A NNALES DE LA FACULTÉ DES SCIENCES DE T OULOUSE
A LEX B IJLSMA
An elliptic analogue of the Franklin-Schneider theorem
Annales de la faculté des sciences de Toulouse 5
esérie, tome 2, n
o2 (1980), p. 101-116
<http://www.numdam.org/item?id=AFST_1980_5_2_2_101_0>
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AN ELLIPTIC ANALOGUE OF THE FRANKLIN-SCHNEIDER THEOREM
Alex Bijlsma (1)(*)(**)
Annales
Faculté
des Sciences Toulouse VolI I,1980,
P.101 à 116(1)
lnstitut des Hautes EtudesScientifiques,
35 route de Chartres - 91440Bures-sur-Yvette,
France.
Resume : Soit p une fonction
elliptique
de Weierstrass d’invariantsg2
etg3 algébriques.
Soient aet b des nombres
complexes
tels que ni a ni ab soitparmi
lespoles
de p. On donne une minoration pourI’approximation
simultanée dep(a),
b etp(ab)
par des nombresalgébriques, exprimée
dansleur hauteur et leur
degré.
Par uncontre-exemple,
on montrequ’une
certainehypothèse
sur lesnombres
j8 qui approximent
b est nécessaire.Summary :
Let p be a Weierstrasselliptic
function withalgebraic
invariantsg2
andg3.
Let a andb be
complex
numbers such that a and ab are not among thepoles
of p. A lower bound isgiven
for the simultaneous
approximation
ofp(a),
b andp(ab) by algebraic numbers, expressed
intheir
heights
anddegrees. By
acounterexample
it is shown that a certainhypothesis
on the num-bers
~3 approximating
b is necessary.1- INTRODUCTION
If a ~ 0 and b are
complex numbers,
the numbers a, b anda
cannotsimultaneously
be
approximated by algebraic
numbers in such a way that the totalapproximation
error is smallin terms of the
heights
anddegrees
of thesealgebraic numbers,
except in the case where all the(*) Research supported by the Netherlands Organisation for the Advancement of Pure Research (Z.W.O.).
(**) Present address : Technische Hogeschool Eindhoven, Onderafdeling der Wiskunde, Postbus 513 - - 5600 MB Eindhoven, The Netherlands.
102
numbers used to
approximate
b can be chosen rational. This result is known as the Franklin-Schneider
theorem ;
itssharpest
version up until now is due to M. Waldschmidt[12],
whoproved
the
following :
there exists a numberC, effectively computable
in terms of a, b and the determina-tion of the
logarithm
of a used indefining ab,
such that for alltriples
ofalgebraic
numberswith
~i
irrationalwhere
log2
meansloglog,
D is thedegree
of the fieldover Q
andH ee
is a bound forthe
heights of a, ~i
and 7. The condition that is crucial : in[2],
thepresent
authorproved that,
even ifb ~ Q,
no bound of thetype (1 )
exists if~i
is allowed to be rational.The purpose of this paper is to prove
elliptic analogues
of both these statements. We fix thefollowing
Notation. Let
w1, W2
becomplex numbers, linearly independent
overIR ;
let 03A9 denote the set~ m 1 w 1
+m~ w2
andp
the Weierstrasselliptic
function withperiod
latticeS~.
Then p satisfies a differential
equation
of thetype
with
g~ ~ ;
we shall assumeeverywhere
thatg~ go
arealgebraic. By el’ e~, e~
we denotethe roots of the
equation 4X~ - g~X - g~
=0 ;
K will denote the field ofcomplex multiplica-
tion of
p,
that is to say,K
=Q( ~ / c~)
if~~ / o~
is aquadratic irrationality,
K= Q
other-wise.
We now state the
analogue
of(1)
we propose to prove.THEOREM 1.
Suppose
a, b ~ ~ such that a and ab arenot poles
of p. Then there exists an effec-tively computable
C e)R, depending
onp,
a andb,
such that notriple (u,03B2,v)
e~
satisfiesp(v) algebraic, 03B2 ~
K andwhile
p(u), a, p(v)) ;
D andp(u)),H(~i~,H( p(v)))
H. °The
proof
of this theoremdepends
on a result on linear forms inalgebraic points
of p(see
Lemma 1below) ;
in the caseof complex multiplication,
i.e. whenQJt
would also have beenpossible
to deduce aslightly
lesssharp
version of Theorem 1 from the results of M. Anderson announcedin [9] .
