C OMPOSITIO M ATHEMATICA
K UNRUI Y U
Linear forms in p-adic logarithms. II
Compositio Mathematica, tome 74, n
o1 (1990), p. 15-113
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Linear forms in p-adic logarithms
IIDedicated to the memory of Professor
Loo-keng
HuaKUNRUI YU*
Institute of Mathematics, Academia Sinica, Beijing 100080, China Received 20 March 1989; accepted 17 August 1989
Compositio
© 1990 Kluwer Academic Publishers. Printed in the Netherlands.
0. Introduction and results
0.1 The present paper is a continuation of the
study
in Yu[20]
and[21],
wherea brief
history
of thetheory
of linear forms inp-adic logarithms
wasgiven,
andprecise
resultssubject
to a Kummer condition wereproved.
In this paper we shall remove the Kummercondition, thereby establishing
thep-adic analogue
ofa celebrated theorem of Baker on linear forms in
logarithms
ofalgebraic
numbers
(i.e.
Theorem 2 of Baker[2])
and thep-adic analogue
of Baker’s well- knownSharpening
II(i.e.
Baker[1]).
Let 03B11,..., an be
n(, 2)
non-zeroalgebraic
numbers and let K be the field ofdegree d generated by
ai , ... , an over the rationals Q. We denoteby
p aprime
number and
by p
anyprime
ideal of thering
ofintegers
in K,lying
above p.We shall establish estimates for
where
b1, ... , bn
are non-zero rationalintegers
andord
denotes the exponentto
which p
divides theprincipal
fractional idealgenerated by
theexpression (assumed non-zero)
inparentheses.
Our result will be in terms of real numbersh1,...,hn satisfying h1
...hn
andwhere
log
03B1j has itsimaginary
part in the interval(- n, n]
andh(03B1)
denotes the*Forschungsinstitut für Mathematik, ETH Zentrum, CH-8092 Zürich, Switzerland.
logarithmic
absoluteheight
of a. This is definedby
where m is the
degree
of a, a is theleading
coefficient of the minimalpolynomial
of a over the rational
integers 7L,
and03B1(1),
... , OE(m) are theconjugates
of oc. Thenas a
simple
consequence of our main result(see
Section0.2),
we havewhere B is the maximum of
the 1 bj (1 j n)
andwith d’ =
max(d, 2)
and h’ =max(h",1).
Whenordp bn
= minordp bj,
h’ can bereplaced by max(hn-1, 1).
This is thep-adic analogue
of Baker’s[2]
Theorem 2.As a second
corollary, analogous
with Baker’s[1] Sharpening II,
we suppose that the above condition onordp bn
is satisfied and h’ is modifiedaccordingly;
then for
any £5
with 0 03B41,
we haveThus we have overcome all the difficulties associated with the work of
[14] -
seethe discussion in our earlier papers
[20], [21] -
and except for the minorreplacement
of pby p2
in the case d =1,
we have established andstrengthened
all the main assertions
(Theorems 1, 3
and4) given
there.In order to overcome the essential
probem
inapplying
the Kummertheory
tothe final descent in the
p-adic
case, we introduce a newingredient
into theanalytic
part of ourproof.
It is anirreducibility
criterion for thepolynomial xrk -
a, where r is aprime
number(see
Lemma1.8),
and it is obtained as a consequence of theVahlen-Capelli
Theorem(see Capelli [6]
and Rédei[15]).
This enables us to construct a new
auxiliary
function(see
theproof
of Lemma2.1),
and both the
extrapolation
and the passage from the Jth step to the(J
+l)th step
in theproof
of the main inductive argumentdepend strongly
on thiscriterion
(see
theproof
of Lemmas2.3,
2.4 and2.5).
The research of this paper was
partly
done at the Max-Planck-Institut für Mathematik Bonn and at the ETH Zürich. The author would like to express hisgratitude
to the Alexander vonHumboldt-Stiftung,
the Max-Planck- Institut für MathematikBonn,
theForschungsinstitut
für Mathematik and theMathematikdepartement
of the ETH Zürich for their support. The author isalso very
grateful
to Professors A.Baker,
R.Tijdeman,
M. Waldschmidt and G. Wüstholz for valuable discussions.Finally,
the author expresses hisgratitude
to his
wife,
DehuaLiu,
for her assistance in many respectsduring
the courseof the work.
0.2 Detailed statements of the main results. Let 03B11, a,, be
n( 2)
non-zeroalgebraic
numbers andLet p be a
prime
number. SetLet K be an
algebraic
number field ofdegree
D over Q such thatDenote
by
aprime
ideal of thering
ofintegers
in K,lying
above p. For03B1~KB{0},
writeord ,
a for the exponentof p
in theprime
factorization of the fractional ideal(a);
defineordfi 0
= oo. Denoteby e
the ramification index of andbye
the residue classdegree of p.
