R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
A LAIN E SCASSUT
The equation y
0= f y in C
pwhen f is quasi-invertible
Rendiconti del Seminario Matematico della Università di Padova, tome 86 (1991), p. 17-27
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Cp
ALAIN
ESCASSUT (*)
SUMMARY - Let K be a
complete algebraically
closed extension ofCp .
Let D be aclopen
bounded infraconnected set inK,
let H(D) be the Banachalgebra
ofthe
analytic
elements on D,let f E
H(D) and let be the space of the solu- tions of theequation y’ = fy
in H(D). We construct such a set Dprovided
witha T-filter lX such that there exists a
quasi-invertible f
E H(D) such that8(f)
has non zero
elements g
whichapproach
zeroalong T.
Inextending
this con-struction we show that for every t E N, we can make a set D and
an f E
H(D) such that8(f)
has dimension t. That answersquestions suggested
inprevious
articles.
I. Introduction and theorems.
Let K be an ultrametric
complete algebraically
closedfield,
of char-acteristic zero and residue
characteristic p #
0.Let D be an infraconnected bounded
clopen
set in K and letH(D)
bethe Banach
algebra
of theAnalytic
Elements on D(i. e. , H(D)
is thecompletion
of thealgebra R(D)
for the uniform convergence norm onD) [E1 , E2 , E3 , K1 , K2, R].
Recall that a set D in K is said to be infraconnected it for every a E D the
mapping a )
has animage
whose adherence in R is an inter-val ;
thenH(D)
has noidempotent
different from 0 and 1 is andonly
if Dis infraconnected
[E2]
On the otherhand,
an open set D is infraconnect- ed if andonly if f’ = 0 implies f = ct
for everyf E H(D) [E6 ].
Letf E H(D);
we denoteby 8(f)
the differentialequation y’ = fy (where
y E
H(D))
andby
the space of the solutions of8(f ).
In
[E 7 ]
we saw that has dimension 1 as soon as it contains (*) Indirizzo dell’A.: Universite Blaise Pascal (ClermontII), D6partement
de
Math6matiques Pures,
63177 Aubi6reC6dex,
France.a g invertible in
H(D).
IfH(D)
has no divisor of zero,8( f)
doesn’thave dimension
greater
than one.In
[E ]
we saw that if the residue characteristic of K is zero, thennever has dimension
greater
than one.But when the residue characteristic p is different from zero, in
[E ]
we saw that there does exist infraconnected
clopen
bounded sets with aT-filter
F[E4]
and an elementf
annulledby Y such
that the solutions of8( f )
are also annulledby M
Thanks to suchT-filters,
for every n E Nwe could construct infraconnected
clopen
bounded sets D withf E H(D)
such that
S(f)
has dimension n, and we even constructed sets D withf E H(D)
such that isisomorphic
to the space of the sequences of limit zero.Thus
[Eg ] suggested
that a situation where the solutions ofwere not invertible in
H(D)
should be associated to a nonquasi-invert-
ible element
f,
and so should be spacesS(f)
of dimensiongreater
thanone.
(Recall that f is
said to bequasi-invertible
inH(D)
if it factorizes in the formP(x) g(x)
where P is apolynomial
the zeros of which are in Dand g is an invertible element of
H(D)) [E i , E2 , E3 , E4 ].
Here we will prove this connection does not hold in
constructing
aninfraconnected
clopen
bounded set D with a T-filter Y and aquasi-in-
vertible element
f E H(D)
such that8(f)
has solutionsstrictly
annulledby M
Next,
for every fixedinteger t,
an extension of that construction willprovide
us with a set D and aquasi-invertible f E H(D)
such thatdim S(f) = t.
