R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
A LBERTO F ACCHINI
On the ring of quotients of a noetherian commutative ring with respect to the Dickson topology
Rendiconti del Seminario Matematico della Università di Padova, tome 62 (1980), p. 233-243
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Ring
Ring with Respect
tothe Dickson Topology.
ALBERTO FACCHINI
The aim of this paper is to
investigate
the structure of thering
of
quotients ~3)
of a commutative Noetherianring
.R withrespect
tothe Dickson
topology
Ð. Inparticular
westudy
under which condi- tions R =Ra)
orR’J)
is the totalring
of fractions of R(§ 2),
the structure ofRen
when .R is a GCD-domain and when .I~ is local and satisfies con-dition
S2 (§ 3),
and theendomorphism ring
of the R-module(~ 4).
1. Preliminaries.
The
symbol
.R will be usedconsistently
to denote a commutative Noetherianring
with anidentity
element.Let Ð be the Dickson
topology
onI~,
that is the Gabrieltopology
on I~
consisting
of the ideals I of R such that is an Artinianring,
i.e. the ideals I of .R which contain the
product
of a finite number of maximal ideals of R(see [12], Chap. VIII, §2).
For every R-module~VI we
put X(M) = (z
E = 0 for some I E~~. X(lVl)
is a sub-module of M said the D-torsion submodule of M. The functor X has been studied
by
E. Matlis([6]).
Let us define(*) Indirizzo dell’A. : Istituto di
Algebra
e Geometria - Via Belzoni 7 - 35100 Padova.where the direct limit is taken over the downwards directed
family
1‘~.M5)
is called the module ofquotients
of .lVl withrespect
to thetopo- logy
D. It is known thatRÐ
becomes aring
in a natural way and that.MU)
becomes anRD-module.
We shall
always
suppose that R has no D-torsion. This isequivalent
to
request
that every maximal ideal of R bedense,
that is torequest
that(since R
isNoetherian)
every maximal ideal of R contain aregular
element. Under such a
hypothesis
R is asubring
ofRp
and1~~
is asubring
ofQ,
the totalring
of fractions of .R. A more convenientdescrip-
tion of
Rp
is thatRa)
_Ix
EQ IxI ç R
for some I E Moreprecisely
we have that
1.1. LEMMA.
If R
possesses maximal idealso f grade 1,
then =- ~x I
there exist maximal ideals ... , in Rof grade
1 suchthat ...
A,, C 1~~.
OtherwiseRÐ
= R.PROOF. It is clear that the
products
of a finite number of maximal ideals of .R form a basis for ~. ThereforeI
there exist maximal idealsvtL1,..., vtLn
in .R such that1~~.
Henceit is sufficient to prove that if ... ,
7 A,,,
are maximal idealsof
R,
gr(~~) ~ 1,
thenzfli
... Forthis it is
enough
to show thatif y
EQ,
is a maximal ideal ofR,
gr
( ~ ) ~
1and y
vtL çR, then y
E R.Now y
= s-lr for some r, s ER,
s
regular.
Hence from it follows that that is A Cr).
By
themaximality
of(Rs:r)
or = R. In the firstcase gr
(fl) =
~. Therefore R _ i.e. y = s-1r ER.From lemma 1.1 we
immediately
have acomplete description
ofthe
grade
of all ideals in a localring R
such that .R ~ For sucha
ring
it is easy to prove that gr(I)
= 1 for everyregular
ideal I ofR,
and
gr(J)
= 0 for everynon-regular
ideal J. From lemma 1.1 it is also clear that thestudy
ofR5)
isequivalent
to thestudy
of the maximalideals of
grade
1 in R.It is also easy to describe the elements of
1~~
in relation to theprimary decomposition
in .Z~. In fact let x EQ.
Then x ERp
if andonly
if x is of the forms-lr,
where s is aregular
element ofR,
and ifRs = ... r1
Q~
is a normalprimary decomposition
of the idealRs,
where
Qi
is associated to aprime
non-maximal ideal for i =1,
..., t and to a maximal ideal for i = t+ 1,
... , n, ... r1Qt .
Now let us « count» the number of the
generators
of the ideals in the localizationRp.
We havealways supposed
that thering
Ris Noetherian. Of course we cannot
hope
that thisimplies
thering
of
quotients Rp
is Noetherian. There existrings R
such thatRp
pos-sesses ideals which cannot be
generated by
a finite number of elements.Nevertheless there exists an upper bound for the number of elements needed to
generate
any ideal ofR5).
