J.P. B ÉZIVIN
A. B OUTABAA
Decomposition of p-adic meromorphic functions
Annales mathématiques Blaise Pascal, tome 2, n
o1 (1995), p. 51-60
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DECOMPOSITION OF
p-ADIC
MEROMORPHIC FUNCTIONSJ.P. Bézivin and A. Boutabaa
Ann. Math. Blaise Pascal, Vol. 2, N° 1, 1995, pp.51- 60
Abstract. Given a
meromorphic function,
we are interested in how many ways it can beexpressed
as acomposite
of othermeromorphic
functions. Is thisalways possible? And,
when it is the case, what about
unicity
of such anexpression?
Etc...In the case of
polynomials,
the discussion isrelatively
easy and has been studied in detailsby
Ritt[13].
As soon as one passes to rational functions theproblem
becomes muchmore difficult .
Various aspects of the
problem concerning
thecomplex meromorphic
functions have been studiedby
many authors[3], [7], [8], [9], [10], [11], [12]
and[14].
In this
work,
we deal withp-adic meromorphic
functions. We will see that one neithercan
always
use the same methods that those of thecomplex
case nor obtain the sameresults. This is due to the fact that the distribution of the
singularities
is not the samein the two cases. For
instance,
in C one can consider entire functions whose all zeroslie on a
single straight
line(real
numbers forexample).
Then many resultsconcerning
such functions and based upon a theorem of Edrei
[6]
are obtained(11J, [12].
This result of Edreiplays
animportant
role in thecomplex decomposition theory
and there is nopossible p-adic
version of it.1991 Mathematics
subject classification:
:Primary 12H25; Secondary 46S10.
I.
p-adic meromorphic
functionsWe note
A( Cp)
thering
of entire functions inCp
andM(Cp)
the field ofmeromorphic
functions in all
Cp.
Definition 1.1. Let
f ; f l, ... , f n
eM(Cp)
such that :f ‘f10~2~...p fn.
We say that this is a
decomposition
off
and thatf i, ... fn
are factors off
.Given a rationnal function
R(,r)
= whereP x x
arepolynomials of p[x]
relatively prime,
we calldegree
ofR( x)
the numberdeg
R =max(deg P; deg Q).
Definition 1.2. A function
f
EM(Cp)
is said to beindecomposable if,
in alldecompo-
sition of
f,
allfactors, except
at most one, are rational functions ofdegree
one.Definition 1.3. A function
f
EM(Cp)
is said to bepseudo-indecomposable
if in alldecomposition,
allfactors,
except at most one, are rational functions.Remark 1.4. Let
f
be an entire function. We havenaturally
a notion ofindecompos- ability (resp. pseudo-indecomposability) depending
on whether weregard f
as an elementof
A( p )
or ofp ).
We will see however that( Corollary 2.4.) f
isindecompos-
able
(resp. pseudo-indecomposable)
in if andonly
if it isindecomposable (resp.
pseudo-indecomposable)
inJI~I{~,).
This result is false in theComplex
case(Remark 2.5.).
Definition 1.5. Two
decompositions
of h E h =f1 o
... ofn
= gl o ... o gn are said to beequivalent
if there exist rational functions ofdegree
one,L1; L2 ;
... ;Ln-1
suchthat :
,fl
= 91 0Lli
lf2
=L-11 o g2
0L2; ... fn-I
=L-1n-2
0 gn-1 o Ln-1 ;f n
=We will now recall some basic definitions and results on
p-adic meromorphic
functions.Let
f(x)
=V anxn
be an entire function in ~ p. For all R >0,
we note, n>oI f I (R)
=maxn~0 | an | Rn,
the maximum modulus function off.
. This is extended tomeromorphic
functions h== 2014 by h ~ (R)
= .Proposition
1.6. Letf
be an entirefunction
in ~ p. Then thefunction
(R
~ 8f(log R)
=log f ‘ (R))
is a convexpolygonal
curve. Moreover The numberof
zerosof f
in the closed(resp. open)
discof
radius p and center 0 isgiven by
theright-derivative 03B8+f (log p) (resp le f t
derivative03B8-f(log 03C1)) of
8f(u)
at thepoint log
p.For the
proof
see~1~
forexample :
:Proposition
1.7. LetF, G,
H be threemeromorphic functions
.Suppose
that H is not arational
function
and that F = H o G. Then G is entire.53
Proof. We have H( x) = A(x) B(x) where A( x) and B( x) are entire functions without common
zeros. The fact that
H ( x)
is not rationalimplies
that there exist at most one value 03C90 Ep
such that
A( x)
- is apolynomial.
