beta_{\ell}}}\text{ such that $\alpha_{i},\beta_{j}\in\mathbb{N}$ and $c\sum_{i=1}^{k}\alpha_{i}+a\sum_{j=1}^{\ell-1}\beta_{j}=a\beta_{\ell}$}.

(4)

Consider any element from (4). Since $\gcd(a,c)=1$ , $a\mid\sum_{i=1}^{k}\alpha_{i}$ . So, $\sum_{i=1}^{k}\alpha_{i}=qa$ for some $q\in\mathbb{N}^{+}$ and $\alpha_{k}=qa-\sum_{i=1}^{k-1}\alpha_{i}$ . We then have

\beta_{\ell}=\frac{c\sum_{i=1}^{k}\alpha_{i}}{a}+\sum_{j=1}^{\ell-1}\beta_{j}=\frac{(c\sum_{i=1}^{k-1}\alpha_{i})+c\alpha_{k}}{a}+\sum_{j=1}^{\ell-1}\beta_{j}=\frac{(c\sum_{i=1}^{k-1}\alpha_{i})+cqa-c(\sum_{i=1}^{k-1}\alpha_{i})}{a}+\sum_{j=1}^{\ell-1}\beta_{j}=cq+\sum_{j=1}^{\ell-1}\beta_{j}.

We have

	$\displaystyle\frac{x_{1}^{\alpha_{1}}\cdots x_{k}^{\alpha_{k}}y_{1}^{\beta_{1}}\dots y_{\ell-1}^{\beta_{\ell-1}}}{y_{\ell}^{cq+\sum_{j=1}^{\ell-1}\beta_{j}}}$	$\displaystyle=\left(\frac{x_{1}}{x_{k}}\right)^{\alpha_{1}}\dots\left(\frac{x_{k-1}}{x_{k}}\right)^{\alpha_{k-1}}\left(\frac{x_{k}^{\sum_{i=1}^{k}\alpha_{i}}}{y_{\ell}^{cq}}\right)\left(\frac{y_{1}}{y_{\ell}}\right)^{\beta_{1}}\dots\left(\frac{y_{\ell-1}}{y_{\ell}}\right)^{\beta_{\ell-1}}$
		$\displaystyle=\left(\frac{x_{1}}{x_{k}}\right)^{\alpha_{1}}\dots\left(\frac{x_{k-1}}{x_{k}}\right)^{\alpha_{k-1}}\left(\frac{x_{k}^{qa}}{y_{\ell}^{cq}}\right)\left(\frac{y_{1}}{y_{\ell}}\right)^{\beta_{1}}\dots\left(\frac{y_{\ell-1}}{y_{\ell}}\right)^{\beta_{\ell-1}}$
		$\displaystyle=t_{1}^{\alpha_{1}}\dots t_{k-1}^{\alpha_{k-1}}u^{q}v_{1}^{\beta_{1}}\dots v_{\ell-1}^{\beta_{\ell-1}}.$

This shows that every element from (4) is an element of ${\rm\bf k}(t_{1},\dots,t_{k-1},u,v_{1},\dots,v_{\ell-1})$ and consequently (3) is true. Since $\operatorname{{\rm Frac}}{B_{(y_{\ell})}}=K$ , it follows from (3) that $K={\rm\bf k}(t_{1},\dots,t_{k-1},u,v_{1},\dots,v_{\ell-1})$ . Since $\operatorname{{\rm trdeg}}_{\rm\bf k}K=k+\ell-2$ , it now suffices to show that $u\in{\rm\bf k}(t_{1},\dots,t_{k-1},v_{1},\dots,v_{\ell-1})$ .

Let $M=\big{\{}\,(\mu_{1},\dots,\mu_{k})\in\mathbb{N}^{k}\,\mid\,\sum_{i=1}^{k}\mu_{i}=a\,\big{\}}$ and let $N=\big{\{}\,(\nu_{1},\dots,\nu_{\ell})\in\mathbb{N}^{\ell}\,\mid\,\sum_{i=1}^{\ell}\nu_{i}=c\,\big{\}}$ . Write

g(X_{1},\dots,X_{k})=\sum_{(\mu_{1},\dots,\mu_{k})\in M}a_{m}X_{1}^{\mu_{1}}\dots X_{k}^{\mu_{k}}\text{ and }h(Y_{1},\dots,Y_{\ell})=\sum_{(\nu_{1},\dots,\nu_{\ell})\in N}b_{n}Y_{1}^{\nu_{1}}\dots Y_{\ell}^{\nu_{\ell}}

where $m=(\mu_{1},\dots,\mu_{k})\in M$ , $n=(\nu_{1},\dots,\nu_{\ell})\in N$ and the coefficients $a_{m}\in{\rm\bf k}$ (resp $b_{n}\in{\rm\bf k}$ ) are not all zero. Then

	$\displaystyle 0$	$\displaystyle=g(x_{1},\dots,x_{k})+h(y_{1},\dots,y_{\ell})=\sum_{(\mu_{1},\dots,\mu_{k})\in M}a_{m}x_{1}^{\mu_{1}}\dots x_{k}^{\mu_{k}}+\sum_{(\nu_{1},\dots,\nu_{\ell})\in N}b_{n}y_{1}^{\nu_{1}}\dots y_{\ell}^{\nu_{\ell}}$
	$\displaystyle 0$	$\displaystyle=\sum_{(\mu_{1},\dots,\mu_{k})\in M}\frac{a_{m}x_{1}^{\mu_{1}}\dots x_{k-1}^{\mu_{k-1}}x_{k}^{\mu_{k}}x_{k}^{\sum_{i=1}^{k-1}\mu_{i}}}{x_{k}^{\sum_{i=1}^{k-1}\mu_{i}}y_{\ell}^{c}}+\sum_{(\nu_{1},\dots,\nu_{\ell})\in N}\frac{b_{n}y_{1}^{\nu_{1}}\dots y_{\ell}^{\nu_{\ell}}}{y_{\ell}^{c}}$
		$\displaystyle=\sum_{(\mu_{1},\dots,\mu_{k})\in M}\frac{a_{m}x_{1}^{\mu_{1}}\dots x_{k-1}^{\mu_{k-1}}x_{k}^{a}}{x_{k}^{\sum_{i=1}^{k-1}\mu_{i}}y_{\ell}^{c}}+\sum_{(\nu_{1},\dots,\nu_{\ell})\in N}\frac{b_{n}y_{1}^{\nu_{1}}\dots y_{\ell}^{\nu_{\ell}}}{y_{\ell}^{c}}$
		$\displaystyle=u\sum_{(\mu_{1},\dots,\mu_{k})\in M}a_{m}t_{1}^{\mu_{1}}\dots t_{k-1}^{\mu_{k-1}}+\sum_{(\nu_{1},\dots,\nu_{\ell})\in N}b_{n}v_{1}^{\nu_{1}}\dots v_{\ell-1}^{\nu_{\ell-1}}$

from which it follows that

u=\frac{-\sum_{(\nu_{1},\dots,\nu_{\ell})\in N}b_{n}v_{1}^{\nu_{1}}\dots v_{\ell-1}^{\nu_{\ell-1}}}{\sum_{(\mu_{1},\dots,\mu_{k})\in M}a_{m}t_{1}^{\mu_{1}}\dots t_{k-1}^{\mu_{k-1}}}

noting that the denominator is non-zero because not all the $a_{m}$ are zero and $t_{1},\dots,t_{k-1}$ are algebraically independent over ${\rm\bf k}$ . It follows that $u\in{\rm\bf k}(t_{1},\dots,t_{k-1},v_{1},\dots,v_{\ell-1})$ which completes the proof.

∎

3.7 Lemma.

Let $a,c\in\mathbb{N}^{+}$ and suppose $\gcd(a,c)=1$ . If $m,n\in\mathbb{N}$ satisfy $ma+nc=ac$ , then either $m=0$ or $n=0$ .

Proof.

Assume $ma+nc=ac$ . Then $ma=(a-n)c$ so $c\mid ma$ . Since $\gcd(a,c)=1$ , $c\mid m$ . So $m=0$ or $m=c$ , in which case $n=0$ . ∎

3.8 Corollary.

Let $a,c\in\mathbb{N}^{+}$ be such that $\gcd(a,c)=1$ and let $B=\mathbb{C}_{c,c,a,a}[X_{0},X_{1},X_{2},X_{3}]/\langle f\rangle$ where $f$ is irreducible and homogeneous of degree $ac$ . Then $\operatorname{{\rm Proj}}B$ is rational over $\mathbb{C}$ .

Proof.

By Lemma 3.7, $f=g(X_{0},X_{1})+h(X_{2},X_{3})$ where each of $g(X_{0},X_{1})$ and $h(X_{2},X_{3})$ is either zero or is homogeneous of degree $ac$ . We write $B=\mathbb{C}[x_{0},x_{1},x_{2},x_{3}]$ where $x_{i}$ is the canonical image of $X_{i}$ in $B$ . Let $g=g(X_{0},X_{1})$ , $h=h(X_{2},X_{3})$ and let $\bar{f},\bar{g}$ and $\bar{h}$ denote the images of $f,g$ and $h$ in $B$ .

If $\bar{g}=0$ , then $f\mid g$ so $f=g$ . Since $f$ is irreducible, $f=\lambda_{0}X_{0}+\lambda_{1}X_{1}$ where $\lambda_{0},\lambda_{1}\in\mathbb{C}$ are not both 0. We then obtain that $\operatorname{{\rm Proj}}B\cong\mathbb{P}(c,a,a)\cong\mathbb{P}(c,1,1)$ which is rational. The same argument shows that $\operatorname{{\rm Proj}}B$ is rational if $\bar{h}=0$ . Assume now that $\bar{g}\neq 0$ and $\bar{h}\neq 0$ . Then $g(X_{0},X_{1})$ and $h(X_{2},X_{3})$ are nonzero and the result follows from Proposition 3.6. ∎

We state Lemma 7.1 of [12]. Note that the last line should read $A_{-(k-\alpha)}$ instead of $A_{-k-\alpha}$ as in [12].

