Families of Thue equations associated with a rank one subgroup of the unit group of a number field

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by  Claude Levesque

Families of Thue equations associated
with a rank one subgroup of the
unit group of a number field
by
Claude Levesque and Michel Waldschmidt
7
1 Abstract. Twisting a binary form F (X,Y) ∈ Z[X,Y] of degree d ≥ 3
0
0 by powers υa (a ∈ Z) of an algebraic unit υ gives rise to a binary form
2
F (X,Y) ∈ Z[X,Y]. More precisely, when K is a number field of degree
a
n d, σ ,σ ,...,σ the embeddings of K into C, α a nonzero element in K,
a 1 2 d
J a ∈ Z, a > 0 and
0 0

5 d
(cid:89)
F (X,Y) = a (X −σ (α)Y),
0 0 i
]
T i=1
then for a ∈ Z we set
N
d
(cid:89)
. F (X,Y) = a (X −σ (αυa)Y).
h a 0 i
t i=1
a
Given m ≥ 0, our main result is an effective upper bound for the solutions
m
(x,y,a) ∈ Z3 of the Diophantine inequalities
[

0 < |F (x,y)| ≤ m
1 a
v for which xy (cid:54)= 0 and Q(αυa) = K. Our estimate involves an effectively
0
3 computable constant depending only on d; it is explicit in terms of m, in
2 terms of the heights of F and of υ, and in terms of the regulator of the
0
1
number field K.
0
.
1
0 Keywords: Families of Thue equations, Diophantine equations
7
MSC: 11D61, 11D41, 11D59
1
:
v
i
X 1 Introduction and the main results
r
a
Let d ≥ 3 be a given integer. We denote by κ ,κ ,... positive effectively
1 2
computable constants which depend only on d.
Let K be a number field of degree d. Denote by σ ,σ ,...,σ the em-
1 2 d
beddings of K into C and by R the regulator of K. Let α ∈ K, α (cid:54)= 0, and
let a ∈ Z, a > 0, be such that the coefficients of the polynomial
0 0
d
(cid:89)(cid:0) (cid:1)
f (X) = a X −σ (α)
0 0 i
i=1
1
are in Z. Let υ be a unit in K, not a root of unity. For a ∈ Z, define the
polynomial f (X) in Z[X] and the binary form F (X,Y) in Z[X,Y] by
a a
d
f (X) = a (cid:89)(cid:0)X −σ (αυa)(cid:1)
a 0 i
i=1
and
d
F (X,Y) = Ydf (X/Y) = a (cid:89)(cid:0)X −σ (αυa)Y(cid:1).
a a 0 i
i=1
Define
d d
(cid:89) (cid:89)
λ = a max{1,|σ (α)|} and λ = max{1,|σ (υ)|}.
0 0 i i
i=1 i=1
Let m ∈ Z, m > 0. We consider the family of Diophantine inequalities
(1) 0 < |F (x,y)| ≤ m,
a
where the unknowns (x,y,a) take their values in the set of elements in Z3
such that xy (cid:54)= 0 and Q(αυa) = K. It follows from the results in [4] that
the set of solutions is finite. However, the proof in [4] relies on Schmidt’s
subspace theorem, which is not effective. Here we give an effective upper
boundformax{|x|,|y|,|a|}intermsofm,R,λ andλ,byusinglowerbounds
0
for linear forms in logarithms.
For x ∈ R, x > 0, we stand to the notation log(cid:63)x for max{1,logx}.
Here is our main result.
Theorem 1. There exists an effectively computable constant κ > 0,
1
depending only on d, such that any solution (x,y,a) ∈ Z3 of (1), which
verifies xy (cid:54)= 0 and Q(αυa) = K, satisfies
|a| ≤ κ λd2(d+2)/2(R+logm+logλ )Rlog(cid:63)R.
1 0
Under the assumptions of Theorem 1, with the help of the upper bound
H(F ) ≤ 2dλ λ|a|
a 0
for the height of the form F , it follows from the bound (3.2) in [1, Theorem
a
3] (see also [2, Th. 9.6.2]) that
logmax{|x|,|y|} ≤ κ(cid:0)R+log(cid:63)m+|a|logλ+logλ (cid:1)R(log(cid:63)R)
0
with
κ = 3r+27(r+1)7r+19d2d+6r+15.
