Algebraic independence
overpar PETER BUNDSCHUH et KUMIKO NISHIOKA
RÉSUMÉ. Soit
f(x)
une série entière03A3n>1
où(e(n))
est une suite récurrente linéaire d’entiers
naturels, strictement
croissante, et
(03B6(n))
une suite de racines de l’unité dansQp,
quisatisfait à une
hypothèse technique
convenable. Alors nous nous sommesparticulièrement
intéressés à caractériserl’indépendance algébrique
surQp
des élémentsf(03B11),..., f(03B1t)
deCp
en fonctiondes 03B11, ... , at ~
Qp,
deux à deuxdistincts,
avec 0|03B103C4|p
1pour 03C4 = 1, ... , t. Une
application remarquable
de notre résultatprincipal
dit que, dans le case(n)
= n,l’ensemble {f(03B1)|03B1
~Qp,
0|03B1|p 1}
estalgébriquement indépendant
surQp,
si(03B6(n))
satisfait à"l’hypothèse technique".
Nous terminerons parune conjecture portant sur des suites
(e(n)) plus générales.
ABSTRACT. Let
f(x)
be a power series03A3n~1 03B6(n)xe(n),
where(e(n))
is astrictly increasing
linear recurrence sequence ofnon-
negative integers, and(03B6(n))
a sequence of roots ofunity
inQp
satisfying
an appropriate technical condition. Then we aremainly
interested in
characterizing
thealgebraic independence
overQp
ofthe elements
f(03B11), ... , f(03B1t)
fromCp
in terms of the distinct03B11,..., 03B1t ~
Qp satisfying
0|03B103C4|p
1 for 03C4 =1, ... , t. A strik-ing
application
of our basic result saysthat,
in the casee(n)
= n,the
set {f(03B1)| 03B1
~Qp,
0|03B1|p 1}
isalgebraically independent
over
Qp
if(03B6(n))
satisfies the "technical condition". We close witha conjecture concerning more
general
sequences(e(n)).
1. Introduction and results
Let p be a fixed
prime, Qp
be thep-adic completion
ofQ,
letQp
be thealgebraic
closure ofQp,
andCp
be thep-adic completion
ofQp,
which isan
algebraically
closedcomplete
field with a valuationuniquely
extendedfrom
Qp.
Manuscrit requ le 21 mars 2003.
This work was done during the second-named author’s stay at the University of Cologne supported by the Alexander von Humboldt Foundation. Both authors are very grateful to the foundation for giving them the opportunity for collaboration.
The
question
of transcendence oralgebraic independence
of elementsfrom
Qp
or even fromCp over Q
is rather wellinvestigated
in the litera- ture. In contrast to thissituation,
thecorresponding question
forCp
overQp
has been studied in thepast only occasionally. Seemingly,
the first suf- ficient criterion(of Liouville-type)
for transcendence was statedimplicitly by
Amice[1, p.74~
as an exercise.Concerning algebraic independence,
thereis a first article
by Lampert [8],
who usedp-adic
series of the formwith infinite sequences
(rk)
ofpositive non-integral
rational numbers toanswer two
questions
of Koblitz[6, p.75]
about the transcendencedegrees
of
Cp
overc;nr
and ofc;nr
overQp.
Herec;nr
denotes thep-adic
closureof in
Cp,
whereQ;nr
is the maximal unramified extension field ofQp.
A few years
later,
Nishioka[9],
based on anapproximation type
criterion foralgebraic independence,
gave moreexplicit examples
foralgebraically independent
elements fromCp
overCun, (and
fromCpn,
overQp
aswell)
than
Lampert
did.Whereas the ak in
Lampert’s
series(1)
were certain roots ofunity
inCp,
the first-named author and Chirskii
[2], [3]
veryrecently proved
avariety
ofresults
giving
sufficient conditions for thealgebraic independence
overQp
of numbers from
Cp, again
definedby
infinite series oftype (1),
but nowwith coefficients from the
ring Zp
:=Ix
E1}
ofp-adic integers.
In the
present
paper, we will bemainly
interested in thealgebraic
inde-pendence
off (al), ... , f (at)
overQp,
where al, ... , at E~p
:=Qp B 101
and
f (x)
is a power serieswith a
strictly increasing
sequence(e(n))n=1,2,...
ofnon-negative integers.
Our motivation to consider series of
type (2) originates
from Gouv6a’s book[4, p.165],
where the transcendence of overQp
isproved
underassumptions
which are muchstronger
than our’s will be in thesequel.
