C OMPOSITIO M ATHEMATICA
G. R. E VEREST
A “Hardy-Littlewood” approach to the S-unit equation
Compositio Mathematica, tome 70, n
o2 (1989), p. 101-118
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101
A "Hardy-Littlewood" approach
tothe S-unit equation
G.R. EVEREST Compositio
© 1989 Kluwer Academic Publishers. Printed in the Netherlands.
School of Mathematics, University of East Anglia, Norwich, Norfolk NR4 7TJ, England
Received 17 March 1988; accepted in revised form 20 October 1988
This paper is concerned with the
quantitative theory
of the S-unitequation.
Suppose
K is analgebraic
numberfield,
finite dimensional overQ,
withring
of
algebraic integers
denoted OK. We define aprime
to be anequivalence
class of non-trivial valuations on K. The infinite
primes
are those which contain archimedean valuations and will be denotedby S~.
It is usual tonormalise the valuations so that the
product
formulaholds;
where the
product
extends over all the valuations.Suppose
S is a finite setof valuations of K which contains the archimedean
valuations,
S ~S~.
Write
US
for the set of all a E K* with the propertyThe set
Us
forms a group undermultiplication
and the structure ofUs
is wellknown;
Here T denotes a
(finite)
torsion group,consisting
of the rootsof unity
inside K, also1 SI
= s + 1. We aregoing
to assumethroughout
the paper that ifv E S extends the valuation
corresponding
to the finite(rational) prime
p then all of the w which extend p are also contained in S.Let co, ... , cn denote non-zero
algebraic
numbers. There is a consider- able amount of interest in the values takenby
theexpression
where
the Xi
are allowed to runthrough
the group of S-units of K. Several authors[4-7, 9, 10]
have considered thequalitative theory,
thatis,
where one sets the
expression (4)
to zero and considers the number ofpossible
solutions. It becomes clear that one must forbid scalermultiples
of(4). Thus,
it is convenient to work inside theprojective
spaceP"(K).
This isthe set of all vectors x =
(xo,..., xn),
xi E K, where two such are identified if one is a non-zeromultiple
of the other.In
[4],
it wasproved
that theequation
has
only
a finite number ofsolutions x
=(xo,
... ,xn )
ePn(K),
whereeach Xi ~ US, provided
there are no proper,vanishing
subsums. Thatis,
for all proper subsets
{i0,
... ,ir}
c{0,
... ,n}.
In the
quantitative theory
a veryimportant
role isplayed by
thefollowing
definition.
Given x
=(xo,..., xn)
EPn(K),
define theprojective height
ofx to be
This is well-defined on
P"(K)
because it isindependent
of scalarmultiples by
theproduct
formula(1).
Observe that wheneach x;
EUs,
one may rewrite(7)
moresimply
asThe aim of the
theory
is to make acomparison
between theexpression
and the
height
functionH(x)
in(8).
Tosimplify
thepresentation
of thispaper suppose that co =
1, ci
EZ, i
=1,
... , n(thus,
theproduct
in(9)
can run over the valuations of
K).
It follows from Evertse’s Theorem 2 in[4] that, given
6 >0,
there is a constantc(ë)
> 0 such that for all x -(xo,..., xn )
Epn(K), xi E US ,
whichsatisfy
condition(6),
we haveIn this paper we will introduce a new
approach,
which could be seen as arefinement of the result
(10).
It contains atechnique
not dissimilar from theHardy-Littlewood
method. Thetechnique employs (10),
also the theorem of Baker on linear forms inlogarithms.
The reader should note that(10) depends
upon Schlickewei’sgeneralisation
of theSubspace
Theorem ofW. Schmidt.
The
techniques
we wish to introduce canperfectly
well be exhibited in thecase n = 2. Given the
projective
nature of theproblem
let x =(1,
xl ,x2 ),
Xi
EUs,
so that(9) becomes,
For the
complex
variable z, definewhere the "click" indicates that those x have been
omitted,
whichgive
rise to a
vanishing
subsum. The result(10)
guarantees thatonly finitely
many x have
N(x)
= 1 sothey
also can be omitted withoutharming
the kind of result we seek. A similar
assumption
willprevail throughout
the paper.
