on some problems in transcendental number theory and ... › ... › 30350 › 4 ›...

On some problems in Transcendental numbertheory and Diophantine approximation

Ngoc Ai Van Nguyen

Thesis submitted to the

Faculty of Graduate and Postdoctoral Studies

in partial fulfillment of the requirements

for the Doctorate in Philosophy degree in Mathematics1

Department of Mathematics and Statistics

Faculty of Science

University of Ottawa

c© Ngoc Ai Van Nguyen, Ottawa, Canada, 2014

1The Ph.D. Program is a joint program with Carleton University, administered by the Ottawa-Carleton Institute of Mathematics and Statistics

Abstract

In the first part of this thesis, we present the first non-trivial small value estimate that

applies to an algebraic group of dimension 2 and which involves large sets of points.

The algebraic group that we consider is the product C×C∗, of the additive group Cby the multiplicative group C∗. Our main result assumes the existence of a sequence

(PD)D≥1 of non-zero polynomials in Z[X1, X2] taking small absolute values at many

translates of a fixed point (ξ, η) in C×C∗ by consecutive multiples of a rational point

(r, s) ∈ (Q∗)2 with s 6= ±1. Under precise conditions on the size of the coefficients

of the polynomials PD, the number of translates of (ξ, η) and the absolute values of

the polynomials PD at these points, we conclude that both ξ and η are algebraic

over Q. We also show that the conditions that we impose are close from being best

possible upon comparing them with what can be achieved through an application of

Dirichlet’s box principle.

In the second part of the thesis, we consider points of the form θ = (1, θ1, . . . , θd−1, ξ)

where {1, θ1, . . . , θd−1} is a basis of a real number field K of degree d ≥ 2 over Q and

where ξ is a real number not in K. Our main results provide sharp upper bounds

for the uniform exponent of approximation to θ by rational points, denoted λ(θ),

and for its dual uniform exponent of approximation, denoted τ(θ). For d = 2, these

estimates are best possible thanks to recent work of Roy. We do not know if they

are best possible for other values of d. However, in Chapter 2, we provide additional

information about rational approximations to such a point θ assuming that its ex-

ponent λ(θ) achieves our upper bound. In the course of the proofs, we introduce

new constructions which are interesting by themselves and should be useful for future

research.

ii

Acknowledgements

First of all, I would like to express my deepest gratitude to my supervisor Professor

Damien Roy for his direction, support, patience, and understanding.

I would like to thank the professors of the Ottawa-Carleton Institute of Mathematics

and Statistics for teaching me and thank the staff of the Faculty of Science for helping

and giving me a nice environment to study and work in.

I also sincerely thank to my former supervisors Professor Michel Waldschmidt (Uni-

versity of Paris VI) and Professor Bui Xuan Hai (University of Science Ho Chi Minh

City) for providing me with the opportunity to study here and for having encouraged

me for these years.

I am deeply grateful to all the Vaillancourt family and to Thu Huong Nguyen for

giving me warmth, help and encouragement during my staying with them.

Last but not least, I would like to thank my family, especially my grandmother, par-

ents, and my husband. Without their love, I would not have been able to complete

this thesis.

Ottawa, September 2013

Ngoc Ai Van Nguyen

iii

Dedication

First and foremost, I dedicate this work to my father to fulfill my last promise to him.

Daddy, you left a void never to be filled in my life, but your memory always gave

me strength whenever I was weak. I wish you could know that I am always proud of

being your daughter.

Mama, although I cannot fill the void Dad left in you, I dedicate this work to you

with hope that it will make you happier.

I also dedicate this to my grandmother, who is illiterate, but taught us the value of

studying and worked hard to provide us with the opportunities to study.

iv

Introduction

This thesis has two parts. In the first part, which is Chapter 1, we prove a new small

value estimate for the group C × C∗. This result provides necessary conditions for

the existence of certain sequences of non-zero polynomials with integer coefficients

taking small absolute values at points of C× C∗. In the second part, divided in two

chapters, we prove two new results of Diophantine approximation.

Part I.

We present the first non-trivial small value estimate that applies to an algebraic group

of dimension 2 and which involves large sets of points. The algebraic group that we

consider here is the product C × C∗. Our main result shows that if there exists a

sequence (PD)D≥1 of non-zero polynomials in Z[X1, X2] taking small absolute values

at many translates of a fixed point (ξ, η) in C × C∗ by multiples of a rational point

(r, s) ∈ (Q∗)2 with s 6= ±1, then both ξ and η are algebraic over Q. More precisely,

for each integer D ≥ 1, we request that PD has degree at most D and norm at most

eDβ

for some fixed number β > 0. The translates at which we evaluate PD are points

of the form γi

= (ξ, η) + i(r, s) with 0 ≤ i < 3bDσc where σ > 1 is fixed. We request

that

|PD(γi)| ≤ e−D

ν

(0 ≤ i < 3bDσc) (1)

where ν is fixed. The conclusion that ξ and η are algebraic is then obtained by

assuming that the parameters β, σ and ν satisfy the conditions

1 ≤ σ < 2, β > σ + 1, ν > max

{β + 2− σ +

(σ − 1)(2− σ)

β − σ + 1, σ + 2

}. (2)

v

An application of Dirichlet’s Box principle shows that, given (ξ, η), (r, s) ∈ C×C∗,there always exists such a sequence (PD)D≥1 satisfying condition (1) if 0 ≤ σ < 2, β >

σ + 1 and ν < β + 2− σ.

Since (σ − 1)(2− σ)/(β − σ + 1) ≤ 1/8, the main lower bound that we impose on

ν is weaker than

ν ≥ (β + 2− σ) +1

8.

We do not know if the conditions (2) can be improved but this shows that if it is not

best possible, the largest saving that we could achieve is no more than 1/8. Therefore,

in a sense, it is close to be best possible.

We also show that, in order to reach the conclusion ξ, η ∈ Q, we need the param-

eter σ to be at least 1. Assuming that σ < 1, β > 2σ, we show the existence of a

point (ξ, η) with algebraically independent coordinates for which there is a sequence

(PD)D≥1 satisfying (1) for any ν > 0. This is a consequence of a construction of

Khintchine–Philippon.

The proof of our main result is an adaption of the argument of D. Roy in [21]. In

this paper, the author proves a similar result. He also considers a sequence (PD)D≥1

of polynomials in Z[X1, X2] of degree ≤ D and norm ≤ eDβ. The difference is that,

these polynomials PD are assumed to have the absolute values at most e−Dν

at one

point (ξ, η) in C × C∗ together with their derivatives with respect to the operator

D = ∂∂X1

+X2∂

∂X2up to order 3bDτc− 1, while in our work, the polynomials PD have

absolute values at most e−Dν

at 3bDτc translates of (ξ, η). The constraints on the

parameters τ, β, ν in [21] are almost the same as (1) (where τ replaces σ and β > τ

replaces β > σ + 1). In both cases, the conclusion is that ξ, η ∈ Q.

To prove our result, we apply elimination theory in the form developed by M. Lau-

rent and D. Roy in [14] in terms of height of a Q-cycle relative to a convex body.

More precisely, as in [19], we consider some homogenization of the polynomials PD

and for each D ≥ 1, we define an appropriate convex body CD. Then using elimi-

nation theory, we obtain a zero-dimensional Q-subvariety ZD whose height hCD(ZD)

relative to CD is very small (negative). Up to this, the argument is very similar to

[21]. The rest of the proof is different since we deal with several points.

In order to reach the conclusion, we need to analyze the distance from the points

vi

of ZD and the points γi = (1, ξ + ir, ηsi) (i ∈ Z). This analysis is complicated and

involves a new interpolation estimate as well as a diophantine analysis of the ideal

of homogeneous polynomials of C[X0, X1, X2] vanishing on all the points γi with

0 ≤ i < bDσc. We refer readers to the precise outline of the proof given in Chapter

1. Despite this big difference in the proof of our main result, it is surprising that we

reach the same conclusion ξ, η ∈ Q by asking constraints on σ, β, ν which are almost

the same as those in [21] for τ, β, ν.

In [17], D. Roy made a statement in the form of a small value estimate and

prove that it is equivalent to Schanuel’s conjecture, one of the main open problems in

transcendental number theory. In this paper, the author considers a certain sequence

(QD)D≥1 of polynomials in Z[X1, X2] with partial degree ≤ Dt1 in X1 and partial

degree ≤ Dt2 in X2 and norm ≤ eD. He requests that the polynomials QD take

the absolute values ≤ e−Du

with their derivatives up to order Ds1 at all the points

m1Υ1 + · · ·+m`Υ` (0 ≤ mi ≤ Ds2) where Υi = (ξi, ηi) (0 ≤ i ≤ `) are fixed points of

the algebraic group C× C∗ such that ξ1, . . . , ξ` are linearly independent over Q.

Assuming that

max{1, t1, 2t2} < min{s1, 2s2}, max{s1, s2 + t2} < u <1

2(1 + t1 + t2),

he shows that

tr.degQ(ξ1, . . . , ξ`, η1, . . . , η`) ≥ `.

Our present result implies that if tr.degQ(ξ, η) ≥ 1, then for each (r, s) ∈ Q∗2 with

s 6= ±1, and for each triple (σ, β, ν) satisfying (2), there exist infinitely many integers

D for which any non-zero polynomial P of Z[X1, X2] of degree ≤ D and norm ≤ eDβ

satisfies

max0≤<3bDσc

|P ((ξ, η) + i(r, s))| > e−Dν

.

This is a modest step in the direction of the Schanuel conjecture, but it improves on

previously known results.

vii

Part II.

The second part of the thesis deals with the two most basic problems of Diophantine

approximation. One of them consists in finding good rational approximations to

a given real point (θ1, . . . , θn). The other consists in finding small linear integral

combination of 1, θ1, . . . , θn. In their precise form both problems request to solve

some systems of linear inequations. In the first case, we look for non-zero integral

solutions x = (x0, . . . , xn) to the system

|x0| ≤ X, |x0θ1 − x1| ≤ X−λ, . . . , |x0θn − xn| ≤ X−λ (3)

where λ > 0 is fixed and X goes to infinity. If x = (x0, . . . , xn) is a solution of the

system with X large enough, then x0 6= 0 and the point (x1/x0, . . . , xn/x0) provides

a rational approximation to (θ1, . . . , θn). In the second case, we look for non-zero

integral solutions x = (x0, . . . , xn) of the system

|x0 + x1θ1 + · · ·+ xnθn| ≤ X−τ , |x1| ≤ X, . . . , |xn| ≤ X (4)

where τ > 0 is fixed and X goes to infinity. The two problems are dual of each other

and the geometry of numbers provides remarkable connections between them. In this

thesis, we are interested in the so-called uniform exponents of approximation attached

to each problem. Following a convention introduced by Bugeaud and Laurent in

[2], we denote by λ(1, θ1, . . . , θn) (resp. by τ(1, θ1, . . . , θn)) the supremum of all real

numbers λ > 0 (resp. τ > 0) such that the system (3) (resp. (4)) has a non-zero

integer solution for each sufficiently large X. An application of Minkowski’s first

convex body theorem shows that, if θ := (1, θ1, . . . , θn) has Q-linearly independent

coordinates, then λ(θ) ≥ 1/n and τ(θ) ≥ n.

It came as a surprise when it was shown in [18], some ten years ago, that there

exist real points θ with coordinates in a field of transcendence degree 1 for which

at least one of these exponents (and in fact both of them) strictly exceed the above

lower bounds. In [2] and [22], Bugeaud, Laurent and Roy produced more examples

of such points. However, in all cases, these points lay on an algebraic curve in R3

defined by an irreducible homogeneous polynomial of Q[x0, x1, x2] of degree 2. For

transcendental points on algebraic curves of higher degree (defined over Q), we only

viii

have upper bounds on their exponents of approximation. For example, Davenport

and Schmidt showed in [6] that λ(1, θ, θ2, θ3) ≤ 1/2 for any real number θ which is

not an algebraic number of degree ≤ 3. This upper bound was improved by Roy to

about 0.4245 in [19], but at present an optimal upper bound is not known. More

recently Lozier and Roy showed in [15] that λ(1, θ, θ3) ≤ 2(9 +√

11)/35 ' 0.7038 for

any real number θ such that 1, θ, θ3 are linearly independent over Q.

Let α be a quadratic real number. It is shown in [22] that, for any ξ ∈ R \Q(α),

we have λ(1, α, ξ) ≤ (√

5 − 1)/2 ' 0.618, with equality for a countable set of real

numbers ξ. The proof of the upper bound in this case is simpler than the estimate

λ(1, ξ, ξ2) ≤ (√

5−1)/2 proved by Davenport and Schmidt for non-quadratic irrational

real numbers ξ in [6]. This motivated us to establish upper bounds for the uniform

exponents of approximation to points of the form

θ := (1, θ1, . . . , θd−1, ξ)

where {1, θ1, . . . , θd−1} is a basis of a real number field K of degree d ≥ 2 over Q and

where ξ ∈ R \K. In a simplified form, our main result in Chapter 2 says that such a

point satisfies

λ(θ) ≤ λd <1

d− 1− 1

d2(d− 1)(5)

where λd is the unique positive real root of the polynomial (d−1)d−1xd+· · ·+(d−1)x2+

x−1. This improves on the trivial upper bound λ(θ) ≤ λ(1, θ1, . . . , θd−1) = 1/(d−1).

Similarly, our main result in Chapter 3 is that

τ(θ) ≤ τd :=1 +√

5

2(d− 1) + 1. (6)

Following the pioneer work of Davenport and Schmidt in [6], the proofs of both results

are based on an analysis of the sequences of so-called minimal points attached to θ,

in relation to the problem under consideration.

Our main contribution in Chapter 2 is a careful study of the heights of the sub-

spaces spanned by consecutive minimal points. It leads to an inequality relating the

norms of properly chosen minimal points. It took us much work to discover and prove

this result but with its help, the proof of (5) goes relatively easily.

ix

Our analysis of the sequence of minimal points attached to the other problem

is quite different. In Chapter 3, we assume that θ = (1, α, . . . , αd−1, ξ) where α is

a primitive element of the field K. Then we combine several linearly independent

minimal points to construct polynomials in α with small non-zero absolute values

and then we use Liouville’s inequality to bound from below these absolute values.

This yields inequalities relating the corresponding minimal points. These estimates

and others coming from geometry of numbers lead to the proof of (6).

A more complete outline of each proof is given in the corresponding chapter. In

both chapters we also give alternative proofs of some of our results when they are

obtained through non-explicit constructions based on Diriclet’s box principle or on

geometry of numbers. These alternative arguments are based on the construction of

explicit auxiliary polynomials adapted to our problem. In Chapter 2, we also present

the construction of a point (1, 3√

2, 3√

4, ξ) with surprising Diophantine properties.

x

Contents

Abstract ii

Acknowledgements iii

Dedication iv

Introduction v

1 A new small value estimate 1

1.1 Introduction and results . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.1 Dimension and degree of algebraic subsets of Pm(C) . . . . . . 8

1.2.2 Basic results in Elimination Theory . . . . . . . . . . . . . . . 9

1.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 Outline of the proof of Theorem 1.1.5 . . . . . . . . . . . . . . . . . . 15

1.5 An interpolation estimate for homogeneous polynomials . . . . . . . . 19

1.6 Decomposition of polynomials in I(T ) . . . . . . . . . . . . . . . . . . 25

1.7 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.8 Construction of Q-subvarieties of dimension 0 . . . . . . . . . . . . . 35

1.9 Proof of the main theorem 1.1.5 . . . . . . . . . . . . . . . . . . . . . 48

2 On approximation by rational points 60

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

2.1.1 Statement of the results . . . . . . . . . . . . . . . . . . . . . 60

xi

2.1.2 Proofs of the corollaries . . . . . . . . . . . . . . . . . . . . . 66

2.1.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

2.2 Construction of minimal points . . . . . . . . . . . . . . . . . . . . . 69

2.3 Construction of sequences of vector spaces . . . . . . . . . . . . . . . 73

2.4 On the norms of minimal points . . . . . . . . . . . . . . . . . . . . . 77

2.5 Proof of the main theorems . . . . . . . . . . . . . . . . . . . . . . . 79

2.5.1 Proof of Theorem 2.1.3 . . . . . . . . . . . . . . . . . . . . . . 79

2.5.2 Proof of Theorem 2.4.3 . . . . . . . . . . . . . . . . . . . . . . 81

2.6 The polynomials ϕ and Φ . . . . . . . . . . . . . . . . . . . . . . . . 85

2.7 The morphism Ψ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

2.8 An explicit construction of a point with exponent of approximation

≥ 1/3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3 On the dual Diophantine problem 107

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

3.2 Sequences of minimal points associated to Tθ . . . . . . . . . . . . . . 110

3.3 The set I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

3.4 Proof of Theorem 3.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 120

3.5 Alternative approach using polynomials . . . . . . . . . . . . . . . . 122

Bibliography 129

xii

Chapter 1

A new small value estimate for the

group C× C∗

1.1 Introduction and results

The theory of transcendental numbers started with Liouville’s memoir of 1844. There,

he investigated a class of numbers x, now called Liouville numbers, for which there

exists a rational number p/q such that |x − p/q| ≤ 1/qn for any positive integer n,

and showed that these are transcendental.

In 1873, Hermite showed that e is transcendental. This is the first number proven

transcendental but not constructed to be transcendental.

In 1882, Lindemann proved that e to any non-zero algebraic number power is

transcendental. As a consequence, π is transcendental. This yields the negative

answer for the squaring circle problem, proposed by ancient Greek geometers.

Generalizing the method of Lindemann, Weierstrass established a result, named

for both of them.

Theorem 1.1.1. (Lindemann-Weierstrass) If α1, . . . , αn are algebraic numbers which

are linearly independent over Q then eα1 , . . . , eαn are algebraically independent over

Q.

1

CHAPTER 1. A NEW SMALL VALUE ESTIMATE 2

In 1934, Gel’fond and Schneider proved independently that if α and β are alge-

braic numbers with α 6= 0 and β /∈ Q, then for any choice of logα 6= 0, the number

αβ = eβ logα is transcendental.

A basis tool in transcendental number theory consists of the construction of aux-

iliary functions taking small values at many points of an algebraic group. If these

values are integers < 1, then they all vanish and we can apply a zero estimate to

conclude. If these values are algebraic, we can instead apply Liouville’s inequality

and hopefully conclude that these values are zero, such as in the proof of Gel’fond-

Schneider Theorem. When the field generated by these values has transcendence

degree 1 over Q, a substitute for Liouville’s inequality is given by Gel’fond’s criterion

in [10]. When the transcendence degree of this field is higher, one can use Philippon’s

criterion (Theorem 2.11 of [16]). We recall these criterions below.

Gel’fond criterion. Let ξ ∈ C. Assume that there exist real numbers

β > 1, ν > β + 1

and a sequence of non-zero polynomials (PD)D≥1 ⊂ Z[X] such that

degPD ≤ D, ‖PD‖ ≤ eDβ

, |PD(ξ)| ≤ e−Dν

where ‖PD‖ denotes the norm of polynomial PD, i.e. the largest absolute value of its

coefficients. Then PD(ξ) = 0 for all sufficiently large integers D ≥ 1. In particular,

ξ ∈ Q.

Philippon’s criterion. Let θ = (1, θ1, . . . , θm) ∈ Cm+1, let θ denote the correspond-

ing point of Pm(C), and let k be an integer with 0 ≤ k ≤ m. Moreover, let (Dn)n≥1

be a non-decreasing sequence of positive integers, and let (Tn)n≥1 and (Vn)n≥1 be non-

decreasing sequences of positive real numbers such that

lim supn→∞

Vn(Dn + Tn)Dk

n

=∞.

Suppose also that for each n ≥ 2 there exists a non-empty family Fn consisting of

homogeneous polynomials in Z[X0, X1, . . . , Xm] which satisfy the following two prop-

erties.


(i) For each P ∈ Fn, we have

deg(P ) = Dn, h(P ) ≤ Tn and |P (θ)| ≤ e−Vn‖P‖ ‖θ‖Dn .

(ii) The polynomials of Fn have no common zero α in Pm(C) with

dist(θ, α) ≤ e−Vn−1 .

Then we have k < m and the transcendence degree over Q of the field Q(θ1, . . . , θm)

is ≥ k + 1.

Using his criterion, Gel’fond proved in [9] the following result.

Theorem 1.1.2. If α and β are algebraic numbers with α 6= 0 and [Q(β) : Q] = 3,

then for any choice of logα 6= 0, the numbers eβ logα and eβ2 logα are algebraically

independent over Q.

Applying Philippon’s criterion, G. Diaz established the following result in [8].

Theorem 1.1.3. Let α and β be algebraic numbers with α 6= 0 and [Q(β) : Q] = d.

Then, for any choice of logα 6= 0, we have

tr.degQQ(eβ logα, . . . , eβd−1 logα) ≥

⌊d+ 1

2

⌋.

For future progress in Transcendence and Algebraic Independence, it is desirable

to study situations where the values are not small enough so that we can apply

Philippon’s criterion.

D. Roy presented in [21] such a situation and showed an improvement on a di-

rect application of Philippon’s criterion. More precisely, he established the following

result.

Theorem 1.1.4. Let (ξ, η) ∈ C× C∗ and let τ, β, ν ∈ R with

1 ≤ τ < 2, β > τ, ν > max

{β + 2− τ +

(τ − 1)(2− τ)

β − τ + 1, τ + 2

}.


Suppose that, for each sufficiently large positive integer D, there exists a non-zero

polynomial PD ∈ Z[X1, X2] of degree ≤ D and norm ≤ exp(Dβ) such that

max0≤i<3bDτ c

|DiPD(ξ, η)| ≤ e−Dν

where D =∂

∂X1

+X2∂

∂X2

.

Then, we have ξ, η ∈ Q and moreover DiPD(ξ, η) = 0 (0 ≤ i < 3bDτc) for each

sufficiently large integer D.

In this chapter, we adapt the approach of D. Roy in [21] to establish the following

result.

Theorem 1.1.5. Let (ξ, η) ∈ C × C∗ and (r, s) ∈ Q∗2 with s 6= ±1. Let σ, β, ν ∈ Rsuch that

1 ≤ σ < 2, β > σ + 1, ν > max

{β + 2− σ +

(σ − 1)(2− σ)

β − σ + 1, σ + 2

}.

Suppose that, for each sufficiently large positive integer D, there exists a non-zero

polynomial PD ∈ Z[X1, X2] such that


, max0≤i<3bDσc

|PD(ξ + ir, ηsi)| ≤ e−Dν

. (1.7)

Then we have ξ, η ∈ Q.

For any (ξ, η), (r, s) ∈ C2, Dirichlet’s Box principle ensures the existence of a

sequence of polynomials satisfying (3.3) when the condition

ν > max{β + 2− σ +(σ − 1)(2− σ)

β − σ + 1, σ + 2}

is replaced by ν < β + 2 − σ. So we are not able to conclude anything in this case.

More precisely, we have the following result.

Proposition 1.1.6. Let (ξ, η), (r, s) ∈ C2. Let σ, β, ν ∈ R such that

0 ≤ σ < 2, β > σ + 1, ν < β + 2− σ.

Then, for each D � 1, there exists 0 6= PD ∈ Z[X1, X2] such that


, max0≤j<3bDσc

|PD(ξ + jr, ηsj)| ≤ e−Dν

.


Proof. Fix a large integer D. Put S = 3bDσc. Let UD be the set of polynomials

in Z[X1, X2]≤D with non-negative integer coefficients and norm ≤ eDβ. Consider the

map

f : UD −→ RS

P 7−→ (P (ξ + jr, ηsj))0≤j<S

We have

CardUD ≥ exp

(Dβ

(D + 2

2

))≥ exp

(1

2Dβ+2

).

Moreover, for each 0 ≤ j < S, we have

|P (ξ + jr, ηsj)| ≤(D + 2

2

)eD

β

max{1, |ξ + jr|, |ηsj|}D ≤ e4Dβ

since β > σ + 1. So (P (ξ + jr, ηsj))0≤j<S belongs to S-cube[−e4Dβ , e4Dβ

]S.

On the other hand, the interval[−e4Dβ , e4Dβ

]can be covered by a union of at

most 1 + 2e4Dβ+Dν subintervals of length e−D

ν. Hence the S-cube

[−e4Dβ , e4Dβ

]Sis

covered by at most (3e4Dβ+Dν )S ≤ exp(16Dmax{β,ν}+σ) smaller S-cubes of edges of

length e−Dν. Since σ < 2, and ν < β + 2− σ, we find that UD has a cardinal greater

than the number of such small S-cubes.

By Dirichlet’s Box Principle, there exist two distinct polynomials QD, Q′D in UD

mapping to the same small S-cube. This means that

|(QD −Q′D)(ξ + jr, ηsj)| ≤ e−Dν

for all 0 ≤ j < S. Since QD and QD′ have coefficients in [0, eDβ], the polynomial

PD = QD −Q′D is non-zero and has norm ‖PD‖ ≤ eDβ. Thus it satisfies the required

properties.

The above Proposition implies that, we cannot reduce the lower bound on ν in

Theorem 1.1.5 by more than

(σ − 1)(2− σ)

β − σ + 1<

(σ − 1)(2− σ)

2≤ 1

8.

Now we will explain why we need σ ≥ 1. This follows from a result of Khintchine

revisited by Philippon in [16, Appendix].


Theorem 1.1.7. (Khintchine - Philippon) Let ψ : N −→ (0, 1) be a decreasing

function. Then there exists (ξ, η) ∈ R× R∗ with the following properties

• ξ and η are algebraically independent over Q,

• for each D ≥ 1, there exists a non-zero linear form LD ∈ Z[X1, X2] such that

‖LD‖ ≤ D, |LD(ξ, η)| ≤ ψ(D).

Corollary 1.1.8. Let (r, s) ∈ Q×Q∗. Let σ, β, ν ∈ R such that

0 ≤ σ < 1, β > 2σ, ν > 0.

Then there exists (ξ, η) ∈ R× R∗ with the following properties

• ξ, η are algebraically independent over Q,

• for each D � 1, there exists a non-zero polynomial PD ∈ Z[X1, X2]≤D such that


, max0≤j<3bDσc

|PD(ξ + jr, ηsj)| ≤ e−Dν

.

Proof. From theorem 1.1.7, we deduce the existence of (ξ, η) ∈ R × R∗ with the

following properties

• ξ and η are algebraically independent over Q,

• for each D ≥ 1, there exists a non-zero linear form LD ∈ Z[X1, X2] such that

‖LD‖ ≤ D, |LD(ξ, η)| ≤ exp(−Dν −Dβ).

Set

PD(X1, X2) =∏

0≤j<3bDσc

djLD(X1 − jr, s−jX2)

where d is a positive integer such that ds−1, dr ∈ Z. Assuming D large enough, we

have

degPD = 3bDσc ≤ D since σ < 1.


Moreover, we get

‖PD‖ ≤ 33Dσ max0≤j<3bDσc

‖djLD(X1 − jr, s−jX2)‖3Dσ

≤ 33Dσ(d3D

σ‖LD‖(1 + 3Dσ|r|+ |s−1|3Dσ))3Dσ

≤(3 d3D

σ

D(1 + |r|+ |s−1|)3Dσ)3Dσ

� eDβ

(since β > 2σ),

and

|PD(ξ + jr, ηsj)| = d9D2σ |LD(ξ, η)|

∏j′ 6=j

0≤j′<3bDσc

|LD(ξ + (j′ − j)r, ηsj′−j)|

≤ d9D2σ

e−Dν−Dβ

(‖LD‖

(1 + |ξ|+ 3Dσ|r|+ |η|(|s|+ |s−1|)3Dσ

) )3Dσ≤ e−D

ν−Dβd9D2σ(D(1 + |ξ|+ |r|+ (|η|+ 1)(|s|+ |s−1|)

)3Dσ )3Dσ� e−D

ν

(since β > 2σ).

This result shows that Theorem 1.1.5 does not hold if we replace the condition

1 ≤ σ < 2 by 0 ≤ σ < 1. Indeed, for such σ, the pair (ξ, η) constructed by Corollary

1.1.8 satisfies all the hypotheses of the theorem (for any choice of β > σ + 1 and

ν > 0) but it does not satisfy the conclusion.

1.2 Preliminaries

In this section, we introduce the results of dimension theory and elimination theory

that we will need in the proof of our main result (Theorem 1.1.5).

Letm be a positive integer. We denote by C[X] the ring of polynomials in variables

X0, . . . , Xm with coefficients in C. For each integer D ≥ 0, we denote by C[X]D its

homogeneous part of degree D.


1.2.1 Dimension and degree of algebraic subsets of Pm(C)

Let S be a subset of C[X] consisting of homogeneous polynomials. We denote by

Z(S) the set of common zeros in Pm(C) of the polynomials of S. Then Z(S) = Z(I)

where I is the homogeneous ideal generated by S.

Given a subset Z of Pm(C), we say that Z is an algebraic subset of Pm(C) if

Z = Z(I) for some homogeneous ideal I of C[X]. If the corresponding ideal is prime,

we say that Z is an irreducible algebraic subset of Pm(C).

By a Q-subvariety of Pm(C), we mean the zero set in Pm(C) of a homogeneous

prime ideal of Q[X0, X1, . . . , Xm] distinct from the ideal 〈X0, . . . , Xm〉. Such a set is

non-empty but may not be irreducible as an algebraic subset of Pm(C).

Let Z be an algebraic subset of Pm(C). We say that Z has dimension t and write

dim(Z) = t if there exists a chain of irreducible algebraic subsets

∅ = Z0 · · · Zt+1 ⊆ Z,

but no longer chain.

Example 1.2.1. (i) dim(Pm(C)) = m.

(ii) dim(∅) = −1.

(iii) dim(Z(P )) = m− 1 if P is a non-zero homogeneous polynomial of C[X].

