VG1 .- Conjecture of Lehmer, Conjecture of Schinzel-Zassenhaus

1 / 49 Obj. : to study the problem of the minoration of the Mahler measure M(α) of a nonzero algebraic number α, which is not a root of unity, in higher dimension

the minoration of the height h(P) of a rational point P of an algebraic variety, by means of the dynamical zeta function ζβ (z) of the β-shift.

2 / 49 1.– The Mahler measure of a polynomial

To study solutions of Diophantine equations it is natural to try to associate to such solutions a “size”, a ”complexity” or a ”measure”. Any computation on partial sets of solutions then takes advantage of some finiteness properties when this size lies below some bound. The first step consists in attributing a real value to an algebraic number, to solve the problem of finding such a “size” for algebraic numbers. An algebraic number α is described uniquely by its minimal polynomial Pα (X), or say f ∈ Z[X]. It has integer coefficients, which are relatively prime and by convention its dominant coefficient is positive. It is natural to establish several notions of “size” on polynomials.

3 / 49 Let f ∈ C[X] be a nonzero polynomial with complex coefﬁcients of degree d :

d d d−1 f = a0X + a1X + ...ad = a0 ∏(X − αi ). i=1 Several notions of height for f ; at least : (i) the Mahler measure of f :

d M(f ) = |a0| ∏ max{1,|αi |}, i=1 (ii) the usual height of f :

H(f ) = max{|a0|,|a1|,...,|ad |}, (iii) the Euclidean norm of f :

Z 1 1/2 2 2 2 1/2 2iπt 2 L2(f ) = (|a0| + |a1| + ... + |ad | ) = |f (e )| dt , 0

4 / 49 (iv) the sup norm on the unit disc (or on the unit circle, which is the same by the maximum modulus principle) :

|f |1 = sup |f (z)| = sup |f (z)|, |z|≤1 |z|=1

(v) the length of f

L(f ) = |a0| + |a1| + ... + |ad |,

(vi) Bombieri’s norm of f (for factorization algorithms) :

!1/2 d |a |2 [f ] = i ∑ d i=0 i Inequalities :

−1/2 d (d + 1) M(f ) ≤ H(f ) ≤ L2(f ) ≤ |f |1 ≤ L(f ) ≤ 2 M(f ).

5 / 49 and, from the symmetric functions of the roots

d |a | ≤ M(f )(Norm inequality). i i

[recall a = (−1)d−i a ... ] i 0 ∑ αi1 αid−i i1,...,id−i ⊂{1,...,d}

If f is monic and has roots only in D(0,1), then

M(f ) ≤ [f ] since [f ] ≥ 1. The same holds if all the roots of f are outside D(0,1). So if we want [f ] ≤ M(f ) some roots should be inside some outside D(0,1).

6 / 49 Theorem (Degot´ - Jenvrin, ’98)

Let f ∈ C[X] be monic, deg(f ) = d decomposed as d = n1 + n2 + n3, where n1 = number of roots inside D(0,α), 0 < α < 1, n2 = number of roots outside D(0,β), β > 1, and where n1,n2,n3,d,α,β satisfy :

2 n1 1 n1 1 + α n2 1 + αβ n3 1 + α (1 + αβ)n1 (1 + )n2 (1 + β)n3 × + + β 2 d 1 + αβ d 1 + β 2 d 1 + β

n 1 + α n 1 + β n 2 n3 × 1 + 2 + 3 ≤ 1, d 1 + αβ d 1 + β 2 d 1 + β then [f ] ≤ M(f ).

7 / 49 Theorem (Jensen’s formula) For any f ∈ C[X], f 6≡ 0,

Z 2π M(f ) = exp Log|f (eit )|dt 0

Pf. The logarithm is additive. It sufﬁces to consider f (z) = z − α. 3 cases : |α| > 1 : the function Log|f (z)| is harmonic in a neighbourhood of the unit circle ; hence Log|f (0)| = Log|α|. |α| < 1 : g(z) = 1 − αz has no zero on |z| = 1, and Log|g(z)| is harmonic in a neighbourhood of |z| = 1. We have : |f (z)| = |g(z)| on |z| = 1, hence

8 / 49 Pf. |α| = 1 : it comes from

1 Z 2π 1 Z 2π Log|eit − α|dt = Log|eit − 1|dt = 0. 2π 0 2π 0

Deﬁnition The logarithmic Mahler measure of an algebraic number α is LogM(α), and denoted by m(α).