3.9 Lemma.

Let $X=X_{f_{1},\dots,f_{r}}\subseteq\mathbb{P}(w_{0},\dots,w_{n})$ be a well-formed quasismooth weighted complete intersection. Let $A$ be the graded ring $\mathbb{C}_{w_{0},\dots,w_{n}}[X_{0},\dots,X_{n}]/\langle f_{1},\dots,f_{r}\rangle$ . Then,

H^{i}(X,\operatorname{\mathcal{O}}_{X}(k))\cong\begin{cases}A_{k}&\text{if }i=0\\ 0&\text{if }1\leq i<\dim X\\ A_{-(k-\alpha)}&\text{if }i=\dim X\\ \end{cases}

for all $k\in\mathbb{Z}$ .

3.10 Proposition.

Let $X=X_{f_{1},\dots,f_{r}}\subseteq\mathbb{P}(w_{0},\dots,w_{n})$ be a well-formed quasismooth weighted complete intersection of dimension $q$ that is not contained in a hyperplane. If $\alpha$ belongs to the submonoid $\langle w_{0},\dots,w_{n}\rangle$ of $(\mathbb{N},+)$ , then $h^{q}(X,\operatorname{\mathcal{O}}_{X})>0$ . Moreover, $X$ is not uniruled and not rational.

Proof.

Let $A$ be the graded ring $\mathbb{C}_{w_{0},\dots,w_{n}}[X_{0},\dots,X_{n}]/\langle f_{1},\dots,f_{r}\rangle$ and let $\tilde{X}$ be a resolution of singularities of $X$ . Suppose $\alpha\in\langle w_{0},\dots,w_{n}\rangle$ . If $\alpha=0$ , then $A_{\alpha}=\mathbb{C}$ . Otherwise, $\alpha>0$ and since $X$ is not contained in a hyperplane, $A_{\alpha}$ contains a nonzero monomial of $A$ . We obtain

0\neq A_{\alpha}\cong H^{q}(X,\operatorname{\mathcal{O}}_{X})\cong H^{q}(\tilde{X},\operatorname{\mathcal{O}}_{\tilde{X}}),

the first isomorphism by Lemma 3.9, the second since $X$ has rational singularities. This shows that $0<h^{q}(X,\operatorname{\mathcal{O}}_{X})=h^{q}(\tilde{X},\operatorname{\mathcal{O}}_{\tilde{X}})$ . Since $\tilde{X}$ has non-zero genus, it has non-negative Kodaira dimension, and hence is not uniruled. It follows that $X$ is not uniruled and not rational. ∎

3.11 Example.

Proposition 3.10 may fail if $X$ is contained in a hyperplane. Let $f_{1}=X_{0}^{3}+X_{1}^{3}+X_{2}^{7}+X_{3}^{7}$ , $f_{2}=X_{4}$ and let $X=X_{f_{1},f_{2}}\subset\mathbb{P}(7,7,3,3,1)$ . Then $X$ is quasismooth and well-formed, the amplitude $\alpha=1$ belongs to the submonoid $\langle 7,7,3,3,1\rangle$ of $\mathbb{N}^{+}$ but $X\cong\operatorname{{\rm Proj}}B_{3,3,7,7}$ is rational by Proposition 3.6.

3.12 Remark.

In Proposition 2.13 (ii) of [16], the authors prove that if $X=X_{f_{1},\dots,f_{r}}\subseteq\mathbb{P}(w_{0},\dots,w_{n})$ is a quasismooth well-formed weighted complete intersection, $\alpha>0$ and $w_{i}\mid\alpha$ for all $i\in\{0,\dots,n\}$ , then $X$ is not uniruled. They then remark that they do not know whether the assumption that $w_{i}\mid\alpha$ for all $i$ is necessary for their result to hold. Proposition 3.10 shows that their assumption can often be relaxed.

3.13 Corollary.

Suppose $X=X_{f}\subset\mathbb{P}(w_{0},w_{1},w_{2},w_{3})$ is a well-formed quasismooth weighted hypersurface of degree $d\in\mathbb{N}^{+}$ .

(a)

If $\alpha<0$ , then $X$ is rational.
(b)

If $\alpha$ belongs to the submonoid $\langle w_{0},w_{1},w_{2},w_{3}\rangle$ of $(\mathbb{N},+)$ then $h^{2}(X,\operatorname{\mathcal{O}}_{X})>0$ and $X$ is not rational.
(c)

If $d=\sum_{i=0}^{3}w_{i}$ or $d\geq\max\{2L,\sum_{i=0}^{3}w_{i}\}$ where $L=\operatorname{{\rm lcm}}(w_{0},w_{1},w_{2},w_{3})$ , $X$ is not rational.

Proof.

For (a), it is well-known that $X$ is a del Pezzo surface with quotient singularities (see p.790 of [2]) and so $X$ is rational. For (b), we note that if $\alpha\in\langle w_{0},w_{1},w_{2},w_{3}\rangle$ then $\alpha\geq 0$ and so $X$ is not contained in a hyperplane and the result follows from Proposition 3.10. Part (c) follows from Theorem A(i) and part (b). ∎

3.14 Theorem.

(a)

$X$ is rational,
(b)

$h^{2}(X,\operatorname{\mathcal{O}}_{X})=0$ ,
(c)
one of the following holds
1. (i)
  
  $n=1$ , $w_{0}=w_{1}$ , $w_{2}=w_{3}$ and $\gcd(w_{0},w_{2})=1$ ;
2. (ii)
  
  $nL-\sum_{i=0}^{3}w_{i}<0$ .

Proof.

Let $\Gamma=\langle w_{0},w_{1},w_{2},w_{3}\rangle$ , let $X=X_{f}$ , and recall that $\alpha=nL-\sum_{i=0}^{3}w_{i}$ . Since $X$ is well-formed, we have $\gcd\big{\{}\,w_{i}\,\mid\,i\in I\,\big{\}}=1$ for every subset $I$ of $\{0,1,2,3\}$ of cardinality $3$ .

That (a) implies (b) follows from the fact that $X$ has rational singularities. Assume (b) holds. Then Corollary 3.13 (b) implies that $\alpha\notin\Gamma$ , so either $\alpha<0$ or $\alpha\in\mathbb{N}\setminus\Gamma$ . If $\alpha<0$ , (ii) holds. If $\alpha\in\mathbb{N}\setminus\Gamma$ , Theorem A(ii) implies that $n=1$ and $|\{w_{0},w_{1},w_{2},w_{3}\}|=2$ , so (i) holds. This proves that (b) implies (c). We show that (c) implies (a). If (i) holds, then $X$ is rational by Corollary 3.8. If (ii) holds, then $X$ is rational by Corollary 3.13 (a).

∎

3.15 Remark.

As discussed in Remark 2.3, Proposition 3.10 and Theorem 3.14 further motivate the study of numerical semigroups generated by well-formed tuples.

3.16.

Let $X=X_{f}\subseteq\mathbb{P}(w_{0},\dots,w_{n})$ be a well-formed quasismooth hypersurface. We already noted that $X_{f}$ is a normal projective variety with cyclic quotient singularities. Since $X$ is Cohen-Macaulay, its canonical sheaf coincides with its dualizing sheaf, so Theorem 3.3.4 of [8] implies that $\operatorname{\mathcal{O}}_{X}(K_{X})\cong\operatorname{\mathcal{O}}_{X}(\alpha)$ , where $\alpha$ is the amplitude of $X$ . Note that the affine cone $C_{X}$ is a hypersurface of $\mathbb{A}^{n}$ with only one singular point; so $C_{X}$ and its coordinate ring which we denote by $B$ , are normal. Since $X$ is well-formed, Proposition 3.5 implies that $B$ is saturated in codimension one and Lemma 3.4 implies that there exists an ample integral divisor $D\in\operatorname{{\rm Div}}(X)$ such that $\operatorname{\mathcal{O}}_{X}(nD)\cong\operatorname{\mathcal{O}}_{X}(n)$ for all $n\in\mathbb{Z}$ . This implies that $\operatorname{\mathcal{O}}_{X}(\alpha D)\cong\operatorname{\mathcal{O}}_{X}(\alpha)\cong\operatorname{\mathcal{O}}_{X}(K_{X})$ . It follows that $\lfloor\alpha D\rfloor$ is a canonical divisor of $X$ ; since $D$ is integral, $\lfloor\alpha D\rfloor=\alpha D$ , so $\alpha D$ is a canonical divisor of $X$ .

3.17 Corollary.

Let $w_{0},w_{1},w_{2},w_{3}\in\mathbb{N}^{+}$ where $w_{0}\leq w_{1}\leq w_{2}\leq w_{3}$ and let $L=\operatorname{{\rm lcm}}(w_{0},w_{1},w_{2},w_{3})$ . Suppose $X_{f}\subset\mathbb{P}(w_{0},w_{1},w_{2},w_{3})$ is a well-formed quasismooth hypersurface of degree $nL$ for some $n\in\mathbb{N}^{+}$ . The following are equivalent:

(a)

$X_{f}$ is rational and has ample canonical divisor;
(b)

$n=1$ , $w_{0}=w_{1}$ , $w_{2}=w_{3}$ , $\gcd(w_{0},w_{2})=1$ , and $L>2w_{0}+2w_{2}$ .

Proof.

