R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
L UIGI S ALCE
P AOLO Z ANARDO
Some cardinal invariants for valuation domains
Rendiconti del Seminario Matematico della Università di Padova, tome 74 (1985), p. 205-217
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LUIGI SALCE - PAOLO ZANARDO
Introduction.
The condition on a valuation domain h of
being maximal,
whichgoes back to Krull
[8],
and the very close condition ofbeing
almostmaximal,
due toKaplansky [7],
wereextensively investigated by
many
authors,
on account of theirimportance
for the consequencesderiving
for thering
structure of .R and for many classes of R-modules.On the
contrary,
theproblem
ofmeasuring
in some way the fai- lure of themaximality
did not receive too much attention up to now;the
only
contributions in this direction knownby
the authors areby
Brandal
[1],
andby
Facchini and Vamos[2].
As any valuation domain .R is a
subring
of a maximal valuation domain~S,
which is an immediate extension ofit,
it is natural totry
to measure the failure of the
maximality
for .I~by looking
for cardinalinvariants which measure,
roughly speaking,
howlarge
is S over .R.In this paper,
given
any ideal I ofR,
we will introduce two cardi- nal invariants associated with I : thecompletion defect at I,
denotedby
and the totaldefect
atI,
denotedby
theirdefinition,
which seems very
technical,
raised upnaturally
in theinvestigation
of
indecomposable finitely generated
modules in[13].
The
completion
defect measures howlarge
is the com-(*)
Indirizzodegli
AA.: L. SALCE : Istituto di Matematica, Informatica eSistemistica, Università di Udine; P. ZANARDO: Istituto di
Algebra
e Geome- tria, Università di Padova.Lavoro svolto con il contributo del Ministero della Pubblica Istruzione
e nell’ambito del GNSAGA del C.N.R.
pletion
ofRaj1
in the idealtopology,
and the total defect measu-res how
large
is8jI8
overR/I;
recall that both and arecontained in up to canonical
isomorphisms.
It is
noteworthy
that the invariants that we areintroducing
donot
depend
on thering
structure which is notunique
up to iso-morphism,
butonly
on the R-module structure ofS,
which is a pure-injective
hull of .R(see [7]
and[12]).
In the first section we introduce the breadth ideal
(of
non maxima-Zity)
of the valuation domainR,
aconcept originally
due to Bran-dal
[1],
and the breadth ideal o a unit ofS,
aconcept
defined in asligh- tly
different wayby Ostrowsky [11], Kaplansky [7]
and Nishi[10].
The breadth ideals of the units of S are used in the second section to define the
completion
defect and the total defect at an ideal I of .R.The main result in this section is an
inequality
which relates the total defect at an ideal I with thecompletion
defects at the idealscontaining
I.This
inequality
however is ingeneral strict,
as isshown,
for aspecial
class of discrete valuation
domains, by
Facchini and the second author in[3].
In section 3 we compare the total defect at an ideal I with the Goldie dimension of
8jI8
as anR/I-module; they
turn out to beequal
if I is a
prime ideal,
while in thenon-prime
case the total defect becomesgenerally larger.
We remark that the invariants of the valuation domain .R that
we
investigate
hereplay a
relevant role in thestudy
of many classes of R-modules: besidesfinitely generated
modules(see [13]),
theR-modules
JSIIS,
where I > J are fractional ideals ofR(see [4]);
indecomposable injective
R-modules(see [2])
and torsion-free R- modules of finire rank(see [5]).
1. The breadth.
I~ will
always
denote a valuationdomain,
P its maximal ideal andQ
its field ofquotients.
Recall that .R is maximal if it islinearly compact (in
the discretetopology);
.R is almost maximal if every proper factor of it islinearly compact.
A valuation domain 8
containing
.R as asubring
is an immediateextension of .R if
(i)
every ideal of S is of the formIS,
where I is an ideal of.R,
and
(ii) S/PS
isnaturally isomorphic
to.R/P
or,equivalently, S = PS -~- .R.