9 / 49 Deﬁnition The Mahler measure of an algebraic number α is the Mahler measure of its minimal polynomial Pα (X) ∈ Z[X]. If

P(X) = a0(X − α1)(X − α2)...(X − αn) = n n−1 a0X + a1X + ... + an−1X + an ∈ C[X], a0an 6= 0 then M(P) := |a0| ∏ |αi | and M(P) ≥ |a0|. i,|αi |>1 The smallest Mahler measures to be considered when P has integer coefﬁcients : for |a0| = 1, i.e. over algebraic integers.

10 / 49 Properties : (i) multiplicativity : P = P1 × P2 × ... × Pm ⇒ M(P) = M(P1) ...M(Pm).

M(α1 ...αm) = M(α1)...M(αm),

(ii) if P∗(X) := X deg(P)P(1/X) is the reciprocal polynomial of P, then

M(P∗) = M(P), in particular, for any nonzero α algebraic number

M(±α±1) = M(α),

q d q q(d+1) 1/q (iii) if Pq(z) = a0 ∏j=1(z − αj ) = (−1) ∏ζ,ζ d =1 P(z ζ), then, for any norm on C[X]q, 1/q lim kPqk = M(P), q→∞ (iv) for α any root of unity M(α) = 1.

11 / 49 Theorem (Kronecker, 1857) Let α be an algebraic integer. Then

M(α) = 1 ⇐⇒ α is a root of unity, or α = 0.

Pf.

=⇒ : let f be the minimal polynomial of α. Since M(fq) = 1, the (q) coefﬁcient aj of fq satisﬁes (from the Norm inequality) :

d |a(q)| ≤ . j j

Thus the set of such fq is ﬁnite, and the set of all roots of all fq is ﬁnite. Consequently, αk = αr for some k > r. We deduce αk−r = 1 or α = 0.

12 / 49 Deﬁnition β is a Pisot number if it is an algebraic integer (real) > 1 and if all its (Galois) conjugates |β (i)| < 1. β > 1 is a Salem number if |β (i)| ≤ 1 with at least one conjugate on |z| = 1. β and 1/β are conjugated with pairs of complex conjugated roots of modulus 1. A Perron number β ≥ 1 is an algebraic integer for which either β = 1 or |β (i)| < β if β > 1.

Denote PPerron := {Perron}. The set PPerron is dense in [1,∞).

The notations are usual : S:= {Pisot},T:= {Salem}, Theorem (Adler Marcus, ’79 - Memoirs of the AMS Vol.20, No 219) For any algebraic number α, the Mahler measure M(α) is a Perron number.

13 / 49 “Topological entropy and equivalence of dynamical systems”, uses the theory of Perron-Frobenius. ... in which the topological entropy of an automorphim of the d-dimensional torus = LogM(P)...

Inclusions : S ⊂ PPerron

T ⊂ PPerron

{M(α) | α alg. number} ⊂ PPerron,

{M(P) | P ∈ Z[X]} ⊂ PPerron. The last two inclusions are strict (Dubickas 2004, Boyd 1981) : for example, for m > 3, the Perron number β of minimal polynomial −1 − x + xm is not a Mahler measure.

14 / 49 Theorem (Dubickas, ’04) If β is a Perron number then nβ is a Mahler measure for some n ∈ N. Proof : Assume that β has minimal polynomial P(x) of degree d with ∗ leading coefficient b ∈ N and with na¨ıve height H(β). Let β be the maximal modulus of all roots of P different from β (if d = 1, set β ∗ = 0). Since β is a Perron number, β ∗ < β. There exists infinitely many pairs of prime numbers q 6= t for which β ∗ < q/(tb) < β. Consider such a pair (q,t) such that q > H(β). Then we claim that nβ, where n = qd−1tb, is the Mahler measure of tbβ/q. Indeed tbβ/q is the root of the polynomial P(qx/(tb)). Furthermore, the polynomial td bd−1P(qx/(tb)) has integer coefficients. Its two extreme coefficients are qd and td bd−1P(0). These two integers are relatively prime. Thus td bd−1P(qx/(tb)) is an integer irreducible polynomial. All its roots, except for tbβ/q lie in the unit circle. Then

M(tbβ/q) = qd tbβ/q = qd−1tbβ = nβ.

