The Norm of Gaussian Periods

The norm of Gaussian periods

P. Habegger

Departement Mathematik und Informatik Preprint No. 2016-22 Fachbereich Mathematik November 2016 Universit¨atBasel CH-4051 Basel www.math.unibas.ch THE NORM OF GAUSSIAN PERIODS

P. HABEGGER

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 provide a rate of convergence in a case of Myerson’s Conjecture for periods of arbitrary odd length. Our method involves a result of Bombieri, Masser, and Zannier 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 1beanintegerandp aprimenumber.Weareinterestedintheasymptotic behavior of the norm of

a1 a2 a 2⇡p 1/p (1) ⇣ + ⇣ + + ⇣ f where ⇣ = e ··· as a1,...,af Z and p vary. These cyclotomic2 integers appear naturally in algebraic number theory. We identify the Galois group of Q(⇣)/Q with Fp⇥ = Fp r 0 .Saya1,...,af represent the elements { } of a subgroup G Fp⇥ of order f.Thenthesum(1)isthetraceof⇣ relative to the ✓ subfield of Q(⇣)fixedbythesaidsubgroupanditiscalledaGaussianperiod.Ithas degree k =[Fp⇥ : G]=(p 1)/f over Q,werefertoChapter10.10ofBerndt,Evans, and Williams’ book [3] for these and other facts. A Gaussian period can be expressed in terms of p and f for small values of k.Indeed,ifk =1thenG = Fp⇥ and the Gaussian 2 p 1 period is of course ⇣ + ⇣ + + ⇣ = 1. Gauss evaluated the sum if k =2andthe minimal polynomial of a Gaussian··· period has been computed if k 4. The absolute norm of a Gaussian period appears in combinatorial problems, cf. My- erson’s work [18, 19]. Let A1,...,Ak Fp⇥ denote a complete set of representatives of 2 Fp⇥/G.Thenthecardinalitysatisfies 1 p 1 (2) # (x ,...,x ) Gk : A x + + A x =0 f k = { 1 k 2 1 1 ··· k k }p p where = ⇣tg ; g G ! t Fp⇥/G 2Y X2 t note that ⇣ is well-defined for t Fp as is the sum. If A1x1 + + Akxk were to 2 ··· attain all values of Fp equally often then would vanish. As Myerson [18] observed, 2010 Mathematics Subject Classification. Primary: 11R06, secondary: 05A15, 11J71, 11K60, 37A45. Key words and phrases. Gaussian period, Mahler measure, Diophantine Approximation, Ergodic Theory and Number Theory.

1 THENORMOFGAUSSIANPERIODS 2 this linear form attains non-zero values equally often. It is tempting to interpret (2) tg as an error term. But note that the trivial estimate g G ⇣ f leads to the upper 2  bound (1 p 1)f k for the modulus of (2). This bound exceeds p 1f k for all p 3. In P this paper we will improve on the trivial bound if the length f of the Gaussian period is a ﬁxed prime and p is large. When well-deﬁned, the logarithmic absolute norm of the Gaussian period is

p 1 1 1 (3) log = log ⇣tg . k | | p 1 t=1 g G 2 X X Our Theorem 1 below determines the asymptotic behavior of this value as p when f =#G is a ﬁxed prime. Before stating our ﬁrst result, we survey what is!1 known for groups G of order f. Certainly (3) vanishes if G is trivial. If f =2,thenG = 1 and p is odd. Note 1 1 2 {± } that ⇣ + ⇣ = ⇣ (⇣ +1)isaunitinQ(⇣). So (3) is again zero. The value of , i.e. its sign, can be computed using the Kronecker symbol. Already the case f = 3 is more involved. It requires the logarithmic Mahler measure

1 1 2⇡p 1x1 2⇡p 1xn m(P )= log P e ,...,e dx1 dxn, 0 ··· 0 ··· Z Z ⇣ ⌘ 1 1 of a non-zero Laurent polynomial P C[X1± ,...,Xn± ]; for the fact that this integral converges and other properties we refer2 to Chapter 3.4 in Schinzel’s book [23]. If p 1 ⌘ (mod 3), then Fp⇥ contains an element ✓ of order 3. Myerson, cf. Lemma 21 [19], proved

p 1 1 2 (4) log ⇣t + ⇣t✓ + ⇣t✓ =m(1+X + X )+o(1) p 1 1 2 t=1 X as p .ThelogarithmicMahlermeasureof1+ X + X was evaluated by Smyth !1 1 2 [26] and equals L0( 1,)where is the non-trivial character modulo 3 and L( ,)is its associated Dirichlet L-function. Duke [8] gave a new proof of (4) which extended· to 3 2 alargerclassofvectorsinFp containing the exponent vector (1,✓,✓ ). Moreover, he provided a rate of convergence. Due to a fortunate factorization, the case f =4issimilartoorder2.Indeed,Theorem 6[19]implies= 1ifp 1 (mod 4) and also determines the sign. So the limit in question is again zero.± ⌘ For higher order, the approaches in [19] and [8] break down. But Myerson’s Conjecture [18] predicts convergence of (3) as p and the limit point. The full conjecture is !1 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, Schmidt, and Verbitskiy [15] 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 (Cr 0 )n.Theycomputedthelimitofthisaverageforcertainsequencesofgroupsof roots{ of} unity. THENORMOFGAUSSIANPERIODS 3

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 [12] 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 [21]. Our approach is quantitative in the sense that it can provide a rate of convergence. The following theorem is the special case of our main result when #G is an odd prime. We refer to Corollary 21 below for a more general statement. This corollary contains the case q = p of Myerson’s Conjecture.

Theorem 1. Suppose f is an odd prime. For a prime p with p 1 (mod f) let Gp Fp⇥ denote the subgroup of order f. Then ⌘ ✓

p 1 1 p tg 1 2⇡ 1 p 5(f 1)2 (5) log e =m(1+X1 + + Xf 1)+O p p 1 ··· t=1 g Gp X X2 ⇣ ⌘ as p ; in particular, the logarithm is well-defined for all suciently large p. !1 1 We have the estimate m(1 + X1 + + Xf 1) 2 log f by Corollary 6 in Chapter 3.4 [23]. This justifies the treatment of (2)··· as an error term if f =#G is a prime and p is large. The value m(1 + X1 + + Xf 1)isnon-zeroiff 3. This implies an amusing corollary of Theorem 1 on··· Gaussian periods that are units. Corollary 2. Suppose f is an odd prime. There are at most finitely many primes p with p 1(modf) such that ⌘ 2⇡p 1 g e p g G X2 is an algebraic unit where G Fp⇥ denotes the subgroup of order f. ✓ Gaussian periods and their generalizations were investigated by Duke, Garcia, and Lutz [9] 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 [1]. Our main technical result is Theorem 20 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 dicult when f =#G 5isthattheintegrand 1+X1 + + Xf 1 ··· in 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. As in Duke’s work [8] we start with a basic observation; to simplify notation THENORMOFGAUSSIANPERIODS 4

n we set n = f 1. Let a =(a1,...,an) Z ,then 2 p 1 1+⇣ta1 + + ⇣tan ··· t=1 Y is the product of P =1+Xa1 + + Xan a ··· evaluated at all roots of unity of order p.Ifan >an 1 > >a1 > 0, then Pa is a ··· p 1 monic polynomial and the product above is the resultant of Pa and 1 + X + + X . If this resultant is non-zero, then by symmetry properties we ﬁnd that (3) equals···

log(n +1) 1 d (6) + log ↵p 1 . p 1 p 1 | i | i=1 X 1 p where ↵1,...,↵d are the roots of Pa. Comparing p 1 log ↵i 1 to the local contribution | | log max 1, ↵i of the logarithmic Mahler measure m(Pa)isacrucialaspectofthe problem{ at| hand;|} see Section 2 for details on the Mahler measure. Indeed, by a result of Lawton [14] the value m(Pa)convergestowardsthelogarithmicMahlermeasureofa multivariate polynomial as in Theorem 1 if a for a in suciently general position. In a self-contained appendix, we provide a| quantitative|!1 version of Lawton’s Theorem, see Theorem 24. Baker’s theory on linear forms in logarithms yields a lower bound for non-zero values p of ↵i 1 . But the current estimates are not strong enough to directly establish Theorem 1,| see Duke’s| comment after the proof of this Theorem 3 [8]. However, as we shall see p p below, strong lower bounds for ↵i 1 are available if ↵i =1.Indeed, ↵i 1 ↵ p 1 ↵ 1 .If ↵ =1wewilluseanoldresultofMahlerontheseparation| | | |6 | | || i| ||| i| | | i|6 of distinct roots of a polynomials to bound ↵i 1 from below. If ↵i lies in µ ,the || | | p 1 set of all roots of unity in C,thenasucientlystronglowerboundfor ↵i 1 follows from simpler considerations. | | This estimate gives us sucient control on each term in the sum (6) subject to the 1 1 condition that Pa does not have any root in S r µ ,hereS is the unit circle in C.But it seems unreasonable to expect this hypothesis to1 hold for all a.Toaddressthisconcern 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.Weexploitthissymmetry by using a result of Bombieri, Masser, and Zannier2 [5] on unlikely intersections. This in combination with Dirichlet’s Theorem in diophantine approximation allows us to assume 1 that Pa has no roots on S rµ after a suitable transformation as described above. Here the parity assumption on f in1 Corollary 21 below is used. At this point we have a sucient lower bound for each term of the sum (6). However, the method cannot proceed if too many terms are close to this bound. Duke [8] already p use the following basic principle. Suppose that for some ↵ = ↵i the distance ↵ 1 is 1 | | small, i.e. at most a fixed power of p .Thanonecanexpectthat↵ is close to some ⇣ µp where µp is the set of roots of unity in C of order dividing p. As Pa(↵)=0we find2 that (7) 1+⇣a1 + + ⇣an | ··· | THENORMOFGAUSSIANPERIODS 5 is small. Myerson [20] proved a lower bound for non-vanishing sums of roots of unity if 1 n =1, 2, and 3. His estimates are are polynomial in p and are strong enough to imply p Duke’s result. However, for fixed n 4onlyexponentialboundssuchas(n +1) are known to hold in general. They are not good enough for our purposes. p If ↵i 1 is small for many i,weareabletoshowthat(7)issmallformanyp-th roots| of unity | ⇣. 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 [12] on diophantine approximations on definable sets in an o-minimal structure. Theorem 3. Let n 1. For all ✏>0 there exist constants c = c(n, ✏) 1 and = (n, ✏) 1 with the following property. If p is a prime and ⇣1,...,⇣n µp satisfy 1+⇣ + + ⇣ =0, then 2 1 ··· n 6 t t 1 ✏ (8) # t Fp : 1+⇣1 + + ⇣n

