C OMPOSITIO M ATHEMATICA
P IETER L. C IJSOUW
M ICHEL W ALDSCHMIDT
Linear forms and simultaneous approximations
Compositio Mathematica, tome 34, n
o2 (1977), p. 173-197
<http://www.numdam.org/item?id=CM_1977__34_2_173_0>
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173
LINEAR FORMS AND SIMULTANEOUS APPROXIMATIONS
Pieter L.
Cijsouw
and Michel WaldschmidtNoordhoff International Publishing
Printed in the Netherlands
1. Introduction
We shall
improve
andgeneralize
several results on transcendence and on simultaneousapproximations
of numbers associated with theexponential
function. We deduce these results from thefollowing
lower bound of an
inhomogeneous
linear form inlogarithms
ofalgebraic
numbers withalgebraic
coefhcients :THEOREM 1: Let n > 1 and d > 1 be
given integers.
There exists aneffectively computable
numberC, depending only
on n andd,
with thefollowing
property. LetIl, ...,
ln becomplex
numbers such that a, ==el1, ...,
an = e ln arealgebraic
numbersof degrees
at mostd; for
1j
n,let Aj >
6 be an upper boundfor
theheight of aj and for elljl.
Further, let03B20, 03B21, ..., 03B2n be algebraic
numbersof degrees
at most d andheights
at mostB
(>6) ;
putand
define
Then, if
A #0,
we haveFrom this
point
on, we shall writelog
ai instead ofli
so that(1)
takesthe form
then the theorem says,
that,
if A ~ 0,(2)
holds where C isindependent
of the branches of the
logarithms.
We
remark,
that it is known that the linear form A in(3)
is non-zero when 03B11,..., an are non-zero and eitherlog
03B11,...,log
an not all zeroand 03B21, 03B22, ..., 03B2n linearly independent
over Q, or03B20 ~
0.For earlier results in this
direction,
see inparticular
A. Baker’spapers
[2]
and[3].
It wa spointed
out to usby
A.J. van der Poortenthat
by
minorchanges
we could havereplaced
the boundexp
1- Cf2 (Log n)(Log
B +Log 03A9)} by
where
03A9’ = (Log A1) ... (Log An-1).
We shalldevelop
this remark atthe end of the present paper. We understand
that, independently,
A.Baker
recently
obtained a similar bound for the linear form(1),
which is somewhat more refined than our bound.However,
these refine- ments are not essential for ourapplications,
and moreover, ourproof
is
simpler.
The actual
proof
of theorem 1 will followclosely
theproof given
in[3].
Nevertheless asimplification
is introduced here. In earlier similarproofs,
theextrapolation
isperformed by
means of aninduction,
and then a further inductive argumentpermits
one to decrease the numberof terms in the
auxiliary
function. In ourproof,
there is no need toincrease the number of zeros, and we therefore avoid the induction in the
extrapolation
argument. Thissimplification
allows us to reducethe main part of the
proof
from a chain of lemmas to asingle
unified argument.Moreover,
in a certain part of theproof
where aprime
p has to be used whichordinarily
must besufficiently large,
we can dealwith an
arbitrary prime
p, even with p = 2.We want to express our
gratitude
to A. Baker and A. J. van derPoorten for some very useful discussions at the conference on
transcendental number
theory during early
1976 atCambridge,
wherethis paper was
put
into its final form.2. Simultaneous
approximations
We shall
apply
theorem 1 togive
bounds for the simultaneousapproximability
of certain numbers.First,
let a EC with a ~0,
a ~1,
bEC,
E > 0 and d E N begiven.
We consider the number of solutions inalgebraic
numbers a,j3
and y, ofdegrees
at mostd,
of theinequality
where H is a bound for the
heights
of a,03B2
and y and where ab= eb log a for some fixed value of thelogarithm.