Fix an algebraic subset Z of Pm(C) of dimension d. Denote by I(Z) the ideal gen-

erated by all homogeneous polynomials of C[X] vanishing on Z. Then C[X]/I(Z) is a

graded C[X]-module whose homogeneous part of degree t is denoted by (C[X]/I(Z))t.

It is well-known that there exists a polynomial HZ(t) ∈ Q[t], called the Hilbert poly-

nomial of Z, such that

HZ(t) = dimC(C[X]/I(Z))t

for each sufficiently large integer t. More precisely, HZ(t) is a polynomial of degree d

of the form

HZ(t) = a0

(t

d

)+ a1

(t

d− 1

)+ · · ·+ ad

(t

0

)


where a0, a1, . . . , ad are integers.

If Z 6= ∅, we have d ≥ 0, and we define the degree of Z to be deg(Z) = a0. This

is a positive integer.

Example 1.2.2. We have

HPm(C)(t) = dimC(C[X0, . . . , Xm]t) =

(t+m

m

),

and so deg(Pm(C)) = 1.

To establish our result, we will work with Q-subvarieties of Pm(C) of dimension

0. Note that, if Z is a Q-subvariety of Pm(C) of dimension 0, then Z is finite,

more precisely, deg(Z) = |Z| and if (α0, α1, . . . , αm) is a representative in Cm+1 of

a point of Z with at least one coordinate equal to 1, then Z consists of the points

(σ(α0) : σ(α1) : . . . : σ(αm)) ∈ Pm(C) where σ runs through all embeddings of

Q(α0, . . . , αm) into C.

1.2.2 Basic results in Elimination Theory

In our work, we will use consequences of the following result, which derives from [5,

Lemma 3].

Theorem 1.2.3. Assume that Z is an algebraic subset of Pm(C) of dimension d ≥ 1.

Let P be a non-constant homogeneous polynomial of C[X] such that Z(P ) does not

contain any irreducible component of Z (over C). Then the intersection Z ∩ Z(P )

has dimension d− 1 and degree at most deg(Z) · deg(P ).

Moreover, if Z is d-equidimensional, i.e., if every component of its decomposition

into irreducible algebraic subsets of Pm(C) has dimension d, then Z∩Z(P ) is (d−1)-

equidimensional.

In fact, Lemma 3 of [5] shows that deg(Z ∩ Z(P )) = deg(Z) deg(P ) if P has no

multiple factor so that the ideal 〈P 〉 is reduced.

If Z = Pm(C) then we have deg(Pm(C)) = 1 and so, by the theorem, we get the

following result.


Corollary 1.2.4. Let P be a non-constant homogeneous polynomial of C[X]. Then

deg(Z(P )) ≤ deg(P ).

If Z = Z(Q) where Q is a non-constant homogeneous polynomial of C[X], then

Z(Q) is (m− 1)-equidimensional. In particular, if P and Q belong to Q[X] and have

no common factor in Q[X], then they also have no common factor in C[X]. Therefore,

we obtain the following result.

Corollary 1.2.5. Assume that P and Q are non-zero homogeneous polynomials of

C[X] (resp. Q[X]) which have no common factor in C[X] (resp. Q[X]). Then Z(P,Q)

is (m− 2)-equidimensional and has degree deg(Z(P,Q)) ≤ deg(P ) deg(Q).

We now introduce the main tool used in our work, the Chow form of Q-subvarieties

Z of Pm(C). We start with the definition of resultant, which is the Chow form of

Pm(C) as we will see below.

Let D ∈ N∗. For each ν = (ν0, . . . , νm) ∈ Nm+1, we define Xν = Xν00 · · ·Xνm

m . Let

Ui =∑

ν∈Nm+1

|ν|=D

ui,νXν , i = 0, . . . ,m

be m + 1 generic homogeneous forms in X0, . . . , Xm of degree D, i.e. homogeneous

forms in X with indeterminate coefficients.

As is well-known, there is a polynomial in ui,ν with integer coefficients, called the

resultant, denoted ResD(U0, . . . , Um), such that

• ResD(U0, . . . , Um) is irreducible over C,

• ResD(U0, . . . , Um) is homogeneous of degree Dm in (ui,ν)|ν|=D for each index

i = 0, . . . ,m and it has total degree (m+ 1)Dm,

• viewing the resultant as a polynomial map ResD : C[X]m+1D −→ C, we have

ResD(P0, P1, . . . , Pm) = 0 iff Z(P0, . . . , Pm) 6= ∅

for any tuple (P0, P1, . . . , Pm) ∈ C[X]m+1D .


(See [23, Chapter XI] for more details.)

We now define the Chow form of Q-subvarieties of Pm(C). Assume that Z is a

Q-subvariety of Pm(C) of dimension t. The first section of [16] shows the existence of

a polynomial F (U0, . . . , Ut) ∈ Z[ui,ν ; 0 ≤ i ≤ t, |ν| = D] with the following properties

• F (U0, . . . , Ut) is irreducible over Z,• F is homogeneous of degree Dt deg(Z) in (ui,ν)|ν|=D for each i = 0, . . . , t and it

has total degree (t+ 1)Dt deg(Z),

• viewing F (U0, . . . , Ut) as a polynomial map F : C[X]t+1D −→ C, we have

Z(F ) = {(P0, . . . , Pt) ∈ C[X]t+1D ; Z(P0, . . . , Pt) ∩ Z 6= ∅}.

For given Z and D, such a polynomial is unique up to multiplication by ±1. We

call it the Chow form of Z in degree D.

We define the (logarithmic) height h(Z) of Z as the logarithm of norm of its Chow

form in degree 1.

By the definition, when Z = Pm(C), the corresponding Chow form is simply the

resultant in the same degree. For D = 1, this is ± det

((ui,ν)0≤i≤m

|ν|=D

)which has

non-zero coefficients ±1. Thus we have h(Pm(C)) = 0.

In the case where Z is a Q-subvariety of Pm(C) of dimension 0, the corresponding

Chow form F in degree 1 is a homogeneous polynomial of degree deg(Z) in m + 1

variables. Viewing it as a polynomial map F : C[X]1 −→ C, we have

Z(F ) = {L ∈ C[X]1; Z(L) ∩ Z 6= ∅}.

Note that, for any point of such Z with representative α = (α0, α1, . . . , αm) in Pm(C)

with at least one coordinate equal to 1, Z consists of the deg(Z) points

(σ(α0) : σ(α1) : . . . : σ(αm)) ∈ Pm(C)

where σ runs through all embeddings of Q(α0, . . . , αm) into C. Therefore, writing F


as a polynomial in X0, . . . , Xm, F has the form

ad∏i=1

(σi(α0)X0 + σi(α1)X1 + · · ·+ σi(αm)Xm

), a ∈ Z.

Let C be a compact subset of C[X]D with non-empty interior. We call it a convex

body of C[X]D if we have aP + bQ ∈ C for any P, Q ∈ C and for any a, b ∈ C with

0 ≤ |a|+ |b| ≤ 1. Then all the polynomials of C[X]D of norm ≤ 1 form a convex body

of C[X]D. We call it the unit convex body of C[X]D.

For a Q-subvariety Z of Pm(C) of dimension t and its corresponding Chow form

F in degree D, we define the height of Z relative to convex body C of C[X]D to be

hC(Z) = hC(F ) = log ‖F‖C

where ‖F‖C = sup{|F (P0, . . . , Pt)|;P0, . . . , Pt ∈ C}. We also use the same notation

‖F‖C not only for the Chow form but also for any polynomial map F : C[X]t′D −→ C

with t′ ≥ 1.

Given t ∈ {0, . . . ,m}, we define a Q-cycle of dimension t in Pm(C) to be a formal

linear combination of distinct Q-subvarieties Z1, . . . , Zs of Pm(C) of dimension t

Z = m1Z1 + · · ·+msZs

for some positive integers m1, . . . ,ms. Such Q-subvarieties Z1, . . . , Zs are called the

irreducible components of Z.

We extend to cycles the notions of degree, height and height relative to a convex

body by writing

deg(Z) =s∑i=1

mi deg(Zi), h(Z) =s∑i=1

mih(Zi)

and

hC(Z) =s∑i=1

mihC(Zi)

where C stands for an arbitrary convex body of C[X]D for some D ∈ N∗.

By the definition, we get the following result.


Corollary 1.2.6. Let Z be a Q-cycle of Pm(C) and C be a convex body of C[X]D.

Assume that

hC(Z) ≤ ah(Z) + b deg(Z)

for some a, b ∈ R. Then there exists an irreducible component Z ′ of Z such that

hC(Z′) ≤ ah(Z ′) + b deg(Z ′).

In the proof of our main result, we construct a certain Q-subvariety of dimension

0, obtained as an irreducible component of a certain Q-cycle of P2(C) of dimension

0. To derive estimates relative to such a Q-subvariety, we use the following lemmas,

taken from the paper [21] of D. Roy (see also [5]).

The first lemma compares the height of a Q-cycle Z with its height relative to the

unit convex body of C[X]D.

Lemma 1.2.7. [19, Lemma 2.1] Let D be a positive integer and let B be the unit

convex body of C[X]D. Then, for any integer t ∈ {0, 1, . . . ,m} and any Q-cycle Z of

Pm(C) of dimension t, we have

|hB(Z)−Dt+1h(Z)| ≤ (t+ 4)(t+ 1) log(m+ 1)Dt+1 deg(Z).

In particular, we have hB(Pm) ≤ (m+ 4)(m+ 1) log(m+ 1)Dm+1.

The second lemma provides estimates for the intersection of such a Q-cycle with

a certain type of hypersurface.

Lemma 1.2.8. [19, Proposition 2.2] Let D be a positive integer, let C be a convex

body of C[X]D, and let Z be a Q-subvariety of Pm(C) of dimension t > 0. Suppose

that there exists a polynomial P ∈ Z[X]D ∩ C such that Z(P ) does not contain Z.

Then there exists a Q-cycle Z ′ of Pm(C) of dimension t− 1 which satisfies:

(i) deg(Z ′) = D deg(Z);

(ii) h(Z ′) ≤ Dh(Z) + deg(Z) log ‖P‖+ 2(t+ 5)(t+ 1) log(m+ 1)D deg(Z);

(iii) hC(Z′) ≤ hC(Z) + 2t log(m+ 1)Dt+1 deg(Z).


The third lemma deals with the case where the Q-cycle Z has dimension 0.

Lemma 1.2.9. [19, Proposition 2.3] Let D be a positive integer, let C be a convex

body of C[X]D, and let Z be a Q-subvariety of Pm(C) of dimension 0, and let Z be

a set of representatives of the points of Z by elements of Cm+1 of norm 1. Then, we

have ∣∣∣∣∣∣hC(Z)−Dh(Z)−∑α∈Z

log sup{|P (α)|;P ∈ C}

∣∣∣∣∣∣ ≤ 9 log(m+ 1)D deg(Z).

Moreover, if there exists a polynomial P ∈ Z[X]D ∩ C which does not belong to I(Z),

then we have hC(Z) ≥ 0 and

0 ≤ 7 log(m+ 1)D deg(Z) +Dh(Z) +∑α∈Z

log |P (α)|.

We will also need the following result, which is a special case of Proposition 3.7

in [14].

Lemma 1.2.10. Let D, s ∈ N∗. Assume that F1, . . . , Fs are non-zero multi-homogeneous

polynomial maps from C[X]m+1D to C and that F = F1 · · ·Fs has multi-degree (d0, . . . , dm).

Let C be a convex body of C[X]D. Then we have

‖F‖C ≤s∏i=1

‖Fi‖C ≤(D + 2

2

)2(d0+···+dm)

‖F‖C.

1.3 Notation

In this chapter, the letters i, j, k always denote non-negative integers.

We fix (ξ, η) ∈ C× C∗ and (r, s) ∈ Q∗2 with s 6= ±1.

For each i, set γi = (1 : ξ + ir : ηsi) ∈ P2(C), then γi = (1, ξ + ir, ηsi) is a

representative of γi in C3. For each integer T , we put

ST = {γi ; 0 ≤ i < T}

and

CT =

{‖γ0‖+ |r|T if |s| < 1,

|r|T + |s|T‖γ0‖ if |s| > 1.


So we have ‖γi‖ ≤ CT for 0 ≤ i < T .

In this chapter, for any ring R, we denote by R[X] the polynomial ring in the

variables X0, X1, X2 with coefficients in R. For any ν = (ν0, ν1, ν2) ∈ N3, we denote

by Xν the monomial Xν00 X

ν11 X

ν22 and set |ν| = ν0 + ν1 + ν2.

We define the norm ‖P‖ of a polynomial P ∈ C[X] as the largest absolute value

of its coefficients and define the length L(P ) as the sum of all absolute values of its

coefficients.

Let τ denote the map

τ : C3 −→ C3

(x, y, z) 7−→ (x, y + rx, sz)

and let τ denote the induced map from P2(C) to P2(C). Viewing C×C∗ as a subset of

P2(C) under the standard embedding mapping (y, z) to (1 : y : z), the map τ restricts

to translation by (r, s) in the group C× C∗.Let Φ denote the C−algebra isomorphism on C[X] which sends a homogeneous

polynomial P (X0, X1, X2) ∈ C[X]D to P (X0, X1 +rX0, sX2) ∈ C[X]D. Then we have

Φj(P )(τ i(z)) = Φi+j(P )(z) for all z ∈ C3.

Now for each integer T ≥ 0, we denote by I(T ) the ideal of C[X] generated by

all homogeneous polynomials in C[X] vanishing on ST and denote by I(T )D its homo-

geneous part of degree D which consists of 0 and all polynomials in I(T ) which are

homogeneous of degree D. For any α ∈ P2(C) with representative α in C3 of norm

1, we also define

|I(T )D |α = sup{|P (α)|; P ∈ I(T )D , ‖P‖ ≤ 1}.

For any subset W of P2(C), we write W to denote an arbitrary set of representa-

tives of points of W by points of C3 of norm 1.

1.4 Outline of the proof of Theorem 1.1.5

We provide here an outline of the proof of our main result. The strategy is similar

to the one of D. Roy in [21]. The difference is that, in [21], the author considers


polynomials whose derivatives are small up to a large order at one point while in our

work, we consider polynomials taking small values at a large number of points which

are translates of a fixed point (ξ, η) by multiples of a rational point (r, s). Despite

this difference, it is surprising that we obtain a so similar looking result.

Arguing by contradiction, as in [21], we first replace PD by an appropriate ho-

mogenization PD of PD such that PD(γi) is equal to PD(ξ + ir, ηsi) up to a product

of powers of η and s, and such that X0 - PD, X2 - PD. The last condition ensures

that the polynomials Φi(PD) with i ∈ Z are relatively prime.

For each degree D ≥ 1, we define a convex body CD consisting of polynomi-

als of C[X]D of bounded norm taking small values at γ0, . . . ,γTD where TD =

bDσc. The precise condition defining CD ensures that CD contains all the polyno-

mials c′2DTDΦj(PD) with 0 ≤ j < 2TD for an appropriate positive integer c′.

The first crucial property which we prove is that the height hCD(P2(C)) of P2(C)

relative to CD is a very small negative number. Recall that this height is the logarithm

of the supremum of the absolute values of the resultant at triples of polynomials from

CD. A result of [21] implies that the resultant vanishes up to order TD at each triple of

homogeneous polynomials vanishing at all points γi of STD . The problem is that the

polynomials of CD may not vanish on STD . However, they take small values at each

point of STD . In Section 1.5, we prove an interpolation estimate which shows that,

for each polynomial of CD, there exists a homogeneous polynomial of the same degree

and small norm which takes the same values at each point γi of STD . Therefore, each

triple of polynomials of CD is close to a triple of polynomials vanishing on STD . As the

resultant vanishes at the modified triples up to a very large order, an application of

Schwarz’s lemma implies that the resultant takes very small absolute values at triples

in C3D. This means that hCD(P2(C)) is a very small negative number.

Based on this, we adapt the argument in [21] to construct a Q-subvariety ZD of

dimension 0 contained in Z(Φj(PD); 0 ≤ j < 2TD) whose height hCD(ZD) relative to

CD is very small (negative). The existence of this Q-subvariety ZD is based on the

lemmas of Section 1.2 which are not proved in the thesis. However, for the convenience

of the reader, we give here some explanation on how ZD is obtained (for details, see

[14] and [21]). First of all, we observe that the divisor of PD is a Q-cycle of dimension


1 whose height relative to CD is very small (negative) since PD ∈ CD and hCD(P2(C))

is very small (negative). Then we choose an irreducible component Z ′ of this Q-cycle

whose height relative to CD is smallest compared to the standard height and degree

of Z ′. Since X0 - PD, X2 - PD, there exists a polynomial Φi(PD) with 0 ≤ i < D not

vanishing on Z ′. The intersection of Z ′ with the divisor of Φi(PD) is a Q-cycle of

dimension 0 whose height relative to CD is very small (negative). Then, we take for

ZD an irreducible component of this Q-cycle in a similar fashion as we did for Z ′.

The rest of the argument is new and differs a lot from the argument in [21]

although the same idea is to reach a contradiction by intersecting (a translate of) ZD

with the divisor of a polynomial of the form Φi(PD) for a smaller degree D′. Such a

descent argument is typical in algebraic independence and is crucial for example in the

proof of Philippon’s criterion for algebraic independence [16]. To put this in practice,

we first note that, by the penultimate lemma of Section 1.2, the height hCD(ZD) is

essentially equal to ∑α∈ZD

log sup{|P (α)|; P ∈ CD},

where ZD denotes an arbitrary set of representatives of points of ZD by points of C3

of norm 1. We also note that

sup{|P (α)|; P ∈ CD} ≥ |I(TD)D |α

for each α ∈ ZD with representative α ∈ C3 of norm 1. We show in Section 1.7 that,

for each α ∈ P2(C), we also have

log dist(α,STD) ≤ cT 2D + log |I(TD)

D |α

for some constant c > 0, where dist(α,STD) denotes the smallest distance between α

and a point of STD (we use the projective distance defined in Section 1.7). Putting

all the estimates together, we conclude that

Θ =∑α∈Z0

D

log dist(α,STD)

is small (negative), where Z0D is a subset of ZD obtained by extracting the points of

ZD which are far from any points of STD .


For each α ∈ Z0D, we choose an integer tα ∈ {0, 1, . . . , TD−1} for which γtα ∈ STD

is closest to α. Then we have

Θ =∑α∈Z0

D

log dist(α, γtα).

For each pair of integers (m,n) with 0 ≤ m < n ≤ TD, we define

Θ(m,n) =∑α∈Z0

Dm≤tα<n


We also define recursively a sequence of pair (mk, nk) with k ∈ N starting with

(m0, n0) = (0, TD) such that nk −mk is essentially TD/2k, and Θ(mk, nk) is at most

nk −mk

TDΘ. We show that there exists a largest integer k such that

τ−mk(ZD) ⊂ Z(Φi(PDk); 0 ≤ i < 2TDk), (1.8)

where Dk is the smallest integer satisfying nk−mk ≤ TDk . We show that Dk tends to

infinity with D. Based on (1.8), we deduce upper bounds for the degree and height of

τ−mk(ZD) in terms of Dk. Similar upper bounds then follow for the degree and height

of Z ′D := τ−mk+1(Z) because |mk −mk+1| ≤ TD. Now we put D′ = Dk+1 where Dk+1

is defined similarly as we did for Dk. Because of the choice of k, there exists an integer

i0 with 0 ≤ i < 2TD′ such that the polynomial P := Φi0(PD′) does not vanish on Z ′D.

Using a lemma of Section 1.2, this implies a lower bound for∑

α∈ZDlog |P (α)| in

terms of the height and degree of Z ′D.

Define WD to be the set of α ∈ Z0D such that mk+1 ≤ tα < nk+1, and for

each α ∈ WD, define α′ = τ−mk+1(α). Then we obtain a similar lower bound for∑α∈WD

log |P (α′)| where α′ denotes a representative in C3 of α′.

For each α ∈ WD, the point α′ is close to τ−mk+1(γtα) = γ`α where `α = tα−mk+1

is an integer in the range 0 ≤ `α < nk+1 −mk+1 ≤ TD′ . Moreover, we have

|P (α′)| ≤∣∣∣∣P ( γ`α

‖γ`α‖

)∣∣∣∣+DL(P ) dist(α′, γ`α)

≤ e−12D′ν +D′L(P ) dist(α′, γ`α).


If, for some α0 ∈ WD, we have |P (α′0)| < 2e−12D′ν , then this is easily found

to contradict the lower bound for∑

α∈WDlog |P (α′)|. We are thus reduced to the

case where |P (α′)| is essentially bounded above by dist(α′, γ`α) or equivalently by

dist(α, γtα). This give an upper bound for∑

α∈WDlog |P (α′)| in terms of∑

α∈WD

log dist(α, γtα) = Θ(mk+1, nk+1).

Again, this contradicts the lower bound on∑

α∈WDlog |P (α′)|.

1.5 An interpolation estimate for homogeneous poly-

nomials

In this section, we establish an upper bound for the length of an arbitrary homoge-

neous polynomial of C[X]L in terms of the values which it takes at the points of SMwhere M =

(L+22

). This implies that any polynomial in C[X]L is determined uniquely

by its values on SM . We will use this result to construct interpolation polynomials in

the next section.

Lemma 1.5.1. Let L ∈ N and put M =(L+22

). Then there exists a constant c =

c(r, s, ξ, η) ≥ 3 such that any Q ∈ C[X]L has length satisfying

L(Q) ≤

cL2 · max

0≤i<M|Q(γi)| if |s| > 1,

cL3 · max

0≤i<M|Q(γi)| if |s| < 1.

(1.9)

Consequently, the linear map

φ : C[X]L −→ CM

Q 7−→ (Q(γi))0≤i<M

is bijective.

Proof. Note that the estimate (1.9) implies that the linear map φ is injective. Then,

since dim C[X]L = dim CM , this yields that φ is bijective.


It remains to show the first assertion. The result is clear for L = 0 since then

Q ∈ C. Assume that L > 0. We note that, for each (j, k) ∈ N2 with j + k = L, the

polynomial

Qjk(X) = XL−j−k0

j−1∏i=0

(X1 − (ξ + ir)X0)

has degree j in X1. Hence, for each k ≤ L, the polynomials Qj,k with j = 0, . . . , L−kare linearly independent. This implies that the polynomials

(η−1X2)kQj,k ((j, k) ∈ N2, j + k = L)

are linearly independent, and so form a basis of C[X]L since their cardinal is M =

dimC[X]L.

Fix Q ∈ C[X]L. We write

Q(X) =∑j+k≤L

cjk(η−1X2)

kQj,k(X)

for some cjk ∈ C. We have

L(Q) ≤∑j+k≤L

|cjk| |η|−kj−1∏i=0

(1 + |ξ|+ i|r|)

≤∑j+k≤L

|cjk| |η|−k(1 + |ξ|+ |r|)j−1∏i=1

i(1 + |ξ|+ |r|)

≤2M maxj+k≤L

{|cjk| |η|−k(1 + |ξ|+ |r|)j.j!

}.

To find an upper bound for |cjk|, we set P (X) = Q(X0, X1 + ξX0, ηX2). We have

P (X) =∑j+k≤L

cjkXL−j−k0 X1(X1 − rX0) · · · (X1 − (j − 1)rX0)X

k2 .

For each (i, k) ∈ N2, put

u(j,k)i =

{i(i− 1) · · · (i− j + 1)rjsik if j > 0,

sik if j = 0,

and, for each, k ∈ N define a sequence u(j,k) by u(j,k) = (u(j,k)i )i∈N. Set

u =∑j+k≤L

cjku(j,k).


Then

ui =∑j+k≤L

cjk u(j,k)i = P (1, ir, si) = Q(γi).

Let τ denote the linear operator on CN which sends a sequence (xn)n∈N to the shifted

sequence (xn+1)n∈N. For each (j′, k′) ∈ N2 satisfying j′ + k′ ≤ L, we will construct a

polynomial Fj′,k′ ∈ C[T ] of degree < M such that

(Fj′,k′(τ)(u(j,k))

)0

=

{1 if (j′, k′) = (j, k),

0 else.(1.10)

If we take this for granted, then cjk =(Fjk(τ)(u)

)0. Moreover, since degFjk < M ,

we have

|cjk| = |(Fjk(τ)(u)

)0| ≤ L(Fjk) max{|(τ i(u))0| ; 0 ≤ i < M}

≤ L(Fjk) max{|ui| ; 0 ≤ i < M}

≤ L(Fjk) max{|Q(γi)| ; 0 ≤ i < M}

So we also need an upper bound for L(Fjk) to estimate |cjk|.Fix (j0, k0) ∈ N2 such that j0 +k0 ≤ L. We now proceed to construct Fj0k0 . We claim

that

(τ − sk)m(u(j,k)) =

{j(j − 1) · · · (j −m+ 1)(rsk)mu(j−m,k) if m ≤ j

0 if m > j(1.11)

Indeed, for m = 1, and j ≥ 1, we have

((τ − sk)(u(j,k)))i = (i+ 1)i · · · (i− j + 2)rjs(i+1)k − i(i− 1) · · · (i− j + 1)rjsik+k

= (rsk)((i+ 1)− (i− j + 1)

)i · · · (i− j + 2) rj−1sik

= (rsk)ju(j−1,k)i .

So by induction on m, we find that (1.11) is true for m ≤ j. Since u(0,k)i = sik, we

also have (τ −sk)(u(0,k)) = 0. From this, we deduce that (2.30) is also true for m > j.

Now using (1.11) and u(j,k)0 = δ0j, we get

((τ − sk0)j0(u(j,k0)))0 = (rsk0)j0j0!δj0j. (1.12)


In particular, we have

(τ − sk0)L−k0+1(u(j0,k0)) = 0 since j0 + k0 ≤ L.

Since this holds for any (j0, k0) with j0 + k0 < L, we deduce that∏k′ 6=k0

k′=0,...,L

(τ − sk′)L−k′+1(u(j,k)) = 0 when k 6= k0, j + k ≤ L. (1.13)

By [19, Lemma 3.2], there exists a unique polynomial aj0,k0(Y ) ∈ C[Y ] of degree

≤ L− j0 − k0 such that

aj0,k0(Y )∏k′ 6=k0

k′=0,...,L

(1− Y

sk′ − sk0

)L−k′+1

≡ 1 mod Y L−j0−k0+1 (1.14)

and it satisfies

L(aj0,k0) ≤(M − j0 − 1

L− j0 − k0

)maxk′ 6=k0

k′=0,··· ,L

{1,

1

|sk′ − sk0|

}L−j0−k0

≤

2M max{

1, 1|s|−1

}Lif |s| > 1,

2M max{

1, 1|s|L(1−|s|)

}Lif |s| < 1

≤

{2McL1 if |s| > 1

2McL2

2 if |s| < 1

where c1 =|s||s| − 1

and c2 =1

|s|(1− |s|). Replacing Y by T − sk0 in (1.14), we get

aj0,k0(T − sk0)∏k′ 6=k0

k′=0,...,L

(T − sk′

sk0 − sk′)L−k′+1

≡ 1 mod (T − sk0)L−j0−k0+1. (1.15)

This yields the following congruence modulo (X − sk0)L−k0+1

(T − sk0)j0aj0,k0(T − sk0)∏k′ 6=k0

k′=0,...,L

(T − sk′

sk0 − sk′)L−k′+1

≡ (T − sk0)j0 .


Now take

Fj0,k0(T ) =1

(rsk0)j0j0!(T − sk0)j0aj0,k0(T − sk0)

∏k 6=k0

k=0,...,L

(T − sk

sk0 − sk

)L−k+1

Then Fj0,k0 has degree < M and from (1.13) and (1.15) we get

Fj0,k0(τ)(u(j,k)) =

0 if k 6= k0,

1(rsk0 )j0j0!

(τ − sk0)j0(u(j,k0)) if k = k0.

By (1.12), we get (1.10) as required.

Now, it remains to find an upper bound for L(Fj0,k0). We have

L(Fj0,k0) ≤1

|rsk0|j0j0!(1 + |s|k0)j0L(aj0,k0)(1 + |s|k0)L−j0−k0

∏k 6=k0

k=0,...,L

(1 + |s|k

|sk − sk0|

)L−k+1

≤ L(aj0,k0)

|rsk0|j0j0!(1 + |s|k0)L−k0

∏k 6=k0

k=0,...,L

(1 + |s|k

||s|k − |s|k0|

)L−k+1

.

In the case where |s| > 1, we have

L(Fj0,k0) ≤2McL1|rsk0 |j0j0!

(2|s|k0)L−k0∏k 6=k0

k=0,...,L

(2|s|k

|s|k−1(|s| − 1)

)L−k+1

≤2M+LcL1|r|j0j0!

|s|k0(L−k0−j0)∏k 6=k0

k=0,...,L

(2c1)L−k+1

≤(4c1|s|)M+L

|r|j0j0!

upon noting that k0(L− k0) ≤ L2/4 ≤M .


In the case where |s| < 1, we have

L(Fj0,k0) ≤L(aj0,k0)

|r|j0j0!(1 + |s|k0)L−k0|s|k0(L−k0)

∏k 6=k0

k=0,...,L

(1 + |s|k

||s|k − |s|k0 |

)L−k+1

≤ 2McL2

2

|r|j0j0!

L∏k=0

(2

|s|L(1− |s|)

)L−k+1

≤ 4McL(M+L)2

|r|j0j0!≤ (2c2)

4L3

|r|j0j0!.

Now we have

L(Q) ≤ 2M maxj+k≤L

{(1 + |ξ|+ |r|)jj!|η|−k|cjk|

}≤ 2M max

j+k≤L

{(1 + |ξ|+ |r|)jj!|η|−kL(Fjk)|

}max0≤i<M

|Q(γi)|.

Take c′ = 1 + 1+|ξ|+|r||r| + |η|−1 we get

L(Q) ≤

2Mc′L(4c1|s|)M+L · max

0≤i<M|Q(γi)| if |s| > 1,

2Mc′L(2c2)4L3 · max

0≤i<M|Q(γi)| if |s| < 1.