2. Notation The supremum norm on Rn is for any n 1. We have already seen the deﬁnition |·| 1 1 of the logarithmic Mahler measure m(P )ifP C[X1± ,...,Xn± ] r 0 .TheMahler measure of P is M(P )=em(P ). 2 { } The absolute logarithmic Weil height, or just height, of an algebraic number ↵ with minimal polynomial P in Z[X]andleadingtermp0 1is

1 1 h(↵)= m(P )= log 0p0 max 1, z 1 [Q(↵):Q] [Q(↵):Q] { | |} z C B P Y(z2)=0 C B C @ A where the second equality follows from Jensen’s Formula. We write H(↵)=eh(↵). Moreover, we set H(↵ ,...,↵ )=max H(↵ ),...,H(↵ ) if ↵ ,...,↵ are algebraic. 1 n { 1 n } 1 n 3. Algebraic Numbers Close to the Unit Circle An algebraic number ↵ Cr 1 of degree D =[Q(↵):Q]canbeboundedaway from 1 using Liouville’s Inequality,2 { Theorem} 1.5.21 [4], log ↵ 1 D log 2 Dh(↵). | | The modulus ↵ = p↵↵ is again an algebraic number, here and below denotes complex conjugation. Its| | height satisﬁes · 1 h( ↵ ) h(↵↵) h(↵) | |  2  THENORMOFGAUSSIANPERIODS 6 since h(↵)=h(↵). If ↵ is real, then clearly Q( ↵ )=Q(↵). However, for D 2weonly | | have [Q( ↵ ):Q] D(D 1) and equality is possible. So Liouville’s Inequality applied to ↵ gives| |  | | log ↵ 1 D(D 1) log 2 D(D 1)h(↵) || | | if ↵ =1.WewillusearesultofMahlertoimproveonthedependencyinD in front of log| 2.|6

Theorem 4 (Mahler). Let P Z[X] be a polynomial with D =degP 2.Ifz,z0 C are distinct roots of P , then 2 2 (D+2)/2 (D 1) z0 z > p3D M(P ) . | | Proof. We may assume that P has no multiple roots over C after replacing it by its squarefree part. The estimate then follows from Theorem 2 [16] as the absolute value of the new discriminant is at least 1. ⇤ Lemma 5. Let ↵ C be an algebraic number of degree D =[Q(↵):Q].If ↵ =1then 2 | |6 log ↵p 1 log ↵ 1 1 (D +1)log(2D) 2(2D 1)Dh(↵) | | || | | for all integers p 1. p p p 1 Proof. The first inequality follows from ↵ 1 ↵ 1 = ↵ 1 ↵ + +1 ↵ 1 .Toprovethesecondinequalitywemayassume| ||| | | || ↵| |·||1/2,| in particular··· | || | | | | ↵ =0.LetP Z[X]denotetheminimalpolynomialof↵.WewillapplyMahler’s 6 2 D 1 Theorem to F = P (X)P (1/X)X Z[X]. Observe that F (↵)=F (↵ )=0and 1 2 (2D+2)/2 (2D 1) deg F =2D.Therefore,↵ ↵ > p3(2D) M(F ) since ↵ = 1. As M(P (1/X)XD)=M(P )andsincetheMahlermeasureismultiplicative,wefind,after| | | |6 multiplying with ↵ = ↵ ,that | | | | 2 (D+1) 2(2D 1) ↵ 1 > p3 ↵ (2D) M(P ) . | | | | Observe that log M(P )= Dh(↵)and ↵ 1 = ↵ 2 1 /( ↵ +1).Therefore, || | | || | | | | ↵ (D+1) 2(2D 1) p3 (D+1) 2(2D 1)Dh(↵) ↵ 1 > p3 | | (2D) M(P ) (2D) e || | | ↵ +1 3 | | using ↵ 1/2. We conclude the proof by taking the logarithm. | | ⇤ 4. AFirstEstimate n Let n 1andp 2beintegers.Fora =(a1,...,an) Z we define 2 p 1 ta1 tan 2⇡p 1/p (9) (a)= 1+⇣ + + ⇣ where ⇣ = e . p | ··· | t=1 Y a1 an If p is a prime, then p(a)istheQ(⇣)/Q norm of the cyclotomic integer 1+⇣ + +⇣ up-to sign. We attach to a the lacunary Laurent polynomial ···

a1 an 1 (10) Pa =1+X + + X Z[X± ]. ··· 2 Say e =max 0, a ,..., a 0, then { 1 n} (11) XeP = Xe + Xe+a1 + + Xe+an a ··· THENORMOFGAUSSIANPERIODS 7 is a polynomial with integral coecients, non-zero constant term, and degree d = max ai aj :0 i, j n 2 a ,wherea0 =0.Wewrite↵1,...,↵d C for the zeros{ of XeP with multiplicity, } | i.e.| XeP = p (X ↵ ) (X ↵ )where2p 1 a a 0 1 ··· d 0 is the leading term of Pa.Wenotethat↵i =0, 1foralli. Our goal in this section is to bound 6 1 (12) log (a) m(P ) p 1 p a e d from above, here m(Pa)=m(X Pa)=logp0 + log max 1, ↵i is the logarithmic i=1 { | |} Mahler measure of Pa. P As p(a) is essentially a resultant we can rewrite it as a product over the roots ↵i. This will allow us to express the di↵erence (12) in terms of these roots. n In the next 4 lemmas we obtain several statements on the roots ↵i in terms of a Z . 2 p 1 d p Lemma 6. We have (a)=p (n +1) ↵ 1 . p 0 i=1 | i | Proof. AvariantofthiscalculationcanalsobefoundintheproofofDuke’sTheorem3Q [8]. We have p 1 d p 1 d p t p 1 t p 1 1 ↵i (a)= P (⇣ ) = p ⇣ ↵ = p . p | a | 0 | i| 0 1 ↵ t=1 i=1 t=1 ! i=1 i Y Y Y Y The lemma follows since p d (1 ↵ )=P (1) = n +1. 0 i=1 i a ⇤ Each ↵i is an algebraic numberQ with Di =[Q(↵i):Q] d whose height is bounded by the next lemma.  Lemma 7. Let i 1,...,d , then D h(↵ ) m(P ) log(n +1). 2{ } i i  a  Proof. Recall that ↵i is a root of the polynomial in (11) which, by the Gauss Lemma, is e divisible by the minimal polynomial Q in Z[X]of↵i.SoDih(↵i) = m(Q) m(X Pa)=  m(Pa)asthelogarithmicMahlermeasureisadditiveandnon-negativeonZ[X] r 0 . { } By Corollary 6, Chapter 3.4 [23] the Mahler measure M(Pa)isatmosttheeuclidean norm of the coecient vector of P .Thisgivesm(P ) log(n +1). a a  ⇤ We now come to a lower bound for ↵p 1 which is independent of p under the | i | assumption that ↵i lies o↵the unit circle or is a root of unit of order not divisible by p. Lemma 8. Suppose a =0, let i 1,...,d , and let p 1 be an integer. 6 p 2{ } (i) If ↵i =1, then log ↵i 1 18 log(n +1)a log(2 a ). (ii) If ↵| |is6 a root of unity| and |↵p =1, then log ↵|p | 1 | | 2log(2a ). i i 6 | i | | | Proof. Observe that ↵ =1impliesn 2. According to Lemmas 5 and 7 we have | i|6 log ↵p 1 1 (D +1)log(2D ) 2(2D 1) log(n +1) | i | i i i (D +1)log(2D ) 4D log(n +1). i i i Now D 2 a by (11). The first part of the lemma follows from i  | | (2 a +1)log(4a )+8a log(n +1) 18 log(n +1)a log(2 a ). | | | | | |  | | | | p The second part is more elementary. Let m 2bethemultiplicativeorderof↵i .If m 3, then ↵p 1 sin(2⇡/m) 2/m.Thebound ↵p 1 2/m certainly also holds | i | | i | THENORMOFGAUSSIANPERIODS 8 for m =2.Itiswell-knownthatEuler’stotientfunction' satisfies '(m) m/2. As p 2 p '(m)=[Q(↵i ):Q] Di we find m 2Di . Hence log ↵i 1 2logDi 2log(2a ), as D 2 a .  | |p | | i  | | ⇤ p This last lemma allows us to compare log ↵i 1 with the corresponding contribution p log max 1, ↵ in the logarithmic Mahler| measure. | { | i|} Lemma 9. Suppose a =0, let i 1,...,d , and let p 1 be an integer. 6 2{ } (i) If ↵ =1, then | i|6 (13) log ↵p 1 p log max 1, ↵ 18 log(n +1)a log(2 a ). | i | { | i|}  | | | | p (ii) If ↵i is a root of unity and ↵i =1, then 6 log ↵p 1 2log(2a ). | i |  | | p Proof. For the proof of part (i) let us first assume ↵i < 1. Then ↵i 1 2and p | | | | Lemma 8(i) yields log ↵i 1 18 log(n +1)a log(2 a ), as desired. | | 1 ||  | | | | If ↵i > 1weusethat↵i is a root of P a.Weobtainthesameboundasbeforefor | |p p log ↵i 1 = log ↵i 1 p log ↵i and this completes the proof of (i). To| prove (ii)| we argue| as in| the case| |↵ < 1abovebutuseLemma8(ii). | i| ⇤ Suppose for the moment that all ↵ satisfy ↵ =1andforsakeofsimplicityalso i | i|6 p0 =1.ByLemma6theboundgiveninpart(i)ofthelastlemmaleadstothebound d a log(2 a ) 2 a 2 log(2 a ) (14) | | | | | | | | p  p 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 ↵i into two parts depending on a parameter 1. The first part (15) p p B = B(p, a, )= i : ↵ < 1and ↵ 1 1and ↵ 1 1weproceedsimilarlyandobtain log ↵p i i | | | THENORMOFGAUSSIANPERIODS 9