After someearly
results of Ricci
[10]
and Franklin[8],
Schneider[11] proved
that ifb ~
Q, then there areonly finitely
many solutions(a, (3, y)
with03B2~Q
when K > 5. Smelev
[12] improved
this to K >4,
and Bundschuh[5]
reached K >
4,
without the restriction to theirrationality
of03B2. Note,
that b E Qimplies
thatj3
= b for Hlarge enough
and that in this caseinfinitely
many solutions arepossible
when a is a suitable U-number.The next theorem
shows,
that there areonly finitely
many solutions(a, (3, y)
with03B2~
Q even when K > 3.THEOREM 2: Let a E C with a ~
0,
letlog
a be anarbitrary
valueof
the
logarithm of a
withlog
a # 0. Letb ~ C,
E> 0 and d E N begiven.
There are
only finitely
manytriples (a,,6, y)
withB~
Qof algebraic
numbers withdegrees
at most dfor
whichwhere H is a bound
f or
theheights o f 03B1, 03B2
and y and where a b = eb log a.PROOF: Let
(a, 03B2, y)
be atriple
as indicated in the statement of thetheorem,
and let H besufliciently large.
Since a ~ 0 andab ~ 0,
wehave a~ 0 and
03B3~ 0. Further,
we know that for suitable deter- minations of thelogarithms
and
moreover,
log
03B1~ 0 sincelog
a ~ 0.Combining
bothinequalities
andusing
that03B2
is close tob,
we see tbatHowever,
in both caseslog
condition
03B2 ~ Q
that the linear formj6 log a -log
y does not vanish.We obtain from theorem 1 the
inequality
so that we have a contradiction for H
large enough. Hence, H
mustbe bounded and the theorem follows. D
Wallisser
[15]
stated a theoremsaying essentially
that foralgebraic
03B11, ..., an and certain numbers
b 1,...,
bn which can beapproximated
very well
by algebraic numbers,
the number03B1b11... 03B1bnn
must be transcendental.Meyer [9] improved
this and also consideredebo03B1b11
...03B1bnn;
in both cases he achievedessentially
the bestpossible
result. Bundschuh
[6] recently showed,
that for fixedalgebraic
03B11, ..., an, the numbers
b 1, ..., bn, al’
...an-
cannot beapproximated simultaneously
very well(up
to exp1-(Log H)n2+2n+2+~}) by infinitely
many
(n
+1)-tuples
ofalgebraic
numbers of boundeddegrees;
heused
only
naturalhypotheses.
The next theoremgives
animportant improvement
of Bundschuh’s main theorem for aspecial
case.In theorem 4 we state an
analogous
result in whichonly
naturalconditions are used.
THEOREM 3: Let a,,..., an be non-zero
algebraic numbers,
letlog
aibe an
arbitrary
valueof
thelogarithm
withlog
03B1j ~ 0(j
= 1, ...,n),
letb1,...,
bn becomplex numbers,
E > 0 and dEN. There areonly finitely
many
(n
+1)-tuples (03B21,..., (3n, 03B3) of algebraic
numbers with 1,03B21, ..., 03B2n linearly independent
over 0 andof degrees
at most dfor
which
where H denotes a bound
for
theheights of 03B21,
...,03B2n,
y and wherePROOF: Use the linear form
of which ai, ..., an, y and
log 03B11, ..., log 03B1n
are non-zero.By
theassumption
on the linearindependence
of1, 03B21, ..., 03B2n
this linearform is non-zero. Use theorem 1 and
proceed
as in theproof
oftheorem 2. 0
THEOREM 4: Let 03B11, ..., 03B1n,
log
03B11,...,log
03B1n,b1,
...,bn,
E and d beas in theorem 3. Let bo E C with bo =1= 0.