Let $X=X_{f}$ . By 3.16, there exists an ample integral divisor $D\in\operatorname{{\rm Div}}(X)$ such that $K_{X}=\alpha D$ where $\alpha=nL-\sum_{i=0}^{3}w_{i}$ . Suppose (a) holds. Then $K_{X}=\alpha D$ is ample, so $\alpha=nL-\sum_{i=0}^{3}w_{i}>0$ , so condition (ii) of Theorem 3.14 is not satisfied. Since $X$ is rational, condition (i) of Theorem 3.14 must be satisfied, so (b) holds. Conversely, if (b) holds then $X$ is rational by Theorem 3.14, and $\alpha=L-2w_{0}-2w_{2}>0$ , so $K_{X}=\alpha D$ is ample. ∎

3.18 Example.

For any $n\geq 2$ , we can construct a normal rational projective $\mathbb{C}$ -variety of dimension $n$ with quotient singularities and ample canonical divisor. Indeed, let ${\rm\bf k}=\mathbb{C}$ and choose some $f$ satisfying the assumptions of Proposition 3.6 such that $X_{f}=\operatorname{{\rm Proj}}(S/\langle f\rangle)\subset\mathbb{P}(c,\dots,c,a,\dots,a)$ is well-formed and quasismooth, and such that the amplitude $\alpha=ac-kc-\ell a>0$ . By 3.16, the weighted hypersurface $X_{f}$ is a normal projective variety with quotient singularities and ample canonical divisor and by Proposition 3.6, $X_{f}$ is rational. In particular, the variety $X_{f}=\operatorname{{\rm Proj}}B_{3,3,7,7}\subset\mathbb{P}(7,7,3,3)$ is one such example. By Theorem 8.1 in [12], one can produce many such examples.

3.19 Remark.

In Section 5 of [14], Koll $\rm{\acute{a}}$ r defines a family of weighted hypersurfaces $H(a_{1},\dots,a_{n})\subseteq\mathbb{P}(a_{1},\dots,a_{n})$ , now known as Koll $\acute{a}$ r hypersurfaces; some of these (see Theorem 39 in [14]) are rational with ample canonical divisor. As far as we know, they represent the first family of rational weighted projective hypersurfaces with ample canonical divisor, with Example 3.18 providing another such family. (Currently, we only know of these two families with this property.) Note the special case of 2-dimensional Koll $\rm{\acute{a}}$ r hypersurfaces is studied in detail in [19].

In view of Theorem 3.14 and Remark 3.19, we note that we know of no examples of two-dimensional non-rational quasismooth weighted hypersurfaces $X$ such that $h^{2}(X,\operatorname{\mathcal{O}}_{X})=0$ . Hence we ask the following question:

3.20 Question.

Does there exist a non-rational two-dimensional well-formed quasismooth hypersurface $X$ such that $h^{2}(X,\operatorname{\mathcal{O}}_{X})=0$ ? Theorem 3.14 implies that if such an $X$ exists and is a hypersurface, the degree of $f$ is strictly positive and not divisible by $L$ .

4 The Rational Affine Pham-Brieskorn Threefolds

Let $B_{a_{0},\dots,a_{n}}=\mathbb{C}[X_{0},\dots,X_{n}]/\langle X_{0}^{a_{0}}+X_{1}^{a_{1}}+\dots+X_{n}^{a_{n}}\rangle$ . The ring $B_{a_{0},\dots,a_{n}}$ is called a Pham-Brieskorn ring and $\operatorname{{\rm Spec}}B_{a_{0},\dots,a_{n}}$ is called an affine Pham-Brieskorn variety. We apply the results of the previous sections to answer the following question of R.V. Gurjar:

4.1 Question.

For which 4-tuples $(a_{0},a_{1},a_{2},a_{3})$ is $\operatorname{{\rm Spec}}B_{a_{0},a_{1},a_{2},a_{3}}$ a $\mathbb{C}$ -rational variety?

4.2.

Let $n\geq 2$ , $(a_{0},\dots,a_{n})\in(\mathbb{N}^{+})^{n+1}$ and $f=X_{0}^{a_{0}}+X_{1}^{a_{1}}+\dots+X_{n}^{a_{n}}\in\mathbb{C}[X_{0},\dots,X_{n}]$ . Let $L=\operatorname{{\rm lcm}}(a_{0},\dots,a_{n})$ and let $\deg(X_{i})=w_{i}=L/a_{i}$ for each $i\in\{0,\dots,n\}$ . Then $f$ is a homogeneous irreducible element of the graded ring $\mathbb{C}_{w_{0},\dots,n_{n}}[X_{0},\dots,X_{n}]$ , $B_{a_{0},\dots,a_{n}}=\mathbb{C}_{w_{0},\dots,w_{n}}[X_{0},\dots,X_{n}]/\langle f\rangle$ is a graded ring and $\deg(x_{i})=w_{i}$ for each $i\in\{0,\dots,n\}$ where $x_{i}\in B_{a_{0},\dots,a_{n}}$ is the canonical image of $X_{i}$ .

4.3 Proposition.

Let $B=B_{a_{0},a_{1},a_{2},a_{3}}$ . Then $\operatorname{{\rm Spec}}B$ is rational over $\mathbb{C}$ if and only if $\operatorname{{\rm Proj}}B$ is rational over $\mathbb{C}$ .

Proof.

Let $K$ be the function field of $\operatorname{{\rm Proj}}B$ and recall that the function field of $\operatorname{{\rm Spec}}B$ is isomorphic to $K^{(1)}$ . Assume $\operatorname{{\rm Spec}}B$ is rational over $\mathbb{C}$ . Then $K^{(1)}\cong\mathbb{C}^{(3)}$ and so the function field of $\operatorname{{\rm Proj}}B$ is stably rational, hence unirational. By Castelnuovo’s Theorem, $K$ is rational. The converse is clear. ∎

4.4 Definition.

Given a tuple $S=(a_{0},\dots,a_{n})\in(\mathbb{N}^{+})^{n+1}$ , we define $S_{i}=(a_{0},\dots,a_{i-1},\hat{a}_{i},a_{i+1},\dots,a_{n})$ and $L_{i}=\operatorname{{\rm lcm}}S_{i}$ . We define $\operatorname{{\rm cotype}}(S)=|\big{\{}\,i\in\{0,\dots,n\}\,\mid\,L_{i}\neq L\,\big{\}}|$ .

4.5.

Let $B=B_{a_{0},\dots,a_{n}}$ be Pham-Brieskorn ring. Lemma 3.1.14 in [3] shows that there exists a Pham-Brieskorn ring $B^{\prime}=B_{b_{0},\dots,b_{n}}$ such that $\operatorname{{\rm cotype}}(b_{0},\dots,b_{n})=0$ and $\operatorname{{\rm Proj}}B\cong\operatorname{{\rm Proj}}B^{\prime}$ . Moreover, given a Pham-Brieskorn ring $B$ , it is straightforward to determine $B^{\prime}$ (or equivalently the tuple $(b_{0},\dots,b_{n})$ ) explicitly; in particular, the proof of Lemma 3.1.14 in [3] shows that $b_{i}\leq a_{i}$ for all $i\in\{0,\dots,n\}$ . This is demonstrated in Example 4.8 below. Consequently, to answer Question 4.1, it suffices to consider the special case where $\operatorname{{\rm cotype}}(a_{0},a_{1},a_{2},a_{3})=0$ .

4.6 Proposition.

[3, Proposition 2.4.25] Let $f=X_{0}^{a_{0}}+\dots+X_{n}^{a_{n}}$ where $n\geq 2$ and $a_{i}\geq 1$ for all $i$ . Then, the weighted hypersurface $X_{f}\subset\mathbb{P}=\mathbb{P}(w_{0},\dots,w_{n})$ is a well-formed quasismooth hypersurface if and only if $\operatorname{{\rm cotype}}(a_{0},\dots,a_{n})=0$ .

4.7 Theorem.

(a)

$a_{0}=a_{1}$ , $a_{2}=a_{3}$ and $\gcd(a_{0},a_{2})=1$ ;
(b)

$\frac{1}{a_{0}}+\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}>1$ .

Proof.

Let $X=\operatorname{{\rm Proj}}B_{a_{0},a_{1},a_{2},a_{3}}$ . Let $L^{\prime}=\operatorname{{\rm lcm}}(a_{0},a_{1},a_{2},a_{3})$ and let $L=\operatorname{{\rm lcm}}(w_{0},w_{1},w_{2},w_{3})$ . Since $w_{i}=L^{\prime}/a_{i}$ for each $i\in\{0,1,2,3\}$ , we obtain that $L^{\prime}=nL$ for some $n\in\mathbb{N}^{+}$ . Proposition 4.6 shows that $X$ is a well-formed quasismooth hypersurface of degree $nL$ so $X$ satisfies the hypotheses of Theorem 3.14. It is easy to see that condition (a) is equivalent to condition (i) in Theorem 3.14. Since

\frac{1}{a_{0}}+\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}>1\iff nL-\sum_{i=0}^{3}w_{i}<0,

condition (b) is equivalent to condition (ii) in Theorem 3.14. These two equivalences prove the theorem.

∎

4.8 Example.

We show that both $\operatorname{{\rm Proj}}B_{3,7,14,87}$ and $\operatorname{{\rm Spec}}B_{3,7,14,87}$ are rational. Note that $\operatorname{{\rm cotype}}(3,7,14,87)=2$ so we cannot directly apply Theorem 4.7. By Proposition 5.2 in [4], we find $\operatorname{{\rm Proj}}B_{3,7,14,87}\cong\operatorname{{\rm Proj}}B_{3,7,14,3}\cong\operatorname{{\rm Proj}}B_{3,7,7,3}\cong\operatorname{{\rm Proj}}B_{3,3,7,7}$ . Since $\operatorname{{\rm cotype}}(3,3,7,7)=0$ , $\operatorname{{\rm Proj}}B_{3,3,7,7}$ is rational by Theorem 4.7. Consequently, $\operatorname{{\rm Proj}}B_{3,7,14,87}$ is rational and hence $\operatorname{{\rm Spec}}B_{3,7,14,87}$ is rational by Proposition 4.3.

We also obtain the following. Note there is no assumption on the cotype.

