C OMPOSITIO M ATHEMATICA
D. W. M ASSER
The transcendence of certain quasi-periods associated with abelian functions in two variables
Compositio Mathematica, tome 35, n
o3 (1977), p. 239-258
<http://www.numdam.org/item?id=CM_1977__35_3_239_0>
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239
THE TRANSCENDENCE OF CERTAIN
QUASI-PERIODS
ASSOCIATED WITH ABELIAN FUNCTIONS IN TWO VARIABLES
D. W. Masser Noordhoff International Publishing
Printed in the Netherlands
1. Introduction
Let A be a lattice in the
complex
spaceC2.
An Abelian function withrespect
to A is ameromorphic
functionf (z)
of the variablez =
(zi, Z2)
such thatfor all w in A. The
totality
of such functions form afinitely generated field %
whose transcendencedegree
over C is well-known not to exceed 2(see
forexample [9],
p.78). Throughout
this paper we shallassume that this transcendence
degree
isexactly
2. Hence there existsa vector A =
(A,, A2)
ofalgebraically independent
Abelian functionsA1
1 andA2,
and a function Balgebraic over C(A),
such thatFurther,
from(1)
it is clear that the differentialoperators dlazl, aI az2
take the
field C(A, B)
into itself.The
object
of this paper is to prove the transcendence of certain numbers associatedwith J,
andconsequently
we shallsuppose S
defined with reference to the field A of
algebraic
numbers in thefollowing
sense.First,
the function B isalgebraic
overA(A),
andsecondly
theoperators al az1, laldZ2
take the fieldA(A, B)
into itself.We then say
that J
isalgebraically
defined withrespect
to its generators A and B.It is easy to construct
examples
of such Abelian function fields. Animportant example
for this paper arises fromtaking
A as the Car-tesian
product
of latticesAI, A2
each in C. IfPi(z), P2(z)
are theWeierstrass
elliptic
functions associated with these lattices the cor-responding
Abelian function field isgiven by
theright-hand
side of(2)
withFurthermore if the invariants of
VI(z), P2(z)
arealgebraic
numbersthe above
generators provide
analgebraic
definition of this function field.Our main concern in this paper is with a
slightly
different class offunctions. We define a
quasi-periodic
function withrespect
to A as ameromorphic
functionf (z)
such that for each tu in A the differenceis constant. This
plainly implies
thataf/az1, afl az2
lie inC(A, B).
In addition we say thatf
isalgebraically
defined if these derivatives lie inA(A, B). Clearly
any linear functionalz, + a2Z2
isquasi-periodic,
and if a,, a2 are
algebraic
numbers it is alsoalgebraically
defined. The existence of less trivialexamples,
at least in the Cartesian casementioned
above,
can be seen as follows. IfCl(z), (2(Z)
are theWeierstrass zeta functions associated with
VI(z), P2(z) respectively,
then any combination such as
is
quasi-periodic;
alsof
isalgebraically
defined if a,, a2,03B21, 03B22
arethemselves
algebraic
numbers. Notice that if 03C91, W2 areperiods
ofA1, ll2 respectively, and n1,
172 are thecorresponding quasi-periods,
thequasi-period
attached tof
and theperiod w
=03C91, Cù2)
of A isgiven by
More
generally quasi-periodic
functions may be constructed fromintegrals
of the second kind(see
p. 121 of[8])
orby logarithmically differentiating
theta functions(see (7) below).
Let now A be a lattice such that the
corresponding
Abelian func-tion
field %
isalgebraically
defined with respect to certaingenerators
A and B. Fix a non-zero
period w = (03C91, W2)
ofA,
and let H be aquasi-periodic
function withrespect
toA, algebraically
defined withrespect
to the samegenerators.
In this paper we shall determine when thequasi-period n(H, tu)
is transcendental.Partial results on this
problem
were first obtained in 1941by
Schneider[8].
Histheorems,
which are stated underslightly
differenthypotheses
for Abelian functions onCn, imply
thefollowing
whenn = 2. If H is not itself an Abelian
function,
and 03C91, (d2 aresuitably
chosen
periods
ofA,
then at least one ofn(H, tu,), n(H, W2)
istranscendental.
In this connexion there is an assertion in
[6]
whichought
to becorrected. On
taking H(z)
= zi, say, in this result of Schneider we seethat the first components of wl, w2 cannot both be
algebraic. Similarly
for the second components; from the
generalization
in C" Schneider deduced his well-known theorem on values of the beta function. The resultquoted
in[6]
andmistakenly
attributed toSchneider,
that thecomponents
of asingle period
cannot all bealgebraic,
appears to have been firstproved by Lang
in[3].
