A corollary to a theorem of Laurent Mignotte Nesterenko

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1 ACTA ARITHMETICA XXXVI (1998 A corollary to a theorem of aurent Mignotte Nesterenko by M Mignotte (Strasbourg 1 Introduction For any algebraic number α of degree d on Q, whose minimal polynomial over Z is a d i=1 (X α(i the roots α (i are complex numbers, we define the absolute logarithmic height of α by h(α = 1 ( d log a log max(1, α (i d i=1 et α 1, α be two non-zero algebraic numbers, and let log α 1 and log α be any values of their logarithms We consider the linear form Λ = b log α b 1 log α 1, b 1 and b are positive integers Without loss of generality, we suppose that α 1 and α are 1 Put The main result of [MN] is: D = [Q(α 1, α : Q]/[R(α 1, α : R] Theorem 1 et K be an integer, an integer, and R 1, R, S 1, S integers > 0 et ϱ be a real number > 1 Put R = R 1 R 1, S = S 1 S 1, N = K, g = 1 4 N 1RS, b = ((R 1b (S 1b 1 ( K 1 /(K k! K et a 1, a be positive real numbers such that for i = 1, Suppose that (1 k=1 a i ϱ log α i log α i Dh(α i, Card{α r 1α s : 0 r < R 1, 0 s < S 1 }, Card{rb sb 1 : 0 r < R, 0 s < S } > (K Mathematics Subject Classification: Primary 11J86 [101]

2 10 M Mignotte and that ( K( 1 log ϱ (D 1 log N D(K 1 log b g (Ra 1 Sa > 0 Then { Se Λ ϱ K1/ with Λ S Λ /(b } = Λ max, ReR Λ /(b1 b b 1 In the case when the numbers α 1 and α are multiplicatively independent we shall deduce from Theorem 1 the following result, which is a variant of Théorème of [MN] Theorem 15 Consider the linear form Λ = b log α b 1 log α 1, b 1 and b are positive integers Suppose that α 1 and α are multiplicatively independent Put D = [Q(α 1, α : Q]/[R(α 1, α : R] et a 1, a, h, k be real positive numbers, and ϱ a real number > 1 Put λ = log ϱ and suppose that ( ( b1 h D log b ( log λ f(k 00, a a 1 (4 a i max{1, ϱ log α i log α i Dh(α i } (i = 1,, (5 and f(x = log (1 x 1 x x 1 Then we have the lower bound log Λ λk a 1 a a 1 a λ, x x 1 log x 6x(x 1 log 4 log x 1 = [h/λ], K = 1 [ka 1 a ] max{λ( 05log(( / k max{a 1, a }, D log }, provided that k satisfies with ku V k W 0 U = ( 1λ h, V = /, W = 1 ( 1 1 a 1 a a 1 a

3 Theorem of aurent Mignotte Nesterenko 10 Remark 1 Put = V 4UW The condition on k implies k k 0 k0 = V, k 0 = V V U 4U = V U W U V V U U 4W U with V U = 1 λ (h λ 1 λ 1 (h λ (h λ (h λ = λ, since (V/U/ < 0 and (1 h/λ, and W = 1 ( 1 1 a 1 a a 1 a (1, a 1 a so that W U a 1 a 1 λ (h λ 4 λ 1 a 1 a λ 1, 4 since a 1 a λ Hence k 4/(9λ and ka 1 a kλ 9 4 ( (1 = ψ( (say Clearly ψ increases with and computation gives ψ( > 6 Proof of Theorem 15 We suppose that α 1 and α are multiplicatively independent, and we apply Theorem 1 with a suitable choice of the parameters The proof follows the proof of Théorème of [MN] For the convenience of the reader we keep the numbering of formulas of [MN], except that formula (5i in [MN] is here formula (i; moreover, when there is some change the new formula is denoted by (i Put (1 Recall that = [h/λ], K = 1 [ka 1 a ], R 1 = 1 [ a /a 1 ], S 1 = 1 [ a 1 /a ], R = 1 [ (K 1a /a 1 ], S = 1 [ (K 1a 1 /a ] a i ϱ log α i log α i Dh(α i for i = 1, By the iouville inequality, log Λ D log Db 1 h(α 1 Db h(α D log 1 (b 1a 1 b a = D log 1 b a 1 a, b = b 1 a b a 1