1 p log ↵i log(2p). If ↵i =1,thenbyhypothesis↵i is a root of unity whose order| does| not|  divide p,soLemma9(ii)implies| | log ↵p 1 2log(2a ) 2log(2p). | | i ||  | |  We recall d 2 a and sum over 1,...,d r B to find  | | { } d (17) log ↵p 1 p log max 1, ↵ (d #B)max log(2p), 2log(2p) | i | { | i|}  { } i=1 Xi B 62 4 a log(2p).  | | If i B then ↵ = 1. We apply Lemma 9(i) and use a p.Thebound(13)holds 2 | i|6 | | for #B roots and therefore log ↵p 1 p log max 1, ↵ 18 log(n +1)a log(2p)#B. | i | { | i|}  | | i B X2 We combine this bound with (17) to obtain d (18) log ↵p 1 p log max 1, ↵ 18 log(n +1)a log(2p)( +#B). | i | { | i|}  | | i=1 X The logarithmic Mahler measure of P is log p + d log max 1, ↵ where p 1 a 0 i=1 { | i|} 0 is the leading term of Pa. We use Lemma 6 and apply the triangle inequality to find 1 P that p 1 log p(a) m(Pa) is at most | | 1 pp 1 d log 0 (p 1) log p + log ↵p 1 (p 1) log max 1, ↵ p 1 n +1 0 p 1 | | i | { | i|}| ✓ ◆ i=1 X log(n +1) a log(2p) 1 +18log(n +1)| | ( +#B)+ m(P )  p 1 p 1 p 1 a where we used (18). From Lemma 7 we deduce m(P ) log(n +1). So(16)holds a  true. ⇤ If we ignore for the moment all logarithmic contributions, then we have traded in d in (14) for +#B in (16). Before continuing we make two elementary, but important, observations on symmetry properties of p(a)whenp is a prime. Lemma 11. Let p be a prime and let a Zn be arbitrary. n 2 (i) If a0 Z with a a0 (mod p), then p(a)=p(a0). 2 ⌘ (ii) If t Z with p - t, then p(a)=p(ta). 2 Proof. Part (i) follows using the definition (9) and ⇣p =1.Forpart(ii)observethat ⇣ ⇣t permutes the factors in (9) since p - t. 7! ⇤ n Using this lemma we will transform a0 = ta b with p - t and b pZ such that a0 is small compared to p.ThiswillbedoneusingDirichlet’sTheoremfromdiophantine 2 | | approximation. A theorem of Bombieri-Masser-Zannier leads to a criterion that rules out that Pa0 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 suciently large but fixed .Thisiswherecountingrationalpointsclosetoadefinablesetscomesintoplay. THENORMOFGAUSSIANPERIODS 10

5. Lacunary Polynomials with Roots off the Unit Circle n Say n 2. In this section we investigate a condition on a Z such that Pa,as 2 defined in (10), does not vanish at any point of S1 r µ .Itwillproveusefultorestrict n 1 a to a subgroup ⌦ Z and we will introduce a condition on ⌦that ensures that Pa ✓ does not have any roots in S1 r µ for a ⌦ingeneralposition. Our condition is based on a theorem1 of Bombieri,2 Masser, and Zannier [5] on unlikely intersections in the algebraic torus. We also make use of (very rudimentary) tropical geometry. Say K is the field of Puiseux series over C;itisalgebraicallyclosedandequippedwithasurjectivevaluation ord : K Q .Let be an irreducible subvariety defined over K of the algebraic n! [{1} X n torus Gm.ThetropicalvarietyTrop( )of is the closure in R of X X (ord(x ),...,ord(x )) : (x ,...,x ) (K) . { 1 n 1 n 2X } We need the following two basic facts. First, if is an irreducible subvariety defined over K,thenTrop( ) Trop( ); this followsY directly✓X from the definition. Y ✓ X Second, if r =dim 1andafterpermutingcoordinates,theprojectionofTrop( ) X X to the first r coordinates of Rn contains Qr.Weprovethisusingbasicalgebraicgeom- r etry. After permuting coordinates the projection ⇡ : (K) Gm(K)ontothefirstr X r ! coordinates contains a Zariski open and dense subset of Gm.Thereexistsapolynomial 1 1 r P K[X1± ,...,Xr± ] r 0 such that any point of Gm(K)outsideofthezerolocusof 2 { } r P lies in the image of ⇡.Let(a1,...,an) Q be arbitrary. For suciently general r 2 a1 ar r (c1,...,cr) Gm(C)thepolynomialP does not vanish at (c1T ,...,crT ) Gm(K). This point has2 a pre-image under ⇡ in (K) and the valuation of its first r coordinates2 X are a1,...,ar.Thisyieldsourclaim. Einsiedler, Kapranov and Lind’s [10] Theorem 2.2.5 on the structure of Trop( )can be used instead of this second property in the proof of Lemma 12 below. X We write , for the standard scalar product on Rn. If⌦is a subgroup of Zn,then we set h· ·i n ⌦? = a Z : a, ! =0forall! ⌦ . { 2 h i n2 } We write e1,...,en for the standard basis elements of R augmented by e0 =0. Lemma 12. Let ⌦ Zn be a subgroup of rank m 2 for which the following property ✓ holds. If (↵,) Z2 r 0 and if i, j, k, l 0,...,n are pairwise distinct with v = 2 { } 2{ } ↵(ei ej)+(ek el), then ⌦ (vZ)?. Then there exist finitely many subgroups 6✓ N ⌦1,...,⌦N ⌦ of rank at most m 1 such that if a ⌦ r ⌦i then Pa does not ✓ 2 i=1 vanish at any point of S1 r µ . 1 S n Proof. We consider an irreducible component Gm of the zero set of X✓ 1 1 (19) 1 + X + + X and 1 + X + + X . 1 ··· n 1 ··· n 1 1 Then dim = n 2asthesetwopolynomialsarecoprimeinC[X1± ,...,Xn± ]. X 1 a1 an Say a =(a1,...,an) ⌦suchthatPa has a root z S rµ .Then1+z + +z = 0 by definition and after2 applying complex conjugation2 we find1 ···

a1 an 1+z + + z =0. ··· So za =(za1 ,...,zan ) (C)foroneoftheirreduciblecomponentsabove. 2X THENORMOFGAUSSIANPERIODS 11

a n But z also lies in an algebraic subgroup of Gm of dimension 1. According to Bombieri, Masser, and Zannier’s Theorem 1.7 [5] there are two cases. Either za lies in a ﬁnite set a n that depends only on ,andhenceonlyonn,orz (C)where Gm is an irreducible componentX of an algebraic subgroup with 2H H✓