Before
writing
down theprecise
statement of ourprincipal result,
it isuseful to
give
theDefinition. A finite subset
{al, ... , atl
ofcCp
:=Cp B 101
is called(e(n))-
dependent
if there exist a E(CPX
I roots ofunity (i,..., (t,
and numberscCp
such that thefollowing
conditions hold:(i)
aT =Ta
forr= 1,...,
t ..
for any
large
n.We are now in a
position
to formulate our main result.Theorem 1. Let
(e( n) )n=I,2,...
be astrictly increasing
sequenceof
non-negative integers forming
a linear recurrence. Assume(~(n))~-1,2 ".
to bea sequence
of
rootsof unity,
whose orders are allprime
to p,satisfying
~(n) ~ ~p(~(1), ... , ((n - 1)) for
anylarge
n E N‘ :=11, 2,...}.
Putand let al, ... , at E
Q~
be distinct withlarlp
1for
T =1, ... ,
t. Thenf (al), ... , f (at)
arealgebraically dependent I if
andonly if
there exists anon-empty
subsetof { aI, ... , at}
which is(e(n))-dependent.
Remark 1. An
example
of a(-sequence
as in Theorem 1 is thefollowing.
Let ~ be a
prime #
p, and let((n)
be aprimitive
f’1-th root ofunity
forany
Since, by
a Vandermondeargument,
nonon-empty
finite subset ofC§
can be
(n)-dependent,
the casee(n)
= nyields
our firstapplication.
Corollary
1. Let(((n))
be as in Theorem1,
anddefine
Then the set E
Q;, lalp 11
isalgebraically independent.
Corollary
2. Let(((n))
be as in Theorem1,
anddefine
for fixed dEN B f 11. Suppose
a 1, ... , at E~p
to be distinct withI a, lp
1for
T= 1,
... , t. Thenfd(al),
... ,fd(at)
arealgebraically dependent if
andonly if
there exists anon-empty (dn)-dependent
subsetof f a,,
... ,at I -
With
regard
toCorollary
2 we can even go onestep
furtherby allowing
dto vary over
N B f 11.
Thecorresponding result,
which we state as Theorem2,
is not a direct consequence of Theorem1,
but itsproof
is rather similar.Theorem 2. Assume
(((n))
to be as in Theorem1,
andfd(x)
as inCorollary
2.Suppose
al, ... , at EQpx
to be distinct withI a, lp
1for
T
=1, ... ,
t. Then the EN B f 11,
T =l, ... , t}
isalgebraically dependent if
andonly if
there exist ado
E~1, ~1~
and anon-empty
subsetof
... ,at I
which is(don) - dependent.
An immediate consequence of Theorem 2 is
1 In our paper "algebraically dependent or independent" means always "over Qp" .
Corollary
3.Suppose
that no(1
T T’t)
is a rootof unity.
Then the E
I~, ~1}, T
=l, ..., t}
isalgebraically independent.
To conclude this introduction we propose the
following
Conjecture.
Let be astrictly increasing
sequenceof non-negative integers,
and let(C(n))nEN
be as in Theorem 1.Suppose f
to bedefined by (3),
and let al, ... , at EQp’
be distinct withlaTlp
1for
T =1,
... , t.Then
f (al),
... , arealgebraically dependent if
andonly if
there existsa
non-empty (e(n))-dependent
subsetof {aI, ... , atl.
2. Theorem 1: The
if-part
andpreparation
of theonly-if-part
Let be an
(e(n))-dependent
subset of{al, ... , at}
CQpx.
By definition,
there exist a EC;,
roots ofunity (I,
... ,~t~,
and numbersðl,
EC;
such thatFrom
(i)
we see a EQp ,
and from(ii)
we may suppose61
... , EQp,
not all zero, compare the
argument
in[11, p.83].
To prove now the
if-part,
let a,, ... , atsatisfy
thehypotheses
of our The-orem
1,
and let{c~i,..., at, )
be an(e(n))-dependent
subset offal7 ....
Then, using (i)
and(ii),
we findwith
61 , ... , 6t,
EQp,
not all zero. This lastequation
shows that1, I(al), ...
f (at,)
arelinearly dependent
overQp,
and thealgebraic dependence
off (al), ... , f( at’ ) (over Qp)
follows.Clearly,
thisreasoning
isindependent
of the fact that
(e(n~~
is a recurrence sequence.To
begin
with theonly-if-part,
we supposef (al), ... , f (at)
to bealge- braically dependent whereas, w.l.o.g.,
every t-1 of these numbers arealge- braically independent.