Then there are
only finitely
many(n + 2)-tuples (03B2o, (3l,..., I3n, y) of algebraic
numbersof degrees
at most dfor
whichwhere H denotes a bound
for
theheights of 03B2o, 03B21, ..., 03B2n,
y and where;
P ROOF : Use the linear form
03B2o
+03B21 log
« 1 + · · ·+ 03B2n log an - log
y which is nonzero for Hlarge enough,
since then the condition bo # 0implies 03B2o
~ 0. 0Pemark,
that(4)
enables us toreplace (Log Log H)1+E by (Log Log H)’
in theorem 3 and theorem 4.In his paper
[6],
Bundschuh remarked that in theorem 3 and theorem4,
one should like toreplace
the aiby complex
numbers aiwhich can be
approximated
very wellby algebraic
numbers. Thefollowing
theoremsgive
such results. In theorem5,
of which the second part isquite
an extension of theorem2,
weagain
need acondition on linear
independence;
in theorem 6 this is not the case.THEOREM 5: Let al, ..., an be non-zero
complex numbers,
letlog
ai denote anarbitrary
valueof
thelogarithm of
ai such thatlog
ai 7-’ 0.Let
bi,...,
bn becomplex numbers,
let E > 0 and d ~ N.Firstly,
suppose that there existinfinitely
many(2n)-tuples (al,
..., an,(31, ..., (3n) of algebraic
numbers with 1,03B21, ..., {3n linearly independent
over 0 andof degrees
at mostd, for
whichwhere H is a bound
for
theheights of
03B11, ..., an,03B21,..., {3n.
Thenarl
...abnn (where abji
= ebjlog aJ)
is transcendental.Secondly,
there areonly finitely
many(2n
+1)-tuples (al,..., an,
03B21, ..., 03B2n, y) of algebraic
numbers with 1,03B21,..., {3n linearly
in-dependent
over 0 andof degrees
at mostd, for
whichwhere H is a bound
for
theheights of
PROOF : Use the linear form
03B21 log
a 1 + · · ·+ J3n log
03B1n -log
an -log
y with y fixed for the first assertion and free for the second one, andproceed
as in theproof
of theorem 3. DTHEOREM 6: Let al, ..., an,
log
a,, ...,log
an,b i,
...,bn,
E and d beas in theorem 5 and let bo E C with bo 94- 0.
Firstly,
suppose that there existinfinitely
many(2n
+1)-tuples (al,
..., an,03B2o, 03B21, ..., (3n) of algebraic
numbersof degrees
at mostd, for
whichwhere H is a bound
for
theheights of
03B11, ..., an,03B2o,..., 03B2n.
Theneboab11
...ann (where abjj
=ebJ log al)
is transcendental.Secondly,
there areonly finitely
many(2n
+2)-tuples (a,, ...,
an,03B2o, 03B21,
...,(3n, y) of algebraic
numbersof degrees
at mostd, for
whichwhere H is a bound
for
theheights of
ai,..., an,03B2o, 03B21, ..., 03B2n,
y and whereafj = ehJ log aJ.
PROOF: As the
proof
of theorem 5. DWe would like to remark
here,
that at thepresent
state of de-velopment
of Baker’smethod,
thedependence
on thedegrees
of thealgebraic
numbers involved seems to become moreinteresting (for
some old
results,
see[7]). Among
manypossibilities,
one could try toinvestigate
theorems like the above ones, in which thedependence
onthe
heights
as well as on thedegrees
isexplicitly given.
3.
Preliminary
lemmasWhen a is an
algebraic number,
we denoteby la the
maximum ofthe absolute values of the
conjugates
of a,by H(a)
theheight
of aand
by den(a)
the denominator of a.For elements a,
j6
of a number field K ofdegree
D, we have thefollowing inequalities
where Ci,...,C4
depend only
onD;
from the firstinequality,
wededuce,
for a 760,
We shall also use the fact that for any
algebraic
number a and anypositive integer
q,We shall use the notation
Log
for theprinciple
value of thelogarithm.
LEMMA 1: Let K be a number
field of degree
D over Q. Let ai,i(1 i n, 1 j m)
bealgebraic integers
inK,
whoseconjugates
arebounded in absolute value
by
aninteger
A.If
n >mD,
there exists a non-trivial solution(xl, ..., xn) of
the systemin rational
integers
xi, boundedby
PROOF: This is lemma 1.3.1 of
[14].
DLEMMA 2: Let
f
be ananalytic function
in a discIz 1 -
Rof
thecomplex plane.
Let E be afinite
subsetof
the discIzl:5
r with r --5R/2,
consisting of
kpoints
which arelying
on astraight
line and have amutual distance at least 03B4 with 03B4 min
(r/2, 1).