We deduce that there exists c = c(r, s, ξ, η) > 1 satisfying (1.9).

Now we will give an example which shows that the estimate (1.9) established in

Lemma 1.5.1 is a good upper bound for L(Q).

Example 1.5.2. Take r = 1, ξ = 0, η = 1. Since φ is an isomorphism, there exists

Q ∈ C[X]L such that (Q(1, i, si))0≤i<M = (0, . . . , 0, 1).

Write

Q(X) = c0LXL2 +

∑j+k≤Lk 6=L

cjkXL−j−k0 X1(X1 − rX0) · · · (X1 − (j − 1)rX0)X

k2 .

Note that, in the proof of Lemma 1.5.1, if j+ k = L then we have ajk(X) = 1. Thus,

the polynomial

F0L(T ) =L−1∏k=0

(T − sk

sL − sk

)L−k+1


satisfies c0L = [F0L(τ)(u)]0 ≤ L(Q) where ui = 0 for all i < M−1 and uM−1 = 1. Since

degF0L =∑L−1

k=0 (L− k+ 1) = M − 1 and (τ i(u))0 = 0 for i < M − 1, [τM−1(u)]0 = 1,

we deduce that [F0L(τ)(u)]0 is exactly the leading coefficient of F0L(T ). So

[F0L(τ)(u)]0 =L−1∏k=0

(1

sL − sk

)L−k+1

.

Take s = 1/2, by induction on L, we can check

|[F0L(τ)(u)]0| =L−1∏k=0

(2L+k

2L − 2k

)L−k+1

≥ 216L3

.

So

216L3 ≤ L(Q) ≤ cL

3

= cL3 ·max{|Q(γi)|, 0 ≤ i < M}.

1.6 Decomposition of polynomials in I(T )

In general, given an arbitrary homogeneous ideal J in C[X], we cannot expect that

JN ⊂ 〈JD〉 when N ≥ D where JN , JD denote the homogeneous parts of J of

respective degrees N,D. In this section, we consider the ideal I(T ) = I(ST ), defined

in section 1.3. We will show that

I(T )N ⊂ 〈I(T )D 〉

when N ≥ D and T ≤(bD/2c+2

2

). More precisely, for any polynomial Q ∈ I(T )N , we

will prove that

Q =∑ν∈N3

|ν|=N−D

XνQν

for some Qν ∈ I(T )D with an upper bound for∑L(Qν). Assuming that N = T, this

will lead to an upper bound for |I(T )T |α in terms of |I(T )D |α, valid for any α ∈ P2(C).

Lemma 1.6.1. Let K,L,N, T ∈ N such that

N + L ≤ 2K ≤ 2N ≤ 3K + 2 , T ≤(L+ 2

2

)


and let Q ∈ I(T )N . Then we can write

Q =2∑j=0

XN−Kj Qj

for some Qj ∈ I(T )K satisfying

2∑j=0

L(Qj) ≤

{3cL

3cK1 T

KL(Q) if |s| < 1,

3cL2cTK1 L(Q) if |s| > 1

where c1 = 1 + |ξ|+ |η|+ |r|+ |s| and c is as in Lemma 1.5.1.

Proof. Since N > 3(N −K − 1), for each triple ν in S := {ν ∈ N3; |ν| = N}, there

exists at least one coordinate ≥ N −K. For each t = 0, 1, 2, set

St = {ν = (ν0, ν1, ν2) ∈ S ; νt ≥ N −K}.

Then

S = S0 ∪ S1 ∪ S2.

Fix Q =∑ν∈N3

|ν|=N−D

cνXν ∈ I(T )N . Then we have Q =

∑2j=0X

N−Kj Pj where

P0 =∑ν∈S0

cνXν

XN−K0

,

P1 =∑

ν∈S1\S0

cνXν

XN−K1

,

P2 =∑

ν∈S2\(S0∪S1)

cνXν

XN−K2

are polynomials in C[X]K . We find that L(Q) =∑2

j=0 L(Pj).

Applying Lemma 1.5.1 with M =(L+22

), we get that, for each 1 ≤ j ≤ 2, there

exists Rj ∈ C[X]L such that

Rj(γi) =

{Pj(γi) if 0 ≤ i < T,

0 if T ≤ i < M.


and

L(Rj) ≤

cL3

max0≤i<T

|Pj(γi)| if |s| < 1,

cL2

max0≤i<T |Pj(γi)| if |s| > 1.

For each 0 ≤ i < T, we have

|Pj(γi)| ≤ L(Pj) ·max{1, |ξ + ir|t1|ηsi|t2 ; t1 + t2 ≤ K}

≤

{cK1 T

K · L(Pj) if |s| < 1,

cTK1 · L(Pj) if |s| > 1.

Hence

L(Rj) ≤

{cL

3cK1 T

K · L(Pj) if |s| < 1,

cL2cTK1 · L(Pj) if |s| > 1.

Since 2K − L ≥ N, and N ≥ K ≥ L, the polynomials

Q0 = P0 +X2K−L−N0 (XN−K

1 R1 +XN−K2 R2),

Q1 = P1 −XK−L0 R1,

Q2 = P2 −XK−L0 R2

belong to C[X]K . By construction, we have Q1, Q2 ∈ I(T ). Since Q =∑2

j=0XN−Kj Qj

belongs to I(T ), we deduce that XN−K0 Q0 ∈ I(T ), hence Q0 ∈ I(T ). Moreover we have

2∑j=0

L(Qj) ≤ 2L(R1) + 2L(R2) +2∑j=0

L(Pj)

≤

{2cL

3cK1 T

K · (L(P1) + L(P2)) + L(Q) if |s| < 1,

2cL2cTK1 · (L(P1) + L(P2)) + L(Q) if |s| > 1.

≤

{3cL

3cK1 T

KL(Q) if |s| < 1,

3cL2cTK1 L(Q) if |s| > 1.

Proposition 1.6.2. Let D,N, T be positive integers with N ≥ D and T ≤(bD/2c+2

2

).

Then any Q ∈ I(T )N can be written in the form

Q =∑ν∈N3

|ν|=N−D

XνQν


for a choice of polynomials Qν ∈ I(T )D satisfying

∑ν∈N3

|ν|=N−D

L(Qν) ≤

{cN

3c2N1 T 2N · L(Q) if |s| < 1,

cN2c2NT1 · L(Q) if |s| > 1,

(1.16)

where c1, c are as in Lemma 1.6.1 .

Proof. We will proceed by induction on N . The result is clear for N = D. When

N > D, we consider two cases.

Case 1: 2N ≤ 3D

Take K = D and L = 2D −N . Then we have

N + L = 2K ≤ 2N ≤ 3K

and L ≥ D/2 (since 2N ≤ 3D), so

T ≤(bD/2c+ 2

2

)≤(L+ 2

2

).

Lemma 1.6.1 ensures the existence of Q0, Q1, Q2 ∈ I(T )D such that Q =∑2

j=0XN−Dj Qj

and

2∑j=0

L(Qj) ≤

{3cL

3cD1 T

D · L(Q) if |s| < 1,

3cL2cDT1 · L(Q) if |s| > 1,

and (1.16) is satisfied since D ≤ N , L < N and c ≥ 3.

Case 2: 2N > 3D

Take K = N −bN/3c, L = bN/3c. Since N/3 ≥ D/2, we have L ≥ bD/2c and so

T ≤(bD/2c+ 2

2

)≤(L+ 2

2

).

On the other hand, we have

N + L = N + bN/3c ≤ N + (N − 2bN/3c) = 2K

≤ 2N ≤ 2N + (N − 3bN/3c) = 3K.


Lemma 1.6.1 ensures the existence of Q0, Q1, Q2 ∈ I(T )K such that Q =∑2

j=0XN−Kj Qj

and

2∑j=0

L(Qj) ≤

{3cL

3cK1 T

K · L(Q) if |s| < 1,

3cL2cKT1 · L(Q) if |s| > 1.

If K ≤ D then (1.16) is satisfied since L,D < N . Otherwise, applying the induction

hypothesis, for each 0 ≤ j ≤ 2, we can write

Qj =∑ν′∈N3

|ν′|=K−D

Xν′Qjν′

for a choice of polynomials Qjν′ ∈ I(T )D satisfying

∑ν′∈N3

|ν′|=K−D

L(Qjν′) ≤

{cK

3c2K1 T 2K · L(Qj) if |s| < 1,

cK2c2KT1 · L(Qj) if |s| > 1.

So

Q =2∑j=0

XN−Kj

∑ν′∈N3

|ν′|=K−D

Xν′Qjν′

with

2∑j=0

∑ν′∈N3

|ν′|=K−D

L(Qjν′) ≤

{cK

3c2K1 T 2K · 3cL3

cK1 TK · L(Q) if |s| < 1,

cK2c2KT1 · 3cL2

cKT1 · L(Q) if |s| > 1,

≤

{cN

3c2N1 T 2N · L(Q) if |s| < 1,

cN2c2NT1 · L(Q) if |s| > 1,

using K3 + L3 < (K + L)3 = N3, K2 + L2 < (K + L)2 = N2 and c ≥ 3.

Applying the above proposition with N = T , we obtain the following result.

Corollary 1.6.3. Let D,T be positive integers with D ≤ T ≤(bD/2c+2

2

). Let c and

c1 be as in Lemma 1.6.1. For any α ∈ P2(C) , we have

|I(T )T |α ≤

{cT

3c2T1 T 2T3D · |I(T )D |α if |s| < 1,

cT2c2T

2

1 3D · |I(T )D |α if |s| > 1.


Proof. Let α ∈ C3 be a representative of α of norm 1 and let Q ∈ I(T )T with ‖Q‖ ≤ 1.

Write Q =∑ν∈N3

|ν|=T−D

XνQν as in Proposition 1.6.2. Then

|Q(α)| ≤∑ν∈N3

|ν|=T−D

|Qν(α)| ≤∑ν∈N3

|ν|=T−D

‖Qν‖ · |I(T )D |α

≤∑ν∈N3

|ν|=T−D

L(Qν) · |I(T )D |α

≤

{cT

3c2T1 T 2TL(Q) · |I(T )D |α if |s| < 1,

cT2c2T

2

1 L(Q) · |I(T )D |α if |s| > 1.

The conclusion follows since L(Q) ≤ 3D‖Q‖ ≤ 3D and c > 3.

1.7 Distance

For any points u, v ∈ P2(C) with representatives u = (u0, u1, u2), v = (v0, v1, v2)

in C3, we define the projective distance between u and v by

dist(u, v) =‖u ∧ v‖‖u‖ ‖v‖

.

This is independent of the choice of u and v. The projective distance from v to a

finite subset S of P2(C) is defined by

dist(v,S) = min{dist(v, γ); γ ∈ S}.

Recall that γi is the point of P2(C) with homogeneous coordinates γi = (1, ξ +

ir, ηsi) and that ST is the set of points γi with 0 ≤ i < T . Recall also that

CT =

{‖γ0‖+ |r|T if |s| < 1,

|r|T + |s|T‖γ0‖ if |s| > 1.

In this section, we establish some estimates for the projective distance which are

crucial for the proof of the main result.


Lemma 1.7.1. Let α ∈ P2(C) with representative α = (α0, α1, α2) of norm 1.

(i) If dist(α, γi) <1

2‖γi‖for some i ≥ 0, then |α0| >

1

2‖γi‖.

(ii) For any positive integer T , there exists at most one non-negative integer i < T

such that

dist(α, γi) <|r|

4C2T

.

Proof. (i) Assume that dist(α, γi) < 1/(2‖γi‖). Then we have

max{|α0(ξ + ir)− α1|, |α0ηsi − α2|} ≤ ‖α ∧ γi‖ < 1/2.

This implies that

|α0| >max{|α1|, |α2|} − 1/2

‖γi‖,

which yields the required estimate for α0 since ‖α‖ = 1.

(ii) Assume that dist(α, γi) <|r|

4C2T

for some i < T . Since CT ≥ max{|r|, ‖γi‖}, we

find that dist(α, γi) < (2‖γi‖)−1. We conclude from part (i) that |α0| > (2‖γi‖)−1 >(2CT )−1. For any integer j with j 6= i, we have

dist(α, γi) + dist(α, γj) ≥‖α ∧ γi‖+ ‖α ∧ γj‖

CT

≥‖α ∧ (γi − γj)‖

CT

≥ |α0(i− j)r|CT

≥ |r|2C2

T

.

Using the assumption, we conclude that dist(α, γj) ≥ |r|(4C2T )−1.

Proposition 1.7.2. Let D,T be as in Corollary 1.6.3. Then there exists a constant

c2 = c2(r, s, ξ, η) > 1 such that

dist(α,ST ) ≤

{cT

3

2 |I(T )D |α if |s| < 1,

cT2

2 |I(T )D |α if |s| > 1

(1.17)

for any α ∈ P2(C).


Proof. Let α ∈ P2(C) with representative α in C3 of norm 1. Note that, for each i,

we have

dist(α, γi) = max{|Li(α)|, |L′i(α)|, |L′′i (α)|}

where

Li = ‖γi‖−1((ξ + ir)X0 −X1),

L′i = ‖γi‖−1(ηsiX0 −X2),

L′′i = ‖γi‖−1(ηsiX1 − (ξ + ir)X2).

Let Mi ∈ {Li, L′′i , L′′′i } such that dist(α, γi) = Mi(α). Then we have ‖Mi‖ ≤ 1 and

Mi(γi) = 0. We conclude that the polynomial

Q =T−1∏i=0

Mi

belongs to I(T )T and has length L(Q) ≤ 2T . Applying Corollary 1.6.3, we get

T−1∏i=0

dist(α, γi) = |Q(α)| ≤ L(Q)|I(T )T |α ≤

{cT

3c2T1 T 2T6T · |I(T )D |α if |s| < 1,

cT2c2T

2

1 6T · |I(T )D |α if |s| > 1.

By Lemma 1.7.1 (ii), we also have

T−1∏i=0

dist(α, γi) ≥ (4C2T )1−T · |r|T−1 · dist(α,ST ).

Hence

dist(α,ST ) ≤

{cT

3c2T1 T 2T62TC

2(T−1)T · |r|1−T · |I(T )D |α if |s| < 1,

cT2c2T

2

1 62TC2(T−1)T · |r|1−T · |I(T )D |α if |s| > 1.

So there exists a constant c2 > 1 which depends only on r, s, ξ, η and satisfies (1.19).

Proposition 1.7.3. Let µ, α ∈ P2(C) with representatives µ,α of norm 1 and let

P ∈ C[X]D. Then

|P (α)| ≤ |P (µ)|+DL(P ) dist(α, µ).


Proof. Without loss of generality, we can assume that µ = (1, µ1, µ2). Write P (X) =∑i+j≤D ci,jX

D−i−j0 X i

1Xj2 . Then

P (α)− αD0 P (µ) =∑i+j≤D

ci,j(αD−i−j0 αi1α

j2 − αD0 µi1µ

j2)

=∑i+j≤D

ci,j

(i∑t=1

(α1 − α0µ1)αD−i−j+t−10 αi−t1 αj2µ

t−11

+

j∑t=1

(α2 − α0µ2)αD−j+t−10 αj−t2 µi1µ

t−12

)So

|P (α)| ≤ |αD0 P (µ)|+∑i+j≤D

|ci,j|D dist(α, µ) ≤ |P (µ)|+DL(P ) dist(α, µ).

Recall that the map τ : C3 −→ C3 sends (x, y, z) to (x, y + rx, sz) and induces

τ : P2(C) −→ P2(C).

Lemma 1.7.4. Let α, γ be points of P2(C) and t be an integer. Then there exists a

constant c3 which depends only on s and C[X] such that

| log dist(τ t(α), τ t(γ))− log dist(α, γ)| ≤ c3|t|. (1.18)

Proof. Let α = (α0, α1, α2) and γ = (γ0, γ1, γ2) be representatives of α and γ in C3

of norm 1. We may assume that one of the coordinates of α is 1. We have

dist(α, γ) = max{|α1γ0 − α0γ1|, |α2γ0 − α0γ2|, |α1γ2 − α2γ1|}

and

dist(τ t(α), τ t(γ)) =‖τ t(α) ∧ τ t(γ)‖‖τ t(α)‖ · ‖τ t(γ)‖

where τ t(α) = (α0, α1 + tα0, stα2) and τ t(γ) = (γ0, γ1 + tγ0, s

tγ2). We find that

‖τ t(α) ∧ τ t(γ)‖ = max{|α1γ0 − α0γ1|, |(α2γ0 − α0γ2)s

t|,

|(α1 + trα0)γ2 − α2(γ1 + trγ0)| · |s|t}.


Since

|(α1 + trα0)γ2 − α2(γ1 + trγ0)| ≤ |(α2γ1 − α1γ2)|+ |tr(α2γ0 − α0γ2)|,

we deduce that

dist(τ t(α), τ t(γ)) ≤(1 + (|tr|+ 1)|s|t

) dist(α, γ)

‖τ t(α)‖ · ‖τ t(γ)‖. (1.19)

Note that

max{|α0|, |1 + trα0|} ≥(|tr|+ 1) ·max{|α0|, 1− |trα0|}

|tr|+ 1,

=max{|α0|(|tr|+ 1), 1 + |tr|(1− |α0|(|tr|+ 1))}

|tr|+ 1

≥ 1

|tr|+ 1.

Since one of the coordinates of (α0, α1, α2) is equal to 1, as we assumed before, then

we find that

‖τ t(α)‖ = max{|α0|, |α1 + trα0|, |stα2|}

≥

|α0| = 1 if α0 = 1,

max{|α0|, |1 + trα0|} if α1 = 1,

|st| if α2 = 1,

≥ min{|st|, (|tr|+ 1)−1}.

and so1

‖τ t(α)‖≤ max{|s−t|, |tr|+ 1}.

Similarly,1

‖τ t(γ)‖≤ max{|s−t|, |tr|+ 1}.

From (1.19), this yields the existence of c3 which only depends on |s| and |r| satisfying

(1.18).


1.8 Construction of Q-subvarieties of dimension 0

In this section, we define a general convex body C of C[X]D which is adapted

to our problem. Then, we construct a Q-subvariety Z of P2(C) of dimension 0 and

provide estimates for hC(Z), deg(Z), h(Z) (in this order).

We first recall the following result.

Theorem 1.8.1. [19, Theorem 5.6] Let Σ be a non-empty finite subset of G = C× C∗

and let S be a positive integer. Denote by I the ideal of C[X] generated by the homo-

geneous polynomials P satisfying

DiP (1, γ) = 0 for each γ ∈ Σ and each i = 0, . . . , S − 1. (1.20)

Suppose that there exists a finite subset Σ1 of G and an integer S1 ≥ 0 such that

D < (S1 + 1) min{|π1(Σ1)|, |π2(Σ1)|}, (1.21)

(S + S1)|Σ + Σ1| <(D + 2

2

), (1.22)

where Σ+Σ1 = {γ+γ1; γ ∈ Σ, γ1 ∈ Σ1} denotes the sumset of Σ and Σ1 in G. Then,

the resultant in degree D viewed as a polynomial map

ResD : C[X]3D −→ C

vanishes up to order S|Σ| at each point of (ID)3.

Here, π1 : G → C and π2 : G → C∗ are the projections from G to its first and

second coordinates.

Let T and D be positive integers. Set

Σ = {(ξ + ir, ηsi); 0 ≤ i < T}, Σ1 = {(ξ + ir, ηsi); 0 ≤ i ≤ D}

and set S = 1, S1 = 0. Note that if T ≤(D+12

), then the condition (1.21) and (1.22)

are satisfied because |π1(Σ1)| = |π2(Σ1)| = D + 1 and |Σ + Σ1| = T + D. Moreover,

by definition, I(T ) is the ideal of C[X] generated by all the homogeneous polynomials

vanishing at all points (1, γ) with γ ∈ Σ. Therefore, the above theorem has the fol-

lowing consequence.


Lemma 1.8.2. Let T and D be positive integers such that T ≤(D+12

). Then the

resultant in degree D viewed as a polynomial map

ResD : C[X]3D −→ C

vanishes up to order T at each triple (P,Q,R) of elements of I(T )D .

We now introduce the convex body C that is relevant to our problem and estimate

the corresponding height of P2(C). Recall that, for any convex body C of C[X]D, the

height of P2(C) relative to C is

hC(P2(C)) = hC(ResD) = log ‖ResD‖C (1.23)

where ‖ResD‖C = sup{ResD(P0, P1, P2); P0, P1, P2 ∈ C}.

Proposition 1.8.3. Let D,T be positive integers with T ≤(D+12

)and let Y, U be

positive real numbers such that

Y ≥

{2T log c if |s| > 1

3T 3/2 log c if |s| < 1

with c as in Lemma 1.5.1. Then, for the choice of convex body

C = {P ∈ C[X]D; ‖P‖ ≤ eY , max0≤i<T

|P (γi)| < e−U},

we have

hC(P2(C)) ≤ −TU + 3Y D2 + 21 log(3)D3.

Proof. By Lemma 1.2.7, we get

hB(ResD) = hB(P2(C)) ≤ 18 log(3)D3 (1.24)

where B is the unit convex body of C[X]D.

As C is compact, there exist P0, P1, P2 ∈ C such that ‖ResD‖C = |ResD(P0, P1, P2)|and so, by (1.23), we have

hC(P2(C)) = log |ResD(P0, P1, P2)|.


Let L denote the smallest non-negative integer such that T ≤(L+22

). Since

T ≤(D+12

), we have L < D. Moreover, we have L2 < 2

(L+12

)< 2T, so L3 < 3T 3/2.

Set M =(L+22

). For each j = 0, 1, 2, Lemma 1.5.1 ensures the existence of a unique

polynomial Qj ∈ C[X]L such that

Qj(γi) =

{Pj(γi) if 0 ≤ i < T

0 if T ≤ i < M

and

‖Qj‖ ≤ L(Qj) ≤

cL2 ·max0≤i<M |Qj(γi)| if |s| > 1

cL3 ·max0≤i<M |Qj(γi)| if |s| < 1

≤

c2T e−U if |s| > 1

c3T3/2e−U if |s| < 1

≤ eY−U .

We also have Pj − XD−L0 Qj ∈ C[X]D, and (Pj − XD−L

0 Qj)(γi) = 0 for 0 ≤ i < T .

Hence Pj −XD−L0 Qj ∈ I(T )D . According to Lemma 1.8.2, the polynomial

f(z) = ResD(P0 − (1− z)XD−L0 Q0, . . . , P2 − (1− z)XD−L

0 Q2) ∈ C[z]

vanishes up to order at least T at z = 0. Then we can write f(z) = zTg(z) for some

g(z) ∈ C[z]. Using the Maximum Modulus Principle, we find that

|f(1)| = |g(1)| ≤ ‖g‖R ≤ R−T‖f‖R

for any R ≥ 1. Choosing R = eU , we get

exp(hC(ResD)) = |ResD(P0, P1, P2)| = |f(1)| ≤ e−TU‖f‖eU . (1.25)

Note that, since ‖Qj‖ ≤ eY−U , for any |z| ≤ eU , we get

‖Pj − (1− z)XD−L0 Qj‖ ≤ ‖Pj‖+ (1 + eU)‖Qj‖ ≤ 3eY .


Therefore, we have

‖f‖eU = sup{|f(z)|; |z| ≤ eU}

≤ sup{|ResD(R0, R1, R2)|;Rj ∈ C[X]D, ‖Rj‖ ≤ 3eY , 0 ≤ j ≤ 2

}≤ (3eY )3D

2

sup {|ResD(R0, R1, R2)|;Rj ∈ C[X]D, ‖Rj‖ ≤ 1, 0 ≤ j ≤ 2}

= (3eY )3D2

exp(hB(ResD)). (1.26)

where the penultimate inequality follows from the fact that ResD is homogeneous of

total degree 3D2. Combining inequalities (1.24), (1.25) and (1.26), we get

exp(hC(ResD)) ≤ e−TU(3eY )3D2

exp(hB(ResD))

≤ e−TU(3eY )3D2

318D3

≤ exp(−TU + 3Y D2 + 21 log(3)D3).

Recall that

Φ : C[X] −→ C[X]

P (X0, X1, X2) 7−→ P (X0, X1 + rX0, sX2)

is a C−algebra isomorphism which preserves homogeneity and degree of polynomials

of C[X].

The construction of Z needs the two following results.

Lemma 1.8.4. Let D ∈ N, let i ∈ Z and let P ∈ Q[X]D be irreducible in Q[X].

Then Φi(P ) is also irreducible in Q[X]. Moreover, if i 6= 0, then P divides Φi(P ) if

and only if P is a constant multiple of either X0 or X2.

Proof. Fix i ∈ Z. Let P ∈ Q[X]D be irreducible in Q[X]. Since Φi is a Q−algebra

isomorphism, it preserves irreducibility. So Φi(P ) is irreducible in Q[X].

Now, suppose that i 6= 0, we will prove the last statement.

Assume that P divides Φi(P ). We will show that P is a constant multiple of either

X0 or X2. The converse is obvious.


Since P is irreducible, it is enough to prove that P is divisible by either X0 or X2.

To this end, we assume that P is not divisible by X0 and show that P is divisible by

X2.

Since degP = deg Φi(P ) and P |Φi(P ), there exists a constant c ∈ Q such that

P = cΦi(P ), i.e.

P (X0, X1, X2) = cP (X0, X1 + irX0, siX2). (1.27)

Write P (X0, X1, X2) =∑

j+k≤D

cjkXD−j−k0 Xj

1Xk2 with cjk ∈ Q. Substituting X0 = 0

into (1.27), we obtain∑j+k=D

cjkXj1X

k2 = c

∑j+k=D

cjkXj1(siX2)

k = c∑

j+k=D

cjk sikXj

1Xk2 . (1.28)

So cjk = c cjk sik for each (j, k) ∈ N2 with j + k = D. As X0 does not divide P , there

exists a pair (j, k) with j + k = D such that cjk 6= 0. Since s 6= ±1, we deduce that

there exists a unique (j0, k0) ∈ N2 with j0 + k0 = D such that cj0k0 6= 0 and, for this

choice of (j0, k0), we have c = s−ik0 . Now we have

P (X0, X1, X2) =∑

j+k<D

cjkXD−j−k0 Xj

1Xk2 + cD−k0,k0X

D−k01 Xk0

2 (1.29)

In the case where k0 6= 0, substituting X2 = 0 into (1.27) and using (1.29), we

find ∑0≤j<D

cj0XD−j0 Xj

1 = c∑

0≤j<D

cj0XD−j0 (X1 + irX0)

j

Suppose that there exists an integer j with 0 ≤ j < D such that cj0 6= 0 and let j′ be

the largest one. We deduce that

cj′0XD−j′0 Xj′

1 = c cj′0XD−j′0 Xj′

1 ,

so cj′0 = c cj′0. Since c = s−ik0 6= 1, this is a contradiction. We conclude that cj0 = 0

for all j < D. Thus X2 divides P .

In the case where k0 = 0, we have c = 1 and cD,0 6= 0. Replacing X2 by 0 into

(1.27) and using (1.29), we get∑0≤j<D

cj0XD−j0 Xj

1 + cD0XD1 =

∑0≤j<D

cj0XD−j0 (X1 + irX0)

j + cD0(X1 + irX0)D


Then

cD−1,0X0XD−11 = cD−1,0X0X

D−11 + cD0DirX0X

D−11 .

Since cD0 6= 0, this is a contradiction.

Lemma 1.8.5. Let D be a positive integer and let P ∈ Q[X]D with X0 - P and

X2 - P . Then the polynomials P,Φ(P ), . . . ,ΦD(P ) have no common irreducible factor

in Q[X]. Moreover, there exist a1, . . . , aD ∈ Z in the range 0 ≤ ai ≤ D which are not

all 0 and for which

Q =D∑i=1

aiΦi(P )

is relatively prime to P .

Proof. Write P = P e11 · · ·P

ekk as a product of irreducible factors in Q[X]. Then

Φi(P ) = Φi(P1)e1 · · ·Φi(Pk)

ek is also a decomposition of Φi(P ) into irreducible factors.

Suppose that P,Φ(P ), . . . ,ΦD(P ) have a common irreducible factor, say P1. Then

for each i ∈ {0, . . . , D}, there exists ji ∈ {1, . . . , k} such that P1 and Φi(Pji) are con-

stant multiples of each other. Since D ≥ k, then there exist two distinct indices

i1, i2 ∈ {0, . . . , D} such that ji1 = ji2 =: j′. So both Φi1(Pj′) = Φi1−i2(Φi2(Pj′)) and

Φi2(Pj′) are constant multiples of P1, this mean that they divide each other. Lemma

1.8.4 implies that Φi2(Pj′) is a constant multiple of either X0 or X2. So P1 is also a

constant multiple of either X0 or X2. This is a contradiction since X0 - P and X2 - P .

We deduce that P,Φ(P ), . . . ,ΦD(P ) have no common irreducible factor in Q[X].

To prove the last statement, we fix ξ1, ξ2 ∈ C which are algebraically independent

over Q and consider the canonical maps

ϕi : Q[X] −→ Q[X]/〈Pi〉

Q 7−→ Q := Q+ 〈Pi〉

for i = 1, . . . , k. By Normalization Theory, for each i, there exist Yi, Y′i ∈ Q[X]

algebraically independent over Q(Pi) such that Q[X]/〈Pi〉 is integral over Q[Y i, Y ′i

].

Since ξ1, ξ2 ∈ C are algebraically independent over Q, for each i, there exists an


embedding ϕi of Q[Y i, Y ′i

]into C which sends Yi to ξ1 and sends Y ′i to ξ2. Since

Q[X]/〈Pi〉 is integral over Q[Y i, Y ′i

], there exists an embedding ϕi of Q[X]/〈Pi〉 into

C which extends the embedding ϕi.

Let U1, . . . , UD be indeterminates over C. Set

R(U1, . . . , UD) =k∏i=1

(D∑j=1

Uj ϕi(ϕi(Φj(P )

))).