The condition
~i ~ ~
that was necessary for(1 )
isreplaced
in Theorem 1by ~i ~
OC.The
necessity
of the latterassumption
follows fromTHEOREM 2. For every function g :
~12 -~
~ there exist a E~, 1 OC,
such that a and ab arenot
poles
ofp
and such that for every C E R there existinfinitely
manytriples
E~3
sa-tisfying p(u), ~i, p(v) algebraic
andwhile
[~~P~u)~ ~~P~~)) ~ ~~ ~
Dandmax(H(p(u)), H~P~~))) ~
H.2 - PROOF OF THEOREM 1
For a set K in the
complex plane, K°
denotes the interior of K.By
the size of analge-
braic
complex number,
we mean the maximum of its denominator and the absolute values of itsconjugates.
LEMMA 1. For every
compact
subset K of ~B
St there exists aneffectively computable
C Ept, depending only
on p andK,
with thefollowing property.
Let u,(3,
v E~ satisfy
u, v EK°, p(u), (3, p(v) algebraic
and~i ~
UC. Letee
be an upper bound formin(H( p(u)), H( p(v))) ;
; letA2 > A~
be an upper bound formax(H( p(u)), H( p(v))).
Let B > e be an upper bound forand take D :_
[~(p(u),(3,p(v)) : ~]
. ThenProof. I. Let
(u,Av)
be atriple satisfying
the conditions of the lemma.By cl’ c2,...
we shall denoteeffectively computable
real numbersgreater
than 1 thatdepend only
on p and K. Let x be somelarge
realnumber ;
further conditions on x will appear at laterstages
of theproof.
Put B’ : = xDBlog A2’
E : =4D1 j21og1 ~2 A
and assumeFrom this we shall obtain a
contradiction,
which proves(3).
Define
in case
H(p(u)) A1.lf H(p(u))
>A~,
the definitions ofL1
andL2
should beinterchanged. By
~
we denote aprimitive
element forQ( p(u), p, p(v))
of the formm1 p(u)
+ +m3 p(v),
where
ml’ m2, m3
E IN - 1 andm~
+m2
+m~ D2 (cf. [3] ,
Lemma1).
Consider the auxilia-ry function
where the
p(03BB1,03BB2, s )
are rationalintegers
to be determined later. For t EIN -1,
s ~IN such thatsu ~ ~, st3u
we haveAlso define
where e : = v. For t E
)N - 1,
s E IN such thatsu ~ 03A9, sv ~
03A9 we haveNow
put
As in
[7] JX, §
1 anapplication
of the boxprinciple
shows that we may assume there is a subset Vof{1,...,S}
suchthat # V c-11 S and su, sv ~ Ko + 03A9
for all s ~ V.Moreover,
it is no restriction to assume that thepoints
su : s e V are all distinct.Indeed,
this caneither be
brought
aboutby interchanging
u and v, or there exists~, s~ ~ ~ 1,...,S }
andmi,...,m~E 2,
their absolute values boundedby
such thatHowever,
in that caseand
(3)
then follows from Fel’dman’s result[6] .
Putand consider the
system
of linearequations
in the
p(A1,A2,S ).
To solve thissystem,
we shall use a method devisedby
M.Anderson.Lemma 5.1 of
[10]
states that for every w EC,
there existpolynomials
03A6w, 03A6*w ~ Z [X1,...,XS] ,
theirheights
anddegrees
boundedby
an abso)uteconstant,
such thatas a
meromorphic
function in z, whileLemma 7.1 of
[8] ,
which remains valid withoutcomplex multiplication,
states thatfor every s E
IN,
there existcoprime polynomials ~ , degree
at mosts2
such thatas a
meromorphic
function in z ; the coefficients~Y~
are themselvespolynomials
ing /4,
~ ~ s s
2
2g3,
with adegree
at most s and rationalinteger
coefficients notlarger
thanc 3 .
Now defineand
then for s E V we have
Therefore we have found a solution of
(7)
if we choose thep(A~,~2,8 )
in such a way thatThe number of
equations
in(9)
is at mostwhile the number of unknowns is
According
to Lemma 5.2 of[10],
theexpression
can be written as a
polynomial
inp(u)
andp’(u)
ofdegree
at most in each of the varia-bles ; the
coefficients of thispolynomial belong
toQ( p( w1/2), p"( w1 /2)),
and their size is boun- dedby s2
+ tlog t)). Moreover, they
have a common denominator of the formmn,
where m
depends only
on p and nc (~ s2
+ tlog t).
Thus the coefficients of thesystem
of linearequations (9)
lie in a field ofdegree
at mostcgD
and their size and common denominatorare bounded
by
According
to Lemma 1.3.1of [11 ] , if x >
thisimplies
the existence not all zero, such that(9)
andthereby (7) hold,
whileII. Put T’ : =
[x2T] .