THEOREM 1. There exist an
infraconnected clopen
bounded set Dquasi-invertible
elementsf E H(D)
such that8(f)
has solutions
strictly
annulledby
T andS(f)
has dimension 1.More
precisely,
we willconcretely
construct such a set D andf E H(D)
inProposition
B.THEOREM 2. Let t E N. There exist an
infraconnected clopen
bounded set D and
quasi-invertible
elementsf E H(D)
such thatdim
(S( f ))
= t.Theorem 2 will also be proven
by
a concrete construction.REMARK. We are not able to construct an infraconnected
clopen
bounded set D with a
quasi-invertible f E H(D)
such that8(f)
has infinite dimension.By then,
thefollowing conjecture
seems to belikely.
CONJECTURE.
If f
isquasi-invertible, 8(f)
hasfinite
dimen-sion.
The
following Proposition
A will demonstrate Theorem 1by
show-ing
how to obtain the setD,
the T-filter~,
and the elementf.
PROPOSITION A. Let N be a sequence in d -
(0, 1)
such thatI bm I I bm + 1 I,
and let(Pm)m E
N be a sequenceof integers
in theform p
qmwhere qm is a sequence
of integers satisfying
Let R be ,
1,
and let D =d(O,
let
For each m E N*
Then the sequence converges in
H(D)
to a limit h that isstrictly
annulled
by
theincreasing Tlilter ff of
center 0of
diameter1,
andh E
S(,T).
00The
series Y- pj/(bm - x)
converges isH(D)
to a limitf quasi-in-
m=1 1
vertibLe in
H(D)
and h is a solutionof 8(f).
II. The
proof
ofProposition
AThe
proof
ofproposition
will use thefollowing
Lemma B.LEMMA B. Let q and n be two
integers
such that C n ~pq.
ThenPROOF. If n is a
multiple
of somep ,
thenpq - n
isobviously
mul-tiple of ph.
Let b thebijection
from{1, ..., n}
onto{(pq- n+ 1), ..., pq }
defined
by b( j) _ ~q - j +
1.By
the lastsentence, when j
is divisibleby ph, b( j + 1)
is also divisibleby p~
hence therefore!(p,-l)(p,-2)...(p~-~+l)!~!(~-l)!! I
andfinally IC;q
PROOF OF PROPOSITION A. Since we have Thus we can
easily
define a sequence ofintegers 1,,,
such that andWe
put
Then we haveAs the holes of D are in the form d - it is
easily
seenthat
Let us consider We have
and in the same way hence
Now let us consider
hm+ 1 (x) - (x) when I x ~ Ibm
and let usput
Then it is clear that
lu(x)1 I
and then forwe obtain
by
Lemma B.we see that and then
every term is upper bounded
by
therefore
u(x) ~ ~
whenever x E D nbm 1).
Finally by (6)
we see that hence thesequence
hm
converges inH(D)
to theconvergent
infiniteproduct
By (3)
andby
the definition of D it iseasily
seen that theincreasing
filter T of center
0,
of diameter1,
is a T-filter and it is theonly
one T-filter on
D [E4 ].
On the other
hand, by (5)
we have wheneverx
E DBd - (o, ~ Ibm I)
and therefore h isclearly
annulledby £
and it isstrictly
annulledby T (because .1-r
is theonly
T-filter onD),
andh(x)
= 0whenever X E hence h E
ðo (Y).
Now let us consider the series Since
(4)
we see that series series converge to alimit f E H(D). Moreover,
it iseasily
seen that forevery j
EN* , hence, by (2),
we have , hence
f
is not annulledby M
Since Y is the
only T-filter, f
is thenquasi-invertible.
At
last,
weshortly verify
that h is solution of8(f ).
By Corollary of [E6 ]
we know thath’ E ,. H(D)
and the sequenceh:n
converges to h’ in
H(D)’ .
On the otherhand,
it iseasily
seen thathence
and therefore h is a solution of
8(f),
and that ends theproof
ofProposi-
tion A.
III. The
proof
of Theorem 2.LEMMA C. Let q, n be
integers
such that 0 n q. Then|q!/n!| p1-(q-n)/p.