Then R(U1, . . . , UD) is a homogeneous polynomial in U1, . . . , UD of degree k with

coefficients in C. We claim that R(U1, . . . , UD) 6= 0. Otherwise, there exists i with

1 ≤ i ≤ k such thatD∑j=1

Uj ϕi(ϕi(Φj(P )

))= 0.

This implies that ϕi (ϕi (Φj(P ))) = 0 for all j = 1, . . . , D. Thus, ϕi (Φ

j(P )) = 0, i.e.,

Pi |Φj(P ) for all j = 1, . . . , D, which is impossible since P,Φ(P ), . . . ,ΦD(P ) have no

common factor.

Since R(U1, . . . , UD) is a non-zero homogeneous polynomial in U1, . . . , UD of degree

k, there exist integers a1, . . . , aD ∈ {0, 1, . . . , k} such that

R(a1, a2, . . . , aD) 6= 0.

Hence, for each i = 1, . . . , k, we get

ϕi

(ϕi

(D∑j=1

aj Φj(P )

))=

D∑j=1

aj ϕi(ϕi(Φj(P )

))6= 0.

This means that ϕi

(∑Dj=1 aj Φj(P )

)6= 0 for all i = 1, . . . , k. So

∑Dj=1 aj Φj(P ) is

relatively prime to P . Note that k ≤ deg(P ) = D so a1, . . . , aD ≤ D as required.

Let c′ be the smallest positive integer such that c′r, c′s ∈ Z. Then, for any

polynomial P ∈ Z[X]D, we have c′iDΦi(P ) ∈ Z[X]D for all i ∈ N. In the proof

of the main result, we use the following proposition to construct a Q-subvariety of

dimension 0 of small height relative to an appropriate convex body.


Proposition 1.8.6. Let D,T, Y, U and C be as in Proposition 1.8.3 and set A =

TU/(D2Y ). Suppose that

5 ≤ A, D < 2T, 25 log(3)D ≤ Y.

Suppose that there exists a non-zero polynomial P ∈ Z[X]D∩C such that X0 - P, X2 -P and c′iDΦi(P ) ∈ C for all i = 1, . . . , 2T − 1.

Then there exists a Q-subvariety Z ⊂ Z(Φi(P ); 0 ≤ i < 2T ) of dimension 0 with

hC(Z) ≤ −A′(Dh(Z) + Y degZ)

where A′ = (A− 5)/6.

Proof. We have dim(P2(C)) = 2, deg(P2(C)) = 1, h(P2(C)) = 0 and P ∈ Z[X]D ∩ C.Applying Lemma 1.2.8 with Z = P2(C), we deduce that there exists a Q-cycle Z ′ of

P2(C) of dimension 1 and degree D with

h(Z ′) ≤ Y + 42 log(3)D ≤ 3Y,

hC(Z′) ≤ hC(P2(C)) + 4 log(3)D3.

By Proposition 1.8.3, we get

hC(Z′) ≤ −TU + 3D2Y + 25 log(3)D3

≤ −(A− 4)D2Y

≤ −A− 4

6D(Dh(Z ′) + 3Y deg(Z ′)

).

Lemma 1.2.6 ensures the existence of an irreducible component Z1 of Z ′ with

hC(Z1) ≤ −A− 4

6D(Dh(Z1) + 3Y deg(Z1)

).

By Lemma 1.8.5, the polynomials P,Φ(P ), . . . ,ΦD(P ) have no common factor. We

deduce that there exists 0 ≤ i ≤ D such that Φi(P ) /∈ I(Z1). Since c′iDΦi(P ) ∈C ∩ Z[X]D, Lemma 1.2.8 implies that there exists a Q-cycle Z ′′ of dimension 0 and

degree D deg(Z1) satisfying

h(Z ′′) ≤ Dh(Z1) + deg(Z1) log ‖c′iDΦi(P )‖+ 24 log(3)D deg(Z1)

≤ Dh(Z1) + 2Y deg(Z1),


and

hC(Z′′) ≤ hC(Z1) + 2 log(3)D2 deg(Z1)

≤ −A− 4


)+

2

25DY deg(Z1)

= −A− 4


)−(A− 4

6− 2

25

)DY deg(Z1)

≤ −A− 5

6(Dh(Z ′′) + Y deg(Z ′′)).

Similarly, by linearity of degree and height, we deduce that there exists a subvariety

Z ⊂ Z ′′ such that

hC(Z) ≤ −A− 5

6(Dh(Z) + Y deg(Z)).

So hC(Z) < 0. From Lemma 1.2.8, we deduce that

I(Z) ⊃ (C ∩ Z[X]D) ⊃ {c′jDφj(P ); 0 ≤ j < 2T}.

Therefore, Z ⊂ Z(φj(P ); 0 ≤ j < 2T ).

In the proof of our main result, we need to consider translates of a Q-subvariety

of dimension 0 and we need to estimate their heights. The next proposition fulfills

this purpose.

Proposition 1.8.7. Let Z be a Q-subvariety of P2(C) of dimension 0 with Z * Z(X0)

and let t be an integer. Then

|h(τ t(Z))− h(Z)| � |t| deg(Z)

where the constant involved in the symbol � depends only on r, s.

Proof. Let F and G be Chow forms of Z and τ t(Z) in degree 1. Since Z * Z(X0),

the variety Z contains a point α ∈ P2(C) with representative (1, α1, α2) ∈ Q3. As Z

is a Q-subvariety of dimension 0, we have

Z = {(1 : σ(α1) : σ(α2)); σ ∈ G}


where G is the set of all embeddings of Q(α1, α2) into C (see the preliminaries in

Section 1.2). By definition of τ , this implies that

τ t(Z) = {(1 : σ(α1 + tr) : σ(stα2)); σ ∈ G}.

Therefore, we have

F (X) = a∏σ∈G

(X0 + σ(α1)X1 + σ(α2)X2),

G(X) = b∏σ∈G

(X0 + (σ(α1) + tr)X1 + stσ(α2)X2),

where |a| and |b| are the smallest positive integers such that the above products belong

to Z[X]. Let c be a common positive denominator of r, s and s−1 and set

P (X) = F (c|t|(X0 + trX1), c|t|X1, c

|t|stX2).

Then P belongs to Z[X] and since degF = deg(Z), we get

P (X) = ac|t|deg(Z)∏σ∈G

(X0 + (σ(α1) + tr)X1 + stσ(α2)X2).

So P is a constant multiple of G. Since G is irreducible over Z, we deduce that G

divides P in Z[X]. Therefore, by the definition of P , we obtain

‖G‖ ≤ ‖P‖ ≤ c|t| deg(Z)(|st|+ |tr|+ 1)deg(Z)‖F‖ ≤ (c(|s|+ |r|+ 1))|t| deg(Z)‖F‖.

Since h(Z) = log ‖F‖ and h(τ t(Z)) = log ‖G‖, this implies that

h(τ t(Z)) ≤ h(Z) + c′|t| deg(Z)

where c′ = log(c(|s| + |r| + 1)). Since Z = τ−t(τ t(Z)) and deg(Z) = deg(τ(Z)), this

result applied with Z replaced by τ t(Z) and t replaced by −t implies in turn that

h(Z) = h(τ−t(τ t(Z))) ≤ h(τ t(Z)) + c′|t| deg(Z).

The conclusion follows.

The last proposition provides upper bound estimates for the degree and the height

of Q-varieties of dimension 0 contained in the zero set of families of polynomials of

the form Φi(P ) with P ∈ Z[X]D fixed.


Proposition 1.8.8. Let D,T ∈ N∗. Let P ∈ Z[X]D with X0 - P and X2 - P and let

Y ∈ R. Suppose that

max{

25 log(3)D, log ‖P‖, log ‖c′D2

Φ(P )‖, . . . , log ‖c′D2

ΦD(P )‖}< Y

and that W = Z(Φi(P ); 0 ≤ i < T + D) is not empty. Then W has dimension 0.

Moreover, any Q-subvariety Z of P2(C) contained in W has dimension 0 with

degZ ≤ D2

Tand

T−1∑i=0

h(τ i(Z)) ≤ 3DY.

In particular, we have h(τ i(Z))� DY

T+D2 for each 0 ≤ i < T .

Proof. Since X0 - P and X2 - P , Lemma 1.8.5 implies that there exist integers

a1, . . . , aD not all 0, with 0 ≤ ai ≤ D, such that

Q =D∑i=1

ai(c′D2

Φi(P )) ∈ Z[X]

is relatively prime to P . Then dimZ(P,Q) = 0 and W ⊂ Z(P,Q). Since W 6= ∅, we

deduce that dimW = 0.

Let Z ⊂ W be an arbitrary Q-subvariety of P2(C), so dimZ = 0 and Z is finite.

Then, for each 0 ≤ i < T and for each representative z in C3 of a point z of Z, we

have

Φj(P )(τ i(z)) = Φi+j(P )(z) = 0

when 0 ≤ j ≤ D since z ∈ W = Z(Φi(P ); 0 ≤ i < T +D), and so

Q(τ i(z)) = c′D2

D∑j=1

aiΦj(P )

(τ i(z)

)= 0.

Thus τ i(z) ∈ Z(P,Q) for all i < T . Since Z is irreducible, so is τ i(Z) for all i < T .

The above observation then implies that Z, τ(Z), . . . , τT−1(Z) are disjoint irreducible

components of Z(P,Q). Since they all have dimension 0 as Z(P,Q), we have

T−1∑i=0

deg(τ i(Z)) =T−1∑i=0

∣∣τ i(Z)∣∣ ≤ |Z(P,Q)| = deg(Z(P,Q)) ≤ D2 (1.30)


where the last inequality follows from Corollary 1.2.5. Since deg(Z) = deg(τ i(Z)) for

all i, we deduce that

deg(τ i(Z)) = degZ ≤ D2

T. (1.31)

Now we will prove that∑T−1

i=0 h(τ i(Z)) ≤ 3DY .

Consider the polynomial map

F : C[X]D −→ C

L 7−→ ResD(P,Q, L).

Since ResD is homogeneous of degree D2 in each of its polynomial arguments, we

conclude that the polynomial underlying F is homogeneous of degree D2.

For each 0 ≤ i < T, denote by Fi a Chow form of the Q-subvariety τ i(Z) in degree

D (viewed as a polynomial map from C[X]D to C). We have

hB(τ i(Z)) = hB(Fi) = log sup{|Fi(L)|;L ∈ B} (1.32)

where B = {R ∈ C[X]D; ‖R‖ ≤ 1}. So, for such i, Lemma 1.2.7 gives

Dh(τ i(Z)) ≤ 4 log(3)D deg(τ i(Z)) + hB(τ i(Z))

= 4 log(3)D deg(Z) + hB(Fi), (1.33)

using (1.32) and deg(τ i(Z)) = deg(Z).

Since τ i(Z) ⊂ Z(P,Q), we get Z(Fi) ⊂ Z(F ). So Fi|F for all i < T . Since

Z, τ(Z), . . . , τT−1(Z) are disjoint irreducible varieties, the polynomials F0, . . . , FT−1

are non-associate irreducible polynomials. Hence∏T−1

i=0 Fi divides F as polynomials

over Q. Moreover, since F, Fi have coefficients in Z and Fi is irreducible over Z, we

deduce∏T−1

i=0 Fi divides F as polynomials over Z. Hence there exists a polynomial G

with coefficients in Z such that F = G∏T−1

i=0 Fi. Then we have 1 ≤ ‖G‖ ≤ ‖G‖B and

so hB(G) ≥ 0. This implies that

T−1∑i=0

hB(Fi) ≤ hB(G) +T−1∑i=0

hB(Fi)

≤ hB(F ) + 2D2 log

(D + 2

2

)(by Lemma 1.2.10)

≤ hB(F ) + 2D3 log(3). (1.34)


Since ResD(P,Q, L) is homogeneous of degree D2 in both P and Q and since we have

‖P‖ < eY and ‖Q‖ ≤∑D

i=1D‖c′D2Φi(P )‖ < D2eY , we find that

|ResD(P,Q, L)| ≤∣∣∣∣ResD

(P

‖P‖,Q

‖Q‖, L

)∣∣∣∣ (eY )D2

(D2eY )D2

for any L ∈ C[X]D. Therefore, we get

hB(F ) = log sup{|ResD(P,Q, L)|; ‖L‖ ≤ 1}

≤ log sup

{∣∣∣∣ResD

(P

‖P‖,Q

‖Q‖, L

)∣∣∣∣ ; ‖L‖ ≤ 1

}+ 2D2Y + 2D2 logD

≤ hB(P2) + 2D2Y + 2D2 logD (since hB(P2) = log ‖ResD‖B)

≤ 18 log(3)D3 + 2D2Y + 2D2 logD (by Lemma 1.2.7)

≤ 19 log(3)D3 + 2D2Y. (1.35)

Since deg(τ i(Z)) = deg(Z), it follows from (1.33) that

DT−1∑i=0

h(τ i(Z)) ≤ 4 log(3)DT degZ +T−1∑i=0

hB(Fi)

≤ 4 log(3)D3 + (hB(F ) + 2 log(3)D3) (using (1.34))

≤ 25 log(3)D3 + 2D2Y (using (1.35))

≤ 3D2Y.

So we haveT−1∑i=0

h(τ i(Z)) ≤ 3DY .

To show the last inequality, we note that, for each index j with 0 ≤ j < T , we

haveT−1∑i=0

h(τ i(Z)) = Th(τ j(Z)) +T−1∑i=0

(h(τ i(Z))− h(τ j(Z))

).

This implies that

|h(τ j(Z))| ≤ 1

T

(T−1∑i=0

h(τ i(Z)) +T−1∑i=0

∣∣h(τ i(Z))− h(τ j(Z))∣∣)

� 1

T

(3DY +

T−1∑i=0

|i− j| deg(τ j(Z))

)(by Proposition 1.8.7)

� DY

T+D2 (by (1.31)).


1.9 Proof of the main theorem 1.1.5

Proof. Suppose on the contrary that (1 : ξ : η) /∈ P2(Q). We will show that this leads

to a contradiction.

Fix a positive integer D. In the computations below, we assume that D is suf-

ficiently large so that all the inequalities marked with a star (i.e. ≤∗ or ≥∗) are

satisfied.

Step 0. Reduction to the case where |s| > 1.

Suppose that |s| < 1. We set

β′ = β +1

3ε , ν ′ = ν − 1

2ε

where ε = ν −max

{β + 2− σ +

(σ − 1)(2− σ)

β − (σ − 1), σ + 2

}so that

ν ′ > max

{β′ + 2− σ +

(σ − 1)(2− σ)

β − (σ − 1), σ + 2

}and β′ > σ + 1.

Moreover, for each integer D, the polynomial

P ∗D(X1, X2) = c′DT∗DPD(X1 + T ∗Dr, s

T ∗DX2), (with T ∗D = 3bDσc − 1)

belongs to Z[X1, X2]≤D and satisfies

max0≤i<3bDσc

{|P ∗D(ξ + i(−r), η(s−1)i)|} = max0≤i<3bDσc

{c′DT ∗D |PD(ξ + ir, ηsi)|}

≤ c′3D1+σ

e−Dν

≤∗ e−Dν′

and

‖P ∗D‖ ≤ c′DT∗D3D‖PD‖(1 + T ∗D|r|)D since |s| < 1

≤ c′3D1+σ

3DeDβ

(1 + |r|)3D1+σ

since ‖PD‖ ≤ eDβ

≤∗ eDβ′

.


So if we replace r by −r, s by s−1, ν by ν ′, β by β′, and PD by P ∗D, then all the

hypotheses of the theorem still hold. Therefore, it is enough to consider the case

where |s| > 1. We therefore assume from now on that |s| > 1.

Step 1. Construction of a convex body.

For each D ∈ N, we put

TD = bDσc, YD = 2Dβ, UD = 12Dν

and define a convex body of C[X]D by

CD = {Q ∈ C[X]D; ‖Q‖ ≤ eYD , max0≤i<TD

|Q(1, ξ + ir, ηsi)| ≤ e−UD}.

We also denote by PD the homogeneous polynomial of Z[X]D determined by the con-

dition

PD(1, X1, X2) = XaD1 X−bD2 PD(X1, X2)

where bD stands for the largest integer b such that Xb2 divides PD, and where aD =

D − deg(PD) + bD. Then, by construction, PD is not divisible by neither X0 nor X2,

moreover, ‖PD‖ = ‖PD‖.

By the definition of c′ given just before Proposition 1.8.6, we have

c′2DTDΦj(PD) ∈ Z[X]D

for any positive integer j < TD. We claim that, for any sufficiently large D, we have

c′2DTDΦj(PD) ∈ CD ∩ Z[X]D

for all integers j with 0 ≤ j < 2TD. In particular, this means that

Φj(PD) ∈ CD

for all integers j with 0 ≤ j < 2TD since c′ > 1.


To prove the claim, fix an integer j with 0 ≤ j < 2TD. Since β > σ + 1, we have

‖c′2DTDΦj(PD)‖ ≤ c′2DTD(D + 2

2

)‖PD‖ · max

t1+t2≤D{(1 + |jr|)t1|sj|t2}

≤ c′2Dσ+1

3DeDβ

(1 + |jr|+ |s|j)D

≤ 3Dc′2Dσ+1

eDβ

(1 + |r|+ |s|)2Dσ+1

≤∗ eYD .

Moreover, for each 0 ≤ i < TD, we have

|c′2DTDΦjPD(1, ξ + ir, ηsi)| ≤∣∣∣c′2Dσ+1

PD(1, ξ + (i+ j)r, ηsi+j

)∣∣∣= c′2D

σ+1|ξ + (i+ j)r|aD · |ηsi+j|−bD · |PD(ξ + (i+ j)r, ηsi+j)|

≤ c′2Dσ+1

(i+ j)D(|ξ|+ |r|)D|η|−bDe−Dν

≤∗ e−12Dν

since ν > σ + 1.

Step 2. Construction of a Q-subvariety of dimension 0.

Since ν > β − σ + 2, 1 ≤ σ < 2, and β > σ, the hypotheses of Proposition 1.8.6

hold for T = TD, Y = YD, U = UD and the convex body CD for each sufficiently large

D. So there exists a Q-subvariety ZD of P2(C) contained in Z(Φj(PD); 0 ≤ i < 2TD)

with dimZD = 0 and

hCD(ZD) ≤ − 1

25Dν−β+σ−2(2Dβ deg(ZD) +Dh(ZD)).

By Lemma 1.2.9, we get∑α∈ZD

log sup{|Q(α)|;Q ∈ CD} ≤ hCD(ZD)−Dh(ZD) + 9 log(3)D deg(ZD)

≤ hCD(ZD) + 9 log(3)D deg(ZD) (since h(ZD) ≥ 0)

≤ − 1

25Dν−β+σ−2(Dβ deg(ZD) +Dh(ZD)) (1.36)

since 9 log(3)D ≤∗ (1/25)Dν+σ−2. For any α ∈ P2(C) with representative α ∈ C3 of

norm 1, we have

sup{|Q(α)|;Q ∈ CD} ≥ sup{|Q(α)|;Q ∈ I(TD)D , ‖Q‖ ≤ 1} = |I(TD)

D |α


since I(TD)D ⊂ CD. Proposition 1.7.2 gives

|I(TD)D |α ≥ c

−T 2D

2 dist(α, STD) ≥ c−D2σ

2 dist(α, STD). (1.37)

Put

Z0D = {α ∈ ZD; dist(α, STD) < (4CTD)−1}

where CTD = |r|TD + |s|TD‖γ0‖. For α ∈ ZD\Z0D, we get

|I(TD)D |α ≥ c−D

2σ

2

1

4CTD,

and so we have

0 ≤∗ log |I(TD)D |α + log(2c2)D

2σ

because logCTD � TD ≤ Dσ. For the other points α ∈ Z0D, the inequality (1.37)

gives

log dist(α, STD) ≤ log |I(TD)D |α + log(c2)D

2σ.

We conclude that∑α∈Z0

D

log dist(α,STD)

≤∑α∈Z0

D

(log |I(TD)

D |α + log(c2)D2σ)

+∑

α∈ZD\Z0D

(log |I(TD)

D |α + log(2c2)D2σ)

≤∑α∈ZD

log |I(TD)D |α + log(2c2)D

2σ deg(ZD) (since |Z0D| ≤ |ZD| = deg(ZD))

≤∑α∈ZD

log sup{|Q(α)|;Q ∈ CD}+ log(2c2)D2σ deg(ZD)

≤ − 1

25Dν−β+σ−2(Dβ deg(ZD) +Dh(ZD)) + log(2c2)D

2σ deg(ZD)

≤∗ − 1

30Dν−β+σ−2(Dβ deg(ZD) +Dh(ZD)).

where the penultimate estimate uses (1.36) and the last estimate uses the fact that

ν > σ + 2 .

Step 3. Subsets of the Q-subvariety.

For each α ∈ P2(Q), we denote by tα the smallest non-negative integer with

0 ≤ tα < TD such that dist(α,STD) = dist(α, γtα).


For each m,n ∈ N with 0 ≤ m ≤ n < TD, define

Θ(m,n) =∑α∈Z0

Dm≤tα<n


Step 2 provides an upper bound for Θ(0, TD). Our goal is to construct subsums

of Θ(0, TD) of the form Θ(m,n) which are small compared to the number n −m of

values of tα that they involve. In fact, we construct recursively a finite sequence of

subsums such that each subsum is computed over an interval that is essentially half

of the one of the preceding subsum. More precisely, it is its first half or its second half

of this interval depending on which gives the smaller subsum compared to its length

(which may vary by ±1). Technically, we define recursively a finite sequence of pairs

(mj, nj) by putting

(m0, n0) = (0, TD)

and

(mj+1, nj+1) =

{(mj, kj) if Θ(mj, kj) ≤ kj−mj

nj−mj Θ(mj, nj),

(kj, nj) else,

where kj = b(mj + nj)/2c as long as nj −mj ≥ 2.

When nj −mj ≥ 2, we have mj < kj < nj and

Θ(mj, nj) = Θ(mj, kj) + Θ(kj, nj).

We deduce that

Θ(mj+1, nj+1) ≤nj+1 −mj+1

nj −mj

Θ(mj, nj).

By induction, this yields Θ(mj, nj) ≤nj −mj

TDΘ(0, TD). Using the upper bound for

Θ(0, TD) computed in Step 2, we deduce that

Θ(mj, nj) ≤nj −mj

DσΘ(0, TD)

≤ −nj −mj

30

(Dν−2 deg(ZD) +Dν−β−1h(ZD)

)(1.38)

for all pairs (mj, nj) of our sequence.


Step 4. Selection of a particular subset

Define D0 = D and Dj = d(nj −mj)1σ e so that TDj ≥ nj −mj and thus

{γ0, . . . , γnj−mj−1} ⊂ {γ0, . . . , γTDj } = STDj (1.39)

Then by the hypothesis, the functions ΦiPDj with 0 ≤ i < 2TDj take small absolute

values at those points .

Note that, since m0 = 0, we have

τ−m0(ZD) = ZD ⊂ Z(ΦiPD0 ; 0 ≤ i < 2TD0).

So, for fixed D, there exists a largest non-negative integer k such that nk −mk ≥ 2

and

τ−mk(ZD) ⊂ Z(ΦiPDk ; 0 ≤ i < 2TDk). (1.40)

Note that the set Z(ΦiPDk ; 0 ≤ i < 2TDk) is finite.

We claim that Dk goes to infinity with D.

Indeed, suppose on the contrary that Dk is bounded above by some positive integer

D∗ independently of the choice of D. Then τ−mk(ZD) is contained in the set

D∗⋃N=1

Z(ΦiPNk ; 0 ≤ i < 2TNk),

which is finite and independent of D. By equation (1.38) and the fact that Θ(mk, nk)

involves at most deg(ZD) terms, there exists (for sufficiently large D) a point α ∈ Z0D

with mk ≤ tα < nk such that

log dist(α, γtα) ≤ −nk −mk

30Dν−2 ≤ − 1

15Dν−2.

Then we find

log dist(τ−mk(α),STD∗ ) ≤ log dist(τ−mk(α),STDk ) since Dk ≤ D∗

≤ log dist(τ−mk(α), γtα−mk) by (1.39)

≤ log dist(α, γtα) + c3mk by Lemma 1.7.4

≤ − 1

15Dν−2 + c3D

σ

� −Dν−2 since ν > σ + 2.


Thus, as D goes to infinity, the distance between τ−mk(α) and STD∗ tends to

zero. However, it is not equal to zero since STD∗ ∩ P2(Q) = ∅. So the points

τ−mk(α) make an infinite sequence in ∪D∈N τ−mk(ZD). This contradicts the finite-

ness of ∪D∈N τ−mk(ZD). Now the claim is verified.

Step 5. The conclusion

Put D′ = Dk+1. Since nk+1 −mk+1 � nk −mk, we have

D′ =⌈(nk+1 −mk+1)

1/σ⌉�⌈(nk −mk)

1/σ⌉� Dk,

and so D′ and TD′ � TDk go to infinity with D.

Put

Z ′D = τ−mk+1(ZD).

Note that, since ZD is a Q-subvariety of P2(C) of dimension 0, so is Z ′D. Set

WD = {α ∈ Z0D; mk+1 ≤ tα < nk+1}.

Since WD ⊂ Z0D ⊂ ZD, we have

|WD| ≤ |ZD| = deg(ZD), τ−mk+1(WD) ⊂ Z ′D.

For any α ∈ WD, we set

α′ := τ−mk+1(α) ∈ Z ′D,

`α := tα −mk+1,

then we have

(α′, γ`α) = (τ−mk+1(α), τ−mk+1(γtα)), (1.41)

and

0 ≤ `α < TD′

since tα−mk+1 ≤ nk+1−mk+1 ≤⌈(nk+1 −mk+1)

1/σ⌉σ

. Note that, there is no reason

to conclude that `α = tα′ .

Consider

S =∑α∈WD

log dist(α′, γ`α).


We will find an upper bound and a lower bound for S in terms of h(Z ′D) and deg(ZD).

This will lead to the desired contradiction.

By (1.41), Lemma 1.7.4 gives

S ≤ Θ(mk+1, nk+1) + c3mk+1|WD|

≤ Θ(mk+1, nk+1) + c3Dσ deg(ZD)

since mk+1 ≤ TD ≤ Dσ and |WD| ≤ deg(ZD). Using (1.38), we find that

S ≤ −nk+1 −mk+1

30(Dν−2 deg(ZD) +Dν−β−1h(ZD)) + c3D

σ deg(ZD).

Since Dν−2 > Dσ and since nk+1 −mk+1 � D′σ goes to infinity with D, we deduce

that

S � −D′σ(Dν−β−1h(ZD) +Dν−2 deg(ZD)).

By Proposition 1.8.7, there exists a constant c′′ > 0 such that

h(ZD) ≥ h(Z ′D)− c′′mk+1 deg(ZD) ≥ h(Z ′D)− c′′Dσ deg(ZD)

since mk+1 ≤ TD ≤ Dσ. We conclude that

S � −D′σDν−β−1h(Z ′D)−D′σ(Dν−2 − c′′Dν−β−1+σ) deg(ZD)

≤∗ −D′σDν−β−1h(Z ′D)− 1

2D′σDν−2 deg(ZD)) (1.42)

where the last inequality follows from the fact that Dν−2 ≥∗ 2c′′Dν−β−1+σ since

β > σ + 1. This gives an upper bound for S in terms of h(Z ′D) and deg(ZD). Now

we search for a lower bound.

By Step 4, Dk goes to infinity with D, so we have nk −mk ≥∗ 2. By the choice of

k, we deduce that

Z ′D 6⊂ Z(ΦiPD′ ; 0 ≤ i < 2TD′).

So there exists an integer i0 with 0 ≤ i0 < 2TD′ such that Φi0PD′ does not vanish

on Z ′D.


For any α ∈ WD, Proposition 1.7.3 gives

|Φi0PD′(α′)| ≤ ‖γ`α‖

−D′ |Φi0(PD′)(γ`α)|+D′L(Φi0(PD′)) dist(α′, γ`α)

≤ |Φi0(PD′)(γ`α)|+D′2D′‖Φi0(PD′)‖ dist(α′, γ`α)

where α′ is a representative of α′ in C3 of norm 1 and where the last inequality uses

‖γ`α‖ ≥ 1.

By Step 1, we get

‖Φi0PD′‖ ≤∗ e2D′β,

and

|Φi0(PD′)(γ`α)| ≤∗ e−(1/2)D′ν

since 0 ≤ `α ≤ TD′ . This implies that

|Φi0PD′(α′)| ≤∗ e−(1/2)D′ν +

1

2e3D

′βdist(α′, γ`α) (1.43)

for any α ∈ WD.

By Step 1 and the fact that 0 ≤ i0 < 2TD′ , we also have

c′2D′TD′Φi0(PD′) ∈ (CD′ ∩ Z[X])\I(Z ′D).

Applying Lemma 1.2.9 to this polynomial, we obtain

0 ≤ 7 log(3)D′ deg(Z ′D) +D′h(Z ′D) +∑α∈Z′D

log∣∣∣c′2D′TD′Φi0(PD′)(α)

∣∣∣Since |Z ′D| = deg(Z ′D) = deg(ZD), this implies that∑

α∈Z′D

log |Φi0PD′(α)|

≥ −7 log(3)D′ deg(ZD)−D′h(Z ′D)− 2 log(c′)D′TD′ deg(ZD)

≥∗ −D′h(Z ′D)− 3 log(c′)D′1+σ deg(ZD),

using 7 log(3) ≤∗ TD′ ≤ D′σ. Since log ‖Φi0PD′‖ ≤ 2D′β, we find

log |Φi0PD′(α)| ≤ D′ log(3) + log ‖Φi0PD′‖ ≤ 4D′β


for any α ∈ C3 of norm 1. Hence, for any non-empty subset W ⊂ Z ′D, we have∑α∈W

log |Φi0PD′(α)| ≥∑α∈Z′D

log |Φi0PD′(α)| − 4D′β deg(ZD)

≥ −D′h(Z ′D)− 3 log(c′)D′1+σ deg(ZD)− 4D′β deg(ZD)

≥∗ −D′h(Z ′D)− 5D′β

deg(ZD) (1.44)

since D′β ≥∗ 3 log(c′)D′1+σ.