Let t be apositive integer.
ThenPROOF: Without loss of
generality,
we may assume that rR/4.
Consider the circles T:
|03B6| = R and
Tx :|03B6 - x| = 03B4/2
forx E E,
des-cribed in the
positive
sense; defineQ(z)
=IIxEE (z - x)t.
We useHermite’s
interpolation
formula:Finally,
we useand
LEMMA 3: For any
positive integer k,
letv(k)
be the least commonmultiple of 1, 2,...,
k.Define, for
z E C,and
For any
integers
1 > 0, m > 0, k -0,
and any z E C,Moreover,
let q be apositive integer,
and let x be a rational numbersuch that qx is a
positive integer.
Thenis a
positive integer,
and we havev(k):5
4k.PROOF: The second part is lemma TI of
[13].
The first one follows from thefollowing
estimates(see [13]):
LEMMA 4: For any
positive integers k,
R and L with k >R,
thepolynomials (L1(z + r; k)Y (r = 0, 1, ..., R - 1;
l = 1, ...,L),
where,A (z; k)
has been introduced in lemma3,
arelinearly independent.
PROOF: Follows
immediately by
induction on L from thefollowing
result
(see
lemma 2 of[1]):
If P is apolynomial
withdegree n
> 0 and with coefficients in a fieldK, then,
for anyinteger m
with 0 m n,the
polynomials
P(x),
P(x
+1),...,
P(x
+m)
and1,
x, ..., xn-m-1 arelinearly independent
over K. DLEMMA 5: Let M and T be
non-negative integers,
let a, b and E-,(m = 0, 1,..., M ; t
= 0, 1, ...,T)
becomplex
numbers.If
PROOF: First we prove that a certain linear combination of the left hand side of
(6)
for t =0, 1,..., n just gives
the left hand side of(5)
with t = n, when 0 S n S T. We prove:Namely,
the left hand side of(7)
isequal
towhich, by replacing
t - Tby
t andapplying
the binomialformula,
isequal
to theright
hand side of(7).
We now prove the lemma
by
induction. For t = 0,(5)
and(6)
are thesame.
Suppose (6)
has beenproved
for t = 0, 1, ..., n - 1 with n :5 T.Then from
(7)
it follows that(6)
is also true for t = n. ElLEMMA 6: Let ai, ..., an be non-zero elements
of
analgebraic
number
field
K and leta:/p,
...,a,,"
denotefixed p-th
rootsfor
someprime
p. Further let K’ =K(a:/p,..., an’p,).
Then eitherK’(an"p)
is anextension
of
K’of degree
p or we havefor
some y in K and someintegers
PROOF: This is lemma 3 of
[4].
DIn
fact,
in order to prove theorem 1 we need lemma 6only
f orp = 2. We remark that for this
particular
case the arguments of[4]
could be
considerably simplified.
LEMMA 7:
Suppose
that a,{3
are elementsof
analgebraic
numberfield
withdegree
D and thatfor
somepositive integer
q we havea =
3 q. If
aa is analgebraic integer for
somepositive
rationalinteger
a and
if
b is theleading coefficient
in the minimaldefining polynomial
of {3,
then b - aD/q.PROOF: This is lemma 4 of
[4].
0LEMMA 8: Let al, ..., an be
algebraic
numbersof degrees
at most dand
of heights
at mostA; if Log al, ..., Log an
are0-linearly
de-pendent,
then there exist rationalintegers bl,..., bn,
not all zero,of
absolute values at most
such that
PROOF: This is lemma 2 of
[3].
D4. A
special
case of theorem 1PROPOSITION 1: Let n ± 1, D > 1 and po ± 2 be
given integers.