Now n(H, 03C9)
canquite easily
be analgebraic number,
for if H is itself an Abelian function thisquasi-period
is zero. Also if a,tu, + a2(1)2 = 0 foralgebraic
ai, a2, wehave 7q (H, tau)=
0 whenAccordingly
we call wspecial
if itscomponents
arelinearly
de-pendent
over A. Our theoremimplies
that the above are theonly examples
ofalgebraic quasi-periods.
THEOREM:
If
H is not an Abelianfunction
and (d is notspecial
then,q(H, w) is
transcendental.If w is
special
we can use results of someprevious
papers toinvestigate
the transcendence of thequasi-periods.
For then Theorem II of[6]
shows that A isessentially
a Cartesianproduct,
and it is notdifficult to deduce that any
quasi-period
isgiven by
theexpression
onthe
right
of(5).
It follows from Baker’s theorem on two Weierstrasszeta functions
[1] that n(H, (d)
is either zero or transcendental.Finally
thegeneralization
of[1] proved
in[7]
enables us to determineexactly
whichquasi-periods
can vanish.We recover Theorem 1 of
[6] simply by making
the choice(6)
in our present theorem. In fact we shall establish our theoremby extending
the methods used in
[6].
Under theassumption
that71(H, tu)
is analgebraic number,
we construct for eachlarge integer
k a non-zeropolynomial
in H and A with many zeros. If H is not inJ,
thisauxiliary
function is notidentically
zero. The maindifficulty
incarrying
out the
extrapolation
concerns certain division values associated with H.By using
a well-known but notentirely
trivial lower bound forpolynomials
inalgebraic
numbers we obtain estimates for these division valueswhich, although they
areprobably
far from bestpossible, just
sufHce for our limited purposes.We then borrow an elimination
procedure
from[7]
to remove thequasi-periodic part
of theauxiliary
function. For each k thisyields
anon-zero
homogeneous polynomial
inA[x]
with small absolute valueat x = to. We conclude from the familiar transcendence criterion of Gelfond that if
o), 0
0 then03C92/03C91
must bealgebraic; i.e.,
theperiod
isspecial.
Following Schneider,
the result of this paper, as well as those of[6],
may be stated in terms of Jacobians. LetF(x, y)
be apolynomial
with
algebraic
coefficients such that theequation F(x, y )
= 0 re-presents
a curve C of genus 2. We call (6special
if there exists arational map from Cf6 onto an
elliptic
curve. It is not hard to see that 16is
special
if andonly
if its Jacobianvariety splits
into aproduct,
andso the
study
of such curves may be reduced to the consideration of twoelliptic
curves as in[6].
Hence we shall suppose that % is notspecial.
Now on W there exist non-zero diff erentials of the first kind
having
the form cp =
R(x, y) dx
whereR(x, y )
is a rational function withalgebraic
coefficients. Let Y be any closedpath
on the associatedRiemann surface of C. The theorems of
[6] imply
that theintegral
ofcp around Y is transcendental if L is not trivial in the sense of
homology.
Similarly
there exist differentials o, of the second kind on (6 also definedby
rational functions withalgebraic
coefficients. The theorem of the present paper shows that if a is not the differential of a rationalfunction,
theintegral
of cr around a non-trivial closedpath L
istranscendental.
As an
example
consider the curve Cf6corresponding
toF(x, y)
=y2 + x6 - X.
Twolinearly independent
differentials of the first kind aregiven by
Their
periods
can berapidly
calculatedby applying
the endomor-phisms
x --->,Ex,y -->,E’y (e5 = 1)
of(6,
and it followsby inspection
thatC is not
special.
The differentialsare of the second
kind;
furthermore if a,, a2,03B21, 03B22
are not all zerothen
is not the differential of a rational
function,
since not all of itsperiods
vanish. From a familiar
argument
one of theseperiods
7q isgiven by
the formula
181
in which y assumes
only non-negative
values. Hence if the coeffi- cients ah a2,03B21, (32
arealgebraic numbers,
theperiod 7q
is a tran-scendental number. This leads to a result on the beta function
B(x, y)
as follows.