4 104 M Mignotte We consider two cases: b λk or b > λk In the first case, iouville s inequality implies log Λ D log λk a 1 a and Theorem 15 holds Suppose now that b > λk Then max{b 1 /a, b /a 1 } > λk, hence b 1 > λ k (K 1a /a 1 or b > λ k (K 1a 1 /a Since k 4/(9λ and, we have λ k > 1, which proves that Card{rb sb 1 : 0 r < R, 0 s < S } = R S and, by the choice of R and S, this is > (K 1 Moreover, since α 1 and α are multiplicatively independent we have Card{α r 1α s : 0 r < R 1, 0 s < S 1 } = R 1 S 1 This ends the verification of condition (1 of Theorem 1 Remark The condition b > kλ implies by Remark 1 and λ/d h/d (log(kλ log λ f(k ( log(ψ( log 4 > 881, Suppose that ( holds Then Theorem 1 implies Notice that log Λ Kλ λ/, { Se Λ S Λ /(b } = Λ max, ReR Λ /(b1 b b 1 R = R 1 R 1 a /a 1 (K 1a /a a ka 1 (1/ ka 1 (1/ ka, A = max{a 1, a } and, in the same way, This shows that S = S 1 S 1 1 (1/ ka max{r, S} (1/ k A

5 Theorem of aurent Mignotte Nesterenko 105 As we may, suppose that log Λ λk a 1 a 4 Then { R Λ max, S Λ } (11 k a 1 a e λk a 1 a 4 b b 1 ( a 1 a e 4 a 1 a /(9λ 4, λ since k 4/(9λ and λk a 1 a > 1 The last term is an increasing function of λ, thus for λ 1, { R Λ max, S Λ } ( a 1 a e 4 a 1 a /9 4 < 01 b b 1 since a 1 a 4 For λ 1, { R Λ max, S Λ } b b 1 and, since a 1 a λ, we get { R Λ max, S Λ } b b 1 In all cases, which implies ( ( a 1 a e 4 a 1 a /(9λ 4 λ e 4 λ/9 4 < e 4 /9 4 < 01 Λ Λ ( (1/ k max{a 1, a }, log Λ λk a 1 a λ( 05 log(( / k max{a 1, a } and Theorem 15 follows Now we have to verify that condition ( is satisfied: we have to prove that Φ 0 = K( 1 log ϱ (D 1 log N D(K 1 log b g (Ra 1 Sa > 0, when b > λk We replace this condition by the two conditions Φ > 0, Θ > 0, Φ 0 Φ Θ The term Φ is the main one, Θ is a sum of residual terms As indicated in [MN], the condition Φ > 0 leads to the choice of the parameters (1, as Θ > 0 is a secondary condition, which leads to assuming some technical hypotheses on h and a 1, a Here, we follow the advice given in [MN]: for some applications one can modify these technical hypotheses As in [MN] (emme 8 we get (17 log b log ( b1 a b a 1 h D 00 D log λ log(πk/ e f(k K 1 log(πk/ e, K 1