2
CombiningthisupperboundwithourTheorem1providesaneffectiveupper
bound for max{|x|,|y|,|a|}.
For i = 1,...,d, set υ = σ (υ) and assume
i i
|υ | ≤ |υ | ≤ ··· ≤ |υ |.
1 2 d
Our proof actually gives a much stronger estimate for |a|, see Theorem 2,
which involves some extra parameter µ > 1 defined by

λ if |υ | = |υ | or |υ | = |υ |,
 1 d−1 2 d

 (cid:26)|υ | |υ |(cid:27)
 d−1 d
min , if |υ | < |υ | = |υ | < |υ |,
µ = |υ | |υ | 1 2 d−1 d
1 2

|υd−1| if |υ | < |υ |.
 2 d−1
|υ |
2
Notice that the condition |υ | = |υ | means |υ | = |υ | = ··· = |υ |
1 d−1 1 2 d−1
and that the condition |υ | = |υ | means |υ | = |υ | = ··· = |υ |; using
2 d 2 3 d
Lemma 11, we deduce that each of these two conditions implies that d is
odd, hence that the field K is almost totally imaginary (namely, with a
single real embedding) – compare with [9].
Theorem 2. There exists a positive effectively computable constant κ ,
2
depending only on d, with the following property. Let (x,y,a) ∈ Z3 satisfy
xy (cid:54)= 0, [Q(αυa) : Q] = d and 0 < |F (x,y)| ≤ m.
a
Then
logλ (cid:18) (logλ)2(cid:19)
(2) |a| ≤ κ (R+logm+logλ +logλ)Rlog R .
2 0
logµ logµ
On the one hand, using Lemma 12 (§3.5), we will prove in §5 that
logµ ≥ κ λ−d2(d+2)/2(logλ)2,
3
which will enable us to deduce Theorem 1 from Theorem 2. On the other
hand, thanks to (5), we have µ ≤ λ2. In general, we expect µ to be as
large as λκ4 (which is therefore the maximum possible), in which case the
conclusion of Theorem 2 becomes
(3) |a| ≤ κ (R+logm+logλ +logλ)R(logR+log(cid:63)log(cid:63)λ)
5 0
with a positive effective constant κ depending only on d. In §2, we give a
5
few examples where this last bound is valid.
In Theorem 1, the hypothesis that υ is not a root of unity cannot be
omitted. Here is an example with α = a = m = 1. Let Φ (X) be the
0 n
cyclotomic polynomial of index n and degree ϕ(n) (Euler totient function).
3
Let ζ be a primitive n–th root of unity. Set f = Φ and u = ζ . For a ∈ Z
n 0 n n
with gcd(a,n) = 1, the irreducible polynomial f of ζa is nothing else than
a n
f . Hence, if the equation
0
F (x,y) = ±1
0
has a solution (x,y) ∈ Z2 with xy (cid:54)= 0, then for infinitely many a ∈ Z
the twisted Thue equation F (x,y) = ±1 has also the solution (x,y), since
a
F = F . For instance, when n = 12, we have Φ (X) = X4 −X2 +1 and
a 0 12
the equation
x4−x2y2+y4 = 1
has the solutions (1,1), (−1,1), (1,−1), (−1,−1).
The main result of [5], which deals only with non totally real cubic equa-
tions, is a special case of Theorem 2; the “constants” in [5] depend on α
and υ, while here they depend only on d. The main result of [6] deals with
Thue equations twisted by a set of units which is not supposed to be a group
of rank 1, but it involves an assumption (namely that at least two of the
conjugates of υ have a modulus as large as a positive power of υ) which we
do not need here. Our Theorem 2 also improves the main result of [7]: we
remove the assumption that the unit is totally real (besides, the result of [7]
is not explicit in terms of the heights and regulator). We also notice that
the part (iii) of Theorem 1.1 of [8] follows from our Theorem 2. The main
result of [9] does not assume that the twists are done by a group of units of
rank 1, but it needs a strong assumption which does not occur here, namely
that the field K has at most one real embedding.