4.9 Corollary.

Let $B_{a_{0},a_{1},a_{2},a_{3}}$ be a Pham-Brieskorn ring and consider the following statements:

(a)

$\operatorname{{\rm Spec}}B_{a_{0},a_{1},a_{2},a_{3}}$ has a rational singularity at the origin;
(b)

$\frac{1}{a_{0}}+\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}>1$ ;
(c)

$\operatorname{{\rm Proj}}B_{a_{0},a_{1},a_{2},a_{3}}$ and $\operatorname{{\rm Spec}}B_{a_{0},a_{1},a_{2},a_{3}}$ are rational.

Then $(a)\iff(b)\Rightarrow(c)$ .

Proof.

The equivalence of (a) and (b) is stated in Example 2.21 of [9] so it suffices to prove (b) implies (c). By 4.5, there exists a tuple $(b_{0},b_{1},b_{2},b_{3})$ such that $\operatorname{{\rm Proj}}B_{a_{0},a_{1},a_{2},a_{3}}\cong\operatorname{{\rm Proj}}B_{b_{0},b_{1},b_{2},b_{3}}$ , $\operatorname{{\rm cotype}}(b_{0},b_{1},b_{2},b_{3})=0$ and $b_{i}\leq a_{i}$ for all $i\in\{0,1,2,3\}$ . Then $\frac{1}{b_{0}}+\frac{1}{b_{1}}+\frac{1}{b_{2}}+\frac{1}{b_{3}}\geq\frac{1}{a_{0}}+\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}>1$ . By Theorem 4.7 (b), $\operatorname{{\rm Proj}}B_{b_{0},b_{1},b_{2},b_{3}}$ is rational and so $\operatorname{{\rm Proj}}B_{a_{0},a_{1},a_{2},a_{3}}$ and $\operatorname{{\rm Spec}}B_{a_{0},a_{1},a_{2},a_{3}}$ are rational as well. ∎

5 Proof of Theorem A

This section provides the proof of Theorem A. Recall the definition of a well-formed tuple from Definition 2.1.

Theorem A.

Suppose $(d_{1},d_{2},d_{3},d_{4})\in(\mathbb{N}^{+})^{4}$ is well-formed, let $\Gamma=\langle d_{1},d_{2},d_{3},d_{4}\rangle$ and let $L=\operatorname{{\rm lcm}}(d_{1},d_{2},d_{3},d_{4})$ .

(i)

If $N\geq\max\{2L-\sum_{m=1}^{4}d_{m},0\}$ then $N\in\Gamma$ .
(ii)

If $n\in\mathbb{N}^{+}$ and $n\cdot L-\sum_{i=1}^{4}d_{i}\in\mathbb{N}\setminus\Gamma$ , then $n=1$ and $|\{d_{1},d_{2},d_{3},d_{4}\}|=2$ .

The proof of Theorem A is given in paragraphs 5.2–5.16. We preserve the following notation until the end of proof.

5.1 Notation.

Given $d_{1},d_{2},d_{3},d_{4}\in\mathbb{N}^{+}$ , we define $L=\operatorname{{\rm lcm}}(d_{1},d_{2},d_{3},d_{4})$ and $\Gamma=\langle d_{1},d_{2},d_{3},d_{4}\rangle$ . Given $i,j\in\{1,2,3,4\}$ , we define $L_{i,j}=\operatorname{{\rm lcm}}(d_{i},d_{j})$ and $g_{i,j}=\gcd(d_{i},d_{j})$ .

5.2 Proposition.

(a)

Let $d_{1},d_{2}\in\mathbb{N}^{+}$ and $n\in\mathbb{N}$ . If $n>L_{1,2}-d_{1}-d_{2}$ and $g_{1,2}\mid n$ then $n\in\langle d_{1},d_{2}\rangle$ .
(b)

If $d_{1},d_{2},d_{3}\in\mathbb{N}^{+}$ are relatively prime then $F(d_{1},d_{2},d_{3})\leq L_{1,2}+L_{1,3}-d_{1}-d_{2}-d_{3}$ .

Proof.

The case $g_{1,2}=1$ of (a) is a well-known result of Frobenius; the general case follows. For (b), note that $F(d_{1},d_{2},d_{3})\leq L_{1,2}+g_{1,2}d_{3}-d_{1}-d_{2}-d_{3}$ by the $k=3$ case of statement (6) in the introduction of [1]. Since $\gcd(g_{1,2},g_{1,3})=1$ , we get $g_{1,2}g_{1,3}\mid d_{1}$ , implying that $g_{1,2}d_{3}$ divides $\frac{d_{1}d_{3}}{g_{1,3}}=L_{1,3}$ ; so $g_{1,2}d_{3}\leq L_{1,3}$ and (b) follows. ∎

5.3 Assumption.

We assume from this point onward that $d_{1},d_{2},d_{3},d_{4}\in\mathbb{N}^{+}$ and $(d_{1},d_{2},d_{3},d_{4})$ is well-formed.

5.4 Lemma.

If $i,j,k,l$ are such that $\{i,j,k,l\}=\{1,2,3,4\}$ , then $g_{k,l}\mid\frac{L}{L_{i,j}}$ .

Proof.

Let $m=L/L_{i,j}$ . Since $g_{k,l}$ is relatively prime to $d_{i}$ and also to $d_{j}$ , it is relatively prime to $L_{i,j}$ . Since $g_{k,l}$ divides $L=mL_{i,j}$ and is relatively prime to $L_{i,j}$ , we obtain $g_{k,l}\mid m$ , as desired. ∎

Proof of Theorem A(i). We may assume that $d_{1}\leq d_{2}\leq d_{3}\leq d_{4}$ . We may also assume that $d_{1}\geq 2$ , otherwise the result is trivial.

Consider first the case where $L_{1,3}=L$ . Since $d_{2}$ and $d_{4}$ divide $L_{1,3}$ and (by Lemma 5.4) $g_{2,4}=1$ , we get $d_{2}d_{4}\mid L_{1,3}\mid d_{1}d_{3}$ , so $d_{2}d_{4}\leq d_{1}d_{3}$ . Since $d_{1}\leq d_{2}$ and $d_{3}\leq d_{4}$ it follows that $d_{1}=d_{2}$ and $d_{3}=d_{4}$ . Note that $d_{1}d_{3}-d_{1}-d_{3}>0$ , because $d_{1},d_{3}\geq 2$ and $g_{1,3}=\gcd(d_{1},d_{2},d_{3},d_{4})=1$ . We have $N\geq 2L-\sum_{m=1}^{4}d_{m}=2(d_{1}d_{3}-d_{1}-d_{3})>d_{1}d_{3}-d_{1}-d_{3}$ , so Proposition 5.2(a) gives $N\in\Gamma$ , as desired.

Next, consider the case where $L_{1,3}\neq L$ . Proposition 5.2(b) gives $F(d_{1},d_{3},d_{4})\leq L_{1,3}+L_{1,4}-d_{1}-d_{3}-d_{4}$ , so

	$\displaystyle\textstyle N-F(d_{1},d_{3},d_{4})$	$\displaystyle\textstyle\geq 2L-\sum_{m=1}^{4}d_{m}-L_{1,3}-L_{1,4}+d_{1}+d_{3}+d_{4}$
		$\displaystyle=(L-L_{1,3})+(L-L_{1,4})-d_{2}.$

We have $(L-L_{1,3})+(L-L_{1,4})\geq L-L_{1,3}=L_{1,3}(\frac{L}{L_{1,3}}-1)\geq L_{1,3}$ , so

\textstyle N-F(d_{1},d_{3},d_{4})\geq L_{1,3}-d_{2}.

(5)

We claim that $L_{1,3}-d_{2}>0$ . Indeed, $L_{1,3}-d_{2}\geq d_{3}-d_{2}\geq 0$ ; if $L_{1,3}-d_{2}=0$ then $L_{1,3}=d_{2}=d_{3}$ , so $d_{1}\mid d_{2}$ and $d_{1}\mid d_{3}$ , so $\gcd(d_{1},d_{2},d_{3})\geq d_{1}>1$ , a contradiction. This shows that $L_{1,3}-d_{2}>0$ , so (5) gives $N>F(d_{1},d_{3},d_{4})\geq F(\Gamma)$ as required. $\square$

5.5 Assumption.

Let $\alpha=L-\sum_{i=1}^{4}d_{i}$ and assume that $\alpha\in\mathbb{N}\setminus\Gamma$ .

5.6 Remark.

In order to finish the proof of Theorem A, it suffices to prove that Assumption 5.5 implies that $|\{d_{1},d_{2},d_{3},d_{4}\}|=2$ . (This follows from part (i) of Theorem A.) Assumption 5.5 remains in effect until the end of the proof of Theorem A.

5.7.

Assumptions 5.5 immediately imply:

(a)

$\alpha>0$
(b)

$\min(d_{1},d_{2},d_{3},d_{4})\geq 2$
(c)

$\sum_{i=1}^{4}d_{i}<L$ , so in particular $L\notin\{d_{1},d_{2},d_{3},d_{4}\}$ .
(d)

$|\{d_{1},d_{2},d_{3},d_{4}\}|\geq 2$
(e)

If $i,j,k\in\{1,2,3,4\}$ are distinct, then $\gcd(g_{i,j},d_{k})=1$ .

5.8 Lemma.

If there exist distinct $i,j\in\{1,2,3,4\}$ such that $d_{i}\mid d_{j}$ , then $|\{d_{1},d_{2},d_{3},d_{4}\}|=2$ .

Proof.