6 106 M Mignotte which follows from the condition h D(log b log λ f(k 00 emme 9 of [MN] gives (18 Put g(ra 1 Sa 1 / (K 1a 1 a / a 1 a 1 (a 1 a / a 1 a 6(1 K 1 (1 and ( Φ = K( 1λ Kh / (K 1a 1 a / a 1 a (a 1 a Θ = 00(K 1 h / ( a 1 a πk 6(1 K 1 D log e (D 1 log(k By (17 and (18 we see that Φ 0 Φ Θ, ka 1 a < K 1 ka 1 a, hence Φ > ka 1 a (( 1λ h a 1 a k which implies / a 1 a Φ a 1 a > ku V k W (a 1 a, This proves that Φ > 0 provided that ku V k W 0 To prove that Θ 0, rewrite ( as Θ = Θ 0 (D 1 Θ 1, ( π Θ 0 = log(λb f(k log log, e ( π Θ 1 = 00K log K log log e log(λb f(k / a 1 a 6(1 K 1 We conclude by proving that Θ 0 and Θ 1 are both positive Since b > kλ, by Remark 1 we have log(λb > ψ(, which shows that Θ 0 is positive

7 Thus, Theorem of aurent Mignotte Nesterenko 107 Notice that, by the proof of Remark, / a 1 a = a 1 a 1 ha 1 a /λ 1 h > 1 (log(ψ( f(k 00 = φ(k (say ( 16π Θ 1 00K log K log 9 f(k e φ(k (1 K 1 and an elementary numerical verification shows that Θ 1 is positive for K 4, which holds by Remark 1 A corollary of Theorem 15 Now we can apply Theorem 15 to get a result closer to Théorème of [MN] Theorem Consider the linear form Λ = b log α b 1 log α 1, b 1 and b are positive integers Suppose that α 1 and α are multiplicatively independent Put D = [Q(α 1, α : Q]/[R(α 1, α : R] et a 1, a, h, k be real positive numbers, and ϱ a real number > 1 Put λ = log ϱ, χ = h/λ and suppose that χ χ 0 for some number χ 0 0 and that ( ( b1 h D log b ( log λ f( K 0 00, a a 1 (4 a i max{1, ϱ log α i log α i Dh(α i } (i = 1,, (5 f(x = log (1 x 1 x x 1 and K 0 = 1 ( χ0 λ Put a 1 a λ, x x 1 log x 6x(x 1 log 4 log x 1 (1 χ 0 λ ( 1 1 4λ χ 0 9 a 1 a a 1 a a 1 a v = 4χ 4 1/χ, m = max{ 5/ (1 χ /, (1 χ 5/ /χ}

8 108 M Mignotte Then we have the lower bound log Λ 1 ( v λ 6 1 v 9 4λv ( 1 max{λ(15 χ a 1 1 a 8λm a 1 a a 1 a log((( χ / ( χ k A ( χ, D log }, A = max{a 1, a } and k = 1 λ ( 1 χ χ 1 ( λ χ (1 χ1/ χ 4 Proof of Theorem We apply Theorem 15 with k = k 0 First we estimate certain quantities of the form k 0 α The formula x x α λx (λ h = = x α 1 ((α 1λx α(λ h (λx (λ h λx α 1 ((α 1x α(1 χ (λx (λ h shows that the functions α /U are non-increasing in the interval I = [1 χ, χ] for α Hence, U (1 χ λχ = 4χ 4 1/χ λ = v λ Moreover, the previous formula also shows that the function α /U is unimodular for all α, which implies 5/ U 1 { } ( χ 5/ λ max (1 χ5/, 1 χ χ = 1 λ max{5/ (1 χ /, (1 χ 5/ /χ} = m λ These remarks imply λk 0 a 1 a 1 ( v λ 6 1 v 9 4vλ ( 1 1 8λm a 1 a a 1 a a 1 a Besides, k 0 V U W U