WriteK
for thecompletion
of K withrespect to the
(additive)
valuationord;
and thecompletion
oford
will bedenoted
again by ord.
Now let1
be analgebraic
closureof Qp. Write Cp
forthe
completion of E
with respect to the valuation of03A3,
which is theunique
extension of the
valuation ||p
ofQu.
Denoteby ordp
the additive form of the valuation onCp. According
to Hasse[9],
pp.298-302,
we can embedK
intoCp:
there exists a
Q-isomorphism t/J
from Kinto 03A3
such thatK
isvalue-isomorphic to Qp(03C8(K)),
whence we canidentify K
withQp(03C8(K)). Obviously
Let be the set of
non-negative
rationalintegers
and defineSet
LK:= { l~|el
EK}.
For 1 Eft7K
definewhere
h(a)
denotes thelogarithmic
absoluteheight
of analgebraic
number a(see,
for
example, Lang [10], Chapter IV).
LetVl , ... , Vn
be real numberssatisfying
and
where and in the
sequel log 03B1j = log|03B1j| +
i arg 03B1j with - TC arg 03B1j n(1 j n).
Letb 1, ... , bn
EZ,
not all zero, and let B,B 1, ... , Bn
bepositive
numbers such that
Set
Define
THEOREM 1.
Suppose
thatand
Then we have
where
COROLLARY 1.
Suppose
that(0.13)
and(0.14)
hold. Thenwhere
THEOREM 2.
Suppose
that(0.13), (0.14)
hold andLet
with
Let Z =
03C903A6/ V
withThen for any j
with 1j
n andany ô
with 0 03B4Zf(log p)/D,
we haveWhen 03B11,...,03B1n are n
( 2)
non-zero rationalnumbers,
thehypothesis (0.13)
in Theorems
1,
2 andCorollary
1 may be omitted. Forexample,
Theorem 1 hasthe
following
COROLLARY 2.
Suppose
that(0.14)
holds andLet
A1,
... ,An
be real numbers such thatA1 ··· An
andSet A =
An-1 if ordp bn
=min1jn ord, bj
orlog
03B1n islinearly dependent
on ni,log
a 1, ... ,log
03B1n-1, and set A =An
otherwise. LetThen we have
In the
general
case, thehypothesis (0.13)
can also be removed. Thefollowing
Theorems l’and 2’ are the version in terms of the additive valuation on
Ko
=Q(03B11,..., 03B1n) and
withoutassuming (0.13).
Denoteby po
anyprime
idealof the
ring
ofintegers
inKo, lying
above p. Letord,*o,
eo,fo
be defined withrespect
to the fieldKo.
SetLet
Va,..., V"
be real numberssatisfying V1 ··· Vn
andand let B,
B 1, ... , Bn
and V be definedby (0.10)
and(0.11 ).
THEOREM l’.
Suppose
that(0.14)
holds. Then we havewhere
and
THEOREM 2’.
Suppose
that(0.14)
and(0.15)
hold. Letwith 03C1’ = 1.0752 if p > 2 and 03C1’ =
1.1114if p
= 2. LetThen
for any j
with 1j
n and any 03B4 with 0 03B41 403C903A6/(D0 Vj),
we havewhere co is
given by (0.17).
1. Preliminaries
For the basic facts about
p-adic exponential
andlogarithmic
functions inCp,
we refer to Hasse
[9],
pp.262-274,
or Section 1.1 of Yu[21].
We assume thatthe variable z takes values from
Cp.
Ifordp
z0,
we say that z isintegral.
Thefollowing
concepts of normal series and functions are due to Mahler[13].
A
p-adic
power serieswhere zo E
Cp
isintegral,
is called a normalseries,
ifA
p-adic function,
which is definableby
a normal series in aneighborhood
of anintegral point
inCp,
is called a normal function. For the fundamentalproperties
of normal
functions,
we refer to Mahler[13].
LEMMA 1.1. Let KE7L be
defined by
and set
then
Proof.
This is Lemma 1.2 of Yu[21].
For later
references,
note thatby (1.1)
and(1.2)
we haveand
LEMMA 1.2.
Suppose
that 0 > 0 is a rationalnumber,
q is aprime
number with q =1= p, and M >0,
R > 0 are rationalintegers
withq 1 R. Suppose further
thatF(z)
is a
p-adic
normalfunction
andThen
for
all rationalintegers k,
we haveREMARK. Here
log R
andlog p
denote the usuallogarithms
forpositive
numbers.
Proof.
This is Lemma 1.4 of Yu[21].