Thereexists an
effectively computable
number CI >0, depending only
on n,D and po, with the
following
property. Let a,, ..., an be non-zeroalgebraic numbers ; for 1 j
n, letlog
ai be any determinationof
thelogarithm of
ah and letA; >
6 be an upper boundfor
theheight of
aiand
for
expIlog ail. Further,
let{3o,..., {3n-l
bealgebraic
numbersof heights
at most B(- 6);
let K be a numberfield containing
ah ..., an,(30, ..., {3n-l
and assume that thedegree of
K over 0 is at most D. Putand
define
Suppose
that there exists aprime
p _ po such that thedegree of
thefield K(al"P,...,a,,’P)
over K isequal
topn,
wherea/lp =
exp
((1lp ) log ai), (1 j n).
ThenPROOF:
First,
wegive
some notations and discuss a certainauxiliary function; then,
wegive
theproof
itself in three steps.The numbers Ct,..., c4 were
already
introduced.By
cs, c6, ..., C15 we shall denotepositive
numbers that can bespecified
in termsof n,
D and po
only.
Let p be aprime satisfying
thesupposition
ofproposition
1. Let v be asufliciently large integer.
Without loss ofgenerality,
we may assume that fi and B are at leastC2,
where C2 islarge
incomparison
to v(otherwise,
wereplace f2 by
max(f2, C2),
or B
by
max(B, C2)).
We show that, in this case, CI can be chosen as2n +5
V .
We define the
integers Lo, LI, ..., Ln,
S and Tby
For
abbreviation,
we denoteby
U the numberf2 (Log fi)(Log
B +Log fi).
We suppose thatand we shall arrive at a contradiction.
To introduce our
auxiliary function,
we choose R =[v(Log B + Log il)],
and wesignify by A,o(z), (Ao = 0, ..., Lo)
the functions(L1(z + r; R))’, (r = 0,
1, ..., R -1;1 = 1,2, ..., (Lo + 1)/R).
Remark that it follows from the lemmas 3 and 4 that these functions
L1Ào, (Ào
=0, 1, ..., Lo)
arelinearly independent,
thatfor 1 z 1:::;
6S and 0 t T - 1, and that for allintegers
q and s with 1 s q T and 0 s s sqS,
the numbers~(t)03BB0(s/q)
(Ào = 0, 1,..., Lo ; t = 0, 1, ...,T - 1)
are rational numbers with a common denominator which is at mostWe shall use the
auxiliary
functionRemark that the exponent is
equal
toFor
brevity,
we shall writeaj
foreZ log a,;
we use2(A)
instead ofEkj=o ... Ik,=o
andp (A )
instead ofp(Ào,
...,An ).
We shall compute the derivatives of F; for this purpose we
introduce the
polynomials
The derivatives of F
satisfy
where 17-1
= t stands for To >_ 0, ..., Tn-l >_ 0 and To + ... + Tn-l = t and where the summation is extended over alltuples (To, ..., Tn_1) E Zn
with this property;
further,
where
We shall also consider the entire functions
Step
1. Constructionof
the p(03BB)
There exist rational
integers p(À),
not all zero, bounded in absolute valueby
exp(C7V2n+3 U),
such that thecorresponding functions 03A6(T) defined
abovesatisfy 4>(T)(S)
= 0for
all(n
+1)-tuples (To,..., Tn-l, s)
E Zn+l with0 ITI
T - 1 and 0:5 s S - 1.be a common denominator of the numbers ,r 0 = 0, 1, ..., To). Consider the system
of less than
linear
homogeneous equations
withunknowns
P (03BB),
and with coefficients which areintegers
of K, ofconjugates
boundedby
Using
lemma 1, we obtainStep
1.Step
2. Induction. When J is anon-negative integer satisfying pJ
T - 1, we show that there exist rationalintegers
not all zero, bounded in absolute value
by
exp(C7p2n+3 U),
such thatthe
functions
PROOF: For
J = 0,
we choosep(O)(À) = p(A),
thanks toStep
1.Assume that the assertion in
Step
2 is correct for someinteger
J with1 pJ (T - 1)lp.
Definewhere the sum is for
The
following
relation betweenfJ,(T)
and CPJ,(T) will be an essential tool in thesequel.
PROOF OF LEMMA 9: Consider the formula
Clearly,
we haveby
theinequality
On the other
hand,
we bound thelogarithm
of the absolute value ofusing (8), by
Consequently
This proves lemma 9.
