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The Quarterly Journal of Mathematics
, Volume 69 (1) – Mar 1, 2018

30 pages

/lp/ou_press/the-norm-of-gaussian-periods-jPSgPn98Wd

- Publisher
- Oxford University Press
- Copyright
- © 2017. Published by Oxford University Press. All rights reserved. For permissions, please email: journals.permissions@oup.com
- ISSN
- 0033-5606
- eISSN
- 1464-3847
- D.O.I.
- 10.1093/qmath/hax028
- Publisher site
- See Article on Publisher Site

Abstract Gaussian periods are cyclotomic integers with a long history in number theory and connections to problems in combinatorics. We investigate the asymptotic behavior of the absolute norm of a Gaussian period and prove a case of Myerson’s Conjecture for periods of arbitrary odd length and provide a rate of convergence. Our method involves a result of Bombieri et al. on unlikely intersections in the algebraic torus as well as work of the author on the diophantine approximations to a set definable in an o-minimal structure. In the appendix, we make a result of Lawton on Mahler measures quantitative. 1. Introduction Let f≥1 be an integer and p a prime number. We are interested in the asymptotic behavior of the norm of ζa1+ζa2+⋯+ζafwhereζ=e2π−1/p (1)as a1,…,af∈Z and p vary. These cyclotomic integers appear naturally in algebraic number theory. We identify the Galois group of Q(ζ)/Q with Fp×=Fp⧹{0}. Let f be a divisor of p−1 and say a1,…,af represent the elements of the subgroup G⊆Fp× of order f. Then the sum (1) is the trace of ζ relative to the subfield of Q(ζ) fixed by the said subgroup and it is called a Gaussian period. It has degree k=[Fp×:G]=(p−1)/f over Q, we refer to Chapter 10.10 of Berndt et al. book [4] and Chapter 10 of Konyagin and Shparlinski’s book [14] for these and other facts. A Gaussian period can be expressed in terms of p and f for small values of k. Indeed, if k=1, then G=Fp× and the Gaussian period is of course ζ+ζ2+⋯+ζp−1=−1. Gauss evaluated the sum if k=2 and the minimal polynomial of a Gaussian period has been computed if k≤4. The absolute norm of a Gaussian period appears in combinatorial problems, cf. Myerson’s work [20, 21]. Let A1,…,Ak∈Fp× denote a complete set of representatives of Fp×/G. Then #{(x1,…,xk)∈Gk:A1x1+⋯+Akxk=0}−1pfk=p−1pΔ, (2)where Δ=∏t∈Fp×/G(∑g∈Gζtg);note that ζt is well defined for t∈Fp as is the sum. If A1x1+⋯+Akxk were to attain all values of Fp equally often, then Δ would vanish. As Myerson [20] observed, this linear form attains non-zero values equally often. It is tempting to interpret (2) as an error term. Note that the trivial estimate ∣∑g∈Gζtg∣≤f leads to the upper bound (1−p−1)fk for the modulus of (2), which exceeds p−1fk for all p≥3. This upper bound was improved by Konyagin and Shparlinski [14] who showed, for fixed f, that limsupp→∞1klog∣Δ∣ is at most the Mahler measure of a certain multivariate polynomial, see (10.4) loc. cit.; here the limits run over primes p with p≡1(modf). If f is an odd prime, we will see that the limit superior is at most 12logf and Δ is justifiably an error term. The aim of this paper is to investigate the asymptotics of the error term. We show that 1klog∣Δ∣ converges to a Mahler measure when the length f of the Gaussian period is a fixed odd number and p→∞. When well defined, the logarithmic absolute norm of the Gaussian period is 1klog∣Δ∣=1p−1∑t=1p−1log∣∑g∈Gζtg∣. (3)Before stating our first result, we survey what is known for small f=#G. Certainly, (3) vanishes if G is trivial. If f=2, then G={±1} and p is odd. Note that ζ+ζ−1=ζ−1(ζ2+1) is a unit in Q(ζ). So (3) is again zero. The value of Δ, that is its sign, can be computed using the Kronecker symbol. Already the case f=3 is more involved. It requires the logarithmic Mahler measure m(P)=∫01⋯∫01log∣P(e2π−1x1,…,e2π−1xn)∣dx1…dxn,of a non-zero Laurent polynomial P∈C[X1±1,…,Xn±1]; for the fact that this integral converges and other properties we refer to Chapter 3.4 in Schinzel’s book [26]. If p≡1(mod3), then Fp× contains an element θ of order 3. Myerson, cf. [21, Lemma 21], proved 1p−1∑t=1p−1log∣ζt+ζtθ+ζtθ2∣=m(1+X1+X2)+o(1) (4)as p→∞. The logarithmic Mahler measure of 1+X1+X2 was evaluated by Smyth [29] and equals L′(−1,χ), where χ is the non-trivial character modulo 3 and L(·,χ) is its associated Dirichlet L-function. Duke [9] gave a new proof of (4) which extended to a larger class of vectors in Fp3 containing the exponent vector (1,θ,θ2). Moreover, he provided a rate of convergence. Due to a fortunate factorization, the case f=4 is similar to order 2. Indeed, [21, Theorem 6] implies Δ=±1 if p≡1(mod4) and also determines the sign. So the limit in question is again zero. For higher order, the approaches in [9, 21] break down. But Myerson’s Conjecture [20] predicts convergence of (3) as p→∞ and the limit point. The full conjecture is more general as it also covers subgroups of Fq×, where q is a fixed power of p. Myerson’s Conjecture has an ergodic flavor. Indeed, using methods from ergodic theory, Lind et al. [17] proved convergence in the following setting. They suitably averaged the value of the logarithm of the modulus of a polynomial evaluated at a finite subgroup of roots of unity. Their polynomials are required to satisfy an intersection theoretical property with respect to the maximal compact subgroup of (C⧹{0})n; see the related property in work of Agler et al. [1]. Our 1+X1+⋯+Xn has this property for n≥2 and it is implicitly used in the proof of Lemma 5.1. Lind et al. computed the limit of this average for all equidistributed sequences of groups of roots of unity. In this paper, we concentrate on the special case when G has fixed odd order. We prove that (3) converges and compute the limit. Our method is based on a recent result [13] of the author on diophantine approximation of sets definable in a polynomially bounded o-minimal structure. It counts strong rational approximations to a definable set and is related to the Pila–Wilkie Counting Theorem [24]. Our approach is quantitative in the sense that it can provide a rate of convergence. At the moment, it is not clear to the author whether the methods from o-minimality can be made effective in our case; however, Gal Binyamini has recently announced progress towards an effective Pila–Wilkie Theorem. The following theorem is the special case of our main result when #G is an odd prime. We refer to Corollary 9.2 below for a more general statement. This corollary contains the case q=p of Myerson’s Conjecture. Theorem 1.1 Suppose f is an odd prime. For a prime p with p≡1(modf), let Gp⊆Fp×denote the subgroup of order f. Then 1p−1∑t=1p−1log∣∑g∈Gpe2π−1tgp∣=m(1+X1+⋯+Xf−1)+O(p−15(f−1)2), (5)as p→∞; in particular, the logarithm is well defined for all sufficiently large p. We have the estimate m(1+X1+⋯+Xf−1)≤12logf for all f≥2 by Corollary 6 in Chapter 3.4 [26]. This is asymptotically sharp by work of Myerson and Smyth [23, 29] who show that the Mahler measure is 12log(f)−γ2+o(1) as f→∞ and where γ is Euler’s constant. In fact, m(1+X1+⋯+Xf−1) is non-zero for all f≥3. This implies an amusing