THEOREM:
The function N(z)
isanalytic
in thehalf-plane Re(z)
>2s,
where|S|
= s + 1 ; it admitsmeromorphic
continuation toRe(z)
> 2s - 2 where it isanalytic
apartfrom simple poles
at z = 2s and 2s - 1. Thepole
atz = 2s has residue
where
t/11
is apositive
rational number(a
combinatorial constant,independent of
theci)
and OJK’RK
denote(respectively)
the numberof
rootsof unity
in Kand the
regulator of
K. Thepole
at z = 2s - 1 has residueThis is not the first time that results from
Diphantine Approximation
have beenused to
give
information aboutsuitably
definedcomplex
series. Forexample,
one can
guarantee
convergence of the Hasse-Weil03B6-function
for anelliptic
curve
using
theHasse-Davenport
bound for the number of solutions ofequations
over finite fields.Our ideas are
being presented
in this way for three reasons. The first is that results such as these translate into results aboutcounting
functions viaan
appropriate
Tauberian theorem. One such is theDelange-Ikahara
Tauberian Theorem
(see [8]).
One may deduce thefollowing.
COROLLARY: Let N(q) =
#(x:
condition(6)
holds andN(x) q}.
Then thefollowing asymptotic formula
is valid,where
03C85
is a constant which isindependent
of the Ci.The second is
that,
ingeneral application
of themethod,
the "secondpole"
(in
this case, thepole
at z = 2s -1) usually
contains information of anarithmetic nature. This can be of
quite
amysterious kind,
see theexample
2 in
§3.
The third reasonis;
we believe that the functionN(z) provides
asuitable context for
studying
much harder and moregeneral questions
aboutthe values taken
by N(x).
Forexample,
it is folklore that one does not expectN(x)
to take pure power values veryoften,
up to asuitably
defined notion ofdegeneracy.
Our methodsprovide
a context in whichasymptotic
resultsmight
begiven.
The
proof
of the theoremdepends,
in part, upon acomparison
madebetween
N(z)
and the series definedby
Write
H’(z)
for the sub-series of those x which do notgive
rise to avanishing
sub-sum of
(5).
Note that the sub-series definedby
where x
runs over those x which dogive
rise to avanishing subsum,
isanalytic
in thehalf-plane Re(z)
> s. Thusthey
are, in a veryprecise
sense,negligible.
Theproof
of this will become apparent when we compare the series(16), (17)
with suitableintegrals.
See the remark after theproof
ofCorollary
1.Note also that we could have introduced the refinement of
letting
x1, x2run over fixed
subgroups
ofUs. Indeed,
this is much closer to theapproach
in
[3].
This too would have created notational difficulties but it does have anamusing
effect upon the residue at z = 2s - 1. Thefactor |c1c2|v
becomesweighted according
to the relative ranks of thesubgroups
ofUs.
The
layout
of the paper is as follows. In§ 1
a sequence of Lemmas isprovided
which expressgeometric
results in terms ofcomplex
functions. In§2,
someasymptotic
formulae areobtained,
based upon thegeometric
considerations of
§ 1
and the results from thetheory
ofDiophantine Approximation,
to which reference hasalready
been made. In§3,
a shortdiscussion is
presented
of how modifications to thisgeneral technique
canbe
applied
to some other classes ofproblems
whichgive
rise to an S-unitequation.
The letters01, 03B82, ...
will be used to denote a sequence ofpositive
constants.
§1. Suppose Lv11,
...,Lls, Lv12, ..., LZS
are linear forms on R8 withLJ" , j = 1,
... , slinearly independent,
for fixed i. DefineLî +1 by
Extend
Lvji
to V = !Rs 0 [Rsby setting,
Define:
Mj(y)
= maxwhere the
integral
is taken over V= !R2s - B, B denoting
an open ball about theorigin.