We shall use also the
following
arithmetic property of the numbers’PJkr)(S/ p).
PROOF OF LEMMA 10: The number
’PJ,(T)(S/P)
is analgebraic
numberwith a denominator
which, using (9),
is at mostis a common denominator of all numbers
Further,
theconjugates
of this number’PJkr)(S/P)
have absolute values at most exp(C12V2n+3U).
Since thedegree
of’PJ,(Tls/p)
over a is notmore than
Dp" Dp’ ,
it follows that theheight
of it is not more thanexp
(C13v2n+3U).
Hence, either’PJ,(T)(S/P)
= 0 orand lemma 10 has been
proved.
We now
proceed
to prove the assertion ofStep
2 for J + 1.By
lemma 9, we obtain from the induction
hypothesis
the functions
f,,(,, satisfy
the differentialequations
It follows from
(10)
thatfor all values of
(T),
m, s with0 :5ITI 2p -JT,
0 m -2p -JT,
0 spJS -1
and(s,p)
= 1. Weapply
lemma 2 to the above indicated functionsf,,(,)
with t =[2p-JT]
and!pJS :5
k :5pJS, using
R = 6r, r =pJS,
8 = 1. We have forIzl
= RIt follows with
(8)
thatLemma 2 and
inequalities (11) give
Hence,
for all s e Z with
0spj+1S
and all(T)
withO:5ITI!p-JT.
Wecompare
fj,r>(slp)
with’PJ,(T)(S/ p );
once more lemma 9gives
for 0 s :5
pJ+l S and 0 :5ITI 1 2p-JT.
By
lemma 10, we see that’PJkrls/p)
= 0 for0 s [T[ p -J-lT,
0:5 spJ+l S.
From thispoint
on, weonly
use these numbers for so far(s, p) = 1. Using
ourassumption [K(a:/p,..., a,,"): K]
=pn,
we canexpress
’PJ,(T)(S/P)
on the basis1(a’,"p ... «nn’p), (1 1,
...,ln )
E{0,1,...,p - 1}"}.
We remark that the coefficients in thisexpression
consist of those contributions of
2(x)
for whichAi =-À 0 (mod p),
(j
=1, ..., n).
Since each of these coefficients must be zero, we haveThis is
possible by
the inductionhypothesis.
We denoteby
For fixed r,, ..., Tn-1 and s, write
into
the formSince this
expression
is zero for all values of To with 0 ToP -J-l T - (Tl
+... +Tn-1),
it f ollows f rom lemma 5 thatis zero for the same To. Hence, the numbers
are zero, where
We
proceed
to prove that the numbersare zero for the same values of
(T)
and s.Since
we have
so that each number
’PJ+l,(T)(S)
can be written as a linear combination of the numbersÇJ+l,(T)(S); hence, they
are zero for the mentioned values of(T)
and s.Step
3. Contradiction. Let Jo be theinteger
such thatpjo
T - 1pJo+l.
Sincebecomes a
polynomial;
it has zeros for z = s, 0 S s SpJoS - 1, (s, p) =
1. The number of these zeros exceeds
ST/(2p )
which is more thanthe
degree
of ’PJo,(O); so ’PJo,(O) must beidentically
zero. But thepolyno-
mialsA4(A,,
= 0, 1,...,Lo -1)
arelinearly independent;
hence allp(JO> (Ào,
0,...,0)
are zero, in contradiction to their construction.5. Proof of theorem 1
We first use the arguments of
[3],
§4 to prove thefollowing
result.PROPOSITION 2: Let n > 1, D > 1 and p > 2 be
given integers,
with pprime.
There exists aneffectively computable
number C3 >0, depen-
ding only
on n, D and p, with thefollowing
property. Let K be anumber
field, containing
thepth
rootsof unity, of degree
at most Dover Q; let al, ..., an be nonzero elements
of K; for
1:5 j :5
n, letlog
aibe any determination
of
thelogarithm of
ai, and letAj 2::
6 be an upper boundfor
theheight of
aj andfor
expIlog ail.
