J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
A
RT ¯
URAS
D
UBICKAS
The mean values of logarithms of algebraic integers
Journal de Théorie des Nombres de Bordeaux, tome
10, n
o2 (1998),
p. 301-313
<http://www.numdam.org/item?id=JTNB_1998__10_2_301_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
The
meanvalues of
logarithms
of
algebraic
integers
par ART016ARAS DUBICKAS
RÉSUMÉ. Soit 03B11 = 03B1, 03B12,..., 03B1d l’ensemble des
conjugués
d’unentier
algébrique
03B1 dedegré
d,
n’étant pas une racine de l’unité. Dans cet article on propose de minorer$$Mp(03B1)=p~1 d03A3i=1|log|03B1i||p
où p>
1.ABSTRACT. Let 03B1 be an
algebraic integer
ofdegree
d withconju-gates 03B11 = 03B1, 03B12, ..., 03B1d. In the paper we
give
a lower bound forthe mean value
$$Mp(03B1)=p~1 d/d03A3i=1|log|03B1i||p
when 03B1 is not a root of unity and p > 1.
1. INTRODUCTION.
Let a be an
algebraic
number ofdegree
d > 2 withas its minimal
polynomial
over Z and adpositive. Following Mahler,
the Mahler measure of a is definedby
The house of an
algebraic
number is the maximum of the modulus of itsconjugates:
- --Put also
for
the "symmetric
deviation" ofconjugates
from the unit circle. Denotefor p > 0
Our main concern here is the lower bound for this mean value when a is an
algebraic integer
(ad
= 1)
which is not a root ofunity.
In
1933,
D.H. Lehmer[8]
asked whether it is true that for everypositive
e there exists an
algebraic
number a for which 1M(a)
1 + c. In itsstrong
form Lehmer’sproblem
has been reformulated as whether it is truethat if a is not a root
unity
thenM(a) >
ao = 1.1762808... where cxo is the root of thepolynomial
In
1971,
C.J.Smyth
[16]
proved
that if a is anon-reciprocal
algebraic
integer
thenM(a)
> 8 = 1.32471... where 0 is the real root of thepoly-1. This result reduces Lehmer’s
problem
to the caseof
reciprocal algebraic integers
(those
with minimalpolynomial
satisfying
the
identity
P(x) -
xdP(1/x)).
P.E.Blanksby
andH.L.Montgomery
[2]
used Fourier
analysis
to prove thatM(a)
> 1 +1/52d
log(6d).
In1978,
C.L. Stewart[18]
proved
thatM(a)
> 1 +1/104d
logd.
Although
this result is weaker than theprevious
one, the method used has become veryimportant
and led to furtherimprovements. Recently
M.Mignotte
andM. Waldschmidt
[12]
obtained Stewart’s result via theinterpolation
deter-minant.In
1979,
E. Dobrowolski[4]
obtained a remarkableimprovement
of theseresults
showing
that for each £ >0,
there exists an effectived(é)
such thatfor d >
d(~)
D.C. Cantor and E.G. Straus
[3]
in 1982 introduced theinterpolation
determinant tosimplify
Dobrowolski’sproof
and toreplace
the constant 1 - êby
2 - ê.Finally,
R. Louboutin[9]
was able toimprove
this constant to9/4 -
E. M.Meyer
[11]
obtained Louboutin’s resultusing
a version ofSiegel’s
lemma due to Bombieri and Vaaler.Recently
P. Voutier[19]
showed thatinequality
(1)
holds for all d > 2 with the weaker constant1/4
instead of 1- E.In
1965,
A. Schinzel and H. Zassenhaus[13]
conjectured
that there existsan absolute
positive constant -y
such thatïar
> 1+ ~y/d
whenever a is not awe have
where d > In
fact,
bothinequalities
(1), (2)
and therespective
con-jectures
can be considered in terms of the lower bound forMp(a).
Indeed,
notice that
Therefore,
forlao I 2::
2,
If
laol
=1,
thenLouboutin’s result can be written as follows
Taking
p =oo, we can write the
inequality
(2)
in thefollowing
formThe function p --~
Mp(a)
isnondecreasing.
Hence theinequality
cp where 1 p oo and cp > 0 lies between theconjecture
of Lehmer p = 1 andthe "symmetric"
form of theconjecture
of Schinzeland Zassenhaus p = oo
(see
also[1]
for aproblem
which lies betweenthese two
conjectures).
We have noticed above that theconjectural
value for ci is 2log
aa. It would be of interest to find out whether it is truethat
d(a)
> Theequality
holds for thepolynomial
2. Weconjecture
that the answer to the abovequestion
isaffirmative,
so thatCoo =
log
2. In thispaper, we take up the
interpolation
determinantagain
(see
[3],[5],[9],[10],
[19])
and fill the gap betweeninequalities
(3)
and(4)
(Theorem 2).
One can also consider the mean value ofconjugates
of analgebraic
integer
The lower bound for
ml (a)
where a is atotally
positive integer
was consi-deredby
I. Schur[14],
C.L.Siegel
[15],
C.J.Smyth
[17].
In1988,
M.Langevin
[7]
solved Favard’sproblem proving
thatt(a)
:= maxi,jlai-aj
I
> 2 - c for analgebraic integer
of asufhciently large degree.
The author[6]
proved
thatt2 (a)
>4 e -
c. Theproblem
offinding
an upper boundfor
:=1/
aj )
is known as aseparation
problem.