An immediate
extension 8
of R is maximalif, given
any valuation domain S’containing S
as a propersubring,
either(i)
or(ii)
fails for S’.It is well known
(see [7]
or[12])
that every valuation domain is contained in a maximal immediateextension S,
which is a maxi-mal valuation domain. However S is
uniquely
determined up toR-isomorphism only,
and not as aring,
unless R is almostmaximal,
in which case S is the
completion
of .R(in
the valuationtolopogy).
.R coincides with S if and
only
if it is maximal.Brandal considered in
[1]
thefamily
of ideals of Rand he showed that either Y =
W,
or there exists aprime
ideal Zof I~ such that
Fixed a maximal immediate extension S of
.R,
we reformulate this resultby introducing
thefollowing
subset of.R,
called the breadtho f
I~(with
a moremeaningful
term we could call it the breadthof
nonmaximality of R):
Notice that
B(.R) _ 0
whenever S = R for all a thishappens exactly
if ~‘ =.R,
i.e. if .R ismaximal;
thus from now on weshall assume that R is a valuation domain not
maximal,
soB(R)
isan ideal of R.
PROPOSITION 1.1. Let .R be a valuation domain not maximal.
Then its breadth
B(R)
is aprime
ideal of R such that:PROOF. Assume
that a, b
Then ~S = .R-~-
aS = 1~+
+ bs,
and bS =+ as) implies S
= .R+ + aS)
= R+
Thereforeab 0 B(R),
so thatB(.R)
isprime.
If
.R/I
islinearly compact,
then8118
in a natural way, therefore S = .R+ IS;
thus It follows thatB(R)
is contained inr’1 f I: R/I
islinearly compact}. Conversely,
ifI > B(R),
thenS = .R -~- IS,
thus8118
islinearly compact. Being B(R) prime,
eitherB(R)
=P,
in which case the firstequality
istrivial,
or is the intersection of the ideals
properly containing it,
thusthe first
equality
is obvious. The secondequality
can beproved
ina similar way.
///
From
Proposition
1.1 it follows that does notdepend
on thechoice of
S,
and that it coincides with the ideal Lquoted
in the Bran-dal’s result. Notice that .R is almost maximal
(and
notmaximal)
if and
only
ifB(R)
= 0.The valuation domain
.R/B(.R)
isalways
almostmaximal;
Bran-dal
gives examples
in[1] showing
that can be maximal or not.Let us denote
by U(S)
and( U.R) respectively
themultiplicative
groups of the units of S and 1~.
Every
0 ~ x e S can be written ina
unique
way, up to units ofE,
in the form x = Er, with 8 EU(S)
and r E
R; by
this reason we will confine ourselves to consider units of S in thefollowing
discussion.Given any E E consider the ideal of
I~,
called the breadthof
8REMARK. Our definition of breadth of a unit of S is
essentially
the same as the one
given by
Nishi[10],
which is aslight
modifica-tion of the
original
definition of breadth of apseudoconvergent
setof elements of .l~
given by Kaplansky [7],
andoriginally
due to Ostrow-sky [11].
The definition of breadth(of
nonmaximality)
of R isoriginated by
the two above definitions.From the definitions of and
B(E)
ittrivially
follows thatB(E) ~ B(R). Conversely,
let a EB(.R) ;
then S > R+ as,
thus thereexists 8 E such
that E E R -E- aS,
therefore aeB(E);
we haveproved
PROPOSITION 1.2. Let .R be a valuation domain not maximal.
Then
B(R)
= u~B(E) :
E eThe
following
result will be useful in the nextsection;
it is similarto
[10, Prop. 6].
LEMMA 1.3. Let R be a valuation domain not maximale and e E E If 1t E
U(R)
and 0 ~ r EP,
thenB(u + ~~~) - rB(,e).
PROOF. if and
only
if and thisobviously
isequivalent
to u+
R+
raS.///
We introduce the
following
notation:given
letbe the canonical
surjection;
then theimage fIll
of R is asubring
ofSIIS isomorphic
to itscompletion,
wheneverR/I
is
Hausdorff,
is denotedby
Noticethat, being
pure in and8118 complete,
we have thefollowing
inclusions:The
topology
consideredabove,
as in thefollowing proposition,
on the factor
ring
is the idealtopology »,
which has as a basisof
neighborhoods
of 0 the ideals a EBBI.