We assume thatal, ..., an are
arranged
in such a way that A1 A2:5 ... :5 An. Fur-ther,
let81, - - -, {3n
be elementsof
Kof heights
at most B withB 2::
Log
An. Assume that the linearform
does not vanish.
Then there exist an
integer
v, 1 v :5 n + 1, non-zero elementsa1,..., a l, 0 fi, ..., 03B2 v of
K, and determinationslog a 1, ..., log
a lof
thelogarithms of
a ( ,..., a 1,
such that(a)
A, isequal
to(1 J v),
hasdegree
pV over K,(c)
theheight of a
and expIlog ail
are at mostA; 3n-v,
where(in
caseof v
= n+ 1)
Ao should be read as 6, and theheights of
thenumbers ’ (j
= 1,...,v)
are at most BCBREMARK: When
{3],..., {3n
arerational,
we will find{3,..., {3
rational. When
log
a i, ...,log
an are theprinciple
values of thelogarithm,
we will findlog
a, ...,log
a lprinciple
valued too.PROOF: The numbers C16,..., C23 will
depend only
on n, D and p.Let h > 1 be the greatest
integer
such that exp(p -hi7r)
EK ;
thenph c 16.
Note that exp(p -hi7T)
is analgebraic
number ofheight
1;note also that exp
(P-(h+l)i7T)
generates an extension of K ofdegree
p since K contains thep th
roots ofunity.
Itclearly
sufices to prove theproposition
for linear forms Ai in which a 1= exp(P-hi7T)
andlog
al 1=p-hi7T, using
Al 1 =6,
when we show that v:5 n instead of v :5 n + 1 and(in
case of v =n )
that theheight
of a’ and expIlog ail
are at most 6c3.We shall prove this
by
induction. For n =1,
our statement is satisfiedby
our choice of al. Let n 2::: 2 and suppose that the statement is true for all linear forms of the indicated typehaving n - 1 logarithms (we
shall refer to thisassumption
as "the inductionhypo-
thesis on
n ").
Let m be the greatestinteger
with the property that, forIL = 1,..., m,
a’/
generates an extension ofK’ = K(al/p, ... a’pl)
ofdegree
p, where K! should be read asK;
remark that m > 1by
ourchoice of a 1. For m = n, the statement holds
trivially. Consequently,
we may assume that m n, and moreover, we may suppose that the statement is true for all linear forms of the indicated type,
having n logarithms,
and with mreplaced by
m’ with m m’ n(we
shallrefer to this
assumption
as "the inductionhypothesis
onm "). By
lemma6,
we havefor some y E K and some
integers
ri, ..., rm withm).
We construct, as far aspossible,
a séquence’elements in K such that
where the
integers
r,,jsatisfy
0 ri,j p.Clearly,
we havewhere the sn; are
integers
with 0 s,,jp’.
The denominator ofso
that, by
lemma7,
theleading
coefficient in the minimalpolynomial
of qyi is at most
A’-D+,.
Since eachconjugate
ofhas absolute value at most
we obtain that the
height
of yi does not exceedRemark,
that it follows thatWe deduce from
(12)
the existence of somesl,l
E Z such thatWe have from
(13):
in
particular
Define
We
distinguish
two cases,according
as the sequence yl, y2, ... ter- minates at yl for some 1 withp’ :5
H(case 1),
or it does not(case 2).
Case 1. Assume that yi,
y2, ... terminates at yl
withp’
H. From(13), Ai
isequal
to a linear form A2 inlog
al, ...,log
an, wherelog
am+lis
replaced by Log
y,, and where the coefficients arealgebraic
num-bers in K
having heights
at mostBC20; namely
where
£i = Q
+Qm+is l,i
andCi
=(3;
+{3m+lS"j (2 j ~ m). By
the sup-position
that the sequence terminates at yi, we deduce from lemma 6 thaty’p
generates an extension ofK(a:’p,..., amp)
ofdegree
p. We recall that theheight of yi
and expILog yi are
at most A211.Hence,
replacing
Am+iby Am+,, Aj by
max(Aj, A21,) (m
+1 S j S n)
and Bby
max
(BC20,
C18Log Am+,),
the inductionhypothesis
on mgives
thedesired result
(remark,
that a 1 remainsunchanged,
and that theinduction on m is
finite,
since 1 = m =n ).