We conclude this §1 with some more definitions and properties.
When f is a polynomial in one variable of degree d with coefficients in Z
and leading coefficient c > 0, the (usual) height H(f) of f is the maximum
0
of the absolute values of the coefficients of f, while the Mahler measure of
f is
d
(cid:89)
M(f) = c max{1,|γ |},
0 i
i=1
where γ ,γ ,...,γ are the roots of f in C.
1 2 d
Let us recall1 that the logarithmic height h(γ) of an algebraic number γ
ofdegreedis 1 logM(γ)whereM(γ)istheMahlermeasureoftheirreducible
d
polynomial of γ. We have
√
(4) M(f) ≤ d+1H(f) and H(f) ≤ 2dM(f)
(see [12], Annex to Chapter 3, Inequalities Between Different Heights of a
Polynomial, pp. 113–114; see also [2, §1.9]). The second upper bound in (4)
could be replaced by the sharper one
(cid:18) (cid:19)
d
H(f) ≤ M(f),
(cid:98)d/2(cid:99)
1Our h is the same as in [2], it corresponds to the logarithm of the h in [1].
4
but we will not need it.
Let υ be a unit of degree d and conjugates υ ,...,υ with
1 d
|υ | ≤ |υ | ≤ ··· ≤ |υ |,
1 2 d
so that υ = |υ |. Let λ = M(υ) and let s be an index in {1,...,d−1} such
d
that
|υ | ≤ |υ | ≤ ··· ≤ |υ | ≤ 1 ≤ |υ | ≤ ··· ≤ |υ |.
1 2 s s+1 d
We have
λ = M(υ) = |υ ···υ | ≤ |υ |d−s ≤ |υ |d−1
s+1 d d d
and
M(υ−1) = |υ ···υ |−1 = M(υ) = λ
1 s
with
λ ≤ |υ |−s ≤ |υ |−(d−1).
1 1
Therefore we have
(5) λ1/(d−1) ≤ |υ | ≤ λ and λ−1 ≤ |υ | ≤ λ−1/(d−1).
d 1
2 Examples
The lower bound µ ≥ λκ4 quoted in section 1 is true
• when d = 3 and the cubic field K is not totally real;
• for the simplest fields of degree 3 (see [8]), and also for the simplest fields
of degrees 4 and 6;
• when −υ is a Galois conjugate of υ (which means that the irreducible
polynomial of υ is in Z[X2]), and more generally when |υ | = |υ | and
1 2
|υ | = |υ | with d ≥ 4.
d−1 d
Here is an example of this last situation. Let (cid:15) be an algebraic unit, not
a root of unity, of degree (cid:96) ≥ 2 and conjugates (cid:15) ,(cid:15) ,...,(cid:15) . Let h ≥ 2 and
1 2 (cid:96)
let d = (cid:96)h. For a ∈ Z, define
(cid:96)
(cid:89)
(6) F (X,Y) = (Xh−(cid:15)aYh).
a i
i=1
Let R be the regulator of the field Q((cid:15)1/h).
From Theorem 2 we deduce the following corollary.
Corollary 3. Let m ≥ 1. If the form F in (6) is irreducible and if there
a
exists (x,y) ∈ Z2 with xy (cid:54)= 0 and |F (x,y)| ≤ m, then
a
|a| ≤ κ (R+logm+log (cid:15))Rlog(cid:63)(Rlog (cid:15)).
6
5
Proof. Without loss of generality, assume |(cid:15) | ≤ |(cid:15) | ≤ ··· ≤ |(cid:15) |, so
1 2 (cid:96)
that|(cid:15) | = (cid:15). Letζ beaprimitiveh-throotofunity. Letυ = (cid:15)1/h. Weapply
(cid:96)
Theorem 2 with α = ζ, a = 1, λ = 1, λ ≤ (cid:15)(cid:96), F (X,Y) = (Xh−Yh)(cid:96) and
0 0 0
υ = ζj−1(cid:15)1/h (0 ≤ i ≤ (cid:96)−1, 1 ≤ j ≤ h).