PROPOSITION 1.4. Let R be a valuation
domain,
and Then.R/.I
is Hausdorff and noncomplete
if andonly
if I =B(E)
for someE E
PROOF. In order to show that is
Hausdorff,
it isenough
to prove
B(E) implies pa 0 B(E)
for somep E P.
So let E EER+aB;
thenBut ~S = R -~- PS implies
that s
= t + ps’,
for some t ER, p E P
and s’ ES ;
therefore weget:
E = r
--~- at + aps’ E
R+ paS,
as we want.Clearly
E+ B(E) S 0 0 fB(,,)R,
but it is the limit of aCauchy
net of elements offor,
given implies
that there exists suchthat E - ur E
rS;
thus E-E-- B(E) S
is the limit of theCauchy
net{Ur +
So we have
proved
that is notcomplete.
Conversely, assuming
thatR/I
is Hausdorff and notcomplete,
from the inclusions
(1)
weget
an element E E such that cis the limit of a
Cauchy
net-~-
IS :r ~ l)
inf I R.
So E+
rSif and
only
if therefore1 = B(E). 1/1
From the
proof
of thepreceding proposition
we deduce thefollowing
COROLLARY 1.5.
and I R. Then ê + IS E
E
/7.R
if andonly
if if andonly
ifA
particular
case is whenI = 0 ;
then the elements of thecomple-
tions
jS
of .R areexactly
those x = E EU(S))
suchthat
B(E)
= 0. It was shownby
Nishi[10]
that1~
is the centerZ(A)
of the
ring
A =EndR E(RjP),
which isisomorphic
to sowe have the inclusions:
2. The
completion
defect and the total defect.We introduce now a new
concept,
which firstappeared
in[13].
Let .R be a valuation domain not
maximal,
and S a maximal imme- diate extension of1~;
let 8,, ..., En EU(S)
andI P;
we say that theEi’s
areu-independent
over I ifimplies
for all i.Conversely,
if(2)
holds for someU(.R)
the
Eils
are saidu-dependent
over I.LEMMA 2.1.
(i) lT(S)ER,
then 8 isu-independent
overB(E).
(ii)
If ~1, ... ,U(S)
areu-independent
overI P,
thenPROOF.
(i)
If then if andonly
ifal ft P and,
in this case, 8 E-~- .R,
which is absurd.(ii) If,
forsome j, 8j E R,
then(2)
holds with ao = E~,aj = -1 and a,
= 0 for0 =1= i
~j.
Assume now thatB(Ej)
I for somej.
Then Ej
+ IS
= u+
IS for some u EU(l~),
so(2)
holds with ao = u, aj = -1and ai = 0
for0 # i
#j. ///
We say that a
family
of units of ~S not in .R is u-inde-pendent
over an idealI C P,
if any finite subset of it isu-indipendente
over
I;
so theu-independence
is aproperty
of finitecharacter,
andmaximal families of units with this
property
do exist.Having
fixed the ideal we consider all the families of units of S which areu-independent
overI,
such that = Ifor all Let be the minimal cardinal such that >
+ I
for all these families.’
We call the
completion defect of R at I ; clearly
is an inva-riant of R not
depending
on the choice of~S, being
theu-indepen-
dence defined
by linearity.
Obviously
B is almost maximal if andonly
if = 1 for all non-zero ideals I.
The
following
result compares thecompletion
defects at isomor-phic
ideals.PROPOSITION 2.2. Let
I ~ J
beisomorphic
ideals of R contained in P. Then =c~(J).
PROOF. It is
enough
to showthat, given
afamily (si :
which is
u-independent
over1,
where I = for all 2 EIl.,
thereexists a
family
which isit-independent
overJ,
whereJ = for all h
Being
there exists a such that either J = aI oraJ = I ;
we can assume otherwise J = I and the claim is trivial. IfJ = aI,
let qz =1-~
asi for all ThenB(i7,a)
= J for E ll follows from Lemma 1.3. Assume nowthat
then
implies
thatthus, by
theu-independence
of thecAl s,
we deduce that/ n B
and
+ j£
ia2
EP;
it follows that au E P too.Conversely,
assume thatI (a
EP).