15 / 49 Usual notations : ∗ A polynomial P ∈ C[X] such that P = P is said reciprocal.

The minimal polynomial of a Salem number is monic and reciprocal i.e. P∗ = P. By convention, a Salem polynomial denotes the product of cyclotomic polynomials by the minimal polynomial of a Salem number.

The minimal polynomial of a Pisot number is nonreciprocal. A Pisot polynomial is the minimal polynomial of a Pisot number ; it is monic, irreducible and nonreciprocal. A Perron polynomial is the minimal polynomial of a Perron number.

Open question (explicit algebraicity of the Mahler measure) : For any algebraic number α what is the minimal polynomial of M(α) ?, i.e. what is the Perron polynomial N−1 N g(X) = b0 + b1X + ... + bN−1X + X ∈ Z[X] such that g(M(α)) = 0?

16 / 49 What are the relations between N, d = deg(α), the bj s and the coefficients ai of the minimal polynomial of α ? If α is a Pisot number or a Salem number, then M(α) = α, N = d, and the coefficient vector of g is exactly the coefficient vector of the minimal polynomial. Open question (“inverse problem for Mahler measures”) : For any Perron number β ∈ PPerron is β the Mahler measure M(α) of some algebraic number α, i.e.

β = M(α)?

[answer : effective method found by Dixon and Dubickas ’04] If yes, how many (ﬁnitely many) predecessors ?

β = M(α1) = M(α2) = ... = M(αs)? [partial answers : Staines, ’12 ]

17 / 49 Adler and Marcus’s questions : m ∗ Q1 : Is the dominant root of z − z − 1 a Mahler measure for every m > 3 ?

∗ Q2 : If α is reciprocal, then must M(α) be reciprocal ? answer to Q1 (Boyd) : no. answer to Q2 (Boyd) : no. There are inﬁnitely many reciprocal α with M(α) non-reciprocal.

M. Staines PhD 2012 (Univ. of East Anglia) “On the Inverse Problem for Mahler measures”

18 / 49 Open limit problems on the derived sets (adherence) : what are the respective sets of limit (accumulation) points of

M(P) for P ∈ Z[X]?

M(α) for α ∈ Q? T (Salem numbers)?

Theorem (Salem, ’44) The set S of Pisot numbers is closed in [1,∞) for the usual topology, i.e. S = S. The smallest Pisot number is Θ = 1.3247..., real root of the trinomial X 3 − X − 1 by a Theorem of Siegel (’44).

19 / 49 Conjecture (Boyd) (i) S ∪ T is closed, (ii) the ﬁrst derived set T (1) of the set of Salem numbers T satisﬁes :

T (1) = S.

Is 1 in the adherence of {M(α) | α ∈ Q}?

Conjecture (Lehmer’s Conjecture, ’33)) There exists c > 0 such that

M(α) ≥ 1 + c for any algebraic number α 6= 0 which is not a root of unity, i.e. the open interval (1,1 + c) ∩ PPerron is deprived of any value of Mahler measure of any algebraic number.

20 / 49 − > discontinuity at 1. Lehmer’s problem (1933) : is it possible to ﬁnd integer polynomials with arbitrarily close to 1 Mahler measure. Absolute logarithmic, Weil height of α : LogM(α) h(α) := d fact : M(α) = M(α−1) ;M(α) = α if α ∈ S (set of Pisot numbers) or T = set of Salem numbers). M(α) = 1 if α is a root of unity. absolute heights, local heights 2.– Minorations of the Mahler measure

Fundamental Question : minoration of the Mahler measure M(α) of a nonzero algebraic integer α which is not a root of unity ?

Theorem (Smyth, ’71) Let f ∈ Z[X], f (0) 6= 0 and f assumed not reciprocal. Then

M(f ) ≥ Θ = 1.3247....

... the Conjecture of Lehmer is true for the subset of algebraic numbers which are not algebraic integers, \ , and for the subset of Q OQ algebraic integers whose minimal polynomials are nonreciprocal.

21 / 49 Corollary (Cassels’s Theorem, ’66) d Let f (z) = ∏j=1(z − αj ) ∈ Z[z] where the roots satisfy

LogΘ |α | < 1 + 1 ≤ j ≤ d. j d Then : f = ±f ∗ (reciprocal or antireciprocal).

