R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
G. D E M ARCO
On the algebraic compactness of some quotients of product groups
Rendiconti del Seminario Matematico della Università di Padova, tome 53 (1975), p. 329-333
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Algebraic Compactness
of Some Quotients of Product Groups.
G. DE MARCO
(*)
It is well known
that, given
a countablefamily (Gn)neN of
abeliangroups, y the group is
algebraically compact (see [B]
and[H]).
This result has beengeneralized by
Fuchs([F],
Theorem42.1).
Fuchs’s
generalization suggests
thefollowing enquiry
about the si- tuation which arises when the «countability » assumption
is removed from thehypotheses.
In this direction O. Gerstner hasproved
thatthe direct
product
ofinfinitely
manycopies
of theintegers
modulothe
corresponding
direct sum isalgebraically compact only
when theindex set is countable
([G],
Satz2).
This result shall be extended here to
arbitrary
nonalgebraically compact
groups. Related theorems shall also be obtained.Let us fix some notations. Let X be a
set, q
a(perhaps improper)
filter on
X,
A an abelian group; denoteby
AX the group of all functions from X toA,
withpointwise addition, by
thesubgroup
of allfunctions in AX which vanish on some element
of 92 (hence E,(92)
= Agiff T
isimproper), by AT
the groupAlso,
letT*
be the filtergenerated by
the intersections of countable subsets of 99. Thusg*
is proper
iff 92
has the countable intersectionproperty (c.i.p.).
thatis,
every countable subset
of g
hasnon-empty
intersection.Moreover, [F].
Theorem
42.1,
shows that isalgebraically compact.
THEOREM 1. Let 1p
be afilter
thatproperly
containsis
algebraically compact,
then A isalgebraically compacct.
(*) Indirizzo
dell’A.: Istituto di MatematicaApplicata,
via Belzoni3~
35100 Padova
(Italy).
330
PROOF. Let
be a
system
of linearequations
inA,
and assume that every finite sub-system
of(1)
is solvable in A. Pick V E and let oci denote thefunction which is ai on and 0 on
V,
andput ai =
oci + In this way we obtainlinear
system
in~~ ( y~ ) /~d ( g~ ) . Clearly (2)
isfinitely solvable,
and hencesolvable, by
thehypothesis.
Then there exist(yeN)
suchthat the sets
belong to q;
for everyi E N. Since is not contained in V.
Take ~
EZO’V.
Then is a solution of
( 1 )
in A.Plainly,
allsubgroups
are pure in All. Then :COROLLARY
1. Let 99
be afilter
onX,
with the countable intersectionproperty.
ThenA~ =
isalgebraically compact iff
A isalge- braically compact.
Observe that the direct sum A(Z) is the group where is the filter of cofinite subsets of X. Then :
COROLLARY 2. Let X be uncountable. Then is
algebraically compact iff
A isalgebraically compact.
Stronger
results may be obtained for certain torsion-free groups.First we need a
simple
lemma.LEMMA. Let 99 be a
filter
onX,
closed under countableintersection,
p a
prime.
Thenpro A
= 0(resp.
A 1=0 ) implies
0(resp:
A~
=0).
.PROOF. Assume that
f 0
is of infinitep-height
moduloE,(92).
Then for every kEN there exists
f k
E Ax such thatZk
={0153
E X :f (x)
==Thus Since
f 0 -r,(92),
thereis ~
EZo
such that
f (~) ~ 0. Clearly
Theproof
for A 1 is almost identical.REMARK.
If g
is not closed under countableintersection,
and Ais
unbounded,
thenA; ¥=
0. Thus thehypothesis
on g cannot be weakened.THEOREM 2. Let A be a
(necessarily torsion- f ree)
group such thatpwA
== qW A
= 0for at
least twodifferent primes
p,q. Let
g~ be afilter
on X.Then every
algebraically compact subgroup o f AZI1:Â(lfJ)
is con-tained in which is then the
largest algebraically compact subgroup of A9,.
PROOF. What we have to prove amounts to show that
A-,*
= contains no non-zero
algebraically compact subgroup.
And in
fact, by
thelemma, 0,
whichimplies
that nonon-zero r-adic module can be a
subgroup
ofA~* ,
for noprime
r.THEOREM 3. Let A be a countable reduced torsion
free group. Let
99 be afilter
on X. Then thelargest algebraically compact subgroup of A9,
isPROOF.