corollary of Theorem 1.1 on Gaussian periods that are units. Corollary 1.2 Suppose f is an odd prime. There are at most finitely many primes p with p≡1(modf)such that ∑g∈Ge2π−1gpis an algebraic unit, where G⊆Fp×denotes the subgroup of order f. Gaussian periods and their generalizations were investigated by Duke et al. [10] from several points of view. In Theorem 6.3, they prove that the Galois orbit of a Gaussian period becomes equidistributed in a suitable sense. Our average (3) involves the logarithm whose singularity at the origin often makes applying equidistribution directly impossible, see for example Autissier’s example [2]. Our main technical result is Theorem 9.1 below. It essentially amounts to a convergence result when averaging over groups of roots of unity of prime order. It is used to deduce the theorem and corollary above. The main difficulty when f=#G≥5 is that the integrand 1+X1+⋯+Xf−1in the logarithmic Mahler measure (5) has singularities along a positive dimensional real semi-algebraic set. Our approach requires new tools and we now give a brief overview of the proof of Theorem 1.1. As in Duke’s work [9], we start with a basic observation; to simplify notation, we set n=f−1. Let a=(a1,…,an)∈Zn, then ∏t=1p−1(1+ζta1+⋯+ζtan)is the product of Pa=1+Xa1+⋯+Xanevaluated at all roots of unity of order p. If an>an−1>⋯>a1>0, then Pa is a monic polynomial and the product above is the resultant of Pa and 1+X+⋯+Xp−1. If this resultant is non-zero, then by symmetry properties we find that (3) equals −log(n+1)p−1+1p−1∑i=1dlog∣αip−1∣, (6)where α1,…,αd are the roots of Pa. Comparing 1p−1log∣αip−1∣ with the local contribution log max{1,∣αi∣} of the logarithmic Mahler measure m(Pa) is a crucial aspect of the problem at hand; see Section 2 for details on the Mahler measure. Indeed, by a result of Lawton [16], the value m(Pa) converges towards the logarithmic Mahler measure of a multivariate polynomial as in Theorem 1.1 if ∣a∣→∞ for a in sufficiently general position. In a self-contained appendix, we provide a quantitative version of Lawton’s Theorem, see Theorem A.2. Baker’s theory on linear forms in logarithms yields a lower bound for non-zero values of ∣αip−1∣. But the current estimates are not strong enough to directly establish Theorem 1.1, see Duke’s comment after the proof of this [9, Theorem 3]. However, as we shall see below, strong lower bounds for ∣αip−1∣ are available if ∣αi∣≠1. Indeed, ∣αip−1∣≥∣∣αi∣p−1∣≥∣∣αi∣−1∣. If ∣αi∣≠1, we will use an old result of Mahler on the separation of distinct roots of a polynomials to bound ∣∣αi∣−1∣ from below. If αi lies in μ∞, the set of all roots of unity in C, then a sufficiently strong lower bound for ∣αip−1∣ follows from simpler considerations. This estimate gives us sufficient control on each term in the sum (6) subject to the condition that Pa does not have any root in S1⧹μ∞, here S1 is the unit circle in C. But it seems unreasonable to expect this hypothesis to hold for all a. To address this concern, we use symmetry in (3). Indeed, this mean is invariant under translating a by an element of pZn and also by replacing a by ta with t∈Z coprime to p. We exploit this symmetry by using a result of Bombieri et al. [6] on unlikely intersections. This in combination with Dirichlet’s theorem in diophantine approximation allows us to assume that Pa has no roots on S1⧹μ∞ after a suitable transformation as described above. Here the parity assumption on f in Corollary 9.2 below is used. At this point, we have a sufficient lower bound for each term of the sum (6). However, the method cannot proceed if too many terms are close to this bound. Duke [9] already uses the following basic principle. Suppose that for some α=αi, the distance ∣αp−1∣ is small, that is at most a fixed power of p−1. Then one can expect that α is close to some ζ∈μp, where μp is the set of roots of unity in C of order dividing p. As Pa(α)=0, we find that ∣1+ζa1+⋯+ζan∣ (7)is small. Myerson [22] proved a lower bound for non-vanishing sums of roots of unity if n=1,2 and 3. His estimates are polynomial in p−1 and are strong enough to imply Duke’s result. However, for fixed n≥4, only exponential bounds such as (n+1)−p are known to hold in general. They are not good enough for our purposes. If ∣αip−1∣ is small for many i, we are able to show that (7) is small for many pth roots of unity ζ. This situation can be analyzed using the following theorem that counts small sums of roots of unity of prime order. Its proof requires recent work of the author [13] on diophantine approximations on definable sets in an o-minimal structure. Theorem 1.3 Let n≥1. For all ϵ>0, there exist constants c=c(n,ϵ)≥1and λ=λ(n,ϵ)≥1with the following property. If p is a prime and ζ1,…,ζn∈μp, then #{t∈Fp:∣1+ζ1t+⋯+ζnt∣<c−1p−λ}≤cpϵ. (8) Igor Shparlinksi has pointed out to the author that 1+ζ1+⋯+ζn≠0 for all ζ1,…,ζn∈μp if p is large in terms of n; indeed, this follows from Mann’s [19, Theorem 1]. As this paper was being finished up, Dimitrov [8] announced an extension of Lind et al.’s work to arbitrary integral polynomials but Cartesian power subgroups. His approach used ideas from diophantine approximation and is independent from ours. We hope to expand the connection between counting points approximating a definable set and questions related to ergodic theory in future work. The author thanks Igor Shparlinksi and the anonymous referee for helpful comments. 2. Notation The supremum norm on Rn is ∣·∣ for any n≥1. We have already seen the definition of the logarithmic Mahler measure m(P) if P∈C[X1±1,…,Xn±1]⧹{0}. The Mahler measure of P is M(P)=em(P). The absolute logarithmic Weil height, or just height, of an algebraic number α with minimal polynomial P in Z[X] and leading term p0≥1 is h(α)=1[Q(α):Q]m(P)=1[Q(α):Q]log(p0∏z∈CP(z)=0max{1,∣z∣}),where the second equality follows from Jensen’s Formula. We write H(α)=eh(α). Moreover, we set H(α1,…,αn)=max{H(α1),…,H(αn)}, if α1,…,αn are algebraic. 3. Algebraic numbers close to the unit circle An algebraic number α∈C⧹{1} of degree D=[Q(α):Q] can be bounded away from 1 using Liouville’s inequality, Theorem 1.5.21 [5] log∣α−1∣≥−Dlog2−Dh(α).The modulus ∣α∣=αα¯ is again an algebraic number, here and below ·- denotes complex conjugation. Its height satisfies h(∣α∣)≤12h(αα¯)≤h(α),since h(α¯)=h(α). If α is real, then clearly Q(∣α∣)=Q(α). However, for D≥2, we only have [Q(∣α∣):Q]≤D(D−1) and equality is possible. So Liouville’s inequality applied to ∣α∣ gives log∣∣α∣−1∣≥−D(D−1)log2−D(D−1)h(α),if ∣α∣≠1. We will use a result of Mahler to improve on the dependency in D in front of log2. Theorem 3.1 (Mahler) Let P∈Z[X]be a polynomial with D=degP≥2. If z,z′∈Care distinct roots of P, then ∣z′−z∣>3D−(D+2)/2M(P)−(D−1). Proof We may assume that P has no multiple roots over C after replacing it by its squarefree part. The estimate then follows from [18, Theorem 2] as the absolute value of the new discriminant is at least 1.