dim a max 1, dim +dim n +1 =max 1, dim 1 . z X\H { X H } { H } Moreover, in the second case comes from a finite set that depends only on n,cf.the first paragraph of the proof ofH Theorem 1.7 [5] on page 26. In the first case we claim that a must lie in one of finitely many subgroups of ⌦of 1 rank 1 m 1. Indeed, we may assume a =0.Ifz0 S r µ is a root of some Pa 1 0  a a0 6 2 with a0 ⌦andz = z0 ,thena and a0 are linearly dependent as z0 is not a root of unity. Our2 claim follows as there are only finitely many possible za in this case. We add these rank 1 subgroups to our collection of ⌦i. In the second case there is an irreducible component of positive dimension at least dim 1thatcontainsza.InthiscasenY✓3.X\ WeH recall that algebraic Hn n subgroups of Gm are in bijection with subgroups of Z ,seeTheorem3.2.19[4]for details. Let ⇤ Zn be a subgroup from a finite set depending only on n with rank r = n dim such✓ that is contained in the algebraic subgroup defined by all H H (20) Xb1 Xbn 1where(b ,...,b ) ⇤. 1 ··· n 1 n 2 a,b Hence zh i =1forallb ⇤. As z is not a root of unity we find a ⇤?. 2 2 We would like to add ⌦ ⇤? to our list of ⌦i. 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 [⌦:⌦ ⇤?]⌦ ⇤? and hence ⌦ ⇤? because \ ✓ ✓ ⇤? is primitive. We now derive a contradiction from this situation by analyzing ⇤. Since all monomials (20) vanish on we find that Trop( )liesin⇤?R,thevectorsub- n H H space of R generated by ⇤?.WealsoneedtostudyTrop( ). Say (x1,...,xn) (C). Since it is a zero of the first polynomial in (19) and by theX ultrametric triangle2 inequal-X ity the minimum among 0, ord(x1),...,ord(xn)isattainedtwice.Thesameargument applied to the second polynomial in (19) shows that the maximum is attained twice. Hence Trop( )iscontainedinthefiniteunionofthecodimension2vectorsubspacesof X Rn defined by relations ord(x )=ord(x ), ord(x )=ord(x )with#i, j, k, l =4 i j k l { } and ord(x )=ord(x ), ord(x )=0 with # i, j, k =3. i j k { } By the discussion before this lemma, Trop( ) Trop( ) Trop( ) Trop( ) ⇤?R. Y ✓ X \ H ✓ X \ Moreover, the projection of Trop( )tosomechoiceofdim distinct coordinates of Rn dim Y Y contains Q Y .So⇤?R intersected with one of the codimension 2 subspaces mentioned above must have dimension at least dim dim 1=n r 1. Therefore, Y H there exists (↵,) Z2 r 0 and pairwise distinct i, j, k, l 0,...,n with v = 2 { } 2{ } ↵(e e )+(e e ) ⇤. Recall that ⌦ ⇤?.So⌦liesintheorthogonalcomplement i j k l 2 ✓ of v,whichcontradictsthehypothesisofthelemma. ⇤ THENORMOFGAUSSIANPERIODS 12

Suppose n 2, let ⌦ Zn be a subgroup of rank at least 2 and let a Zn.Fora prime p we deﬁne ✓ 2

(21) ⇢p(a;⌦)=inf ! : ! ⌦ r 0 and !, a 0(modp) {| | 2 { } h i⌘ } this is a well-deﬁned real number as the set is non-empty.

Proposition 13. Let ⌦ Zn be a subgroup of rank m 2 that satisfies the following ✓ hypothesis. If (↵,) Z2 r 0 and if i, j, k, l 0,...,n are pairwise distinct with 2 { } 2{ } v = ↵(ei ej)+(ek el), then ⌦ (vZ)?. Then there exists a constant c = c(⌦) 1 with the following property. Say a 6✓ ⌦ and let p be a prime with ⇢ (a;⌦) c. Then 2 p there exist t Z with p - t and ! ⌦ such that a0 = ta p! =0, 2 1 1/m 2 6 (i) we have a0 cp , and | | 1 1 (ii) the Laurent polynomial Pa Z[X± ] does not vanish at any point of S r µ . 0 2 1 Proof. The proposition follows from combining Lemma 12 with Dirichlet’s Theorem from diophantine approximation. Indeed, we let ⌦1,...,⌦N be the subgroups of ⌦from this lemma. If N =0weset⌦1 = 0 .Wewillseehowtochoosec below. We fix a basis (! ,...,! )oftheabeliangroup{ } ⌦. Thena = ⌫ ! + + ⌫ ! , 1 m 1 1 ··· m m where ⌫1,...,⌫m R are unique. If p 3, then2 Dirichlet’s Theorem, cf. Theorem 1B [24], applied to ⌫ /p, . . . , ⌫ /p, 1 m and p 1 > 1 yields t, ⌫10 ,...,⌫m0 Z such that 1 t p 1and t⌫i/p ⌫i0 1/m 2   m | | (p 1) . The same conclusion holds for p =2.Weset! = i=1 ⌫i0!i ⌦. Then 1/m 1/m 2 a0/p ( !1 + + !m )(p 1) cp where a0 = ta p! for c large enough. This| | yields | part| ··· (i). | |  P As each ⌦ has rank at most m 1wehave⌦ (⌦ )? =0foralli.Foreachi we i \ i 6 fix a non-zero !⇤ ⌦ (⌦ )? of minimal norm. If !⇤,ta p! =0,then !⇤,a 0 2 \ i h i h i⌘ (mod p), since p - t.So!⇤ ⇢p(a;⌦) c by hypothesis. We can avoid this outcome | | by fixing c large in terms of the ⌦ (⌦ )?.Thus !⇤,a0 =0.Thisimpliesa0 ⌦ for \ i h i6 62 i all 1 i N and in particular a0 =0.Part(ii)followsfromtheconclusionofLemma   6 12. ⇤

6. Rational Points Close to a Definable Set In this section we prove Theorem 3. To do this we temporarily adopt the language of o-minimal structures. Our main reference is van den Dries’ book [27] and his paper with Miller [28]. 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 [12] 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 n height that are close to a subset of R 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✏>0wedefine (Z,✏)tobethesetof N y Rn for which x y <✏for some x Z.WerecallthattheheightH( )wasdefined in2 Section 2. | | 2 · THENORMOFGAUSSIANPERIODS 13

n Theorem 14 (Theorem 2 [12]). Let Z R be a closed set that is deﬁnable in Ran and ✓ 1 let ✏>0. There exist c = c(Z,✏) 1 and ✓ = ✓(Z,✏) (0, 1] such that if ✓ then 2 n alg ✓ ✏ # q Q r (Z ,T ):H(q) T and there is x Z with x q 0and>0dependonlyonZ.Wemayassume c c < 1, so 1 1 (23) q˜ x

t 2⇡p 1(tq tq ) 2⇡p 1x 1+ ⇣ e j b j c e j0 2⇡(n 1) q˜ x0 . j   | t | j J j J 2 2 X X Hence there is a constant c2 > 0dependingonlyonn with

t ✓ 1+ ⇣j

Recall #J n 1. Now suppose c0 > 0and0 > 0aretheconstantsfromthis theorem applied by induction to the terms in J.Wearefreetoassumethat satisﬁes ✓/2 ✓ ✓/2 ✓/2 1 2 c c0 and ✓ 20.Thenc p c 2 p c0 p 0 as p 2. So 2 2  2 

t 1 0 1+ ⇣j

1 t t t 1 1+ (⇣ ⇣ ) = 1+⇣ + + ⇣

2⇡p 1# Lemma 15. Suppose ↵ = re with r (0, 1] and # [0, 1).Ifp 2 is an integer with ↵p 1 1/2, then there exists t 02,...,p such that2 | | 2{ } t 1 (24) # ↵p 1 . p  p | | 4 2 n If in addition a Z and Pa(↵)=0 , then 2 2⇡p 1t/p p (25) P e 5n a ↵ 1 . a  | || | ⇣ ⌘ Proof. Observe that z 1 z 1/2 z/ z 1 for all z Cr 0 ,seeLemma11.6.1[22]. p| | |p| | | p/|2 2|⇡p 1#p 2 { } p p p We substitute z = ↵ to ﬁnd ↵ 1 r e 1 . Now 1 r r 1 ↵ 1 , so rp 1 ↵p 1 1/2byhypothesis.Weﬁnd| | | | | || | | | 2⇡p 1#p p (26) e 1 p2 ↵ 1 .  | | Let t be an integer with #p t 1/2. Then t 0,...,p as # [0, 1). So | | 2{ } 2 2⇡p 1#p 2⇡p 1(#p t) (27) e 1 = e 1 4 #p t | | by elementary geometry. Combining (26) with (27) and dividing by p 2yields(24). THENORMOFGAUSSIANPERIODS 15

2⇡p 1t/p To prove the second claim we set ⇠ = e and estimate 2⇡p 1# 2⇡p 1# 2⇡p 1(# t/p) ↵ ⇠ = re ⇠ r 1 + e ⇠ = r 1 + e 1 | | | || | | | | | | | (2 8 ) ↵p 1 +2⇡ # t/p (1 + ⇡/p8) ↵p 1 ;( | | | | | | where we used r 1 rp 1 ↵p 1 . | || e || | We ﬁx e Z such that X Pa(X)isapolynomialwithnon-zeroconstantterm.Then 2 n P (⇠) = ⇠eP (⇠) ↵eP (↵) ⇠ak+e ↵ak+e | a | | a a | | | Xk=0 here a =(a1,...,an)anda0 =0.Someak + e vanishes and we use ⇠ =1and ↵ 1 to ﬁnd | | | |