If the
positive integers
m, n varysubject only
to m ~- n modulo5,
the values
B(1/5)m, in)
span a vector space over A of dimension at most 5. Our theoremimplies,
after asimple calculation,
that this dimension isexactly 5; equivalently
the numbersB (1/2, 1/5m ) (1
ms5)
arelinearly independent
over A. It would bepleasant
to include the number 7Tby removing
the restriction on m and n, but all 1 can prove is the weaker resultthat 7q
+ 7T is either zero or transcendental.Finally
1 am much indebted to Michel Waldschmidt forcorrecting
amisconception
of mine about Lemma 4.1.2. Normalization and
preliminary
lemmasSuppose 8
isalgebraically
defined withrespect
to thegenerators
A andB,
and let H be analgebraically
definedquasi-periodic
function.By multiplying
Bby
a function in thering A[A]
if necessary, we mayassume that B is
integral
over thisring.
Also thepartial
derivatives ofH, A,
B can beexpressed
as rational functions ofA,
B with acommon denominator C in
A[A].
It followseasily
that theoperators al az1, dl’aZ2
take H into thering A[A, B, C-1]
and also take thisring
into itself.
Our first task is to find a convenient
representation
of all these functions asquotients
of entire functions.By
a theta function withrespect
to the lattice A we mean an entire functionO(z)
such that for each w in A there exist constants a i,a2, b
such thatThis
implies (cf.
Lemma 1 of[6])
that for some cindependent
of zwe have the estimate
where
More
generally,
if 0 is an entire functionsatisfying (8)
and(9)
weshall say that 0 has
growth
order at most 2.LEMMA 2.1: There is a non-zero theta
function
0 such thatOH, 0Ai, OA2, OB, OC-1
are entirefunctions of growth
order at most 2.PROOF: Let us
temporarily
denote the operatoral aZi by ai (i
=1, 2).
Since
a1H, a2H
lie inJ,
we can find(e.g., [9],
p.91)
a non-zero theta function 0 such that the functionsare entire. We
proceed
toshow, using
theunique
factorizationproperty ([9],
p.11)
of thering +
of power seriesconvergent
near theorigin,
that 8H is also entire.Let a be an
arbitrary point of C2 .
There existmutually
non-associate non-unit
primes
p,,..., p., q 1, ..., qn of Z and a unit u of :rtogether
withpositive integers
1£,, ..., ILm, VI,..., Pn such thatfor z near 0. Thus for i =
1, 2
theexpressions
Because
O(z
+a)aiH(z
+a)
isholomorphic
at z =0,
we deduce thatQ
divides8(z
+a)Pi.
If nowQ
does not divide8(z
+a),
neither doesqkk+1
forsome k,
andhence qk
divides eachPi.
It follows from(10) that qk
divides eachaiqk.
But this iseasily
seen to beimpossible
sinceqk is not a unit. Therefore
Q
does indeed divide8(z + a),
and so8(z + a)H(z + a)
isholomorphic
at z = 0. Onrecalling
that a isarbitrary,
we conclude that cp = 8H is entire.To calculate the
growth
order of ~, let z be apoint of C2 ,
andsuppose for the moment that H is
holomorphic
at z. If tu is aperiod
of A nearest to z, the
point
u = z - w lies in somecompact
setindependent
of z. Hence if the c’s denotepositive
constants alsoindependent
of z, we have from Lemma 1 of[6]
with R
given by (9).
ThusNow is bounded and
plainly
whence
This
inequality
continues to hold when the restriction on z isremoved,
and so ço hasgrowth
order at most 2. Since the otherfunctions
8A¡, 8A2, 8B, OC-’
are themselves thetafunctions,
thiscompletes
theproof
of the lemma.Let J denote the Jacobian determinant of A with entries
aAlaz¡
(i, j = 1, 2).
Since J is notidentically
zero([9],
p.77),
there exists apoint
a at which none ofC, 0,
J vanishes. Then A isholomorphic
at a, andby slightly shifting
a we can ensure thatA(a)
hasalgebraic
coordinates. The effect of
translating
the variable zby
a is to takea = 0 in the
preceding remarks; supposing
this done it follows thatA 1, A2, B, C-1
areholomorphic
at theorigin
and takealgebraic
valuesthere,
while
0(0), J(0)
are both non-zero. Furthermore H isholomorphic
atthe
origin
and itsquasi-periods
areunchanged. They
remain so even ifwe subtract an
arbitrary
constant fromH ;
thus we may assume thatH(0)
= 0.Plainly
none of theseoperations
affects thealgebraic
definition
of J
or H.The
non-vanishing
ofJ(0)
has thefollowing
useful consequence.LEMMA 2.2:
If P(x) is a
non-zeropolynomial of
totaldegree
atmost
L,
thefunction P (A(z))
cannot have a zeroof
order greater than L at theorigin.