ih+j i+1
From |υ | = |υ | = |(cid:15) |1/h < 1 and |υ | = |υ | = |(cid:15) |1/h we deduce
1 2 1 d−1 d (cid:96)
(cid:12)(cid:12)(cid:15)(cid:96)(cid:12)(cid:12)1/h (cid:12)(cid:12)υd(cid:12)(cid:12)
µ = (cid:12) (cid:12) = (cid:12) (cid:12)
(cid:12)(cid:15) (cid:12) (cid:12)υ (cid:12)
1 1
and using (5) we conclude
2
logµ ≥ logλ. (cid:50)
d−1
A variant of this proof is to take α = 1, λ = 1, F (X,Y) = (X −Y)d,
0 0
and to use the fact that ζa is also a primitive h-th root of unity since F is
a
irreducible.
3 Auxiliary results
3.1 An elementary result
For the convenience of the reader, we include the following elementary result
–similarargumentsareoftenusedwithoutexplicitmentionintheliterature.
Lemma 4. Let U and V be positive numbers satisfying U ≤ V log(cid:63)U.
Then U < 2V log(cid:63)V.
Proof. IflogU ≤ 1,theassumptionisU ≤ V andtheconclusionfollows.
√
Assume logU > 1. Then logU ≤ U, hence the hypothesis of the lemma
√
implies U ≤ V U and therefore we have U ≤ V2. We deduce
logU ≤ 2logV,
hence
U ≤ V logU ≤ 2V logV. (cid:50)
3.2 Diophantine tool
In this section only, the positive integer d is not restricted to d ≥ 3.
The main tool is the following Diophantine estimate ([6, Proposition 2],
[12, Theorem 9.1] or [2, Th. 3.2.4]), the proof of which uses transcendental
number theory.
6
Proposition 5. Let s and D be two positive integers. There exists an
effectively computable positive constant κ(s,D), depending only upon s and
D, with the following property. Let η ,...,η be nonzero algebraic numbers
1 s
generating a number field of degree ≤ D. Let c ,...,c be rational integers
1 s
and let H ,...,H be real numbers ≥ 1 satisfying
1 s
H ≥ h(η ) (1 ≤ i ≤ s).
i i
Let C be a real number with C ≥ 2. Suppose that one of the following two
statements is true:
(i) C ≥ max |c |
1≤j≤s j
or
(ii) H ≤ H for 1 ≤ j ≤ s and
j s
(cid:26) (cid:27)
H
j
C ≥ max |c | .
j
1≤j≤s Hs
Suppose also ηc1···ηcs (cid:54)= 1. Then
1 s
|ηc1···ηcs −1| > exp{−κ(s,D)H ···H logC}.
1 s 1 s
The statement (ii) of Proposition 5 implies the statement (i) by per-
muting the indices so that H ≤ H for 1 ≤ j ≤ s; however, we find it
j s
more convenient to use the part (i) so that we can use the estimate without
permuting the indices.
We will use Proposition 5 several times. Here is a first consequence.
Corollary 6. Let d ≥ 1. There exists a constant κ , which depends only
7
on d, with the following property. Let K be a number field of degree d. Let
α , α , υ , υ be nonzero elements in K and let a be a nonzero integer. Set
1 2 1 2
γ = α υa and γ = α υa. Let λ and λ satisfy
1 1 1 2 2 2 0
max{h(α ),h(α )} ≤ logλ , max{h(υ ),h(υ )} ≤ logλ
1 2 0 1 2
and assume γ (cid:54)= γ . Define
1 2
(cid:18) (cid:26) log(cid:63)λ (cid:27)(cid:19)
χ = (log(cid:63)λ )(log(cid:63)λ)log(cid:63) |a|min 1, .
0 log(cid:63)λ
0
Then
|γ −γ | ≥ max{|γ |,|γ |}e−κ7χ.
1 2 1 2
Proof. By symmetry, without loss of generality, we may assume
|γ | ≥ |γ |. Set
2 1
υ α
1 1
s = 2, η = , η = , c = a, c = 1,
1 2 1 2
υ α
2 2
(cid:26) (cid:26) (cid:27)(cid:27)
H
H = 2log(cid:63)λ, H = 2log(cid:63)λ , C = max 2,|a|min 1, 1 ·
1 2 0
H
2
The conclusion of Corollary 6 follows from Proposition 5 (via part (i) if
H ≥ H , via part (ii) otherwise), thanks to the relation
1 2
|ηc1ηc2 −1| = |γ |−1|γ −γ |. (cid:50)
1 2 2 1 2
7
3.3 Lower bound for the height and the regulator
Fortherecord,wequoteKronecker’sTheoremanditseffectiveimprovement.