To
justify
the introduction of thesedefinitions,
first observe that thefunction
H,
in(8),
is invariant under the action of T in thefollowing
sense;Thus,
in thesequel,
we will work modulo T. Nowwrite V.
for theinteger
points
in V. Choose bases for the S-units xl , x2 . Then thelogarithms
of theabsolute values of the units are linear forms in the
exponents
of the units.Identify
these forms with theLvji
asabove, (so
thatequation (18)
becomesa restatement of the
product
formula(1)).
Thenlog H(x)
becomes identifiedwith
M(x), x
EVz.
We will tend to abuse this fact and switch between the two; aliberty granted by
the decision to work modulo T.LEMMA 1:
The function I(z)
isanalytic for Re(z)
> 2s with ameromorphic
continuation to the whole
plane.
Here it isanalytic apart from simple poles
atz =
1, 2,
..., 2s. The residueoJ
thepole
at z = 2s ist/11/Ri.
Proof.-
This istrivial,
theintegral
can be evaluatedexplicitly.
The resi-due arises in the evaluation of the
integral,
as the Jacobian of asimple
transformation.
Note that the
dependence
upon the choice of B is of a very trivial kind. If B is altered then the resultantchange
inI(z)
is to add to it an entire function.COROLLARY 1: The
function N(z)
isanalytic
in thehalf-plane Re(z)
> 2s.Proof.- By
the result(10)
it is sufficient to prove that the functionH(z)
isanalytic
forRe(z)
> 2s. But this follows at once from Lemma 1by
theintegral
test. Infact, H(z) (and
thereforeN(z))
forms anabsolutely
con- vergent series which isuniformly
convergent on compact subsets of thehalf-plane Re(z)
> 2s. Theanalyticity
follows from the usualtheory
aboutsuch series.
Note that the remark in the introduction
concerning
thehalf-plane
ofconvergence for the series
(17)
iseasily
verified. Avanishing
subsum canoccur
only
in a"diagonal"
way, thatis,
when x, =± X2.
Thus theintegral
test
gives
acomparison
between(17)
and anintegral
over RsLEMMA 2:
The function w-2KH(Z) - lez)
isanalytic for Re(z)
> 2s - 1.Proof.
Write Cx for the unit cube with center x. ThenNow
apply
the mean-value theorem to each of theintegrals. Also,
observethe
following quasi-linearity
property of M:This property
(22)
follows because the coefficients of theforms,
also thelength
ofbx
are alluniformly
boundedquantities. Thus,
with our earlieridentifications,
Expand
the inner bracketby
the binomial theorem. Theleading
termscancel and the lower order terms guarantee,
together
with the known convergenceproperties
ofH(z),
thatWK2 H(z) - I(z)
isanalytic
forRe(z)
> 2s - 1.COROLLARY 2:
H(z)
has ameromorphic
continuation toRe(z)
> 2s - 1 withonly a simple pole
at z =2s,
residue03C81(wK/RK)2.
Next we will show that there is a massive
non-uniformity
in the distribution of the units. Asusual, write x
=(1,
xl ,x2 )
for a(projective)
vector, whereXI, x2 E
US .
For v E S, writeAlso,
letH*v(x) (respectively H**v(x))
denote the second(third) largest
member of the set in
(23). Repetitions
are not excluded so wemight
haveHv(x) = H*(x)
etc. For each v E S suppose constantsAv
>0, Bv
> 1 aregiven.
DefineUo = Uo (Av , Bv) by
LEMMA 3:
H(z) - Ho(z) is analytic
in thehalf-plane Re(z)
> 2.s - 1.The reader should notice the coincidence between the
expression
in(24)
andthe form of
(38)
in the statement of Theorem A.Here, and in the
sequel,
we will use thefollowing terminology. If f(03C3)
andg(Q)
are two function of the real variable 6 which takepositive,
real values then we will say that gmajorises f if f(03C3)
=0 (g(u»,
with the usualbig
0notation.