□ Lemma 3.2 Let α∈Cbe an algebraic number of degree D=[Q(α):Q]. If ∣α∣≠1, then log∣αp−1∣≥log∣∣α∣−1∣≥−1−(D+1)log(2D)−2(2D−1)Dh(α),for all integers p≥1. Proof The first inequality follows from ∣αp−1∣≥∣∣α∣p−1∣=∣∣α∣−1∣·∣∣α∣p−1+⋯+1∣≥∣∣α∣−1∣. To prove the second inequality, we may assume ∣α∣≥1/2, in particular α≠0. Let P∈Z[X] denote the minimal polynomial of α. We will apply Mahler’s Theorem to F=P(X)P(1/X)XD∈Z[X]. Observe that F(α)=F(α¯−1)=0 and degF=2D. Therefore, ∣α−α¯−1∣>3(2D)−(2D+2)/2M(F)−(2D−1) since ∣α∣≠1. As M(P(1/X)XD)=M(P) and since the Mahler measure is multiplicative, we find, after multiplying with ∣α∣=∣α¯∣, that ∣∣α∣2−1∣>3∣α∣(2D)−(D+1)M(P)−2(2D−1).Observe that logM(P)=Dh(α) and ∣∣α∣−1∣=∣∣α∣2−1∣/(∣α∣+1). Therefore, ∣∣α∣−1∣>3∣α∣∣α∣+1(2D)−(D+1)M(P)−2(2D−1)≥33(2D)−(D+1)e−2(2D−1)Dh(α)using ∣α∣≥1/2. We conclude the proof by taking the logarithm.□ 4. A first estimate Let n≥1 and p≥2 be integers. For a=(a1,…,an)∈Zn, we define Δp(a)=∏t=1p−1∣1+ζta1+⋯+ζtan∣whereζ=e2π−1/p. (9)If p is a prime, then Δp(a) is the Q(ζ)/Q norm of the cyclotomic integer 1+ζa1+⋯+ζan up-to sign. We attach to a the lacunary Laurent polynomial Pa=1+Xa1+⋯+Xan∈Z[X±1]. (10)Say e=max{0,−a1,…,−an}≥0, then XePa=Xe+Xe+a1+⋯+Xe+an (11)is a polynomial with integral coefficients, non-zero constant term, and degree d=max{ai−aj:0≤i,j≤n}≤2∣a∣,where a0=0. We write α1,…,αd∈C for the zeros of XePa with multiplicity, that is XePa=p0(X−α1)⋯(X−αd), where p0≥1 is the leading term of Pa. We note that αi≠0,1 for all i. Our goal in this section is to bound ∣1p−1logΔp(a)−m(Pa)∣ (12)from above, here m(Pa)=m(XePa)=logp0+∑i=1dlog max{1,∣αi∣} is the logarithmic Mahler measure of Pa. As Δp(a) is essentially a resultant, we can rewrite it as a product over the roots αi. This will allow us to express the difference (12) in terms of these roots. In the next four lemmas, we obtain several statements on the roots αi in terms of a∈Zn. Lemma 4.1 We have Δp(a)=p0p(n+1)−1∏i=1d∣αip−1∣. Proof A variant of this calculation can also be found in the proof of Duke’s [9, Theorem 3]. We have Δp(a)=∏t=1p−1∣Pa(ζt)∣=p0p−1∏i=1d(∏t=1p−1∣ζt−αi∣)=p0p−1∏i=1d∣1−αip1−αi∣.The lemma follows since p0∏i=1d(1−αi)=Pa(1)=n+1.□ Each αi is an algebraic number with Di=[Q(αi):Q]≤d whose height is bounded by the next lemma. Lemma 4.2 Let i∈{1,…,d}, then Dih(αi)≤m(Pa)≤log(n+1). Proof Recall that αi is a root of the polynomial in (11) which, by the Gauss Lemma, is divisible by the minimal polynomial Q in Z[X] of αi. So Dih(αi)=m(Q)≤m(XePa)=m(Pa) as the logarithmic Mahler measure is additive and non-negative on Z[X]⧹{0}. By Corollary 6, Chapter 3.4 [26], the Mahler measure M(Pa) is at most the euclidean norm of the coefficient vector of Pa. This gives m(Pa)≤log(n+1).□ We now come to a lower bound for ∣αip−1∣ which is independent of p under the assumption that αi lies off the unit circle, or is a root of unity of order not divisible by p. Lemma 4.3 Suppose a≠0, let i∈{1,…,d}, and let p≥1be an integer: If ∣αi∣≠1, then log∣αip−1∣≥−18log(n+1)∣a∣log(2∣a∣). If αiis a root of unity and αip≠1, then log∣αip−1∣≥−2log(2∣a∣). Proof Observe that ∣αi∣≠1 implies n≥2. According to Lemmas 3.2 and 4.2, we have log∣αip−1∣≥−1−(Di+1)log(2Di)−2(2Di−1)log(n+1)≥−(Di+1)log(2Di)−4Dilog(n+1).Now Di≤2∣a∣ by (11). The first part of the lemma follows from (2∣a∣+1)log(4∣a∣)+8∣a∣log(n+1)≤18log(n+1)∣a∣log(2∣a∣). The second part is more elementary. Let m≥2 be the multiplicative order of αip. If m≥3, then ∣αip−1∣≥sin(2π/m)≥2/m. The bound ∣αip−1∣≥2/m certainly also holds for m=2. It is well known that Euler’s totient function φ satisfies φ(m)≥m/2. As φ(m)=[Q(αip):Q]≤Di we find m≤2Di2. Hence log∣αip−1∣≥−2logDi≥−2log(2∣a∣), as Di≤2∣a∣.□ This last lemma allows us to compare log∣αip−1∣ with the corresponding contribution plog max{1,∣αi∣} in the logarithmic Mahler measure. Lemma 4.4 Suppose a≠0, let i∈{1,…,d}, and let p≥1be an integer: If ∣αi∣≠1, then ∣log∣αip−1∣−plogmax{1,∣αi∣}∣≤18log(n+1)∣a∣log(2∣a∣). (13) If αiis a root of unity and αip≠1, then ∣log∣αip−1∣∣≤2log(2∣a∣). Proof For the proof of part (i), let us first assume ∣αi∣<1. Then ∣αip−1∣≤2 and Lemma 4.3(i) yields ∣log∣αip−1∣∣≤18log(n+1)∣a∣log(2∣a∣), as desired. If ∣αi∣>1, we use that αi−1 is a root of P−a. We obtain the same bound as before for ∣log∣αi−p−1∣∣=∣log∣αip−1∣−plog∣αi∣∣ and this completes the proof of (i). To prove (ii) we argue as in the case ∣αi∣<1 above but use Lemma 4.3(ii).□ Suppose for the moment that all αi satisfy ∣αi∣≠1 and for the sake of simplicity also p0=1. By Lemma 4.1, the bound given in part (i) of the last lemma leads to the bound d∣a∣log(2∣a∣)p≤2∣a∣2log(2∣a∣)p (14)for (12) up-to a factor depending only on n. However, this estimate is not strong enough for our aims due to the contribution ∣a∣2/p. To remedy this, we begin by splitting up the roots αi into two parts depending on a parameter λ≥1. The first part B=B(p,a,λ)={i:∣αi∣<1and∣αip−1∣<p−λ}∪{i:∣αi∣>1and∣αi−p−1∣<p−λ} (15)corresponds to those roots whose pth power is excessively close to 1. The second part is the complement {1,…,d}⧹B.Later, we will bound the cardinality of B. Proposition 4.5 Let λ≥1, let a∈Zn⧹{0}, and let p≥1be an integer satisfying ∣a∣≤p. Suppose that the only roots of Pathat lie on the unit circle are roots of unity of order not dividing p. Then Δp(a)≠0and 1p−1logΔp(a)=m(Pa)+O(∣a∣log(2p)p(λ+#B)), (16)where the implied constant depends only on n. Proof We recall Lemma 4.1, it implies Δp(a)≠0 under the given circumstances. Let α1,…,αd be the roots of Pa with multiplicities, as above. Say first i∈{1,…,d}⧹B. If ∣αi∣<1, then ∣αip−1∣≥p−λ. Thus ∣log∣αip−1∣∣≤λlog(2p) since λlog(2p)≥log2. For ∣αi∣>1, we proceed similarly and obtain ∣log∣αip−1∣−plog∣αi∣∣≤λlog(2p). If ∣αi∣=1, then by hypothesis αi is a root of unity whose order does not divide p, so Lemma 4.4(ii) implies ∣log∣αip−1∣∣≤2log(2∣a∣)≤2log(2p). We recall d≤2∣a∣ and sum over {1,…,d}⧹B to find ∑i=1i∉Bd∣log∣αip−1∣−plogmax{1,∣αi∣}∣≤(d−#B)max{λlog(2p),2log(2p)}≤4∣a∣log(2p)λ. (17) If i∈B, then ∣αi∣≠1. We apply Lemma 4.4(i) and use ∣a∣≤p. The bound (13) holds for #B roots and, therefore, ∑i∈B∣log∣αip−1∣−plogmax{1,∣αi∣}∣≤18log(n+1)∣a∣log(2p)#B.We combine this bound with (17) to obtain ∑i=1d∣log∣αip−1∣−plogmax{1,∣αi∣}∣≤18log(n+1)∣a∣log(2p)(λ+#B). (18) The logarithmic Mahler measure of Pa is logp0+∑i=1dlog max{1,∣αi∣}, where p0≥1 is the leading term of Pa. We use Lemma 4.1 and apply the triangle inequality to find that ∣1p−1logΔp(a)−m(Pa)∣ is at most 1p−1log(p0pn+1)−(p−1)logp0+1p−1∑i=1d∣log∣αip−1∣−(p−1)logmax{1,∣αi∣}∣≤log(n+1)p−1+18log(n+1)∣a∣log(2p)p−1(λ+#B)+1p−1m(Pa),where we used (18). From Lemma 4.2, we deduce m(Pa)≤log(n+1). So (16) holds true.□ If we ignore for the moment all logarithmic contributions, then we have traded in d in (14) for λ+#B in (16). Before continuing, we make two elementary, but important, observations on symmetry properties of Δp(a) when p is a prime. Lemma 4.6 Let p be a prime and let a∈Znbe arbitrary: If a′∈Znwith a≡a′(modp), then Δp(a)=Δp(a′). If t∈Zwith p∤t, then Δp(a)=Δp(ta). Proof Part (i) follows using the definition (9) and ζp=1. For part (ii), observe that ζ↦ζt permutes the factors in (9) since p∤t.□ Using this lemma, we will transform a′=ta−b with p∤t and b∈pZn such that ∣a′∣ is small compared with p. This will be done using Dirichlet’s Theorem from diophantine approximation. A theorem of Bombieri–Masser–Zannier leads to a criterion that rules out that Pa′ has roots on the unit circle of infinite order. This opens the door to applying the previous proposition. In a final step, we will need a strong upper bound for B for a sufficiently large but fixed λ. This is where counting rational points close to a definable sets comes into play. 5. Lacunary polynomials with roots off the unit circle Say n≥2. In this section, we investigate a condition on a∈Zn such that Pa, as defined in (10), does not vanish at any point of S1⧹μ∞. It will prove useful to restrict a to a subgroup Ω⊆Zn and we will introduce a condition on Ω that ensures that Pa does not have any roots in S1⧹μ∞ for a∈Ω in general position. Our condition is based on a theorem of Bombieri et al. [6] on unlikely intersections in the algebraic torus. We also make use of (very rudimentary) tropical geometry. Say K=⋃m≥1C((T1/m))is the field of Puiseux series over C. It is algebraically closed and equipped with a valuation ord:K→Q∪{∞}, where ord(T)=1. Let X be an irreducible subvariety defined over K of the algebraic torus Gmn. The tropical variety Trop(X) of X is the closure in Rn of {(ord(x1),…,ord(xn)):(x1,…,xn)∈X(K)}. We need the following two basic facts. First, if Y⊆X is an irreducible subvariety defined over K, then Trop(Y)⊆Trop(X); this follows directly from the definition. Secondly, if r=dimX≥1 and after permuting coordinates, the projection of Trop(X) to the first r coordinates of Rn contains Qr. We prove this using basic algebraic geometry. After permuting coordinates, the projection π:X(K)→Gmr(K) onto the first r coordinates contains a Zariski open and dense subset of Gmr. There exists a polynomial P∈K[X1±1,…,Xr±1]⧹{0} such that any point of Gmr(K) outside of the zero locus of P lies in the image of π. Let (a1,…,an)∈Qr be arbitrary. For sufficiently general (c1,…,cr)∈Gmr(C), the polynomial P does not vanish at (c1Ta1,…,crTar)∈Gmr(K). This point has a pre-image under π in X(K) and the valuation of its first r coordinates are a1,…,ar. This yields our claim. Einsiedler et al.’s [11] Theorem 2.2.5 on the structure of Trop(X) can be used instead of this second property in the proof of Lemma 5.1 below. We write ⟨·,·⟩ for the standard scalar product on Rn. If Ω is a subgroup of Zn, then we set Ω⊥={a∈Zn:⟨a,ω⟩=0forallω∈Ω}. We write e1,…,en for the standard basis elements of Rn augmented by e0=0. Lemma 5.1 Let Ω⊆Znbe a subgroup of rank m≥2for which the following property holds. If (α,β)∈Z2⧹{0}and if i,j,k,l∈{0,…,n}are pairwise distinct with v=α(ei−ej)+β(ek−el), then Ω⊄(vZ)⊥. Then there exist finitely many subgroups Ω1,…,ΩN⊆Ωof rank at most m−1such that if a∈Ω⧹⋃i=1NΩi, then Padoes not vanish at any point of S1⧹μ∞. Proof We consider an irreducible component X⊆Gmn of the zero set of 1+X1+⋯+Xnand1+X1−1+⋯+Xn−1. (19)Then dimX=n−2 as these two polynomials are coprime in C[X1±1,…,Xn±1]. In particular, 1+X1+⋯+Xn is atoral as in [17], cf. Lemma 2.6 loc. cit. Say a=(a1,…,an)∈Ω such that Pa has a root z∈S1⧹μ∞. Then 1+za1+⋯+zan=0 by definition and after applying complex conjugation, we find 1+z−a1+⋯+z−an=0.So za=(za1,…,zan)∈X(C) for one of the irreducible components above. But za also lies in an algebraic subgroup of Gmn of dimension 1. According to Bombieri et al. [6, Theorem 1.7], there are two cases. Either za lies in a finite set that depends only on X, and hence only on n, or za∈H(C), where H⊆Gmn is an irreducible component of an algebraic subgroup with dimzaX∩H≥max{1,dimX+dimH−n+1}=max{1,dimH−1}.Moreover, in the second case, H comes from a finite set that depends only on n, cf. the first paragraph of the proof of [6, Theorem 1.7, p. 26]. In the first case, we claim that a must lie in one of the finitely many subgroups of Ω of rank 1≤m−1. Indeed, we may assume a≠0. If z′∈S1⧹μ∞ is a root of some Pa′ with a′∈Ω and za=z′a′, then a and a′ are linearly dependent as z′ is not a root of unity. Our claim follows as there are only finitely many possible za in this case. We add these rank 1 subgroups to our collection of Ωi. In the second case, there is an irreducible component Y⊆X∩H of positive dimension at least dimH−1 that contains za. In this case, n≥3. We recall that algebraic subgroups of Gmn are in bijection with subgroups of Zn, see [5, Theorem 3.2.19] for details. Let Λ⊆Zn be a subgroup from a finite set depending only on n with rank r=n−dimH such that H is contained in the algebraic subgroup defined by all X1b1⋯Xnbn−1where(b1,…,bn)∈Λ. (20)Hence z⟨a,b⟩=1 for all b∈Λ. As z is not a root of unity, we find a∈Λ⊥. We would like to add Ω∩Λ⊥ to our list of Ωi. However, we must ensure that its rank is at most m−1. Once this is done, our proof is complete. Suppose the rank does not drop, then [Ω:Ω∩Λ⊥]Ω⊆Λ⊥ and hence Ω⊆Λ⊥ because Λ⊥ is primitive. We now derive a contradiction from this situation by analyzing Λ. Since all monomials (20) vanish on H, we find that Trop(H) lies in Λ⊥R, the vector subspace of Rn generated by Λ⊥. We also need to study Trop(X). Say (x1,…,xn)∈X(C). Since it is a zero of the first polynomial in (19) and by the ultrametric triangle inequality the minimum among 0,ord(x1),…,ord(xn) is attained twice. The same argument applied to the second polynomial in (19) shows that the maximum is attained twice. Hence Trop(X) is contained in the finite union of the codimension 2 vector subspaces of Rn defined by relations ord(xi)=ord(xj),ord(xk)=ord(xl)with#{i,j,k,l}=4and ord(xi)=ord(xj),ord(xk)=0with#{i,j,k}=3. By the discussion before this lemma, Trop(Y)⊆Trop(X)∩Trop(H)⊆Trop(X)∩Λ⊥R. Moreover, the projection of Trop(Y) to some choice of dimY distinct coordinates of Rn contains QdimY. So Λ⊥R intersected with one of the codimension 2 subspaces mentioned above must have dimension at least dimY≥dimH−1=n−r−1. Therefore, there exists (α,β)∈Z2⧹{0} and pairwise distinct i,j,k,l∈{0,…,n} with v=α(ei−ej)+β(ek−el)∈Λ. Recall that Ω⊆Λ⊥. So Ω lies in the orthogonal complement of v, which contradicts the hypothesis of the lemma.□ Suppose n≥2, let Ω⊆Zn be a subgroup of rank at least 2, and let a∈Zn. For a prime p we define ρp(a;Ω)=inf{∣ω∣:ω∈Ω⧹{0}and⟨ω,a⟩≡0(modp)}, (21)this is a well-defined real number as the set is non-empty. Proposition 5.2 Let Ω⊆Znbe a subgroup of rank m≥2that satisfies the following hypothesis. If (α,β)∈Z2⧹{0}and if i,j,k,l∈{0,…,n}are pairwise distinct with v=α(ei−ej)+β(ek−el), then Ω⊄(vZ)⊥. Then there exists a constant c=c(Ω)≥1with the following property. Say a∈Ωand let p be a prime with ρp(a;Ω)≥c. Then there exist t∈Zwith p∤tand ω∈Ωsuch that a′=ta−pω≠0, we have ∣a′∣≤cp1−1/m, and the Laurent polynomial Pa′ ∈Z[X±1]does not vanish at any point of S1⧹μ∞. Proof The proposition follows from combining Lemma 5.1 with Dirichlet’s Theorem from diophantine approximation. Indeed, we let Ω1,…,ΩN be the subgroups of Ω from this lemma. If N=0 we set Ω1={0}. We will see how to choose c below. We fix a basis (ω1,…,ωm) of the abelian group Ω. Then a=ν1ω1+⋯+νmωm, where ν1,…,νm∈R are unique. If p≥3, then Dirichlet’s Theorem, cf. [27, Theorem 1B], applied to ν1/p,…,νm/p, and p−1>1 yields t,ν1′,…,νm′∈Z such that 1≤t≤p−1 and ∣tνi/p−νi′∣≤(p−1)−1/m. The same conclusion holds for p=2. We set ω=∑i=1mνi′ωi∈Ω. Then ∣a′/p∣≤(∣ω1∣+⋯+∣ωm∣)(p−1)−1/m≤cp−1/m, where a′=ta−pω for c large enough. This yields part (i). As each Ωi has rank at most m−1 we have Ω∩(Ωi)⊥≠0 for all i. For each i, we fix a non-zero ω*∈Ω∩(Ωi)⊥ of minimal norm. If ⟨ω*,ta−pω⟩=0, then ⟨ω*,a⟩≡0(modp), since p∤t. So ∣ω*∣≥ρp(a;Ω)≥c by hypothesis. We can avoid this outcome by fixing c large in terms of the Ω∩(Ωi)⊥. Thus, ⟨ω*,a′⟩≠0. This implies a′∉Ωi for all 1≤i≤N and in particular a′≠0. Part (ii) follows from the conclusion of Lemma 5.1.□ 6. Rational points close to a definable set In this section, we prove Theorem 1.3. To do this, we temporarily adopt the language of o-minimal structures. Our main reference is van den Dries’ book [30] and his paper with Miller [31]. We work exclusively with the o-minimal structure Ran of restricted analytic functions. It contains the graph of any function [0,1]n→R that is the restriction of an analytic function Rn→R. The main technical tool in this section is a result of the author [13] which we cite in a special case below. We retain much of the notation used in the said reference. Roughly speaking, the result gives an upper bound for the number of rational points of bounded height that are close to a subset of Rn that is definable in Ran. Note that Ran is a polynomially bounded o-minimal structure as required by this reference. For any subset Z⊆Rn, we write Zalg for the union of all connected, semi-algebraic sets that are contained completely in Z. For ϵ>0, we define N(Z,ϵ) to be the set of y∈Rn for which ∣x−y∣<ϵ for some x∈Z. We recall that the height H(·) was defined in Section 2. Theorem 6.1 [13, Theorem 2] Let Z⊆Rnbe a closed set that is definable in Ranand let ϵ>0. There exist c=c(Z,ϵ)≥1and θ=θ(Z,ϵ)∈(0,1]such that if λ≥θ−1then #{q∈Qn⧹N(Zalg,T−θλ):H(q)≤Tandthereisx∈Zwith∣x−q∣<T−λ}≤cTϵfor all T≥1. Proof of Theorem 1.3 For n=1, the theorem follows with λ=1 and taking c sufficiently large. Our proof is by induction on n and we suppose n≥2. We will choose c≥1 and λ≥1 in terms of n and ϵ during the argument. Let Z={(x1,…,xn)∈[0,1]n:1+e2π−1x1+⋯+e2π−1xn=0},which is compact and definable in Ran. Let us write ζj=e2π−1qj for j∈{1,…,n} with qj∈1pZ∩[0,1). Say t is as in the set (8). For convenience, we identify it with its representative in {0,…,p−1}. Then ∣1+ζ1t+⋯+ζnt∣<c−1p−λ, (ζ1,…,ζn) has precise order p, and t≠0. We claim that q˜t=(tq1−⌊tq1⌋,…,tqn−⌊tqn⌋)∈1pZn∩[0,1)n lies close to Z. Indeed, a suitable version of Łojasiewicz’s inequality, see 4.14.(2) [31], implies that the distance of this point to Z is at most c1∣1+e2π−1tq1+⋯+e2π−1tqn∣δ≤c1c−δp−λδ≤c1c−δ,where c1>0 and δ>0 depend only on Z. We may assume c1c−δ<1, so ∣q˜t−x∣<p−λδforsomex∈Z. (22) The vectors q˜0,…,q˜p−1 are pairwise distinct. So it is enough to bound the number of q˜t with (22). If λ is sufficiently large, then λδ≥θ−1 where θ is provided by Theorem 6.1 applied to Z and ϵ. So there are at most cpϵ many q˜t that are not in the p−θλδ-tube around Zalg. To prove (8), we need to consider only those t with ∣q˜t−x′∣<p−θλδ for some x′=(x1′,…,xn′)∈Zalg. The algebraic locus Zalg is well understood; see Ax’s work [3] and how it is applied for example in the proof of [13, Theorem 7]. There exists ∅≠J⊊{1,…,n} such that 1+∑j∈Je2π−1xj′=0. We subtract this from the partial sum over the coordinates of t(q1,…,qn) and get 1+∑j∈Jζjt=∑j∈J(e2π−1tqj−e2π−1xj′).Thus ∣1+∑j∈Jζjt∣≤∑j∈J∣e2π−1(tqj−⌊tqj⌋)−e2π−1xj′∣≤2π(n−1)∣q˜t−x′∣.Hence, there is a constant c2>0 depending only on n with ∣1+∑j∈Jζjt∣<c2p−θλδ. Recall #J≤n−1. Now suppose c′>0 and λ′>0 are the constants from this theorem applied by induction to the terms in J. We are free to assume that λ satisfies 2θλδ/2≥c2c′ and θλδ≥2λ′. Then c2p−θλδ≤c22−θλδ/2p−θλδ/2≤c′−1p−λ′ as p≥2. So ∣1+∑j∈Jζjt∣<c′−1p−λ′.By induction, there are at most cpϵ possibilities for t.□ 7. Counting small sums of roots of unity Let Pa be a lacunary Laurent polynomial as in (10) with a∈Zn, where n≥1. The goal of this section is to bound the number of roots α of Pa coming from B=B(p,a,λ) defined in (15). If αp is close to 1, then α is close to a root of unity ζ with ζp=1. So ∣Pa(ζ)∣ will be small. Thus, 1+ζa1+⋯+ζan is small in modulus, where a=(a1,…,an). This is where the counting result proved in Section 6 comes into play. We make the first part of this approach precise in the next lemma. Lemma 7.1 Suppose α=re2π−1ϑwith r∈(0,1]and ϑ∈[0,1). If p≥2is an integer with ∣αp−1∣≤1/2, then there exists t∈{0,…,p}such that ∣ϑ−tp∣≤142∣αp−1∣. (23)If in addition a∈Znand Pa(α)=0, then ∣Pa(e2π−1t/p)∣≤5n∣a∣∣αp−1∣. (24) Proof Observe that ∣z−1∣≥∣z∣1/2∣z/∣z∣−1∣ for all z∈C⧹{0}, see [25, Lemma 11.6.1]. We substitute z=αp to find ∣αp−1∣≥rp/2∣e2π−1ϑp−1∣. Now 1−rp≤∣rp−1∣≤∣αp−1∣, so rp≥1−∣αp−1∣≥1/2 by hypothesis. We find ∣e2π−1ϑp−1∣≤2∣αp−1∣. (25) Let t be an integer with ∣ϑp−t∣≤1/2. Then t∈{0,…,p} as ϑ∈[0,1). So ∣e2π−1ϑp−1∣=∣e2π−1(ϑp−t)−1∣≥4∣ϑp−t∣ (26)by elementary geometry. Combining (25) with (26) and dividing by p≥2 yields (23). To prove the second claim, we set ξ=e2π−1t/p and estimate ∣α−ξ∣=∣re2π−1ϑ−ξ∣≤∣r−1∣+∣e2π−1ϑ−ξ∣=∣r−1∣+∣e2π−1(ϑ−t/p)−1∣≤∣αp−1∣+2π∣ϑ−t/p∣≤(1+π/8)∣αp−1∣, (27)where we used ∣r−1∣≤∣rp−1∣≤∣αp−1∣. We fix e∈Z such that XePa(X) is a polynomial with non-zero constant term. Then ∣Pa(ξ)∣=∣ξePa(ξ)−αePa(α)∣≤∑k=0n∣ξak+e−αak+e∣here a=(a1,…,an) and a0=0. Some ak+e vanishes and we use ∣ξ∣=1 and ∣α∣≤1 to find ∣Pa(ξ)∣≤nmax0≤k≤n{ak+e}∣ξ−α∣≤2n∣a∣∣ξ−α∣.We recall (27) to obtain (24).□ Next we show that many elements in B will lead to many different roots of unity as given by the lemma above. The reason for this is that roots of lacunary polynomials are nearly angularly equidistributed by a result of Hayman, known already to Biernacki. Lemma 7.2 Let a∈Znand suppose p≥2is an integer with ∣a∣≤p. If λ≥1, then #B≤12n#{ζ∈μp:∣Pa(ζ)∣<5n∣a∣p−λ}. Proof We may assume a≠0. Let us partition B=B(p,a,λ) into B<1={i∈B:∣αi∣<1} and B>1={i∈B:∣αi∣>1}. We construct a map ψ<1:B<1→{0,…,p}in the following manner. If i∈B<1, then αi=∣αi∣e2π−1ϑ for ϑ∈[0,1), which depends on i, and we have ∣αip−1∣<p−λ≤1/2. Lemma 7.1 yields t∈{0,…,p} with ∣ϑ−t/p∣≤∣αip−1∣/32<p−λ/32. We set ψ(i)=t. We define ψ>1:B>1→{0,…,p} in the same spirit. For if i∈B>1, then ∣αi∣>1 and there is ϑ∈[0,1) such that αi=∣αi∣e−2π−1ϑ. We apply the said lemma to αi−1 and P−a to obtain ψ(i)∈{0,…,p} with ∣ϑ−ψ(i)/p∣<p−λ/32. For i∈B<1, we get ∣Pa(ξψ(i))∣<5n∣a∣p−λwith ξ=e2π−1/p. If i∈B>1, then the same bound holds for ∣P−a(ξψ(i))∣=∣Pa(ξ−ψ(i))∣. Now #B≤2#B<1 or #B≤2#B>1. We assume the former, the latter case is dealt with similarly. Let us fix e∈Z such that Q=XeP(X) is a polynomial with non-zero constant term. Then degQ≤2∣a∣ and Q has at most n+1 terms. Elements of B<1 that are in a fiber of ψ come from roots of Q that are in an open sector of the complex plane with angle 4πp−λ/32=πp−λ/2. By [25, Proposition 11.2.4], such an open sector contains at most deg(Q)22pλ+n+1≤2∣a∣22pλ+n+1≤∣a∣2p+n+1≤12+n+1≤3nroots of Q, counting multiplicities; here we used λ≥1 and ∣a∣≤p. Therefore, #B<1≤3n#{0≤t≤p:∣Pa(ξt)∣<5n∣a∣p−λ}.We must compensate for the fact that we may be counting 1=ξ0=ξp twice, so #B<1≤6n#{0≤t≤p−1:∣Pa(ξt)∣<5n∣a∣p−λ}.The lemma now follows from #B≤2#B<1.