PROOF: See Lemma 2 of
[6].
Since we intend to construct
auxiliary polynomials
inH, Ai, A2
weneed to know when these functions are
algebraically independent.
The next lemma shows the obvious necessary condition also to be sufficient.
LEMMA 2.3:
If
H isnot in J
then thefunctions H, A 1, A2
arealgebraically independent.
PROOF: If
H, Ai, A2
arealgebraically dependent
let m denote thedegree
of H overS.
Thus there are functionsfi, ..., f m in J
such thatFor i =
1,
2 weapply
the operatordl,9zi
to(11), obtaining
where go, ..., gm-1 also lie in
J.
Hence this must be a trivialequation
for H. In
particular
its first coefficientvanishes for
i = 1, 2
and so H diff ers f rom - m-’ f by only
a constant.Thus in fact m = 1 and H lies in
J.
The second stage of normalization involves the introduction of an
algebraic
number field. Let K be analgebraic
number fieldcontaining Ai(0), A2(0), B (0)
such that B isintegral
overK[A],
C lies inK[A],
and the operatorsalaz1, al az2
take H intoK[A, B, C-1]
and thisring
into itself. If I is the
ring
ofintegers
ofK,
we may even suppose that B isintegral
overI[A],
C lies inI[A],
and these operators take H into R =I[A, B, C-11
as well as 9î intoitself;
for if necessary wereplace
B and Cby integral multiples
of themselves. The action of differential operators on thering R[H)
is nowparticularly
easy to examine in detail. For the rest of this section c, cl, ... will denotepositive
constantsdepending only
on the differentialequations
im-plied by
the above remarks. Also d denotes thedegree
of B overI[A].
LEMMA 2.4: Let P be a
polynomial
inI [xo,
x, y,z]
withdegree
atmost
Lo
in xo and totaldegree
at most L.If
a is adifferential
operatorof
order notexceeding
k we havewhere
Q
inI[xo,
x, y,z]
hasdegree
at mostLo
in xo and totaldegree
atmost L + ck. Furthermore
if
thecoefficients of
P have sizes notexceeding S,
thoseof Q
have sizes notexceeding (cL)k k! S.
PROOF: A
straightforward
induction onk ;
compare Lemma 3 of[6]
and Lemma 5 of
[7].
Note that thedegree
of P in the variable xocorresponding
to the function H does not increase on differentiation.This remark is the basis of the elimination
procedure
of[7],
which ismodified for our present situation in the
following
lemma.LEMMA 2.5: Let P be a
polynomial
inI [xo, x] of degree
at mostLo
in xo and total
degree
at most L.Suppose
thefunction
is not
identically
zero. Then there exists a non-zeropolynomial Q
inI[x] of
totaldegree
at mostcLoL
with thefollowing property.
Thefunction
--- ,has a zero at each
point
where H and A areholomorphic
and 0 has azero
of
order at least2Lo
+ 1. Furthermoreif
thecoefficients of
P havesizes not
exceeding S,
thoseof Q
have sizes notexceeding (CL)cLOLscLo.
PROOF:
Using
Lemma 6 of[7]
we deduce the existence of apolynomial R
inI [x, y, z]
of totaldegree M-C,L0L
such that the functionis not
identically
zeroyet
has therequired
zeros. Also the coefficients of R have sizes at most(c,L) c,LOLSc,Lo .
To obtain e we would like toclear of denominators and take a suitable norm; this can be
formally
carried out as follows. The function g =
CMf
lies inI[A, B],
and wecan write
for
hr,s
inI[A].
From theseequations
we construct the characteristicpolynomial of g
overI[A].
Since this is a power of thecorresponding
minimal
polynomial,
its constant term 03A8 is notidentically
zero. Onthe other hand 03A8 vanishes at all zeros
of g
where A isholomorphic.
The lemma follows on
noting
that up tosign 03A8
is the determinant of the functionsh,,S (0 -
r, s s d -1),
and its coefficients arereadily
estimated from
(12).
3.
Heights
of division valuesIn this section we fix a non-zero
period to
of A andstudy
thevalues of Abelian and
quasi-periodic
functions at thepoints rwl q
forcoprime integers q - 1,
r. Forsimplicity
we shall sometimesrestrict q
to the set 21
consisting
of all oddintegers
q ~ 3 such that0(rwlq) 0
0for all r
prime
to q. This conditionimplies
that the functionsH, A, B, C-1
areholomorphic
at thecorresponding points r03C9lq.
Since0(0) ~
0the function
03B8(z03C9)
is notidentically
zero, and so it is clear from(7)
that 9 contains all
sufficiently large
oddintegers.
We retain the normalizations introduced in the
previous section,
and forbrevity
we denote the fieldK(A, B) by J(K).
Thefollowing simple
result is very useful.LEMMA 3.1: The
coefficients
in the power seriesexpansions of
Habout the
origin
lie in K. Furthermore afunction in J holomorphic
atthe
origin
has thisproperty if
andonly if
itbelongs
toB(K).
PROOF: If P lies in
I[xo, x,
y,z]
thenby putting
z = 0 in Lemma 2.4we see that
P (H, A, B, C-1)
lies inJ(K)
thering
of power seriesconvergent
near 0 with coefficients in K. Hence so does H and anyfunction in
?j(K) holomorphic
at 0.Conversely
any functionin %
andX(K)
can beexpressed
asfor
coprime polynomials Po,..., Pd. Normalizing Pd
to have some coefficientunity,
we conclude as in theproof
of Lemma 6.2 of[4]
thatPo,..., Pd
lie inK[x].
We shall
frequently
make use of the deviceanalogous
toL’Hôp-
ital’s rule
employed
on p. 505 of[5]
and p. 102 of[6]
to ensure that certain denominators do not vanish. Thepositive
constants c, cl, ...will
depend only
on theperiod w
and the differentialequations underlying
the normalizations of section 2. The first three lemmas of this section deal with division values of functions in?’j(K).
LEMMA 3.2:
Suppose f
in?’j(K)
can be written as thequotient gt!g2 of
non-zerofunctions
gb g2 inI[A, B] of
totaldegrees
at most L in A,B,
and let S be an upper boundfor
the sizesof
thecoe ffi cients
in gi, 92-Let q:t-- 3,
r becoprime integers
such thatf
isholomorphic
atrwlq.
Then the
number f (rwlq) is
analgebraic
numberof degree
at mostcq4
and
height
at most(cqLS)Cq4.
PROOF: This estimate is
implicit
in theproof
of Lemma 7 of[6],
andso we
give
no more than a sketch of theargument.
We know thatf (z)
is a rational function of
A(z), B(z)
of totaldegree
at most L. Fromthe
multiplication
formulafor % given
as Lemma 4 of[6]
the func- tionsA1(qz), A2(qz)
can beexpressed
as rational functions ofA(z), B(z)
of totaldegrees
at mostclq 2.
LetP(x, y)
be apolynomial
withundetermined coefficients whose total
degree
in x isformally CLq2
and whose
degree
in y isformally Cq4.
If C c2 is asufficiently large integer,
we can use a linear forms lemma to choose the coefficients of P asintegers
ofI,
not all zero, such thatFurther, by keeping
track of themagnitudes
of all thealgebraic quantities occurring,
we find that the sizes of the coefficients of P canbe
supposed
not to exceed(c,q’S)c3". Finally
the substitution z =rtulq
in(13)
afterusing
theL’Hôpital
deviceyields
analgebraic equation for 03BE = f (r03C9/q)
over I. We derive anequation
over therational field Q almost
immediately,
and with it therequired
estimatesfor the
degree
andheight
ofe.
One consequence of this lemma will be that for
each q
in 2 there exists analgebraic
number fieldKq
ofdegree
at mostcq4
whichcontains all the numbers
f (rwlq)
as r andf
varysubject
to theconditions of the lemma. To see
this,
we fix rprime to q
andtemporarily
writeKq,,
for the fieldgenerated
over Kby
the numbersf (r03C9/q)
asf
varies over all functions infj(K) holomorphic
atr03C9/q.
LEMMA 3.3:
For q in 21
and rprime to q
thefield Kq,r
isgenerated
over K
by A1(r03C9/q), A2(r03C9/q), B(r03C9/q),
and it isindependent of
r.PROOF:
Using
theL’Hôpital
device if necessary to evaluate quo-tients,
weimmediately verify
thatKq,r
has thegenerators
indicated.Next let r, r’ be
integers prime
to q, and denoteby s
anyinteger
suchthat r ~ sr’ modulo q. Then if
f (z)
is any ofA1, A2,
B the functionf’(z) = f (sz)
lies inJ(K) by
Lemma3.1,
and solies in
Kq,r,. Consequently Kq,r
is contained inKqr,
and thisby
symmetry implies
the last assertion of the lemma.For q in 9 we denote the fields
Kq,r by Kq,
and we writedq
for thedegree
ofKq.
The next lemma summarises all werequire
on divisionvalues of Abelian functions. It
gives
a rather curiousdecomposition
of the division value a into constituents which need in no sense be
’simpler’
than a. Such a reduction seems to be a necessaryprecaution
before
using
the lower bound of Lemma 4.2.LEMMA 3.4:
For q
in 2 we havedq cq4.
Let r beprime
to q, and supposef
inJ(K)
isholomorphic
atr03C9/q.
Thenfor
some constant Cwhich may
depend
onf
we can writewhere a,, a2 are
algebraic
numbersof Kq with degrees exactly dq
andheights
at mostqcq4.
PROOF : Let K be an
algebraic
number such that K =Q(K).
Thepreceding
lemma showsthat,
as in theproof
of Lemma 1.6 of[4],
wecan find
integers m¡,
m2, n with absolute values at mostqcl
such thatthe number
generates
Kq
=Kq,r
over Q. Since y is the value atrtolq
of thefunction
it follows from Lemma 3.2 that its
degree
does not exceedcq4.
Sincef (rwlq)
lies inKq,
there issimilarly
a non-zerointeger
k of absolute value at mostq c2
such that the numberalso generates
Kq
over Q. Thus(14)
is satisfied withand these numbers have
degrees dq.
The formulae(15), (16) present
them as the values atr03C9/q
of the functionskg(z), f (z) - kg(z) respectively;
hence Lemma 3.2gives
the desired estimate for theirheights.
Finally
we turn to the division values associated with thequasi- periodic
function H normalized as in section 2. In the Cartesian situation described in section1,
these are related to certain values of thecorresponding
Weierstrass zeta functions. Such numbers werefirst studied in detail
by
Baker in his paper[1],
and his methodssuggest
theapproach
we take here. Thefollowing decomposition
issimilar to that of the
preceding
lemma butslightly
more elaborate.for
some p S q, where 03B11, ..., a2p arealgebraic
numbersof Kq
withdegrees exactly dq
andheights
at mostqCq4,
and m ~0,
m,, ..., m2p are rationalintegers
with absolute values at most 2q.PROOF: We consider the function
Plainly
this is an Abelianfunction,
and from Lemma 3.1 it lies in?j(K).
We have for anypositive integer p
in which every term is
holomorphic
atriulq
because q isodd;
since q a 3 this last condition alsoimplies
that we canchoose p s
q such that 2p = 1+ kq
for someinteger
k ~ 0.Hence, writing q = q (H, 03C9)
we obtain
and
(18)
evaluated atr03C9/q yields
The lemma now follows on
applying
theprevious
lemma withf
= Dto
split
up the division valuesD(2’rwlq).
The
decomposition (17)
is somewhat delicate. We cannot, forexample,
absorb theintegers
mi into thealgebraic
numbers a;, as thismight give
rise to numbers ofheight cq5.
In fact it isimplicit
inLemma 3.5 that
H(rwlq) - rn/q
is analgebraic
number ofKq
whoseheight
does not exceedqCq5. Upper
bounds of the formc qK
with someabsolute constant K - 6 are much easier to
obtain,
and these sufhce toprove the result on ~ + 03C0 mentioned in section 1. For the present
proof, however,
we need K16/3.
In any case we shall not use such estimatesexplicitly,
since weprefer
toapply (17) directly
to theauxiliary
function.4. Additional lemmas on
algebraic
numbersWe
require
two results well-known in thetheory
of transcendental numbers.Unfortunately they
areusually
stated with reference to thering Z
of rationalintegers.
For convenience we shall use instead theirgeneralizations
to thering
I ofintegers
of analgebraic
number fieldK. In either case there is no
difficulty
inproving
the desired extensionby
the obviousprocedure
of’taking
the norm’.Throughout
this section thepositive
constants c, cl, ... willdepend only
on K. AlsoIIPII
will denote the maximum of the absolute values of the coefficients of apolynomial
P. The first result is a lower bound for aparticular type
ofalgebraic number,
and is due to Feldman. Itgives
asubstantial,
and for our purposesindispensable, improvement
on the usual norm estimate. We suppose 03B11,..., am are
algebraic
numbers of exact
degrees di, ..., dm
andheights
at mostHl,..., Hm respectively,
and we denoteby
D thedegree
of the fieldQ(a, ..., 03B1m).