Case 2. Let 1 be the least
integer
withp’ > H.
Sincelog
al, ...,log
am+1,Log
y, arelinearly dependent by (13),
we know that alsoLog al, ..., Log am+t, Log-y,
arelinearly dependent.
Wededuce from lemma 8 that there are rational
integers bo, b°, b2,
...,bm+1,
not all zero and of absolute values at mostH,
such thatIt follows that there exists a rational
integer
bl withfor which
Eliminating Log
yl from(13)
and(14),
we findWe see that
We claim that at least one
b;, (2:5 j :5 m + 1)
is not zero; infact,
if b’-+, =0,
we deduce from its definition that(because p
>H ~ 1 b,,,I)
bm+, = bo =
0;
in this caseclearly b;
=p’bj, (1 j m),
and the claim followsby
the observation that now b2 =0,
..., bm = 0imply b?
= 0.Let
jo,
with 2 -jo 5 m
+1,
be such thatbio -:j; O. Using (15),
we express the linear form A, aswhere
{3j = {3j - ({3jo/bjo)bj, (1 j
:5 m +1)
and{3j
=f3i (m
+2:5 j :5 n).
Since
{3 jo = 0,
and since theheight
of the numbers{3 j, (1 j
:5n)
is notmore than
BC23, the
statement followsby
the inductionhypothesis
onn.
Remark,
that a 1 remainsunchanged
and that the order remains intact. In both cases, our statement has beenproved.
Thiscompletes
the
proof
ofproposition
2.Finally,
wegive
theproof of
Theorem 1. We may assume without loss ofgenerality
that A 1 _ A 2 ~···~ An :5eB,
and thatA:F {3o.
The num-bers c24, ..., C28 will
depend only
on n and d. Let p be any fixedprime,
say p =2,
and take K =Q(j8o, j3i,..., {3n,
al,..., an,Ç) where {:F 1
is ap t’’
root ofunity.
Then K hasdegree
D :5d2n+lp ~
c24. Wefirst use
proposition
2 with the linear form lll = A -{3o,
to see that forsome v with 1 v n + 1 there exist non-zero elements a’, ...,
a, {3, ..., {3
inK,
and determinations oflog a’, ..., log a v,
such thatand such that the field
K((a )l’p, ..., (a v)l’p)
hasdegree pV
overK;
moreover, the
height
ofa
and expIlog ail
are at mostAi,25,,-,,,
where(in
case of v = n
+ 1)
Ao should be read as6,
and theheights
of thenumbers
{3o, j3i,..., (3
are at most B C26.Remark,
that theproduct
ofthe
logarithms
of the boundsAi,25,,-, (j
=1,..., v)
is at most C27[J.Subsequently,
weapply proposition
1 to the linear formin which the coefficients have
heights
at most BC28. Thisgives
a lowerbound of the desired form for
1-({3)-lA 1. As jS’,
is non-zero, we have1{3 I
>(BC28
+1)-1,
so thatinequality (2)
can be deduced from the above lower boundby multiplying by lo’,I.
6. A final remark
As we said
already
in theintroduction,
it ispossible
toreplace
thefactor
Log {l by Log
f2’ in the conclusion of theorem 1. To do so, oneonly
has to make somechanges
inproposition
1 and itsproof (§4).
Namely,
use the new parameters:Further,
in the induction of step 2, the number J shouldsatisfy p J :5
L’ n + 1. At the end of theinduction,
we arrive at functions ’PJo,{-r) for whichLnJ°’
= 0. Wereplace
step 3by
usual arguments(see
forexample [2], beginning
of§4).
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(Oblatum 18-111-1976 & 2-VI-1976) Pieter L. Cijsouw
Department of Mathematics of the Technological University Eindhoven
The Netherlands Michel Waldschmidt
Université P. et M. Curie (Paris VI) 4, Place Jussieu
75230 Paris Cedex 05 France
Added in
proof
As