We shall provethat,
for ourspecial
choice of the),
we have
For s E
V,
tE ~ 0,...,T-1 ~ , comparison
of(5)
and(6) yields
Now define
where
here a is the
sigma-function
of Weierstrasscorresponding
to S~ . Then G is an entire func-tion and
.
as I I
T ! +T )), by Cauchy’s inequality,
substitution of(11 )
shows that
Lemma 2 of
[5]
states thatFor the factors in the
right
hand member of(14),
we possess thefollowing
estimates :Substitution in
(14) yields
and
thus, by Cauchy’s inequality,
Fix s e V and let t be the smallest number in
{ 0,...J’-1 ~
such thatfs t
~=0. From(8)
it isthen clear that
)(su)
= 0 for T =1,...,t.
For all terms with r ~ 0in (12)
we have an es-timate of the form
while the left hand member is bounded
by (16).
This shows thatNow the definition of
V, together
with Lemma 7.1 of[8], implies
thatand therefore
adjusting
the value ofc29
ensures that(17)
also holdsfor I (su) I
. ThusAccording
to(8),
we havethus, by
the choice of t,Therefore
However,
onceagain by
Lemma 5.2 of[10] ,
is analgebraic
number ofdegree
at mostc36D
and of size at most
by
formula(1.2.3)
of[11 ],
thisimplies
so if x >
c3Sc40
we obtain a contradiction with(18).
This shows thatwhich, by (8), implies (10).
III. To obtain the final contradiction we shall use an
argument involving
resultants. It has beenbrought
to the author’s attention that this method is due to W.D. Brownawell and D.W.Masser,
who willpublish
a detailed account of it in[4] .
Putand
Suppose
P is notidentically
zero. LetP*(X,Y)
be anarbitrary
non-constant irreducible factor ofP(X,Y)
withalgebraic
coefficients and putThen,
for every s GV,
where
Px
andPy
denote the derivatives of P* with respect to the first and second variable respec-tively.
ThusPut
then
(20)
may,by
the differentialequation
for p, be written asWe consider the case where neither
Pv
norPy
isidentically
zero ; from(19)
and(21 )
weget
where R denotes the resultant of P* and
Q
withrespect
to the second variable. LetLi, L2
denotethe
degree
of P* withrespect
toX,
Yrespectively ;
then R is apolynomial
ofdegree
at mostwe find that either
or R is
identically
zero.Suppose (23)
is notsatisfied,
so R isidentically
zero. As P* isirreducible,
it followsthat P* divides
Q.
Let A denote the resultant of P* andPy
withrespect
to Y. If A were identi-cally
zero, it would follow from theirreducibility
of P* that P* dividesP*,
thus thatP*
isidentically
zero ; therefore there is someB
S~with A(
=~= 0. I nparticular
wemay choose
zi
in such a way that is transcendental.Clearly
there existsa ~
withP*( p(z 1 ), ~’ )
= 0. If( p(z 1 ), ~ ) - 0,
it would followthat 0 ( p(z 1 )) - 0 ;
thusP Y * ( p(Z 1 ), ~’ )
~ 0.The
implicit
function theorem now states that there exists aholomorphic
functionh,
defined on aneighbourhood
U ofz1,
such thath(z1 ) _ ~ ,
whileP*( p(z), h(z))
= 0 andPy(p(z), h(z)) ~
0 for z E U. Differentiation shows that for z E U we havethus
and
On the other
hand, Q(p(z),h(z)) = 0,
so we find thatfor z E U. Because
p(z~)
istranscendental, h(z~ ) - ~’ ~ ~ e~,e2,e ~
andthus, by (24),
h’(z~ ) ~ 0 ;
it is no restriction to assumeh’(z)
~= 0 for all z E U. Differentiation of(24)
nowyields
an
equation
that shows that the coefficientshn
of theTaylor development
of h aroundzl
sa-tisfy
It is clear that at most two
analytic
functions h on U cansimultaneously satisfy (25)
andBut
as ~ ~ ~ e~,e2,e3 ~ ,
there arez2,z2
such thatThe functions z 1-+
and
z 1-+ aredifferent, yet they
bothsatisfy (25)
and(26) ;
therefore either
h(z)
=for
z E U orh(z)
= for z E U. It is no restriction toassume that the first
equality
holds. ThusP*( p(z),
= 0 for z EU,
andby analytic
conti-nuation for every z that is not a
pole
of eitherelliptic
function. Here we obtain a contradiction with thealgebraic independence
over ~ of theelliptic
functionsinvolved,
and we haveproved
that(22)
holds for every irreducible factor P* of P with theproperty
that neitherP*X
norP*Y
areidentically
zero. IfPy
isidentically
zero butPx
is not, it followsimmediately
from(22)
and thedistinctness modulo S~ of the
points
su thatIf
Px
isidentically
zero butPy
isnot,
it likewise follows from(22)
thatnote that the number of P*
fitting
this lastdescription
is at mostL2.