Pf. Apply LogΘd M(f ) < 1 + ≤ eLogΘ = Θ d and Smyth’s Theorem.

22 / 49 Proof of Smyth’s Theorem

Let α be any complex number and z − α B (z) = . α 1 − αz

If |α| < 1, then Bα (z) is holomorphic on an open disk containing {|z| ≤ 1} with the property : |Bα (z)| = 1 for |z| = 1. Proposition

Let α1,α2,...,αr be a r-tuple of complex numbers of modulus < 1. Let

r ( ) = ( ) = + + 2 + ... B z ∏ Bαj z c0 c1z c2z j=1

be the Blaschke function associated with the r-tuple. Then

2 2 2 1 = |c0| + |c1| + |c2| + ....

23 / 49 Deduced from

Z 2π Z 2π 1 it 2 1 i(k−l) 1 = |B(e )| dt = ∑ck cl e dt. 2π 0 2π k,l 0

Consider f a real polynomial without zeros on |z| = 1, such that f (0) 6= 0. Let B and Bb the Blaschke function associated with the zeroes of f and f ∗ inside the open unit disk, respectively (multiplicities are allowed). Then B/f has no zeroes, its poles being the roots of f outside the closed unit disk and the 1/α, where α runs through the roots of f in the open unit disk. The same for Bb/f ∗ so that B Bb = . f f ∗ ∗ Let ck , dk and ak the Taylor coefﬁcients of B, Bb and f /f at z = 0 respectively. Assume that the constant and dominant coefﬁcient of f are 1.

Then c = |c0| = |d0| = 1/M(f ) and a0 = ±1.

24 / 49 If, furthermore, f /f ∗ is not constant, then there exists a smallest k ≥ 1 such that ak 6= 0, with the property :

ck = a0dk + ak d0, i.e. ck − (±dk ) = ak d0 6= 0.

We deduce |dk | ≥ |ak d0|/2 = |ak |c/2 or |ck | ≥ |ak |c/2. Hence, from the preceding Proposition applied to B or Bb,

|a |2 1 ≥ c2 + k c2, 4 thus 1/2 |a |2 M(f ) ≥ 1 + k . 4 Assume now that f is a monic integer irreducible nonreciprocal polynomial, with f (0) = 1. f has no roots on the unit circle ; if not, f ∗ would also have the same roots on the unit circle, and everywhere, but it would mean that f = f ∗ that is excluded.

25 / 49 1/2 r |a |2 5 Then |a | ≥ 1 and thus : M(f ) ≥ 1 + k ≥ = 1.118.... Following k 4 4 C. Smyth, the sharpening of the lower bound is a consequence of the following Lemma. Lemma

1. For all real numbers x0,x1,...,xn the inequality holds

2 2 2 2 2 (c0x0) + (c0x1 + c1xn) + ... + (c0xn + ... + cnx0) ≤ x0 + ... + xn ,

2. for   cn cn−1 cn−2 ... c0  cn−1 cn−2 ... c0 0    A =  c ... c 0 0   n−2 0  . .c0 the matrices 1 + A and 1 − A are symmetric positive and deﬁnite,

26 / 49 3. in particular the following inequalities hold

2 1 ≥ c0 + |cn|,

2 2 2 cn 2 cn −(1 − c0 − ) ≤ c2n ≤ 1 − c0 − , 1 + c0 1 − c0 2 1 ≥ d0 + |dn|, 2 2 2 dn 2 dn −(1 − d0 − ) ≤ d2n ≤ 1 − d0 + − . 1 + d0 1 − d0

Pf. n 1. We have : |B(z)| = 1 on |z| = 1. For any p(z) = x0 + x1z + ... + xnz ,

Z 2π n 1 it 2 2 |p(e )| dt = ∑(xj ) , 2π 0 j=0 and, 27 / 49 n Z 2π Z 2π 2 1 it it 2 1 it 2 ∑(c0xj + ... + cj x0) = |B(e )p(e )| dt ≤ |p(e )| dt. j=0 2π 0 2π 0 2. From Cauchy-Schwarz inequality :

±xt Ax ≤ |x||Ax| ≤ |x|2.

3. Let ε = ±1. Since 1 + εA > 0, in particular,   1 + εc2n εcn εc0 1 + εcn εc0 det > 0, det εcn 1 + εc0 0  > 0. εc0 1 εc0 0 1 and the same inequalities from Bb and the dk s.

28 / 49 We had : max{|cn|,|dn|} ≥ |c|/2. From the above inequality in 3., 1 − c2 ≥ c/2, =⇒ M(f )2 − M(f )/2 − 1 ≥ 0.

Hence √ 1 + 17 M(f ) ≥ = 1.28.... 4 Without loss of generality, we can assume M(f ) ≤ 4/3 and B(0) > 0 and ∗ k l l+1 Bb(0) > 0 . We have f /f = 1 + ak z + al z + O(z ) and

k l l+1 B = (1 + ak z + al z + O(z )Bb, with c := c0 = d0 = 1/M(f ), ak 6= 0, al 6= 0. Furthermore

cj = dj , 0 ≤ j < k

ck = dk + c0ak

ck+1 = dk+1 + d1ak

cl−1 = dl−1 + dl−k−1ak

cl = dl + dl−k ak + c0al .

29 / 49 We deduce : |ak | = 1 = |al | and |ck | + |dk | = c.

∗ Indeed, if |ak | ≥ 2, since max{|ck |,|dk |} ≥ c, and by 3. in the Lemma, we would have 1 − c2 ≥ c that is M(f )2 − M(f ) − 1 ≥ 0. Contradiction since M(f ) is assumed less than 4/3, 2 ∗ if |al | ≥ 2, since max{|cl |,|dl |,|dl−k |} ≥ 3 c, and by 3. in the Lemma, we 2 2 would have 1 − c ≥ 3 c. Contradiction since M(f ) is assumed less than 4/3,

∗ we have : |ck | + |dk | ≥ |ck − dk | ≥ c. If we assume a strict inequality, then c = |ck | − |dk | or c = −|ck | + |dk | and max{|ck |,|dk |} ≥ c. Contradiction. 2 cases : • 2k ≤ l (− > proof), • 2k > l (exercise).

30 / 49 Case 2k ≤ l :

We can assume 2k < l. Otherwise, a2k = al is equal to ±1, and we interchange f and f ∗ by k 2k −1 k 2 2k (1 + ak z + a2k z + ...) = 1 − ak z + (ak − a2k )z + ... 2 with ak − a2k = 0 (the value 2 for a2k does not ﬁt). Let us apply the inequalities in 3. of the Lemma : 2 2 2 ck 2 ck −(1 − c − ) ≤ c2k ≤ 1 − c − , 1 + c0 1 − c0 2 2 2 dk 2 dk −(1 − c − ) ≤ −d2k ≤ 1 − c + − . 1 − d0 1 + d0 Adding both : d2 c2 d2 c2 −2(1 − c2) + k + k ≤ c − d = d a ≤ 2(1 − c2) + − k − k . 1 − c 1 + c 2k 2k k k 1 + c 1 − c Let x2 (c − x)2 ρ(x) := 2(1 − x2) − + . 1 + c 1 − c

31 / 49 Since |ak | = 1 and that |ck | + |dk | = c, we deduce

|dk | ≤ max{ρ(|ck |),ρ(|dk |)}.

But, from 3. in the Lemma (and the same with ck and dk interchanged), 2 2 1 − c ≥ |dk | ≥ c − |ck | ≥ c + c − 1. If I = [c + c2 − 1,1 − c2], then

c2 + c − 1 ≤ maxρ(x). x∈I The function x → ρ(x) takes its maximum at x = (1 + c)/2, with (1 + c)/2 ≥ 1 − c2, is increasing on I and thus

c2 (c − 1 + c2)2 c2 + c − 1 ≤ ρ(1 − c2) = 2(1 − c2) − − , 1 + c 1 − c i.e. −c3 − c2 + 1 ≥ 0, that is

M(f )3 − M(f ) − 1 ≥ 0 =⇒ M(f ) ≥ Θ.

32 / 49 Theorem (Dobrowolski, ’79) For any reciprocal algebraic integer α of degree d,

LogLog d 3 M(α) > 1 + (1 − ε) , d > d (ε), Log d 1 where 1 − ε can be replaced by 1/1200 in an effective version (n ≥ 2).

Theorem (Voutier, ’96) For any reciprocal algebraic integer α of degree d,

1 LogLog d 3 M(α) > 1 + , d ≥ 2. 4 Log d

33 / 49 Theorem (Garza, ’07) d Soit P(X) = c ∏i=1(X − αi ) le polynomeˆ minimal d’un nombre algebrique´ α = α1, de degre´ d ≥ 1. Soit

d 1/d 1/d 1/d H(α) := |c| ∏ max{1,|αi |} = M(α) i=1 la hauteur absolue de α. Soit r le nombre de racines reelles´ de P(X). On pose R = r/d. Alors

 q R/2 1− 1 1− 1 2 R + 4 R + 4 H(α) ≥   . (1) 2 elementary proof : Hoehn, Int. J. Number Theory, (2011) 2 pages. (Pf. : uses absolute Mahler measures and local heights). [Notations : do not confuse Na¨ıve height H and absolute height H].

34 / 49 Pf. 0.– Show that H(α) = H(α−1). 1.– Let 0 < a < 1/2 and

a √ |1 − z2| f (z) = z . |z|

f (z) Show that the maximum of z → max{1,|z|} is

a MC = 2 on C ; and that it has

1 − 2a1/4 1 a/2 M = (4a)a R 1 + 2a 1 − 4a2 as maximum on R.

35 / 49 2.– We assume that α is an algebraic integer, i.e. c = 1. Show that

d 1/2−a a ∏f (αi ) = |P(0)| |P(1)P(−1)| ≥ 1 i=1 and H(α) ≥ M−RMR−1. R C 1 1/R −1/2 Deduce, with a = 2 (1 + 4 ) , the result (1). 3.– We now assume that α is an algebraic number which is not an algebraic integer, i.e. c ≥ 2. The previous result can now be extended as follows.

36 / 49 Let K = Q(α) be the number ﬁeld generated by α over Q. The notation v denotes a place of K and |.|v is the normalized corresponding absolute value of K. The local degree is denoted dv = [Kv : R] and αv is the image of α by a Q-automorphism σi (Galois) of the algebraic closure (normal extension) of K, which satisﬁes : dv /d dv /d |α|v = |αv | = |αi | where αi = σi (α). If γ 6= 0 denotes an arbitrary element of K, the normalized height of γ is equal to H(γ) = ∏v max{1,|γ|v }, and we recall the product formula : ∏|γ|v = 1. v

37 / 49 By chosing well γ, show that

−1 a dv /d (|αv − αv | ) −1 1 1 ≤ × (max{1,|αv |}max{1,|α |}) 2 . ∏ −1| dv ∏ v varchim (max{1,|αv |}max{1,|αv }) 2d vnon archim By using the fact that the function

|x − x−1|a g(x) = (max{1,|x|}max{1,|x−1|})1/2 satisﬁes g(x) = g(1/x), deduce

1 ≤ MRM1−R × H(α)1/2H(α−1)1/2. R C Then deduce (1).

38 / 49 4.– Show that the function f (z) introduced above is optimal in the family

{|z|u|1 − z2|w | 0 < u,w < 1/2} to obtain the best minoration in (1). 5.– Show that for R = 1 the lower bound given by (1) is optimal. What is its value ?

These bounds (Dobrowolski, Voutier, Garza) do not take into account the Conjecture of Lehmer by the fact that, when d tends to inﬁnity, the limit minorants are equal to 1.

39 / 49 Conjecture (Schinzel - Zassenhaus, ’65) There exists a constant C > 0 such that, for any algebraic integer α 6= 0, deg[α) = d, not being a root of unity, such that the house α := max{|α(i)|} of α satisﬁes :

C α > 1 + . d

S-Z’s ﬁrst result : if 2s conjuguates of α are not reals, then :

α > 1 + 4−s−2.

Theorem (Dubickas, ’93) 64 LogLog d 3 1 α > 1 + − ε , d > d (ε) π2 Log d d 2

40 / 49 3.– Heights of algebraic numbers

Assume that α = p/q is a rational number, gcd(p,q) = 1. It is the root of its minimal polynomial qX − p. We deﬁne its height by

H(α) := max{|p|,|q|}.

This measures the size, or complexity, of α, and the number of information we need to write down α (number of digits). In addition, {α | H(α) < C} is ﬁnite for any real number C > 0. The height function H possesses a decomposition into local factors.

41 / 49 To each prime number p we can associate the valuation |.|p of Q deﬁned by

−n |α|p = p where pn is the exact power of p in the prime decomposition of α. A valuation of a ﬁeld K is a function v : K \{0} → R+ such that v(α) = 0 if and only if α = 0 which satisﬁes :

v(αβ) = v(α)v(β)

v(α + β) ≤ v(α) + v(β) Two valuations v and w are said equivalent if there is a real number s > 0 such that v(α) = v(α)s for all α ∈ K . By a Theorem of Ostrowski we know all classes of equivalence of valuations : v is either equivalent to a |.|p for some prime number p or to the usual absolute value which will be denoted by |.|∞.

42 / 49 Denote by Hp(α) := max(1,|α|p) the local height of α at the prime number p (ﬁnite places). Let us extend the notation for p = ∞ by H∞(α) := max(1,|α|∞) (inﬁnite place). Theorem For any rational number α 6= 0,

H(α) = max(1,|α|∞) ∏ max(1,|α|p) p<∞ i.e. ∗ H = H∞ × ∏ Hp on Q p<∞

Pf. Prime factor decomposition of the numerator and the denominator of α 6= 0.

43 / 49 Let us generalize this notion of local height to algebraic numbers. For an algebraic number α of minimal polynomial

d f = a0 ∏(X − αj ), j=1 set the absolute height as

1/d " d # H(α) := |a0| ∏ max{1,|αj |} , j=1 so that H(α) = M(f )1/d . The Weil height h is deﬁned as its logarithm :

LogM(α) h(α) = d

44 / 49 The normalizing power 1/d is usually inserted to have a decomposition formula of the height function in local contributions which do not depend upon the ﬁeld from which the valuations are taken.

A class of equivalence of valuations of a number ﬁeld K is called a place of K or a prime of K . The set of places of K is denoted by PK . ∞ The notation PK is reserved for the set of archimedean places of K , i.e. the set of equivalence classes of valuations which extend the usual absolute value on Q (up to equivalence).

Usual normalization : the representatives |.|v for the places v of K can be chosen in a unique way that one has :

∏ |α|v = |NK /Q(α)| (norm) ∞ v∈PK and ∏ |α|v = 1. (product formula) v∈PK

45 / 49 Theorem Let K be a number ﬁeld and α ∈ K . Then

1/[K :Q] H(α) = ∏ max{1,|α|v } . v∈PK

(the value of the rhs does not depend upon the ﬁeld K )

Pf.

(i) K = Q(α) and α integral : in this case |α|v ≤ 1 for all ﬁnite places of K and

r s 2 ∏ max{1,|α|v } = ∏max{1,|σj (α)} ∏ max{1,|σj (α)| } v∈PK j=1 j=r+1

(ii) general case :

1 1/[K :Q[α)] |a0| = ∏ = ∏ ∏max{1,|α|v } p ﬁnite |a0|p p ﬁnite v|p

46 / 49 Properties

(h Weil height) : for algebraic numbers α1 and α2,

∗ h(α1α2) ≤ h(α1) + h(α2),

∗ h(α1 + α2) ≤ Log2 + h(α1) + h(α2), ∗ for any n ∈ Z and any algebraic number α 6= 0, h(αn) = |n|h(α).

Lehmer’s Conjecture - general form : nonzero algebraic number α not being a root of unity

LogM(α) c0 h(α) = ≥ for some c0 > 0. deg(α) deg(α)

Ref. : M. Waldschmidt, Diophantine Approximation on Linear Algebraic Groups, Transcendence properties of the Exponential Function in Several Variables, Springer (2000), Chap. 3.

47 / 49 4.– Limit Mahler measures

The set of limit points of {M(P) | P(X) ∈ Z[X]} is obtained by the following useful Theorem of Boyd (’81) and Lawton (’83) which correlates Mahler measures of univariate polynomials to Mahler measures of multivariate polynomials : Theorem

let P(x1,x2,...,xn) ∈ C[x1,...,xn]) and r = (r1,r2,...,rn), ri ∈ N>0. Let r r r Pr (x) := P(x 1 ,x 2 ,...,x n ). Let

n n q(r) := min{H(t) | t = (t1,t2,...,tn) ∈ Z ,t 6= (0,...0), ∑ tj rj = 0}, j=1

where H(t) = max{|ti | | 1 ≤ j ≤ n}. Then

lim M(Pr ) = M(P). q(r)→∞

48 / 49 example :

lim M(−1 + x + xn) = M(−1 + x + y). n→∞

49 / 49