By the choice of k, we have

τ−mk(ZD) ⊂ Z(ΦiPDk ; 0 ≤ i < 2TDk)

⊂ Z(ΦiPDk ; 0 ≤ i < TDk +Dk)

Recall that, by construction, we have X0 - PDk , X2 - PDk . Moreover, the estimates of

Step 1 give

max{

25 log(3)Dk, log ‖PDk‖, log ‖c′D2kPDk‖, . . . , log ‖c′D2

kΦDkPDk‖}< YDk = 2Dβ

k .

Applying Proposition 1.8.8 to τ−mk(ZD), we get

degZD = deg(τ−mk(ZD)) ≤ D2k

TDk� D2−σ

k � D′2−σ (1.45)

and

h(τ−mk(ZD))� DkYDkTDk

+D2k � Dβ−σ+1

k � D′β−σ+1

since β > σ + 1. By Proposition 1.8.7, this implies that

h(Z ′D) = h(τmk−mk+1(τ−mk(ZD)))

� h(τ−mk(ZD)) + (mk+1 −mk) deg(ZD)

� D′β−σ+1 + TD′D′2−σ

� D′β−σ+1

(1.46)

since 0 ≤ mk+1 −mk ≤ nk −mk ≤ TDk � TD′ ≤ D′σ and β > σ + 1.

Combining (1.44), (1.45) and (1.46), we obtain∑α∈W

log |Φi0PD′(α)| � −D′β−σ+2. (1.47)


For each α ∈ WD, applying the above inequality to W = {α′} where α′ is a

representative of α′ = τ−mk+1(α) in C3 of norm 1, we get

|Φi0PD′(α′)| ≥ 2e−(1/2)D

′ν

when D′ is sufficiently large, i.e., when D is sufficiently large, since ν > β − σ + 2.

By (1.43), we conclude that

2e−12D′ν ≤ |Φi0PD′(α

′)| ≤ e−12D′ν +

1

2e3D

′βdist(α′, γ`α) when D � 1.

So, for such points, when D is large, we have

e−12D′ν ≤ 1

2e3D

′βdist(α′, γ`α),

and thus

|Φi0PD′(α′)| ≤ e3D

′βdist(α′, γ`α).

This means that

log dist(α′, γ`α) ≥ log |Φi0PD′(α′)| − 3D′β

for any α ∈ WD. This yields that

S =∑α∈WD

log dist(α′, γ`α) ≥∑α∈WD

log |Φi0PD′(α′)| − 3D′β|WD|

≥∑α∈WD

log |Φi0PD′(α′)| − 3D′β deg(ZD).

Applying (1.44) to W = τ−mk+1(WD) ⊂ Z ′D, we obtain∑α∈WD

log |Φi0PD′(α′)| ≥ −D′h(Z ′D)− 5D′β deg(ZD).

So we conclude that

S ≥ −D′h(Z ′D)− 8D′β deg(ZD).

We just found a lower bound for S in the term of h(Z ′D) and deg(ZD). Combining

it with the upper bound given by (1.42), we get

D′h(Z ′D) + 8D′β deg(ZD) ≥ λ

(D′

σDν−β−1h(Z ′D) +

1

2D′

σDν−2 deg(ZD)

)


for some constant λ > 0 which is independent of the choice of D. This implies that

(D′ − λD′σDν−β−1)h(Z ′D) ≥(λ

2D′

σDν−2 − 8D′

β

)deg(ZD)

≥∗ λ4D′

σDν−2 deg(ZD) > 0

where the last estimate follows from the fact that

λ

4D′

σDν−2 ≥ λ

4D′

σ+ν−2 ≥∗ 8D′β

since ν > max{β + (2− σ), 2}. Note that

0 ≤ h(Z ′D)� D′β−σ+1, and deg(ZD) ≥ 1.

We conclude that

D′ >∗ λD′σDν−β−1

and

D′σDν−2 � D′h(Z ′D)� D′β−σ+2.

This implies that

D′σ−1 � Dβ+1−ν and Dν−2 � D′β−2σ+2. (1.48)

Since σ ≥ 1, this implies that

D(ν−2)(σ−1) � D′(β−2σ+2)(σ−1) � D(β+1−ν)(β−2σ+2),

and thus

(ν − 2)(σ − 1) ≤ (β + 1− ν)(β − 2σ + 2).

This means that

ν ≤ β + 2− σ +(σ − 1)(2− σ)

β − (σ − 1).

which contradicts the hypothesis on ν. We conclude that ξ, η ∈ Q.

Chapter 2

On approximation by rational

points

2.1 Introduction

2.1.1 Statement of the results

Let θ = (1, θ1, . . . , θd) ∈ Rd+1. We say that a real number λ ≥ 0 is a uniform

exponent of approximation to θ if there exists a constant c = c(θ) > 0 such that

|x0| ≤ X, max1≤i≤d

|x0θi − xi| ≤ cX−λ (2.1)

admits a non-zero solution (x0, x1, . . . , xd) ∈ Zd+1 for each X ≥ 1. We denote by

λ(θ) the supremum of all these exponents.

Note that for θ ∈ Cd+1 \ Rd+1, this definition would give λ(θ) = 0. Indeed,

WLOG, suppose that Im(θ1) 6= 0. Then |x0θ1 − x1| > |Im(θ1)| > 0 for any integer

x0 6= 0. Hence, if (x0, x1, . . . , xd) ∈ Zd+1 is a solution of (2.1) with X large enough

and some fixed λ > 0 then the inequality |x0θ1 − x1| ≤ cX−λ implies that x0 = 0,

and so we have xi = 0 for all i ≤ d. This is impossible.

The following lemma gathers several properties of the exponent λ.

60

CHAPTER 2. ON APPROXIMATION BY RATIONAL POINTS 61

Lemma 2.1.1. Let θ = (1, θ1, . . . , θd) ∈ Rd+1.

(i) We have λ(1, θ1, . . . , θm) ≥ λ(θ) if m ≤ d.

(ii) Let {1, e1, . . . , et} be a basis of the vector space 〈1, θ1, . . . , θd〉Q. Then λ is a

uniform exponent of approximation to θ if and only if λ is a uniform exponent

of approximation to (1, e1, . . . , et). In particular, we have

λ(1, e1, . . . , et) = λ(θ).

Proof. The assertion (i) is clear from the definition.

Assume that λ is a uniform exponent of approximation to θ′ = (1, e1, . . . , et), i.e.

that there exists a constant c > 0 such that the inequalities

|x0| ≤ X, max1≤i≤t

|x0ei − xi| ≤ cX−λ (2.2)

admit a solution in Zt+1 \ {0} for any sufficiently large value of X.

We will show that λ is also a uniform exponent of approximation to θ.

Let M ∈ Mat(Q) be the (t+ 1)× (d+ 1) matrix with coefficients in Q such that

θ = θ′M . The first column of M is t(1, 0 . . . , 0). Let m ∈ N such that mM ∈ Mat(Z).

Suppose that x ∈ Zt+1 is a solution of (2.2) for someX > 1. Then the point y = mxM

belongs to Zd+1 and upon writing y = (y0, . . . , yd), we have

|y0| = |mx0| ≤ mX,

max1≤i≤d

|y0θi − yi| = ‖mx0θ′M −mxM‖ � ‖x0θ′ − x‖ � X−λ

with implied constants depending on M . Thus λ is also a uniform exponent of

approximation to θ.

Conversely, assume that λ is a uniform exponent of approximation to θ. WLOG,

we may assume that {1, θ1, . . . , θt} is a basis of 〈1, θ1, . . . , θd〉Q = 〈1, e1, . . . , et〉Q. It

follows from the definition that λ is also a uniform exponent of approximation to

(1, θ1, . . . , θt) and so, by the above, it is also a uniform exponent of approximation to

(1, e1, . . . , et) since {1, θ1, . . . , θt} is a basis of 〈1, e1, . . . , et〉Q.

Throughout this chapter, we restrict to points θ = (1, θ1, . . . , θd) ∈ Rd+1 with

Q-linearly independent coordinates.


Lemma 2.1.2. Let θ = (1, θ1, . . . , θd) ∈ Rd+1 with Q-linearly independent coordi-

nates. Then

(i) λ(θ) ≥ 1/d;

(ii) λ(θ) = 1/d if θ1, . . . , θd are algebraic over Q.

Proof.

(i) If λ = 1/d and c = 1, then the volume of the convex body defined by (2.1) is

2d+1 for any X > 0. From Minkowski’s First Convex Body Theorem, it follows that

for any X > 0, this convex body contains a non-zero point of Zd+1. This shows that

1/d is a uniform exponent of approximation to θ and so λ(θ) ≥ 1/d.

(ii) Suppose that λ(θ) 6= 1/d. From (i), we have λ(θ) > 1/d, i.e., there exist ε > 0

and c > 0 such that the inequalities

|x0| ≤ X, max1≤i≤d

|x0θi − xi| ≤ cX−1/d−ε

have a non-zero solution x = (x0, . . . , xd) in Zd+1 for each X > 1. This in turn implies

the existence of ε′ > 0 such that

max1≤i≤d

|x0θi − xi| < ‖x‖−1/d−ε′

admits a non-zero solution x ∈ Zd+1 for each X � 1. Such solution x satisfies

|M0(x)M1(x) · · ·Md(x)| < ‖x‖−dε′

where M0(x) = x0,M1(x) = x0θ1−x1, . . . ,Md(x) = x0θ1−xd are Q-linearly indepen-

dent linear forms with algebraic coefficients. By Schmidt’s Subspace Theorem, these

points lie in a finite number of proper subspaces of Rd+1 defined over Q. However,

these points converge to θ projectively as X goes to infinity. So θ must belong to

one of these proper subspaces. This is impossible since θ has linearly independent

coordinates.


The following are results that apply to points of the form θ = (1, θ, . . . , θd) where

θ is either transcendental or algebraic of degree > d.

It is well-known that λ(1, θ) = 1.

In 1969, H. Davenport and W. M. Schmidt proved in [6] that

λ(1, θ, . . . , θd) ≤

1/γ ' 0.618 if d = 2,

1/2 if d = 3,

bd/2c−1 if d ≥ 4,

where γ =1 +√

5

2denotes the golden ratio.

In the case d = 2, it is shown in [18, 2004] that the above upper bound is best

possible. More precisely, there exist real non-quadratic irrational numbers θ such that

1/γ is a uniform exponent of approximation to (1, θ, θ2). Any such θ is transcendental

over Q, and the set of these numbers is countable.

Nevertheless, no optimal upper bound for λ(1, θ, . . . , θd) is known for d ≥ 3 when

θ is transcendental. Note that, when θ is algebraic of degree > d, then it follows from

Lemma 2.1.2 that λ(1, θ, . . . , θd) = 1/d.

In [13, 2003], M. Laurent proved that λ(1, θ, . . . , θd) ≤ dd/2e−1 if d ≥ 3 which

improves the result of H. Davenport and W. M. Schmidt for odd d ≥ 5.

In [19, 2008], D. Roy sharpened the estimate in case d = 3, by showing that

λ(1, θ, θ2, θ3) ≤ 1

2

(2 +√

5−√

7 + 2√

5

)' 0.4245.

In this chapter, we will consider points of the form θ = (1, θ1, . . . , θd−1, ξ) where

{1, θ1, . . . , θd−1} is a basis of a real field extensionK ofQ of degree d ≥ 2 and ξ ∈ R\K.

These conditions ensure that the coordinates of θ are Q-linearly independent. By

noting that K = Q(α) for some algebraic number α ∈ R of degree d, we deduce from

Lemma 2.1.1 (ii) that

λ(1, θ1, . . . , θd−1, ξ) = λ(1, α, . . . , αd−1, ξ).


From the two previous lemmas , we deduce that

1

d≤ λ(θ) ≤ λ(1, α, . . . , αd−1) =

1

d− 1. (2.3)

In the case d = 2, it is shown in [22] that 1/γ is an optimal upper bound for

λ(1, α, ξ). More precisely, D. Roy proved in [22] that this value is the largest exponent

of approximation achieved by points with Q-linearly independent coordinates on any

real conic defined over Q.

The main result of this chapter applies to any integer d ≥ 2. We establish a

general upper bound for λ(θ) which reduces to 1/γ when d = 2. Moreover, this

upper bound is strictly smaller than1

d+

1

d2=

1

d− 1− 1

d2(d− 1), which is a notable

improvement on (2.3).

Theorem 2.1.3. Let K be a real number field of degree d ≥ 2 with basis {1, θ1, . . . , θd−1}over Q and let ξ ∈ R \K. Let c and λ be positive real numbers. Suppose that for any

sufficiently large value of X, the inequalities

|x0| ≤ X

|x0θ1 − x1| ≤ cX−λ

· · ·|x0θd−1 − xd−1| ≤ cX−λ

|x0ξ − xd| ≤ cX−λ

(2.4)

admit a non-zero solution x = (x0, x1, . . . , xd) ∈ Zd+1. Then λ ≤ λd where λd is the

unique positive real root of the equation

x+ (d− 1)x2 + · · ·+ (d− 1)d−1xd = 1. (2.5)

The following corollary provides an estimate for λd. However, we do not know if

our upper bound λd is optimal for d ≥ 3.

Corollary 2.1.4. Under the notation of Theorem 2.1.3, we have

λ(1, θ1, . . . , θd−1, ξ) ≤

1/γ if d = 2,

λ3 ' 0.40527 if d = 3,

λd <1d−1 −

1d2(d−1) if d ≥ 2.


Applying the main result of Y. Bugeaud and M. Laurent in [3] to our main result,

we obtain the following consequence.

Corollary 2.1.5. Let the notation be as in Theorem 2.1.3. Assume that λ > λd.

Then, for any η ∈ R, there are arbitrarily large values of X such that the inequalties|x0 + x1θ1 + · · ·+ xd−1θd−1 + xdξ + η| ≤ X−1/λ

‖x‖ ≤ X(2.6)

have a solution x = (x0, . . . , xd) in Zd+1.

The proof of our main theorem follows the approach of Davenport and Schmidt

in [6]. Its details occupy four sections. In the notation of the theorem, let θ =

(1, θ1, . . . , θd−1, ξ). In Section 2.2, we construct a canonical sequence of primitive

integer points (xi)i∈N converging to θ projectively. We call it a sequence of minimal

points for θ and establish some basic properties of this sequence, similarly as it is done

in [6]. Section 2.3 however is novel and provides the key to the proof of our theorem.

In this section, we look at subspaces of Rd+1 spanned by consecutive minimal points

and study how their dimension varies with the length of the sequence, i.e. as a

function of the first and last minimal point in the spanning set. In this way, based on

the choice of a fixed “initial” minimal point xi0 , we construct, for each k = 1, . . . , d,

families of subspaces (Ukt )k−1≤t≤d−1 of dimension k and (V k+1

t )k−1≤t≤d−1 of dimension

k + 1. Based on properties of these spaces under sum and intersection, we obtain

some strong inequality linking their heights. This is done by descending induction

on k, using a theorem of Schmidt. Then, our key inequality is obtained by taking

k = 1. The latter relates the heights of some explicit set of minimal points associated

with the choice of xi0 . Section 2.4 provides the last tool that we need, a very general

upper bound for the norm of any minimal point in terms of the norm of the next

minimal point. It is obtained through geometry of numbers, by observing that the

first d coordinates of any minimal point provide a good rational approximation to

the point (1, θ1, . . . , θd−1). The proof of Theorem 2.1.3 is presented in Section 2.5,

based on all these estimates. Thus, we have λ(θ) ≤ λd. In the same section, we also

consider the hypothetical situation where λ(θ) = λd is precisely the uniform exponent


of approximation to θ. In that case, we show that there exists a subsequence (yi)i∈N

of the sequence of minimal points which satisfies very rigid growth estimates and

which “explains” the fact that λ(θ) = λd. Moreover, any d+ 1 consecutive points in

this sequence are linearly independent and their determinant is bounded from above.

This situation is very similar to that of the extremal numbers in [18], except that

here we don’t know if extremal points θ exist. The precise construction is delicate

and obtained by “pasting” together the finite sequences of minimal points defined in

Section 2.3.

Section 2.6 provides an alternative proof of the main estimate of Section 4. In

this section, we construct a multi-linear symmetric polynomial map Φ from (Rd+1)d

to R defined over Q, whose restriction ϕ(x) = Φ(x, . . . ,x) to the diagonal is closely

connected to the norm map from K to Q. Looking at the values of ϕ at minimal

points and showing that they are non-zero, we obtain an alternative proof for the

main estimate of Section 2.4. In Section 2.7, we use Φ to construct a polynomial map

Ψ : (Rd+1)d → Rd+1. Then, we study its properties and use them to provide algebraic

relations between the points of the sequence (yi)i∈N constructed in Section 2.5 in the

case where λ(θ) = λd. We view Ψ as an analog of the bracket operator of [18, §2].

We were not able to go further on these lines of investigation but hope that the above

mentioned result will be useful for further study in this topic.

The last section of this chapter derives from several unsuccesful trials to construct

points θ of the above form, for which the uniform exponent of approximation λ(θ)

is greater than the trivial lower bound 1/d provided by the box principle. In this

section, we fix d = 3, set α = 3√

2, and construct a transcendental number ξ ∈ Rsuch that λ(1, α, α2, ξ) ≥ 1/3. However, the sequence of minimal points attached

to (1, α, α2, ξ) have their norms growing in the fastest possible way, something that

cannot be achieved by an application of the box principle.

2.1.2 Proofs of the corollaries

Proof of Corollary 2.1.4. By direct computation, we get λ2 = 1/γ and λ3 ' 0.40527.

We will verify that λd <d+1d2

for d ≥ 2. Set y = (d − 1)x. The equation (2.5) is


equivalent to

y + y2 + · · ·+ yd = d− 1.

Set f(y) = 1 + y + y2 + · · ·+ yd − d. Then f(y) is an increasing function on R+. We

need to prove that the unique positive real zero of f is less than y0 = (d2 − 1)/d2. It

is enough to show that f(y0) > 0.

We have

f(y0) =1− yd+1

0

1− y0− d = d2 − d− d2

(d2 − 1

d2

)d+1

The inequality f(y0) > 0 is equivalent to

d2 − d > d2(d2 − 1

d2

)d+1

,

which is equivalent to (d

d− 1

)d>

(d+ 1

d

)d+1

.

This is true since the function

(d

d− 1

)dis decreasing for d > 1.

The proof of Corollary 2.1.5 is based on the main result of Y. Bugeaud and M.

Laurent in [3]. In order to state this result, we first introduce the following notation.

For any positive integer n, and for any point x ∈ Rn, we denote by

{x} = miny∈Zn‖x− y‖

the distance from x to a closest integer point.

Let A be a real n × m matrix. For any columm vector η in Rn, we denote by

ω(A,η) the supremum of the real numbers ω for which, for arbitrarily large real

numbers X, the inequalities

{A · x + η} ≤ X−ω, ‖x‖ ≤ X (2.7)

have a non-zero solution x = (x0, . . . , xd) in Zm. We denote by ω(A,η) the supre-

mum of the real numbers ω for which, for all sufficiently large real numbers X, the

inequalities (2.7) have a solution x in Zm. We define furthermore two homogeneous

exponents ω(A) and ω(A) by setting ω(A) := ω(A, 0) and ω(A) := ω(A, 0).

The main result in [3] is reads as follows.


Theorem 2.1.6 (Y. Bugeaud, M. Laurent, 2005). For any real n×m matrix A, and

any column vector η ∈ Rn, we have the lower bounds

ω(A,η) ≥ 1

ω(At)and ω(A,η) ≥ 1

ω(At)(2.8)

with equality in (2.8) for almost all η with respect to the Lebesgue measure on Rn.

Based on this, we can now prove Corollary 2.1.5.

Proof of Corollary 2.1.5. Set θ = (1, θ1, . . . , θd−1, ξ), η = (η, 0, . . . , 0) ∈ Rd+1 and set

Aθ =

0 θ1 . . . θd−1 ξ

0 −1 . . . 0 0...

.... . .

......

0 0 . . . −1 0

0 0 . . . 0 −1

.

Then, for any x ∈ Zd+1, we have

Atθxt = (0, x0θ1 − x1, . . . , x0θd−1 − xd−1, x0ξ − xd)t

and

{Aθ · xt + ηt} = {x1θ1 + · · ·+ xd−1θd−1 + xdξ + η}.

From definition, we conclude that λ(θ) = ω(Atθ). By (2.8), we get

ω(Aθ,η) ≥ 1

ω(Atθ)=

1

λ(θ)≥ 1

λd.

Since λ > λd, this implies that ω(Aθ,η) > 1/λ. This means that there are arbitrarily

large real numbers X such that the system

{x1θ1 + · · ·+ xd−1θd−1 + xdξ + η} ≤ X−1/λ, ‖x‖ ≤ X/2

has a solution in x = (x0, . . . , xd) ∈ Zd+1. If X is large enough then the system (2.6)

also has a solution in Zd+1.


2.1.3 Notation

.

In this chapter, for any point x in Rd+1, we denote by x− the point of Rd whose

coordinates are the first d coordinates of x.

For any point θ = (θ0, θ1, . . . , θd) in Rd+1, we define the function Lθ : Rd+1 → Rby

Lθ(x) = max1≤i≤d

|x0θi − θ0xi|

for each x = (x0, . . . , xd) ∈ Rd+1.

2.2 Construction of minimal points

In this section, we fix a point θ = (1, θ1, . . . , θd) ∈ Rd+1 with Q-linearly indepen-

dent coordinates.

To study the problem of uniform aproximation to θ, we follow the approach of

Davenport and Schmidt in [6] by first defining a certain sequence of primitive points

(xi)i∈N in Zd+1, called minimal points, which converges projectively to θ.

First of all, for each real X > 1, we consider the set of integer points x =

(x0, . . . , xd) with

1 ≤ x0 ≤ ‖x‖ ≤ X, |x0θ1 − x1| ≤1

2, . . . , |x0θd − xd| ≤

1

2.

Since 1, θ1, . . . , θd are linearly independent over Q, there is a unique point among

them for which Lθ(x) has its least value, and we call this point the minimal point

corresponding to X.

It is clear that the minimal points are primitive. Moreover, if x is the minimal

point corresponding both to X ′ and to X ′′, it is also the minimal point corresponding

to any X between X ′ and X ′′. Hence there is a sequence of integers X1 < X2 < . . .

such that the same minimal point xi corresponds to all X in the range Xi ≤ X < Xi+1

but to no X outside this range.

So for fixed θ, the sequences of minimal points (xi)i∈N are uniquely determined

up to the choice of their first points.


Now fix such a sequence (xi)i∈N. Then (xi)i∈N has the following properties

(a) xi is primitive for each i,

(b) the norms Xi = ‖xi‖ form a strictly increasing sequence,

(c) the positive real numbers Li = Lθ(xi) form a strictly decreasing sequence,

(d) if a non-zero point x ∈ Zd+1 satisfies Lθ(x) < Li for some i ≥ 1 then

‖x‖ ≥ Xi+1.

The following lemma shows that one can compute λ(θ) directly from a sequence

of minimal points.

Lemma 2.2.1. A positive real number λ is a uniform exponent of approximation to

θ if and only if Li � X−λi+1 for each index i.

Proof. Assume that λ is a uniform exponent of approximation to θ, i.e. that there

exists a constant c = c(θ) > 0 such that

|x0| ≤ X, Lθ(x) ≤ cX−λ (2.9)

admits a non-zero solution x = (x0, x1, . . . , xd) ∈ Zd+1 for each X ≥ 1. Then there

exists a constant c′ = c′(θ) > 0 such that any such solution x satisfies

|x0| ≤ ‖x‖ ≤ c′|x0|.

Now fix a sufficiently large index i so that c′−1Xi+1 > 1. For any real number X with

1 < X < c′−1Xi+1, a solution x ∈ Zd+1 to (2.9) satisfies

‖x‖ ≤ c′|x0| < Xi+1, Lθ(x) ≤ cX−λ.

From the property (d), we deduce that

Li ≤ Lθ(x) ≤ cX−λ.

Since we can choose X arbitrarily close to c′−1Xi+1, we conclude that

Li ≤ c(c′−1Xi+1)−λ.

The reverse implication is easy to prove.


We also note that, any two consecutive minimal points xi,xi+1 are Q-linearly

independent since they are primitive and distinct with positive first coordinates. We

have the following estimate for the height of the vector spaces generated by two such

points.

Lemma 2.2.2. Let Vi be the R-vector space generated by xi and xi+1. Then, we have

Vi ∩ Zd+1 = 〈xi,xi+1〉Z and H(Vi) � Xi+1Li.

Proof. Assume that Vi∩Zd+1 6= 〈xi,xi+1〉Z. Then there exist r, r′ ∈ R with |r|, |r′| ≤12

such that

z = rxi + r′xi+1 ∈ Zd+1 \ {0}.

So we get

‖z‖ ≤ |r|‖xi‖+ |r′|‖xi+1‖ < Xi+1,

Lθ(z) ≤ |r|Li + |r′|Li+1 < Li.

This is a contradiction.

So Vi ∩ Zd+1 = 〈xi,xi+1〉Z and thus

H(Vi) = ‖xi ∧ xi+1‖ � Xi+1Li.

Indeed, it is obvious that ‖xi ∧ xi+1‖ � Xi+1Li. We will clarify that

‖xi ∧ xi+1‖ � Xi+1Li. (2.10)

Set u = xi−xi,0θ and v = xi+1−xi+1,0θ. Write u = (u0, . . . , ud) and v = (v0, . . . , vd).

Then Li = |ut| for some positive integer 1 ≤ t ≤ d. Moreover, there exists 0 ≤ l ≤ d

such that |ul− vl| = ‖u−v‖ = Lθ(xi+1−xi) ≥ Li since ‖xi+1−xi‖ < Xi+1. We have

‖xi ∧ xi+1‖ ≥ |xi,0vt − xi+1,0ut| ≥ xi+1,0Li − xi,0Li+1 ≥ (xi+1,0 − xi,0)Li.

In the other hand, we have

‖xi ∧ xi+1‖ ≥ |xi,0vl − xi+1,0ul| = |xi,0(vl − ul) + ul(xi,0 − xi+1,0)|

≥ xi,0Li − Li(xi+1,0 − xi,0).

This yields

3‖xi ∧ xi+1‖ ≥ xi,0Li + (xi+1,0 − xi,0)Li = xi+1,0Li.

so we get (2.10).


In the proof of Theorem 2.1.3, we will need another property of the above sequence

of vector spaces (Vi)i∈N. We state it below in a very general form.

Lemma 2.2.3. Let (Vi)i∈N be a sequence of subspaces of Rd+1 of dimension t ≤ d

defined over Q. Suppose that the union of these spaces contains a sequence of non-

zero points (yi)i∈N in Rd+1 converging to θ = (1, θ1, . . . , θd) projectively. Then the

sequence (Vi)i∈N contains infinitely many distinct vector spaces.

Proof. Assume by contradiction that there exist only finitely many distinct vector

spaces among the sequence (Vi)i∈N. Then there exists one of them Vi0 containing an

infinite subsequence (zn)n∈N of (yi)i∈N. Since (yi)i∈N converges to θ projectively, so

does any of its subsequences. We deduce that θ ∈ Vi0 .Since the vector space Vi0 is defined over Q and dimVi0 ≤ d, there exists a non-

zero vector u = (u0, . . . , ud) ∈ Qd+1 orthogonal to Vi0 . So it is also orthogonal to θ.

Then we have

u0 + u1θ1 + · · ·+ udθd = 0.

This contradicts the hypothesis that 1, θ1, . . . , θd are linearly independent over Q.

For the last result of this section, we drop the condition that θ = (1, θ1, . . . , θd)

has Q−linearly independent coordinates. In fact, this result provides a criterion for

the coordinates of θ to be linearly independent over Q.

Lemma 2.2.4. Let θ = (1, θ1, . . . , θd) be an arbitrary point in Rd+1. Assume that

there exists a sequence of points yn = (yn,0, . . . , yn,d) in Zd+1 with n ∈ N∗ such that

(i) det(yn,yn+1, . . . ,yn+d) 6= 0 for infinitely many integers n,

(ii) limn→∞ Lθ(yn) = 0.

Then 1, θ1, . . . , θd are Q–linearly independent.

Proof. Assume on the contrary that 1, θ1, . . . , θd are Q–linearly dependent. WLOG,

we may write θd = a0 + a1θ1 + . . .+ ad−1θd−1 for some a0, a1, . . . , ad−1 in Q.


For each n ∈ N, we have

xn,0θd − xn,d = xn,0(a0 + a1θ1 + · · ·+ ad−1θd−1)− xn,d= a1(xn,0θ1 − xn,1) + · · ·+ ad−1(xn,0θd−1 − xn,d−1) +R(n)

where

R(n) = a0xn,0 + a1xn,1 + · · ·+ ad−1xn,d−1 − xn,d.

Then we get

|R(n)| ≤ |xn,0θd − xn,d|+ |a1(xn,0θ1 − xn,1)|+ · · ·+ |ad−1(xn,0θd−1 − xn,d−1)|

� Lθ(xn).

Since Lθ(xn) converges to 0 when n tends to infinity, so does R(n). Since R(n) ∈Z[a0, a1, . . . , ad−1] with ai ∈ Q for all i = 1, . . . , d − 1, we deduce that R(n) = 0,

namely

xn,d = a0xn,0 + a1xn,1 + · · ·+ ad−1xn,d−1,

when n is sufficiently large. This implies that det(xn,xn+1,xn+2,xn+3) = 0 when n is

sufficiently large. This is a contradiction. So 1, θ1, . . . , θd are Q–linearly independent.

