BRAUER{THRALLFORTOTALLYREFLEXIVEMODULESLARSWINTHERCHRISTENSEN,DAVIDA.J - PDF document

BRAUER{THRALLFORTOTALLYREFLEXIVEMODULESLARSWINTHERCHRISTENSEN,DAVIDA.JORGENSEN,HAMIDREZARAHMATI,JANETSTRIULI,ANDROGERWIEGANDAbstract.LetRbeacommutativenoetherianlocalringthatisnotGoren-stein.Itisknownthatthecategoryoftotallyre exivemodulesoverRisrepresentationinnite,providedthatitcontainsanon-freemodule.Themaingoalofthispaperistounderstandhowcomplexthecategoryoftotallyre exivemodulescanbeinthissituation.Localrings(R;m)withm3=0arecommonlyregardedasthestructurallysimplestringstoadmitdiversecategoricalandhomologicalcharacteristics.Forsuchringsweobtainconclusiveresultsaboutthecategoryoftotallyre exivemodules,modeledontheBrauer{Thrallconjectures.Startingfromanon-freecyclictotallyre exivemodule,weconstructafamilyofindecomposabletotallyre exiveR-modulesthatcontains,foreveryn2N,amodulethatisminimallygeneratedbynelements.Moreover,iftheresidueeldR=misalgebraicallyclosed,thenweconstructforeveryn2Naninnitefamilyofindecomposableandpairwisenon-isomorphictotallyre exiveR-modules,eachofwhichisminimallygeneratedbynelements.Themodulesinbothfamilieshaveperiodicminimalfreeresolutionsofperiodatmost2.Contents1.Introductionandsynopsisofthemainresults22.Totallyacycliccomplexesandexactzerodivisors43.Familiesofindecomposablemodulesofdierentsize74.Brauer{ThrallIovershortlocalringswithexactzerodivisors125.Exactzerodivisorsfromtotallyre exivemodules146.Familiesofnon-isomorphicmodulesofthesamesize177.Brauer{ThrallIIovershortlocalringswithexactzerodivisors218.Existenceofexactzerodivisors249.Shortlocalringswithoutexactzerodivisors|Anexample2910.Familiesofnon-isomorphicmodulesofinnitelength32Acknowledgments33References33 Date:October1,2010.2010MathematicsSubjectClassication.Primary:16G10;13D02.Secondary:16G60;13C14.Keywordsandphrases.Brauer{Thrallconjectures,exactzerodivisor,Gorensteinrepresenta-tiontype,maximalCohen{Macaulaymodule,totallyre exivemodule.ThisresearchwaspartlysupportedbyNSAgrantH98230-10-0197(D.A.J.),NSFgrantDMS0901427(J.S.),andbyaUNLFacultyDevelopmentFellowship(R.W.).1 2L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGAND1.IntroductionandsynopsisofthemainresultsTherepresentationtheoreticpropertiesofalocalringbearpertinentinformationaboutitssingularitytype.AnotableillustrationofthistenetisduetoHerzog[10]andtoBuchweitz,Greuel,andSchreyer[3].TheyshowthatacompletelocalGorensteinalgebraisasimplehypersurfacesingularityifitscategoryofmaximalCohen{Macaulaymodulesisrepresentationnite.Amodulecategoryiscalledrepresentationniteifitcomprisesonlynitelymanyindecomposablemodulesuptoisomorphism.TypicalexamplesofmaximalCohen{MacaulaymodulesoveraCohen{Macaulaylocalringarehighsyzygiesofnitelygeneratedmodules.OveraGorensteinlocalring,allmaximalCohen{Macaulaymodulesariseashighsyzygies,butoveranarbitraryCohen{Macaulaylocalringtheymaynot.Totallyre exivemodulesareinnitesyzygieswithspecialdualityproperties;theprecisedenitionisgivenbelow.Onereasontostudythesemodules|infact,theonediscoveredmostrecently|isthattheyaordacharacterizationofsimplehypersurfacesingularitiesamongallcompletelocalalgebras,i.e.withoutanyaprioriassumptionofGorensteinness.Thisextensionoftheresultfrom[3,10]isobtainedin[5].Itisconsonantwiththeintuitionthatthestructureofhighsyzygiesisshapedpredominantlybythering,andthesameintuitionguidesthiswork.Thekeyresultin[5]assertsthatifalocalringisnotGorensteinandthecategoryoftotallyre exivemodulescontainsanon-freemodule,thenitisrepresentationinnite.Themaingoalofthispaperistodeterminehowcomplexthecategoryoftotallyre exivemodulesiswhenitisrepresentationinnite.Ourresultssuggestthatitisoftenquitecomplex;Theorems(1.1)and(1.4)belowaremodeledontheBrauer{Thrallconjectures.ForanitedimensionalalgebraA,therstBrauer{ThrallconjectureassertsthatifthecategoryofA-modulesofnitelengthisrepresentationinnite,thenthereexistindecomposableA-modulesofarbitrarilylargelength.Thesecondcon-jectureassertsthatiftheunderlyingeldisinnite,andthereexistindecomposableA-modulesofarbitrarilylargelength,thenthereexistinnitelymanyintegersdsuchthatthereareinnitelymanyindecomposableA-modulesoflengthd.TherstconjecturewasprovedbyRoter(1968);thesecondconjecturehasbeenver-ied,forexample,foralgebrasoveralgebraicallyclosedeldsbyBautista(1985)andBongartz(1985).Inthissection,Risacommutativenoetherianlocalring.Thecentralquestionsaddressedinthepaperare:Assumingthatthecategoryoftotallyre exiveR-modulesisrepresentationinniteandgivenanon-freetotallyre exiveR-module,howdoesoneconstructaninnitefamilyofpairwisenon-isomorphictotallyre- exiveR-modules?And,canonecontrolthesizeofthemodulesinthefamilyinaccordancewiththeBrauer{Thrallconjectures?AnitelygeneratedR-moduleMiscalledtotallyre exiveifthereexistsaninnitesequenceofnitelygeneratedfreeR-modulesF:


�!F1�!F0�!F�1�!


;suchthatMisisomorphictothemoduleCoker(F1!F0),andsuchthatbothFandthedualsequenceHomR(F;R)areexact.ThesemoduleswererststudiedbyAuslanderandBridger[1],whoprovedthatRisGorensteinifandonlyifeveryR-modulehasatotallyre exivesyzygy.OveraGorensteinring,thetotallyre exive BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES3modulesarepreciselythemaximalCohen{Macaulaymodules,andthesehavebeenstudiedextensively.IntherestofthissectionweassumethatRisnotGorenstein.Everysyzygyofanindecomposabletotallyre exiveR-moduleis,itself,indecom-posableandtotallyre exive;aproofofthisfolkloreresultisincludedinSection2.Thusifoneweregivenatotallyre exivemodulewhoseminimalfreeresolutionisnon-periodic,thenthesyzygieswouldformthedesiredinnitefamily;thoughonecannotexerciseanycontroloverthesizeofthemodulesinthefamily.Inpractice,however,thetotallyre exivemodulesthatonetypicallyspotshaveperiodicfreeresolutions.Toillustratethispoint,considertheQ-algebraA=Q[s;t;u;v]=(s2;t2;u2;v2;uv;2su+tu;sv+tv):Ithassomeeasilyrecognizabletotallyre exivemodules|A=(s)andA=(s+u)forexample|whoseminimalfreeresolutionsareperiodicofperiodatmost2.Italsohasindecomposabletotallyre exivemoduleswithnon-periodicfreeresolutions.However,suchmodulesaresignicantlyhardertorecognize.Infact,whenGasharovandPeeva[8]didso,itallowedthemtodisproveaconjectureofEisenbud.ThealgebraAabovehasHilbertseries1+4+32.Inparticular,Aisalocalring,andthethirdpowerofitsmaximalidealiszero;informallywerefertosuchringsasshort.Fortheserings,[14]givesaquantitativemeasureofhowchallengingitcanbetorecognizetotallyre exivemoduleswithnon-periodicresolutions.Shortlocalringsarethestructurallysimplestringsthataccommodateawiderangeofhomologicalbehavior,and[8]and[14]arebuttwoarmationsthatsuchringsareexcellentgroundsforinvestigatinghomologicalquestionsinlocalalgebra.Fortherestofthissection,assumethatRisshortandletmbethemaximalidealofR.Notethatm3iszeroandsete=dimR=mm=m2.AreadersoinclinediswelcometothinkofastandardgradedalgebrawithHilbertseries1+e+h22.Thefamiliesoftotallyre exivemodulesconstructedinthispaperstartfromcyclicones.Overashortlocalring,suchmodulesaregeneratedbyelementswithcyclicannihilators;HenriquesandSega[9]calltheseelementsexactzerodivisors.Theubiquityofexactzerodivisorsinshortlocalalgebrasisalong-standingempiri-calfact.ItstheoreticalunderpinningsarefoundinworksofConca[7]andHochsterandLaksov[11];weextendtheminSection8.ThemainresultsofthispaperareTheorems(1.1){(1.4).ItisknownfromworkofYoshino[19]thatthelengthofatotallyre exiveR-moduleisamultipleofe.InSection4weprovetheexistenceofindecomposabletotallyre exiveR-modulesofeverypossiblelength:(1.1)Theorem(Brauer{ThrallI).IfthereisanexactzerodivisorinR,thenthereexistsafamilyfMngn2Nofindecomposabletotallyre