Let E be an
algebraic
numberfield, ’
be aprime
ideal of thering
ofintegers
in E,
lying
above theprime
number p. Letordji’,
e’,f/t’
be defined in the sameway as in Section 0.2. For a
polynomial P,
denoteby L(P)
itslength,
i.e. the sumof the absolute values of its coefficients.
LEMMA 1.3.
Suppose
thatP(x1,..., xm)
E7L [x 1 , ... , xm] satisfies
if 03B2m
E E andP(03B21,..., Pm) =1= 0,
thenProof.
This is Lemma 2.1 of Yu[21].
LEMMA 1.4.
Suppose
that rx =1= 0 is analgebraic
number in K and b EZB{0}.
If rxb =1= 1,
thenREMARK. Note that here K may be chosen to be any
algebraic
number fieldcontaining
a.Proof.
Iford03B1 ~ 0,
then it iseasily
seen thatord(03B1b - 1) 0;
and when a isa root of
unity,
wehave, by
Lemma1.3,
Thus we may assume that
ord
a = 0 and a is not a root ofunity.
Let s be the leastpositive integer
such thatThen
By
an argument similar to that in theproof
of Lemma 1.1(see
Yu[20],
p.418)
we see that if
PECp
satisfiesordp(03B2 - 1) 1/e,
thenwhere 03BA ~ Z is defined
by
theinequality pk-1(p - 1) % e p pk(p-1),
whenceOn
applying (1.7)
to as, weget
Note that
03B1sp03BA ~ 1,
since a is not a root ofunity. By
the basicproperties
of thep-adic exponential
andlogarithmic
functions(see,
forexample, Yu[21], §1.1)
and
by Lemma 1.3, (1.6), (1.8),
we obtainOn
noting
theinequality ef D,
the lemma follows at once.LEMMA 1.5. Let
(03B21,...,03B2r~K. Suppose
that(not
allzero) satisfy
Write
and
If
n > mD, then there exist y,, ... , Yn E 7L withsuch that
Proof.
This is Lemma 2.2 of Yu[21].
Define for z e C
and for
l,
m~NFor every
positive integer k,
letv(k)
be the least commonmultiple
ofl, 2, ... ,
k.LEMMA 1.6. For any z~C and any
integers
k1,
11,
m0,
we haveLet q be a
positive integer,
and let x be a rational number such that qx is apositive integer.
Thenand we have
Finally, for
anypositive integers k,
R and L with kR,
thepolynomials (0(z
+ r;k))l(r
=0, 1,...,
R -1;
1 =1,..., L)
arelinearly independent.
Proof. (1.11)
is aslight improvement
of Lemma 2.4 of Waldschmidt[18]
and Lemma 2.3 of Yu
[21],
and will beproved
below.(1.12)
isjust
Lemma Tlof
Tijdeman [17].
For aproof of (1.13),
see theproof of
Lemma 2.3 of Yu[21].
The last assertion of the lemma
is just
Lemma 4of Cijsouw
and Waldschmidt[7].
To prove
(1.11),
we may assume m kl. Thuswhere the summation is over all
selections j1,...,jm
of mintegers
from the set1,..., k repeated
1 times. Now(1.14) implies
thatHence it sufHces to show that
(1.15)
isobviously
true for k =1,
and we mayassume k
2. WriteIt is easy to see that the
polynomial f(x) - g(x)
hasnon-negative
coeffi-cients,
whence so does thepolynomial ( f(x))1 - (g(x))l,
because off 1 - gl
=By
this observation we getThus to prove
(1.15)
it sufHces to show thatFor x
0,
we haveFrom this and the
inequality
k ! >(203C0k)1/2kke-k (see
Yu[21],
Lemma2.7)
weobtain,
for x 0 and k2,
This
is just (1.16),
whence theproof
of(1.11)
iscomplete.
Let
B’, Bn
bepositive numbers, T, L 1, ... , Ln (n 2)
bepositive integers.
SetL’ = max1jnLj.
LEMMA 1.7.
Suppose
thatb1,..., bn,03BB1,...,03BBn,03C41,...,03C4n-1
are rationalintegers satisfying
Then
Proof.
This is Lemma 2.4 of Yu[21],
which is aslight improvement
of aLemma in
Loxton, Mignotte,
van der Poorten and Waldschmidt[12].
For a field E and a
positive integer h,
writeEh
={ah|a~E}.
LEMMA 1.8. Let r be a
prime number,
k apositive integer,
and Ea field.
Whenr =
2,
we supposefurther
that -1E E2. If
a~E andaf/Er,
then thepolynomial
is irreducible in
E[x].
Proof.