Proof of
Lemma 3:Apply
theintegral
test to the sum which definesH(z) - Ho(z).
Use the notation of Lemma 1 and write a = Re(z).
DefineM*
andMj**
tosatisfy
; for the set
If one of the
inequalities
in(24)
is violated then(in "log space"),
Thus the
integral
ismajorised by
a finite sum ofintegrals
oversubregions
ofV defined
by inequalities (26).
There are three cases to consider:For
(b),
writeNow make the
change
ofvariables;
The
region
becomeslarger
if wereplace Mj by 2Nj.
The(linear) change
ofvariables maps the open ball about the
origin
into another open ball about theorigin.
We canenlarge
the new open ball at the cost ofadding
an entirefunction. Similar remarks hold in case
(c).
Thus we compare with theintegral
of the function(03A3Nj)-03C3
over theregion
(The
lowerinequalities
guarantee that an open ball has been removed about theorigin).
109 To facilitate the
integration
write N* =N*j
+03A3i~jNj.
Now theregion
becomes
larger
when wewrite,
for some03B85,
To see
this, replace successively N by N*j
+04 (log N )
on theright
hand sideof
(27).
The result converges and we may take05 = 203B84.
Perform the
N integral
first togive,
Expand
the first bracket and cancel theleading
terms to obtain0, (log N*)N* -6
+ lower order terms.Replace log
N*by 03B86N*03B5,
8 > 0. Theremaining integrations
arestraight-
forward and
yield
a functionanalytic
in thehalf-plane 03C3
> 2s - 1 + s.a
COROLLARY 3:
H0(z)
admitsmeromorphic
continuation toRe(z)
> 2s - 1.In this
haif-plane,
theonly singularity
is asimple pole
at z = 2s, with residueLEMMA 4: The
function H(x)
hasmeromorphic
continuation toRe(z)
> 2s - 2 where it isanalytic
apartfrom simple poles
at z =2s,
2s - 1.
Proof.
The idea is to reconsider theexpression
As
before,
we abuse notation andidentify x
with xe n
andconsequently log H(x)
withM(x). (It
will be convenient to alternate between thetwo).
Change
the variables in theintegral
in(28);
write c for the unit cubewith centre
origin.
ThenFor x e Uo,
use the mean-value theorem togive,
where the
a(x)
areuniformly
boundedquantities. By
Lemma3,
this series converges back toRe(z)
> 2s - 2.For the sum
03A3x~U0,
use the fact thatfor x
EUo
withlog H(x) sufficiently large,
and for all w E c,(and not just M(x)
+O(1)
as in(23)).
In otherwords,
the same linear form isbeing
used to definey(x
+w), y(w)
andM.(x).
Thus
Expand
the inner bracket and cancel theleading
terms to obtain+
something analytic
forRe(z)
> 2s - 2.By
Lemma3, Ho (z
+1)
hasmeromorphic
continuation toRe(z)
> 2s - swith
only
asimple pole
at z = 2s - 1. ~COROLLARY 4: The residue
of
thepole of H(z)
at z = 2s - 1 lookslike,
Proof:
This arisesby putting together
the formulae(29)
and(31 ).
The firstterm appears in
Corollary
2. The second comes fromCorollary
2 and theevaluation of the
integral
inside thecurly
bracket in formula(31).
This
non-uniformity
of distributionpersists. Given,
asbefore, Av
> 0,Bv
> 1 for v E S, writeÛo
for the set ofall x
suchthat,
This
gives
rise to a seriesLEMMA 5:
H(z) - Ho(z) is analytic for Re(z)
> 2s - 2.Pro of.
Use the notation of theproof
of Lemma 3. Write a =Re(z).