Lemma 7. (a) If a nonzero algebraic integer α has all its conjugates in
the closed unit disc {z ∈ C | |z| ≤ 1}, then α is a root of unity.
(b) More precisely, given d ≥ 1, there exists an effectively computable
positive constant κ , depending only on d, such that, if α is a nonzero alge-
8
braic integer of degree d satisfying h(α) < κ , then α is a root of unity.
8
Proof. Voutier (1996) refined an earlier estimate due to Dobrowolski
(1979) by proving that the conclusion of the part (b) in Lemma 7 holds with
(cid:40)
log2 if d = 1,
κ =
8 2d(logd)3 if d ≥ 2.
See for instance [2, Prop. 3.2.9] and [12, §3.6]. (cid:50)
Lemma 8. There exists an explicit absolute constant κ > 0 such that
9
the regulator R of any number field of degree ≥ 2 satisfies R > κ .
9
Proof. According to a result of Friedman (1989 – see [2, (1.5.3)]) the
conclusion of Lemma 8 holds with κ = 0.2052. (cid:50)
9
3.4 A basis of units of an algebraic number field
Here is Lemma 1 of [1]. See also [2, Proposition 4.3.9]. The result is essen-
tially due to C.L. Siegel [11].
Proposition 9. Let d be a positive integer with d ≥ 3. There exist
effectively computable constants κ ,κ ,κ depending only on d, with the
10 11 12
following property. Let K be a number field of degree d, with unit group
of rank r. Let R be the regulator of this field. Denote by ϕ ,ϕ ,...,ϕ a
1 2 r
set of r embeddings of K into C containing the real embeddings and no pair
of conjugate embeddings. Then there exists a fundamental system of units
{(cid:15) ,(cid:15) ,...,(cid:15) } of K which satisfies the following:
1 2 r
(cid:89)
(i) h((cid:15) ) ≤ κ R;
i 10
1≤i≤r
(ii) max h((cid:15) ) ≤ κ R;
i 11
1≤i≤r
(iii) The absolute values of the entries of the inverse matrix of
(log|ϕ ((cid:15) )|)
j i 1≤i,j≤r
do not exceed κ .
12
8
The next result is [10, Lemma A.15].
Lemma 10. Let (cid:15) , (cid:15) , ..., (cid:15) be an independent system of units for K
1 2 r
satisfying the condition (ii) of Proposition 9. Let β ∈ Z with N (β) =
K K/Q
m (cid:54)= 0. Then there exist b ,b ,...,b in Z and β˜ ∈ Z with conjugates
1 2 r K
β˜ ,β˜ ,...,β˜ , satisfying
1 2 d
β = β˜(cid:15)b1(cid:15)b2···(cid:15)br
1 2 r
and
|m|1/de−κ13R ≤ |β˜ | ≤ |m|1/deκ13R for j = 1,...,d.
j
The conclusion of Lemma 10 can be written
(cid:12) (cid:16) (cid:17)(cid:12)
(cid:12)log |m|−1/d|β˜ | (cid:12) ≤ κ R for j = 1,...,d.
(cid:12) j (cid:12) 13
3.5 Estimates for the conjugates
Lemma 11. Let γ be an algebraic number of degree d ≥ 3. Let γ ,
1
γ , ..., γ be the conjugates of γ with |γ | ≤ |γ | ≤ ··· ≤ |γ |.
2 d 1 2 d
(a) If |γ | < |γ | and γ ∈ R, then |γ | < |γ |.
1 2 2 2 3
(b) If |γ | < |γ | and γ ∈ R, then |γ | < |γ |.
d−1 d d−1 d−2 d−1
Proof. (a) The conditions |γ | < |γ | ≤ |γ | for 3 ≤ i ≤ d imply that γ
1 2 i 1
is real and that −γ is not a conjugate of γ . Hence the minimal polynomial
1 1
of γ is not a polynomial in X2. Assume |γ | = |γ |. Since −γ is not a
2 3 2
conjugate of γ , we deduce γ (cid:54)∈ R, hence d ≥ 4. We may assume γ = γ .