Pa(⇠) n max ak + e ⇠ ↵ 2n a ⇠ ↵ . 0 k n | |   { }| | | || | We recall (28) to obtain (25). ⇤ Next we show that many elements in B will lead to many di↵erent 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 16. Let a Zn and suppose p 2 is an integer with a p.If 1, then 2 | | #B 12n# ⇣ µ : P (⇣) < 5n a p .  { 2 p | a | | | } Proof. We may assume a =0.LetuspartitionB = B(p, a, )intoB = i B : 6 <1 { 2 ↵i < 1 and B>1 = i B : ↵i > 1 . | We| construct} a map{ 2 | | } : B 0,...,p <1 <1 !{ } 2⇡p 1# in the following manner. If i B<1,then↵i = ↵i e for # [0, 1), which p 2 | | 2 depends on i,andwehave ↵i 1 1 : B>1 0,...,p in the same spirit. For if i B>1,then ↵i > 1 !{ } 2⇡p 1# 2 | | 1 and there is # [0, 1) such that ↵i = ↵i e .Weapplythesaidlemmato↵i 2 | | and P a to obtain (i) 0,...,p with # (i)/p 1, the same bound holds for P a(⇠ ) = Pa(⇠ ) . 2 | | | | Now #B 2#B<1 or #B 2#B>1.Weassumetheformer,thelattercaseisdealt with similarly.  Let us fix e Z such that Q = XeP (X)isapolynomialwithnon-zeroconstantterm. 2 Then deg Q 2 a and Q has at most n +1terms.ElementsofB<1 that are in a fiber of come from | roots| of Q that are in an open sector of the complex plane with angle 4⇡p /p32 = ⇡p /p2. By Proposition 11.2.4 [22], such an open sector contains at most deg(Q) 2 a a 1 + n +1 | | + n +1 | | + n +1 + n +1 3n 2p2p  2p2p  p2p  p2  THENORMOFGAUSSIANPERIODS 16 roots of Q,countingmultiplicities;hereweused 1and a p.Therefore, | | t #B 3n# 0 t p : P (⇠ ) < 5n a p <1    | a | | | We must compensate for the fact that we may be counting 1 = ⇠0 = ⇠p twice, so t #B 6n# 0 t p 1: P (⇠ ) < 5n a p . <1    | a | | | The lemma now follows from #B 2#B<1.  ⇤ Proposition 17. For all ✏>0 there exist constants c = c(n, ✏) 1 and = (n, ✏) 1 n with the following property. Let a Z and let p be a prime with a p such that Pa 2 | | does not vanish at any point of µp. Then #B(p, a, ) cp✏.  Proof. Say c0 and 0 are from Theorem 3. We apply Lemma 16 to a fixed 0 +1 2 1 +1 that satisfies 2 0 5nc0 and hence p 5nc0p 0 . 2⇡p 1/p Say ⇠ = e and a =(a1,...,an). Now any ⇣ µp with Pa(⇣) < 5n a p equals t 2 | | | | ⇠ for some t Fp and 2

a1t ant 1 a 1 1+⇠ + + ⇠ < 5n a p c0 | |p 0 c0 p 0 , ··· | |  p 

a1 an as a p.Recallthat1+ ⇠ + + ⇠ = Pa(⇠) =0fora as in the hypothesis. So by Theorem| | 3 the number of possible···t is at most cp✏6 for c suciently large in terms of n and ✏. ⇤

8. Main Technical Result Suppose n 2, let ⌦ Zn be a subgroup of rank at least 2, and let a Zn.Weset ✓ 2 (29) ⇢(a;⌦)=inf ! : ! ⌦ r 0 and !, a =0 {| | 2 { } h i } and recall (21).

Proposition 18. Let ⌦ Zn be a subgroup of rank m 2 that satisﬁes the hypothesis in Proposition 13. There exists✓ a constant c = C(⌦) 1with the following property. Say a ⌦ and let p be a prime with ⇢p(a;⌦) c. There exists a0 ⌦r 0 with a0 = at p!, 2 1 21/m { } where t Z is coprime to p and ! ⌦, such that a0 cp , p(a0)=p(a) =0, 2 2 | | 6 1 1 (30) ⇢(a0;⌦) ⇢ (a;⌦), and log (a0)=m(P )+O p 2m ; p p 1 p a0 ⇣ ⌘ here the constant implicit in O( ) depends only on ⌦. · Proof. Let !1 ⌦ r 0 ,then p!1,a 0(modp), hence ⇢p(a;⌦) p !1 .By increasing c we2 may assume{ } that hp is largeri⌘ than a prescribed constant. Say |c | 1is 1 the constant from Proposition 13; we may suppose c max 3,c1 .Thereist Z not { } 1 1/m 2 divisible by p and ! ⌦suchthata0 = ta p! satisﬁes 0 < a0 c p and the 2 | | 1 only roots of Pa0 on the unit circle are roots of unity. We may assume a0 < (p 1)/2 by increasing c. | | Suppose !0,a0 =0with!0 ⌦ r 0 .Then !0,a 0(modp)ast and p are h i 2 { } h i⌘ coprime. So ! ⇢ (a0;⌦) byhypothesis. Thisimpliestheinequalityin(30). | 0| p THENORMOFGAUSSIANPERIODS 17

e e Let e Z such that X Pa0 is a polynomial with non-zero constant part, then deg(X Pa0 ) 2 e  2 a0

Let A Matmn(Z)beamatrixwithentriesaij.Wedeﬁne 2 n a1j a2j amj 1 1 (31) PA =1+ X1 X2 Xm Z[X1± ,...,Xm± ]. ··· 2 j=1 X n If we consider a Z as a 1 n matrix and identify X1 with X,thenthedeﬁnitions (10) and (31) coincide.2 ⇥ Now suppose that ⌦ Zn is a subgroup of rank m 1andassumethattherows ✓ of A are a basis of ⌦. Then the value m(PA)isindependentofthechoiceofbasis. Indeed, if the rows of B Matmn(Z) constitute another basis of ⌦, then A = UB 2 with U GLm(Z). The Mahler measure is known to be invariant under a change of 2 coordinates by U,i.e.m(PA)=m(PUB)=m(PB), cf. Corollary 8 in Chapter 3.4 [23]. So m(PA)dependsonlyon⌦andwewrite

(32) m(⌦) = m(PA) for any A as before.

Theorem 19. Let ⌦ Zn be a subgroup of rank m 2 that satisfies the hypothesis in ✓ Proposition 13 and say ✏>0. Suppose a ⌦ and p is a prime such that ⇢p(a;⌦) is suciently large in terms of ⌦. Then (a)2=0and p 6 1 1 +✏ (33) log (a)=m(⌦)+O ⇢ (a;⌦) 4n p 1 p p ⇣ ⌘ as ⇢ (a;⌦) where the implicit constant depends only on ⌦ and ✏. p !1 Proof. If ⇢ (a;⌦) issucientlylarge, thenbyProposition18wehave (a) =0and p p 6 a0 ⌦withthestatedproperties. 2 Let us fix a basis !1,...,!m of ⌦and write A Matmn(Z)forthematrixwithrows 2 !1,...,!m.Wefixatupleofindependentelements(!1⇤,...,!m⇤ )in⌦andaninteger d 1with !⇤,! =0ifj = l and !⇤,! = d for all 1 j, l m. h j li 6 h j ji   It remains to check that m(Pa0 ) converges to m(PA)=m(⌦).Observe

⌫1 ⌫2 ⌫m Pa0 = PA(X ,X ,...,X ) m where ⌫ =(⌫1,...,⌫m) Z is determined by a0 = ⌫1!1 + + ⌫m!m. 2 ··· THENORMOFGAUSSIANPERIODS 18

Our quantitative version of Lawton’s Theorem, Theorem 24 in the self-contained m m n appendix, relies on ⇢(⌫; Z ). Say (1,...,m) Z r 0 has norm ⇢(⌫; Z )with m m 2 { } ⌫ =0.Then !⇤,a0 =0andhence j=1 j j h j=1 j j i m P P m ⇢p(a;⌦) ⇢(a0;⌦) j!j⇤ ⇢(⌫; Z )( !1⇤ + + !m⇤ )    | | ··· | | j=1 X where we used (30). The theorem follows asPA has at most n+1 andatleast 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(⌦) was defined in (32). 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 1 for a prime p taking the value p at p. Theorem 20. Let K be a number field, let ↵ ,...,↵ K, and define the subgroup 1 n 2 n ⌦= (b1,...,bn) Z : b1↵1 + + bn↵n =0 ? { 2 ··· } of rank m. Let ✏>0. We suppose m 2 and that the following hypothesis holds. The 0=↵0,↵1,...,↵n are pairwise distinct and (↵i ↵j)/(↵k ↵l) Q for all pairwise 62 distinct i, j, k, l 0,...,n . Let p be a prime, v0 a place of K that extends the p-adic 2{ } n absolute value, and e(v0) the ramification index of K/Q at v0. Suppose (a1,...,an) Z 2 with ai ↵i v0 < 1 for all i 1,...,n , then (34) | | 2{ } p 1 1 1 ta1 tan +✏ 2⇡p 1/p log 1+⇣ + + ⇣ =m(⌦)+O p 4n[K:Q]e(v0) where ⇣ = e p 1 ··· t=1 X ⇣ ⌘ as p and the implicit constant depends only on ↵1,...,↵n, and ✏; in particular, the logarithm!1 is well-defined for all large p. n Proof. Observe that ⇤= (b1,...,bn) Z : b1↵1 + + bn↵n =0 is a primitive n { 2 ··· } subgroup of Z of rank r = n m.Ifr 1andifM Matrn(Z) is a matrix whose rows are a basis of ⇤, then 2 inclusion multiplication by M 0 p⌦ pZn pZr 0 ! ! ! ! n is a short exact sequence. By the Snake Lemma we find that the image of ⌦in Fp equals n r the kernel of multiplication by M taken as an endomorphism Fp Fp. n ! Say a =(a1,...,an) Z is as in the hypothesis. Then the left-hand side of (34) 1 2 n equals p 1 log p(a). This mean is invariant under translating a by a vector in pZ ,cf. Lemma 11(i). For all (b ,...,b ) ⇤wefind 1 n 2 a1b1 + + anbn v = (a1 ↵1)b1 + +(an ↵n)bn v max ai ↵i v < 1. 0 0 1 i n 0 | ··· | | ··· |    | | By the previous paragraph we may assume, after adding an element of pZn to a,i.e. without loss of generality, that a ⌦. The current theorem will follow from Theorem 19. 2 THENORMOFGAUSSIANPERIODS 19

