Ordinal length and the canonical topology

0.19 MB·English

by  Hans Schoutens

ORDINAL LENGTH AND THE CANONICAL TOPOLOGY
HANSSCHOUTENS
3
1
0 ABSTRACT. Weextendtheclassicallengthfunctiontoanordinal-valuedinvariantonthe
2 classofallfinite-dimensionalNoetherianmodules.Weshowhowtocalculatethiscombi-
n natorialinvariantbymeansofthefundamentalcycleofthemodule,thuslinkingthelattice
a ofsubmodulestohomologicalpropertiesofthemodule.Usingthis,wedefineonamodule
J itscanonicaltopology,inwhicheverymorphismiscontinuous.

8
2

] 1. INTRODUCTION
C
A Thepurposeofthispaperistolaythefoundationsofanew,ordinal-valuedinvariantin
CommutativeAlgebra:the(ordinal)lengthlen(M)ofaNoetherianmoduleM,measuring
.
h (by meansof an ordinal)the longestdescendingchain of submodulesin M. Recall that
t
a anordinalisanisomorphismclassofatotalwell-order(=admittingthedescendingchain
m condition),andtheclassofordinals,orderedbytheinitialsegmentrelation,isagainawell-
[ order;eachnaturalnumberisanordinalbyidentifyingitwithachainofthatlength;theor-

derof(N,≤)isdenotedωandisequaltothesupremumofalln∈N;theorder-typeofthe
1
lexicographicalorderon Nd isdenotedωd. Apartfromtheusual(non-commutative)sum
v
7 α+β,wealsoneedthe(commutative)shuffleornaturalsumα⊕β givenbycoefficient-
5 wiseadditionintheCantornormalform(fordetails,see §2). Sincedescendingchainsin
4 a Noetherian module M are well-ordered with respect to the reverse inclusion, they are
6
ordinals,andwedefinelen(M)asthesupremumofallordinalsobtainedthisway.
.
1 In[8]weshowedthat,whereaslengthcannolongerbeadditiveonexactsequences,it
0 isstillsemi-additive:
3
1 Theorem(Semi-additivity). If0→N →M →Q→0isanexactsequenceofNoether-
: ianR-modules,then
v
i (1) len(Q)+len(N)≤len(M)≤len(Q)⊕len(N).
X
r Moreover,ifthesequenceissplit,thenthelastinequalityisanequality.
a
As an application, we obtain that a Noetherian ring has length ωd if and only if it
is a domain of Krull dimension d. In this paper, we prove the remarkable fact that this
combinatorial invariant can be described in terms of homological invariants: recall that
the local multiplicity e (M) at a prime ideal p is defined as the length of the p-torsion
p
submoduleofM (whenceisnon-zeroifandonlyifpisanassociatedprimeofM).
p
MainTheorem. ThelengthofM isequalto e (M)ωdim(R/p).
Lp p
Combiningthiswithsemi-additivity,leadstomanyinterestingapplications([8,6,7]).
Inthispaper,weapplythetheorytodefineacanonicaltopologyoneverymodule:theopen
submodulesinthistopologyarepreciselythosethathavethesamelengthasM. Weshow
Date:January29,2013.
Key words and phrases. Commutative Algebra; ordinal length; local cohomology; Chow group; local
multiplicity.
1
2 HANSSCHOUTENS
thatanymorphismbetweenmodulesiscontinuousinthistopology. Anyopensubmodule
is essential, and in [7], we will characterize those modules for which the converse also
holds. In the last section, as an applicationof this material, we discuss the phenomenon
of degradation: how source and target of a module may force the morphism to become
(almost)zero.Forinstance,weshow
Theorem. If an endomorphism on M factors through a module that has no associated
primesincommonwithM,thenitmustbenilpotent.
2. ORDINALS AND ORDINAL LENGTH
Apartialorderingiscalleda (partial)well-order ifithasthedescendingchaincondi-
tion, that is to say, any descendingchain must eventuallybe constant. A total orderis a
well-orderifandonlyifeverynon-emptysubsethasaminimalelement.Anordinalisthen
anequivalenceclass,uptoanorder-preservingisomorphism,ofatotalwell-order.Theset
of ordinals is a transfinite extension of the set of natural numbers N, in which the usual
inductionisreplacedbytransfiniteinduction.Wesaythatα≤β ifαcanbeembeddedas
atotalorderinβ. Anyboundedsubsetofordinalshasthenaninfimumandasupremum.
The(Cantor)sumα+β istheordinalcorrespondingtothewell-orderon α⊔β obtained
bylettinganyelementofβ belargerthananyelementofα. Thus,1+ωisthesameasω,
whenceinparticulardifferentfrom ω+1. Wewillnotneedarbitraryordinalmultiplica-
tion,butonlyproductsoftheformnαwithn ∈N,simplydefinedasthesumofncopies
of α (be aware that logicianswould use the more awkwardnotation α·n forthis). The
supremumofthenωisdenotedω2andistheorder-typeofthelexicographicalorderingon
N2. Theωdaresimilarlydefined,andtheirsupremumisdenotedωω.
LetObethecollectionofordinalsstrictlybelowωω. Anyα ∈ OhasauniqueCantor
normalform
(2) α=a ωd+···+a ω+a
d 1 0
with a ∈ N, called its Cantor coefficients. The support of α, denoted Supp(α) ⊆ N,
n
consistsof all i forwhich a 6= 0. The maximumandminimumof the supportof α are
i
called respectivelyits degree and order. An ordinalis a successor ordinal(=of the form
α+1forsomeordinalα) ifandonlyif itsorderis zero. Thesumofall a iscalledthe
i
valenceofα. Theordering≤onordinalscorrespondstothelexicographicalorderingon
thetuplesofCantorcoefficients(a ,...,a ). Letussaythatαisweakerthanβ,denoted
d 0
α (cid:22) β, if a ≤ b foralli, where,likewise, theb aretheCantorcoefficientsof β. Note
i i i
that≤extendsthepartialorder(cid:22)toatotalorder.
Apart from the usual ordinal sum, we make use of the natural or shuffle sum α⊕β
giveninCantornormalformas (a +b )ωd +···+a +b . Notethattheshufflesum
d d 0 0
iscommutative,andα+β willin generalbesmallerthan α⊕β. Infact, weshowedin
[8, Theorem 7.1] that the shuffle sum is the largest possible ordinal sum one can obtain
fromwritingbothordinalsasasumofsmallerordinalsandthenrearrangingtheirterms.
Inparticular,Oisclosedunderbothadditions.Moreover,sinceα(cid:22)β ifandonlyifthere
existsγ suchthatα⊕γ =β,wemayview(O,⊕,(cid:22))asapartiallyorderedcommutative
semi-group.
3. LENGTHAND SEMI-ADDITIVITY
Allringswillbe commutative,Noetherian,of finiteKrulldimension,andallmodules
will be finitely generated. Throughout, if not specified otherwise, R denotes a (finite-
dimensional, Noetherian) ring and M denotes some (finitely generated) R-module. By
ORDINALLENGTHANDTHECANONICALTOPOLOGY 3
Noetherianity,thecollectionofsubmodulesofM orderedbyinverseinclusionisapartial
well-order,calledtheGrassmanianofM anddenotedGr (M)(orjustGr(M)). Inpar-
R
ticular,anychaininGr (M)is(equivalentto)anordinal. Thesupremumofallpossible
R
ordinalsarisingasachaininthiswayiscalledthelengthofM andisdenotedlen (M),or
R
whenthebaseringisclear,simplybylen(M). NotethatthelengthofM asanR-module
isthesameasthatofanR/Ann (M)-module,andsowemayassume,ifnecessary,that
R
M isfaithful. ItfollowsfromtheJordan-Ho¨ldertheorythat thisordinallengthcoincides
withtheusuallengthformodulesoffinitelength. Thelengthofaringisthatofamodule
overitself. Inotherwords,len(R)isthelongestdescendingchainofidealsinR.
Itisusefultohavealsoatransfinitedefinitionoflength:wedefineaheightrankl(·)on
Gr (M),asfollows. Putl(M) := 0. GivenasubmoduleN ⊆ M,atasuccessorstage,
R
wesaythatl(N) ≥ α+1, if thereexistsa submoduleN′ ⊆ M containingN suchthat
l(N′) ≥ α. Ifλisalimitordinal(thatistosay,notasuccessorordinal),thenwesaythat
l(N)≥λ,providedforeachα<λ,thereexistsasubmoduleN ⊆M containingN with
α
l(N )≥α. Finally,wesaythatl(N)=αifl(N)≥αbutnotl(N)≥α+1.Weprovein
α
[8,Theorem3.10]thattheheightrankofthezeromoduleisthelengthofM.Infact,more
generally,foranysubmoduleN ⊆M,itsheightrankequalsitsco-length,thatistosay,
(3) l(N)=len(M/N).
Notethatheightranksatisfiesthefollowingcontinuityproperty:l(N)islessthanorequal
tothesupremumofalll(W)+1,forallW strictlycontainingN. Usingsemi-additivity
(seeintroduction)weshowedin[8]:
3.1.Theorem(Dimension). LetM beafinitelygeneratedmoduleoverafinite-dimensional
NoetherianringR. Thenthedegreeoflen(M)isequaltothedimensionofM. Inpartic-
ular,Risad-dimensionaldomainifandonlyiflen(R)=ωd. (cid:3)
The order of a module is by definition the order of its length, and will be denoted
ord(M); the valence val(M) is the valence of its length. We will calculate these two
invariantsinCorollary4.5below. Letuscallanexactsequencestronglyequilateralifwe
haveequalityatbothsidesof(1);ifweonlyhaveequalityattheright,wecallthesequence
equilateral. Beingstronglyequilateralisreallyapropertyofordinals: α+β = α⊕β if
andonlyifthedegreeofβ isatmosttheorderofα,andhence
3.2.Corollary. AnexactsequenceoffinitelygeneratedR-modules
0→N →M →Q→0,
isstronglyequilateralifandonlyifdimN ≤ord(Q). (cid:3)
4. LENGTH ASA COHOMOLOGICAL RANK
Ascustomary,wedefinethedimensionofaprimeidealp⊆R,denoteddim(p),asthe
KrulldimensionoftheresidueringR/p.Wedenotethecollectionofallassociatedprimes
(= prime ideals of the form Ann(a) with a ∈ M) by Ass(M); it is always a finite set.
We willmakefrequentuse, fora shortexactsequence 0 → N → M → Q → 0, ofthe
followingtwoinclusions(see,forinstance,[2,Lemma3.6])
(4) Ass(N)⊆Ass(M);
(5) Ass(M)⊆Ass(N)∪Ass(Q)
4 HANSSCHOUTENS
We define the the (zero-th) local cohomology1of M at p, denoted H0(M), as the p-
p
torsionofM ,thatistosay,thesubmoduleofM consistingofallelementsthatarekilled
p p
bysomepowerofp. AsH0(M)isamoduleoffinitelengthoverR ,wedenotethislength
p p
bye (M) and call itthe localmultiplicity ofM atp (see, forinstance, [2, p. 102]). An
p
alternativeformulationisthroughthenotionofthefinitisticlengthofamoduleM,defined
asthesupremumlenfin(M)ofalllen(N)withN ⊆Mandlen(N)<ω.ByNoetherianity,
M hasalargestsubmoduleH offinitelength,andhencelenfin(M) = len(H). Withthis
notation,wehave
(6) e (M)=lenfin(M ),
p p
foranyprimeidealp,andthisisnon-zeroifandonlyifpisanassociatedprimeofM. We
nowdefinethecohomologicalrankofamoduleM as
coh(M):=Mep(M)ωdim(p).
p
It is instructiveto view this fromthe pointof view ofChow cycles. Let A(R) be the
ChowringofR, definedasthefreeAbeliangroupon Spec(R). AnelementD ofA(R)
will be called a cycle, and will be represented as a finite sum a [p ], where [p] is the
P i i
symboldenotingthefreegeneratorcorrespondingtotheprimeidealp. Thesumofalla
i
iscalledthedegreedeg(D)ofthecycleD. WedefineapartialorderonA(R)bytherule
thatD (cid:22)E,ifa ≤b ,foralli,whereE = b [p ].Inparticular,denotingthezerocycle
i i P i i
simplyby0,wecallacycleD effective,if0 (cid:22) D,andweletA+(R)bethesemi-group
of effective cycles. This allows us to define a map from effective cycles to ordinals by
sendingtheeffectivecycleD = a [p ]totheordinal
Pi i i
o(D):=Maiωdim(pi).
i
Clearly,ifD andE areeffective,theno(D+E) = o(D)⊕o(E). Moreover,ifD (cid:22) E,
theno(D)(cid:22)o(E),sothatwegetamap(A+(R),+,(cid:22))→(O,⊕,(cid:22))ofpartiallyordered
semi-groups.
ToanyR-moduleM,wecanassignitsfundamentalcycle,bytherule
(7) cycR(M):=Xep(M)[p]
p
Immediately from (6) we get o(cyc(M)) = coh(M). Our main result now links this
cohomologicalinvarianttoourcombinatoriallengthinvariant:
4.1.Theorem. ForanyfinitelygeneratedmoduleM overafinite-dimensionalNoetherian
ringR,wehavelen(M)=o(cyc(M))=coh(M).
Beforewegivetheproof,wederivetwolemmas. Itisimportanttonoticethatthefirst
oftheseisnottrueatthelevelofcycles.
4.2.Lemma. IfM →QisapropersurjectivemorphismofR-modules,then
coh(Q)<coh(M).
Proof. Let N be the (non-zero) kernel of M → Q, and let d be its dimension. If p ∈
Ass(M) but not in the support of N, then M ∼= Q , so that they have the same local
p p
cohomology. This holds in particular for any p ∈ Ass(M) with dim(p) > d, showing
thatcoh(Q) and coh(M) can only start differingat a coefficientof ωi for i ≤ d. So let
1Thisdeviatesslightlyfromthepracticein[2,App.4]aswealsolocalize.
ORDINALLENGTHANDTHECANONICALTOPOLOGY 5
p∈Ass(M)∩Supp(N)havedimensiond.Ingeneral,localcohomologyisonlyleftexact,
but by Lemma 4.3 below, we have in fact an exact sequence (8). Since e (N) 6= 0, we
p
mustthereforehavee (Q)<e (M). Itnoweasilyfollowsthatcoh(Q)<coh(M). (cid:3)
p p
4.3.Lemma. Givenanexactsequence0 → N → M → Q → 0,ifpisaminimalprime
ofM,then
(8) 0→H0(N)→H0(M)→H0(Q)→0
p p p
isexact.
Proof. Itiswell-known(see [2, App.4])thatH0(·)isleftexact, sothatweonlyneedto
p
proveexactnessat the final map. By assumption, M has finite length and hence N =
p p
H0(N). Choose n high enoughso that pnN = 0. Suppose a¯ ∈ H0(Q), that is to say,
p p p
pma¯=0inQ ,forsomem. Leta∈M beapre-imageofa¯underthesurjectionM →Q.
p
Therefore,pma∈N ,whencepm+na=0inM ,showingthata∈H0(M). (cid:3)
p p p
4.4.Corollary. Givenanexactsequence0→N →M →Q→0,ifM hasnoembedded
primes,thenwehaveanequalityofcycles
(9) cyc (M)+D =cyc (N)+cyc (Q)
R R R
whereDisaneffectivecyclesupportedonAss(Q)\Ass(M).
Proof. Let D be the cycle given by (9), so that D has supportin Ass(M)∪Ass(Q) by
(5). WeneedtoshowthatD iseffectiveandsupportedonAss(Q)\Ass(M). Sinceany
associatedprimepofM isminimal,DisnotsupportedinpbyLemma4.3. Ontheother
hand,anyassociatedprimeofQnotinAss(M)appearswithapositivecoefficientin D,
showingthatthelatteriseffective. (cid:3)
Proof of Theorem 4.1. Let us first prove len(M) ≤ coh(M) by transfinite induction
on coh(M), where the case coh(M) = 0 corresponds to M = 0. Let N be any non-
zero submoduleof M. By Lemma4.2, we have coh(M/N) < coh(M), and hence our
inductionhypothesisappliedtoM/N yieldslen(M/N)≤coh(M/N)<coh(M). Since
len(M/N) = l(N)by(3),continuitythereforeshowsthatlen(M) = l(0)canbeatmost
coh(M),asweneededtoshow.
Toprovetheconverseinequality,weinductonthelengthof M. Chooseanassociated
primepofM ofminimaldimension,say,dim(p) = e. Byassumption,thereexistsm ∈
M such that Ann (m) = p. Let H be the submodule of M generated by m. Clearly,
R
coh(H) = ωe,andsobywhatwealreadyproved,len(R/p) ≤ ωe. ByTheorem3.1,this
thenisanequality. SowemayassumethatQ := M/H isnon-zero. ByLemma4.3, we
gete (M)=e (Q)−1. Bysemi-additivity,wehaveaninequality
p p
len(Q)+len(H)≤len(M)
andtherefore,byinduction
(10) coh(Q)+ωe ≤len(M)
LetgbeanyassociatedprimeofM differentfromp.Byminimalityofdimension,gcannot
containp. In particular, M ∼= Q , whence e (M) = e (Q). Expandingcoh(Q) in its
g g g g
Cantornormalform b ωi,letλ := b ωi. Puttingtogetherwhatweprovedsofar,
P i Pi≥e i
wecanfindanordinalαwithord(α)≥e(stemmingfromprimesassociatedtoQbutnot
toM),suchthatλ⊕ωe =coh(M)⊕α. Sincecoh(Q)+ωe =λ⊕ωe,weget,from(10)
andthefirstpart,inequalities
coh(M)⊕α≤len(M)≤coh(M)
6 HANSSCHOUTENS
whichforcesα=0andallinequalitiestobeequalities. (cid:3)
4.5.Corollary. The orderofa moduleisthesmallestdimensionofanassociatedprime,
anditsvalenceisthedegreeofitsfundamentalcycle. (cid:3)
By[1,Proposition1.2.13],overalocalring,wehave
(11) depth(M)≤ord(M).
Thisinequalitycanbestrict: forexample,atwo-dimensionaldomainwhichisnotCohen-
Macaulay,hasdepthonebutordertwobyTheorem3.1. Ournextresultgivesaconstraint
onthepossiblelength(nottobeconfusedwithitsheightrank)ofasubmodule,whichwas
exploitedin[6],tostudybinarymodules.
4.6.Theorem. IfN ⊆M,thenlen(N)(cid:22)len(M). Conversely,ifν (cid:22)len(M),thenthere
existsasubmoduleN ⊆M oflengthν.
Proof. ThefirstassertionisimmediatefromTheorem 4.1,inclusion(4),andthefactthat
(8) is always left exact. For the second assertion, let µ := len(M). We induct on the
(finite collection) of ordinals ν weaker than µ to show that there exists a submodule of
thatlength. The case ν = 0 being trivial, we may assume ν 6= 0. Let i be the orderof
ν andwrite ν = θ⊕ωi forsome θ (cid:22) ν. Sincethenθ (cid:22) µ, thereexistsasubmoduleof
length θ by induction. Let H ⊆ M be maximalamong all submodulesof length θ. By
Theorem4.1, thereexistsan i-dimensionalassociatedprimepofM,suchthatH0(H)is
p
strictlycontainedinH0(M). Hencewecanfindx∈M outsideH suchthatpx⊆H. Let
p
N :=H +Rxandletx¯betheimageofxmoduloH,sothatN/H =Rx¯. Sincepx¯=0,
the length of Rx¯ is at mostωi. By semi-additivityapplied to the inclusion H ⊆ N, we
have an inequality len(N) ≤ θ ⊕ len(Rx¯), and hence len(N) ≤ ν. Maximality of H
yieldsθ <len(N). Ontheotherhand,sincelen(N)(cid:22)µbyourfirstassertion,minimality
ofithenforceslen(N)=ν,asweneededtoshow. (cid:3)
4.7. Remark. In fact, if N ⊆ M, then cyc(N) (cid:22) cyc(M), so that the fundamental
cycle map is a morphism Gr(M)◦ → A(R) of partially ordered sets, where Gr(M)◦
is the opposite order given by inclusion. On the other hand, by (3) and Theorem 4.1,
the map Gr(M) → A(R) given by N 7→ cyc(M/N) factors through the length map
Gr(M) → O,butthereisnonaturalorderingon A(R)forwhichthisbecomesamapof
orderedsets.
Wemayimprovethelowersemi-additivitybyreplacing≤by(cid:22):
4.8.Corollary. If0→N →M →Q→0isexact,thenlen(Q)+len(N)(cid:22)len(M).
Proof. Thisisreallyjustafactaboutordinals: withµ,ν,θbeingtherespectivelengthsof
M,N,Q, semi-additivitygivesθ+ν ≤ µ ≤ θ⊕ν, whereasTheorem4.6givesν (cid:22) µ,
and we now show that these inequalities imply that θ +ν (cid:22) µ. Write µ = ν ⊕α and
letdbethedimensionofν. Foran arbitraryordinal β, we havea uniquedecomposition
β = β+ ⊕ β− with β− of degree strictly less than d and β+ of order at least d. By
assumption,ν+ = aωd forsomea,andhencethesemi-additivityinequalitiesatdegreed
and higherbecome θ+ ⊕aωd ≤ α+ ⊕aωd ≤ θ+ ⊕aωd, showing that θ+ = α+. By
definitionofordinalsum,θ+ν =θ+⊕ν andso
(θ+ν)⊕α− =θ+⊕ν⊕α− =α+⊕α−⊕ν =α⊕ν =µ
provingthatθ+ν (cid:22)µ. (cid:3)
ORDINALLENGTHANDTHECANONICALTOPOLOGY 7
Recall that a moduleM is called unmixed of dimension d if all its associated primes
havedimensiond. Theorem4.1thenshowsthatM isunmixedifandonlyiford(M) =
dimM,inwhichcaselen(M)=ℓgen(M)ωd,wherewedefinethegenericlengthofM as
thesumofthelocalmultiplicitiesatalld-dimensionalprimes. Corollary3.2yields:
4.9.Corollary. LetRbead-dimensionalNoetherianringand0 → N → M → Q → 0
a short exact sequence. If Q is unmixed of dimension d, then this sequence is strongly
equilateral. (cid:3)
Recall the dimension filtration D (M) ⊆ D (M) ⊆ ··· ⊆ D (M) = M of a d-
0 1 d
dimensional finitely generated R-module M defined by Schenzel in [5], where D (M)
i
is the submodule of all elements of dimension at most i, where we define the dimen-
sion of an element x ∈ M as the dimension of the module it generates, that is to say,
dim(R/Ann (x)). Equivalently,D (M)isthelargestsubmoduleofM ofdimensionat
R i
mosti.
4.10.Proposition. Givenad-dimensionalmoduleM,theexactsequence
0→D (M)→M →M/D (M)→0
i i
is strongly equilateral, for each i. Moreover, len(D (M)) is obtained from len(M) by
i
omittingthemonomialsofdegreebiggerthani.
Proof. Itfollowsfrom[5,Corollary2.3]thattheassociatedprimesofM/D (M)arepre-
i
cisely the associated primes of M of dimension strictly largerthan i. By Corollary 4.5,
thismeansthatlen(M/D (M))hasorderatleasti+1. Sincelen(D (M))hasdegreeat
i i
mostibyTheorem3.1,theresultfollowsfromCorollary3.2. (cid:3)
Anotherwaytoformulatethisresultisasthefollowingformulaforcalculatinglength
d
(12) len(M)=Mlen(Di(M)/Di−1(M)),
i=0
andeachnon-zeroD (M)/D (M)isunmixedofdimensioniandoflengtha ωi,where
i i−1 i
a isitsgenericlength.
i
4.11.Example. LetRbethecoordinateringofaplanewithanembeddedlineinsidethree
dimensionalspaceoverkgivenbytheequationsx2 =xy =0inthethreevariablesx,y,z.
Using Theorem 4.1, one easily calculates that len(R) = ω2 +ω, where the associated
primesarep = (x)andq = (x,y). Theidealsoflengthω areexactlythosecontainedin
p. Theidealsoflengthω2+ω(theopenidealsintheterminologyfromthenextsection),
arepreciselythosethatcontainanon-zeromultipleofxandanon-zeromultipleofy(this
follows,forinstance,from[6,Proposition4.10]). Finally,theremaining(non-zero)ideals
oflengthω2,arethosecontainedinqbutdisjointfromp(notethatifI isnotcontainedin
q,thenIR =R andIR =R ,sothatI mustbeopen).
p p q q
5. OPEN SUBMODULES
By semi-additivity, len(N) is at most len(M), and, in fact, len(N) (cid:22) len(M) by
Theorem4.6. IfM hasfinitelengthandN isapropersubmodule,thenobviouslyitslength
must be strictly less, but in the non-Artinian case, nothing excludes this from being an
equality. So,wecallN ⊆M open,iflen(N)=len(M). ImmediatelyfromTheorem4.1
andTheorem4.6,weget
8 HANSSCHOUTENS
5.1.Corollary. AsubmoduleN ⊆M isopenifandonlyifH0(N)=H0(M)forallp,if
p p
andonlyifcyc (N)=cyc (M),ifandonlyifval(N)=val(M). (cid:3)
R R
5.2.Remark. Inparticular,thelongexactsequenceoflocalcohomologyyieldsthatN ⊆
M is open if and only if the canonicalmorphism H0(M/N) → H1(N) is injective, for
p p
every(associated)primep(ofM).
5.3. Proposition. Given an exact sequence 0 → N → M → Q → 0, if dim(Q) <
ord(M),thenN isopen. Inparticular,anynon-zeroidealinadomainisopen,andmore
generally,anyidealinanunmixedringcontainingaparameterisopen.
Proof. Letν = a ωi,µ= b ωi,andθ = c ωi betherespectivelengthsofN,M
P i P i P i
and Q. By semi-additivity,µ ≤ θ ⊕ν, whence b ≤ a +c , forall i. By assumption,
i i i
c = 0wheneverb 6= 0,sothatinfactb ≤ a ,foralli,thatistosay,µ ≤ ν. Sincethe
i i i i
otherinequalityalwaysholds,N isopen. Toprovethelastassertion,letxbeaparameter
inad-dimensionalunmixedringR,sothatR/xRhasdimensionstrictlylessthand. Since
ord(R) = d byTheorem 4.1, ourfirst assertion showsthatthe ideal (x), andhence any
idealcontainingx,isopen. (cid:3)
Let us call a submodule N ⊆ M equilateral, if 0 → N → M → M/N → 0 is
equilateral, that is to say, if len(M) = len(N)⊕len(M/N). Hence a direct summand
isequilateralbysemi-additivity. ByCorollary3.2,anysubmoduleN suchthatdimN ≤
ord(M/N),isequilateral,buttheconverseneednothold.
5.4.Proposition. Amaximal(proper)submoduleiseitherequilateraloropen. Inpartic-
ular,ifM haspositiveorder,thenanymaximalsubmoduleisopen.
Proof. LetN M bemaximal,sothatQ := M/N issimple,oflengthone. Letν and
µbetherespectivelengthsofN andM. Bysemi-additivity,wehaveν ≤ µ ≤ ν +1. If
theformerinequalityholds,thesubmoduleisopen,andifthelatterholds,itisequilateral.
ThelastassertionnowfollowsfromTheorem4.1,forifM haspositiveorder,itslengthis
alimitordinal,andso,by(3),nomoduleoffiniteco-lengthcanbeequilateral. (cid:3)
Intheringcase,wecanevenprove:
5.5.Proposition. If(R,m)isanon-Artinianlocalring,thenmisopen.
Proof. Let ν andµ be the respectivelengthsof m and R. In view of Proposition5.4, to
ruleoutthatmisequilateral,wemayassumethatitisanassociatedprime. Choosex∈R
withAnn(x) = mandputR¯ := R/xR. SincexRhaslengthone,Corollary3.2applied
totheexactsequences0 → xR → R → R¯ → 0and0 → xR → m → mR¯ → 0yields
µ = len(R¯)+1 and ν = len(mR¯)+1. By induction, len(mR¯) = len(R¯), and hence
ν =µ. (cid:3)
5.6. Remark. As for primary ideals n, they will not be open in general if R has depth
zero. More precisely, suppose len(R) = λ + n with λ a limit ordinal and n ∈ N. If
nH0(R)=0(whichwillbethecaseiflen(R/n)≥n),thenlen(n)=λ. Indeed,thecase
m
n = 0 is trivial, and we may always reduceto this since n is a moduleoverR/H0(m),
R
andthelatterhaslengthλbyProposition5.3.
5.7.Corollary. IfN ⊆M isopen,thenSupp(M/N)isnowheredenseinSupp(M). The
converseholdsifM hasnoembeddedprimes.
ORDINALLENGTHANDTHECANONICALTOPOLOGY 9
Proof. LetQ := M/N. Theconditiononthesupportsmeansthatnominimalprimepof
M liesinthesupportofQ.However,forsuchp,wehaveM =H0(M). HenceN =M
p p p p
byCorollary5.1,showingthatQ =0.SupposenextthatM hasnoembeddedprimesand
p
Supp(Q) is nowheredensein Supp(M). LetD := cyc(N)+cyc(Q)−cyc(M)as
givenbyCorollary4.4,andletpbeanassociatedprimeofQ. SinceSupp(Q)isnowhere
dense,pisnotanassociatedprimeofM.Sincethecyclecyc(M)−cyc(N)hassupport
inAss(M)andisequaltocyc(Q)−D,itmustbethezerocycle,sincecyc(Q)andD
havesupportdisjointfromAss(M). HenceN ⊆M isopenbyCorollary5.1. (cid:3)
TogetherwithProposition5.3,thisproves:
5.8.Corollary. IfN ⊆ M isopenthendim(M/N) < dim(M),andtheconverseholds
ifM isunmixed. (cid:3)
In[6],wecalledamoduleM weaklycompressible,ifM embeds(asasubmodule)in
everyopensubmodule.Thefollowinggeneralizes[6,Corollary6.12]:
5.9.Corollary. IfM hasnoembeddedprimeideals,thenitisweaklycompressible.
Proof. LetN ⊆ M beopenandputQ := M/N. ByCorollary5.7,nominimal(whence
associated)primeofM liesinthesupportofQ. Hencethereexistsx ∈ Ann(Q)outside
any associated prime of M. In particular, xM ⊆ N and multiplication by x gives an
isomorphismM ∼=xM. (cid:3)
6. THECANONICAL TOPOLOGY
Asthenameindicates,thereisanunderlyingtopology.Toprovethis,weneed:
6.1.Theorem. Theinverseimageofanopensubmoduleunderamorphismisagainopen.
Inparticular,ifU ⊆M isopen,andN ⊆M isarbitrary,thenN ∩U isopeninN.
Proof. We start with provingthe secondassertion. Let µ := len(M), ν := len(N) and
α:=len(N ∩U). Wehaveanexactsequence
0→N ∩U →N ⊕U →N +U →0.
SinceN+U isagainopen,semi-additivityyieldslen(N⊕U)≤α⊕µ. Sincetheformer
is equal to ν ⊕µ, we get ν ≤ α, showing that N ∩U is open in N. To prove the first
assertion, let f: M → N be an arbitrary homomorphism and let V ⊆ N be an open
submodule.LetG⊆M ⊕N bethegraphoff andletp: G→M betheprojectiononto
thefirstcoordinate.SinceM ⊕V isopeninM ⊕N bysemi-additivity,(M ⊕V)∩Gis
openinG,bywhatwejustproved.Sincepisanisomorphism,theimageof(M⊕V)∩G
underpisthereforeopeninM. Butthisimageisjustf−1(V),andsowearedone. (cid:3)
We can nowdefine a topologyon M by letting the collectionof open submodulesbe
a basis of open neighborhoodsof 0 ∈ M. This is indeed a basis since the intersection
offinitelymanyopensisagainopenbyTheorem6.1. Anarbitraryopeninthistopology
is then a (possibly infinite) union of cosets x+U with U ⊆ M open and x ∈ M. If a
submoduleN is a unionofcosets x +U of opensubmodulesU ⊆ M, then onesuch
i i i
coset, x+U say, must contain 0, so that U ⊆ N, and hence the inequalitieslen(U) ≤
len(N)≤len(M)areallequalities,showingthatN isindeedopenintheprevioussense.
WecallthisthecanonicaltopologyonM,andTheorem6.1showsthatanyhomomor-
phism is continuous in the canonical topology. Moreover, multiplication on any ring is
continuous: givena ,a ∈ RandanopenidealI suchthata a ∈ I,letJ := a R+I.
1 2 1 2 i i
10 HANSSCHOUTENS
HenceJ isanopenneighborhoodofa andJ ·J ⊆ a a R+I = I. IfM hasdimen-
i i 1 2 1 2
sionzero,thenthecanonicaltopologyistrivial,sinceM isthentheonlyopensubmodule.
ThecomplementofanopenmoduleN istheunionofallcosetsa+N witha ∈/ N,and
henceisalsoopen. Inparticular,anopenmoduleisalsoclosed,andthequotienttopology
on M/N is discrete, whence in generaldifferentfrom the canonical topology. The zero
moduleisclosedifandonlyiftheintersectionofallopensubmodulesis0,thatistosay,
ifandonlyifthecanonicaltopologyisHaussdorf.
6.2.Corollary. Amoduleisnon-Artinianifandonlyifitscanonicaltopologyisnon-trivial.
Proof. OnedirectionisimmediatesinceanArtinianmodulehasnoproperopensubmod-
ules. Fortheconverse,weshow,byinductionon len(M),thatM hasanon-trivial,open
submodule. Assumefirstthatµisalimitordinal. ChooseasubmoduleN ofM ofheight
rank1.By(3),thismeanslen(M/N)=1,andsoN isopenbyProposition5.3. Next,as-
sumeµ=ν+1.LetH beasubmoduleofheightrankν,sothatby(3)again,M¯ :=M/H
haslengthν. Byinduction,wecanfindaproperopensubmoduleofM¯,thatistosay,we
canfindN M containingH suchthatlen(N/H) = ν. Semi-additivityappliedtothe
inclusionH ⊆ N yieldsν+len(H) ≤ len(N). Sincelen(H)6= 0andlen(N) ≤ ν+1,
wegetequality,thatistosay,len(N)=ν+1,whenceN isopen. (cid:3)
6.3. Proposition. The closure of a submodule N ⊆ M is equal to N + D (M). In
0
particular,asubmoduleisclosedifandonlyifithasthesamefinitisticlengthasM.
Proof. Let H := D (M), and let W be an open submodule containing N. By Theo-
0
rem 6.1, the intersection H ∩W is openin H, and since H has finite length(so that its
topologyistrivial),wemusthaveH = W ∩H,provingthatH liesinW. Asthisholds
forallopensW containingN,theclosureofN containsH.
Using Corollary 3.2, oneeasily showsthatthequotienttopologyonM/H isequalto
thecanonicaltopology. Therefore,tocalculatetheclosure,wemaydivideoutH,assume
thatM haspositiveorder,andwethenneedtoshowthatN isclosed. Letx ∈ M beany
elementnotin N andletm bea maximalidealcontaining(N : x). SinceM/mkM has
finitelength,eachmkM isopenbyProposition5.3,whencesoiseachN +mkM. Ifxis
containedineachofthese,thenitiscontainedintheirintersectionW := ∩(N +mkM).
ByKrull’sIntersectiontheorem,thereexistsa∈msuchthat(1+a)W ⊆N. Inparticular,
1+a ∈ (N : x) ⊆ m,contradiction. SoxliesoutsidesomeopenN +mkM,andhence
doesnotbelongtotheclosureofN.
ThelastassertionisnowalsoclearbyTheorem 4.1,sincethefinitistic lengthofM is
equaltothelengthofD (M). (cid:3)
0
6.4. Corollary. A module has a separated canonicaltopology if and only if its order is
positiveifandonlyifanysubmoduleisclosed. (cid:3)
6.5.Example. If (R,m) is localand M has positivedepth, then the canonicaltopology
refines the m-adic topology on M: indeed mkM is then open by Proposition 5.3. We
alreadyshowedthatthis isnolongertruein ringsofpositivedepthin Remark5.6. Also
notethatthecanonicaltopologyonalocaldomainisstrictlyfinerthanitsadictopology,
sinceallnon-zeroidealsareopen.
Theringk[[x,y]]/(x2,xy),oflengthω+1,isnotHaussdorfastheclosureofthezero
ideal is the ideal (x) by Proposition 6.3. It is the only closed, non-openideal, since the
closureofanyidealmustcontainxwhenceisopenwhendifferentfrom(x). Inparticular,
whereasthecanonicaltopologyisnotHaussdorf,theadiconeis.