This is asimple
consequence of thefollowing
VAHLEN-CAPELLI THEOREM: Over a
field
F apolynomial
is reducible
if,
andonly if,
or
( For
aproof
seeCapelli [6] (when
F is a numberfield)
and Rédei[15],
pp. 675- 679 for thegeneral case.)
LEMMA 1.9. Let 03B11,..., 03B1n be non-zero elements
of
analgebraic number , field
K and let
03B11/p1,...,03B11/pn
denotefixed pth
rootsfor
someprime
p. Further let K’ =K(03B11/p1,.. ,03B11/pn-1).
Then either K’(03B11/pn)
is an extensionof
K’of degree
por we have
for
some y in K and someintegers j1,..., jn-1
with 0j,
p.Proof.
This is a lemma of Baker and Stark[4].
LEMMA 1.10. Let a be a non-zero
algebraic integer of degree
d withconjugates
a 1 = rx,rx2,...,rxd. Set
na
=max1jd|03B1j|. If a
is not a rootof unity,
thenProof.
The lemma holds for d =1,
sincelog03B1 log 2
>1. By
a result ofDobrowolski
[8],
which states thatthe lemma is valid for d
21,
sincelog d log
21 > 3. Thus we may assume that 2 d 20 in thesequel. By Smyth [16],
we see that if a is notreciprocal,
thenwhere
03B80
= 1.324... is the real root ofX3 -
x - 1 = 0. For areciprocal
we seethat the lemma holds for d =
2, 4, ... ,16,
in virtue of a result ofBoyd [5].
Itremains to
verify
that the lemma holds for d =18, 20. Obviously
p = 61 is aprime satisfying
On
replacing
6by
5 in theproof
of Dobrowolski[8],
we conclude thatThis
completes
theproof
of the lemma.LEMMA 1.11. Let K be a
number field of degree D
overQ,
andll, ... , lm linearly dependent (over Q)
elementsof 2 K.
Then there existt1,..., tm~Z,
not all zero, such thatand
where
Yl , ... , Vm
arepositive
numberssatisfying
Proof.
This is aslight improvement
of Lemma 4.1 of Waldschmidt[18]. By
virtue of Lemma
1.10,
we mayreplace Co(D)
= 9D2 in theproof
of Lemma 4.1 in[18] by Co(D)
=2D2,
and the lemma follows at once.LEMMA 1.12. Let
K and fit be defined in
Section 0.2.If p = 2 or p ~ 3 (mod 4),
then f
2.Proof. By (0.3)
we may assumeNow the conclusion of the lemma follows
immediately
from Lemma A in theAppendix,
where we takeKo
= Q.We record two
simple inequalities
for later references. For any real numberu > 0 and
integer m 2,
we haveSecondly,
it is easy toverify
that2. Results
subject
to a new Kummer condition.Let p
be aprime number,
K be analgebraic
number field ofdegree
D oversuch that
Denote
by ft
aprime
ideal of thering
ofintegers
in K,lying
above p. Letord,
e,fi,
be defined as in Section 0.2. Wehave, by
Lemma1.12,
Let q,
u, v, ao be defined as followsSuppose
that 03B11,..., rxn e K(n 2)
andVl , ... , Vn, V*-1
are real numbers such thatLet
b1,...,bn~ Z,
not all zero, B,B’, Bn, Bo,
W, W * bepositive
numbers such thatwhere
In this section we shall prove the
following
Theorems and Corollaries.THEOREM 2.1.
Suppose
thatand
then
ord
where a, c are constants
given by
thefollowing
tablesREMARK. Here
03B11/q0,..., 03B11/qn
are fixedqth
roots inCp.
If(2.15)
holds for achoice of
qth
roots inCp,
then it holds for any choice ofqth
roots inCp,
sinceK contains
qth
roots ofunity by (2.1)
and(2.3).
In theproof
of Theorem2.1,
the choice ofqth
roots will be fixedby (2.23)
and(2.25).
THEOREM 2.2. In Theorem
2.1, (2.14)
may be omitted.COROLLARY 2.3.
Suppose
that(2.15)-(2.18)
hold. Then we haveord
where
COROLLARY 2.4. Let
Z’, Z, £5,
W’ bepositive
numberssatisfying
where
a’,
c’ aregiven
inCorollary
2.3 andSuppose
that(2.15H2.18)
hold. ThenWrite
By
Hasse[9],
p. 220 and(2.3), (2.4),
we see thatLet y
be the order towhich q
dividesG,
and letGo, G1
be theintegers
such thatDenote
by (
a fixed Gthprimitive
root ofunity
inKit
such thatand
by 03BE
a fixedqgth primitive
root ofunity in Cp
such thatBy (2.21), (2.22),
we can fix aqth
root03B11/q0
E C such thatBy (2.16)
and Lemma 1.3 of Yu[21],
there existr’1,...,r’n~Z
such thatordp(03B1j03B6r’j - 1) 1/e(1 j n).