Let w be a value different of this eventual one.Then the entire function
A(x)
-03C9B(x)
is transcendental and hence hasinfinitely
many zeros, which we noteSuppose
thatG(x)
has apole
x0. We setG(x)
=(.r 2014 xo)
where
W(x)
is ananalytic
function that has no zeros in an opendisc ~ x -
xa(
R and f.an entire
number ~
1. Forany k
setTk(x)
=W(x)
-ck(x
- For one of these k letus choice p
~]0, R[
suchthat 1 Ck | pl =| W(xo) (.
Hence we have :(W(x4)) [ if0rp
[ if p r R.
On the other
hand,
the fact that W has no zeroin |x -
xo|
Rimplies
that :~W ~(r)=~W(xo)~ [
VrHence we have :
|Tk|(r)={|W(x0)| si 0 r 03C1
| ck|rl si 03C1 r R
Therefore the function
Tk(x)
has at least one zero xk in the circleI
x - xo1=
p. Thus we haveG(Xk)
= ck ; and soF(xk)
= w.consequentely,
sincethe xk are
infinitely
many, the functionF(x)
- w hasinfinitely
many zeros in the discj
x - xo~ R,
which is acontradiction.
HenceG(x)
has nopoles
and so is entire.Remark 1.8. The above
proposition justifies
the factthat, subsequently,
in any decom-position
of ameromorphic
functionf
in the formef = h o g,
wesuppose h meromorphic and g
entire or h rationaland g meromorphic.
The
following
result will enable us to show the existence ofindecomposable
orpseudo- indecomposable
transcendentmeromorphic
functions .For an entire fonction
f
and apositive
number p, we notem(f, p)
the number of zeros off
in the opendisc ~ x ~
p,M( f , p)
the number of zeros off
in the closeddisc I x ~
pand
A( f, p)
the number of zeros off
in the circle‘
x~=
p; each one of these zerosbeing computed
with itsmultiplicity.
Proposition
1.9. Let H and G be two entire and non constantfunctions.
Let pa be apositive
real number such that thefunction | G (r)
isstrictly increasing for
r > po. Letus put F = H o G. Then we have
for
all p > pa :i)
=(
GI (03C1))m(G,03C1)
ii) M(F, p)
=M(H, E G ( ( p))M(G, p)
iii) A(F, p)
=A(H, I G (
+m(H, I
GI (03C1))A(G,03C1)
Proof. From F =
H o G;
we deducethat,
for all r, we have :( F (r) =
Hi (~
G~ (r)).
Since the
function G | (r)
isstrictly increasing
for r > 03C10, we deducethat,
for u >log
pa,we have :
(1) 03B8-F(u)
‘03B8-H(03B8G(u)).03B8-G(u),
and(2)
=03B8+H(03B8G(u)).03B8+G(u).
By proposition 1.6,
we have relationsi)
andii). Substracting i)
fromii),
we obtain rela- tioniii).
Theorem 1.10. Let F E and a a
positive
entire number. We suppose thaton
infinitely
many circlesof
center 0of Cp,
F has a numberof
zeros included between 1 and a. Then alldecomposition of
Fof
theform
F = H o G withH,
G EA(p) implies
that H or G is a
polynomial of degree
between 1 and a.Proof. If any of the functions Hand G is a
polynomial
ofdegree
between 1 and~,
we have for penough large : M(G, p)
> ~ + 1 andm(H, ( G ( p))
> a + 1.On the other
hand,
thehypothesis
of the Theorem and the formulaiii)
of theproposi-
tion 1.9 show that for an
infinity
ofp’s arbitrarily large,
we haveA(H, [
G~ ( p)) #
0or 0. Then for such a p we have
A(F, p)
> a + 1. This is a contradiction with thehypothesis.