Noticethat a 0 1,
becauseJ c P. Being
=I,
there exists anu1
E such that ~a, ==-
u1 -f-
aqi for some r¡;. E S. Without loss ofgenerality,
we can assumethat qi e
U(S) : for,
if ePS,
substituteu~ and qi respectively by
E
U(R)
and 1+ r¡;.
EU(S).
From aJ = I = and fromLemma
1.3,
we deduce that aJ =aB(r¡;.),
so J. Assumenow that
Then
recalling
thatu’, +
=(1 c i ; n),
and that the are u-n
independent
overI,
it follows that acl, ... , an EP;
then an-E- ~
e Jsimplies
ao c P too.///
~Given an ideal
I P,
we introduce now anotherinvariant;
weconsider all the families of units of 8 as in the definition of but we assume
only
that theEa,’s
areu-independent
overI,
without
assuming
that = I for allthus, by
Lemma2.1,
we
only
know that for all A Letdn(1)
be the minimalcardinal such that
IAI -f-
1 for all these families. We calldR(l)
the total
defect of
.R atI ;
here too we notice that is an inavriant of .R notdepending
on the choice of S.The
following
result is an immediate consequence of the defini- tion.LEMMA 2.3.
(i)
If then(ii) dn(l)
= 1 if andonly
if I >B(R)
or I =B(R)
andR/B(R)
is
complete.
(iii)
.R is almost maximal if andonly
if = 1 for everyI ~ 0. ///
Given an R-module M and an ideal
I ~ P,
we say that the elements... , xn E M are
linearly independent
over I if x.-~- 1M,
... , xn-f -
I ~are
linearly independent
elements of theR/I-module
i.e. ifn
aixi e IM(a;
E.R) implies
that ai e I for all i.Obviously
one can1
extend this definition to a
family
of elements of M.Recall
that,
if istorsion-free,
then the rankrkR
if of is thedimension of the
Q-vector
spacewhere Q
is the field of quo- tients ofR,
or,equivalently,
thecardinality
of a maximalsystem
of
linearly independent
elements of M.PROPOSITION 2.4. Let .R be a valuation domain and I a
prime
ideal of .R. Then = and =
PROOF. Given a
family
of elements of whichare
linearly independent
overI,
one can assume, without loss of genera-lity,
thatU(S)
for all and that one ofthem,
say is 1.It follows
trivially
from the definition that(zi: ),
~1}
is afamily
of elements
u-independent
overI,
and Iby Corollary 1.5;
theref ore In a similar way one can see that
rkR/ISjIS
Conversely,
y to prove that(respectively
it is
enough
to showthat, given
afamily fez:
A EA)
ofunits of S with = I for all
(resp.
with whichare
u-independent
overI,
then11,
arelinearly independent
over I. Assume that
if
some ai ft I, let a;
be one of the coefficients not in I with minimal value.By multiplying by c~~ 1,
weget
because
I;
the last relation isabsurd,
because the coefficienti of isequal
to1,
which contradicts theu-independence
of{8A: 2
over I.
///
Vve wish to compare now the total defect at the ideal I with the
completion
defectscR(J)
at the idealsJ ~ I.
LEMMA 2.5. For
every i
=1, ... , n,
let.Ei
be afamily
of unitsof S
u-independent
over the idealsJi,
such thatJi = B(e)
for alls E
Ea .
IfJ1
>J2
> ... >Jn,
and theJ/s
arepairwise
non isomor-phic,
then isu-independent
overJn .
PROOF. We induct on n, the claim
being
trivial for n = 1.So,
assume that n > 1 and that is
u-independent
overJt,
for Letwhere
(0~~)~
for
1 j ~ ~
andnhEEn
forFirst,
notice that ai , ... ,for,
letthen,
for all there existsU(R)
suchthat nh
- vh E
rS. It follows thatand the
u-independence
of the8j’s implies
that a1, ... , Recallnow that is one of the
Ji’s,
for1 c i c n -1,
forall y*;
we willshow that
Being Jn
notisomorphic
toJ1, ... , J n-l, ajB( Bj)
~Jn
forall j ;
letLet for
all j E Al
we can choose an ele-ment uy
EU(.R)
such thatsubstituting
in(3),
weget:
where the sums with indexes in
A2
ate intended to be 0 when-ever
Ai
orÂ2
is void. ,Assume now that
A~ ~ ~b.