Let rl, ... ,rn~Z
be such thatthen, by
Lemma1.1,
where
03BA, 03B8
are definedby (1.1)
and(1.2). By (2.24)
and(2.3)
we see thatwhere the
logarithmic
andexponential
functions arep-adic functions,
are welldefined. Furthermore it is easy to
verify
that there existqth
rootsce 11q,
...,ailq
ECp
such that2.1. The statement of a
proposition
towards theproof
of Theorem 2.1We define
ho,..., h8,03B51,03B52,~ by
thefollowing formulae,
which will be referredas
(2.26).
In Section
2.1-2.5,
we suppose that Co, Cl, C2, C3, C4 are real numberssatisfying
the
following
conditions(2.27)-(2.29):
Set
PROPOSITION 2.1.
Suppose
that(2.14)-(2.18)
hold. Thenordp 0398
U.2.2. Notations
The
following
formulae will be referred as(2.31).
For later covenience we need the
following inequalities (2.32)-(2.47).
(note that, by (2.19), (2.20), Go
=(pfft - 1)/qu.)
In
(2.43)-(2.45), J, k
are rationalintegers
with 0 J[log Ln/log q],
0 k n, 1 isgiven
in(2.26).
Proof of (2.32).
Similar to theproof of (3.12)
of Yu[21].
Note that we use(1.3),
(1.4)
and the factef
D to showProof of (2.33)-(2.37).
Similar to theproof
of(3.13)-(3.17)
of[21].
Proof of (2.38).
Similar to theproof
of(3.18)
of[21J.
vJe use theinequality
c3 160
(see (2.27))
to showProof of (2.39).
From(2.31)
and the definition ofh4 (see (2.26))
we getThus to prove
(2.39)
it sufHces to showsince
log V*n-1 ho.
Nowby (2.30), (2.31)
we haveBy (2.3), (2.2)
it is easy toverify
thatOn
combining
this and(2.27), (2.10), (2.49),
we obtainNow this
inequality
and thefollowing inequality
which can be verified
similarly
as in theproof of (3.19) of [21], yield (2.48)
at once.Proof of (2.40), (2.42)-(2.44), (2.46), (2.47).
Similar to theproof
of(3.20), (3.22)-(3.24), (3.26), (3.27)
of[21].
Proof of (2.41).
Similar to theproof
of(3.21)
in[21].
Here we need(2.7).
Proof of (2.45).
Similar to theproof
of(3.25)
in[21].
We need to use thedefinition
of 1
in(2.26),
from which it follows thatSo far we have established the
inequalities (2.32)-(2.47).
Now we introducesome more notations. For
(J, 03BB-1,...03BBn, 03C40,...,03C4n-1)~ N2n+3
setwhere
A(z; k)
andA(z; k, l, m)
are definedby (1.9)
and(1.10).
In thesequel,
weabbreviate
(03BB-1,...,03BBn)
asÀ, (03C40,...,03C4n-1)
as T and write1 t
= 03C40 + ··· + i" -1- LetBy (2.14)
we can fix a basis of K over 0 of theshape
2.3. Construction of the rational
integers p(À, do, d)
We recall that r1,..., rn are the rational
integers
in(2.24); G, Go, G1
are definedby (2.19), (2.20);
X isgiven
in(2.31); Do, Dl
aregiven
in(2.51).
LEMMA 2.1. For
and A =
(A -
1, ...,An)
in the rangethere exist
p(03BB, do, d)
E 7L withsuch that
for
all(s,
To, ...,03C4n-1)
E Nn+1 satisfying
where SÂ
ranges over(2.54), Id,,,d
ranges over(2.53).
REMARK. In the
sequel
salways
denotes a rationalinteger
and Talways
denotes a
point (03C40,..., 03C4n-1)~
Nn. Theexpression (s, to,... , 03C4n-1)~
N’ + 1 willbe omitted.
Proof.
For t E7L,
defineLet
By (2.20)
we haveDenote
by b
the set of  =(03BB-
1, ... ,Ân)
~ Nn+2satisfying (2.54).
ThenBy (2.59), (2.58), (2.56),
we see that every 03BB~l determinesuniquely
t =t(03BB1,..., 03BBn)~J
and k =k(03BB1,..., 03BBn)~
7L such thatWrite h =
h(Â 1,
...,Ân, do, s)
for the rationalinteger satisfying
By (2.60), (2.21), (2.6), (2.61)
we obtainFor 03BB, d0, d, s, 03C4 with 03BB~, 0 d0 D0,
1d D1,
1s S and (s,q) = 1,
andItl T,
setBy
Lemmas 1.6 and 1.7 we see that eachP03BB,d0,d;s,03C4
is a monomial in xo, xl, ... , x,, with rationalinteger coefhcient,
the absolute value of which is at mostdegxj PÂ,do,d;s,,r P03BASLj
+ D(1 ~ j ~ n).