□ Proposition 7.3 For all ϵ>0, there exist constants c=c(n,ϵ)≥1and λ=λ(n,ϵ)≥1with the following property. Let a∈Znand let p be a prime with ∣a∣≤psuch that Padoes not vanish at any point of μp. Then #B(p,a,λ)≤cpϵ. Proof Say c′ and λ′ are from Theorem 1.3. We apply Lemma 7.2 to a fixed λ≥λ′+1≥2 that satisfies 2λ−λ′−1≥5nc′ and hence pλ≥5nc′pλ′+1. Say ξ=e2π−1/p and a=(a1,…,an). Now any ζ∈μp with ∣Pa(ζ)∣<5n∣a∣p−λ equals ξt for some t∈Fp and ∣1+ξa1t+⋯+ξant∣<5n∣a∣p−λ≤c′−1∣a∣pp−λ′≤c′−1p−λ′,as ∣a∣≤p. Recall that 1+ξa1+⋯+ξan=Pa(ξ)≠0 for a as in the hypothesis. So by Theorem 1.3, the number of possible t is at most cpϵ for c sufficiently large in terms of n and ϵ.□ 8. Main technical result Suppose n≥2, let Ω⊆Zn be a subgroup of rank at least 2, and let a∈Zn. We set ρ(a;Ω)=inf{∣ω∣:ω∈Ω⧹{0}and⟨ω,a⟩=0} (28)and recall (21). Proposition 8.1 Let Ω⊆Znbe a subgroup of rank m≥2that satisfies the hypothesis in Proposition5.2. There exists a constant c=C(Ω)≥1with the following property. Say a∈Ωand let p be a prime with ρp(a;Ω)≥c. There exists a′∈Ω⧹{0}with a′=at−pω, where t∈Zis coprime to p and ω∈Ω, such that ∣a′∣≤cp1−1/m,Δp(a′)=Δp(a)≠0, ρ(a′;Ω)≥ρp(a;Ω),and1p−1logΔp(a′)=m(Pa′)+O(p−12m); (29)here the constant implicit in O(·)depends only on Ω. Proof Let ω1∈Ω⧹{0}, then ⟨pω1,a⟩≡0(modp), hence ρp(a;Ω)≤p∣ω1∣. By increasing c, we may assume that p is larger than a prescribed constant. Say c1≥1 is the constant from Proposition 5.2; we may suppose c≥max{3,c1}. There is t∈Z not divisible by p and ω∈Ω such that a′=ta−pω satisfies 0<∣a′∣≤c1p1−1/m and the only roots of Pa′ on the unit circle are roots of unity. We may assume ∣a′∣<(p−1)/2 by increasing c. Suppose ⟨ω0,a′⟩=0 with ω0∈Ω⧹{0}. Then ⟨ω0,a⟩≡0(modp) as t and p are coprime. So ∣ω0∣≥ρp(a′;Ω) by hypothesis. This implies the inequality in (29). Let e∈Z such that XePa′ is a polynomial with non-zero constant term, then deg (XePa′) ≤2∣a′∣<p−1. Since XePa′ has integral coefficients, we find Pa′ (ζ)≠0 if ζ has order p. Clearly, we also have Pa′(1)=n+1≠0. So Pa′ does not vanish at any point of μp. We apply Proposition 7.3 to ϵ=1/(2m),a′ and p to conclude #B(p,a′,λ)≤c2p1/(2m) for constants c2,λ≥1 that depend only on m and n. We use this bound in the estimate from Proposition 4.5 applied to a′, to get Δp(a′)≠0 and ∣1p−1logΔp(a′)−m(Pa′)∣≤c3∣a′∣logppλ+c2p12m≤c4p−12m,where c3 and c4 depend only on m,n and ϵ. Now recall that a′=ta−pω, so Δp(a′)=Δp(ta)=Δp(a)≠0, by Lemma 4.6, parts (i) and (ii), respectively. This completes the proof.□ Let A∈Matmn(Z) be a matrix with entries aij. We define PA=1+∑j=1nX1a1jX2a2j⋯Xmamj∈Z[X1±1,…,Xm±1]. (30)If we consider a∈Zn as a 1×n matrix and identify X1 with X, then the definitions (10) and (30) coincide. Now suppose that Ω⊆Zn is a subgroup of rank m≥1 and assume that the rows of A are a basis of Ω. Then the value m(PA) is independent of the choice of basis. Indeed, if the rows of B∈Matmn(Z) constitute another basis of Ω, then A = UB with U∈GLm(Z). The Mahler measure is known to be invariant under a change of coordinates by U, that is m(PA)=m(PUB)=m(PB), cf. Corollary 8 in Chapter 3.4 [26]. So m(PA) depends only on Ω and we write m(Ω)=m(PA) (31)for any A as before. Theorem 8.2 Let Ω⊆Znbe a subgroup of rank at least 2 that satisfies the hypothesis in Proposition5.2and say ϵ>0. Suppose a∈Ωand p is a prime such that ρp(a;Ω)is sufficiently large in terms of Ω. Then Δp(a)≠0and 1p−1logΔp(a)=m(Ω)+Oρp(a;Ω)−14n+ϵ (32)as ρp(a;Ω)→∞where the implicit constant depends only on Ω and ϵ. Proof If ρp(a;Ω) is sufficiently large, then by Proposition 8.1 we have Δp(a)≠0 and a′∈Ω with the stated properties. Let m denote the rank of Ω, let us fix a basis ω1,…,ωm of Ω, and write A∈Matmn(Z) for the matrix with rows ω1,…,ωm. We fix a tuple of independent elements (ω1*,…,ωm*) in Ω and an integer d≥1 with ⟨ωj*,ωl⟩=0 if j≠l and ⟨ωj*,ωj⟩=d for all 1≤j,l≤m. It remains to check that m(Pa′) converges to m(PA)=m(Ω). Observe Pa′=PA(Xν1,Xν2,…,Xνm),where ν=(ν1,…,νm)∈Zm is determined by a′=ν1ω1+⋯+νmωm. Our quantitative version of Lawton’s Theorem, Theorem A.2 in the self-contained appendix, relies on ρ(ν;Zm). Say (λ1,…,λm)∈Zm⧹{0} has norm ρ(ν;Zn) with ∑j=1mλjνj=0. Then ⟨∑j=1mλjωj*,a′⟩=0 and hence ρp(a;Ω)≤ρ(a′;Ω)≤∣∑j=1mλjωj*∣≤ρ(ν;Zm)(∣ω1*∣+⋯+∣ωm*∣),where we used (29). The theorem follows as PA has at most n+1 and at least m+1≥2 non-zero terms.□ 9. Applications to Gaussian periods Using our method, we prove the following theorem from which we will deduce two applications. Suppose n≥2. Recall that m(Ω) was defined in (31). A place of a number field is an absolute value on the said number field whose restriction to Q coincides with the standard complex absolute value or the p-adic absolute value for a prime p taking the value p−1 at p. Theorem 9.1 Let K be a number field, let α1,…,αn∈K, and define the subgroup Ω={(b1,…,bn)∈Zn:b1α1+⋯+bnαn=0}⊥.Let ϵ>0. We suppose that Ω has rank at least 2 and that the following hypothesis holds. The 0=α0,α1,…,αnare pairwise distinct and (αi−αj)/(αk−αl)∉Qfor all pairwise distinct i,j,k,l∈{0,…,n}. Let p be a prime, v0a place of K that extends the p-adic absolute value, and e(v0)the ramification index of K/Qat v0. Suppose (a1,…,an)∈Znwith ∣ai−αi∣v0<1for all i∈{1,…,n}, then 1p−1∑t=1p−1log∣1+ζta1+⋯+ζtan∣=m(Ω)+O(p−14n[K:Q]e(v0)+ϵ)whereζ=e2π−1/p (33)as p→∞and the implicit constant depends only on α1,…,αn, and ϵ; in particular, the logarithm is well-defined for all large p. Proof Observe that Λ={(b1,…,bn)∈Zn:b1α1+⋯+bnαn=0} is a primitive subgroup of Zn. Let r denote its rank. If r≥1 and if M∈Matrn(Z) is a matrix whose rows are a basis of Λ, then 0→pΩ⟶inclusionpZn⟶multiplicationbyMpZr→0is a short exact sequence. By the Snake Lemma, we find that the image of Ω in Fpn equals the kernel of multiplication by M taken as an endomorphism Fpn→Fpr. Say a=(a1,…,an)∈Zn is as in the hypothesis. Then the left-hand side of (33) equals 1p−1logΔp(a). This mean is invariant under translating a by a vector in pZn, cf. Lemma 4.6(i). For all (b1,…,bn)∈Λ, we find ∣a1b1+⋯+anbn∣v0=∣(a1−α1)b1+⋯+(an−αn)bn∣v0≤max1≤i≤n∣ai−αi∣v0<1.By the previous paragraph we may assume, after adding an element of pZn to a, that is without loss of generality, that a∈Ω. The current theorem will follow from Theorem 8.2. As Λ⊥⊥=Λ, we find that the hypothesis on Ω implies the hypothesis on Ω in Proposition 5.2. To estimate ρp(a;Ω) say ω=(ω1,…,ωn)∈Ω⧹{0} with ⟨ω,a⟩≡0(modp) and ∣ω∣=ρp(a;Ω). We define Δ=ω1α1+⋯+ωnαn∈K. Observe that Δ≠0 since Λ∩Λ⊥={0}. We estimate ∣Δ∣v0=∣ω1(α1−a1)+⋯+ωn(αn−an)+ω1a1+⋯+ωnan∣v0≤p−1/e(v0). If v is an archimedean place of K, then the triangle inequality yields ∣Δ∣v≤n∣ω∣max{∣α1∣v,…,∣αn∣v}. For any non-archimedean place v of K, we get a stronger bound due to the ultrametric triangle inequality, that is ∣Δ∣v≤max{∣α1∣v,…,∣αn∣v}. We take the product of ∣Δ∣v≠0 over all places with the appropriate multiplicities and use the local bounds above in combination with the product formula to obtain 1≤(n∣ω∣eh(α1)+⋯+h(αn))[K:Q]p−1/e(v0).So (33) follows from (32) as ∣ω∣=ρp(a;Ω).