Hence H or G isnecessarily
apolynomial
ofdegree
between 1 and a.Corollary
1.11. Let F E and .1 an entire > l. We suppose that on aninfinity of
circlesof
center0,
F has a numberof
zeros included between 1 and a and thaton an
infinity of
circlesof
center0,
F has a numberof poles
included between 1 and a.Then any
decomposition of
F in theform
F = H o G with H E and G EA(p) implies
that :G is a
polynomial of degree
between 1 anda,
or H is a rational
function of degree
between 1 and .1.Corollary
1.12. Afunction
Fsatis fying
theassumptions of
theorem l.l 0(resp .
the onesof corollary 1.11)
is :1.
indecomposable
inA(Cp) (resp.
inif
.1=1.2.
Pseudo-indecomposable
inA(p) (resp.
inM(p)) if
03BB ~ 1.Proof. Let H( x) -
H1(x) H2(x)
where H1, H2 ~ A(p) have no common zeros. Hence : F = H1oG H2oG. Then weapply
the theorem 1.10 to each of the functions Hl o G andH2
o G.Proposition
1.13. Let F EM( Cp) possessing
the property to have in aninfinity of
discsof
center 0 and radiusarbitrarily large
aprime
numberof
zeros and in aninfinity of
discsof
center 0 aprime
numberof poles.
Then F isindecomposable
inM(Cp).
Inparticular, if P(x)
Ep[x]
andQ(x)
Ep[x]
havedegrees
which areprime numbers,
the rationalfunction R(x)
= P(x) Q(x) isindecomposable.
55
Proof. We
apply
the relationii)
of theproposition
1.9.Corollary
1.14. Let F E We suppose that in aninfinity of
discsof
center0 and radius
arbitrarily large,
F has a numberof
zerosequal
to theproduct of
twoprime
numbers
(not necessarily distinct). Then,
either F isindecomposable,
or it is acomposite of
twoindecomposable factors.
Proof. We use the
previous proposition.
Corollary
1.15. Let F ECp[z]
and 03BB ~ IN *. We suppose that in aninfinity of
discsof
center 0 and radiusarbitrarily large,
F has a numberof
zerosequal
to theproduct of
aprime
numbers(not necessarily distinct).
Then F is acomposite of
at mostA
indecomposable factors.
Proof. We
proceed gradually using
thecorollary
1.14.Given a
meromorphic function,
it is not easy to know if it isdecomposable
or not.The first of the two next
examples
shows that adecomposable
function may have nonequivalent decompositions.
The secondexample
shows that ameromorphic
function canhave an
infinity
ofprime
factors.Example
1.16. Let q be aprime
number and f apositive
entirerelatively prime
to q.Let be a sequence of elements of
~P
such that :1 an | an+1 |
andlim | an 1=
+00.n~+~
We consider the
following
entire functions :F(x)
=xl (1 - xq an); G(x)
=xl 03A0(1 - x an)l;
H(x)
=xql 03A0 (1- -
and = xq.n~O an
Let rn
=1 an [
Vn > 0. In the disc of center 0 and radius Tn, each of the functions F and G has nq + l zeros. On the other hand the theorem of Dirichletguarantees
thatamong the terms of the arithmetic
progression (nq
+ there is aninfinity
ofprime
numbers. Hence the functions F and G
verify
the conditions of theproposition
1.13,and so are
indecomposable.
The function cp isevidently indecomposable
and we haveThese two
decompositions
areclearly
notequivalent.
Example
1.17. Let(an)n>o
be a sequence of elementsof *p
such that :t an ( I
andlim
n~+~(_
+00.Let
fn(x)
=x(1- an ).
We define the sequence(cpn(x)) by :
= x and ~pn o
fn(x).
Let R > 0. We will show that the sequence of
CPn( x)
is ofCauchy
in the space of functionsanalytic
in the closeddisc a- j
R. We have :03C6n+1(x) = 03C6n(x - x2 an) = 03A3(-x2 an)k03C6(k)n(x) kl.
On the other hand we
have : (R) ~| k! | 03C6n | (R).