Note that
(2.1), (2.3), (2.4)
and(s, q)
= 1imply
thatBy (2.20), (2.21)
we see that(GIs
is a root of xq03BC-u -(03B6qu)s. Thus,
in virtue of Lemma1.8,
it follows that theq03BC-u
elementsare
linearly independent
over K. Oncombining (2.58), (2.59), (2.62)
and the abovefact,
we see that(2.55)
isequivalent
to that for each t~JFor each t E
J,
in(2.63)
there are(1
-Ilq) S(T+nn) equations
and at leastunknowns
p(Â, do, d). By (2.32),
we canapply
Lemma 1.5 to(2.63)
for each t E J(note
thath(03B10)
=0),
and the lemma follows at once.2.4. The main inductive argument
For rational
integers r(J), L(J)j(-1jn)
andp(J)(03BB, do, d)
=p(J)(03BB-1,..., Ân, do, d),
which will be constructed in thefollowing
main inductive argument, setwhere
EÂ
is taken over the set (J) of 03BB =(03BB-1,..., 03BBn) satisfying
Note that
by (2.24),
thep-adic
functionsare normal.
The main inductive argument.
Suppose
that there arealgebraic
numbers03B11, oc, in K and rational
integers bl,..., bn satisfying (2.14)-(2.18),
such thatThen for
every J E 7L with 0 J[log Ln/log q]
+ 1 there exist r(i) E7L, L(J)j~
7L(-1j n)
withand
such that
The main inductive argument will be
proved by
an induction on J. Ontaking
r(0) =
0, LjO)
=Lj(-1 j n), p(0)(03BB, do, d)
=p(03BB, do, d),
which are constructed in Lemma2.1,
we see,by
Lemma2.1,
that the case J = 0 is true. In the rest of Section2.4,
we suppose the main inductive argument is valid for some J with 0 J[log Ln/log q],
and we aregoing
to prove it for J + 1. Wealways keep
the
hypothesis (2.66).
We first prove thefollowing
Lemmas2.2, 2.3, 2.4,
then deduce from Lemma 2.4 the main inductive argument for J + 1.Set
Write
(J)t (t
eZ)
for the set of  =(03BB-1,..., Ân) satisfying
and define
By (2.20),
By (2.65), (2.67), (2.68),
Define
and for t~J(J) define
LEMMA 2.2. For every t E
J(J),
T =(io, ... , 03C4n-1)
with|03C4|
T andy E Q
withy > 0 and
ordp y 0,
we haveProof.
Similar to theproof
of Lemma 3.2 of Yu[21].
LEMMA 2.3. For k =
0,1, ... ,
n we haveProof. By (2.67)-(2.70),
every A E(J)
determinesuniquely
t =t(Â 1, ... , An)
EJ(J)
and k =k(Â 1,
... ,An)t
E 7L such thatLet h =
h(03BB1,..., An’ do, s)
be definedby (2.61).
Thusby (2.75), (2.21), (2.6), (2.61),
we getWe now prove that
(2.74)
isequivalent
to the statement that for every t E J(J) we~Of course, t and k are not necessarily the same as that in (2.60); however we still use these notations, because no confusion will be caused.
have
By
theidentity
(2.77) implies (2.74)
at once.Conversely, by (2.78), (2.76)
and the fact thatare
linearly independent
overK,
which has been established in theproof
ofLemma
2.1,
we see that(2.74) implies (2.77).
ThusIn the
sequel,
let t denote anarbitrarily
fixed element of J(J).By
the maininductive
hypothesis
for J andby (2.79),
we see that(2.77)
with k = 0 is true. Wenow assume
(2.77)
is valid for some k with 0 k n. We shall prove it for k + 1 if k n and include the case k = n for later use. Thus we see,by
Lemma2.2,
thatNote that
by (2.24)
and(2.17),
thep-adic
functionis
normal,
where 0 isgiven by (1.2)
and can be written as 0 =1/m
with1,
mbeing
coprime positive integers,
andp03B8:=03B2l
withp E Cp being
a fixed mth root of p.Further
by (1.14)
and(2.3)
we see thatis a normal
function,
whence so isThus, by (2.72)
we see thatare normal functions. We now
apply
Lemma 1.2 to each function in(2.81), taking
By
an argument similar to theproof of (3.74)
in Yu[21], using
Lemma 2.6 of[21],
we deduce from
(2.80)
thatwhere the second
inequality
follows from(2.42)
and(2.43).