Let and choose foriEA2
all
U(.R)
such thatsubstituting
in(5)
weget :
we will show that
(6)
is absurd. Letjo E A,
be such that has minimalvalue among the
aj’s with j E Å2;
thennotice that in
(7) a701aj E R
forall j E A2,
and thattherefore also the first summand in
(7)
is in .R. But then(7)
isabsurd,
because the coefficient of 8io is 1 and
by
the inductivehypothesis.
Thuswe have
proved
thatA2
=0,
therefore(5)
becomessimply:
by
theu-independence
of ther¡h’S
overJn, (8) gives
thatbl,
... ,bm
EP;
being
from(3)
it follows thatao E P. ///
Given an ideal
J R,
let[J]
denote theisomorphism
class ofJ;
if
I ~ .R
is anotherideal,
then[J] ~ I
means that there exists J’c-[J]
such that
J’ ~ I. By Proposition 2.2,
we can define as the commonvalue where J’ ranges over
[J].
We caneasily
obtain from thepreceding
Lemma 2.5 thefollowing
PROPOSITION 2.6.
The
inequality
inProposition
2.6 is ingeneral strict,
as is shownby
Facchini and the second author in[3]; actually, they
prove a multi-plicative
formularelating
and thecR[J]’s, [J]>I,
for I aprime
ideal of a discrete valuation domain R with
Spec R
well orderedby
the
opposite inclusion; they
alsogive
a realization theorem for these domains withpreassigned completion defects, using
an idea ofNagata [9].
3. - Total defect and Goldie dimension.
Let I be an ideal of the valuation domain
.I~,
and letgR(I )
denotethe Goldie dimension of
S/IS
as anR/I-module.
If I is aprime ideal,
then
gR(I)
=rkRIISjIS.
It follows from the definitionsthat,
for anarbitrary
idealI, g(I) ~ dR(I ),
andProposition
2.4 shows that thisinequality
becomes anequality
if I is aprime
ideal.Recall
that,
if 0 ~ then the subset of Ris a
prime ideal,
which is the union of all the ideals Risomorphic
to I
(see [10]
and[4]).
If I ==0,
we setIll
== 0.LEMMA 3.1. Given any
I R,
PROOF. It is
enough
to showthat, given
E,, ...,~eU(S), they
are
linearly independent
overI#
if andonly
ifthey
arelinearly
inde-pendent
over I.So,
assume thatthey
arelinearly independent over I#
n
and If some let be such that
v(a)
=1
Then
because is an ideal
isomorphic
to1,
hence But thisis a
contradiction,
because some is a unit.Conversely,
let ~i,...,~ belinearly independent
over I and letit
If then for
all
hence for1
all i. If then 0153 == rq,
where q
e andBeing 1#
the union of the ideals
isomorphic
to1,
there exists an ideal such that J=7 for some t c P and r c J. Then tr e1,
thereforen
the
independence
of the over Iimplies
that tai E1
e7=J,
hence for all i.ll/
Recall that an ideal I R is archimedean if
1~
= P. As an imme- diate consequence of thepreceding
lemma weget
COROLLARY 3.2. Given two ideals
7~J,
thenmoreover
gR(I)
= 1 if I is archimedean.PROOF. The first claim follows from the
equality 1#
==J’ ;
the secondequality
follows from theisomorphism ///
Lemma 3.1 and
Proposition
2.4give
thefollowing
COROLLARY 3.4. Given any ideal
COROLLARY 3.5. Given any ideal I
R,
then = 1 if andonly
if either1#
>B(.R),
or1#
=B(R)
andR/B(R)
iscomplete. ///
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L. FUCHS - G. VILJOEN, Onfinite
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Manoscritto