LEMMA 4.1: Let
P in Z[xl,...,xm] ]
havedegree
at mostLi
in xi(1
im),
and put 7T =P(03B11,..., 03B1m ).
Thenif
iro 0 we havePROOF: This is
virtually
Lemma 2 of[2].
The extension to I may be stated as follows.
PROOF: As
suggested above,
we consider the norm of P fromK(x¡, ..., xm)
to0(x,,..., xm).
The other result of this section is due to
Gelfond,
apart from theprecise
constants.LEMMA 4.3:
Let e
be acomplex number,
andfor
someinteger
n let tm tn+lg ... be real numberstending
toinfinity
whose successive ratiostk+dtk
tend to 1.If for
each k > n there exists a non-zeropolynomial Pk
in
Z[x] of degree
at most tk withJIPkll:5 et,
such thatthen e
is analgebraic
number.For later convenience we record the version over I in terms of
homogeneous polynomials.
LEMMA 4.4: Let
el, 03BE2
becomplex numbers,
andfor
someinteger
nlet tm tn+b." be real numbers
tending
toinfinity
whose successive ratiostk+lltk
tend to 1.If for
each k > n there exists a non-zerohomogeneous polynomial Pk
inI[x1, X2] of degree
at mosttk,
withcoefficients of
sizes at mostetk,
such thatthen
Ç1, e2
arelinearly dependent
over A.PROOF:
Straightforward.
5. T he
auxiliary
functionLet %
be analgebraically
defined Abelian function field normalizedas in section
2,
and let H be aquasi-periodic
functionsimilarly
normalized.
Suppose
that H does not lie inJ,
and let £0 =(03C91, ú(2)
bea non-zero
period
of the lattice A. We shall prove our theoremby showing
thatif 7q
=7q(H, £o)
isalgebraic,
theperiod w
isspecial.
Thus we assume 7q is an
algebraic number;
we take the field K introduced in section 2 solarge
as to contain 7q. Let e be a small absolutepositive
constant notexceeding 1/100.
For aninteger k >
1 welet
and
If, - L" J.
In the estimates that
follow,
thepositive
constants c, ci, ... will beindependent
ofk ;
inparticular
we take c solarge
that the ac-companying
calculations are valid whenever k > c.LEMMA 5.1 : There is a non-zero
polynomial
P inZ[xo, x], of degree
at most
Lo
in xo and totaldegree
at most L in x, such that~P~ kc, k
and the
function
has a zero
of
order at least k at eachpoint
s03C9 with s anon-negative integer
notexceeding
h.PROOF: This is an exercise in the use of well-known lemmas on
solutions of
homogeneous
linearequations.
We express the deriva- tives of03A6(z)
in terms ofH(z), A(z), B(z), (C(z))-1 by
means ofLemma 2.4. The essential
point
is thatH(sw) = s7q
for anyinteger
s,and so the
resulting equations
for the coefficients of P have coeffi- cients in K. The calculations are similar to thoseappearing
in theproof
of Lemma 5 of[6].
LEMMA 5.2: The
function
is
entire,
andif
a is adifferential
operatorof
order at most k we havefor
allcomplex
numbers zPROOF: We use
Cauchy’s integral
formula and thegrowth
estimatesof Lemma 2.1
(see
also Lemma 6 of[6]).
To describe the
extrapolation procedure
we recall the set 9- definedin section 3. Also let 03BB 1 be a real number such that whenever
) z ) s À
we haveFor real numbers
Q > 1 , S >
1 we denoteby J (Q, S)
the collection of rationals t = s and t = s +r/q
forintegers q
in9, r, s
withElementary
numbertheory
shows thatJ (Q, S)
contains at leastC4 IQ2 S
distinct numbers t(cf.
p. 103 of[6]).
At thecorresponding points
tca the function 03B8B does not vanish andH,
A,B, C-’
areholomorphic.
The
following
lemmagives
the lower bound which was the soleobjective
of the calculations of section 3.LEMMA 5.3:
If t
lies inJ (Q, S)
and a is adifferential
operatorof
order at most k such
that e
=acp(tCd) :j; 0,
thenPROOF: If t is an
integer
theinequality
is immediate since nodivision values are involved and a
straightforward
norm argument suffices. We can therefore suppose that t = s +r/q
forintegers q
in2,
r, s
satisfying (20)
with q, rcoprime.