On the other
hand, by (2.82), (2.44), (2.45),
we see thatNow we see from
(2.83), (2.84), (2.29)
that eachFJ,t(z,03C4)
in(2.81)
satisfies the condition(1.5)
withR,
Mgiven by (2.82).
Thusby
Lemma 1.2 and(2.81)
we obtainBy
the secondinequality
in(2.83),
we haveObserve that the
right-hand
side of the aboveinequality is, by (2.29),
not less thanthe extreme
right-hand
side of(2.84).
Henceby
Lemma 2.2(note
thatordp(s/q)
0by (2.3)), by
the above observation andby (2.84), (2.82),
we getOn
combining (2.85)
with(2.86),
andutilizing (2.33)
and(2.37),
we obtainFrom now on we assume 0 k n.
On the other
hand, by
Lemma 1.6 and(2.50), (2.76)
we see that for any fixedte $"’(J)
and for1 s qJ+k+1, (s,q) = 1,|03C4| (1-(1-1/q)(k+1)/
with
QJ,t;s,t(x0,x1,..., Xn) being
inZ[x0,x1,..., xn]
andhaving degree
at mostin Xj
(1 j n).
Note thatby
the main inductivehypothesis
for J and Lemmas1.6, 1.7,
wehave,
for  E(J)t,
0 dD0,1
dD1, 1
sqJ+k+ 1 S, (s, q)
=1, |03C4| (1 - ( 1 - llqxk
+1)/(n
+1))q-JT,
thefollowing
estimates:By
the above estimates andby (2.50), (2.88),
thelength of QJ,t;s,03C4(x0,x1,...,Xn) is
at most
Now we assume that there
exist s, s
withsuch that
and we
proceed
to deduce a contradiction.By
Lemma 1.3(note h(oto)
=0), by
thedefinition
of Xo (see (2.31)),
andby (2.88), (2.34)-(2.36), (2.38)-(2.41),
we see thatthe
assumption ~J,t(s, T)
~ 0implies
thatThis
together
with(2.28) implies
thatOn
noting
we see that
(2.89) yields
contradicting (2.87).
This contradiction proves that for any fixed t EJ(J),
This fact and
(2.78) imply (2.74)
for k +1,
and theproof
of the lemma is thuscomplete.
LEMMA 2.4. We have
Proof. By (2.78),
it suffices to show that for any fixed t EJ(J),
we haveWe recall that
(2.87)
holds for k = n, thatis,
ord,
for
For
x, z, z’
ECp
withordpx
>11(p - 1), ordp z 0, ordp z’ 0,
we have thefollowing identity
forp-adic
functions(See
Hasse[9],
p.273.)
Hence wehave, by (2.24), (2.25),
Recall
(2.75)
and let h* =h*(03BB1,
... ,Ân, do, s)
be the rationalinteger satisfying
By (2.75), (2.23), (2.93)
we have for 03BB~(J)t
Now
by
Lemma1.6, (2.50), (2.92)
and(2.94)
we see that for any fixed t E J(J) and for 1 sqJ+1S, (s, q)
=1, |03C4| q-(J+ l)T,
we havewith
Q1,t;S,t(XO’
xl, ... ,Xn) being
inZ[x0,
xl, ... ,xn]
andhaving degree
at mostin xj
(
1j n). By
the main inductivehypothesis
for J andby
Lemmas1.6,1.7,
we
have,
for03BB~(J)t, 0 d0 D0, 1 d D1, 1 s qJ+1S, (s,q) = 1, |03C4|
q-(J+1) T,
thefollowing
estimateswhere
L(J)
=max1jnL(J)j. By
the above estimates andby (2.50), (2.95),
thelength
ofQ1,t;s,t(Xo,
xl, ...,Xn)
is at mostNow we assume that there exist s, t
satisfying 1 s qJ+1S, (s, q)
=1, Itl q-(J+1)T,
such thatand we
proceed
to deduce a contradiction. In Lemma1.3,
let E =K(03B11/q0, 03B11/q1,..., 03B11/qn), ’
be aprime
ideal of thering of integers
inE, lying
above/te
Thusby (2.15),
andNote that
h(03B11/qj)
=(1/q)h(03B1j)
andh(03B11/q0)
= 0.By
Lemma 1.3 and the definition ofXo (see (2.31)),
andby (2.95), (2.34)-(2.36), (2.38)-(2.41), (2.28)
we see that theassumption (2.96) implies
thatcontradicting (2.91).
This contradiction proves(2.90),
whence the lemma follows.LEMMA 2.5. The main inductive argument is true
for
J + 1.Proof.