As ⇤?? =⇤,wefindthatthehypothesison⌦impliesthehypothesison⌦inPropo- sition 13. To estimate ⇢p(a;⌦) say ! =(!1,...,!n) ⌦ r 0 with !, a 0(modp)and ! = ⇢ (a;⌦). Wedefine = ! ↵ + +2! ↵ { }K.Observethath i⌘ =0since | | p 1 1 ··· n n 2 6 ⇤ ⇤? = 0 .Weestimate \ { } 1/e(v0) = ! (↵ a )+ + ! (↵ a )+! a + + ! a p . | |v0 | 1 1 1 ··· n n n 1 1 ··· n n|v0  If v is an archimedean place of K,thenthetriangleinequalityyields v n ! max ↵1 v,..., ↵n v . For any non-archimedean place v of K we get a stronger bound due| to| the ultrametric| | {| | | | } triangle inequality, i.e. max ↵ ,..., ↵ . | |v  {| 1|v | n|v} We take the product of v =0overallplaceswiththeappropriatemultiplicities and use the local bounds above| | in6 combination with the product formula to obtain [K: ] h(↵1)+ +h(↵n) Q 1/e(v0) 1 n ! e ··· p .  | | So (34) follows from (33) as ! = ⇢p(a;⌦). | | ⇤ The following corollary generalizes Theorem 1 to subgroups of Fp⇥ of odd order. Corollary 21. Let f 3 be an odd integer and ✏>0. Say ⇠ is a root of unity of order f and define

f 1 2 f 1 ⌦= (b1,...,bf 1) Z : b1(⇠ 1) + b2(⇠ 1) + + bf 1(⇠ 1) = 0 ? . 2 ··· For any prime p with p 1(modf) let Gp Fp⇥ be the subgroup of order f. Then ⌘ ✓ p 1 1 1 tg 4(f 1)'(f) +✏ 2⇡p 1/p log ⇣ =m(⌦)+O p where ⇣ = e p 1 t=1 g Gp X X2 ⇣ ⌘ as p where the implicit constant depends only on f and ✏; in particular, the logarithm!1 is well-deﬁned for all large p. We will prove this corollary further down. Here we treat the hypothesis on ⌦in Theorem 20 for roots of unity.

Lemma 22. Let ⇣1,⇣2,⇣3,⇣4 be pairwise distinct roots of unity. If (⇣1 ⇣2)/(⇣3 ⇣4) Q, then there exist distinct i, j 1, 2, 3, 4 with ⇣ = ⇣ . 2 2{ } i j 1 1 1 Proof. Set ⌘ = ⇣3⇣4 ,⇠ = ⇣1⇣2 , and ⇣ = ⇣2⇣4 and deﬁne ⇣ ⇣ ⇠ 1 x = 1 2 = ⇣ . ⇣ ⇣ ⌘ 1 3 4 We assume x Q and, after possibly swapping ⇣1 and ⇣2,alsox>0. 2 Note that the number ﬁeld K = Q(⇣1,⇣2,⇣3,⇣4) has only complex embeddings as it contains at least 4 roots of unity. So the K/Q-norm NK/ ( )isnevernegative.Wehave Q · N (⇠ 1) (35) x[K:Q] =N (x)= K/Q . K/Q N (⌘ 1) K/Q We now divide into four cases, depending on whether the orders of ⌘ and ⇠ are prime powers or not. THENORMOFGAUSSIANPERIODS 20

First suppose that neither ⌘ nor ⇠ has order a prime power. Then ⌘ 1and⇠ 1are units and hence x =1by(35).Weobtainthevanishingsumofrootsofunity (36) ⌘ ⇣⇠ + ⇣ 1=0. If a non-trivial subsum vanishes, then ⌘ ⇣⇠ = ⇣ 1=0 or ⌘ + ⇣ = ⇣⇠ 1=0 or ⌘ 1= ⇣⇠ + ⇣ =0. The first and third cases are impossible as ⇣ =1and⌘ =1.Thus⇣⇠ = 1andthis 6 6 implies ⇣1 = ⇣4,asdesired. If no non-trivial subsum vanishes, then Mann’s Theorem 1 [17] implies ⌘6 = ⇠6 = ⇣6 = 1. In the current case, ⌘ and ⇠ must have precise order 6. If ⌘ = ⇠,then(36)implies ⌘ =1or⇣ =1whichcontractsthehypothesis.Hence⌘ = ⇠ and thus ⌘⇠ = 1. When combined with (36) we have 6 ⌘ 1 ⌘⇠ ⇠ 1 ⇠ (37) ⇣⇠ = ⇠ = = = 1, ⇠ 1 ⇠ 1 ⇠ 1 and again ⇣1 = ⇣4. Now suppose that ⇠ has order pe with p aprimeande 1andthattheorderof⌘ [K:Q(⇠)] is not a prime power. Then ⌘ 1remainsaunitbutnowNK/Q(⇠ 1) = p ,so '(pe) e e 1 x = p. As x is rational, we must have '(p )=1,sop =2andthus⇣1⇣2 = ⇠ = 1, which completes this case. Similarly, if ⌘ has prime power order and ⇠ does not, then we apply the argument 1 1 1 from the last paragraph to x = ⇣ (⌘ 1)/(⇠ 1) and conclude ⇣3⇣4 = ⌘ = 1, as desired. Finally, suppose ⌘ has order qe0 and ⇠ has order pe,herep and q are primes and 1/'(pe) 1/'(qe0 ) e e e, e0 1. Now (35) implies x = p q Q.Ifp = q,thenp = q 0 =2andso 2 1/'(pe) 1/6 '(pe0 ) ⌘ = ⇠ = 1, as desired. So say p = q,hencex = p Q and it follows that 2 1 1 Z (p 1)pe 1 (p 1)pe0 1 2 and this entails e = e0.Wefindx =1andarethusbackinthesituationofthefirst case except that ⌘ and ⇠ now have prime power order. We proceed similarly. If a non- trivial subsum in (36) vanishes, then again ⇣1 = ⇣4.OtherwiseMann’sTheoremyields ⌘6 = ⇠6 =1.Thistime⌘ and ⇠ have equal order which is 2 or 3. If the common order is 2thenwearedone.Elsewiseitis3andagainwemusthave⌘ = ⇠ which again implies ⌘⇠ =1.Weconclude⇣ = ⇣ as in (37). 6 1 4 ⇤ Proof of Corollary 21. Let ⇠ be a root of unity of order f.Weset 2 f 1 ↵1 = ⇠ 1,↵2 = ⇠ 1,...↵f 1 = ⇠ 1. Our aim is to apply Theorem 20 to K = Q(⇠)andn = f 1 2. Say ⌦is as in the said theorem. Observe that f 2 ↵1Z + + ↵nZ =(⇠ 1)(Z +(⇠ +1)Z + +(⇠ + + ⇠ +1)Z)=(⇠ 1)Z[⇠] ··· ··· ··· f 1 f 2 as ⇠ = (⇠ + + ⇠ +1).Thisgrouphasrankm = '(f)andtherefore⌦also has rank m 2. ··· THENORMOFGAUSSIANPERIODS 21

We shall show that the hypothesis on ⌦in Theorem 20 is satisfied. Indeed, if i, j, k, l 0,...,n are pairwise distinct, then 2 { } ⇠i ⇠j Q ⇠k ⇠l 2 implies that ⇠ has even order by Lemma 22. This contradicts our hypothesis on f. Finally, observe that p 1(modf)meansthatp splits completely in K.Forsuchp ⌘ there is (a1,...,an)asabove(34)wherev0 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+⇣ta1 + + ⇣tan = ⇣t + ⇣t(1+a1) + + ⇣t(1+an) ··· ··· for all 1 t p 1. ⇤   Proof of Theorem 1. We set ⌦as in the proof of Corollary 21. Its rank is '(f)=f 1 f 1 as f is a prime by hypothesis. Since ⌦is a primitive subgroup of Z we conclude f 1 ⌦=Z .Inthedefinition(32)ofm(⌦)wemaytakeA to equal the (f 1) (f 1) ⇥ unit matrix. The resulting logarithmic Mahler measure is that of 1+X1 + +Xf 1. ⇤ ··· 2⇡p 1g/p Proof of Corollary 2. Observe that g G e is an algebraic integer. In view of 2 Theorem 1 it suces to verify m(1 + X1 + + Xf 1) = 0. The higher dimensional version of Kronecker’s Theorem classifiesP integral··· polynomials 6 whose logarithmic Mahler measure vanishes, see work of Boyd [6], Lawton [13], and Smyth [25]. As our 1 + X1 + + Xf 1 is irreducible and f 3weeasilydeducefromthisclassificationthatits ··· logarithmic Mahler measure is non-zero. ⇤ Appendix On 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 1isaninteger.Recall that the definition of ⇢ is given in (29). 1 1 n Theorem 23 (Lawton, Theorem 2 [14]). Suppose P C[X1± ,...,Xn± ]r 0 .Ifa Z , 2 { } 2 then m(P (Xa1 ,...,Xan )) = m(P )+o(1) as ⇢(a; Zn) . !1 We closely following Lawton’s approach but keep track of estimates to obtain the following refinement. 1 1 Theorem 24. Suppose P C[X1± ,...,Xn± ] r 0 has k 2 non-zero terms and let 2 n { } ✏>0. For a =(a1,...,an) Z we have 2 1 a1 an n 4(k 1) +✏ m(P (X ,...,X )) = m(P )+O ⇢(a; Z ) as ⇢(a; Zn) where the implied constant depends⇣ on n, P, and ✏. ⌘ !1 An important tool is an estimate on the measure of the subset of the unit circle, where apolynomialtakessmallvalues.Weusevol()todenotetheLebesguemeasureonRn. 1 1 · For P C[X1± ,...,Xn± ]andy>0wedefine 2 S(P, y)= x [0, 1)n : P (e(x))