1.2. PROPOSITION. Let R be a
ring, Max(l)(B)
the seto f
all maximal ideals in Rof grade 1, ~
thecardinality of Max(l)(R).
Then is theunion
of a
directedfamily, of cardinality ~o ~ ~ -f- 1, of
Noetheriansubrings of Q. Every
idealof
can begenerated by
at most-~--1)
elements,. In
local,
every idealof
iscountably
gen- erated.PROOF. If
0, by
lemma 1.1~.R~
is therequested family..
Hence let us suppose 0.
Let I be a
regular
ideal of R and setR(I)
= EQ lxl
c..R~~ .
is a
subring
ofQ.
Let s E I be aregular
element. Then if xE Q,
we have that xI C R if and
only
if xs E R and xs E(is :1). Let rl,
... , r,, be a set ofgenerators
of the ideal(I~s :1 )
in R. It follows that ifand
only
if x is a linearcombination
ofs-lr1,
... ,s-1rn
with coefhcients in .R. ThereforeR(,)
= ... , is a Noetherianring.
Now let usconsider the
family
of therings R(j)
where I ranges over the set of allproducts
of a finite number of elements of Thecardinality
of Y is
~ 0 . $.
Y is directed because u ~ andby
lemma 1.1its union is
Finally
if A is an ideal of =U (.R(I~
r1A)
andhence there exists a set of
generators
of A ofcardinality -~- 1).
If R and S are Noetherian
rings
and it mayhappen
that
S’J)
~ This is not the case if ~’ isintegral
over R.1.3. PROPOSITION. Let Noetherian
rings,
S in-tegral
over R. ThenR’J) =
PROOF. First of all note that R and S have the same total
ring
offractions
Q.
Furthermore if JY’ is any maximal idealin S,
N r1 R is amaximal ideal in .R and therefore it contains a
regular
element of R.Hence X contains a
regular
element(of S).
Let us show that Let
x E R’J).
Thenx E Q
andfor suitable maximal ideals
~i
of R. Henceflns C
S.To show that x E
Sp
it is then sufficient to show that == (Jt1S) ... belongs
to the Dicksontopology
of ~’. Hence it is.enough
to prove thatbelongs
to the Dicksontopology
ofS,
andthis is obvious because ~’ is
integral
over R and hence every minimalprime
ideal of AS is a maximal ideal of S.Vice versa let us show that
Stj) ç
Let x ESD.
ThenxN1
...... JY’n ~
~’ for suitable maximal idealsXi
of ~’. It follows thatNow every
Xif1R
is maximal in R because ~’ isintegral
overR;
let rl, ... , rt be a set ofgenerators
ofthe ideal
(JY’1
f1R) ... (Xn
f1R)
of R. Then xrjER’J),
so that there exists an idealAj belonging
to the Dicksontopology
of 1~ such that xr;Aj
C R. From this we have thatx( JY’1
f1R)
...(Xn
f1 ...A, C
Rand the ideal f1
.R)
... f1R) At belongs
to the Dicksontopology
of R. Hence x EThe
preceding proposition
may seem somewhatheavy
due to themany
hypotheses
on R and S. However afterproposition
2.2 we shallbe able to prove that
1.4. PROPOSITION. Let R be a local
ring. I f
=1=R’J) =1= Q,
then there
always
exists a Noetherianring
S ~R5), properly containing
Rand
integral
over R2. The two cases
R’J) ==
R andQ.
Under our
hypotheses (I~
is a Noetherianring
in which every maximal ideal contains aregular element)
we know thatRU) C Q.
The firstproblem
whichnaturally
arises isstudying
under what conditionson R coincides with R and
Q respectively.
The caseQ
ishandled in theorem 2.1 and the case
R’J)
= R in theorem 2.2.2 .1.
THEOREM. Thef ollowing
statements areequivalent : i )
=Q;
.ii)
every maximal idealof
R hasheight 1;
iii)
no proper idealof
is dense inRI);
iv)
Rsatisfies
0-inv condition(see [I1]) ;
is a
1-topology (see [12], Chap. VI, § 6, 1).
Furthermore
if
R is a reducedring
thepreceding
statements are alsoequivalente
to :vi )
is a( V on Neumann- ) regular ring;
vii) R’J)
is asemisimple ring.
PROOF.
i)
=>ii).
Let A be a maximal ideal of .I~. Let se fl, s regular.
Thenby i),
ERD.