If oneof the
inequalities
in(33)
is violated then in"log space"
we obtain aninequality,
forsome j (which might
as well be1),
As in the proof of Lemma 3, replace M* - M** by N*, M, - M** by Nl .
Then the
problem
comes down to thestudy
ofover the
region:
(The
lowerinequalities
gurantee the removal of an open ball about theorigin).
Suppose,
without loss ofgenerality,
thatN2 ... Ns.
ThenThen the
integral
ismajorised by
over the
region
Perform the
N, integral
togive
We may
expand
both brackets(the
second becauseN*1 03B813 log N2 ),
andcancel the
leading
terms togive
lower order terms.
Perform the
N*1 integral
toyield
Observe that
|f(03C3, N2)1 03B814N-03C3+03B52,
e > 0. Thus theintegral
ismajorised by
over the
region 01, N*2,
....N* N3,
....NS.
Perform theN*2,..., N* , N3,
... ,NS integrals
toyield,
Clearly
this is well-defined for a > 2s - 2 + e. Thiscompletes
theproof.
a
COROLLARY 5:
Ho(z)
has ameromorphic
continuation to thehalf-plane Re(z)
> 2s - 2 where it isanalytic apart from simple poles
at z = 2s and2s - 1. The residues are
(respectively)
as stated in Corollaries 2 and 4. ~§2.
THEOREM A:Suppose
K is anumber field
and S isa finite
setof valuations of
K. LetUs
denote the groupof
S-un i tsof
K and suppose 03B11, ..., aS arenon-zero constants
from
K. Thenfor
each v E S there arepositive
constantsA’v, B’v
with the propertyprovided x ~
av, whereH(x)
denotes theprojective height of (7).
The
proof of this
theorem can be puttogether
from the results in[1] and [11 ];
[1]
for the archimedean v E S and[11] for
the non-archimedean v E S. The statement of(38)
isreadily
transformed into one about linear forms inlogarithms.
The left hand sideof (38)
is greater than the sameexpression
butwith
log |x|v (respectively long |03B1v|v) replacing
x(03B1v). Also,
observe thatlog H(x)
is commensurate with theabsolutely largest
exponent of the units inUs.
Note that for each v E S we have six
possibilities
fororderings, Hv(x) H*v(x) Hv**(x).
As we run over v ES,
label allpossibilities
forall
orderings by
i E I.LEMMA 6: Recall the definitions of
Uo
andÛo
from(24)
and(33).
(no vanishing subsum).
Proof.- (i)
Given i E I,where 0;
canrepresent
one of0, log |c1|v
orlog |c2|v, depending
upon which of1, |x1|v
or|x2|v
islargest.
Now(i)
follows from the definition ofUo
andthe fact that
Bv
> 1.(ii)
Given v E S supposethat |c1x1|v |c2x2|v 1,
the other casesbeing entirely
similar. Now we may suppose thatThen
where the last
inequality
follows from Theorem A. Now chooseAv = A’v/2,
Bv = B’v
in(39) (we
may suppose thatB’v
in the statement of Theorem A satisfiesBv
>1).
ThenNow
taking logs
andsumming
over v E Sgives
the formula in(ii). (Note
that the constants
Av, Bv
in the definition(24)
are chosenafter
theapplication
of Baker’s Theorem in the
proof
of Lemma 6.Thus,
the choice of00 is
determined
by
theproof
of thatLemma).
(iii)
This is a directapplication
of the result of Evertse in[4] (see (10)).
a Now we come to the
proof
of the main result.Begin by breaking
up the sumwhich defines
N(z), according
to Lemma 6. Thatis,
Note that
S, requires
further refinementaccording
to Lemma6(i).
Thisdepends
upon thepossible orderings of |x1|v, |x2|v, 1, v ~
S. In factthey
occur with
equal frequency
and we will assume this into the constantOi.
We will switch
freely
between the seriesH(z)
andH’(z) (see (17)) having
observed that
they
differby
a function which isanalytic
in thehalf-plane
Re(z)
> s.Write a =
Re(z).