Then there exists apolynomial
F E7G~,~xl, ... , xt] B 101
with minimal totaldegree
such thatClearly,
with we haveAssuming
and
defining
we
have,
withFrom
(7)
we see~r~!p ~
1 andfor T = and
any k
E N.Using
this and(4)
we deduce from(8)
With
Kk
:=~p(~(1), ... , ~(k))
for everylarge k e N,
letQ(k)
Eid. As we shall
explain
in Remark 2below,
we canextend
a( k)
to an isometricautomorphism
ofCp
overQp,
and therefore wefind from
(9)
Subtracting (9)
from(10)
weget
Since = the last line can be
equivalently
written as
We
put
this maximum exists because of the discreteness
of 1.lp
on If~(1~) _ 0,
then- - _ -/-*,_,, B
Assume
M(k)
>0,
and let l~~ > k be such that.L i ~_i~
Then,
for any n >1~’,
we have. - 1-
Replacing
in(11) k by k’,
andtaking I((k’)u(k’) - ((k’)lp
= 1 intoaccount,
we
get
Combining
this with(13)
wefind, by latip 1,
with 0 ~ := 1.
By (14)
wehave,
inparticular,
/-~B
~
1 ~ -- - !1
Clearly, (14)
and(15)
hold both in theanticipated
caseM(k)
= 0 as well.Remark 2.
Writing
Q for the abovea( k),
we let Q be anautomorphism
of
Qp
overQp.
For each x EQp,
we definelxl*
:=Ixulp. Then 1 . 1*
is anabsolute value on
Qp coinciding
onQp with [ - ip.
andthus
lx,71P
=Ixlp
for any x EQp.
In the above deduction of
(10)
from(9),
we first extend a Eto Q E Then a is
isometric,
and it can be further extended toan isometric
automorphism
ofCp
overQp.
To finish the
proof
of theonly-if-part
of our Theorem 1 we need thefollowing
lemmaconcerning
thequotients aT/at (T
= s +1, ... , t)
fromQp of p-adic
value1,
compare(6).
Lemma 1.
Denoting
q := pfor
p >2, and q
:= 4for
p = 2 there exist distinct rl, ..., qm E 1 + and rootsof unity (,+l, (t
suchthat the
representations
hold with
appropriate p(T)
Ef 1,
... ,mi.
The
proof
followsimmediately
from[4, Corollary 4.3.8]. Clearly,
noquotient 7~,~~y~,~ (,a 54 ~,~~
can be a root ofunity.
3. A
pedagogical example
Before
concluding
theproof
of Theorem 1 in section 4 in thegeneral
situation of linear recurrence sequences
(e(n)),
we treat, forpedagogical
reasons, first the
particular
casee(.)
=g(.) with g
E7G(x~.
To do this weneed
Lemma 2.
Suppose
G EZ[x]
such that(G(V))VEN
is astrictly increasing
sequence
of non-negative integers.
Let ... , be as in Lemma1,
andassume
81, ... , 6m e Cp,
not all zero. ThenProof. W.l.o.g.
we may assume 1for {L
=1,..., m.
Then theSkolem-Lech-Mahler theorem
(for relatively simple proofs
compare, e.g.,[5],
or[10,
Theorem2.5.3]) implies
that there are at mostfinitely
many with2:;=1
=0,
since noquotient (p # ~~,~)
is a root ofunity. Thus, by
ourhypothesis
onG,
there exists vo E N such thatm
and this sum has
p-adic
valuep-v
with rational v > 0.Defining
M :=[vl
ENo
we see~ - 1 E
andthus,
forany ~
E Nwith
appropriate
EN, p~,(~~
EZp.
From this we seefor
any t
EN,
and thereforeThis proves Lemma 2.
To prove the
only-if-part
of Theorem 1 in theparticular
case e = g, g Ewe
write2
n =(p - 1)v
+ r with r E~0, ... , ~ - 21,
thusgetting
g(n) - g(r) (mod(p - 1)). Using
this and Lemma 1 we deduce... r - ,
Defining
for any fixed r E{0,...,
p -2}
the newinteger-valued polynomial Gr by Gr(v)
:=g((p - 1)v
+r),
it isclear,
that everyGr
satisfies the conditions on G in Lemma 2. On the otherhand,
it follows from(12), (15)
and
(16)
that for every r E~0, ... , p - 2}
t ~ ~ ’B
where we defined
2 Obviously, we leave the case p = 2 to the reader.
Using
Lemma 2 we see for any r E~0, ... , p - 2}
This leads to
t’°l’ l-r"’
for any rt E N.
Fix now ~co E
fl,..., m}
such that the set of T EIs
+1, tj
with= po is not
empty.
Then we find for those T withp(T) =
¡toThis,
combined withequation (19) for p
= shows thatpo)
is a
(g(n))-dependent
subset off a 1, - .. ,
Here we recall the fact thatDrF(f(a)) =I
0(T
= s +1, ... , t),
see(5).