PROOF.
q! / n! has q - n
consecutive factors. It iseasily
seen amongthese q - n factors,
the number of them that aremultiple
of p, is at leastInt (q - n)/p)
and thereforev(q! /n!) * Int ((q - n) /p) > (q -
-
n)/p - 1
and that ends theproof
of Lemma C.LEMMA D. Let R ~ I and let
be a Laurent series
convergent for I x
=R,
such that supan IR n
== with
q 0. Then p
does notsatisfy
theinequality
PROOF. We suppose rp satisfies
(1)
and weput By (1)
it iseasily
seen thatIf
q = -1,
relation(2) hence g = 0.
Wewill suppose
q -1
and we will prove that(3) 1 a,,, =
- q -1)!| 1 for n = q + l, q + 2, ..., -2, -1. Indeed,
suppose it has been proven up to therange t with q t -1
and let us prove it at therange t
+ 1.By (2)
we havehence
by (3)
Now
by
Lemma C we know that( - q) ! /( - t) ! ~ ~ 1- ~t - q~~P .
SinceR > ~ro -1 ~p ,
we see thatR t - q > ~ -~t - q)lP ;
henceand Then
by
relation(4)
we haveand therefore
so that relation
(3)
is proven at therange t
+ 1. It is then proven for every n up to -1. Then relation(2)
for n = 0gives
ushence
by (3)
we haveI
and therefore
but we know that hence
(6)
isimpossible.
Lemma D is then proven.
The
following
lemma wasgiven in[55],
inconstructing
the«Produits Bicroulants»
(twice collapsing meromorphic products).
LEMMA E. Let
p, R ’ , R ", R
ER +
with 0 R ’ R " R . There existsequences and with
such
that, if we
denoteby
has an
element p
EH(D) satisfying
PROOF OF THEOREM 2. Let c~1, ..., wt be
points
ind(o,1)
such that= 1 whenever i
=1= j.
Let r E]o,1[
and let(bm)m E N
be asequence in
d - (o, t)
such thatI bm | bm + 1 | 1
and lim = r and letN be a sequence
of integers
such that q1 qm for all m >1,
and Let let
~m =
pom
and letIt is
easily
seen that A admitsa T sequence (Tm , qm ) [S1 ].
Let T bethe
increasing
T-filter of center0,
of diameter r on A. First we will con-struct an infraconnected
clopen
set included ind(o,1),
of diameter1, satisfying
thefollowing
conditions:(1)
(2) 0
has anincreasing
T-filter f1 of center0,
of diameter 1.(3) D
has adecreasing T-filter g
of center0,
of diameterR E ]r, 1[.
(4)
Theonly
T-filters of 0 are1Q ~ g.
(5)
Thereexists p
such thatand
Let
By
Lemma E there exist sequences( ~n )n E N
inr(0, R,1 )
such thatand such that the set
defines an
algebra
that containselements p satisfying
= 1 forg~(r)
= 0 forIxl
= 1. Let usput § =1- ~
and let Q be the setA u
(A "’-d - (0, r)).
Q has
clearly
three T-filter:the filter ~’ on A
the
increasing
filter T of center0,
of diameter 1 thatstrictly
an-nulls p.
the
decreasing filter G
of center0,
of diameter R thatstrictly
an-nulls
Y.
It is
easily
seen these three T-filters are theonly
T-filterson 0,
andQ, p~ ~
are then defined.Then
f1 (x)
=fix)
when x E Q nd(O, R)
andf1 (x)
= 1 whenx E
S2Bd- (0,1).
We can deducethatf1
is aquasi-invertible
element inH(Q). Indeed, by Proposition B, f is
not annulledby T and by f1,
hencef1
is not annulledby 1F
andby either;
on the otherhand,
asf, (x)
= 1when I x
=1, flis
not annulledby ~;
hencefi
is not annulledby
any oneof the three T-filters on S2 so that it is
quasi-invertible
inBy Proposition
B~( fl )
has a solutionNow,
for each let and letdefined
by. In Qj
theequation
has a sol-ution gj
definedby
Let and letf(x) =
Obviously, f(x) when Ix - Cùj
1and f(x)
= 1when Iç- - wll
= 1 forevery 1
=1,
..., t. Each one ofthe.fj
isquasi-invert-
ible in
H(D)
so thatf
is alsoquasi-invertible.