2 3 4 3
Let σ be an automorphism of Q which maps γ to γ ; via σ, let γ be the
2 1 j
image of γ and γ the image of γ . From
3 k 4
γ2 = γ γ
2 3 4
we deduce γ2 = γ γ and |γ |2 = |γ γ |. This is not possible since |γ | > |γ |
1 j k 1 j k j 1
and |γ | > |γ |.
k 1
(b)Wededuce(b)from(a),byusingγ (cid:55)→ 1/γ (orbyrepeatingtheproof,
mutatis mutandis). (cid:50)
Remark. Here is an example showing that the assumptions of Lemma
11 are sharp. The polynomial X4−4X2+1 is irreducible, its roots are
(cid:113) √ (cid:113) √
υ = 2− 3, υ = −υ , υ = 1/υ = 2+ 3, υ = −υ
1 2 1 3 1 4 3
with
υ = |υ | < υ = |υ |.
1 2 3 4
More generally, if h ≥ 2 is a positive integer and (cid:15) is a quadratic unit with
Galois conjugate (cid:15)(cid:48) and if (cid:15)1/h has degree 2h, then it has h conjugates of
absolute value |(cid:15)|1/h and h conjugates of absolute value |(cid:15)(cid:48)|1/h. See also §2.
9
Lemma 12. Let υ be an algebraic unit of degree d ≥ 3. Set λ = M(υ).
Let υ(cid:48) and υ(cid:48)(cid:48) be two conjugates of υ with |υ(cid:48)| < |υ(cid:48)(cid:48)|. Then
|υ(cid:48)(cid:48)|
log ≥ κ λ−(d3+2d2−d+2)/2.
|υ(cid:48)| 14
We will deduce Lemma 12 from Theorem 1 of [3] which2 states the fol-
lowing.
Lemma 13 (X. Gourdon and B. Salvy [3]). Let P be a polynomial of
degree d ≥ 2 with integer coefficients and with Mahler measure M(P). If α(cid:48)
and α(cid:48)(cid:48) are two roots of P with |α(cid:48)| < |α(cid:48)(cid:48)|, then
|α(cid:48)(cid:48)|−|α(cid:48)| ≥ κ M(P)−d(d2+2d−1)/2
15
with √
3(cid:0) (cid:1)−d(d+1)/4−1
κ = d(d+1)/2 .
15
2
Proof of Lemma 12. We apply Lemma 13 to the minimal polynomial
of υ. To conclude the proof of Lemma 12, we use the bounds |υ(cid:48)| ≤ λ and
x |υ(cid:48)(cid:48)|
log(1+x) ≥ for 0 ≤ x ≤ 1 with x = −1. (cid:50)
2 |υ(cid:48)|
4 Proof of Theorem 2
Theorem 2 with the assumption |F (x,y)| ≤ m will be secured if we deal
a
with the equation F (x,y) = m with m (cid:54)= 0.
a
Let (a,x,y,m) ∈ Z4 satisfy m (cid:54)= 0, xy (cid:54)= 0, [Q(αυa) : Q] = d and
F (x,y) = m.
a
Withoutlossofgenerality, wemayrestrict(a,y)toa ≥ 0(otherwise, replace
υ by υ−1) and to y > 0 (otherwise replace F (X,Y) by F (X,−Y)).
a a
The form F˜ (X,Y) = ad−1F (X,Y) has coefficients in Z, and if we set
a 0 a
x˜ = a x, y˜ = y, m˜ = ad−1m we have F˜ (x˜,y˜) = m˜ with (x˜,y˜) ∈ Z2.
0 0 a
Therefore, there is no loss of generality to assume a = 1.
0
Theorem 2 includes the assumption that υ is not a root of unity, hence
λ > 1. More precisely, it follows from the part (b) of Lemma 7 that
logλ ≥ κ .
8
2This reference was kindly suggested to us by Yann Bugeaud.
10