In thisarticle,
weapply
the lower bound forM2 (a)
to estimaterrLp (a)
from below(Theorem 3).
2. STATEMENT OF THE RESULTS.
The notations are the
following.
LetG(x)
be a real valued function in[0; 1]
such thatG(O) =
1,
G(l) =
0. Let also the derivative ofG(x)
be continuous andnegative
in the interval(0; 1).
PutPut also for
brevity
Let a be a
reciprocal algebraic integer,
i.e. d =2m,
m E N,
a2m =
a2m-1 =
1/2(X2, ... , am+1 - l/am
whereIa2l > ... > laml ;2:
1.Suppose
also that a is not a root ofunity.
With thesehypotheses,
our main result is thefollowing:
Theorem 1. For every e > 0 there exists such that we have
whenever d >
duo(£).
The constant
do (e)
and the constantsd1
(e),d2 (e),d3 (e),
d4,
d5 (P)
used below are effective.Taking
G(x) = (1 -
X)2,
weget
I =1/3,
J =1/5,
L =
Corollary
1. For every > 0 there existsd1 (c)
such thatwhenever d >
d1
(e).
This
inequality
obviously implies
Louboutin’s result. On the otherhand,
taking
G(x)
= 1 - we have I = 1 -2/7r,
J =3/2 -
4/-7r,
L =7r2/ 8.
HenceWe can
replace
in theinequality
abovelog
laj
by
laj
1-1,
and so Theorem 1yields
thefollowing Corollary.
Corollary
2. For every - > 0 there existsd2(~)
such thatfor
d >d2(e)
we have
where
Corollary
2implies
theinequality
(2),
since T~ = 1. Thefollowing
theorem fills the gap between
(3)
and(4).
Theorem 2. Let 1 p oo and e > 0. Then there is
d3(E)
such thatfor
d >d3 (e)
we have
-where the constant
bp
isgiven
by
We are not
solving
theproblem
ofcomputing
the maximum in(9)
for afixed p from the interval
(1; oo).
However,
notice that ifG(x) _ (1 -
x)u7
and p = 2 thenby
(5)-(7)
and(9)
weget
b2
> 6.2679.Theorem 3.
If
a is analgebraic
integer
which is not a rootof
unity,
thenfor
every p > 0 there existsd5(p)
such thatfor
d >d5(p)
we haveIn
particular,
Proof of
Theorem 1. Letf (x)
be a continuousnon-negative
function in i=
1
[0; 1]
suchthat j f(z)dz
=1,
and letG(x)
= f f (y)dy.
Puto x
Define
Consider the determinant
where the matrix consists of N -
~ko
+l~1
+ ... +ks) d
columns,
u, =
0,1, ... ,
l~r -1, j -
1, 2, ... ,
d. Here pr is the r-thprime
number(po =
l, pl - 2,p2 =
3,...).
Recall that a isreciprocal
and a2m-1/al, ... , ~m+1 - l/am.
Then see([3],
[5], [9], [10],
[19])
the determinant D isgiven by
where the first
product
is taken overi, j =1, 2, ... ,
m and 0 u v s(if
u = v, then i
j ~ .
The secondproduct
is taken overall i, j
=l, 2, ... ,
m;u, v =
0, l, 2, ... ,
s. Let us denote theseWe first consider
Pl.
We have:Next,
we have for theproduct
P2
Combining
these results we findNow from each term
afu -
in the firstproduct
we takeThis is the
key
point
of ourargument.
Write the determinant D asDenote yl =
y2 -
~3/~2?... ~m-1 =
yrrt =Then D can be
expressed
in the formwhere
p ( y1, ... ,
ym )
is apolynomial
in y2 , ... , The power sj isgiven
by
Using
the maximum modulusprinciple
and theinequalities
1,
j - 1,2,...,m,
we havewhere
Iyrl
=Iygl
]
= ... = = 1. Nowby
Hadamard’sinequality
we find(see
[5])
On the other hand
(see
[9]),
Similarly,
Since
and
For a
sufhciently large
d we haveThis
inequality implies
(8).
DProof of
Theorem 2. If a is notreciprocal,
thenby
Smyth’s
result[16]
2 log 0,
and the theorem follows fromMp(a) 2::
M1(a).
Leta be
reciprocal.
Thenby
(8)
andby
Holder’sinequality
we havewhere
1/p
+llq
= 1.Note first that for a
reciprocal
aFor d
tending
toinfinity
we haveHence
Proof of
Theorem 3. We haveIf a is
reciprocal,
then the inner sumequals
foreven j
and zerofor
odd j.
HenceUtilizing Corollary
3 we haveif d is
large enough
and the statement of Theorem 3 follows.Suppose
now that a is notreciprocal.
Ifthen
for d >
d5 ~p) .
Hence it is sufhcient to consider the case whenI ao
=1. Let a1, a2, ... , ar be theconjugates
of alying strictly
outside the unit circle. PutThen
We shall show now that the last
expression
isgreater
thanwhere 0 =1.32471....
Indeed,
ifthen
Therefore,
the function isincreasing
in the interval(1; oo)
andby
Smyth’s
theoremPut for
brevity
p = zd and r =yd.
We aregoing
to prove thatfor z > 0 and 0 y 1.
Indeed,
g(0)
= 0 andTherefore,
with ourhypotheses
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polynomial. Acta Arith. 34 (1979), 391-401.
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(1996), 81-95. Artfras DUBICKAS Department of Mathematics, Vilnius University, Naugarduko 24, Vilnius 2006, Lithuania