Nowby
Lemma 2.4 the function~03A6(z)
can beexpressed
as apolynomial
inH, A, B, C-’;
thispolynomial
hasdegree
at mostLo
in H and totaldegree
at mostc6k,
while from Lemma 5.1 its coefficients are in I with sizes at most
kC6/i.
Thus
by putting
z = t03C9 andrewriting H(tw)
in terms ofwe obtain a
polynomial F
such thatwhere n is a denominator for 7q and
The
polynomial F
hasdegree
at mostLo
in the firstvariable,
totaldegree
at mostC7k,
and since s :5 S its coefficients are in I with sizesnot
exceeding (c7QS)Lokc7k.
We now use thedecompositions
ofLemmas 3.4 and 3.5 to write
and
for some p S q and
integers
m 00,
m,, ..., m2p of absolute values at most2Q.
The 03B1i,j arealgebraic
numbers ofKq
withdegrees exactly dq
and
heights
at mostQcsQ4.
Thus(21) yields
apolynomial
G such that(22)
1T =(mnq)L°03BE = G(ao,,, ...,
ao,2p, 03B11,,, ...,a4,2)’
This
polynomial
hasdegree
at mostLo
in each of the first2p variables,
anddegree
at mostc9k
in each of the lasteight.
Further-more its coefficients are in I with sizes not
exceeding
W =(cQS),--ok c,k
The lower bound of Lemma 4.2 was needed for
just
this kind of situation. All the ai,j lie inKq
whosedegree
isdq by
definition. It follows that since ir 0 0and q s Q
we haveBecause dqCii Q4
we deduce from thisand the desired
inequality
is now immediate from(22).
We observethat if neither
Q
nor S exceeds a bounded power ofk,
this lowerbound becomes
k -cnQ4(k+LoQ);
and when furtherQ:5 k/Lo
this issimply
k-C14kQ4,
which is asprecise
as the estimate of Lemma 7 of[6].
We alsosee that the weak results on the division values of H are offset
by
thesmallness of
Lo.
After these
preliminaries
we canput
intooperation
the ex-trapolation procedure
as follows.LEMMA 5.4: For any
non-negative integer
n8,E-’-
100 thefunc-
tion
tP(z)
has a zeroof
order at leastk/2n
at eachpoint
tCd with t in:Y(h n/8, h 1+n/4).
PROOF: The result is true for n = 0
by
Lemma 5.1. As in theproof
of Lemma 8 of
[6],
it will suffice to obtain a contradiction from thevalidity
of thepresent
lemma for somenon-negative integer
m8,E-’-
100 and the existence of acounterexample
to the lemma f or n = m + 1. Thus if we write
then t lies in
J( Qm+ l’ Sm+ 1)
and d is a differentialoperator
of order atmost T. From the lemma f or n = m the function
has a zero of order at least T at each
point
ofJ (Qm, Sm),
and we haveseen that there are N >_
cl4 QmSm
of these. We deduce from the maximum modulusprinciple
and Lemma 5.2 thatand since k &
Lh 2 a simple
calculation shows theright
hand side ofthis to be at most
3-NT. If e
is acounterexample
with the order of aminimal a similar
calculation,
as in[6], gives
the upper bound|03BE|
2-NT.
But this contradicts the lower boundarising
from Lemma 5.3 and the observationQm+1 kILo.
The final lemma of this section summarises the information we
need to extract from the
preceding
lemma. Forbrevity
weput
J =J (k 1-15E, k 1-40E).
LEMMA 5.5: The
function tP(z)
has a zeroof
order at least2Lo +
1at each
point
tld with t in JPROOF:
Immediate;
note there is asignificant
reduction in the range of theinteger
s.6. C onclusion of the
proof
We commence
by eliminating
the function H from 0.LEMMA 6.1: There is a non-zero
polynomial Q
inI[x] of
totaldegree
M sc17LoL
withcoefficients of
sizes at mostkC17kLo,
such thatthe
function 1/I’(z)
=Q(A(z))
has a zero at eachpoint
tld with t in 3-.PROOF: From Lemma 2.3 the function 0 is not
identically
zero, andso Lemma 2.5
supplies
therequired polynomial Q.
Next we
put f/J(z) = IP(zto),
and for anon-negative integer m
wedenote the m-th derivative of
f/J(z) by qi(m)(z).
LEMMA 6.2: For each