Now
each gj (1 ; j t)
is a solution of8(f). Indeed,
when|x -
1 we have
gj’ (x) _ , f (x)
= andwhen x - Wj
1=1,
= 0.
On the other
hand,
the gjclearly
havesupports
twoby
twodisjoint- ed,
hencethey
arelinearly independent,
and that shows has di-mension : t.
We will end the
proof
inshowing
that{gl , ... , gt ~ generates S(f).
Log
will denote the reallogarithm
function of base p. Let v be the valuation defined in Kby I
when r # 0 andv(o) _
+ 00.When A is an infraconnected set
containing 0, and f E H(A)
weput
For
each j =1, ..., t,
letDj
= d -1)
n Dand Bj
= d -R);
let D’ =By
definitionof f
we see thatf(x)
= 1 for all x E D’ andd -
(~, 1)
c D’ for every « E D’. Then it is well known that theequation y’ = y
has no solution y inH(d - (~, 1»
but the zero solution. Leth E
s( f ).
For every a ED’,
the restriction of h to d -(a, 1)
is a solution of theequation y’ = y
thatbelongs
toH(d - (a, 1))
hence we see thath(x)
= 0 for all X E D’. Since D’ isequal
to i we seethat
Now let us consider
h(x)
when x EB1.
Since
D, = 0
n d -(0, 1)
the three T-filters1Q £ g of Q are
secant toD1
andthey
are theonly
T-filters onD1.
Then J is theonly
one T-filteron
B1
because Fand S
are not secant tod(O, R).
Thealgebra H(B1 )
hasno divisor of zero. Consider the
restriction/i
of toD1
and the restrictionf,
toB1.
InH(B1 )
the space8( fl )
has dimension oneby
Theorem 3of [E7 ],
hence there existsÀ1 E k
such thath(x) = À1g1 (x)
wheneverx E B1.
Since gl E
ðo (9),
thatimplies h(x)
= 0 whenever x Er(0,
r,R)
hencev(h, -log R) _
+ 00 . We will deduce thatv(h, ,u)
= + 00 whenever[0, -log R].
Indeed,
suppose this is not true. Then h isstrictly
annulledby
anincreasing
T-filter of center0,
ofdiameter > R,
hence h isstrictly
an-nulled
by F.
Since there exists s E]R 1[
such that
On the other
hand,
it iseasily
seen thath(x)
isequal
to a Laurent ser-ies in each annulus
r(0j&~~+i~)
and for every s 1 there exist intervals[r’ , r" ] c ]s,1 [
such that the functionv(h, IA)
isstrictly
decreas-ing
in[ - log r" , -logr’]
and such thath(x)
isequal
to a Laurentseries since
v(h, g)
isstrictly decreasing
in[ - log r" , -log
r’]
thereexists q
0 such that = supan Ipn.
Thenn E Z
h satisfies the
hypothesis
of Lemma D and relation(7)
isimpossible.
But then = +00 for every g E
[0, -log r]
It follows thath(x)
= 0 for every xE r(O, R,1)
because if there existed apoint
aE r(O, R, 1)
with0,
« should be the center of anincreasing
T-filter that would annull h but theunique
T-filter of center a is F and we havejust
seenthat F does not annull h.
Thus we have now proven that
h(x)
= 0 for all X EB1
such that1. Since
gl (x)
= 0 whenever xE r(0, r, 1),
the relationh(x) _
=
xi gi (x)
is then true in allB1.
In the same way, foreach j
=2, ... , t,
wecan show there
exists Aj
E K such thath(x) = Àjgj (x)
for every X EBj
andthen is true in , and of course in
D’,
hence it is true in all D. That finishesproving {gl , ..., gt }
is a base ofS( f ).
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B.,
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