By
the lemma is reduced. SinceA,*
isclearly
torsion-free
(unless
it is0)
we have to prove thatA,*
contains no copy of thep-adic integers,
for noprime
p. Assume thecontrary,
and let C bea copy of
J~ (p-adic integers)
contained inA* ;
C is thecompletion
in the
p-adic topology
of some infinitecyclic subgroup 0153
+Z~(g*)),
Let aI, a2 , ... be an enumeration of the non-zero elements in the range of a. Put
V’i = Then Zo = ( (XBY$)
= sinceSince
g*
is closed under countableintersection, g*
for
some j.
Put V’ =Y~ ,
a = a~ . We prove that thehomomorphism
(o: Z - A
given by co(n)
= na may be extended to anhomomorphism (J
= z-adicintegers),
which isimpossible,
since A is coun-table reduced torsion free. Take a sequence
(xn )
in Z such that Xn+l - E n ! Z for every n E N. Then + is aCauchy
sequence in the z-adictopology
ofC,
which coincides with thep-adic topology
of C. Hence there exist
f E Ag
and such thatf + -~-- ~’d ( ~* ) E C,
for all and such that the setsbelong
tog*
for every n e N. Thus and hence W f1n
0,
sincePick ~
e W r1 V. We havef (~)
- xna =332
=
n!g~(~)
for every n E N. Thenf (~)
is the limit of(xna)
in the z-adictopology
of A.Letting {jj (lim xn )
= lim xn a, we obtain therequired
extension.
Finally,
let A denote anarbitrary
abelian group, 99 a filter on X. Since is a purealgebraically compact subgroup
ofA~,
it is a directsummandofa
and itscomplements
areIn
seeking
informations on thealgebraically compact subgroups
ofwe first observe that
cp*
may be assumed to befree,
i. e.n gg*
_0.
For, otherwise, Alp.
is the direct sum ofA8,
with and ofthe group where y =
(XBS) :
and 1p is closed under countable intersection and free(perhaps improper).
If
T*
is a freefilter,
closed under countableintersection, then,
under the very mild restriction that the
cardinality
of X be non-mea-surable,
the groupAlp*
containsproducts
ofinfinitely
manycopies
ofA,
as we shall now prove.
Thus,
if A hastorsion, Alp*
contains directproducts
ofinfinitely
many finitecyclic
groups, and even unbounded such directproducts,
if the torsionsubgroup
of A is unbounded.PROPOSITION. Let X be a set
of
non measurablecardinality, 99 a free filter
onX,
closed under countable intersection.Then
Alp
contains asubgroup isomorphic
to the directproduct of
coun-tably
manycopies of
A.PROOF.
By
thehypotheses 99
is not an ultrafilter. Then there existsX 1
c X such that neitherXi
norX’XI belong
to 99.Thugs 92
o traces »on a free filter 99,; and 99, is still closed under countable inter-
section,
and free.Again,
there existsX2
c such that neitherX2,
nor is in gg,.
By induction,
we obtain a countablefamily Xi , X2 , ...
ofpairwise disjoint
subsets ofX,
suchthat 99
traces on all ofthem. The
subgroup G
of AXconsisting
of all functions which are con-stant on every
X
and 0 onXi)
isclearly isomorphic
toAN.
i
And since
ZnXi=l=0
forevery Z
E cp and i eN, G
n-r,(99)
= 0.REFERENCES
[B]
S. BALCERZYK, Onfactor
groupsof
somesubgroups of
acomplete
directsum
of infinite cyclic
groups, Bull. Acad. Polon. Sci. Math. Astr.Phys. ,
7
(1959),
pp. 141-142.[F]
L. FUCHS,Infinite
abelian groups, vol. I, New York (1970).[G]
O. GERSTNER,Algebraische Kompaktheit
beiFaktorgruppen
vonGruppen ganzzahliger abbildungen, Manuscripta
Mathematica, 11 (1974), pp. 103-109.[H]
A. HULANICKI, The structureof
thefactor
groupof
an unrestricted sumby the
restricted sumof
abelian groups, Bull. Acad. Pohin.Sci.,
10 (1962), pp. 77-80.Manoscritto