Let N be the first index such that n > Nimplies that I an |>
R. Hence wehave : 03C6n ‘(R) ~|03C6n (R) a .
Consequently
wehave, (R) ==) (R)
for all n > N. It follows that the sequence(cpn(x))
is ofCauchy
in the space of functionsanalytic
in the closeddisc x (
R. Hencethe sequence
(03C6n(x))
converges to an entire function03C6(x).
We can write for all k and n > ~ : :
=
f0
pfi
o ... pfn
0Bn(x),
where
8n(x)
is a function of the same type as and hence will converge to an entire function~(:c).
Hence we will have :~(x)
=fo
oand this relation shows that the
f k( x),
which areindecomposable
are factors of03C6(x).
Hence has
infinitely
manyindecomposable
factors.II.
Comparison
withpolynomials
andcomplex
case.We know that
P(x) being
agiven polynomial,
we have :(1) P(x)
has at least onedecomposition
inindecomposable
factors.(2)
The number ofindecomposable
factors is finiteand,
is the same in anydecomposition of P(x).
(3) P(x)
hasonly
a finite number of nonequivalent decompositions.
Do these
properties
remain true for entire ormeromorphic
functions?Example
1.16 showsthat
(2)
is nolonger
so. Infact,
even if we assume that alldecomposition
off
has a finitenumber of
factors,
we can not show if this number is bounded and if it is the same in anydecomposition.
This is not obvious at all becauseonly
for rational functions thefollowing
result stated
by
Ritt[13]
in 1922 hasonly
beenproved recently [2].
Proposition
2.1.(Ritt).
There exists a rationalfunction
which has twodecomposi-
tions into
indecomposable functions
eachhaving
adifferent
numberof factors.
57
On the other hand Ritt
[13]
had shown that for anydecomposition
of apolynomial
into rational
functions,
there exists anequivalent decomposition
intopolynomials.
Thefollowing
result shows that this is extended top-adic
entire functions.Theorem 2.2. Let F
A(Cp).
Assume that F = G o H withG, H M(Cp).
Then thereexist g, h such that F =
go h. Moreover,
these twodecompositions
areequivalent.
Proof.
1)
IfG ~ Cp(a*),
we have seen(Remark 1.8)
that then JEf must be entire. On the otherhand,
we haveG(a-)
=B ,
whereGi,G2
are without common zerosG2(~)
.
°
So F
= G1(H) G2(H).
We can assume that H is not constant. ThenG2(x)
has no zeros;because if this is not the case and if zo is a zero of
G2(x),
there exists .riCp
suchthat = x0. From which =
G1(x0) ~
0 andG2(H(x1))
=G2(x0)
= 0and F would have a
pole.
Contradiction. HenceG2
has no zeros and is soequal
to aconstant a different from zero. Hence G =
2014 A(Cp).
2)
If then there existrelatively prime
such that=
P(x) Q(x).
So F =P(H) Q(H).
Hbeing
non constant andmeromorphic,
it reaches allvalues of
p
except at most one. Thisimplies
thatQ(x)
has at most one zero. IfQ(x)
hasno zero, we have finished.
Suppose
thatQ(.r)
has a zero a. Set =H1(x) H2(x),
whereH1, H2
6 are without common zeros. Then the function-H’(.r) 2014
a =H1(x) H2(x)-
ahas no zero. Hence
= 03B2 ~
0. So we haveH1
= and H =LoH2,
where L =
03B1x + 03B2 x
. Hence F = ToH2,
with T= P Q
o L. We see that T must be apolynomial;
because if not, F should have apole.
We see also thatH2
= L-1 oH,
where L"~ =
~~.
.Corollary
2.3.If an
entirefunction
F isindecomposable (resp pseudo-indecomposable)
in then F is
indecomposable (resp pseudo-indecomposable)
inM(Cp).
This result is false in
Indeed,
Ozawa[11]
has shown that the functionF(z)
=(~ - l)e~"~
isindecomposable
in~(C),
but that F =fog
wheref
=and g
= This means that inM(Gj),
the function jF is not evenpseudo-decomposable.
III. Common
right
factors of F andF(n).