Lemma 25 (Lawton, Theorem 1 [14]). For k 1 there exists a constant C > 0 with k the following property. If P C[X] is a monic polynomial with k non-zero terms and 2 1/ max 1,k 1 y>0, then vol(S(P, y)) C y { }  k The reference covers the important case k 2. If k =1,then P (e(x)) =1forall n | | x R and the claim is clear with C1 =1. 2 1 1 Lemma 26. Assume P C[X1± ,...,Xn± ] r 0 has k 1 non-zero terms. 2 { } 1/(2 max 1,k 1 ) (i) If for all y (0, 1] then vol(S(P, y)) = O y { } where the constant in O( ) depends2 only on P . q (ii) If q>· 0 then log P (e(x)) dx is well-deﬁned and ﬁnite. [0,1)n | | R Proof. Our proof of (i) is by induction on n.WemayassumethatP is a polynomial. If n =1,thenthelemmafollowsfromLawton’sTheoremafternormalizingP . Say n 2. There is nothing to prove if k =1,sosupposek 2. We may also assume d that P is a polynomial, hence P = P0Xn + + Pd where P0,...,Pd C[X1,...,Xn 1] 1/2 ··· n 1 n 1 2 and P0 =0.Weabbreviate⌃=S(P0,y ) [0, 1) .Ifx0 R and P0(e(x0)) =0, 6 ✓ 2 6 then P (e(x0),X)/P0(e(x0)) is a monic polynomial in X. Fubini’s Theorem implies

vol(S(P, y)) = vol(S(P (e(x0),X),y))dx0 + vol(S(P (e(x0),X),y))dx0 n 1 Z⌃ Z[0,1) r⌃

vol(⌃) + vol(S(P (e(x0),X)/P0(e(x0))),y/P0(e(x0)) )dx0  n 1 | | Z[0,1) r⌃ 1/2 vol(⌃) + vol(S(P (e(x0),X)/P0(e(x0))),y )dx0  n 1 Z[0,1) r⌃ 1/(2k 2) vol(⌃) + C y  k since vol(S(P (e(x0),X),y)) 1andbyLemma25.  By this lemma applied by induction to P0 C[X1,...,Xn 1] r 0 we conclude 1/(2k 2) 2 { } vol(⌃) c(P0)y .Thisyieldspart(i). The statement in (ii) is possible known, we give a proof based on (i). For an integer n q m 0andx R we deﬁne pm(x)=min m, log P (e(x)) 0 which is interpreted 2 { | | || } as m if P (e(x)) = 0. Then pm is a non-decreasing sequence of continuous functions on n [0, 1) .WesetIm = [0,1)n pm(x)dx.BytheMonotoneConvergenceTheoremitsuces to prove that the non-decreasing sequence (Im)m 1 converges. For all suciently large R m1/q n m1/q m we have P (e(x)) e if x [0, 1) .Observethatp equals m on S(P, e ) | | 2 m and that it coincides with pm+1 outside this set. Thus for all large m we have

m1/q m1/q/(2k) Im+1 Im = (pm+1(x) pm(x))dx vol(S(P, e )) = O e S(P,e m1/q )  Z ⇣ ⌘ as pm+1(x) m +1andwhereweused(i),theimpliedconstantisindependentofm.  m1/q/(2k) Since m 1 e < we can use a telescoping sum to show that supm 0 Im < , 1 1 as desired. ⇤ P b n Let N0 denote the non-negative integers. Say b N0 and let g lie in C (R ), the set of 2 real valued functions on Rn whose derivatives exist up-to and including order b and are THENORMOFGAUSSIANPERIODS 23

n continuous. For a multiindex i =(i1,...,in) N0 we set `(i)=i1 + + in.If`(i) b, 2 ···  then we define @ig =(@/@x )i1 (@/@x )in g and 1 ··· n i g b =maxsup @ g(x) C n | | i N0 x Rn | | `(2i) b 2  which is possibly . We now introduce1 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 Cb(R)isnon-decreasingwith(0) = 0,(1) = 1, (x)=0ifx<0, and (x)=1if2 x>1. We ask in addition that @i(0) = @i(1) = 0 for all i 1,...,b .For example, we could take the anti-derivative of xb(1 x)b that attains 02 at{x =0,scaleit} to attain 1 at x =1,andextendby0forx<0andby1for x>1. 2 2 2 Say y (0, 1/2]. We define y as x ((2/y) x 1)/3, which rescales [(y/2) ,y ]to [0, 1], composed2 with .Then 7! 4 i @i( )(t) @i(t) | y | 3y2 ✓ ◆ for all i 0,...,b and all t R. Hence 2{ } 2 2b (38) b = O (y ), | y|C b, here and below the implied constant depends on the quantities appearing in the sub- script. Finally, we define y as 1 (t)logt : t>0, (t)= 2 y y 0:t 0. ⇢  b Then y C (R)andallitsderivativesup-toandincludingorderb vanish outside of ((y/2)2,y2). Thus 2b (39) b = O (y log y ) | y|C b, | | by the Leibniz product rule applied using (38) and since y (0, 1/2]. 2 2 n Say P C[X1,...,Xn] and write g(x)= P (e(x)) ,thisisasmoothfunctionR R and 2 | | !

(40) g b = O (1). | |C b,P,n n We deﬁne fy = y g.IfP (e(x)) =0withx R ,then 6 2 f (x)= ( P (e(x)) 2)= ( P (e(x)) 2)log P (e(x)) y y | | y | | | | and for any x Rn we have 2 0:ifP (e(x)) y/2, f (x)= y log P (e(x)) :if|P (e(x))|y. ⇢ | | | | n b n Moreover, the composition fy : R [0, 1] lies in C (R ). We can bound its norm using the following lemma. !

b b n Lemma 27. Let b 0, C (R), and g C (R ) satisfy Cb < and g Cb < . b(b 1)/2 2 2b | | 1 | | 1 Then g b 2 b max 1, g b . | |C  | |C { | |C } THENORMOFGAUSSIANPERIODS 24

n Proof. The lemma is evident for b =0.Sosayb 1andleti N0 r 0 with n n2 { } ` = `(i) b.Letj N0 be a standard basis vector with i j N0 . Using the chain rule and the Leibniz2 product rule we ﬁnd 2 i i j j i j j ` 1 @ ( g) 0 = @ (@ ( g)) 0 = @ (( 0 g)@ g) 0 2 0 g ` 1 g ` . | |C | |C | |C  | |C | |C i b 1 b(b 1)/2 b 0 b 1 b b b Thus @ ( g) C 2 0 g C g C 2 C max 1, g C ,whereweapplied | |  | | | |  | | { | | } i this lemma by induction to bound 0 g b 1 from above. This upper bound for @ ( | |C | g) 0 continues to hold for i =0anditisthereforeanupperboundfor g b . |C | |C ⇤ This lemma, together with (39) and (40), implies 2b (41) f b = O (y log y ). | y|C b,P,n, | | From now on we suppose b n + 1. In the next three lemmas and if not stated otherwise, P C[X1,...,Xn] r 0 is a polynomial with k 2non-zeroterms. 2 { } Lemma 28. Suppose a Zn and y (0, 1/2], then 2 2 1 log y 1 fy(at)dt = fy(x)dx + Ob,P,,n | 2b | n b n . n y ⇢(a; ) Z0 Z[0,1) ✓ Z ◆ Proof. All implied constants in this proof depend only on b, P, , and n.TheFourier b n coecients fy(m)offy C (R), here m Z ,decayquickly.Indeed,byTheorem 1 2 a 2 a a 3.2.9(b) [11] with s = b and @ fy(m) @ fy C0 @ fy Cb where `(i)=b.We conclude b | || | | | f b log y f (m) = O | y|dC and so f (m) = O | | | y | m b | y | y2b m b ✓ | | ◆ ✓ | | ◆ n log y 1 b b | | for all m Z r 0 by (41). Say H 1, then m H fy(m) = O y2b m H m b , 2 { } | | | | | | | | and hence P ⇣ P ⌘ log y 1b (42) fy(m) = O | 2b | b n | | y H m H ✓ ◆ | X| as b n 1 b Since the Fourier coecients of the continuous function fy are absolutely summable, its Fourier series converges uniformly to fy. Hence 1 1 2⇡p 1 a,m t fy(at)dt = fy(m)e h i dt = fy(0) + fy(m). 0 0 Z m Zn Z m Znr 0 X2 2a,mX=0{ } b b h i b

Now fy(0) equals [0,1)n fy(x)dx,sotheestimateintheassertionfollowsfrom(42)and R b log y 1 fy(m) fy(m) = O | 2b | n b n . ⇤  | | y ⇢(a; Z ) m Znr 0 m ⇢(a;Zn) ✓ ◆ 2a,mX=0{ } | |X h i b b 1 in the reference is the `2-norm |·| THENORMOFGAUSSIANPERIODS 25

Lemma 29. Suppose each non-zero term of P has modulus at least 1.Ifa Zn such 2 that ⇢(a; Zn) is suciently 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| 2 2 1 1 1/(k 1) log P (e(as)) ds = f (as)ds + O y log y | | y | | Z0 Z0 where the implicit constant depends only on P and n.

n Proof. Say a =(a1,...,an)andsupposethat⇢(a; Z )isstrictlylargerthan m m0 for a1 an | | all distinct m, m0 in the support of P .ThenP (X ,...,X )hasthesamecoecients as P .Itisinparticularnon-zeroandthefirstclaimholdstrue.Forthesecondclaim we can apply Lawton’s Lemma 4 [14]. ⇤ Lemma 30. Let ✏>0.Ify (0, 1/2] then 2 1 2(k 1) ✏ (fy(x) log P (e(x)) )dx = O y n | | Z[0,1) ⇣ ⌘ where the implied constant depends only on n, P, and ✏. Proof. Let p>1andfixq>1suchthat1/p+1/q =1.Bydefinitionwegettheequality in

2 (fy(x) log P (e(x)) )dx = (y( P (e(x)) ) 1) log P (e(x)) dx n | | n | | | | Z[0,1) Z[0,1) 2 y( P (e(x)) ) 1 log P (e(x)) dx  n | | | || | || Z[0,1) 1/p 1/q 2 p q y( P (e(x)) ) 1 dx log P (e(x)) dx  n | | | | n | | || ✓Z[0,1) ◆ ✓Z[0,1) ◆ we used the Hölder inequality in the last step; the final integral is finite by Lemma 26(ii). As ( P (e(x)) 2)=1if P (e(x)) y we have y | | | | 2 p 2 p y( P (e(x)) ) 1 dx = y( P (e(x)) ) 1 dx vol(S(P, y)). n | | | | | | | |  Z[0,1) ZS(P,y) Recall k 2. The current Lemma follows from Lemma 26(i) because we may suppose 1/(2p(k 1)) 1/(2k 2) ✏. ⇤ Proof of Proposition 24. Without loss of generality, we may suppose that all non-zero coecients of P have modulus at least 1 and that P is a polynomial. We fix b n +1 suciently large and a suciently small ✏0 > 0with 1 b n (43) + ✏0 + ✏ where = . 2(k 1) 4(k 1) 2b +1/(2k 2) We fix a step function Cb(R)asabove,abbreviateH = ⇢(a; Zn) 1, and set 2 y = H . THENORMOFGAUSSIANPERIODS 26

For H large enough we ﬁnd y 1/2andthat m(P (Xa1 ,...,Xan )) m(P ) equals  | | 1 log P (e(as)) ds log P (e(x)) dx | | n | | Z0 Z[0,1) 1 1 fy(as)ds fy(x)dx + (log P (e(as)) fy(as))ds  n | | Z0 Z[0,1) Z0 + (fy(x) log P (e(x)) )dx . n | | Z[0,1) Then by Lemmas 28, 29, and 30, the ﬁnal one applied to a suciently small ✏0,thissum is in log y 1 1 2(k 1) ✏0 O | | + y y2b Hb n ✓ ◆ where the implied constant here and below depends only on b, P, ,and✏. So the sum is in 2b+n b 2(k 1) +✏0 O (log H)H + H

The proposition follows for⇣ small enough ✏0 as 2b + n b =⌘ /(2k 2), cf. (43). ⇤ References 1. P. Autissier, Sur une question d’équirépartition de nombres algébriques, C. R. Math. Acad. Sci. Paris 342 (2006), no. 9, 639–641. 2. J. Ax, On Schanuel’s conjectures, Ann. of Math. (2) 93 (1971), 252–268. 3. B. C. Berndt, R. J. Evans, and K. S. Williams, Gauss and Jacobi sums, Canadian Mathematical Society Series of Monographs and Advanced Texts, John Wiley & Sons, Inc., New York, 1998. 4. E. Bombieri and W. Gubler, Heights in Diophantine Geometry, Cambridge University Press, 2006. 5. E. Bombieri, D. W. Masser, and U. Zannier, Anomalous Subvarieties - Structure Theorems and Applications, Int. Math. Res. Not. IMRN (2007), no. 19, 1–33. 6. D. W. Boyd, Kronecker’s theorem and Lehmer’s problem for polynomials in several variables,J. Number Theory 13 (1981), no. 1, 116–121. 7. V. Dimitrov, Convergence to the Mahler measure and the distribution of periodic points for algebraic Noetherian Zd-actions, arXiv:1611.04664. 8. W. Duke, A combinatorial problem related to Mahler’s measure, Bull. Lond. Math. Soc. 39 (2007), no. 5, 741–748. 9. W. Duke, S. R. Garcia, and B. Lutz, The graphic nature of Gaussian periods, Proc. Amer. Math. Soc. 143 (2015), no. 5, 1849–1863. 10. M. Einsiedler, M. Kapranov, and D. Lind, Non-Archimedean amoebas and tropical varieties,J. Reine Angew. Math. 601 (2006), 139–157. 11. L. Grafakos, Classical Fourier analysis, third ed., Graduate Texts in Mathematics, vol. 249, Springer, New York, 2014. 12. P. Habegger, Diophantine Approximations on Definable Sets, arXiv:1608.04547. 13. W. M. Lawton, A generalization of a theorem of kronecker, J. Sci. Fat. Chiangmai Unit. (Thailand) 4 (1977), 15–23. 14. , A problem of Boyd concerning geometric means of polynomials, J. Number Theory 16 (1983), no. 3, 356–362. 15. D. Lind, K. Schmidt, and E. Verbitskiy, Homoclinic points, atoral polynomials, and periodic points of algebraic Zd-actions, Ergodic Theory Dynam. Systems 33 (2013), no. 4, 1060–1081. 16. K. Mahler, An inequality for the discriminant of a polynomial, Michigan Math. J. 11 (1964), 257– 262. 17. H. B. Mann, On linear relations between roots of unity, Mathematika 12 (1965), 107–117. THENORMOFGAUSSIANPERIODS 27

18. G. Myerson, A combinatorial problem in finite fields. I, Pacific J. Math. 82 (1979), no. 1, 179–187. 19. , A combinatorial problem in finite fields. II, Quart. J. Math. Oxford Ser. (2) 31 (1980), no. 122, 219–231. 20. , Unsolved Problems: How Small Can a Sum of Roots of Unity Be?, Amer. Math. Monthly 93 (1986), no. 6, 457–459. 21. J. Pila and A. J. Wilkie, The rational points of a definable set, Duke Math. J. 133 (2006), no. 3, 591–616. 22. Q. I. Rahman and G. Schmeisser, Analytic theory of polynomials, London Mathematical Society Monographs. New Series, vol. 26, The Clarendon Press, Oxford University Press, Oxford, 2002. 23. A. Schinzel, Polynomials with special regard to reducibility, Encyclopedia of Mathematics and its Applications, vol. 77, Cambridge University Press, Cambridge, 2000, With an appendix by Umberto Zannier. 24. W.M. Schmidt, Diophantine approximations and Diophantine equations, Lecture Notes in Mathe- matics, vol. 1467, Springer-Verlag, Berlin, 1991. 25. C. J. Smyth, A Kronecker-type theorem for complex polynomials in several variables, Canad. Math. Bull. 24 (1981), no. 4, 447–452. 26. , On measures of polynomials in several variables, Bull. Austral. Math. Soc. 23 (1981), no. 1, 49–63. 27. L. van den Dries, Tame topology and o-minimal structures, London Mathematical Society Lecture Note Series, vol. 248, Cambridge University Press, Cambridge, 1998. 28. L. van den Dries and C. Miller, Geometric categories and o-minimal structures, Duke Math. J. 84 (1996), no. 2, 497–540.

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2016-06 M. Dambrine, I. Greff, H. Harbrecht, B. Puig Numerical solution of the homogeneous Neumann boundary value problem on domains with a thin layer of random thickness 2016-07 G. Alberti, G. Crippa, A. L. Mazzucato Exponential self-similar mixing by incompressible flows

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2016-09 Gabriel A. Dill Effective approximation and Diophantine applications

2016-10 J. Blanc, S. Zimmermann Topological simplicity of the Cremona groups

2016-11 I. Hedén, S. Zimmermann The decomposition group of a line in the plane

2016-12 J. Ballani, D. Kressner, M. Peters Multilevel tensor approximation of PDEs with random data

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Preprints are available under https://math.unibas.ch/research/publications LATEST PREPRINTS

No. Author: Title

2016-13 M. J. Grote, M. Kray, U. Nahum Adaptive eigenspace method for inverse scattering problems in the frequency domain

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2016-16 A. Hyder Conformally Euclidean metrics on ⇥⇤⇤⌅⇧⌃⌥⇤ ⌦⇧⌃ ↵⇤⌃⌃⇤✏⇣⌘ ✓⌃⌘ ◆⇤

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2016-18 R. N. Gantner, M. D. Peters Higher order quasi-Monte Carlo for Bayesian shape inversion

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2016-20 S. Dahlke, H. Harbrecht, M. Utzinger, M. Weimar Adaptive wavelet BEM for boundary integral equations: Theory and numerical experiments

2016-21 A. Hyder, S. Iula, L. Martinazzi Large blow-up sets for the prescribed Q-curvature equation in the Euclidean space

2016-22 P. Habegger The norm of Gaussian periods

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Preprints are available under https://math.unibas.ch/research/publications