2.3 Construction of sequences of vector spaces

In this section, we fix a point of θ = (1, θ1, . . . , θd) ∈ Rd+1 with Q-linearly inde-

pendent coordinates and fix a sequence of minimal points (xi)i≥1 attached to θ. Let

the notation Xi and Li be as in Section 2.2.

Let i0 be a positive integer. For each t = 1, . . . , d− 1, we denote by it the largest

integer such that

dim〈xi0 ,xi0+1, . . . ,xit〉R = t+ 1. (2.11)

The existence of it follows from Lemma 2.2.3. Clearly, the property (2.11) also holds

for t = 0. Moreover, we have

(i) i0 < i1 < . . . < id−1,


(ii) dim〈xi0 ,xi0+1, . . . ,xit+1〉R = t+ 2 for 0 ≤ t ≤ d− 1,

(iii) xit+1 /∈ 〈xi0 ,xi0+1, . . . ,xit〉R for 0 ≤ t ≤ d− 1.

By comparing dimensions, we deduce that

(iv) Rd+1 = 〈xi0 ,xi0+1, . . . ,xid−1+1〉R,

(v) 〈xi0 ,xi0+1, . . . ,xit−1+1〉R = 〈xi0 ,xi0+1, . . . ,xit〉R for 0 < t ≤ d− 1.

For each (t, k) ∈ N2 with 0 ≤ t ≤ d − 1 and 1 ≤ k ≤ t + 1, let s(t, k) be the largest

integer < it + 1 such that

dim〈xs(t,k),xs(t,k)+1, . . . ,xit+1〉R = k + 1.

Then we get

s(t, 1) = it > s(t, 2) > . . . > s(t, t+ 1) ≥ i0

Now we set

V k+1t = 〈xs(t,k),xs(t,k)+1, . . . ,xit+1〉R,

Ukt = 〈xs(t,k),xs(t,k)+1, . . . ,xit〉R.

By definition, we have

dimV k+1t = k + 1, Uk

t ⊂ V k+1t .

Moreover, for such (t, k), we have Ukt ⊂ 〈xi0 ,xi0+1, . . . ,xit〉R. It follows from (iii)

that xit+1 /∈ Ukt . We deduce that

dimUkt = k.

On the other hand, we have xit+1 ∈ V kt by the definition. Thus,

V k+1t = Uk

t + V kt (2.12)

is the sum of two distinct k-dimensional vector spaces.


For 0 < t ≤ d− 1 and 2 ≤ k ≤ t+ 1, it is clear from the definition that Uk−1t is a

(k− 1)-dimensional subspace of the two distinct k-dimensional vector spaces Ukt and

V kt . Hence, for such (t, k), we have

Uk−1t = Uk

t ∩ V kt . (2.13)

For t ≥ 1, note that U t+1t is by definition a (t + 1)-dimensional subspace of

〈xi0 ,xi0+1, . . .xit〉R. Since the latter has dimension t+1 by the choice of it, they coin-

cide. Similarly, V t+1t−1 is a (t + 1)–dimensional subspace of 〈xi0 ,xi0+1, . . . ,xit−1+1〉R =

〈xi0 ,xi0+1, . . . ,xit〉R so they coincide. Hence we have

U t+1t = 〈xi0 ,xi0+1, . . . ,xit−1+1〉R = V t+1

t−1 . (2.14)

The following lemma relates the heights of (Ukj )k≤j≤d−1 and of (V k+1

j )k−1≤j≤d−1.

Lemma 2.3.1. For each k ∈ N with 1 ≤ k ≤ d− 1, we have

H(Ukk )H(Uk

k+1) · · ·H(Ukd−1)� H(V k+1

k−1 )H(V k+1k ) · · ·H(V k+1

d−1 ) (2.15)

Proof. We proceed by descending induction on k.

By (iv), (2.12), and (2.14), we get

Rd+1 = V d+1d−1 = Ud

d−1 + V dd−1.

By (2.13), we have

Ud−1d−1 = Ud

d−1 ∩ V dd−1.

So it follows from Schmidt’s inequality [24, Chap. 1, Lemma 8A] that

H(Ud−1d−1 )� H(Ud

d−1)H(V dd−1).

By (2.14), we have H(Udd−1) = H(V d

d−2). Thus,

H(Ud−1d−1 )� H(V d

d−2)H(V dd−1).

So (2.15) holds for k = d− 1.


Assume that (2.15) is true for some k with 1 < k ≤ d − 1. For each index

t = k − 1, . . . , d− 1, by (2.12) and (2.13), we have

V k+1t = Uk

t + V kt , Uk

t ∩ V kt = Uk−1

t .

Then, it follows from Schmidt’s inequality that

H(V k+1t )� H(Uk

t )H(V kt )

H(Uk−1t )

for each t = k − 1, . . . , d− 1. Combining this with the induction hypothesis, we get

H(Ukk ) · · ·H(Uk

d−1)�H(Uk

k−1)H(V kk−1)

H(Uk−1k−1 )

· · ·H(Uk

d−1)H(V kd−1)

H(Uk−1d−1 )

.

This leads to

H(Uk−1k−1 ) · · ·H(Uk−1

d−1 )� H(Ukk−1)H(V k

k−1) · · ·H(V kd−1).

By (2.14), we have Ukk−1 = V k

k−2, we conclude that (2.15) is true with k replaced by

k − 1. By the induction principle, (2.15) is true for all k = 1, . . . , d− 1.

Proposition 2.3.2. Suppose that λ > 0 is an exponent of approximation to θ. Then

we have

Xi1 · · ·Xid−1� (Xi0+1Xi1+1 · · ·Xid−1+1)

1−λ.

Recall that Xi = ‖xi‖ for each i ≥ 1.

Proof. From the above lemma applied with k = 1, we get

H(U11 )H(U1

2 ) · · ·H(U1d−1)� H(V 2

0 )H(V 21 ) · · ·H(V 2

d−1)

where V 2t = 〈xit ,xit+1〉R and U1

t = 〈xit〉R for t = 0, . . . , d− 1.

For t = 0, . . . , d− 1, we have

H(U1t ) = ‖xit‖ = Xit .

It follows from Lemma 2.2.1 and Lemma 2.2.2 that

H(V 2t ) � ‖xit ∧ xit+1‖ � Xit+1Lit � X1−λ

it+1.

This yields the required inequality.


2.4 On the norms of minimal points

In this section, we fix a real number field K of degree d ≥ 2 with basis {1, θ1, . . . , θd−1}over Q and fix ξ ∈ R\K. Set θ = (1, θ1, . . . , θd−1, ξ). We work with the two functions

L′ : Rd −→ R and L : Rd+1 −→ R defined by

L′(y) = Lθ−(y) for each y ∈ Rd,

L(x) = Lθ(x) for each x ∈ Rd+1.

We fix a sequence of minimal points (xi)i∈N attached to θ. By construction, the

norms Xi = ‖xi‖ form a strictly increasing sequence while the values L(xi) form a

strictly decreasing sequence (see Section 2.2).

In this section, we will show that if λ is a uniform exponent of approximation to

θ, then there exists a constant c > 0 such that

Xi+1 ≤ cX1

(d−1)λ

i for all i ∈ N. (2.16)

To verify this, we first establish the following estimate.

Proposition 2.4.1. For any x ∈ Zd \ {0}, we have

L′(x)� ‖x‖−1d−1 .

Proof. Fix a real number X ≥ 1. Consider the convex body

CX :

|x0 + x1θ1 + · · ·+ xd−1θd−1| ≤ X−(d−1)

|x1|, . . . , |xd−1| ≤ X

and its polar reciprocal parallelepiped

C∗X :

|x0| ≤ Xd−1

max1≤i<d

|x0θi − xi| ≤ X−1.

Suppose first that ν CX contains a non-zero integer point x′ = (x′0, . . . , x′d−1) in Zd,

for some ν > 0. Then we have

‖x′‖ ≤ νCX with C = 1 + |θ1|+ · · ·+ |θd−1|.


Write

y = x′0 + x′1θ1 + · · ·+ x′d−1θd−1.

Since 1, θ1, . . . , θd are Q–linearly independent, we have y 6= 0. Upon choosing m ∈ N∗

such that mθi ∈ OK for i = 1, . . . , d− 1, we get my ∈ OK \ {0} and so NK/Q(my) ∈Z \ {0}. Let σ1, . . . , σd denote the d distinct embeddings of K into C, ordered so that

σ1 is the inclusion of K into R. We find

|NK/Q(my)| = md

d∏i=1

|x′0 + x′1σi(θ1) + · · ·+ x′d−1σi(θd−1)|

≤ mdνX−(d−1)d∏i=2

(C + |σi(θ1)|+ · · ·+ |σi(θd−1)|)νX

≤ (C ′mν)d

where C ′ = max2≤i≤d(C + |σi(θ1)|+ · · ·+ |σi(θd−1)|). Since NK/Q(my) ∈ Z \ {0}, we

conclude that

1 ≤ C ′mν

and so ν ≥ (C ′m)−1. This shows that µ1(CX) ≥ (C ′m)−1.

By Theorem B.4 in [4, Appendix], we have µ1(CX) ·µd(C∗X) � 1. This implies that

µd(C∗X)� 1.

On the other hand, since vol(C∗X) = 2d, it follows from Minkowski’s Second Convex

Body Theorem that1

d!≤ µ1(C∗X) · · ·µd(C∗X) ≤ 1.

Since µ1(C∗X) ≤ µ2(C∗X) ≤ · · · ≤ µd(C∗X), we deduce that

µ1(C∗X) ≥ 1

d! µd(C∗X)d−1> c

for some constant c that is independent of X, with 0 < c < 1.

Now let x = (x0, . . . , xd−1) ∈ Zd \ {0}. We have

|x0| ≤ ‖x‖ = cXd−1 with X = (c−1‖x‖)1d−1 .


Since µ1(C∗X) > c, the point x does not belong to c C∗X and so

L′(x) > cX−1 = cdd−1‖x‖−

1d−1 .

Corollary 2.4.2. Let λ ∈ R+ be a uniform exponent of approximation to θ. We have

Xi+1 � X1

(d−1)λ

i for all i ∈ N.

Proof. Fix an index i. Applying the above proposition to a minimal point xi, we get

L(xi) ≥ L′(x−i )� ‖x−i ‖− 1d−1 ≥ X

− 1d−1

i .

From Lemma 2.2.1, we have L(xi)� X−λi+1. Thus we get

X− 1d−1

i � X−λi+1 ,

from which the result follows.

The previous corollary provides a constraint on the norms of (xi)i∈N which holds

for any uniform exponent λ of approximation to θ. In the special case where λ = λd

is assumed to be an exponent of approximation to θ, we have the following result.

Theorem 2.4.3. Let the notation be as in Theorem 2.1.3. Assume that λd is a uni-

form exponent of approximation to θ = (1, θ1, . . . , θd−1, ξ). Let (xi)i∈N be a sequence

of minimal points attached to θ. Then this sequence admits a subsequence (yn)n∈N∗

such that, for each n ∈ N,

(i) | det(yn,yn+1, . . . ,yn+d)| � 1,

(ii) ‖yn‖ � ‖yn+1‖(d−1)λd ,(iii) L(yn) � L′(y−n ) � ‖yn‖−1/(d−1).

This is a consequence of Theorem 2.1.3 and will be proved in the next section.

2.5 Proof of the main theorems

2.5.1 Proof of Theorem 2.1.3

Fix a sequence of minimal points (xi)i≥1 in Zd−1 attached to θ = (1, θ1, θ2, . . . , θd−1, ξ).

For each i ≥ 1, we define Xi = ‖xi‖ and Li = Lθ(xi) according to the general

convention explained in Section 2.2.


Fix an arbitrary large integer i0. For each t = 1, . . . , d − 1, we denote by it the

largest integer ≥ i0 such that

dim〈xi0 ,xi0+1, . . . ,xit〉R = t+ 1.

By Proposition 2.3.2, we get

Xi1 · · ·Xid−1� (Xi0+1Xi1+1 · · ·Xid−1+1)

1−λ.

The idea is to eliminate successively Xi0+1, Xi1 , Xi1+1, Xi2 , . . . from this equality.

Since i0 + 1 ≤ i1, we first have Xi0+1 ≤ Xi1 , hence the above inequality implies

that

Xi1 · · ·Xid−1� (Xi1Xi1+1 · · ·Xid−1+1)

1−λ (2.17)

which is equivalent to

Xλi1Xi2 · · ·Xid−1

� (Xi1+1 · · ·Xid−1+1)1−λ.

By Corollary 2.4.2, we have Xi1 � X(d−1)λi1+1 . Using this to eliminate Xi1 , we get

X(d−1)λ2i1+1 Xi2 · · ·Xid−1

� (Xi1+1 · · ·Xid−1+1)1−λ. (2.18)

Assume that

X(d−1)λ2+···+(d−1)t−1λt

it−1+1 Xit · · ·Xid−1� (Xit−1+1 · · ·Xid−1+1)

1−λ (2.19)

for some t with 1 < t < d. We just proved (2.19) for t = 2. We will prove that (2.19)

still holds when we replace t by t+ 1.

The inequality (2.19) is equivalent to

X−1+λ+(d−1)λ2+···+(d−1)t−1λt

it−1+1 Xit · · ·Xid−1� (Xit+1 · · ·Xid−1+1)

1−λ. (2.20)

Upon noting that λ ≤ λ(θ−) =1

d− 1(by Lemmas 2.1.1 and 2.1.2), we get

−1 + λ+ (d− 1)λ2 + · · ·+ (d− 1)t−1λt ≤ −1 +t

d− 1≤ 0. (2.21)


So we can use the inequality Xit−1+1 ≤ Xit to eliminate Xit−1+1 from (2.20). This

gives

Xλ+(d−1)λ2+···+(d−1)t−1λt

itXit+1 · · ·Xid−1

� (Xit+1 · · ·Xid−1+1)1−λ

Since Xit � X(d−1)λit+1 , we obtain

X(d−1)λ2+···+(d−1)tλt+1

it+1 Xit+1 · · ·Xid−1� (Xit+1 · · ·Xid−1+1)

1−λ

as required. By the induction principle, we deduce that the inequality (2.19) holds

for all t = 2, . . . , d. Applying (2.19) with t = d, we get

X(d−1)λ2+···+(d−1)d−1λd

id−1+1 � X1−λid−1+1 (2.22)

This implies that λ satisfies

λ+ (d− 1)λ2 + · · ·+ (d− 1)d−1λd ≤ 1.

Since f(λ) = λ + (d− 1)λ2 + · · · + (d− 1)d−1λd is an increasing function on R+, we

deduce that λ ≤ λd.

2.5.2 Proof of Theorem 2.4.3

Note that if there exists a subsequence (yn)n∈N∗ of (xi)i∈N in which any d+ 1 consec-

utive points are linearly independent and if this sequence satisfies the properties (ii),

(iii) of Theorem 2.4.3 for any n ∈ N then we have

1 ≤ | det(yn,yn+1, . . . ,yn+d)| � ‖yn+d‖L(yn) · · ·L(yn+d−1)

� ‖yn+d‖1−λd−(d−1)λ2d−···−(d−1)

d−1λdd = 1,

so the property (i) holds.

Therefore it is enough to establish the existence of a subsequence (yn)n∈N in which

any d + 1 consecutive points are linearly independent and for which the properties

(ii) and (iii) hold. We start with two general observations.


a) Consider an arbitrarily large integer i. For each t = 0, . . . , d − 1, let it be the

largest integer ≥ i such that

dim〈xi,xi+1, . . . ,xit〉R = t+ 1.

In particular, we note that i0 = i. We will show that

Xit−1+1 � Xit � X(d−1)λdit+1 , L(xit) � L(x−it) � X

− 1d−1

it(2.23)

for t = 1, . . . , d− 1.

To this end, consider the proof of Theorem 2.1.3. In our case where λ = λd, we get

an equality in (2.22). Therefore, throughout the proof, we can replace all symbols �and� by �. Otherwise, if in some inequality, one side is much larger than the other,

this would carry to all subsequent estimates, and so we could not have an equality in

(2.22). This uses the fact that for λ = λd, the first inequality in (2.21) is strict for

t = 1, . . . , d− 1. We conclude that

Xit−1+1 � Xit � X(d−1)λdit+1 for all t = 1, . . . , d− 1 (2.24)

where the implied constants do not depend on the choice of i made at the beginning

of Step 1.

Since (d − 1)λd < 1, this means that, for each t = 1, . . . , d − 1, the numbers

Xit−1+1, Xit−1+2, . . . , Xit are about the same size while Xit+1 is much larger. More

precisely, there exists a constant c > 1, independent of the choice of i, such that, if i

is large enough, then

i1 = min{k ∈ N; k > i0, Xk+1 > cXk},

i2 = min{k ∈ N; k > i1, Xk+1 > cXk},

. . .

id−1 = min{k ∈ N; k > id−2, Xk+1 > cXk}.

(2.25)

Moreover, going back to the proof of Corollary 2.4.2, the estimates (2.24) imply that

L(xit) � L′(x−it) � X− 1d−1

it


for all t = 1, . . . , d − 1, where the implied constants are again independent of the

choice of the initial integer i.

b) With the above notation, take j = i1. For each t = 0, . . . , d − 1, let jt be the

largest integer such that

dim〈xj,xj+1, . . . ,xjt〉R = t+ 1.

We will show that jt = it+1 for t = 0, . . . , d− 2 provided that i is large enough. This

is true for t = 0 since j0 = j = i1. Assume that j0 = i1, . . . , jt−1 = it for some t with

1 ≤ t < d− 1. By part a) and the induction hypothesis, we have

jt = min{k ∈ N; k > jt−1, Xk+1 > cXk}

= min{k ∈ N; k > it, Xk+1 > cXk},

and so jt = it+1. By the induction principle, we get

jt = it+1 for t = 0, . . . , d− 2.

Set id = jd−1. Then we have

id = jd−1 = min{k ∈ N; k > jd−2, Xk+1 > cXk}

= min{k ∈ N; k > id−1, Xk+1 > cXk}. (2.26)

We claim that the point xi0 is R–linearly independent from the d points xi1 ,xi2 , . . . ,xid .

To verify this, we first note that

Rd+1 = 〈xi0 ,xi0+1, . . . ,xid−1+1〉R= 〈xi0 ,xi0+1, . . . ,xi1〉R + 〈xi1 ,xi1+1, . . . ,xid−1+1〉R.

Since 〈xi0 ,xi0+1, . . . ,xi1〉R has dimension 2, and since the two points xi0 ,xi1 are prim-

itive with unequal norms, we get

〈xi0 ,xi0+1, . . . ,xi1〉R = 〈xi0 ,xi1〉R.

Moreover, since it+1 = jt for all t = 0, . . . , d− 1, we also have

〈xi1 ,xi1+1, . . . ,xid−1+1〉R = 〈xj0 ,xj0+1, . . . ,xjd−2+1〉R.


By definition of jd−2, this vector space has dimension d. By definition of jd−1, we also

have

〈xj0 ,xj0+1, . . . ,xjd−2+1〉R = 〈xj0 ,xj0+1, . . . ,xjd−1〉R

= 〈xi1 ,xi1+1, . . . ,xid〉R.

Therefore, the (d+ 1)–dimensional vector space

Rd+1 = 〈xi0 ,xi1〉R + 〈xi1 ,xi1+1, . . . ,xid〉R= 〈xi0〉R + 〈xi1 ,xi1+1, . . . ,xid〉R

is the sum of a 1-dimensional vector space and a d–dimensional vector space. We

deduce that xi0 /∈ 〈xi1 ,xi1+1, . . . ,xid〉R. A fortiori, this proves the claim.

c) Construction of (yn)n∈N: Let the constant c be as in part a), and let k0 ∈ Nbe sufficiently large so that (2.25) holds for any choice of i with i ≥ k0. Define

recursively an increasing sequence of integers (kn)n∈N by

kn+1 = min{k ∈ N; k > kn, Xk+1 > cXk},

and put

yn = xkn

for each n ∈ N.

If we apply the construction of parts a) and b) with i = kn for some n ≥ 0, then

the integers i0, i1, . . . , id become kn, kn+1, . . . , kn+d because of (2.25) and (2.26). Note

that the estimates (2.23) of part a) not only hold for t = 1, . . . , d − 1, but also for

t = d because of the construction of id in part b). Considering the cases t = 1 and

t = 2, we have

Xkn+1 � X(d−1)λkn+1+1, Xkn+1+1 � Xkn+2 ,

and

L(xkn+1) � L′(x−kn+1) � X

−1/(d−1)kn+1

.


So we have

‖yn+1‖ � ‖yn+2‖(d−1)λd , L(yn+1) � L′(y−n+1) � ‖yn+1‖−1/(d−1)

for all n ≥ 0, showing that the sequence (yn)n∈N∗ fulfills the properties (ii) and (iii)

of Theorem 2.4.3.

Finally, if we again choose i = kn in part a), then the main result of part b) states

that yn is R–linearly independent of the d successors yn+1,yn+2, . . . ,yn+d. This being

true for all n ≥ 0, it shows that any d+ 1 consecutive points of the sequence (yn)n∈N

are linearly independent. So (yn)n∈N∗ fulfills all the required properties.

2.6 The polynomials ϕ and Φ

Fix a real number field K of degree d ≥ 2. Then there exists an algebraic integer α

of degree d such that K = Q(α). Then {1, α, . . . , αd−1} is a basis of K over Q. Set

θ = (1, α, . . . , αd−1).

Using geometry of numbers, we proved in Proposition 2.4.1 that for any x ∈ Zd,we have

Lθ(x)� ‖x‖−1d−1 . (2.27)

In this section, we provide an algebraic proof of this result based on properties of

a symmetric d-linear form Φ(X1, . . . ,Xd) ∈ Z[X1, . . . ,Xd] which we will construct

below, where Xj = (Xj,0, . . . , Xj,d−1) is a d−tuple of variables for j = 1, . . . , d.

Firstly, we need an auxiliary lemma.

Lemma 2.6.1. There exists a unique choice of a0, a1, . . . , ad−1 in K with ad−1 = 1

such that every embedding τ : K ↪→ C distinct from the inclusion map satisfies

τ(a0) + τ(a1)α + · · ·+ τ(ad−1)αd−1 = 0. (2.28)

These numbers a0, a1 . . . , ad−1 are Q-linearly independent algebraic integers.

Proof. Let F be a normal closure of K over Q. Set G = Gal(F/Q) and H =

Gal(F/K). Any embedding τ of K into C such that τ 6= IdK can be extended


to an automorphism τ of F and then τ ∈ G \ H. Conversely, the restriction to K

of any element σ of G \ H yields an embedding from K into C distinct from the

inclusion. Therefore, condition (2.28) for all τ 6= IdK is equivalent to

σ(a0) + σ(a1)α + · · ·+ σ(ad−1)αd−1 = 0 for all σ ∈ G \H. (2.29)

Since H is a subgroup of G, we have σ ∈ G \H if and only if σ−1 ∈ G \H. Applying

σ−1 to both sides of the equation associated σ, it becomes

a0 + a1σ−1(α) + · · ·+ ad−1σ

−1(αd−1) = 0 for all σ ∈ G \H,

and thus the conditions (2.29) are equivalent to

a0 + a1σ(α) + · · ·+ ad−1(σ(α))d−1 = 0 for all σ ∈ G \H (2.30)

By definition of G and H, the set of numbers τ(α) with τ ∈ G \H consists of all the

conjugates of α but α. Therefore, condition (2.30) is equivalent to asking that the

polynomial

p(T ) = a0 + a1T + · · ·+ ad−2Td−2 + T d−1

has roots α1, . . . , αd−1 where α1, . . . , αd−1 denote all the conjugates of α which are

different from α. Namely, it asks that

p(T ) = (T − α1)(T − α2) · · · (T − αd−1) =min(α,Q)

T − α. (2.31)

Since α ∈ OK , we can write

min(α,Q) = T d + td−1Td−1 + · · ·+ t0

for some integers t0, . . . , td−1. Then (2.31) is equivalent to the following equation

(T d−1 + ad−2Td−2 + · · ·+ a0)(T − α) = T d + td−1T

d−1 + · · ·+ t0.

By comparing coefficients on both sides, we get

ti = ai−1 − aiα for i = 1, . . . , d− 1,


and soad−2 = td−1 + α,

ad−3 = td−2 + ad−2α = td−2 + td−1α + α2,

· · ·

a0 = t1 + t2α + · · ·+ αd−1.

(2.32)

This proves the existence and unicity of a0, . . . , ad−1 in Q(α) satisfying (2.28). Since

α ∈ OK and t0, . . . , td−1 ∈ Z, we get ai ∈ OK for all i = 0, . . . , d − 2. On the other

hand, since α has degree d, the elements 1, α, . . . , αd−1 form a basis of Q(α) over

Q. Moreover, the above formulas show that each ai with i = 0, . . . , d− 1 is a monic

polynomial in α of degree d − i − 1. Hence the elements a0, . . . , ad−1 form another

basis of Q(α) over Q. In particular, they are Q−linearly independent.

Proposition 2.6.2. Let a0, . . . , ad−1 be as in the above lemma. There exists a sym-

metric d-linear form Φ(X1, . . . ,Xd) ∈ Z[X1, . . . ,Xd] such that

(i) Φ(θ,θ,X3, . . . ,Xd) = 0,

(ii) the polynomial ϕ(X1) =1

d!Φ(X1, . . . ,X1) satisfies

ϕ(x) = NK/Q(a0x0 + · · ·+ ad−1xd−1) ∈ Z \ {0}

for all x = (x0, . . . , xd−1) ∈ Zd \ {0}.

Proof.

(i) Let F be a normal closure of K/Q and set G = Gal(F/Q). Then G acts on

F [X1, . . . ,Xd] by

σ(∑

ai1,··· ,idXi11 · · ·X

idd

)=∑

σ(ai1,...,id)Xi11 · · ·X

idd for σ ∈ G,

where Xij denotes the monomial X i0

j,0 · · ·Xid−1

j,d−1 for each i = (i0, . . . , id−1) ∈ Nd.Let τ1, . . . , τd be all the d embeddings of K into C ordered so that τd = IdK . Set

`j(X) = τj(a0)X0 + · · ·+ τj(ad−1)Xd−1 for j = 1, . . . , d,

and set

Φ(X1, . . . ,Xd) =∑ν∈Sd

`ν(1)(X1) · · · `ν(d)(Xd).


It is clear that Φ is a symmetric d-linear form. By the previous lemma, we have

ai ∈ OK for all i = 0, . . . , d − 1, and so Φ ∈ OF [X1, . . . ,Xd]. Moreover, we have

σ(Φ(X1, . . . ,Xd)) = Φ(X1, . . . ,Xd) for any σ ∈ G since σ permutes `1, . . . , `d. We

deduce that

Φ ∈ OGF [X1, . . . ,Xd] = Z[Xi11 . . .X

idd ].

By the same lemma, we have `j(θ) = 0 for j = 1, . . . , d− 1. This implies that

`ν(1)(θ) · `ν(2)(θ) = 0

for all ν ∈ Sd, so we get (i).

(ii) Since a0, . . . , ad−1 are Q- linearly independent algebraic integers, we have

a0x0 + · · ·+ ad−1xd−1 ∈ OK \ {0}

for all x = (x0, . . . , xd−1) ∈ Zd \ {0}. For those points, we conclude that

ϕ(x) =d∏i=1

τi(a0x0 + . . .+ ad−1xd−1)

=NK/Q(a0x0 + . . .+ ad−1xd−1)

is a non-zero integer.

Now we are able to give an alternative proof of Proposition 2.4.1.

Proof of Proposition 2.4.1. Fix x = (x0, . . . , xd−1) ∈ Zd \ {0} and write this point in

the form x = x0θ + ∆ with ∆ ∈ Rd. Then we have Lθ(x) = ‖∆‖. With Φ and ϕ as

in Proposition 2.6.2, we find

ϕ(x) =1

d!Φ(x0θ + ∆, . . . , x0θ + ∆)

=1

d!

(Φ(∆, . . . ,∆) + dx0Φ(θ,∆, . . . ,∆)

)since Φ(x0θ, x0θ,∆, . . . ,∆) = x20Φ(θ,θ,∆, . . . ,∆) = 0. Since ϕ(x) ∈ Z \ {0}, we

conclude that

1 ≤ |ϕ(x)| � ‖x‖ · Lθ(x)d−1 (2.33)

and so Lθ(x)� ‖x‖−1/(d−1).


2.7 The morphism Ψ

As we said before, we don’t know if there exists a point θ ∈ Rd+1 as in Theorem

2.1.3 such that λ(θ) = λd. However, if such θ exists, then there exists a sequence

of primitve points (yn)n∈N∗ satisfying conditions (i)–(iii) of Theorem 2.4.3 for each

n > 0. In this section, we construct explicit algebraic relations between the points of

such a sequence. We hope that these relations will be useful for further study of this

topic.

Let the notation d,K, α, ϕ and Φ be as in Section 2.6. We fix ξ ∈ R \ Q(α) and

set θ = (1, α, . . . , αd−1, ξ).

For any x,y ∈ Rd+1, we define

Ψ(x,y) = ϕ(x−)y − 1

(d− 1)!Φ(x−, . . . ,x−,y−)x ∈ Rd+1.

and set L(x) = Lθ(x).

Theorem 2.7.1. Suppose that there exists a sequence of primitive points (yn)n∈N∗

satisfying conditions (i)–(iii) of Theorem 2.4.3 for each n. Set zn = Ψ(yn+d,yn+d+1)

for each n ∈ N∗. Then we have

(i) (d− 1)! zn ∈ Zd+1 \ {0} for each n > 0,

(ii) ‖zn‖ � ‖yn+d‖d−rd−1 , L(zn)� ‖yn+d‖r−

dd−1 � L(yi) for each n > 0,

(iii) det(yn,yn+1, . . . ,yn+d−1,Ψ(yn+d,yn+d+1)) = 0 when n is sufficiently large.

Condition (i) of Theorem 2.4.3 implies that any d+1 consecutive points of (yn)n∈N∗

form a basis of Rd+1. Therefore, each point yn+d+1 with n > 0 is a linear combination

of its d + 1 predecessors yn, . . . ,yn+d. The following corollary provides us with the

coefficient of yn+d in such a linear combination.

Corollary 2.7.2. Let the assumption and the notation be as in the Theorem 2.7.1.

We have

ϕ(y−n+d)yn+d+1 −1

(d− 1)!Φ(y−n+d, . . . ,y

−n+d,y

−n+d+1)yn+d ∈ 〈yn, . . . ,yn+d−1〉R.


To prove Theorem 2.7.1, we establish the following result.

Proposition 2.7.3. Let x,y ∈ Rd+1 such that

L(y) ≤ L(x) ≤ ‖x‖ ≤ ‖y‖. (2.34)

Then we have

(i) L(Ψ(x,y))� ‖y‖L(x)d,

(ii) ‖Ψ(x,y)‖ � ‖x‖2L(x)d−2L(y) + ‖y‖L(x)d.

Proof. Write x− = x0θ− + ∆x. Then we have

‖∆x‖ ≤ Lθ−(x−) ≤ L(x).

Similarly, write y− = y0θ− + ∆y and so ‖∆y‖ ≤ L(y).

Using the multilinearity of Φ and Proposition 2.6.2 (i), we find that

Φ(x−, . . . ,x−,y−) = Φ(x0θ− + ∆x, . . . , x0θ

− + ∆x, y0θ− + ∆y)

= (d− 1)x0Φ(θ−,∆x, . . . ,∆x,∆y)

+ y0Φ(θ−,∆x, . . . ,∆x) + Φ(∆x, . . . ,∆x,∆y). (2.35)

We deduce from (2.34) that

|Φ(x−, . . . ,x−,y−)| �‖x‖L(x)d−2L(y) + ‖y‖L(x)d−1 + L(x)d−1L(y)

� ‖y‖L(x)d−1. (2.36)

On the other hand, using the definition of ϕ in Proposition 2.6.2 (ii), we find that

ϕ(x−) =1

d!Φ(x−, . . . ,x−) =

1

(d− 1)!x0Φ(θ−,∆x, . . . ,∆x) + ϕ(∆x), (2.37)

and so, by (2.34), we get

|ϕ(x−)| � ‖x‖L(x)d−1. (2.38)


It follows from the definition of Ψ and equalities (2.35), (2.37) that

Ψ(x,y)0 = y0 ϕ(x−)− 1

(d− 1)!x0Φ(x−, . . . ,x−,y−)

= y0 ϕ(∆x)− 1

(d− 2)!x20Φ(θ−,∆x, . . . ,∆x,∆y)

− 1

(d− 1)!x0Φ(∆x, . . . ,∆x,∆y).

Thus we get

|Ψ(x,y)0| � ‖y‖L(x)d + ‖x‖2L(x)d−2L(y) + ‖x‖L(x)d−1L(y)

� ‖y‖L(x)d + ‖x‖2L(x)d−2L(y)

since ‖x‖ ≥ L(x).

On the other hand, we deduce from the inequalities (2.36) and (2.38) that

L(Ψ(x,y)) = L

(ϕ(x−)y − 1

(d− 1)!Φ(x−, . . . ,x−,y−)x

)� |ϕ(x−)|L(y) + |Φ(x−, . . . ,x−,y−)|L(x)

� ‖x‖L(x)d−1L(y) + ‖y‖L(x)d

� ‖y‖L(x)d,

Therefore, we obtain

‖Ψ(x,y)‖ = ‖Ψ(x,y)0θ + (Ψ(x,y)−Ψ(x,y)0θ))‖

� |Ψ(x,y)0|+ L(Ψ(x,y))

� ‖x‖2L(x)d−2L(y) + ‖y‖L(x)d.

Proof of Theorem 2.7.1. (i) By definition of Ψ, we have (d − 1)! zn ∈ Zd+1 for all

n. Since any two consecutive minimal points are linearly independent and since

ϕ(yn+d) 6= 0 for each n ≥ 1 (by Proposition 2.6.2 (ii)), we deduce that zn 6= 0.

(ii) For any n ∈ N∗, the conditions (ii) and (iii) of Theorem 2.4.3 give

‖yn+1‖ � ‖yn‖r, L(yn) � ‖yn‖−1/(d−1),


with r = 1/((d− 1)λd).

Fix a large integer n. We deduce from Proposition 2.7.3 that

L(zn)� ‖yn+d+1‖L(yn+d)d � ‖yn+d‖r−

dd−1 � ‖yn‖(r−

dd−1

)rd .

Since λd is a positive root of (2.5), we get

1

r+ · · ·+ 1

rd= d− 1,

and so

d rd = (d− 1)rd + rd = 1 + r + · · ·+ rd−1 + rd. (2.39)

Thus we have(r − d

d− 1

)rd =

1

d− 1

((d− 1)rd+1 − (1 + r + · · ·+ rd)

)=

1

d− 1

(− 1 + r

((d− 1)rd − (1 + r + · · ·+ rd−1)

))=−1

d− 1.

This leads to

L(zn)� ‖yn+d‖r−dd−1 � ‖yn‖

−1d−1 � L(yn). (2.40)

By Proposition 2.7.3 (ii), we get

‖zn‖ � ‖yn+d‖2L(yn+d)d−2L(yn+d+1) + ‖yn+d+1‖L(yn+d)

d

� ‖yn+d‖2−d−2d−1− rd−1 + ‖yn+d‖r−

dd−1 .

We deduce from the estimates (2.40) that ‖yn+d‖r−dd−1 converges to 0 as n tends to

∞, and since ‖zn‖ ≥ 1, we obtain

‖zn‖ � ‖yn+d‖d−rd−1 .

(iii) Set Dn = det(yn,yn+1, . . . ,yn+d−1, zn) for each n > 0.

By part (i), we get (d − 1)! Dn ∈ Z for all n > 0. So, it is enough to prove that

Dn converges to 0 when n tends to infinity.


From part (ii), we deduce that

Dn � ‖yn+d−1‖L(yn) · · ·L(yn+d−2)L(zn) + ‖zn‖L(yn) · · ·L(yn+d−1)

� L(yn) · · ·L(yn+d−2)(‖yn+d−1‖L(zn) + ‖zn‖L(yn+d−1))

� (‖yn‖ · · · ‖yn+d−2‖)−1d−1

(‖yn+d‖

1r+r− d

d−1 + ‖yn+d‖d−rd−1− 1

(d−1)r

)� ‖yn‖−

1d−1

(1+r+···+rd−2)‖yn+d‖1r+r− d

d−1

� ‖yn‖g(r),

where

g(r) = − 1

d− 1(1 + r + · · ·+ rd−2) + rd

(1

r+ r − d

d− 1

).

By (2.39), we get

g(r) = − 1

d− 1

((d− 1)rd − rd−1

)+

(rd−1 + rd+1 − d

d− 1rd)

= rd−1(r2 − 2d− 1

d− 1r +

d

d− 1

).

Note that 1/d < λd < 1/(d− 1) and so 1 < r < d/(d− 1). We deduce that g(r) < 0.

Since ‖yn‖ grows very fast, Dn converges to 0 when n tends to infinity.

2.8 An explicit construction of a point with expo-

nent of approximation ≥ 1/3

It would be nice to know if the exponent λd given by Theorem 2.1.3 is optimal for

some integer d ≥ 3, namely if there exists a real algebraic number α of degree d and a

real number ξ /∈ Q(α) such that λ(1, α, . . . , αd−1, ξ) = λd. If such numbers exist, then

Theorem 2.4.3 provides us with a sequence of primitive points xn = (xn,0, . . . , xn,d−1)

in Zd+1 satisfying

(i) | det(xn,xn+1, . . . ,xn+d)| � 1,

(ii) ‖xn+1‖ � ‖xn‖r with r = 1/((d− 1)λd),

(iii) L(xn) � L′(x−n ) � ‖xn‖−1/(d−1),where θ = (1, α, . . . , αd−1, ξ) and L = Lθ, L

′ = Lθ− .


Then by (2.33), the property L′(xn) � ‖xn‖−1/(d−1) implies that

(iv) |ϕ(x−n )| � 1

where ϕ is the polynomial associated to α defined in Section 2.6.

In this section, we choose d = 3 and α = 3√

2 and prove the following result.

Theorem 2.8.1. There exist a real number ξ /∈ Q(α) and a sequence of primitive

points (xn)n∈N∗ in Z4 satisfying

(i)’ det(xn,xn+1,xn+2,xn+3) 6= 0,

(ii)’ ‖xn+1‖ � ‖xn‖3/2,(iii)’ L′(x−n ) � ‖xn‖−1/2 and L(xn)� cn‖xn‖−1/2,(iv)’ ϕ(x−n ) = 1,

where c = 180 000 and L = Lθ, L′ = Lθ− with θ = (1, α, α2, ξ).

It is interesting to compare the conditions (i)’–(iv)’ with the conditions (i)–(iv)

for d = 3.

The condition (iv)’ is very restrictive because, as we will see below, it implies that

xn,0α2 + xn,1α+ xn,2 is a unit of Z[α], and these units are sparse since the unit group

of Z[α] has rank 1. However, it is not much more restrictive than condition (iv) which

requests the norm of xn,0α2 +xn,1α+xn,2 to be bounded. So conditions (iv) and (iv)’

are essentially the same.

Consider condition (iii)’. We deduce from the condition (ii)’ on the growth of

‖xn‖ that, for each ε > 0, there exists an integer n0 such that

cn ≤ ‖xn‖ε for all n ≥ n0.

Therefore, conditions (iii)’ and (iii) for d = 3 are also essentially the same. The

condition (i)’ is much weaker than (i) but strong enough to yield that ξ /∈ Q(α) as

we will see below.

The main difference is condition (ii)’ which shows that ‖xn‖ grows much faster

than we would like in comparison with (ii) because for d = 3, we have r ≈ 1.234.


Now let ξ and (xn)n∈N∗ be as in Theorem 2.8.1. Fix ε ∈ R with 0 < ε < 1/6. For

each sufficiently large value of X, condition (ii)’ ensures the existence of a positive

integer n such that

‖xn‖ ≤ X < ‖xn+1‖, cn ≤ ‖xn+1‖ε/3.

Then for x = xn, we have ‖x‖ ≤ X, moreover, it follows from conditions (ii)’ and

(iii)’ that

Lθ(x)� cn‖xn‖−1/2 � cn‖xn+1‖−1/3 ≤ ‖xn+1‖−(1−ε)/3 < X−(1−ε)/3.

Therefore, λ = (1− ε)/3 is a uniform exponent of approximation to θ = (1, α, α2, ξ).

Since ε > 0 can be choosen arbitrarily small, we deduce that λ(θ) ≥ 1/3, a result

which is true for any ξ by the box principle (see Lemma 2.1.2 (i)).

On the other hand, if X is sufficiently large and satisfies

‖xn‖ ≤ X < ‖xn+1‖1−2ε

for some n, then we have

Lθ(x)� ‖xn+1‖−(1−ε)/3 ≤ X−1−ε

3(1−2ε) .

This is meaningful because 1−ε3(1−2ε) > 1/3 and so we cannot construct such a point by

the box principle.

One more thing significant here is that if we could improve the condition (ii)’ and

get ‖xn+1‖ � ‖xn‖r with 1/2 ≤ r < 3/2, then by condition (iii)’, this would give

Lθ(xn)� cn‖xn‖−12 � cn‖xn+1‖−

12r � ‖xn+1‖−(

12r−ε)

for any fixed ε > 0. Then λ = 1/(2r) − ε would be a uniform exponent of approxi-

mation to θ for any ε > 0 and so

λ(θ) ≥ 1

2r> 1/3.


In order to prove Theorem 2.8.1, we will construct a number ξ and a sequence

(xn)n∈N∗ satisfying all the conditions required in the theorem. To do this, we start

with the following observations.

Assume that points xn = (xn,0, . . . , xn,3) ∈ Z4 satisfy the condition (iv)’. By

Proposition 2.6.2, it implies that

NK/Q(a0xn,0 + a1xn,1 + a2xn,2) = 1

with a0, a1, a2 ∈ Q(α) given by (2.32) in Lemma 2.6.1. Since min(α,Q) = T 3 − 2, we

get

a0 = α2, a1 = α, a2 = 1.

So we find

α2xn,0 + αxn,1 + xn,2 ∈ O∗K .

Since K has one real embedding and two complex conjugate embeddings into C,

it follows from Dirichlet’s Unit Theorem that O∗K has rank 1. One can show that

O∗K = {±εm, m ∈ Z}

where ε = α2 + α + 1. So for each n ∈ N, we have

α2xn,0 + αxn,1 + xn,2 = ±εsn

for some sn ∈ Z.Note that for each n ∈ Z, there exists a unique triple (an, bn, cn) in Z3 such that

εn = anα2 + bnα + cn.

Therefore, the points xn must have the form

xn = ±(An, Bn, Cn, yn) (2.41)

for some yn ∈ Z and An = asn , Bn = bsn , Cn = csn .

We need more information about the powers of ε. To derive them, we denote the

three conjugates of α by

α, α1 = ρ3√

2, α2 = α1 = ρ23√

2.


where ρ = e2π3i. Hence the three conjugates of εn are

εn, εn1 = anα21 + bnα1 + cn, εn2 = anα

22 + bnα2 + cn.

Moreover, we have

ε1 = ε−12 eiζ , ε2 = ε−

12 e−iζ

with ζ ≈ −0.5899.

Proposition 2.8.2. For any n ∈ N∗, the number εn has the following properties:

(i) NQ(α)/Q(εn) = 1, gcd(an, bn, cn) = 1,

(ii) |anα− bn| ≤ ε−n2 , |anα2 − cn| ≤

3

2ε−

n2 ,

(iii)1

6εn < an <

1

3εn.

Proof. The property (i) follows immediately from the fact that ε is a unit of Z[α] of

norm 1.

(ii) We have

εn1 − εn2 = an(α21 − α2

2) + bn(α1 − α2)

= (α2 − α1)(anα− bn).

So

|anα− bn| =|εn1 − εn2 ||α2 − α1|

≤ 2ε−n/2

|α2 − α1|≤ ε−

n2 .

For the second inequality, we use

εn1 + εn2 = an(α21 + α2

2) + bn(α1 + α2) + 2cn

= −anα2 − bnα + 2cn

= 2(cn − anα2)− α(bn − anα),

and get

|cn − anα2| = 1

2|(εn1 + εn2 ) + α(bn − anα)| = 1

2

∣∣∣∣(εn1 + εn2 ) + αεn1 − εn2α1 − α2

∣∣∣∣≤ 1

2

(∣∣∣∣1 +α

α1 − α2

∣∣∣∣+

∣∣∣∣1− α

α1 − α2

∣∣∣∣) ε−n/2≤ 3

2ε−n/2.


(iii) We have

εn = anα2 + bnα + cn

= 3anα2 + α(bn − anα) + (cn − anα2).

It is clear by definition that an > 0 for all n > 0. From (ii), we deduce that

εn

6<εn − (α + 3

2)ε−

n2

3α2≤ an ≤

εn + (α + 32)ε−

n2

3α2<εn

3for n > 1.

When n = 1, we have an = 1 so it is clear that1

6ε < an <

1

3ε.

For each n ≥ 1, we request that sn > 0. Since xn has form (2.41), the above

proposition gives

ϕ(x−n ) = 1, ‖x−n ‖ � An � εsn , L′(x−n )� ε−sn/2 � ‖x−n ‖−1/2. (2.42)

By Proposition 2.4.1, we get L′(xn)� ‖x−n ‖−1/2 and so

L′(xn) � ‖x−n ‖−1/2.

Hence (xn)n∈N∗ satisfies the condition (iv)’ and half of (iii)’ in Theorem 2.8.1 for any

choice of yn and sn > 0. We have the freedom of choosing yn and sn such that the

remaining conditions are fulfilled.

Since we want the first coordinate of xn to be positive, we will assume that in

(2.41) we have the sign + so that xn = (An, Bn, Cn, yn)

Note that

L(xn) = max{L′(x−n ), |Anξ − yn|}

holds for any choice of yn and ξ. Moreover, we have

L′(x−n )� ‖x−n ‖−1/2.

So asking that L(xn)� cn‖xn‖−1/2 is equivalent to asking that

|Anξ − yn| � cn‖xn‖−1/2,


which leads to ∣∣∣∣ξ − ynAn

∣∣∣∣� cnA−3/2n . (2.43)

Set

ξn =ynAn

for n ∈ N∗.

Then the conditioon (2.43) implies that

|ξn − ξn−1| ≤ |ξn − ξ|+ |ξn−1 − ξ|

� cnA−3/2n + cn−1A−3/2n−1

� cn−1A−3/2n−1 . (2.44)

To utilise this, we introduce some new notation.

For each n ∈ N∗, we denote by [εn] = (an, bn, cn)T the coordinates of εn in the

basis {1, α, α2} of OK .

For each n ≥ 4, we also denote by Dn the determinant of the matrix

Mn = (xTn xTn−1 xTn−2 xTn−3)

and denote by Dn,i the determinant of the matrix obtained by removing the last row

and the (i+ 1)–th column from matrix Mn for each i = 0, 1, 2, 3.

We have

Dn =

∣∣∣∣∣ [εsn ] [εsn−1 ] [εsn−2 ] [εsn−3 ]

yn − ξn−1An 0 yn−2 − ξn−1An−2 yn−3 − ξn−1An−3

∣∣∣∣∣= (ξn − ξn−1)AnDn,0 + (ξn−2 − ξn−1)An−2Dn,2 + (ξn−3 − ξn−1)An−3Dn,3.

Suppose that xn−1,xn−2,xn−3 have been constructed and that Dn,0 6= 0. We get

|ξn − ξn−1| ≤|Dn|+ |(ξn−2 − ξn−1)An−2Dn,2|+ |(ξn−3 − ξn−1)An−3Dn,3|

|AnDn,0|. (2.45)

We want |ξn − ξn−1| ≤ cn−1A−3/2n−1 . However, as we will see below, for any positive

integers m,n, p with m < n < p, we have∣∣det([εm]T [εn]T [εp]T

)∣∣ ≤ 1√3εp−

m+n2 ,


For each x ∈ K, we have f(x)T = M [x] where

M =

1 α α2

1 α1 α21

1 α2 α22

and where [x] denotes the coordinates of x in the basis {1, α, α2}.

So, for any x, y, z ∈ K, we have

| det(f(x)T , f(y)T , f(z)T )| = |det(M) det ([x], [y], [z])|

= 6√

3 | det([x], [y], [z])|.

In particular, we deduce that

| det([εm], [εn], [εp])| = |A|6√

3

where

A = det(f(εm)T , f(εn)T , f(εp)T

)=

∣∣∣∣∣∣∣∣εm εn εp

εm1 εn1 εp1

εm2 εn2 εp2

∣∣∣∣∣∣∣∣ .Since |ε1| = |ε2| = ε−

12 and m < n < p, we get

|A| ≤ 6|εpεn1εm2 | ≤ 6εp−m+n

2 ,

which proves the upper bound for the absolute value of the determinant in (2.46).

Now for the lower bound, we use

|A| ≥ |εp(εm1 εn2 − εn1εm2 )| − |εn(εm1 εp2 − ε

p1εm2 )| − |εm(εp1ε

n2 − εn1ε

p2)|

≥ εp|(εm1 εm2 )(εn−m2 − εn−m1 )| − 2εn−m+p

2 − 2εm−n+p2

≥ εp−m|2 Im(εn−m1 )| − 2εn−m+p

2 − 2εm−n+p2

≥ 2 |sin((n−m)ζ)|εp−m+n

2 − 2εn−m+p

2 − 2εm−n+p2 .

Since |sin(n−m)ζ| ≥ 1/4 and p > n > m, we obtain that

|A| > εp−n+m

2

(1

2− 2ε−

32(p−n) − 2ε−

32(p−m)

)≥ 1

6εp−

m+n2 ,

so this completes the proof of (2.46).


In view of the above lemma, we ask that

|sin((sn − sn−1)ζ)| > 1/4 (2.47)

for all n > 1 in order to make Dn,0 large.

Note that condition (iii)’ implies that ‖xn‖ � ‖x−n ‖ � εsn . Hence condition (ii)’

requires

sn+1 =3

2sn +O(1). (2.48)

As we will see below, it is easy to construct a sequence (sn)n∈N∗ for which conditions

(2.47) and (2.48) are satisfied.

To construct yn, we consider Dn as a linear form in yn. Since the coefficient of yn

is Dn,0, as we will see below, we can choose yn ∈ Z such that

0 < |Dn| ≤ |Dn,0|. (2.49)

Now we give the details.

Proof of Theorem 2.8.1. We construct the sequence (xn)n∈N∗ and the number ξ so

that they satisfy all the required conditions.

Step 1. We construct recursively a sequence of positive integers (sn)n∈N∗ .

Set s1 = 1. For n > 1, we assume that sn−1 is constructed and choose sn to be

one of the two consecutive integers b(3/2)nc and d(3/2)ne for which

|sin((sn − sn−1)ζ)| > 1/4

is satisfied (see (2.47)). This is possible because of Lemma 2.8.3. Then, for each

n ≥ 1, we find∣∣∣∣sn+1 −3

2sn

∣∣∣∣ ≤∣∣∣∣∣sn+1 −

(3

2

)n+1∣∣∣∣∣+

3

2

∣∣∣∣sn − (3

2

)n∣∣∣∣ < 5

2. (2.50)

(The condition (2.48) is therefore fulfilled.)

Set

x−n = [εsn ]T for each n > 1.


Then, from the previous discussion, we deduce that, for each n > 0, we have

ϕ(x−n ) = 1, L′(x−n ) � ‖x−n ‖−1/2,

‖x−n+1‖ � An+1 � εsn+1 � A3/2n � ‖x−n ‖3/2,

and

1

36√

3εsn−1− 1

2(sn−2+sn−3) ≤ |Dn,0| ≤

1√3εsn−1− 1

2(sn−2+sn−3) if n > 4. (2.51)

Step 2. We construct recursively a sequence of integers (yn)n∈N∗ .

For n = 1, 2, 3, we set yn = An. Assume that yn−1, yn−2, yn−3 have been choosen

for some n > 3. Let t be a real number such that Dn(t) = 0 where

Dn(t) =

∣∣∣∣∣(x−n )T (x−n−1)T (x−n−2)

T (x−n−3)T

t yn−1 yn−2 yn−3

∣∣∣∣∣=− tDn,0 + yn−1Dn,1 − yn−2Dn,2 + yn−3Dn,3.

Set

yn =

t+ 1 if t ∈ Z,

btc if t /∈ Z.

Then we obtain that 0 6= |Dn| ≤ |Dn,0| for all n > 3.

For each n > 0, we define

xn = (x−n , yn).

Step 3. We construct ξ /∈ Q(α) and show that the sequence (xn)n∈N∗ has all the

required properties.

Set ξn = yn/An for each n ≥ 1.

We first prove by induction that the sequence (ξn)n∈N∗ satisfies

|ξn − ξn−1| ≤ cn−1A−3/2n−1 (2.52)

for all n > 1.

This is true when n = 2 or 3 since ξ1 = ξ2 = ξ3 = 1. Assume that

|ξn−1 − ξn−2| ≤ cn−2A−3/2n−2 , |ξn−2 − ξn−3| ≤ cn−3A

−3/2n−3 (2.53)


Similarly, using (2.55), we also find

|(ξn−3 − ξn−1)An−3Dn,3| ≤√

2 cn−2εsn−12(sn−1+sn−2+sn−3).

Substituting the estimates in the upper bound for |ξn − ξn−1| and using the lower

bound for Dn,0 given by (2.51), we deduce that

|ξn − ξn−1| ≤ 35A−3/2n−1 (1 + 72

√6 cn−2εsn−

32sn−1)

< (35 + 6173 cn−2ε5/2)A−3/2n−1 ( by (2.48))

< cn−1A−3/2n−1 (since c = 180 000).

By the induction principle, we conclude that (2.52) holds for any integer n > 1.

This result shows that (ξn)n∈N∗ is a Cauchy sequence, so it converges. Set

ξ = limn→∞

ξn.

By (2.44), we deduce that, for any n ∈ N∗, we have

|ξn − ξ| � cnA−3/2n ,

and thus

|Anξ − yn| � cnA−1/2n . (2.56)

By Step 1, this implies that

‖xn+1‖ � ‖x−n+1‖ � ‖x−n ‖3/2 � ‖xn‖3/2,

and that

L(xn) = max{L′(x−n ), |Anξ − yn|} � cnA−1/2n � cn‖xn‖−1/2.

Note that limn→∞ L(xn) = 0 and Dn 6= 0 for all n > 3. By Lemma 2.2.4, this implies

that ξ /∈ Q(α). So ξ and (xn)n∈N∗ satisfy all the required conditions.

Remark 2.8.5. If we replace 3/2 by a real number r with 1 < r < 3/2 in the

construction of (sn)n∈N∗ in Step 1 and argue as Steps 2, 3, then we obtain a sequence

(xn)n∈N∗ in Z4 and a number ξ /∈ Q(α) satisfying


(i)” det(xn, . . . ,xn+3) 6= 0,

(ii)” ‖xn+1‖ � ‖xn‖r,

(iii)” L′(x−n ) � ‖x−n ‖−1/2 and L(xn)� c′nA−(r−1)n � c′n‖xn‖−(r−1),

(iv)” ϕ(x−n ) = 1,

where the constant c′ only depends on r.

The property (iii)” derives from

|ξn − ξn−1| � A−rn−1

(∣∣∣∣ Dn

Dn,0

∣∣∣∣+ c′n−2εsn−32sn−1

)� A−rn−1,

using |Dn| < |Dn,0|. Therefore, if we could make∣∣∣ DnDn,0

∣∣∣ much smaller than 1 for some

r < 3/2, then we could improve on (iii)” or (iii)’.

Chapter 3

On the dual Diophantine problem

3.1 Introduction

Let θ = (1, θ1, . . . , θd) ∈ Rd+1. We denote by τ(θ) the supremum of the real numbers

τ for which there exists a constant c > 0 such that the convex body

C∗c,X,τ :

|x0 + x1θ1 + · · ·+ xdθd| ≤ cX−τ

|x1|, . . . , |xd| ≤ X(3.1)

contains a non-zero point (x0, x1, . . . , xd) ∈ Zd+1 for any sufficiently large value of X.

Note that

vol(C∗c,X,τ ) = c2d+1Xd−τ .

By Minkowski’s First Convex Body Theorem, we deduce that if τ = d and c = 1,

then the convex body C∗c,X,τ contains a non-zero point in Zd+1 for each X > 0. This

implies that

τ(θ) ≥ d.

The main goal of this chapter is to prove the following result.

Theorem 3.1.1. Let α be an algebraic number of degree d ≥ 2 and let ξ ∈ R \Q(α).

107

CHAPTER 3. ON THE DUAL DIOPHANTINE PROBLEM 108

Let τ, c > 0. Suppose that the inequalities|x0 + x1α + · · ·+ xd−1αd−1 + xdξ| ≤ cX−τ

|x1|, . . . , |xd| ≤ X(3.2)

admit a solution x = (x0, . . . , xd) ∈ Zd+1 \ {0} for any sufficiently large value of X.

Then we have

τ ≤ τd :=1 +√

5

2(d− 1) + 1.

In the notation introduced above, this means that

τ(θ) ≤ τd where θ = (1, α, . . . , αd−1, ξ).

In fact, when d = 2, the estimate τ(1, α, ξ) ≤ τ2 = γ2 can be deduced from the upper

bound λ(1, α, ξ) ≤ λ2 = 1/γ from Chapter 2. Indeed, Jarnık’s transference principle

([11]) gives

τ(1, α, ξ) =1

1− λ(1, α, ξ)≤ 1

1− 1/γ= γ2.

Moreover, it is shown in [22] that, given a quadratic number α, there exists ξ ∈R \ Q(α) such that λ(1, α, ξ) = 1/γ and so τ(1, α, ξ) = γ2. Therefore, the estimate

τ(θ) ≤ τd is optimal for d = 2. We don’t know if it is best possible for d ≥ 3.

Based on the main result of Y. Bugeaud and M. Laurent in [3], arguing as in

Section 3.2, we obtain the following result.

Corollary 3.1.2. Let the notation be as in Theorem 3.1.1. Assume that τ > τd.

Then, for any η = (η0, . . . , ηd) ∈ Rd+1, there are arbitrarily large real numbers X

such that the inequalities|x0αi − xi − ηi| � X−1/τ (1 ≤ i < d),

|x0ξ − xd − ηd| � X−1/τ ,

‖x‖ ≤ X

have a non-zero solution x = (x0, . . . , xd) in Zd+1.


To prove our main result, we start, as in Chapter 2, with the construction of a

sequence of minimal points (xi)i∈N attached to the point θ = (1, α, . . . , αd−1, ξ) or

more precisely to its associated map Tθ(x) = |x · θ| for each x ∈ Zd+1. We then

establish some basic properties of this sequence, assuming that the system (3.2) has

a non-zero integer solution for each sufficiently large real number X and some fixed

τ > 0. This is similar to [6] and occupies Section 3.2. However, by contrast to [6],

it is not so easy to show there exist infinitely many indices i ∈ N such that xi, xi+1

and xi+2 are linearly independent over Q. To prove this, we require that τ > 1. Then

we denote by I the infinite set of all those indices i, and endow it with the natural

ordering of integers.

The proof of the theorem itself uses three main estimates. Two of them are

obtained in a similar way, by working with several linearly independent minimal

points to produce a polynomial in α with small non-zero absolute value and then

by using Liouville’s inequality to bound from below this absolute value. For two

points, this is done in Section 3.2 through an explicit construction. The result is

an upper bound for the norm of any minimal point in terms of the norm of the

preceding point. For three points however, our construction is not explicit as we

obtain it through an application of Dirichlet’s box principle. The resulting estimate

is established in Section 3.3. The triples of points that we use for this purpose are

of the form (xi+1,xj+1,xj+2) for consecutive elements i < j in I, such triples being

linearly independent. The last ingredient that we need uses the fact that, for such

pairs (i, j), we have 〈xi+1,xi+2〉Z = 〈xj,xj+1〉Z and so ‖xi+1 ∧xi+2‖ = ‖xj ∧xj+1‖. A

useful inequality then follows by estimating these norms.

The estimates obtained in Section 3.2 already imply a first upper bound for τ(θ),

namely τ(θ) ≤ 2d− 1. Combining the estimates of Section 3.2 with those of Section

3.3, we prove in Section 3.4 the stronger estimate τ(θ) ≤ τd of Theorem 3.1.1.

In the last section, we obtain one more general estimate through the construction

of explicit non-zero polynomial maps from (Qd+1) to Q which do not vanish simulta-

neously on any triple of linearly independent points in Qd × {0}. Using this estimate

(in place of an estimate constructed in Section 3.3), we obtain an alternative proof

for Theorem 3.1.1 in the case where d = 3.


Notation. For any x = (x0, x1, . . . , xd) in Rd+1, we define x− = (x0, . . . , xd−1)

and x+ = (x1, . . . , xd). For each point θ = (θ0, . . . , θd) in Rd+1, we define a function

Tθ : Rd+1 −→ R by

Tθ(x) = |x · θ| = |x0θ0 + x1θ1 + · · ·+ xdθd|.

3.2 Sequences of minimal points associated to Tθ

Fix a point θ = (1, θ1, . . . , θd) of Rd+1 with Q–linearly independent coordinates.

Replacing Lθ by Tθ in the construction of minimal points in section 2 of chapter 2,

we obtain a sequence of points (xi)i∈N ⊂ Zd+1 such that

(a) xi is primitive for each i ∈ N,

(b) the norms Xi = ‖xi‖ form a strictly increasing sequence,

(c) the positive real numbers Ti = Tθ(xi) form a strictly decreasing sequence,

(d) if a non-zero point x ∈ Zd+1 satisfies Tθ(x) < Ti for some i ≥ 1 then ‖x‖ ≥ Xi+1.

The sequences of minimal points associated to Tθ are uniquely determined up

to the choice of their first points. We fix such a sequence (xi)i∈N and denote by

(xi,0, . . . , xi,d) the coordinates of xi for each i.

Arguing as Daverport and Schmidt in [7, Lemma 2], we also find that, for each

i, the points xi,xi+1 constitute an integral basis for all integer points in the plane

through the origin and these two points. More precisely, we have the following result.

Lemma 3.2.1. For each index i, the two points xi and xi+1 are R-linearly independent

and satisfy

〈xi,xi+1〉R ∩ Zd+1 = 〈xi,xi+1〉Z.

Proof. Since xi and xi+1 are primitive integer points with different norms, they are

R-linearly independent.


Assume that 〈xi,xi+1〉R ∩ Zd+1 6= 〈xi,xi+1〉Z. Then there exists a point of the

form y = rxi + sxi+1 ∈ 〈xi,xi+1〉R ∩Zd+1 for some (r, s) ∈ R \ {0} with |r|, |s| ≤ 1/2.

This implies that

‖y‖ ≤ |r|Xi + |s|Xi+1 < Xi+1, Tθ(y) ≤ |r|Ti + |s|Ti+1 < Ti,

which is impossible because of property (d) of the sequence (xi)i∈N.

The next lemma is also a basic result to which we will refer repeatedly in the

following.

Lemma 3.2.2. Let the notation be as above and let τ, c > 0. Assume that the system|x0 + x1θ1 + · · ·+ xdθd| ≤ cX−τ

|x1|, . . . , |xd| ≤ X(3.3)

admits a solution x = (x0, . . . , xd) ∈ Zd+1 \ {0} for any sufficiently large value of X.

(i) We have

Ti � X−τi+1 for all i ≥ 1.

(ii) If τ > 1, then there exist infinitely many integers i such that the three points

xi,xi+1,xi+2 are linearly independent over R.

As we will see below, the proof of part (i) is quite standard following for example

[6, page 399]. The proof of part (ii) however is more delicate.

Proof. (i) Set c0 = |θ1|+ · · ·+ |θd|+ c+ 1, and choose X0 such that system (3.3) has

a non-zero solution in Zd+1 for each X ≥ X0.

Fix an index i and a real number X with X0 ≤ X < c−10 Xi+1. Let x =

(x0, . . . , xd) ∈ Zd+1 \ {0} be a solution to (3.3). Then we have

|x1|, . . . , |xd| ≤ X, Tθ(x) ≤ cX−τ ,

and so

‖x‖ = max{|x0|, |x1|, . . . , |xd|}

≤ max{|x1θ1|+ · · ·+ |xdθd|+ cX−τ , X}

≤ c0X < Xi+1.


Since the sequence (xi)i∈N has property (d) , we deduce that

Ti ≤ Tθ(x) ≤ cX−τ .

Since we can choose X arbitrarily close to c−10 Xi+1, we conclude that

Ti ≤ c(c−10 Xi+1)−τ .

(ii) Assume on the contrary that there exists an integer n such that xi,xi+1,xi+2 are

R–linearly dependent for all i ≥ n.

For each i ≥ n, set Vi = 〈xi,xi+1,xi+2〉R. Then we have

dimR Vi ≤ 2 for i ≥ n.

Since any two consecutive points of the sequence (xi)i∈N are R–linearly independent,

we deduce that

Vi = 〈xi+1,xi+2〉R = Vi+1 for i ≥ n.

So we find

Vn = Vn+1 = Vn+2 = · · ·

There exists a vector y ∈ Vn \ {0} such that y · θ = 0. Since θ has Q–linearly

independent coordinates, we deduce that y /∈ 〈xn〉R and thus {y,xn} is a basis of Vn.

Fix an index i ≥ n. Since xi ∈ Vi = Vn, we can write

xi = aiy + bixn

for some ai, bi ∈ R with

max{|ai|, |bi|} � ‖xi‖ = Xi,

where the implied constants only depend on y and xn. Then we get

xi ∧ xi+1 =

∣∣∣∣∣ ai bi

ai+1 bi+1

∣∣∣∣∣ (y ∧ xn).

Moreover, we have

Ti = |xi · θ| = |ai(y · θ) + bi(xn · θ)| � |bi|.


We deduce that

1� ‖xi ∧ xi+1‖ � |ai+1bi − aibi+1| � Xi+1Ti +XiTi+1 � Xi+1Ti.

Since Ti � X−τi+1, we get 1 � X1−τi+1 . This is impossible if τ > 1 and if i is large

enough. This shows that there exist infinitely many i such that xi,xi+1,xi+2 are

linearly independent over R.

Considering pairs of consecutive points of (xi)i∈N and applying Liouville’s inequal-

ity, we get the following result.

Lemma 3.2.3. Suppose that θ = (1, α, . . . , αd−1, ξ) where α is an algebraic number

of degree d ≥ 2 and where ξ ∈ R \ Q(α). Let c > 0 and τ > d − 1. Assume that the

system |x0 + x1α + · · ·+ xd−1αd−1 + xdξ| ≤ cX−τ ,

|x1|, . . . , |xd| ≤ X


Then we have τ ≤ 2d − 1 and x−i 6= 0, xi,d 6= 0 for each sufficiently large index i.

Moreover, if τ > d, then we have Xi+1 � Xd−1τ−di .

Proof. We first note, by Lemma 3.2.2 (i), that there exists a constant c1 = c1(α, ξ) > 0

such that

Tθ(xi) = |xi,0 + xi,1α + · · ·+ xi,d−1αd−1 + xi,dξ| ≤ c1X

−τi+1

for all i ∈ N. If x−i = 0, we find

|xi,d| = Xi, |xi,dξ| ≤ c1X−τi+1,

and so i is bounded from above. Hence, there exists an integer N such that x−i 6= 0

for all i ≥ N. By Liouville’s inequality, there exists a constant c2 = c2(α) > 0 such

that

|xi,0 + xi,1α + · · ·+ xi,d−1αd−1| ≥ c2‖x−i ‖−(d−1) ≥ c2X

−(d−1)i for i ≥ N.


Therefore, for any sufficiently large integer i ≥ N , we get

|xi,dξ| ≥ |xi,0 + xi,1α + · · ·+ xi,d−1αd−1| − |xi,0 + xi,1α + · · ·+ xi,d−1α

d−1 + xi,dξ|

≥ c2X−(d−1)i − c1X−τi+1 > 0

since τ > d− 1 and Xi < Xi+1. This means that xi,d 6= 0 when i is sufficiently large.

We deduce that there exists an integer N0 such that

x−i 6= 0, xi,d 6= 0 for each i ≥ N0.

Fix an index i ≥ N0. We have x−i 6= 0, xi,d 6= 0 and x−i+1 6= 0, xi+1,d 6= 0. Since

xi,xi+1 are linearly independent , we deduce that xi+1,dx−i − xi,dx−i+1 6= 0.

Set

Di = (xi+1,dx−i − xi,dx−i+1) · θ

−.

Then

Di = (xi+1,dxi,0 − xi,dxi+1,0) + · · ·+ (xi+1,dxi,d−1 − xi,dxi+1,d−1)αd−1

is a non-zero polynomial in α with integer coefficients of absolute value � XiXi+1.

Applying Liouville’s inequality, we then deduce that

|Di| � (Xi+1Xi)−(d−1).

On the other hand, we have

|Di| = |xi+1,d(xi · θ)− xi,d(xi+1 · θ)| ≤ TiXi+1 + Ti+1Xi � TiXi+1.

Therefore, we get

(Xi+1Xi)−(d−1) � TiXi+1 � X

−(τ−1)i+1 ,

using Lemma 3.2.2 (i), and so

Xτ−di+1 � Xd−1

i .

Since Xi+1 > Xi, this implies that τ − d ≤ d − 1, i.e., τ ≤ 2d − 1. In particular, if

τ > d, then Xi+1 � X(d−1)/(τ−d)i .


3.3 The set I

Fix an integer d ≥ 2 and a point θ = (θ0, . . . , θd) in Rd+1 with Q-linearly independent

coordinates. We fix a sequence of minimal points (xi)i∈N attached to Tθ with norms

Xi and Ti = Tθ(xi)(see section 2). Let τ > 1, and c > 0. Assume that the system|x0 + x1θ1 + · · ·+ xdθd| ≤ cX−τ

|x1|, . . . , |xd| ≤ X


Then it follows from Lemma 3.2.2 (ii) that the set

I = {i ∈ N; xi,xi+1,xi+2 are R-linearly independent }is infinite.

We endow I with the natural ordering of integers. Let i < j be consecutive ele-

ments in I. For each index t with i+1 ≤ t < j, the points xt,xt+1,xt+2 are R-linearly

dependent. Since any two distinct points of (xi)i∈N are R-linearly independent, for

such t, we find that

〈xt,xt+1〉R = 〈xt+1,xt+2〉R.

This means that

〈xi+1,xi+2〉R = · · · = 〈xj,xj+1〉R. (3.4)

Then, Lemma 3.2.1 gives that

〈xi+1,xi+2〉Z = · · · = 〈xj,xj+1〉Z.

On the other hand, it follows from (3.4) that

〈xi+1, . . . ,xj+1〉R = 〈x`,x`′〉R (3.5)

for any `, `′ ∈ N such that i+1 ≤ ` < `′ ≤ j+1. We deduce that xi, x`,x`′ are linearly

independent and so are x`,x`′ ,xj+2 for any `, `′ ∈ N such that i+ 1 ≤ ` < `′ ≤ j + 1.

The following lemma allows us to exploit the properties (3.5).

Lemma 3.3.1. Let y1,y2,y3,y4 be primitive points in Zd+1 and let η > 1 such that


(i) η‖y1‖ < ‖y2‖ ≤ ‖y3‖ < ‖y4‖,

(ii) |y1 · θ| > |y2 · θ| ≥ |y3 · θ| > |y4 · θ|,

(iii) 〈y1,y2〉Z = 〈y3,y4〉Z.

Then we have

‖y2‖|y1 · θ| �η ‖y4‖|y3 · θ|.

Proof. Since y1,y2 are primitive and have distinct norms, they are R-linearly inde-

pendent. So it follows from condition (iii) that

〈y1,y2〉R = 〈y3,y4〉R =: V

has dimension 2. Then there exists a point y ∈ V such that y · θ = 0. Since θ has

Q-linearly independent coordinates, we get y1 · θ 6= 0 and so y /∈ 〈y1〉R. This shows

that

V = 〈y,y1〉R.

For each t = 2, 3, 4, we write

yt = aty + bty1

for some at, bt ∈ R. Then we get

yt · θ = bt(y1 · θ),

and so

|bt| =|yt · θ||y1 · θ|

. (3.6)

Since |y1 · θ| > |yt · θ|, we get |bt| < 1 and thus ‖bty1‖ < ‖y1‖ < ‖yt‖. So we get(1− 1

η

)‖yt‖ ≤ ‖yt‖ − ‖y1‖ ≤ ‖aty‖ ≤ ‖yt‖+ ‖y1‖ ≤ 2‖yt‖

and therefore, we find

|at| �η‖yt‖‖y‖

. (3.7)

Since 〈y1, y2〉Z = 〈y3, y4〉Z, we have

‖y1 ∧ y2‖ = ‖y3 ∧ y4‖.


We also have

y1 ∧ y2 = y1 ∧ (a2y + b2y1) = a2(y1 ∧ y).

and

y3 ∧ y4 = −

∣∣∣∣∣a3 b3

a4 b4

∣∣∣∣∣ (y1 ∧ y)

This implies that

|a2| = |a3b4 − a4b3| ≤ |a3b4|+ |a4b3|.

Using (3.6) and (3.7), we deduce that

‖y2‖‖y1‖

�η‖y3‖‖y1‖

· |y4 · θ||y1 · θ|

+‖y4‖‖y1‖

· |y3 · θ||y1 · θ|

≤ 2‖y4‖‖y1‖

· |y3 · θ||y1 · θ|

,

which proves the required inequality.

The following lemma provides an estimate which can be applied to any triple of

linearly independent points of (xi)i∈N.

Lemma 3.3.2. Suppose that θ = (1, α, . . . , αd−1, ξ) where α is an algebraic number

of degree d and ξ /∈ Q(α). Let x,y, z be linearly independent points in Zd+1. Assume

that

‖x‖ ≤ ‖y‖ ≤ ‖z‖

|x · θ| ≥ |y · θ| ≥ |z · θ|

Then we have

‖x‖ � (‖x‖ · ‖y‖ · ‖z‖)d/2 |x · θ|.

Proof. Put M = 9(‖x‖ · ‖y‖ · ‖z‖)1/2. We consider the set

S =

{(a, b, c) ∈ Z3; 0 ≤ a ≤ M

3‖x‖, 0 ≤ b ≤ M

3‖y‖, 0 ≤ c ≤ M

3‖z‖

}.

Its cardinality is

|S| =(⌊

M

3‖x‖

⌋+ 1

)(⌊M

3‖y‖

⌋+ 1

)(⌊M

3‖z‖

⌋+ 1

)≥ M3

27‖x‖ · ‖y‖ · ‖z‖.


For each (a, b, c) ∈ S, we have

axd + byd + czd ∈ W := {−M,−M + 1, . . . ,M}.

Consider the map

f : S −→ W

(a, b, c) 7−→ axd + byd + czd.

Since M = 9(‖x‖ · ‖y‖ · ‖z‖)1/2, we have

|S| ≥ M3

27‖x‖ · ‖y‖ · ‖z‖= 3M > 2M + 1 = |W |.

We deduce that there exist at least two points of S, (a′, b′, c′) and (a′′, b′′, c′′), which

have the same image under f . Set

(a, b, c) = (a′, b′, c′)− (a′′, b′′, c′′) ∈ Z3 \ {0}.

Then we find

axd + byd + czd = 0. (3.8)

Since x,y, z are linearly independent, we observe that

ax− + by− + cz− 6= 0.

Set

P (T ) = (ax0 + by0 + cz0) + · · ·+ (axd−1 + byd−1 + czd−1)Td−1.

Then we get P ∈ Z[T ]≤d−1 \ {0}. Since 0 ≤ a′, a′′ ≤M/(3‖x‖), we obtain that

|a| = |a′ − a′′| ≤ M

3‖x‖.

Similarly, we also get

|b| ≤ M

3‖y‖, |c| ≤ M

3‖z‖.

This implies that

‖P‖ = ‖ax− + by− + cz−‖ ≤M.


Applying Liouville’s inequality to P (α), we deduce that

M−(d−1) � |P (α)|.

On the other hand, using (3.8), we get

|P (α)| = |(ax− + by− + cz−) · θ− + (axd + byd + czd)ξ|

= |a(x · θ) + b(y · θ) + c(z · θ)|

≤ M

3‖x‖|x · θ|+ M

3‖y‖|y · θ|+ M

3‖z‖|z · θ|

≤ M

‖x‖|x · θ|.

So we conclude that

M−(d−1) � M

‖x‖|x · θ|,

which leads to

‖x‖ �Md|x · θ| � (‖x‖ · ‖y‖ · ‖z‖)d/2|x · θ|.

Applying the above lemmas to points of our sequence (xi)i∈N, we obtain the fol-

lowing result which summarizes the crucial properties that we need for the proof of

Theorem 3.1.1.

Proposition 3.3.3. Suppose that θ = (1, α, . . . , αd−1, ξ) where α is an algebraic

number of degree d and ξ /∈ Q(α). Let i < j be two consecutive indices in I. Then we

have

(i) Xi+1 � (Xi+1Xi+2Xj+2)d/2Ti+1,

(ii) Xi+2Ti+1 � Xj+1Tj if τ > (3/2)d− 1.

Proof. Since i < j are consecutive indices in I, we have

〈xi+1,xi+2〉R = · · · = 〈xj,xj+1〉R := V (see (3.4)).

Moreover, xj,xj+1,xj+2 are R–linearly independent and thus xi+1,xi+2,xj+2 are R–

linearly independent. Applying Lemma 3.3.2 to xi+1,xi+2,xj+2, we get (i).


To prove assertion (ii), we first note that

V ∩ Zd+1 = 〈xi+1,xi+2〉Z = 〈xj,xj+1〉Z,

according to Lemma 3.2.1. Now, suppose that there exist infinitely many i ∈ I such

that

Xi+2 ≤ 2Xi+1.

Then, for such i, applying Lemma 3.3.2 to the R–linearly independent points xi,xi+1,xi+2

and using Xi < Xi+1, Ti ≤ X−τi+1, we get

1� Xd/2−1i X

d/2i+1X

d/2i+2Ti � X

−1−τ+(3/2)di+1 .

Since this holds for infinitely many i, we deduce that τ ≤ (3/2)d − 1. So if τ >

(3/2)d − 1, we get Xi+2 > 2Xi+1 for each sufficiently large i ∈ I. Then we may

apply Lemma 3.3.1 to xi+1,xi+2,xj,xj+1 with η = 2 and this yields the inequality in

(ii).

3.4 Proof of Theorem 3.1.1

Assume that τ > (3/2)d− 1. We will show that τ ≤ τd.

First of all, we fix a sequence of minimal points (xi)i∈N in Zd+1 attached to Tθ.

We know that the corresponding set I is infinite. Set

ρ = inf{r ≥ 1;Xi+2 ≤ Xri+1 for all sufficiently large i ∈ I}.

Lemma 3.2.3 gives

Xi+1 � X(d−1)/(τ−d)i for each i ∈ N.

So we find

1 ≤ ρ ≤ d− 1

τ − d. (3.9)

Now fix a real number ε with 0 < ε < ρ. Then by the definition of ρ, there exist

infinitely many i ∈ I such that

Xi+2 ≥ Xρ−εi+1 . (3.10)


Fix such an index i and let j be the next element in I. By the definition of ρ, we get

Xj+2 ≤ Xρ+εj+1 (3.11)

if i is large enough. Combining Proposition 3.3.3 with the above two inequalities, we

get

1� Xd/2−1i+1 X

d/2i+2X

d/2j+2 Ti+1

� X(d/2−1)(1/(ρ−ε))i+2 X

d/2−1i+2 X

d/2j+2 (Xi+2 Ti+1)

� X(d/2−1)(1/(ρ−ε)+1)i+2 X

d/2j+2 (Xj+1 Tj)

� X(d/2−1)(1/(ρ−ε)+1)i+2 X

(d/2)(ρ+ε)+1−τj+1 (using Tj ≤ X−τj+1),

thus

Xτ−1−(d/2)(ρ+ε)j+1 � X

(d/2−1)(1/(ρ−ε)+1)i+2 .

This holds for infinitely many i ∈ I. Since (Xi)i∈I is strictly increasing and since

j + 1 ≤ i+ 2, we deduce that

τ − 1− d

2(ρ+ ε) ≤

(d

2− 1

)(1

ρ− ε+ 1

).

Since ε can be chosen arbitrarily small, we conclude that

τ − 1− d

2ρ ≤

(d

2− 1

)(1

ρ+ 1

).

This implies that

τ − d

2≤ d

2ρ+

(d

2− 1

)1

ρ

Noting that ρ belongs to [1, (d− 1)/(τ −d)] and that the right hand side of the above

estimate is an increasing function of ρ on [1,∞), we get

τ − d

2≤ d(d− 1)

2(τ − d)+

(d

2− 1

)τ − dd− 1

.

Since τ > (3/2)d− 1 ≥ d, this yields

τ 2 − (d+ 1)τ − d2 + 3d− 1 ≤ 0,

which implies that τ ≤ τd.


3.5 Alternative approach using polynomials

Let α be an algebraic integer of degree d ≥ 3 and let ξ ∈ R \ Q(α). Set θ =

(1, α, . . . , αd−1, ξ). We fix a sequence of minimal points (xi)i∈N attached to Tθ and

set Xi = ‖xi‖ and Ti = Tθ(xi).

Recall that, in Section 3.3, we defined the set

I = {i ∈ N; xi,xi+1,xi+2 are linearly independent}.

We showed that this set is infinite if τ(θ) > 1 (Lemma 3.2.2). Moreover, by using

Dirichlet’s Box principle, we proved that

Xi+1 � (Xi+1Xi+2Xj+2)d/2Ti+1 (3.12)

for any consecutive elements i < j in I. This was the crucial estimate in the proof of

Theorem 3.1.1.

In this section, we construct non-zero polynomial maps from (Qd+1)3 to Q which

are defined over Z and do not simultaneously vanish on any triple of linearly indepen-

dent points ofQd×{0}. Looking at the values of these polynomials at (xi+1,xi+2,xj+2)

where i, j are consecutive elements of I, we will show that

1� Xd−2i+1 X

d−1i+2 Xj+2Ti+1.

Using this estimate instead of (3.12), we will then provide an alternative proof for

Theorem 3.1.1 in the case where d = 3.

To construct the polynomial maps, we first note that, for each j ∈ N, and each

y ∈ Qd × {0}, we have

αj(y · θ) ∈ Q(α).

Fix j ∈ N. Since {1, α, . . . , αd−1} is a basis of Q(α), there exists a unique point

yj ∈ Qd × {0} such that

αj(y · θ) = yj · θ.

In particular, if y ∈ Zd × {0}, then

αj(y · θ) ∈ 〈1, α, . . . , αd−1〉Z


since α is an algebraic integer, and so yj ∈ Zd × {0}.To estimate the norm of yj, we consider the map

Tj : Qd × {0} −→ Qd × {0}

y 7−→ yj.

Since this is a bijective linear map, there exist constants cj, c′j > 0 such that

c′j‖y‖ ≤ ‖yj‖ = ‖Tj(y)‖ ≤ cj‖y‖ for each y ∈ Qd × {0}.

Then, for any y ∈ Qd × {0} and any j ∈ {0, . . . , d− 1}, we have

c′‖y‖ ≤ ‖yj‖ ≤ c‖y‖

where c = max{c0, . . . , cd−1} and c′ = min{c′0, . . . , c′d−1} depend only on α.

We can now construct the desired polynomial maps. They are the determinants

given in the following proposition.

Proposition 3.5.1. Let x,y, z ∈ Zd+1 be linearly independent. Assume that the last

coordinates xd and yd of x and y (respectively) are not both zero. Set

E0 = xdy − ydx and Ei = Ti(E0) for i = 1, . . . , d− 1.

For each j = 0, 1, put

Dj =

{det(x,y, z, E1+j) if d = 3,

det(x,y, z, E1, . . . , Ed−3, Ed−2+j) if d > 3.

Then, D0 and D1 are not both zero. Moreover, if

‖x‖ ≤ ‖y‖ ≤ ‖z‖ and |x · θ| ≥ |y · θ| ≥ |z · θ|,

then we have

1 ≤ ‖x‖d−2‖y‖d−1‖z‖ |x · θ|. (3.13)


Proof. Set

V =

{〈x,y, z〉Q if d = 3,

〈x,y, z, E1, . . . , Ed−3〉Q if d > 3.

Then dimQ V ≤ d. Assume on contrary that D0 = D1 = 0. This implies that Ed−1

and Ed−2 are contained in V . By definition, we have

E0 ∈ 〈x,y〉Q ∩(Qd × {0}

)⊂ V ∩

(Qd × {0}

).

We conclude that

U := 〈E0, . . . Ed−1〉Q ⊂ V ∩(Qd × {0}

). (3.14)

Since x,y are linearly independent and xd, yd are not both zero, we get E0 6= 0. Note

that θ = (1, α, . . . , αd−1, ξ) has Q-linearly independent coordinates. This implies that

E0 · θ 6= 0, and moreover,

E0 · θ, α(E0 · θ), . . . , αd−1(E0 · θ)

are Q-linearly independent. As αi(E0 ·θ) = Ei ·θ for each i = 1, . . . , d−1, we deduce

that E0, . . . , Ed−1 are also linearly independent. Hence, we get dimQ U = d ≥ dimQ V .

By (3.14), this implies that

U = V = Qd × {0}.

Hence we get x,y ∈ Qd × {0}, which is impossible as xd and yd are not both zero.

This contradiction shows that D0 and D1 are not both 0.

Now assume that

‖x‖ ≤ ‖y‖ ≤ ‖z‖ and |x · θ| ≥ |y · θ| ≥ |z · θ|.

By definition, we have

Dj = det

(x · θ y · θ · · · Ed−2+j · θx+ y+ · · · E+

d−2+j

)(j = 0, 1).

By the construction, we get

‖Ei‖ = ‖Ti(E0)‖ � ‖E0‖ � ‖x‖‖y‖


and

|Ei · θ| = |αi(E0 · θ)| � ‖x‖ |y · θ|+ ‖y‖ |x · θ|

� ‖y‖ |x · θ|

for all i = 0, . . . , d− 1. Combining the above estimates, we find that

|Dj| � ‖x‖‖y‖‖z‖(‖x‖‖y‖)d−3(‖y‖|x · θ|) + |x · θ|‖y‖‖z‖(‖x‖‖y‖)d−2

� ‖x‖d−2‖y‖d−1‖z‖|x · θ|.

On the other hand, as α is an algebraic integer, we have Ei ∈ Zd+1for each i =

1, . . . , d− 1, and thus D0, D1 ∈ Z. Since D0 and D1 are not both zero, we get

1 ≤ max{|D0|, |D1|} � ‖x‖d−2‖y‖d−1‖z‖|x · θ|.

As we discussed in Section 3.3, if i < j are consecutive elements of I, then the

points xi+1,xi+2,xj+2 are linearly independent. On the other hand, if the hypothesis

of Lemma 3.2.3 are satisfied, then, for each sufficiently large index i, the last coordi-

nate of xi is not zero. Therefore, we deduce from the above proposition the following

result.

Corollary 3.5.2. Let c > 0 and τ > d− 1. Assume that the system|x0 + x1α + · · ·+ xd−1αd−1 + xdξ| ≤ cX−τ ,

|x1|, . . . , |xd| ≤ X


Then, for any consecutive indices i < j of I, we have

1� Xd−2i+1 X

d−1i+2 Xj+2Ti+1.

We now provide an alternative proof of Theorem 3.1.1 in the case where d = 3.

The argument is the same as the one of the proof in Section 3.4. The difference is

that we use the estimate given in Corollary 3.5.2 instead of Proposition 3.3.3 (i).


Alternative proof of Theorem 3.1.1 in the case where d = 3.

Let (xi)i∈N, I, ρ be as in the proof of Theorem 3.1.1 in Section 3.4 for d = 3. Assume

that τ > (3/2)d− 1 = 7/2. We show that τ ≤ τ3 = 2 +√

5. Fix a real number ε with

0 ≤ ε < ρ. As in the proof of Section 3.4, there exist infinitely many i ∈ I satisfying

(3.10). Fix such an index i and let j be the next element in I. Then, if i is sufficiently

large, the inequality (3.11) holds for this pair (i, j). We now apply Corollary 3.5.2 (in

place of Proposition 3.3.3 (i)) and get

1� Xi+1X2i+2Xj+2 Ti+1

= Xi+1Xi+2Xj+2 (Xi+2 Ti+1)

� X(1/(ρ−ε)+1)i+2 Xj+2 (Xj+1 Tj) by (3.10) and Proposition 3.3.3 (ii)

� X(1/(ρ−ε)+1)i+2 X

(ρ+ε)+1−τj+1 (using Tj ≤ X−τj+1 and (3.11)).

We conclude that

Xτ−1−ρ−εj+1 � X

1/(ρ+ε)+1i+2

holds for infinitely many i ∈ I for each 0 < ε < ρ. So we deduce that

τ − 1− ρ ≤ 1

ρ+ 1.

This implies that

τ ≤ 1

ρ+ ρ+ 2.

Since 1 ≤ ρ ≤ d−1τ−d = 2

τ−3 and since the right hand side of the above estimate is an

increasing function of ρ on [1,∞), we get

τ ≤ τ − 3

2+

2

τ − 3+ 2,

which implies τ ≤ τ3 = 2 +√

5.

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