ForS3
use part(iii)
of Lemma 6 whichgives (log N(x))-1
=O((log H(x))-1).
ThenBy Lemma
5 This converges for 03C3 > 2s -2;
thusS3
defines ananalytic
function in this
half-plane.
For
Sl and S2 expand
the bracketsThen
By
an earlierremark,
we canhappily
remove the "clicks" from each of thesummations;
theonly
costbeing
the addition of a function which isanalytic
in the
half-plane Re(z)
> s.The third term is
analytic
forRe(z)
> 2s - 2by comparison
withH(z
+2).
In the fifth termreplace log log H(x) by (log H(x))E,
e > 0.Then Lemmas 3 and 5
apply
to show that this series isanalytic
forRe(z)
> 2s - 2 + e.Put the first and the fourth terms
together
togive
By Corollary
5 this series has ameromorphic
continuation toRe(z)
> 2s - 2 withsimple poles
at z =2s, 2s -
1. For the second termapply Corollary
3 to obtain themeromorphic
continuation toRe(z)
> 2s -2,
withonly
asimple pole
at z = 2s - 1.To obtain the formulae
(13)
and(14) apply
the Corollaries2, 3,
4 and5, regarding
also theproof
of part(i)
of Lemma 6 whichgives
the form of the constant03A3i~I Oi
as amultiple
oflog (IIvE s |c1c2|v). ~
§3.
Thistechnique
iscapable
of greatapplication.
What follows areexamples
where modifications of thetechnique
willgive
the result stated.1.
Suppose K
is a finite extension of Q withring
ofalgebraic integers 0K
LetT : K ~ Q denote the trace map. One studies
If
0*K
has torsion free rankr(
> 1say)
thenT(z)
is convergent forRe(z)
> rwith a
meromorphic
continuation toRe(z)
> r - 2. There aresimple poles
at z = r, r - 1 with residues:
where
the 03B8i
arepositive
rationals(combinatorial constants)
andwx, Rx
denote
respectively
the number of roots ofunity
and theregulator
of K.More
generally,
one maystudy
the solutions of adegenerate
norm formequation
in this way beexamining
the non-zero traces of the solutions(see [2]
fordetails).
2. Given n a
positive,
square freeinteger,
write L =Q(03B6n)
for the nthcyclotomic
field.Suppose
Q ~ K ~ L with r =Gal(KI Q).
It is knownthat the
conjugates
of a =TL/K(’n)
under r form Z-basis for thering
ofintegers
of K. The set of all suchgenerators
isHere,
one studiesIf rr denotes the torsion free rank of 7Lr* then
N(z)
converges forRe(z)
> r0393 and admits ameromorphic
continuation toRe(z)
> rr - 2with
simple poles
at z = rr, rr - 1. Thepole
at z = rrdepends
upon ronly,
while that at z =rr - 1
looks likewhere
rl , r2
are(messy) group-ring
constants. Here b is apositive integer which- depends
in a rather subtle way upon K. See[3]
for more details.3. Given
K|Q
and S one can alsostudy N(x) = 03A0v~S|1
+xlv
for xE US .
Here there are several
advantages. Firstly,
there is no seriousnon-vanishing
sub-sum condition.
Secondly
there is no need to invoke theSubspace
Theorem so the methods are
entirely
effective.4. One of the motivations for the
study
of S-unitequations
is thetheory
oflinear recurrence sequences
(see [9]).
In thegeneral
case theseprovide
afertile
testing ground
for thetechniques presented
here. Not least because thecoefficients ci
in(4)
are nolonger
constants. This will be thesubject
ofanother paper.
Acknowledgements
This paper was
prepared
while the author was onstudy
leave atMacquarie University, supported
in partby
an SERC grant. He would like to expresshearty
thanks for thehospitality
he receivedduring
his stay.Also,
he wouldlike to thank Professor Alf van der Poorten for
helpful
comments, made onan earlier version of this
manuscript.
References
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