9 thus k 0 1 λ ( 1 χ χ 1 λ ( 1 χ χ Theorem of aurent Mignotte Nesterenko ( ( λ χ a 1 a 1 λ ( χ (1 χ1/ χ a 1 a = k (1 χ1/ χ Since the function f(x is decreasing for x > 1, the last step is to verify that K K 0 (with the notations of Theorem 15 We follow the proof of Remark 1 We have k0 = V V U 4U W U with and so that k0 χ λ V U = 1 / λ( (1 χ χ λ W U ( χ λ a 1 a λ a 1 a (1 χ 9λ ( χ λ a 1 a λ a 1 a and since K = 1 [λk 0 a 1 a ], we get K K 0 [One may verify that K 0 > 4] Remark The number m satisfies { } { } 5/ λ max U I U m = λ max I max λ(4χ 4 1/χ χ I It is possible to simplify some estimates in Theorem without serious loss Consider first the term k given by k = 1 ( 1 χ λ 1 ( χ λ χ (1 χ1/ χ = 1 ( 9 λ 1 h h (1 1 h/λ It is clear that k / λ < 0 Also, k / h < 0 Indeed, k h = ( h λ 1 h h (1 1 h/λ h 1/λ 1 1 h/λ,

10 110 M Mignotte which is < ( 1 h/λ 1 h h λ 1 h/λ = λ(1 h/λ h h λh < 0 1 h/λ Thus, for λ λ 0 and h h 0, we have k 1 ( 1 (1 1 h 0 /λ 0 9 λ 0 h 0 h 0 In particular, when λ log 4 and h 5, we get k 1 Now we consider the term T := log((x x / A x/ log(ax Elementary computation shows that T/ A < 0 and T/ x < 0 When x 4 and A 4 we get T 111 Concerning Theorem, when χ 1, ϱ 4, h 5 and A 4, these remarks imply the simplified estimate log Λ (C 0 c 1 c (λ h a 1 a, C 0 = 1 ( v/ v 9 ( 4λv 1 a 1 1 a 8λm a 1 a λ, (1 χ and λ(15λ h c 1 = (λ h, c = 111λ log(a(λ h a 1 a (λ h a 1 a When a 1 a 0, ϱ 4 and h 5, one can prove that c 004 The formula 15 χ c 1 = (1 χ a 1 a shows that c 1 is a decreasing function of χ and, for example, for χ 15 and a 1 a 0, we have c To summarize, c 1 c < 006 when a 1 a 0, ϱ 4, h 5 and χ 15 Also notice that for χ 1, one has m = 5/ (1 χ / This leads to the following result Corollary Consider the linear form Λ = b log α b 1 log α 1, b 1 and b are positive integers Suppose that α 1 and α are multiplicatively independent Define D, a 1, a, ϱ, λ, h, χ as in Theorem et a 1, a, h, k be real positive numbers, and ϱ a real number > 1 Suppose that ϱ 4 and that { ( ( ( b1 h max 5, 15λ, D log b } log λ , a a 1 (4 a i max{1, ϱ log α i log α i Dh(α i } (i = 1,,

11 Theorem of aurent Mignotte Nesterenko 111 (5 a 1 a max{0, 4λ } et v = 4χ 4 1/χ Then we have the lower bound log Λ (C 0 006(λ h a 1 a, C 0 = 1 ( 1 1 {( λ χ(χ λ ( 1 1 } (1 χ / v a 1 a v a 1 a We apply Theorem After the above preliminaries, we have just to check that the present hypotheses imply K 0 > 8 and use the fact that f(9 < 177 Remark 4 To get a comparison with the estimates of [MN], we can consider the Corollaire of [MN] Thus we suppose also that α 1 and α are both real Then we get log Λ { ( b 1 log D log A ( D 4 max D log A 1 A 1 and A are real numbers > 1 such that { log A i max h(α i, log α i D, 1 D b 006, 1 } log A 1 log A, D This result is obtained with the choice ϱ = 558 in the above Corollary (except that we use the original definitions of c 1 and c, not the estimate c 1 c < 006 In [MN], with (very slightly stronger hypotheses, the constant obtained was 44 } References [MN] M aurent, M Mignotte et Y Nesterenko, Formes linéaires en deux logarithmes et déterminants d interpolation, J Number Theory 55 (1995, 85 1 Département de Mathématique et Informatique Université ouis Pasteur 7, rue René Descartes Strasbourg Cedex, France mignotte@mathu-strasbgfr Received on and in revised form on (101