It follows that for some I ED,
that is I C But then is Artinian and
AIR.-
is aprime
idealin hence a minimal
prime
ideal. Therefore fl is a minimalprime
ideal of Rs in R. Since 8 is
regular,
fl hasheight
1.ii)
=>i).
Let us suppose Then there exists someregular
element s E .R non invertible in
RTJ,
i. e. such that for every ideal I E~,
that isRs 0
~. Therefore is notArtinian,
and henceit has a maximal ideal of
height ~ 1.
It follows that R has a maximal ideal ofheight > 2.
i)
~iii).
Obvious.iii)
~i). Suppose iii)
holds and let us show that if s E R isregular
in R then it is invertible in
(this
will provei)).
Now if s is
regular
inR, s
is, invertiblein Q
and henceregular
in Therefore
Rb - s
is a dense ideal ofBy iii)
=RTJ,
that is s is invertible in
Rp.
ii)
~v).
Let s be aregular
element of 1~. Then every minimalprime
ideal of Rs hasheight l,
and henceby ii)
it is maximal. Therefore Rs E D. It follows that the filter of allregular
ideals is contained in DSince every ideal of D is
regular, D
isexactly
the filter of allregular
ideals of .R. Hence is a
1-topology.
v)
~iv).
Trivial.iv)
=>ii). Suppose R
satisfies Stenstrom’s D-inv condition. Let A be a maximal ideal of R. Then M E D and therefore there exists I E D such that I C A and I is aprojective
ideal. Let I = ... r1Qn
be a normal
primary decomposition
of I. Since I c- 5) the minimalprime
ideals of I areexactly
the maximal ideals of .1~containing
I.Let
Ql
be the~-primary component
of I. Localize withrespect
tothe ideal A. Then is a
projective
ideal(and
hence it isprincipal generated by
aregular
element ofRX),
and = Henceis a
M-primary principal
ideal. From this it follows that theheight
of the ideal inRA
is 1. Hence theheight
of A is 1.Now suppose R is
reduced,
i.e. without non-zeronilpotent
elements.Then
i)
~vii). Trivial,
because the totalring
of fractions of a reduced Noetherianring
isalways semisimple.
vii)
=>vi).
Obvious.vi)
=>i).
Let s be anyregular
element of 1~. Then s = s2x for someelement It follows that 1 = sx, i. e. x = s-1. Therefore s-1 Hence
Q
=R’J).
2.2. THEOREM. The
following
statements areequivalent : i)
R =Rp ;
ii)
’ =0,
where. E is the setof
allregular
elementsof R;
’iii)
No maximal idealof
R is(see [8], § 7);
iv)
No maximal idealof
R is associated to an ideal Rs with sregular
element
of R;
.v)
No maximal idealof
R hasgrade
1.PROOF.
i)
~ii). Suppose
# 0. Then there existssEE
’
s e 27 such that
X(RlRs) 0 0,
that is such thatRjlts
has asimple
sub-module. Let r
--E-
Rs be agenerator
of such a submodule. Then Rs and(Rr -~-- B/A
for some maximal ideal ofR,
and hencerfl c that is R. From this it follows that s-1r E But
s-lr tt R,
for otherwise r e Rs. Hence R =Fii)
=>iii).
Let ~ be a R-reflexive maximal ideal of I~. Then for some with sregular (see [3],
theorem1.5).
Then
(Rr + Rjfl
is asimple
submodule ofIt follows that 0.
’
iii)
~iv).
Let u1L be a maximal ideal of .R associated to the ideal R8 with s aregular
element of Z~. Then A = rad(Rs : r)
for some(see [1],
theorem4.5),
and hence A, C for some natural num-ber n.
Suppose n
is the least for which such relation holds. Then Let 7t ~ (I~s : r).
Then from it follows thati.e. and from it follows that
and so ~ is R-reflexive
([3],
the-orem
1.5).
iv)
=>v). Obvious,
because a maximal ideal ofgrade
1 is associatedto an ideal with s a
regular
element of R.v)
=>i). By
lemma 1.1.Now we are
ready
to proveproposition
1.4:Let u1L be the maximal ideal of .1~. Since
R’J) =F Q, by
theorem 2.1the
height
of ~ is~ 2,
and since is R-reflexiveby
theorem 2.2.Therefore there
exists y E Q
such thatyA C R, y 0
1~. From this it follows thatI~(,~,~ ,
the Noetheriansubring
of defined in theproof
of
proposition 1.2, properly
contains I~. In order to show that S = satisfies the thesis of theproposition
it is sufficient to show that if x EQ
and
xA C: R ,
then x isintegral
over R. Since theheight
of M is>2,
7there exists a
regular prime
ideal P in Rproperly
contained in fl.Now if x
E Q
and xA C:R,
then Let us show that (p.If there would exist p e S such that
xp 0 S.
But xp E.1~,
becausep E A. Let m E Then
xpm 0’5,
a contradiction becauseTherefore Furthermore P is a faithful R-module
(because P
is.regular
in andfinitely generated.
From this it follows that x isintegral
over R.3. The structure of
R5)
for some classes ofrings
R.One of the classes of
rings
for which it ispossible
togive
acomplete description
of the Dickson localization is the class ofGCD-domains,
that is of
integral
domains R such that for everya, b
there existsa
greatest
common divisor[a, (see [4],
page32).
3.1. THEOREM. Let R be a be the
multiplicatively
closed subset
of
Rgenerated by
all elements s E R such that Rs is ac maximal idealof
R. 8-1 R.PROOF. Let us show that in a GCD-domain R every maximal ideal
~ of
grade
1 isprincipal.
If A hasgrade 1,
there exists yE Q
suchthat
yfl C
R andy 0
R.Let y =
with r, s e.R, r, s ~ 0,
and letml , ... , mn be a set of
generators
of A. Then from it followsthat rm for
every i
=1, ... ,
n. Theref ore s divides rm forevery i,
and hence s divides their
greatest
common divisor[rm1,
... ,[ml,
...,mn].
Let d =[ml,
...,mn].
Then Jtt C Rd. Now if d would be a unit in R and hence from it would followsir
in.R,
i.e.y E R,
a contradiction. Since jKj ismaximal, A - Rd,
i.e. Jtt isprin- cipal.
By
lemma1.1, ~x
there exist maximal idealsof
grade
1 such thatxM1
...RI,
from which{x
EQ| I
thereexist
principal
maximal idealsRsi, Rs,,
such that xsl ... Sr ER)
==-
{x
E E R for some 8 E~’~
= S-1.~.3.2. PROPOSITION. be a local
ring
and suppose that at leastone
of
thefollowing four
conditions holds :i)
there exists aregular
non invertible element s E R such that the ideal Rs is a radicalof R;
ii)
Rsatis f ies
condition82 (see [12 ],
Ch.VII, § 6 ) ;
iii)
R is aMacaulay ring;
iv)
R is anintegrally
closed domain.Then if
dim R =1, R~
=Q, and i f
dim R >1, Rp
= R.PROOF.
Suppose
that in .1~ there exists aregular
non-invertible element s such that the ideal Rs is a radical in _R. IfRI) 0 R,
thereexists x E
R5)
such that R and xA cR,
where A is the maximal ideal of .R. Then xs El~,
and hence x = s-1r for some r E R. From thiswe have that rA 9 Rs
and r 0
Rs. Therefore in thering
the max-imal ideal contains
only zero-divisors,
and since thering
isreduced, AIRs
is contained in the union of the minimalprime
idealsof 0 in
RlRs,
and henceAIRs
itself is a minimalprime
ideal of 0 inTherefore ~ is a minimal
prime
ideal of I~s in R and hence it hasheight
1. Thus we haveproved
that if dim R >1,
thenRg)
= R.If dim R =
1,
thenRp
=Q by
theorem 2.1.Next if R satisfies
~2 ,
theprime
ideals ofheight >2
havegrade >2,
from which the thesis follows
by
theorem 2.1 and 2.2.Finally
if R is aMacaulay ring
or anintegrally
closeddomain,
Rsatisfies
~2’
4. The
endonzorphism ring
ofWe shall now
study
the R-module .g = It is a D-torsion R-module(it
is the D-torsion submodule of Therefore(see [6],
theorem
1),
itsplits
in itsA-primary components:
is the submodule of K
consisting
of all elements x E K suchthat
AnnR(x)
contains a power of ~.Now let I be a
regular
ideal. Set I-1={q
EQ Iql ç I~~
and -- It is known that is an ideal of R
canonically
isomorphic
toHomR (HomR (I, R), R),
the .R-bidual of I. Let 93 bea basis of the filter of
neighborhoods
of zero for aring topology
overconsisting
ofregular
ideals of .R. Let us define the bidualtopology
over Ras the
topology having
$-1-1 = E93~
as basis of the filter ofneighborhoods
of zero. Theoriginal topology
is finer than the bidualtopology.
4.1. THEOREM
(see [5],
theorem3.4).
Let R be aring,
maximalideal
of
R. For every natural number n let{x
EKIA-X
=0}.
Then
iii)
The non-zero elementsof
areexactly
the elementsof having
annihilator~;
iv) A(~,),n + 1 /A(,~~,),~
is in a natural way af inite
dimensional vector space over thefield RIA;
v)
iscountably generated.
PROOF.
i)
Theonly
non obvious statement is thatA(A),n - Ext’ (R/An, .R) ;
this isproved by observing
thatA(~,~,~,
=and
applying
the functorHomR (-, .R)
to the short exact sequenceiii)
Obvious.iv)
andv)
Fromiii)
it follows that is a vector space over the field In order to show that it is finite dimensional it isenough
to prove that is afinitely generated
R-module(and
this with
i) immediately gives v)).
To this end weonly
have to showthat the R-submodule of
Q
isfinitely generated.
Let s E
cÂ(;n, S regular
andlet q
E(~At~~)-1. Then qs
E1~, whence q
= s- irfor some r E R. From this we have that
Since the ideal
(Rs:
of R isfinitely generated,
it follows thatis
finitely generated.
Let us consider
End., (.~~, (K) ) .
For everypositive integer
n set=
{f EEndR If (A(,X),.)
=0}.
LetEndR
havethe
topology
definedby
the filtration and letEnd, (K) gz End,
have theproduct topology.
Let us call thistopo- logy
the naturaltopology
ofEnd,, (K).
4.2. PROPOSITION. endowed with its natural
topology
is a
Hausdorff complete topological ring.
PROOF. It is
enough
to prove that everyEndR (XtÂt(K))
iscomplete
and Hausdorff.
Clearly
it is Hausdorff. Now if is aCauchy
sequence in
EndR
let us definef
EEndR
in the fol-lowing
way: if x EA~ jj,~ ,
there exists r e N such that for everyr’, r" E N, -r’, r" ~ r,
we have -fr,,
EH(.Jt),n;
letI(x)
=fr(x).
Theproof
that thisf
is a well-definedhomomorphism
and that it is limit of the sequence(In)
is routine.
Now let .R be a local
ring,
J1L its maximal ideal. Consider the canon-ical
homomorphism q~ :
R -EndR (.g)
which to every element of R associates themultiplication
over Kby
that element. IfEndR (g)
hasits natural
topology
and .I~ has the bidualtopology
of the J1L-adictopo- logy,
it is easy to prove that g~ is a continuousring homomorphism
whose kernel is the closure of 0 in R. Therefore 99 induces a
continuous
EndR (K),
y where R’ is the « Hausdorffized » ofR,
that is .R modulo the closure of 0 in 1~ with thequotient topology.
It is easy to see
that p
is atopological embedding.
Therefore we haveproved
thefollowing
4.3. PROPOSITION. local
ring,
f1 its maximal ideal and let R have the bidualof
the A-adictopology.
Then R modulo the closureof
zero with thequotient topology
is in a natural way atopological ,subring ,of EndR (K)
endowed with its naturaltopology.
Therefore
EndR (_K)
contains the Hausdorffcompletion
of R.The
following
theoremgives
a sufficient condition forEndR (I~)
tobe the Hausdorff
completion
of I~(see [7]).
4.4. THEOREM. Let R be a local
ring,
~ its maximal ideal.Sup-
pose that can be
generated by
two elements. ThenEndR (K)
is theHausdorff completion of
R with the bidualtopology of
the J1L-adictopology.
PROOF: If A-’ =
.R,
then = Z~ for every n. Hence the closure of 0 in .R is R and K = 0.Therefore we may suppose
X-1 =A
R. ThenA(A).,
==(see [7],
lemma2.3,
true also when R is not anintegral domain).
ThusIt follows that if
f
EEndR (K)
and nE N,
then
f
extends to aendomorphism
of and.ae(vtt),n ç Bn,
whereBn
is the submodule ofconsisting
of all elements ofannihilated
by
A,. Butf
coincides overBn
with themultiplication by
an element of R(see [10],
lemma5.11).
It follows that is themultiplication by
an element of R. Henceq(R)
is dense inWe conclude
by proposition
4.2 and 4.3.REFERENCES
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rings, Allyn
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Injective
modules over Noetherianrings,
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[6] E. MATLIS, Modules with
descending
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Algebra,
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