If we nowput
it is
immediately
clear thatHowever, (10)
states thatNs
T’ for s E V and thusIf x >
c4Sc 46’
we obtain a contradiction which shows that P must beidentically
zero ; as thecoefficients
p(A1,A2,b )
are not all zero, thisimplies
the existence of a lineardependence
relationbetween
1, ~ , ~ 2, ... , ~ However, dg ~
= D and so we have obtained the final contradiction thatcompletes
theproof.
Using Lemma 1,
we shall nowgive
aproof
of Theorem 1.By cl’ c2, ...
we shall denoteeffectively computable
real numbersgreater
than 1 thatdepend only
on, a and b.By
C we shalldenote some real number greater than
1 ;
additional restrictions on the choice of C will appear below.Suppose
sometriple
satisfies(2).
First consider the case whereel,e2,e3 ~ .
Thenp’(a) =1= 0 ; according to §
3.3 inChapter
4 of[1 ],
there exists somec 1
suchthatp,
restricted to thedisk I z - a I c 1 1,
has ananalytic inverse,
thusfor I
z - aI c-1. Moreover, there exists a number c3
such that,
for every w with
I
w - p(a) I c ,
theequation p(z)
= w hasexactly
one root in the diskI z - a
Icl1 (ibid.,
Theorem
11).
Choose C solarge
that theright
hand member of(2),
andthereby Ip(a) -p(u) I,
is smaller than
c 31 ;
then there isexactly
one u’with I a - u’ I
andp(u’)
=p(u). By (26)
we now have
In the case where
p(a)
Ee~,e2,e3 , (2)
and Theorem 1.1 of[11]
showthatp(a) =p(u)
if Cexceeds some
c4. Thus p(u’) =p(u)
and(28)
holdtrivially
if we take u’ = a.Similarly,
we find v’with p(v’) =p(v)
andThus, combining (2), (28)
and(29),
Let K be a compact subset of C
1
Qcontaining
a and ab in its interior and letc7
be the constantfrom Lemma 1
corresponding top
and K. Then(30)
contradicts Lemma 1 if C issufficiently large
in terms of
c6
andc~.
3 - PROOF OF THEOREM 2
LEMMA 2. For e very g :
OV 2 -~ ~,
there exist sequences(u ) ~ , (~i ) °°__ , (v ) °°_ , ( E )°°_ ,
,such that for all n ~IN the
following
statements are true : ;Proof. Define =
v1 . - -
w1,
~i1 . _ , E 1 . _ ‘ . Then are algebraic by
5 2 .
Lemma 6.1 of
[8] ,
which remains valid withoutcomplex multiplication.
Now supposef1 ~...~ fN
have been chosen in such a way that(31)
holds forn =
1,...,N
and(32), (33), (34)
hold for n =1,...,N-1,
andproceed by
introduction. Choose 1 1c E
]0,1 [
so small that(32)
and(33)
hold for n = N. Take E]
-, - 3 2[ Q w
13 3
such that
I l
and G]
-, -[ Q Q
such that 0 eThen
(34)
holds for n = N.Finally, put
* thenG ] 1, 3 |03C91 ~ Q 03C91.
4 4
Again according
to Lemma 6,I of[8] ,
, we havealgebraic;
thus(31 )
is satisfiedfor n = N+I ..
LEMMA 3 . /f two sequences of
complex
numbers and( e
ofpositive
real numberssatisfy I l e £ I then I wn I
e§ §£
for almost all n and all m > n . Proof. From e n+1e£
I it follows that limn
= 0. PutAs for all m we know
wm
E it is sufficient to prove that for almost all n and all k > n onehas I k C I Take k > n and z , i.e.
I
z -w I 1/2k+1.
Thenif n is
sufficiently large,
so that in that case z EUsing
Lemmas 2 and3,
we shall nowgive
aproof
of Theorem 2. Construct sequences(u )°°_ , (~i )°° , (v )°°_ ,
as in Lemma 2.According
to Lemma3,
and(j3 )°li
areCauchy
sequences and their l imits a, bsatisfy
1 1 3 3
for almost all n. From
(31 )
it follows thata E [ -, - ] w1, b E [ - , - ] ;
therefore1 3 3 2 4 2
ab E
[
-, -]
We conclude that neither a nor ab arepoles
of p, andthereby (35) implies
4 4
for almost all n, where c does not
depend
on n. In the notation of(32),
theright
hand member of(36)
satisfiesif n is
sufficiently large
in terms of C and c.Finally, (34) implies
the existence ofarbitrarily large
n for which
j8
~=b ;
as,by (33)
and(35),
every~in
is a convergent of the continued fraction ex-pansion
of b and limj3
=b,
it follows that b hasinfinitely
manyconvergents.
Thus b ~ IR B Q andtherefore
b ~ tK. ’
It is clear that in case of
complex multiplication
it makes no essential difference ifwe
replace an ~in w2 /
thus for Theorem 1 the condition is not sufficient and wereally
need~i ~ OC.
116
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