4. The
general
caseTo finish the
proof
of theonly-if-part
of Theorem 1 in thegeneral
case oflinear recurrence sequences we note the existence of
such that
e(n
+h)
=e(n) (mod (p - 1)) (or (mod 2)
if p =2)
holds forany n >
no.Equivalently written,
withe(n)
=e(hv
+r)
=:Er(v)
for anyv E
No,
r E{~o?...
no+ h - 1},
this meansEr (v) * ET(0) (mod (p - 1)) (or (mod 2) if p = 2).
Then we canproceed
as in section 3: Combination of(12)
and(14) yields, by reasoning parallel
to(16),
where
J,(r)
is definedby (17)
withg(r)
nowreplaced by e(r)
=Er(0).
Using
thepostponed
Lemma 3 we deduce from(20)
that the m equa- tions(18),
withg(r) again replaced by e(r)
=ET(0),
hold for any r E(no,
... , no +h - 11 -
Therefore theequations (19)
hold for any n > no withg(n) replaced by e(n),
and from here on theproof
terminates asexplained
at the end of section 3.
Remark 3. For the
proof
of Lemma 3 below we need a few facts on(ho-
mogeneous)
linear recurrence sequences with constantcoefficients,
for shortrecurrence sequences, which the reader may find in the
introductory chapter
"Recurrence
Sequences"
of the book ofShorey
andTijdeman [12].
First,
for ourintegral
recurrence sequence there exists aunique
recurrence relation of minimal order
L,
say,with 0. As it is
easily
seen, the recurrence coefficients ao, ... ,are
rational,
infact, by
a theorem due toFatou, integral. Secondly,
from[12,
TheoremC.la)~
we knowwhere the distinct non-zero
algebraic integers dl,... , di
are the roots oforder pl, ... , pt E N
( pl
+ ... + pt =L)
of thecompanion polynomial
of the recurrence
(21).
Theuniquely determined gx
E(dl, ... ,
havedegrees
pa(A =1, ... , ,~). Thirdly,
we have from[12,
TheoremC.lb)~:
Ifdl, ... , de
E ex aredistinct,
pi,.... Pi EN,
and if one defines ao, ... ,by
and if the gx are any
polynomials
ofdegree
pa(~
=1, ... , ,~),
then thesequence
(e(n))
definedby (22)
satisfies the recurrence relation(21).
When we
applied
earlier Lemma 3 to the sequences(Er(v))"
for fixed r,we must be sure that these are recurrence sequences for every fixed r. But this follows
easily
from our above statement andwhere
D1, ... , Dfl
are the distinct numbers among the powersd1, ... , d~ .
Lemma 3. Let ~yl, ... , ,’)’m E
Cp
be distinct andsatisfy
=:
R~ (~
=1, ... , m).
Let(E(v))
be astrictly increasing
recurrence se-quence
of non-negative integers,
and letbl, ... , ,8m
ECp. If
holds
for
v - oo, with afixed non-negative
real number 791,
then61
=...-b~-p.
Remark 4.
Obviously,
if 1/1- is as in Lemma1,
then11/1- -lip Rp
for anyprime
p.Furthermore,
it should bepointed
outthat,
under the conditionson the
1’S
in Lemma3,
noquotient 1/1-/1/1-’ (p ~ ~c’)
can be a root ofunity,
see
[4, p.154, xiii)].
Proof.
From Remark3,
andrewriting (22)
alittle,
we knowwith distinct non-zero
algebraic integers Dl, ... , De
and with non-zeropoly-
nomials
Ga having
their coefficients in~(Dl, ... ,
Therefore we know0 1
(A
=1, ... , ,~),
and we may assumew.l.o.g.
Denoting
,,-,&. ,,-"" I...&.
we have - 0 as v - oo,
by (24).
Since 1 we maysuppose 1 for every v E
No. Defining
:=for it
=1, ... ,
m, the/31, . 13m
aredistinct,
and theinequalities
11J.t - lip Rp
hold. The sum on the left-hand side of(23)
isW.l.o.g.
we may suppose 1(p
=1, ... , m) .
We assume furtherthat at least one of the
61 , ... , 6m
is non-zero, and derive a contradictionaccording
to the cases t’ = t and t’ t.Case t’ = i. Then
B(v) =
0 andA(v)
=E(v)
for every v.By
theSkolem-Lech-Mahler
theorem,
there is a vo with0,
say,with rational v > 0.
Suppose
now L E Nlarge enough,
to bespecified
later. Then there is anN E N such that for any E
No
thefollowing inequalities
holdThese
imply
where ci > 0
depends only
onG1, ... , G£.
Then for v = vo + Nx weget
by (25).
Here theright-hand
side isp-v
if we take Llarge enough.
Therefore
--- -
for all
large v
of the form Since this contradictshypothesis (23)
of Lemma3,
we haveJi = ... = 8m
= 0 in the first case.Case i’ f. If
B(v) =
0 holdsinfinitely often,
then the infinite setIV
EGx(v)D’
=01
is a union of a finite set and offinitely
many
(at
leastone)
arithmeticalprogressions, by
the Skolem-Lech-Mahler theorem. Thisimplies
the existence ofc, d
E N such thatB(c+ dv)
= 0 foreach P E
No. Putting
we are back to the first case with E.
Therefore we may suppose from now on that 0 for all but
finitely
many v.
Indeed,
we may even assumew.l.o.g.
0 for any v ENo.
Next we transform a little the sum on the left-hand side of
(23)
For every v E
No,
at least one of the sumsis non-zero,
by
a Vandermondeargument,
and since not all81, ... ,8m
van-ish.
Therefore,
for any v ENo,
thereexists j (v)
E~0, ... , ~n -1 }
such that(27)
vanishesfor j
=0, ... , j (v) -1,
but notfor j = j (v).
Thus there existsa
jo E ~0, ... , m -1 ~
such that the sum(27)
vanishes for any v ENo
andj=0,...,jo-1
butfor an
appropriate
vo E N. With some rational w > 0 we defineAs we saw in the first case there exists N E N such that
holds for
every x e No .
Replacing
now the linear recurrence(E(v)) by
the new onewhere
E*(K)
:=E(vo
+NK),
we find from(26)
Using (28)
andI B (vo
+ the lastequation yields
«
for every
large
x. Thisequation,
combined withhypothesis (23)
in Lemma3,
shows that the casejo
= 0 isimpossible.
Hence we know m > 1 andjo E ~1, ... , m -1}.
Sinceholds for any a E
QX
we find from ourprevious
definition ofB(v)
with some real c2 > 1
(independent
ofv). Using this, (29)
and(23)
we findand thus
for every
large x
ENo .
On the other
hand, since t’ t,
there is aAo
with 0I Dxo ip
1. From(30)
followsI D-xol
>1,
thus at least one of theconjugates
ofDxo (which
occur all among the
D1, ... , Dt)
must belarger
than 1 in absolutevalue,
thus
leading
toEstimating
thepositive E*(x)
frombelow,
at least forinfinitely
many x,(31)
leads nowquickly
to the desired contradiction.The
precise
conclusion is as follows. Therepresentation
ofE(v)
fromthe very
beginning
of ourproof
and our definition ofE*(K)
leads us towhere the non-zero
Di,..., D~
are the distinct among theDN D) .
Tothis
representation
of the recurrence weapply
Kubota’s result[7, Corollary 3(ii)],
and we deduce the existence of a constant c6 > 0 such thatholds for
infinitely
many where d :=maxf deg (jA I
= ENo.
Thus Lemma 3 is
completely proved.
5. Sketch of
proof
of Theorem 2Let
dl, ... , d,"i
EI~1 ~ {1~
bedistinct; w.l.o.g.
we may supposedl
> ... >dm
> 1. SinceCorollary
2 allows us to assume m >1,
we may furtherassume
inductively
that thevu’2B--~ / B- I I - I - - I - - - I - - - - /
are
algebraically independent.
For each t’t~
we have to provethat the
f dm (aT) (T
=1,
... ,t’)
arealgebraically independent
over the fieldWe suppose to be
algebraically dependent
overK, whereas, (if t’
>1) every t’
- 1 of these arealgebraically independent.
Then there exists a
polynomial
with minimal total
degree
such thatWe may assume
las,lp
= ~ ~ ’ =lat,lp.
If a E7G~,~fd1(al),... then,
witha(k)
and n > k as in section2,
taking dm-1
>dm
into account.Hence,
we have(11)
withs’, t’
instead ofs, t
and can continue in the same way as in section 2.References
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[11] A.B. SHIDLOVSKII, Transcendental Numbers. De Gruyter, Berlin et al., 1989.
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Peter BUNDSCHUH Mathematisches Institut Universitat zu K61n
Weyertal 86-90
50931 K61n, Germany
E-mail : pb0math.uni-koeln.de
Kumiko NISHIOKA
Mathematics, Hiyoshi Campus
Keio University
4-1-1 Hiyoshi, Kohoku-ku Yokohama 223-8521, Japan E-mail :