Without loss of generality, we may assume that $d_{3}\mid d_{4}$ . Then $g_{1,3}=1=g_{2,3}$ . Let $s=d_{4}/d_{3}\in\mathbb{N}^{+}$ . Define $a=\big{\lceil}\frac{s+1}{g_{1,2}}\big{\rceil}$ and $c=ag_{1,2}-(s+1)\in\mathbb{N}$ . We have $\alpha=L-d_{1}-d_{2}-(s+1)d_{3}\equiv-(s+1)d_{3}\pmod{g_{1,2}}$ and $c\equiv-(s+1)\pmod{g_{1,2}}$ , so $\alpha-cd_{3}\equiv 0\pmod{g_{1,2}}$ . Since $\alpha\notin\Gamma$ and $c\in\mathbb{N}$ , we have $\alpha-cd_{3}\notin\langle d_{1},d_{2}\rangle$ , so Proposition 5.2(a) implies that $\alpha-cd_{3}\leq L_{1,2}-d_{1}-d_{2}$ , i.e.,

0\leq L_{1,2}-d_{1}-d_{2}-\alpha+cd_{3}=L_{1,2}-L+(s+1+c)d_{3}=L_{1,2}-L+ag_{1,2}d_{3}.

(6)

Define $s_{i}=\gcd(s,d_{i})$ for $i=1,2$ . Since $\gcd(s_{1},s_{2})$ divides $\gcd(d_{1},d_{2},d_{4})=1$ , and since $\gcd(s_{1},d_{3})$ divides $\gcd(d_{1},d_{3},d_{4})=1$ and $\gcd(s_{2},d_{3})$ divides $\gcd(d_{2},d_{3},d_{4})=1$ , we have

s_{1}

s_{2}

and

d_{3}

are pairwise relatively prime.

(7)

We claim that

\gcd(s,L_{1,2})=s_{1}s_{2}.

(8)

To see this, first note that $s_{i}\mid\gcd(s,L_{1,2})$ for each $i=1,2$ ; so $s_{1}s_{2}\mid\gcd(s,L_{1,2})$ by (7). Conversely, if a prime power $p^{i}$ divides $\gcd(s,L_{1,2})$ then the fact that $\gcd(s,d_{1},d_{2})=1$ implies that $p^{i}$ divides $s_{1}$ or $s_{2}$ , so $p^{i}\mid s_{1}s_{2}$ , showing that $\gcd(s,L_{1,2})$ divides $s_{1}s_{2}$ . This proves (8). Define

\textstyle\sigma=\frac{\operatorname{{\rm lcm}}(s,L_{1,2})}{L_{1,2}}\in\mathbb{N}^{+}

and note that $\sigma=\frac{\operatorname{{\rm lcm}}(s,L_{1,2})}{L_{1,2}}=\frac{sL_{1,2}}{L_{1,2}\gcd(s,L_{1,2})}=\frac{s}{\gcd(s,L_{1,2})}=\frac{s}{s_{1}s_{2}}$ by (8); so

s=s_{1}s_{2}\sigma.

Since $\gcd(L_{1,2},d_{3})=1$ , we have $L=\operatorname{{\rm lcm}}(L_{1,2},sd_{3})=\frac{L_{1,2}\,sd_{3}}{\gcd(L_{1,2},sd_{3})}=\frac{L_{1,2}\,sd_{3}}{\gcd(L_{1,2},s)}=\operatorname{{\rm lcm}}(L_{1,2},s)d_{3}=L_{1,2}\,\sigma d_{3}$ . So (6) gives $0\leq L_{1,2}-L+ag_{1,2}d_{3}=L_{1,2}(1-\sigma d_{3})+ag_{1,2}d_{3}$ and hence

L_{1,2}(\sigma d_{3}-1)\leq ag_{1,2}d_{3}.

(9)

Since $\gcd(L_{1,2},d_{3})=1$ and $d_{3}>1$ , we have $d_{3}\nmid L_{1,2}$ , so the inequality in (9) is strict; since both sides of this strict inequality are multiples of $g_{1,2}$ , it follows that $L_{1,2}(\sigma d_{3}-1)\leq ag_{1,2}d_{3}-g_{1,2}$ , so

\textstyle\frac{L_{1,2}}{g_{1,2}}\leq\frac{ad_{3}-1}{\sigma d_{3}-1}\,.

(10)

Consider the case $s=1$ . Then $\sigma=1$ , so (10) gives $\big{(}\frac{d_{1}}{g_{1,2}}\big{)}\big{(}\frac{d_{2}}{g_{1,2}}\big{)}=\frac{L_{1,2}}{g_{1,2}}\leq\frac{ad_{3}-1}{d_{3}-1}\leq 2a-1$ , where the last inequality uses $d_{3}\geq 2$ . Recall that $a=\big{\lceil}\frac{s+1}{g_{1,2}}\big{\rceil}=\big{\lceil}\frac{2}{g_{1,2}}\big{\rceil}$ . If $g_{1,2}=1$ then $a=2$ , so $d_{1}d_{2}\leq 3$ , which contradicts 5.7(b). So $g_{1,2}>1$ , in which case we have $a=1$ , so $\big{(}\frac{d_{1}}{g_{1,2}}\big{)}\big{(}\frac{d_{2}}{g_{1,2}}\big{)}\leq 1$ , so $d_{1}=d_{2}$ and $d_{4}=sd_{3}=d_{3}$ , showing that $|\{d_{1},d_{2},d_{3},d_{4}\}|\leq 2$ . In view of 5.7(d), this shows that if $s=1$ then $|\{d_{1},d_{2},d_{3},d_{4}\}|=2$ . So, to complete the proof, it suffices to show that

the case

s\geq 2

does not occur.

(11)

From now-on, we assume that $s\geq 2$ . Since $\gcd(g_{1,2},s)$ divides $\gcd(d_{1},d_{2},d_{4})=1$ , we have

\gcd(g_{1,2},s)=1.

(12)

For each $i\in\{1,2\}$ , define $\delta_{i}=d_{i}/s_{i}\in\mathbb{N}\setminus\{0\}$ . For each $i\in\{1,2\}$ , $g_{1,2}$ divides $d_{i}=\delta_{i}s_{i}$ and is relatively prime to $s_{i}$ , so $g_{1,2}\mid\delta_{i}$ . Thus,

\textstyle\frac{L_{1,2}}{g_{1,2}}=\big{(}\frac{d_{1}}{g_{1,2}}\big{)}\big{(}\frac{d_{2}}{g_{1,2}}\big{)}=\big{(}\frac{\delta_{1}}{g_{1,2}}\big{)}\big{(}\frac{\delta_{2}}{g_{1,2}}\big{)}s_{1}s_{2},\quad\text{where $\big{(}\frac{\delta_{1}}{g_{1,2}}\big{)},\big{(}\frac{\delta_{2}}{g_{1,2}}\big{)}\in\mathbb{N}^{+}$.}

(13)

We shall now consider three cases ( $g_{1,2}=1$ , $g_{1,2}=2$ , $g_{1,2}\geq 3$ ) and show that none of them is possible. Assume that $g_{1,2}\in\{1,2\}$ . Note that $g_{1,2}\mid(s+1)$ , because (12) shows that if $g_{1,2}=2$ then $s$ is odd. So $a=\big{\lceil}\frac{s+1}{g_{1,2}}\big{\rceil}=\frac{s+1}{g_{1,2}}$ . This together with (10) gives

L_{1,2}(\sigma d_{3}-1)\leq g_{1,2}ad_{3}-g_{1,2}=(s+1)d_{3}-g_{1,2}<(s+1)d_{3}.

In view of (13), it follows that

\textstyle\big{(}\frac{\delta_{1}}{g_{1,2}}\big{)}\big{(}\frac{\delta_{2}}{g_{1,2}}\big{)}=\frac{L_{1,2}}{s_{1}s_{2}g_{1,2}}<\frac{(s+1)d_{3}}{s_{1}s_{2}g_{1,2}(\sigma d_{3}-1)}=\big{(}\frac{1}{g_{1,2}}\big{)}\big{(}\frac{s+1}{s}\big{)}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}\,.

(14)

Note that if $\delta_{1}=1=\delta_{2}$ then both $d_{1}=s_{1}$ and $d_{2}=s_{2}$ divide $d_{4}$ , so in fact $L=d_{4}$ , contradicting 5.7(c). So $\delta_{1}\delta_{2}\geq 2$ .

If $g_{1,2}=1$ then (14) gives $2\leq\delta_{1}\delta_{2}<\big{(}\frac{s+1}{s}\big{)}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}\leq 3$ (the last inequality because $s,d_{3}\geq 2$ ), so $\delta_{1}\delta_{2}=2$ . We may assume that $\delta_{1}=1$ and $\delta_{2}=2$ ; then $d_{1}=s_{1}$ and $d_{2}=2s_{2}$ . Since $g_{2,3}=1$ , $d_{3}$ is odd, so $d_{3}\geq 3$ . If $d_{3}\neq 3$ then $d_{3}\geq 5$ and hence $\big{(}\frac{s+1}{s}\big{)}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}<2$ , a contradiction. So $(d_{1},d_{2},d_{3})=(s_{1},2s_{2},3)$ . Since $g_{1,2}=1$ and $g_{1,3}=1$ , $s_{1}$ is odd and $s_{1}\neq 3$ , so $s_{1}\geq 5$ and hence $s\geq 5$ . This implies that $\big{(}\frac{s+1}{s}\big{)}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}<2$ , a contradiction. So the case $g_{1,2}=1$ does not occur.

If $g_{1,2}=2$ then (14) gives $1\leq\big{(}\frac{\delta_{1}}{2}\big{)}\big{(}\frac{\delta_{2}}{2}\big{)}<\big{(}\frac{1}{2}\big{)}\big{(}\frac{s+1}{s}\big{)}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}$ , so $\big{(}\frac{s+1}{s}\big{)}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}>2$ . Since $1=\gcd(g_{1,2},d_{4})=\gcd(2,d_{4})$ , $d_{4}=sd_{3}$ is odd, so $d_{3}$ and $s$ are odd. Hence, $d_{3}\geq 3$ and $s\geq 3$ . Then $2<\big{(}\frac{s+1}{s}\big{)}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}\leq\frac{4}{3}\cdot\frac{3}{2}=2$ , a contradiction. So the case $g_{1,2}=2$ does not occur either.

From now-on, we assume that $g_{1,2}\geq 3$ ; let us show that this leads to a contradicition. The definition $a=\big{\lceil}\frac{s+1}{g_{1,2}}\big{\rceil}$ implies that $a<\frac{s+1}{g_{1,2}}+1$ , so $ag_{1,2}<s+1+g_{1,2}$ and hence $ag_{1,2}\leq s+g_{1,2}$ . So (10) gives $L_{1,2}(\sigma d_{3}-1)\leq ag_{1,2}d_{3}-g_{1,2}\leq(s+g_{1,2})d_{3}-g_{1,2}$ , or equivalently (using (13))

\textstyle\big{(}\frac{\delta_{1}}{g_{1,2}}\big{)}\big{(}\frac{\delta_{2}}{g_{1,2}}\big{)}s_{1}s_{2}g_{1,2}=L_{1,2}\leq\frac{(s+g_{1,2})d_{3}-g_{1,2}}{\sigma d_{3}-1}=\frac{sd_{3}}{\sigma d_{3}-1}+\frac{g_{1,2}(d_{3}-1)}{\sigma d_{3}-1}=\frac{sd_{3}}{\sigma(d_{3}-\frac{1}{\sigma})}+\frac{g_{1,2}(d_{3}-1)}{\sigma(d_{3}-\frac{1}{\sigma})}\,.

Dividing both sides by $s_{1}s_{2}g_{1,2}$ and using $s=s_{1}s_{2}\sigma$ gives

\textstyle 1\leq\big{(}\frac{\delta_{1}}{g_{1,2}}\big{)}\big{(}\frac{\delta_{2}}{g_{1,2}}\big{)}\leq\frac{1}{g_{1,2}}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}+\frac{1}{s}\big{(}\frac{d_{3}-1}{d_{3}-\frac{1}{\sigma}}\big{)}\,.

(15)

If $\sigma\geq 2$ then (using $g_{1,2}\geq 3$ and $s,d_{3}\geq 2$ ) $\frac{1}{g_{1,2}}\big{(}\frac{d_{3}}{d_{3}-\frac{1}{\sigma}}\big{)}+\frac{1}{s}\big{(}\frac{d_{3}-1}{d_{3}-\frac{1}{\sigma}}\big{)}\leq\frac{1}{3}\cdot\frac{2}{3/2}+\frac{1}{2}\cdot\frac{1}{3/2}<1$ , which contradicts (15). So $\sigma=1$ and $s=s_{1}s_{2}$ . Note that $\gcd(s,d_{3})=\gcd(s_{1}s_{2},d_{3})=1$ by (7), and $\gcd(g_{1,2},sd_{3})=\gcd(d_{1},d_{2},d_{4})=1$ ; so

g_{1,2}

s

and

d_{3}

are pairwise relatively prime.

(16)

Substituting $\sigma=1$ in (15) gives

\textstyle 1\leq\big{(}\frac{\delta_{1}}{g_{1,2}}\big{)}\big{(}\frac{\delta_{2}}{g_{1,2}}\big{)}\leq\frac{1}{g_{1,2}}\big{(}\frac{d_{3}}{d_{3}-1}\big{)}+\frac{1}{s}\leq\frac{2}{g_{1,2}}+\frac{1}{s}\,.

(17)

Now $\frac{2}{g_{1,2}}+\frac{1}{s}\geq 1$ is equivalent to $(g_{1,2}-2)(s-1)\leq 2$ , which implies that $(g_{1,2},s)=(3,2)$ (because $g_{1,2}\geq 3$ , $s\geq 2$ and $\gcd(g_{1,2},s)=1$ ). Then (17) gives $1\leq\frac{1}{g_{1,2}}\big{(}\frac{d_{3}}{d_{3}-1}\big{)}+\frac{1}{s}=\frac{1}{3}\big{(}\frac{d_{3}}{d_{3}-1}\big{)}+\frac{1}{2}$ , so $d_{3}\in\{2,3\}$ , which contradicts (16). So the case $g_{1,2}\geq 3$ does not occur. This proves (11), and completes the proof of the Lemma. ∎

5.9 Remark.

In order to finish the proof of Theorem A, it suffices to prove that there exist distinct $i,j\in\{1,2,3,4\}$ such that $d_{i}\mid d_{j}$ . Indeed, if such $i,j$ exist then Lemma 5.8 implies that $|\{d_{1},d_{2},d_{3},d_{4}\}|=2$ , which (by Remark 5.6) completes the proof of the Theorem. Our proof that $i,j$ exist is by contradiction. In other words, we shall prove that Assumptions 5.10 lead to a contradiction.

5.10 Assumptions.

We continue to assume that the conditions of 5.3 and 5.5 are satisfied, and we also assume that

no elements

i,j\in\{1,2,3,4\}

satisfy

i\neq j

and

d_{i}\mid d_{j}

We will derive a contradiction from the above assumption. Without loss of generality, we also assume that the labeling of $d_{1},d_{2},d_{3},d_{4}$ is such that $\max(d_{1},d_{2},d_{3},d_{4})=d_{4}$ . Define positive integers $m,f,g$ by

\textstyle m=\frac{L}{d_{4}},\quad\frac{L}{d_{1}d_{2}d_{3}}=\frac{f}{g}\quad\text{and}\quad\gcd(f,g)=1.

Note that $m\geq 2$ (if $m=1$ then $L=d_{4}$ contradicts 5.7(c)).

5.11 Lemma.

(a)

$g=g_{1,2}\,g_{1,3}\,g_{2,3}$
(b)

$g\mid m$

Proof.

Define $e_{p}=v_{p}\big{(}\frac{L}{d_{1}d_{2}d_{3}}\big{)}$ for each prime number $p$ , and observe that

\textstyle f=\prod\limits_{e_{p}>0}p^{e_{p}}\quad\text{and}\quad g=\prod\limits_{e_{p}<0}p^{-e_{p}}.

If $p$ is a prime factor of $g$ then $v_{p}(d_{1}d_{2}d_{3})>v_{p}(L)\geq\max\limits_{1\leq i\leq 3}v_{p}(d_{i})$ , so $p$ divides exactly two elements of $\{d_{1},d_{2},d_{3}\}$ and consequently there exist distinct $i,j\in\{1,2,3\}$ such that $p\mid g_{i,j}$ . Thus,

every prime factor of

g

is a factor of

g_{1,2}\,g_{1,3}\,g_{2,3}

(18)

Conversely, suppose that $p$ is a prime factor of $g_{1,2}\,g_{1,3}\,g_{2,3}$ . Then there exists exactly one choice of elements $i<j$ of $\{1,2,3\}$ such that $p\mid g_{i,j}$ ; we have $v_{p}(d_{1}d_{2}d_{3})=v_{p}(d_{i}d_{j})=\max(v_{p}(d_{i}),v_{p}(d_{j}))+\min(v_{p}(d_{i}),v_{p}(d_{j}))=v_{p}(L)+v_{p}(g_{i,j})$ , so $e_{p}=v_{p}(L)-v_{p}(d_{1}d_{2}d_{3})=-v_{p}(g_{i,j})<0$ and hence $p\mid g$ . Moreover, $v_{p}(g)=-e_{p}=v_{p}(g_{i,j})=v_{p}(g_{1,2}\,g_{1,3}\,g_{2,3})$ . Hence,

every prime factor

p

g_{1,2}\,g_{1,3}\,g_{2,3}

satisfies

v_{p}(g)=v_{p}(g_{1,2}\,g_{1,3}\,g_{2,3})

(19)

Statements (18) and (19) imply that $g_{1,2}\,g_{1,3}\,g_{2,3}=g$ , which proves (a).

Let $i<j$ be elements of $\{1,2,3\}$ . Then $\gcd(g_{i,j},d_{4})=1$ and $g_{i,j}$ divides $L=md_{4}$ , so $g_{i,j}\mid m$ . Since $g_{1,2},g_{1,3},g_{2,3}$ are pairwise relatively prime, we obtain $(g_{1,2}\,g_{1,3}\,g_{2,3})\mid m$ , proving (b). ∎

5.12 Lemma.

For every choice of distinct $i,j\in\{1,2,3\}$ , we have $gm\nmid d_{i}d_{j}$ .

Proof.

Let $L_{1,2,3}=\operatorname{{\rm lcm}}(d_{1},d_{2},d_{3})$ and $M=\frac{d_{1}d_{2}d_{3}}{g}$ . We will show that

L_{1,2,3}=M.

(20)

Lemma 5.11(a) implies that $\frac{M}{d_{1}}=\frac{d_{2}d_{3}}{g}=\frac{d_{2}}{g_{1,2}\,g_{2,3}}\cdot\frac{d_{3}}{g_{1,3}}\in\mathbb{Z}$ , so $d_{1}\mid M$ . By symmetry, $d_{2},d_{3}\mid M$ , so $L_{1,2,3}\mid M$ . Conversely, consider a prime factor $p$ of $M$ . We can choose $u,v,w$ such that $\{u,v,w\}=\{1,2,3\}$ and $p\nmid d_{w}$ . Then $v_{p}(M)=v_{p}(\frac{d_{u}d_{v}}{g_{u,v}})+v_{p}(\frac{d_{w}}{g_{u,w}\,g_{v,w}})=v_{p}(\frac{d_{u}d_{v}}{g_{u,v}})=v_{p}(L_{u,v})\leq v_{p}(L_{1,2,3})$ , and this shows that $M\mid L_{1,2,3}$ . This shows (20).

Arguing by contradiction, suppose that $i,j,k$ are such that $\{i,j,k\}=\{1,2,3\}$ and $gm\mid d_{i}d_{j}$ . Then $\frac{d_{1}d_{2}d_{3}}{gm}\in d_{k}\mathbb{Z}$ . We have $\frac{d_{1}d_{2}d_{3}}{g}=M=L_{1,2,3}$ by (20), so

\textstyle\frac{d_{4}}{\big{(}\frac{d_{1}d_{2}d_{3}}{gm}\big{)}}=\frac{md_{4}}{\big{(}\frac{d_{1}d_{2}d_{3}}{g}\big{)}}=\frac{L}{L_{1,2,3}}\in\mathbb{Z},

so $\frac{d_{1}d_{2}d_{3}}{gm}$ divides $d_{4}$ . Since $\frac{d_{1}d_{2}d_{3}}{gm}\in d_{k}\mathbb{Z}$ , we have $d_{k}\mid d_{4}$ , which contradicts 5.10. ∎

5.13 Lemma.

Let $i,j,k$ be such that $\{i,j,k\}=\{1,2,3\}$ . If $g_{i,j}=1$ then $\frac{f}{g}\leq\big{(}\frac{m}{m-1}\big{)}\big{(}\frac{1}{d_{i}d_{j}}+\frac{1}{d_{k}}\big{)}$ .

Proof.

Since $g_{i,j}=1$ , Proposition 5.2(a) implies that every integer strictly greater than $d_{i}d_{j}-d_{i}-d_{j}$ belongs to $\langle d_{i},d_{j}\rangle$ and hence to $\Gamma$ . So $\alpha\leq d_{i}d_{j}-d_{i}-d_{j}$ , because $\alpha\notin\Gamma$ . So $md_{4}=L=\alpha+\sum_{\ell=1}^{4}d_{\ell}\leq d_{i}d_{j}-d_{i}-d_{j}+\sum_{\ell=1}^{4}d_{\ell}=d_{i}d_{j}+d_{k}+d_{4}$ and consequently $(m-1)d_{4}\leq d_{i}d_{j}+d_{k}$ . Multiplying both sides by $\frac{m}{(m-1)d_{1}d_{2}d_{3}}$ gives

\textstyle\frac{md_{4}}{d_{1}d_{2}d_{3}}\leq\frac{m}{m-1}\cdot\frac{d_{i}d_{j}+d_{k}}{d_{1}d_{2}d_{3}}=\frac{m}{m-1}\big{(}\frac{1}{d_{k}}+\frac{1}{d_{i}d_{j}}).

Since $\frac{md_{4}}{d_{1}d_{2}d_{3}}=\frac{L}{d_{1}d_{2}d_{3}}=\frac{f}{g}$ , the result follows. ∎

5.14 Lemma.

We have $g_{i,j}>1$ for all choices of distinct $i,j\in\{1,2,3\}$ .

Proof.

Proceeding by contradiction, suppose that there exist $i,j,k$ such that $\{i,j,k\}=\{1,2,3\}$ and $g_{i,j}=1$ . Define $\delta_{i}=\frac{d_{i}}{g_{i,k}}$ , $\delta_{j}=\frac{d_{j}}{g_{j,k}}$ and $\delta_{k}=\frac{d_{k}}{g_{i,k}\,g_{j,k}}$ ; then $\delta_{i},\delta_{j},\delta_{k}\in\mathbb{N}^{+}$ and

\gcd(\delta_{i},\,g_{j,k}\,\delta_{k})=1=\gcd(\delta_{j},\,g_{i,k}\,\delta_{k}),\quad\gcd(\delta_{i},\delta_{j})=1,\quad\text{and}\quad\min(\delta_{i},\delta_{j})\geq 2\,.

(21)

Indeed, $\gcd(\delta_{i},\,g_{j,k}\delta_{k})=\gcd(\frac{d_{i}}{g_{i,k}},\frac{d_{k}}{g_{i,k}})=1$ , $\gcd(\delta_{j},\,g_{i,k}\delta_{k})=\gcd(\frac{d_{j}}{g_{j,k}},\frac{d_{k}}{g_{j,k}})=1$ , and $\gcd(\delta_{i},\delta_{j})=1$ because $g_{i,j}=1$ ; we have $\min(\delta_{i},\delta_{j})\geq 2$ because $d_{i}\nmid d_{k}$ and $d_{j}\nmid d_{k}$ , so (21) is true. We have $g=g_{i,k}\,g_{j,k}$ by Lemma 5.11(a), so Lemma 5.13 gives $\frac{f}{g}\leq\big{(}\frac{m}{m-1}\big{)}\big{(}\frac{1}{d_{i}d_{j}}+\frac{1}{d_{k}}\big{)}=\big{(}\frac{m}{m-1}\big{)}\big{(}\frac{1}{\delta_{i}\delta_{j}g}+\frac{1}{\delta_{k}g}\big{)}$ , i.e.,

\textstyle 1\leq f\leq\big{(}\frac{m}{m-1}\big{)}\big{(}\frac{1}{\delta_{i}\delta_{j}}+\frac{1}{\delta_{k}}\big{)}.

(22)

If $\delta_{k}>3$ then $\delta_{i}\delta_{j}\geq 6$ , so $\frac{1}{\delta_{i}\delta_{j}}+\frac{1}{\delta_{k}}<\frac{1}{6}+\frac{1}{4}<\frac{1}{2}$ ; if $\delta_{k}=3$ then (21) implies that $\delta_{i}\delta_{j}\geq 10$ , so $\frac{1}{\delta_{i}\delta_{j}}+\frac{1}{\delta_{k}}\leq\frac{13}{30}<\frac{1}{2}$ ; since $\frac{m}{m-1}\leq 2$ , (22) gives $1\leq f<1$ in both cases, a contradiction. So $\delta_{k}\in\{1,2\}$ .

If $\delta_{k}=2$ then it follows from (21) that $\delta_{i}\delta_{j}\geq 15$ , so (22) gives $1\leq f\leq\big{(}\frac{m}{m-1}\big{)}\frac{17}{30}$ , which implies that $m=2$ . Then $mg=2g=\delta_{k}g=d_{k}$ divides $d_{i}d_{k}$ , which contradicts Lemma 5.12. So $\delta_{k}=1$ . To summarize,

(d_{i},d_{j},d_{k})=(\delta_{i}g_{i,k},\delta_{j}g_{j,k},g)\quad\text{and}\quad g=g_{i,k}g_{j,k}.

Since $d_{k}\nmid d_{i}$ and $d_{k}\nmid d_{j}$ , we have

g_{i,k}>1\quad\text{and}\quad g_{j,k}>1.

(23)

We have $\delta_{i}\delta_{j}\geq 6$ by (21), so $f\leq\big{(}\frac{m}{m-1}\big{)}\big{(}\frac{1}{\delta_{i}\delta_{j}}+\frac{1}{\delta_{k}}\big{)}\leq\frac{7}{6}\frac{m}{m-1}\leq\frac{7}{6}\cdot 2<3$ , so $f\in\{1,2\}$ .

If $f=2$ then $2\leq\frac{7}{6}\frac{m}{m-1}$ implies that $m=2$ , which is impossible because $g_{i,k}>1$ , $g_{j,k}>1$ and $g_{i,k}\,g_{j,k}=g$ divides $m$ by Lemma 5.11(b). So $f=1$ . Since $\frac{L}{d_{1}d_{2}d_{3}}=\frac{f}{g}$ , this means that $L=\frac{d_{i}d_{j}d_{k}}{g}=\frac{d_{i}d_{j}g_{i,k}g_{j,k}}{g}$ , so

L=d_{i}d_{j}=L_{i,j}.

It follows that $d_{4}$ divides $d_{i}d_{j}=\delta_{i}\delta_{j}g_{i,k}g_{j,k}$ ; since $\gcd(d_{4},g_{i,k}g_{j,k})=1$ , we get $d_{4}\mid\delta_{i}\delta_{j}$ and hence

d_{4}\leq\delta_{i}\delta_{j}.

(24)

Proposition 5.2(b) gives $F(\langle d_{1},d_{2},d_{3}\rangle)\leq L_{k,i}+L_{k,j}-d_{1}-d_{2}-d_{3}$ . Note that $L_{k,i}=\frac{d_{k}d_{i}}{g_{i,k}}=\delta_{i}g$ and $L_{k,j}=\frac{d_{k}d_{j}}{g_{j,k}}=\delta_{j}g$ . Since $\alpha\notin\langle d_{1},d_{2},d_{3}\rangle$ , we get $\alpha\leq F(\langle d_{1},d_{2},d_{3}\rangle)\leq(\delta_{i}+\delta_{j})g-d_{1}-d_{2}-d_{3}$ , so $L\leq(\delta_{i}+\delta_{j})g+d_{4}\leq(\delta_{i}+\delta_{j})g+\delta_{i}\delta_{j}$ by (24). Since $L=d_{i}d_{j}=\delta_{i}\delta_{j}g$ , we obtain $\delta_{i}\delta_{j}(g-1)\leq(\delta_{i}+\delta_{j})g$ , so

\textstyle 1-\frac{1}{g}\leq\frac{1}{\delta_{i}}+\frac{1}{\delta_{j}}.

Condition (23) and $\gcd(g_{i,k},g_{j,k})=1$ imply that $g=g_{i,k}g_{j,k}\geq 6$ , and (21) gives $\frac{1}{\delta_{i}}+\frac{1}{\delta_{j}}\leq\frac{1}{2}+\frac{1}{3}$ ; thus, $\frac{5}{6}\leq 1-\frac{1}{g}\leq\frac{1}{\delta_{i}}+\frac{1}{\delta_{j}}\leq\frac{5}{6}$ . This implies that $g=6$ (so $\{g_{i,k},g_{j,k}\}=\{2,3\}$ ) and that $\{\delta_{i},\delta_{j}\}=\{2,3\}$ . We may assume that $\delta_{i}=2$ and $\delta_{j}=3$ . Then (21) gives $g_{i,k}=2$ and $g_{j,k}=3$ . So (24) implies that $d_{4}\leq\delta_{i}\delta_{j}=6<9=d_{j}$ , contradicting $d_{4}=\max(d_{1},d_{2},d_{3},d_{4})$ . ∎

5.15 Lemma.

$m\leq 6$

Proof.

Without loss of generality, we may assume that $d_{1}\leq d_{2}\leq d_{3}$ ; by 5.10, it follows that $d_{1}<d_{2}<d_{3}<d_{4}$ . Since $\alpha\notin\Gamma$ , we have $\alpha\notin\langle d_{1},d_{2},d_{3}\rangle$ , so Proposition 5.2(b) gives $\textstyle L-\sum_{i=1}^{4}d_{i}=\alpha\leq F(\langle d_{1},d_{2},d_{3}\rangle)\leq L_{1,2}+L_{1,3}-d_{1}-d_{2}-d_{3}$ , so $(1-\frac{1}{m})L=L-d_{4}\leq L_{1,2}+L_{1,3}$ and hence $\frac{1}{L/L_{1,2}}+\frac{1}{L/L_{1,3}}\geq(1-\frac{1}{m})$ . By contradiction, suppose that $m>6$ . Then $\frac{1}{L/L_{1,2}}+\frac{1}{L/L_{1,3}}>\frac{5}{6}$ (where ${L/L_{1,2}}$ and ${L/L_{1,3}}$ are positive integers), so

\textstyle\frac{L}{L_{1,2}}=1\quad\text{or}\quad\frac{L}{L_{1,3}}=1\quad\text{or}\quad\frac{L}{L_{1,2}}=2=\frac{L}{L_{1,3}}.

Also note that $g_{3,4}\mid\frac{L}{L_{1,2}}$ and $g_{2,4}\mid\frac{L}{L_{1,3}}$ by Lemma 5.4. If $\frac{L}{L_{1,2}}=1$ then $d_{3}$ and $d_{4}$ divide $L_{1,2}$ ; since $g_{3,4}=1$ we obtain $d_{3}d_{4}\mid L_{1,2}\mid d_{1}d_{2}$ , which is impossible since $d_{1}d_{2}<d_{3}d_{4}$ . So $\frac{L}{L_{1,2}}\neq 1$ . If $\frac{L}{L_{1,3}}=1$ then $d_{2}$ and $d_{4}$ divide $L_{1,3}$ ; since $g_{2,4}=1$ we obtain $d_{2}d_{4}\mid L_{1,3}\mid d_{1}d_{3}$ , which is impossible since $d_{1}d_{3}<d_{2}d_{4}$ . So $\frac{L}{L_{1,3}}\neq 1$ . It follows that

\textstyle\frac{L}{L_{1,2}}=2=\frac{L}{L_{1,3}}.

If $p$ is a prime factor of $g_{2,4}$ then $p\nmid d_{1}$ and $p\nmid d_{3}$ , so $p\nmid L_{1,3}$ and $p\mid L_{1,2}$ , contradicting $L_{1,2}=L_{1,3}$ . This shows that $g_{2,4}=1$ , and the same argument gives $g_{3,4}=1$ . Thus,

g_{2,4}=1=g_{3,4}.

Let $L_{1,2,3}=\operatorname{{\rm lcm}}(d_{1},d_{2},d_{3})$ . Since $L_{1,2}=L_{1,3}$ , we have $L_{1,2,3}=L_{1,2}=L/2$ , so $v_{2}(L_{1,2,3})+1=v_{2}(2L_{1,2,3})=v_{2}(L)=\max(v_{2}(L_{1,2,3}),v_{2}(d_{4}))$ and hence $v_{2}(d_{4})=v_{2}(L_{1,2,3})+1$ . So $d_{4}$ is even; since $g_{2,4}=1=g_{3,4}$ , $d_{2}$ and $d_{3}$ are odd. Thus, $v_{2}(d_{4})=v_{2}(L_{1,2,3})+1=v_{2}(d_{1})+1$ . So we can write

d_{1}=2^{r}\delta_{1}\quad\text{and}\quad d_{4}=2^{r+1}\delta_{4},\quad\text{where $\delta_{1},\delta_{4}\in\mathbb{N}^{+}$ are odd and $r\in\mathbb{N}$.}

If $p$ is a prime factor of $g_{2,3}$ then $p\nmid d_{1}$ , so $v_{p}(d_{2})=v_{p}(L_{1,2})=v_{p}(L_{1,3})=v_{p}(d_{3})$ . The fact that $v_{p}(d_{2})=v_{p}(d_{3})$ for each prime factor $p$ of $g_{2,3}$ implies that

d_{2}=ab\quad\text{and}\quad d_{3}=ac,\quad\text{where $a,b,c\in\mathbb{N}^{+}$ are odd and pairwise relatively prime.}

Since $d_{2}$ is odd and divides $L=2L_{1,3}$ , we have $d_{2}\mid L_{1,3}$ ; it follows that $b=\frac{d_{2}}{a}$ divides $\frac{L_{1,3}}{a}$ , which divides $\frac{d_{1}d_{3}}{a}=d_{1}c$ ; so $b\mid d_{1}c$ . Since $\gcd(b,c)=1$ , we get $b\mid d_{1}$ . The fact that $\delta_{4}$ divides $\frac{L}{2}=L_{1,2}$ implies that $\delta_{4}\mid d_{1}d_{2}$ ; since $g_{2,4}=1$ , it follows that $\delta_{4}\mid d_{1}$ . So we have $b\mid d_{1}$ , $\delta_{4}\mid d_{1}$ and $\gcd(b,\delta_{4})=1$ (because $g_{2,4}=1$ ); this implies that $b\delta_{4}\mid d_{1}$ . Since $b\delta_{4}$ is odd, it follows that $2^{r}b\delta_{4}\mid d_{1}$ and hence $2^{r}b\delta_{4}\leq d_{1}<d_{4}=2^{r+1}\delta_{4}$ . This implies that $b=1$ . So $d_{2}=a$ divides $d_{3}$ , a contradiction. ∎

5.16.

End of the proof of Theorem A. Lemma 5.15 gives $m\leq 6$ , and we have $g\mid m$ by Lemma 5.11(b). So $g\mid p^{r}q^{s}$ for some prime numbers $p,q$ and some $r,s\in\mathbb{N}$ . Since Lemma 5.11(a) gives $g=g_{1,2}\,g_{1,3}\,g_{2,3}$ where $g_{1,2},g_{1,3},g_{2,3}$ are pairwise relatively prime, it follows that one of $g_{1,2},g_{1,3},g_{2,3}$ is equal to $1$ , contradicting Lemma 5.14. This shows that Assumptions 5.10 lead to a contradiction. By Remark 5.9, this completes the proof of Theorem A.

References

[1] Alfred Brauer. On a problem of partitions. American Journal of Mathematics, 64(1):299–312, 1942.
[2] I. Cheltsov, J. Park, and Constantin Shramov. Exceptional del Pezzo hypersurfaces. Journal of Geometric Analysis, 20:787–816, 2008.
[3] Michael Chitayat. Rigidity of Pham-Brieskorn Threefolds. https://ruor.uottawa.ca/handle/10393/44886, 2023. University of Ottawa, Theses and Dissertations.
[4] Michael Chitayat and Daniel Daigle. On the rigidity of certain Pham-Brieskorn rings. Journal of Algebra, 550:290–308, 2020.
[5] Michael Chitayat and Daniel Daigle. Locally nilpotent derivations of graded integral domains and cylindricity. Transformation Groups, 2022.
[6] Daniel Daigle. Rigidity of graded integral domains and of their Veronese subrings, 2023.
[7] Michel Demazure. Anneaux gradués normaux. In Introduction à la théorie des singularités. II: Méthodes algébriques et géométriques. (Introduction to the theory of singularities. II: Algebraic and geometric methods), pages 35–68. Paris: Hermann, 1988.
[8] Igor Dolgachev. Weighted projective varieties. In James B. Carrell, editor, Group Actions and Vector Fields, pages 34–71, Berlin, Heidelberg, 1982. Springer Berlin Heidelberg.
[9] Hubert Flenner and Mikhail Zaidenberg. Rational curves and rational singularities. Mathematische Zeitschrift, 244(3):549–575, 2003.
[10] Ralf Fröberg, Christian Gottlieb, and Roland Häggkvist. On numerical semigroups. In Semigroup forum, volume 35, pages 63–83. Springer, 1986.
[11] A Hefez and F Lazzeri. The intersection matrix of Brieskorn singularities. Inventiones mathematicae, 25(2):143–157, 1974.
[12] A. R. Iano-Fletcher. Working with weighted complete intersections, page 101–174. London Mathematical Society Lecture Note Series. Cambridge University Press, 2000.
[13] Selmer Martin Johnson. A linear diophantine problem. Canadian Journal of Mathematics, 12:390–398, 1960.
[14] János Kollár. Is there a topological Bogomolov-Miyaoka-Yau inequality? Pure and Applied Mathematics Quarterly, 4, 02 2006.
[15] John Milnor. On the 3-dimensional Brieskorn manifolds $m(p,q,r)$ . Knots, groups, and 3-manifolds, 3:175–225, 1975.
[16] Victor Vladimirovich Przyjalkowski and Constantin Aleksandrovich Shramov. On automorphisms of quasi-smooth weighted complete intersections. Sbornik: Mathematics, 212(3):374, 2021.
[17] José Carlos Rosales, Pedro A García-Sánchez, et al. Numerical semigroups, volume 20. Springer, 2009.
[18] Uwe Storch. Die Picard-zahlen der singularitäten $t_{1}^{r_{1}}+t_{2}^{r_{2}}+t_{3}^{r_{3}}+t_{4}^{r_{4}}=0$ . Journal für die reine und angewandte Mathematik, 350:188–202, 1984.
[19] Giancarlo Urzúa and José Yáñez. Characterization of Kollár surfaces. Algebra & Number Theory, 12(5):1073–1105, 2018.
[20] Etsuo Yoshinaga and Masahiko Suzuki. On the topological types of singularities of Brieskorn-Pham type. Science Reports of the Yokohama National University: Yokohama Kokuritsu Daigaku Rika Kiyō. Dai 1-rui, Sūgaku, Butsurigaku, Kagaku. Mathematics, physics, chemistry. Section I, (21-25), 1974.