Suppose
that ap-adic meromorphic
functionF(z)
and its derivativeF’(z)
have anentire function g as their common
right factor,
it iseasily
shown from F =f o g
andF’ =
h o g
that g must be apolynomial
ofdegree
one. It is nolonger
asimple problem
ofsearching
thepossible
forms of any commonright
factor of F andF~"~. However,
we havethe
following general
result :Theorem 3.1. Let F E
M(Cp) ~
and g E .Suppose
that there exist apositive
entire n and twomeromorphic functions f
and h(which
are notrational functions of degree 1~
such that :F =
f o g
and =h o g
. Then eitherf satisfies
thefollowing equation :
+ ... + + =
0,
whereAa(x)
ECp[x];
or
g’"
= Bo +B1g(x)
+ ... +Bmgm(x),
whereBj
ECp
and m n.We need the
following
results whoseproofs
are in[4]
and(5J :
:Lemma 3.2. Let
ho,
...,hm, Fo,
..., Fm be elementsof M(C p)
such that 0.Let g E
A( Cp)
such thatT(r, ho)
+T(r, h1)
+... +T(r, hm)
=0(Log |
9| (r)).
.We suppose that :
+ ... +
F,n(9)hm(x)
= 0.Then there exist
polynomials Po(x),
...,Pm(x)
withP;(x)
not allequal
to 0 such that :po(9)ho(x)
+ ... +Pm(g)hm(x)
= 0.The function
T(r, .)
above is the Nevanlinna caracteristicfunction.
See[4]
and[5].
Lemma 3.3. Let
P(x,y,y’, ...,y(n))
be adifferential polynomial
iny,y’, ...,y(n)
withcoefficients ai0 i1 ... in(x)
EBCp[x].
Let d =~
0.Suppose
that there exists a transcendentalmeromorphic
solutionf( x) of
thedifferential equatio n
:p(x,y,y’, ...,y(n))
=R(x,y) ;
whereR(x,y)
EBCp(x,y).
Then
R(x, y)
is apolynomial
in yof degree
d.For the
proof
see[5].
Proof of theorem 3.1. From =
h o g,
we have :(i)
"~’ + ... + -hog
=0;
59
where
Dn(g) = g’n; Dn-1(g) = n(n-1) 2g’n-2g"; ...;D1(g) = g(n).
Ingeneral, D=(9)
is ahomogeneous
differentialpolynomial in g
ofdegree
i. Hence weapply
lemma 3.2. withFo
=-h, Fl
=f’,
...,Fn
= andho
= 1,h1
=Ds(g), hn
=Dn(g).
So there existpolynomials
...,P n,l (x)
which are not allequal
to 0 such that :pn,l(9)Dn(g)
+ ... + + = 0Then if
0,
we have :(2) Dn(g)
= +... +Rr,l9)D1 (9)
+R0,1(g)
where the
Rt,l(x)
are rational functions.Multiplying (2) by
andsubstructing
thisfrom
( 1 ) gives
us :(3) [f(n-1)(g)
+ + ... + +R1,1(g)f(n)(g)]D1(g)+
= o
Applying
lemma 3.2. onceagain
we get :(4) Dn-1(g)
=Rn-2,2(g)Dn-2(g)
+ ... +R1,2(g)DICg)
+R0,2(g)
If we suppose that lemma 3.2. does not break
down,
thisfinally gives :
(5) Dn(g)
=(g’)n
=R(g);
whereR(x)
is a rational function. In this case, weapply
lemma 3.3 and
get : (g’)"
= ao + alg + ... +amgm
with m d =2n;
But I (g’)n| I (r) =| amgm | I (r) ~ |gn|(r) rn;
j whichimplies
that m ~ n. If at some stage lemma 3.2 is notapplicable,
theprocedure
breaks down. But this means that one of the functionsoccuring
inequations
similar to(3)
vanishesidentically.
Hence we have :+... + + = 0 where
Ai(x)
ECp[x].
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M. Ozawa - Onuniquely factorizing
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[14]
C.C.Yang - Progress
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Lecture Notes in pure andApplied
Math. 78(1982),
171-192 .Jean-Paul Bezivin
Departement
deMathematiques
etMecanique
Universite de Caen
Esplanade
de la Paix14032 CAEN Cedex France.
bezivin@math.unicaen.fr Abdelbaki Boutabaa Université Blaise Pascal
Laboratoire de
Mathematiques
PuresComplexe scientifique
des Cezeaux63177 AUBIERE Cedex France.