We first show thatFrom
(2.20)-(2.22)
we see thatçG1
is a root of thepolynomial
By (2.15)
we have[K’(03B11/q0): K’]
= q. Thisimplies, by
Abel’s Theorem(see,
forinstance,
Rédei[15],
p.674,
Theorem427),
that03B10~K’q,
whencesince
(p, q)
= 1by (2.3).
Thusby
Lemma1.8, (2.1 ), (2.3)
and(2.100),
thepolynomial
in
(2.99)
is irreducible inK’[x],
thatis,
This
together
with(2.98)
and theidentity
yields (2.97).
Write a =
(03C3-1,..., Un)E Nn+2. By
Lemma 2.4 and(2.50)
we havefor
where (J) is the set of Q =
«(J - 1, ... , (J n) satisfying
Every (u 1,
... ,03C3n) satisfying (2.65)
can beuniquely
written aswith
By
the fact that(G1,q)
= 1(see (2.20))
andby (2.65), (2.102),
we see thatwhere r(J+
1)(03BB*1,
... ,Â*)
is theunique
solution of the congruenceFurther, again by (2.20), (2.65), (2.102),
we see that every (J Erc(J)
determines anunique g
=g(03BB*1,..., À:, 03BB1,..., Àn)E
7L such thatwith
From this and
(2.22)
we getNow on
recalling
theidentity
for x, z, z’ E
Cp
withordp x
>1/(p - 1)
and z, z’being integral (see
Hasse[9],
p.
273),
we see,by (2.24), (2.25), (2.105),
thatOn
combining (2.101)-(2.104), (2.107),
we obtainfor
where
03A303BB1,...,03BBn
is taken over the rangeBy (2.97)
and the fact that( p"s, q)
= 1(see (2.3)),
we see that theqn
elementsare
linearly independent
overK(03BEG1).
By
the main inductivehypothesis
for J, there exists an-tuple (03BB*1,..., À:)
with0 03BB*j q(1 j n),
such that the rationalintegers
are not all zero. Fix this
n-tuple (03BB*1,
... ,03BB*n);
takeset
and define (J+1) to be the set of 03BB =
(03BB-1,..., 03BBn) satisfying
Obviously, by
the choice of then-tuple (03BB*1,..., 03BB*n),(J+1) ~ 0. By (2.110), (2.106),
we obtain from
(2.108)
thatBy
an argument similar to that in theproof
of Lemma 3.5 in Yu[21], utilizing
Lemma 2.6 of
[21],
we conclude from(2.111)
thatThis
completes
theproof
of the lemma.Thus we have established the main inductive argument for
J = 0,1,..., [log Ln/log q]
+ 1.2.5.
Completion
of theproof
ofProposition
2.1The
assumption
thatProposition
2.1 isfalse,
thatis,
there existalgebraic
numbers(03B11,..., a" in K and
bl, ... , bn~
Zsatisfying (2.14)-(2.18),
such thatordp 0398 U,
implies
that the main inductive argument holds forJo
=[log Ln/log q]
+1,
whence we can deduce a contradiction(on utilizing
Lemma 2.5 of Yu[21],
Lemma 1.6 and
(2.46), (2.47);
the argument here iscompletely
the same as inSection 3.5 of
[21]), thereby proving
theProposition.
2.6. Proof of Theorem 2.1
Now this can be reduced to
solving
the system ôfinequalities (2.27)-(2.29).
Wesolve it in the
following
casesrespectively:
Case
(1.a).
p =2, n
8.In this case q =
3,f
2(see (2.2)),
co =17,02
=i.
We have thefollowing
estimates:
It is easy to
verify
thatsatisfy
the system ofinequalities (2.27)-(2.29).
Case(l.b). p = 2, 2 n 7.
In this case q =
3, f 2,
co =9, c2
=158.
We have thefollowing
estimates:It is easy to
verify
thatsatisfy
the system ofinequalities (2.27)-(2.29).
Case
(2.a).
p >2, n
8.In this case q =
2, f 1, D/qu 1 2 (see (2.7)),
Co =17,
C2 = 5. We have thefollowing
estimates:It is easy to
verify
thatsatisfy
the system ofinequalities (2.27)-(2.29).
Case(2.b). p > 2, 2 n 5.
In this case q =
2, f 1, D/qu 1 2,
co =9,02
= 7. We have thefollowing
estimates:
It is easy to
verify
thatsatisfy
thesystem
ofinequalities (2.27)-(2.29).
Case
(2.c).
p >2, n
=6,
7.In this case q =
2, f 1, D/qu 1 2,
co =9,
C2 =27 4.
We have thefollowing
estimates: