C OMPOSITIO M ATHEMATICA
P. L. C IJSOUW
Transcendence measures of exponentials and logarithms of algebraic numbers
Compositio Mathematica, tome 28, n
o2 (1974), p. 163-178
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163
TRANSCENDENCE MEASURES OF EXPONENTIALS AND LOGARITHMS OF ALGEBRAIC NUMBERS
P. L.
Cijsouw
Noordhoff International Publishing Printed in the Netherlands
1. Introduction
Let 6 be a transcendental number. A
positive function f of two integer
variables N and H is called a transcendence measure
of u
iffor all non-constant
polynomials
P ofdegree
at most N and withintegral
coefficients of absolute values at most H.
The purpose of the present paper, which covers a part of the authors thesis
[2],
is togive
transcendence measures for the numbers e"(a
al-gebraic,
03B1 ~0)
andlog
a(a algebraic,
03B1 ~0, 1,
for any fixed value of thelogarithm).
These transcendence measures will be of the formwhere S =
N+ log H,
for aneffectively computable
constant C > 0 and forgiven
constants a,b,
c and d. We try to obtain a small totaldegree
in the exponent in N and S
together,
and to get a minimal contribution of S within this totaldegree.
Such measures areimportant
for certainapplications;
see e.g.[1]
and[14].
On the otherhand,
we do not try to determine the constant C in theexponent
as small aspossible.
Infact,
Cwill be chosen very
large
tokeep
theproof uncomplicated.
As far as we
know,
no transcendence measure for e’ which containsexplicitly
both thedependence
on N and H was everpublished.
Earliertranscendence measures of similar
types
for thespecial
case of the numbere and for
log
a aregiven by
N. I.FEL’DMAN, namely
for e, see
[5 ],
for
log
a, see[3 ],
andfor
log
a, see[4].
Transcendence measures of other types are
published by
several au-thors.
Generally speaking,
in their results theheight plays
a moreimpor-
tant rôle while the
dependence
on thedegree
is notexplicitly given.
However,
in a recent paper,[7],
A. I. GALOÉKINproved
a measure for e’of the form
For more references and
information,
see[2], [8] and [11]. Finally,
weremark that the transcendence of the considered numbers e03B1 and
log
ocwas
proved by
F. LINDEMANN in[9].
2. Formulation of results
We shall prove the
following theorems,
whereagain
S =N+ log H:
THEOREM 1: Let a be a non-zero
algebraic
number. Then there exists aneffectively computable
numberC4
=C4(03B1)
> 0 such thatexp ( - C4N2S}
is a transcendence measure
of
e03B1.THEOREM 2: Let oc be
algebraic,
a 00,
1. Letlog
ce be anyfixed
valueof
thelogarithm of
a. Then there exists aneffectively computable
numberCs = C5 (a)
> 0 such thatexp ( - Cs N2S(1 + log N)2}
is a transcendencemeasure
of log
oc.The method of the
proofs
will be A. O. GEL’FOND’Smethod;
this meth- od was used tooby
N. 1. FEL’DMAN in thequoted
papers. From the na-ture of these
proofs
it is clear that the constantsC4
andC.
are effec-tively computable,
so we will make no further reference to thisaspect.
3. Notations and lemmas For any
polynomial
P withcomplex
coefficientwe call n the
degree
andthe
height
of P. If a is analgebraic number,
then we use thedegree d(a)
and the
height h(a)
as thedegree
andheight
of its minimaldefining poly-
nomial. We call
s(a)
=d(a) + log h(a)
the size of a.Q
will denote the field of the rational numbers. If a is a realnumber,
then[a]
is thegreatest
integer
smaller than orequal
to a.LEMMA 1: Let ex be
algebraic of height h(03B1).
ThenIf moreover
ex 1=0,
then we havePROOF: For the first part, see
[11 ],
Hilfssatz 1. For the second part, take into consideration that ifis the minimal
polynomial
of a, thenis the minimal
polynomial
of03B1-1, apart
from a factor + 1.LEMMA 2: Let ai be
algebraic of degree di
andheight hi (i
=1, ···, n).
Denote
by
d thedegree of Q(03B11, ···, an)
overQ.
Letbe a
polynomial
withintegral coefficients
Pi1 ... j., such that the sumof
theabsolute values
of
thecoefficients
is at most B. ThenP(03B11, ···, an)
= 0 orPROOF: See
[6],
Lemma 2.For convenience we formulate the
following
consequence of Lemma2,
in which
occurring empty
sums should be omitted:LEMMA 3:
Let 03BE
bealgebraic of degree
N and size S. Let n ~ 0 be aninteger
and let 03B1i bealgebraic of degree di
and size si(i
=1, ..., n).
Putd =
[Q(03B11, ···, an) : Q] if
n ~ 1 and d = 1if
n = 0. Letbe a
polynomial
withintegral coefficients
whose sumof
absolute values is at most B. ThenP(03BE,
03B11, ···,an)
= 0 orPROOF :
Apply
Lemma 2 with nreplaced by
n + 1 and 03B11, ···, 03B1nreplaced
by 03BE,
03B11, ···, 03B1n. Use theinequalities
and
LEMMA 4: Let r and s be
positive integers
such that s > 2r. Then any setof
rlinear forms
in s variableswith
complex coefficients
apa such thatlapaI
~ A(p = 1, ···,
r;u =
1,···, s)
hasthe following
property: For everypositive
eveninteger
C there existintegers C1, ···, C,,,
not all zero,with |C03C3| ~
C(03C3
=1, ···, s)
and
PROOF: See
[11 ],
Hilfssatz 28.LEMMA 5: Let
(p
=1, ···,
r; u =1, ···, s)
bepolynomials
withintegral coefficients
p pait ... j., such that the sumof
the absolute valuesof
thecoefficients of
each
polynomial
is at most B. Let oci bealgebraic of degree di
andheight hit (i
=1, ..., n) and put
d =[Q(03B11, ···, an) : Q].
Let C be apositive
even
integer. If
and
then there exist
integers C1, ···, cs,
not all zero,with |C03C3| ~
Cfor
03C3 =
1,...,s,such
thatPROOF: From Lemma 1 we know
that |03B1i| hi+1.
Hence,
Define
Yp
for 03C1 =1, ..., r by
From Lemma 4 we conclude that there exist
integers C1, ··· Cs,
notall zero,
with |C03C3| ~
C for u =1, ···, s and
for p =
1, ···,
r. From(7)
and(9)
it now follows thatfor p =
1, ...,
r.However, Yp
is apolynomial
in al’ ..., an, ofdegree
at most
Ni
in ai and with sum of absolute values of its coefficients at most BsC.Therefore, according
to Lemma2,
theinequality (10) implies
that
Y03C1 = 0
for p =1, ···,
r.LEMMA 6: Let
Ppu(zo,
z1, ···,Zn) for
p =1,
rand 03C3 =1, ···,
s bepolynomials
withintegral coefficients,
such that the sumof
the absolute valuesof
thecoefficients of
eachpolynomial
is at mostB,
and such that thedegree
in ziof
eachpolynomial
is at mostNi (i
=0, 1, ..., n). Let 03BE
bealgebraic of degree
N and size S. Let ai bealgebraic of degree di
and sizesi, i =
1,...,
n. Put d =[Q(03B11, ···, an) : Q] if n ~
1 and d = 1if n
=0,
and let C be apositive
eveninteger. If
and
then there exist
integers Cav (u
=1, ...,
s; v =0, 1, ···,
N -1),
not allzero, such
that |C03C303BD| ~ C for u
=1, ···, s
and v =0, 1, ..., N-l
and such thatPROOF: Define
Ppa,,
for v =0, 1, ···, N- 1 by
Then
Ppa,,
is ofdegree
at mostNo + N-1 in
zo and at mostNi
in zi fori =
1, ..., n.
The sum of the absolute values of the coefficients of eachPpav
is at most B. Theequations (13)
now reduce tofor p = 1, ···, r.
We
apply
Lemma 5 to thepolynomials P,,,;
to this end we re-place
zi , ..., znby
zo, zl, ... , zn, where thedegree
inzo is
at mostNo + N-1;
03B11, ···, anby 03BE,
(Xi, ’ ’ ’, an; sby
Ns and dby
a number thatis at most Nd. For all
positive integers
N we haveHence,
where H denotes the
height of 03BE.
Lethi
be theheight
of ai . We haveFrom this it follows for i =
1, ···, n
thatThe
inequalities (11), (12), (15)
and(16) imply
that conditions(6)
and(7)
with the
appropriate
substitutions are satisfied. Hence, it follows from Lemma 5 that theintegers Cav
can be chosen in therequired
way.LEMMA 7: Let F be an
entire function
and let P and T beintegers
and Rand A be real numbers such that R ~ 2P and A > 2. Put
and
Then
PROOF:
By
the maximum modulusprinciple
we can choose acomplex
number z with
Izl
= R andIF(z)l
=MR.
From the residue theorem ofCauchy
we have thefollowing
well-known consequence:Let p
be one of the numbers0, 1, ’ ’ ’,
P -1 andlet 03BE
be acomplex
numberwith |03B6 - p| = 1 2.
Let qo, q1, ..., qP-1 be the numbers0, 1, ..., P-1,
re-arranged
in such a way thatThen
Hence,
The
inequality (P-1 )!
>(P/3)P
iseasily
checked for P =1, ···,
10. Forhigher
values of P it can beproved by induction, using
theinequality
It follows that
From
(18)
we now obtain the estimateLEMMA 8: Let
be an
exponential polynomial
withcomplex
numbersCkm
and (Ok, such that03C9k ~ 03C9l if
k ~ l. Putand
Let T’and P’ be
positive integers
and putIf
then
for
k =0, 1, ···, K-1
and m =0, 1,
···, M-1.Moreover, if in particular
cok = k03B8(k
=0, 1, ···, K-1) for
some com-plex
number0,
then we mayreplace (20) by
PROOF: See
[13],
Theorem 2.LEMMA 9: Let
~(n, s)
be apositive, f’unction defined for all positive integers
n and all s ~ 1 with the
following properties:
If, for
some transcendental number 0’,for
allalgebraic
numbersç,
where N and S denote thedegree
and the sizeof ç,
thenfor
all non-constantpolynomials
P withintegral coefficients,
where N andH are the
degree
andheight of P
and S = N+log
H.PROOF:
(compare [3 ],
Proof of Theorem2)
If P isirreducible,
it followsby [3 ],
Lemma 5 andby (i)
thatIn the
general
case, write P =aP1 ··· Pm
wherePi
is a non-constant irre- duciblepolynomial
and a > 0 aninteger.
Denotedegree
andheight of Pl by Ni
andHi
andput Si
=Ni + log Hi(i
=1, ···, m).
Thenclearly By
e.g. GEL’FOND’S well-knowninequality
on theheight
of aproduct
ofpolynomials (see [8] p. 135,
LemmaII;
see also[12],
Lemma 3 and[10])
we have
Hi ~ eNH
andthus,
Using (ii)
and(iii)
it follows thatfor i =
1, ···,
m.By multiplying
theseinequalities
we obtain therequired expression (23),
sinceNi
+ ···+Nm
= Nand a ~
1.4. Proof of Theorem 1 First we prove
THEOREM 3: Let 03B1 ~ 0 be an
algebraic
numberof
sizes(a).
Then thereexists an
effectively computable
number81 = S1(03B1)
such thatfor
allalgebraic numbers 03BE of degree
N and size S’ ~81.
PROOF: All estimates
occurring
in thisproof hold for
Ssufficiently large.
Sl
can be chosen as the maximum of thefinitely
many bounds thus ob- tained.Suppose
thatfor some
algebraic number 03BE
ofdegree
N and size S. It will be shown that thisassumption
leads to a contradictionif S ~ Sl .
Observe that(25)
im-plies that |03BE| |e03B1| +1.
Choose the
following positive integers:
We use the
auxiliary exponential polynomial
where the
Ckmy
areintegers
of absolute values at most C. Later we shallspecify
them further.For t =
0, 1, 2, ... and p
=0, 1, 2, ...
we haveDefine
tPtP
for the same tand p by
Then
03A6tp
is analgebraic number, approximating F(t)(p)
veryclosely.
Infact,
for k =
0, 1, ···, K-1 and p
=0, 1, ···, P-1. Hence,
for t =
0, 1, ···, T’ -1 and p
=0, 1, ···, P -1.
We are
going
to choose theintegers Ck.,
such that03A6tp =
0 fort =
0, 1, ···,
T -1 and p =0,
1, ···, P -1. To this end weapply
Lemma 6 with n =
1,
(Xl = a, r = TP and s = KM to thepolynomials
(t=0, 1, ···, T-1; p-0, 1,···, P-1; k=0, 1,
--,K-1;
m =
0, 1, - - -, M-1). Using
the notations of Lemma6,
we have andBy
means of theseinequalities
oneeasily
verifies conditions(11)
and(12)
of Lemma 6.
According
to thislemma,
we now choose theintegers Ckmv,
not all zero, with
|Ckm03BD| ~
C such thatfor t =
0, 1, ···,
T-1and p
=0, 1, ···,
P-1. In this way, our function F iscompletely
fixed.From
(26)
we obtain with(27)
for t =
0, 1, ···,
T -1and p
=0, 1, ···,
P -1. From(1)
it follows that|03B1| ~ es(03B1). Using
thisinequality
we see from the definition of F thatWe
apply
Lemma 7 with R = 2P and A = S. We obtainby (28)
and(29)
Hence,
for t =0,1,..., T’-l and p
=0, 1, ···, P-1
we haveand for the same values of t and p,
using (26),
But
03A6tp
is apolynomial
withintegral
coefficientsin 03BE
and a, ofdegree
at most
(K-1)(P-1)+N-1 ~
KPin 03BE
and less than T’ in a. The sumof the absolute values of its coefficients is not greater than
From Lemma 3 it follows that
Otp
= 0 orfort =
0, 1, ..., T’ -1 and p
=0, 1, ···, P-1.
Since(32) and (33) are
incompatible,
it follows thatçp tp = 0
fort = 0, 1, ···, T’-1
andp =
0, 1, ···,
P-1. From(26)
we now obtain thatfort=
0, 1, ···, T’ - 1 and p
=0, 1, ···, P- 1.
Subsequently
weapply
Lemma 8 to ourexponential polynomial F,
with
and P’ = P.
Using
the notations of Lemma 8 wehave, by |03B1| ~ es(03B1),
theinequality
Thus,
so that condition
(19)
is satisfied.Further, by
Lemma 1 wehave |03B1| ~ es(a)
and
1 oc |
~e-s(03B1).
HenceThus,
it follows from(21)
and(34)
thatfor k
= 0, 1, ···, K+1
and m =0, 1, ’",M-1.
But
is a
polynomial in 03BE
ofdegree
less than N and with sum of the absolutevalues of its coefficients at most NC. It follows from Lemma 3 that
or
for k =
0, 1, ···, K-1
and m =0, 1, ···, M-1. Hence,
for all these k and m.
Since 03BE
isalgebraic
ofdegree N,
the numbers1, 03BE, ···, 03BEN-1
arelinearly independent
overQ. Thus, C’kmy
= 0 fork = 0, 1, ···, K-1, m = 0, 1, ···, M-1 and v = 0, 1, ···, N-1. This
contradicts the choice of the
integers Ckm03BD
andby
this contradiction Theorem 3 isproved.
We now
complete
theproof
of Theorem 1 in thefollowing
way:There are
only finitely
manyalgebraic numbers 03BE
of size SSl.
Sincee03B1 - 03BE ~
0 and since N2S > 0 for all of thesenumbers,
there exists anumber
C6
> 0 such thatfor all
algebraic numbers 03BE
ofdegree N
and size SS1.
ThenC7 =
max(C6, 5.108 e6s(03B1))
will have the propertyfor all
algebraic 03BE
ofdegree
N and size S. From Lemma 9 we obtain that for allpolynomials
P withintegral coefficient,
ofdegree
N andheight H,
with S = N+log
H andC4 = 6C7. By
reasonsof monotony,
the sameinequality
holds forpolynomials
ofdegree at
most N andheight
at mostH.
Hence, exp {- C4 N2S}
is a transcendence measure for e" and Theorem 1 has beenproved.
5. Proof of Theorem 2
Let
log
a be anarbitrary
but fixed value of thelogarithm
of the al-gebraic
number a. N. I. FEL’DMANproved
thefollowing
assertion:(see
Theorem 1 of
[4],
with m =1)
There exists an
effectively computable positive
numberC8 = C8(log oc),
such that
for
allalgebraic numbers 03BE of degree
N andheight H, provided
thatN
(log H)1 4.
From
this,
iteasily
follows thatfor all
algebraic numbers 03BE
ofdegree
N and with sizeS, provided
thatN
(tS)*.
For the
complementary
case N ~(1 2 S)*
we prove thefollowing
theorem:THEOREM 4: Let a :0
0,
1 be analgebraic
numberof size s(03B1)
and letlog
a be an
arbitrary, but fixed,
valueof
thelogarithm of
oc. Then there existsan
effectively computable
numberS2 = S2(log a),
such thatfor
allalgebraic numbers 03BE of degree
N and size S ~S2, provided
thatN ~
(1 2S)1 4.
PROOF: Since the structure of the
proof
is the same as that of Theorem3, only
a shortedproof
isgiven. Suppose
thatfor some
algebraic number ç of degree
N and sizeS,
such that N ~(1 2S)1 4.
We shall derive a contradiction in the case of
large
S.Choose the
integers
Put
where the numbers
Ckmy
areintegers
of absolute values at most C;they
will be
specified
later. We havefor t,
p =0, 1, 2, ···
Put
For t =
0, 1, ···,
T’ -1and p
=0, 1, ···,
P -1 iteasily
follows from theinequality 1 etl
~es(03B1)
thatLet Ptpkm
for t =0, 1, ..., T- 1,
p =0, 1, ..., P-1,
k =0, 1, ...,
K-1and m =
0, 1, ···,
M-1 be thepolynomials (taken
in the obviousway)
such that
We now choose the numbers
Ckm03BD according
to Lemma6, applied
to thepolynomials Ptpkm,
asintegers,
not all zero, of absolute values at mostC,
such thatOtp
= 0 for t =0, 1, ···,
T -1and p
=0, 1, ···, P-1.
Itfollows from
(41)
thatSince
and since N ~
(1 2S)1 4
we see from Lemma 7applied
with R = 2P andA =
Sl/5
thatHence,
and
for t = 0, 1, w, T’ -1 and p = 0, 1, w, P -1.
From Lemma 3 we obtain
03A6tp
= 0 orfor the same values of t
and p. Thus, CP tp
= 0and,
from(41),
for t
= 0, 1, ···, T’-1 and p = 0, 1, ···, P-1.
Condition
(19)
with P’ = Pand 03A9 ~ 103(s(03B1))2| log 03B1|N
is satisfied.Further,
Using
thisinequality
and(46)
we obtain from(21)
for k =
0, 1, ···,
K-1 and m =0, 1, ···,
M-1.From Lemma 3 it now follows that
for all k and m. This
implies
that allintegers Ckm,,
are zero.By
the con-tradiction to the choice of these
integers,
Theorem 4 has beenproved.
We
proceed
to prove Theorem 2. From(39)
it follows that there exists aneffectively computable
numberC9
=C9(log oc)
> 0 such that for allalgebraic numbers 03BE
ofdegree
N and size S with N ~(1 2 S)1 4,
sincethere are
only finitely
manyalgebraic
numbers of size smaller than82.
Taking C10 =
max(2Cg, C9)
we see from(38)
and(48)
thatfor all
algebraic numbers 03BE
ofdegree
N and size S. Asbefore,
theappli-
cation of Lemma 9
completes
theproof.
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