□ The following corollary generalizes Theorem 1.1 to subgroups of Fp× of odd order. Corollary 9.2 Let f≥3be an odd integer and ϵ>0. Say ξ is a root of unity of order f and define Ω={(b1,…,bf−1)∈Zf−1:b1(ξ−1)+b2(ξ2−1)+⋯+bf−1(ξf−1−1)=0}⊥.For any prime p with p≡1(modf)let Gp⊆Fp×be the subgroup of order f. Then 1p−1∑t=1p−1log∑g∈Gpζtg=m(Ω)+O(p−14(f−1)φ(f)+ϵ)whereζ=e2π−1/pas p→∞, where the implicit constant depends only on f and ϵ; in particular, the logarithm is well-defined for all large p. We will prove this corollary further down. Here we treat the hypothesis on Ω in Theorem 9.1 for roots of unity. Lemma 9.3 Let ζ1,ζ2,ζ3,ζ4be pairwise distinct roots of unity. If (ζ1−ζ2)/(ζ3−ζ4)∈Q, then there exist distinct i,j∈{1,2,3,4}with ζi=−ζj. Proof Set η=ζ3ζ4−1,ξ=ζ1ζ2−1, and ζ=ζ2ζ4−1 and define x=ζ1−ζ2ζ3−ζ4=ζξ−1η−1.We assume x∈Q and, after possibly swapping ζ1 and ζ2, also x>0. Note that the number field K=Q(ζ1,ζ2,ζ3,ζ4) has only complex embeddings as it contains at least four roots of unity. So the K/Q-norm NK/Q(·) is never negative. We have x[K:Q]=NK/Q(x)=NK/Q(ξ−1)NK/Q(η−1). (34) We now divide into four cases, depending on whether the orders of η and ξ are prime powers or not. First suppose that neither η nor ξ has order a prime power. Then η−1 and ξ−1 are units and hence x=1 by (34). We obtain the vanishing sum of roots of unity η−ζξ+ζ−1=0. (35) If a non-trivial subsum vanishes, then η−ζξ=ζ−1=0orη+ζ=−ζξ−1=0orη−1=−ζξ+ζ=0.The first and third cases are impossible as ζ≠1 and η≠1. Thus ζξ=−1 and this implies ζ1=−ζ4, as desired. If no non-trivial subsum vanishes, then Mann’s [19, Theorem 1] implies η6=ξ6=ζ6=1. In the current case, η and ξ must have precise order 6. If η=ξ, then (35) implies η=1 or ζ=1 which contracts the hypothesis. Hence η≠ξ and thus ηξ=1. When combined with (35) we have ζξ=η−1ξ−1ξ=ηξ−ξξ−1=1−ξξ−1=−1, (36)and again ζ1=−ζ4. Now suppose that ξ has order pe with p a prime and e≥1 and that the order of η is not a prime power. Then η−1 remains a unit but now NK/Q(ξ−1)=p[K:Q(ξ)], so xφ(pe)=p. As x is rational, we must have φ(pe)=1, so pe = 2 and thus ζ1ζ2−1=ξ=−1, which completes this case. Similarly, if η has prime power order and ξ does not, then we apply the argument from the last paragraph to x−1=ζ−1(η−1)/(ξ−1) and conclude ζ3ζ4−1=η=−1, as desired. Finally, suppose η has order qe′ and ξ has order pe, here p and q are primes and e,e′≥1. Now (34) implies x=p1/φ(pe)q−1/φ(qe′)∈Q. If p≠q, then pe=qe′=2 and so η=ξ=−1, as desired. So say p=q, hence x=p1/φ(pe)−1/φ(pe′)∈Q and it follows that 1(p−1)pe−1−1(p−1)pe′−1∈Zand this entails e=e′. We find x = 1 and are thus back in the situation of the first case, except that η and ξ now have prime power order. We proceed similarly. If a non-trivial subsum in (35) vanishes, then again ζ1=−ζ4. Otherwise Mann’s Theorem yields η6=ξ6=1. This time η and ξ have equal order which is 2 or 3. If the common order is 2, then we are done. Elsewise it is 3 and again we must have η≠ξ which again implies ηξ=1. We conclude ζ1=−ζ4 as in (36).□ Proof of Corollary 9.2 Let ξ be a root of unity of order f. We set α1=ξ−1,α2=ξ2−1,…αf−1=ξf−1−1.Our aim is to apply Theorem 9.1 to K=Q(ξ) and n=f−1≥2. Say Ω is as in the said theorem. Observe that α1Z+⋯+αnZ=(ξ−1)(Z+(ξ+1)Z+⋯+(ξf−2+⋯+ξ+1)Z)=(ξ−1)Z[ξ]as ξf−1=−(ξf−2+⋯+ξ+1). This group has rank m=φ(f) and, therefore, Ω also has rank m≥2. We shall show that the hypothesis on Ω in Theorem 9.1 is satisfied. Indeed, if i,j,k,l∈{0,…,n} are pairwise distinct, then ξi−ξjξk−ξl∈Qimplies that ξ has even order by Lemma 9.3. This contradicts our hypothesis on f. Finally, observe that p≡1(modf) means that p splits completely in K. For such p, there is (a1,…,an) as above (33), where v0 is any place of K extending the p-adic absolute value. Here e(v0)=1. Now 1,1+a1,…,1+an is a complete set of representatives of the subgroup of Fp× of order f=n+1. The corollary follows as ∣1+ζta1+⋯+ζtan∣=∣ζt+ζt(1+a1)+⋯+ζt(1+an)∣for all 1≤t≤p−1.□ Proof of Theorem 1.1 We set Ω as in the proof of Corollary 9.2. Its rank is φ(f)=f−1 as f is a prime by hypothesis. Since Ω is a primitive subgroup of Zf−1, we conclude Ω=Zf−1. In the definition (31) of m(Ω), we may take A to equal the (f−1)×(f−1) unit matrix. The resulting logarithmic Mahler measure is that of 1+X1+⋯+Xf−1.□ Proof of Corollary 1.2 Observe that ∑g∈Ge2π−1g/p is an algebraic integer. In view of Theorem 1.1 it suffices to verify m(1+X1+⋯+Xf−1)≠0. 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Google Scholar CrossRef Search ADS 30 L. van den Dries, Tame Topology and o-Minimal Structures, London Mathematical Society Lecture Note Series vol. 248, Cambridge University Press, Cambridge, 1998. Google Scholar CrossRef Search ADS 31 L. van den Dries and C. Miller, Geometric categories and o-minimal structures, Duke Math. J. 84 ( 1996), 497– 540. Google Scholar CrossRef Search ADS Appendix A. Lawton’s Theorem In this appendix, we provide a rate of convergence for the following theorem of Lawton. The arguments here do not rely on the rest of the paper. Say n≥1 is an integer. Recall that the definition of ρ is given in (28). Theorem A.1 (Lawton [16, Theorem 2]) Suppose P∈C[X1±1,…,Xn±1]⧹{0}. If a∈Zn, then m(P(Xa1,…,Xan))=m(P)+o(1)as ρ(a;Zn)→∞. We closely follow Lawton’s approach, but keep track of estimates to obtain the following refinement. Theorem A.2 Suppose P∈C[X1±1,…,Xn±1]⧹{0}has k≥2non-zero terms and let ϵ>0. For a=(a1,…,an)∈Znwe have m(P(Xa1,…,Xan))=m(P)+O(ρ(a;Zn)−14(k−1)+ϵ)as ρ(a;Zn)→∞where the implied constant depends on n,Pand ϵ. An important tool is an estimate on the measure of the subset of the unit circle, where a polynomial takes small values. We use vol(·) to denote the Lebesgue measure on Rn. For P∈C[X1±1,…,Xn±1] and y>0, we define S(P,y)={x∈[0,1)n:∣P(e(x))∣<y},where e(x)=(e2π−1x1,…,e2π−1xn) for x=(x1,…,xn)∈Rn. Lemma A.3 (Lawton [16, Theorem 1]) For k≥1, there exists a constant Ck>0with the following property. If P∈C[X]is a monic polynomial with k non-zero terms and y>0, then vol(S(P,y))≤Cky1/max{1,k−1} The reference covers the main case k≥2. If k=1, then ∣P(e(x))∣=1 for all x∈Rn and the claim is clear with C1=1. Lemma A.4 Assume P∈C[X1±1,…,Xn±1]⧹{0}has k≥1non-zero terms If for all y∈(0,1], then vol(S(P,y))=O(y1/(2max{1,k−1})), where the constant in O(·)depends only on P. If q>0, then ∫[0,1)n∣log∣P(e(x))∣∣qdxis well defined and finite. Proof Our proof of (i) is by induction on n. We may assume that P is a polynomial. If n=1, then the lemma follows from Lawton’s Theorem after normalizing P. Say n≥2. There is nothing to prove if k=1, so suppose k≥2. We may also assume that P is a polynomial, hence P=P0Xnd+⋯+Pd where P0,…,Pd∈C[X1,…,Xn−1] and P0≠0. We abbreviate Σ=S(P0,y1/2)⊆[0,1)n−1. If x′∈Rn−1 and P0(e(x′))≠0, then P(e(x′),X)/P0(e(x′)) is a monic polynomial in X. Fubini’s Theorem implies vol(S(P,y))=∫Σvol(S(P(e(x′),X),y))dx′+∫[0,1)n−1⧹Σvol(S(P(e(x′),X),y))dx′≤vol(Σ)+∫[0,1)n−1⧹Σvol(S(P(e(x′),X)/P0(e(x′))),y/∣P0(e(x′))∣)dx′≤vol(Σ)+∫[0,1)n−1⧹Σvol(S(P(e(x′),X)/P0(e(x′))),y1/2)dx′≤vol(Σ)+Cky1/(2k−2)since vol(S(P(e(x′),X),y))≤1 and by Lemma A.3. By this lemma applied by induction to P0∈C[X1,…,Xn−1]⧹{0} we conclude vol(Σ)≤c(P0)y1/(2k−2). This yields part (i). The statement in (ii) is possibly known, we give a proof based on (i). For an integer m≥0 and x∈Rn, we define pm(x)=min{m,∣log∣P(e(x))∣∣q}≥0 which is interpreted as m if P(e(x))=0. Then pm is a non-decreasing sequence of continuous functions on [0,1)n. We set Im=∫[0,1)npm(x)dx. By the Monotone Convergence Theorem, it suffices to prove that the non-decreasing sequence (Im)m≥1 converges. For all sufficiently large m we have ∣P(e(x))∣≤em1/q if x∈[0,1)n. Observe that pm equals m on S(P,e−m1/q) and that it coincides with pm+1 outside this set. Thus, for all large m, we have Im+1−Im=∫S(P,e−m1/q)(pm+1(x)−pm(x))dx≤vol(S(P,e−m1/q))=O(e−m1/q/(2k))as pm+1(x)≤m+1 and where we used (i), the implied constant is independent of m. Since ∑m≥1e−m1/q/(2k)<∞ we can use a telescoping sum to show that supm≥0Im<∞, as desired.□ Let N0 denote the non-negative integers. Say b∈N0 and let g lie in Cb(Rn), the set of real valued functions on Rn whose derivatives exist up-to and including order b and are continuous. For a multiindex i=(i1,…,in)∈N0n, we set ℓ(i)=i1+⋯+in. If ℓ(i)≤b, then we define ∂ig=(∂/∂x1)i1⋯(∂/∂xn)ing and ∣g∣Cb=maxi∈N0nℓ(i)≤bsupx∈Rn∣∂ig(x)∣which is possibly ∞. We now introduce a function that equals log∣P(e(·))∣ away from the singular locus, but is continuous on Rn and attains 0 when P(e(·)) does. Say ϕ∈Cb(R) is non-decreasing with ϕ(0)=0,ϕ(1)=1, ϕ(x)=0 if x<0, and ϕ(x)=1 if x>1. We ask in addition that ∂iϕ(0)=∂i(1)=0 for all i∈{1,…,b}. For example, we could take the anti-derivative of xb(1−x)b that attains 0 at x=0, scale it to attain 1 at x=1, and extend by 0 for x<0 and by 1 for x>1. Say y∈(0,1/2]. We define ϕy as x↦((2/y)2x−1)/3, which rescales [(y/2)2,y2] to [0,1], composed with ϕ. Then ϕy takes values in [0,1] and ∣∂i(ϕy)(t)∣≤(43y2)i∣∂iϕ(t)∣for all i∈{0,…,b} and all t∈R. Hence ∣ϕy∣Cb=Ob,ϕ(y−2b), (A.1)here and below the implied constant depends on the quantities appearing in the subscript. Finally, we define ψy as ψy(t)=12ϕy(t)logt:t>0,0:t≤0.Then ψy∈Cb(R) and all its derivatives up-to and including order b vanish outside of ((y/2)2,y2). Thus ∣ψy∣Cb=Ob,ϕ(y−2b∣logy∣) (A.2)by the Leibniz product rule applied using (A.1) and since y∈(0,1/2]. Say P∈C[X1,…,Xn] and write g(x)=∣P(e(x))∣2, this is a smooth function Rn→R and ∣g∣Cb=Ob,P,n(1). (A.3) We define fy=ψy◦g. If P(e(x))≠0 with x∈Rn, then fy(x)=ψy(∣P(e(x))∣2)=ϕy(∣P(e(x))∣2)log∣P(e(x))∣and for any x∈Rn we have fy(x)=0:if∣P(e(x))∣≤y/2,log∣P(e(x))∣:if∣P(e(x))∣≥y.Moreover, the composition fy:Rn→[0,1] lies in Cb(Rn). We can bound its norm using the following lemma. Lemma A.5 Let b≥0,ψ∈Cb(R)and g∈Cb(Rn)satisfy ∣ψ∣Cb<∞and ∣g∣Cb<∞. Then ∣ψ◦g∣Cb≤2b(b−1)/2∣ψ∣Cbmax{1,∣g∣Cb}b. Proof The lemma is evident for b=0. So say b≥1 and let i∈N0n⧹{0} with ℓ=ℓ(i)≤b. Let j∈N0n be a standard basis vector with i−j∈N0n. Using the chain rule and the Leibniz product rule, we find ∣∂i(ψ◦g)∣C0=∣∂i−j(∂j(ψ◦g))∣C0=∣∂i−j((ψ′◦g)∂jg)∣C0≤2ℓ−1∣ψ′◦g∣Cℓ−1∣g∣Cℓ.Thus, ∣∂i(ψ◦g)∣C0≤2b−1∣ψ′◦g∣Cb−1∣g∣Cb≤2b(b−1)/2∣ψ∣Cbmax{1,∣g∣Cb}b, where we applied this lemma by induction to bound ∣ψ′◦g∣Cb−1 from above. This upper bound for ∣∂i(ψ◦g)∣C0 continues to hold for i=0 and it is therefore an upper bound for ∣ψ◦g∣Cb.□ This lemma, together with (A.2) and (A.3), implies ∣fy∣Cb=Ob,P,n,ϕ(y−2b∣logy∣). (A.4) From now on, we suppose b≥n+1. In the next three lemmas and if not stated otherwise, P∈C[X1,…,Xn]⧹{0} is a polynomial with k≥2 non-zero terms Lemma A.6 Suppose a∈Znand y∈(0,1/2], then ∫01fy(at)dt=∫[0,1)nfy(x)dx+Ob,P,ϕ,n(∣logy∣y2b1ρ(a;Zn)b−n). Proof All implied constants in this proof depend only on b,P,ϕ and n. The Fourier coefficients fy^(m) of fy∈Cb(R), here m∈Zn, decay quickly. Indeed, by [12, Theorem 3.2.9(b)] ( ∣·∣ in the reference is the ℓ2-norm) with s=b and ∣∂afy^(m)∣≤∣∂afy∣C0≤∣∂afy∣Cb where ℓ(i)=b. We conclude ∣fy^(m)∣=O(∣fy∣Cb∣m∣b)andso∣fy^(m)∣=O(∣logy∣y2b∣m∣b)for all m∈Zn⧹{0} by (A.4). Say H≥1, then ∑∣m∣≥H∣fy^(m)∣=O∣logy∣y2b∑∣m∣≥H1∣m∣b, and hence ∑∣m∣≥H∣fy^(m)∣=O(∣logy∣y2b1Hb−n) (A.5)as b−n≥1. Since the Fourier coefficients of the continuous function fy are absolutely summable, its Fourier series converges uniformly to fy. Hence ∫01fy(at)dt=∑m∈Zn∫01fy^(m)e2π−1⟨a,m⟩tdt=fy^(0)+∑m∈Zn⧹{0}⟨a,m⟩=0fy^(m).Now fy^(0) equals ∫[0,1)nfy(x)dx, so the estimate in the assertion follows from (A.5) and ∣∑m∈Zn⧹{0}⟨a,m⟩=0fy^(m)∣≤∑∣m∣≥ρ(a;Zn)∣fy^(m)∣=O(∣logy∣y2b1ρ(a;Zn)b−n).□ Lemma A.7 Suppose each non-zero term of P has modulus at least 1. If a∈Znsuch that ρ(a;Zn)is sufficiently large in terms of P. Then ∣P(e(as))∣is non-zero for all s∈[0,1]outside a finite set of points. Moreover, if y∈(0,1/2]then ∫01log∣P(e(as))∣ds=∫01fy(as)ds+O(y1/(k−1)∣logy∣),where the implicit constant depends only on P and n. Proof Say a=(a1,…,an) and suppose that ρ(a;Zn) is strictly larger than ∣m−m′∣ for all distinct m,m′ in the support of P. Then P(Xa1,…,Xan) has the same coefficients as P. It is in particular non-zero and the first claim holds true. For the second claim, we can apply Lawton’s [16, Lemma 4].□ Lemma A.8 Let ϵ>0. If y∈(0,1/2]then ∣∫[0,1)n(fy(x)−log∣P(e(x))∣)dx∣=O(y12(k−1)−ϵ),where the implied constant depends only on n,Pand ϵ. Proof Let p>1 and fix q>1 such that 1/p+1/q=1. By definition, we get the equality in ∣∫[0,1)n(fy(x)−log∣P(e(x))∣)dx∣=∣∫[0,1)n(ϕy(∣P(e(x))∣2)−1)log∣P(e(x))∣dx∣≤∫[0,1)n∣ϕy(∣P(e(x))∣2)−1∣∣log∣P(e(x))∣∣dx≤∫[0,1)n∣ϕy(∣P(e(x))∣2)−1∣pdx1/p∫[0,1)n∣log∣P(e(x))∣∣qdx1/qwe used the Hölder inequality in the last step; the final integral is finite by Lemma A.4(ii). As ϕy(∣P(e(x))∣2)=1 if ∣P(e(x))∣≥y and ∣ϕy(∣P(e(x))∣2)−1∣≤1 in any case, we have ∫[0,1)n∣ϕy(∣P(e(x))∣2)−1∣pdx=∫S(P,y)∣ϕy(∣P(e(x))∣2)−1∣pdx≤vol(S(P,y)).Recall k≥2. The current lemma follows from Lemma A.4(i) because we may suppose 1/(2p(k−1))≥1/(2k−2)−ϵ.□ Proof of Proposition A.2 Without loss of generality, we may suppose that all non-zero coefficients of P have modulus at least 1 and that P is a polynomial. We fix b≥n+1 sufficiently large and a sufficiently small ϵ′>0 with −γ2(k−1)+ϵ′γ≤−14(k−1)+ϵwhereγ=b−n2b+1/(2k−2). (A.6)We fix a step function ϕ∈Cb(R) as above, abbreviate H=ρ(a;Zn)≥1, and set y=H−γ. For H large enough, we find y≤1/2 and that ∣m(P(Xa1,…,Xan))−m(P)∣ equals ∣∫01log∣P(e(as))∣ds−∫[0,1)nlog∣P(e(x))∣dx∣≤∣∫01fy(as)ds−∫[0,1)nfy(x)dx∣+∣∫01(log∣P(e(as))∣−fy(as))ds∣+∣∫[0,1)n(fy(x)−log∣P(e(x))∣)dx∣.Then by Lemmas A.6, A.7 and A.8, the final one applied to a sufficiently small ϵ′, this sum is in O(∣logy∣y2b1Hb−n+y12(k−1)−ϵ′),where the implied constant here and below depends only on b,P,ϕ and ϵ. So the sum is in O(γ(logH)H2bγ+n−b+H−γ2(k−1)+ϵ′γ).The proposition follows for small enough ϵ′ as 2bγ+n−b=−γ/(2k−2), cf. (42).□ © 2017. Published by Oxford University Press. All rights reserved. For permissions, please email: journals.permissions@oup.com

The Quarterly Journal of Mathematics – Oxford University Press

**Published: ** Mar 1, 2018

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