exiveR-moduleswithlengthRMn=neforeveryn.Moreover,theminimalfreeresolutionofeachmoduleMnisperiodicofperiodatmost2.OurproofofthisresultisconstructiveinthesensethatweexhibitpresentationmatricesforthemodulesMn;theyarealluppertriangularsquarematriceswithexactzerodivisorsonthediagonal.Yet,thestrongconversecontainedinTheo-rem(1.2)cameasasurprisetous.Itillustratesthepointthatthestructureoftheringisrevealedinhighsyzygies. 4L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGAND(1.2)Theorem.Ifthereexistsatotallyre exiveR-modulewithoutfreesum-mands,whichispresentedbyamatrixthathasacolumnorarowwithonlyonenon-zeroentry,thenthatentryisanexactzerodivisorinR.Thesetworesults|thelatterofwhichisdistilledfromTheorem(5.3)|showthatexistenceoftotallyre exivemodulesofanysizeisrelatedtotheexistenceofexactzerodivisors.Onedirection,however,isnotunconditional,andinSection9weshowthatnon-freetotallyre exiveR-modulesmayalsoexistintheabsenceofexactzerodivisorsinR.Ifthisphenomenonappearspeculiar,someconsolationmaybefoundinthenexttheorem,whichisprovedinSection8.Foralgebraicallyclosedelds,itcanbededucedfromresultsin[7,11,19].(1.3)Theorem.Letkbeaninniteeld,andletRbeagenericstandardgradedk-algebrawhich(1)hasHilbertseries1+h1+h22,(2)isnotGorenstein,and(3)admitsanon-freetotallyre exivemodule.ThenRhasanexactzerodivisor.IftheresidueeldR=misinnite,andthereisanexactzerodivisorinR,thenthereareinnitelymanydierentones;thisismadepreciseinTheorem(7.6).TogetherwithacoupleofotherresultsfromSection7thistheoremyields:(1.4)Theorem(Brauer{ThrallII).IfthereisanexactzerodivisorinR,andtheresidueeldk=R=misalgebraicallyclosed,thenthereexistsforeachn2NafamilyfMng2kofindecomposableandpairwisenon-isomorphictotallyre exiveR-moduleswithlengthRMn=neforevery.Moreover,theminimalfreeresolutionofeachmoduleMnisperiodicofperiodatmost2.ThefamiliesofmodulesinTheorems(1.1)and(1.4)comefromaconstructionthatcanprovidesuchfamiliesoveragenerallocalring,contingentontheexistenceofminimalgeneratorsofthemaximalidealwithcertainrelationsamongthem.ThisconstructionispresentedinSection2andanalyzedinSections3and6.ToestablishtheBrauer{Thralltheorems,weproveinSections4and7thatthenecessaryele-mentsandrelationsareavailableinshortlocalringswithexactzerodivisors.TheexistenceofexactzerodivisorsisaddressedinSections5,8,and9.InSection10wegiveanotherconstructionofinnitefamiliesoftotallyre exivemodules.Itappliestocertainringsofpositivedimension,anditdoesnotdependontheexistenceofexactzerodivisors.2.TotallyacycliccomplexesandexactzerodivisorsInthispaper,Rdenotesacommutativenoetherianring.ComplexesofR-modulesaregradedhomologically.AcomplexF:


�!Fi+1@Fi+1����!Fi@Fi���!Fi�1�!


ofnitelygeneratedfreeR-modulesiscalledacyclicifeverycycleisaboundary;thatis,theequalityKer@Fi=Im@Fi+1holdsforeveryi2Z.IfbothFandthedualcomplexHomR(F;R)areacyclic,thenFiscalledtotallyacyclic.ThusanR-moduleistotallyre exiveifandonlyifitisthecokernelofadierentialinatotallyacycliccomplex.TheannihilatorofanidealainRiswritten(0:a).Forprincipalidealsa=(a)weusethesimpliednotation(0:a). BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES5Recallfrom[9]thenotionofanexactzerodivisor:anon-invertibleelementx6=0inRiscalledanexactzerodivisorifoneofthefollowingequivalentconditionsholds.(i)Thereisanisomorphism(0:x)=R=(x).(ii)ThereexistsanelementwinRsuchthat(0:x)=(w)and(0:w)=(x).(iii)ThereexistsanelementwinRsuchthat(2.0.1)


�!Rw��!Rx�!Rw��!R�!


isanacycliccomplex.Foreveryelementwasabove,onesaysthatwandxformanexactpairofzerodivisorsinR.IfRislocal,thenwisuniqueuptomultiplicationbyaunitinR.(2.1)Remark.Foranon-unitx6=0theconditions(i){(iii)aboveareequivalentto(iv)ThereexistelementswandyinRsuchthatthesequenceRw��!Rx�!Ry�!Risexact.Indeed,exactnessofthissequenceimpliesthatthereareequalities(0:y)=(x)and(0:x)=(w).Thusthereisanobviousinclusion(y)(w)and,therefore,aninclusion(0:w)(0:y)=(x).Asxannihilatesw,thisforcestheequality(x)=(0:w).Thus(iv)implies(ii)and,clearly,(iv)followsfrom(iii).Startingfromthenextsection,weshallassumethatRislocal.Weusethenotation(R;m)toxmastheuniquemaximalidealofRandthenotation(R;m;k)toalsoxtheresidueeldk=R=m.AcomplexFoffreemodulesoveralocalring(R;m)iscalledminimalifonehasIm@FimFi�1foralli2Z.LetMbeanitelygeneratedR-moduleandletFbeaminimalfreeresolutionofM;itisuniqueuptoisomorphism.TheithBettinumber,Ri(M),istherankofthefreemoduleFi,andtheithsyzygyofMisthemoduleCoker@Fi+1.(2.2)Remark.Thecomplex(2.0.1)isisomorphictoitsowndual,soifitisacyclic,thenitistotallyacyclic.ThusifwandxformanexactpairofzerodivisorsinR,thenthemodules(w)=R=(x)and(x)=R=(w)aretotallyre exive.Moreover,itfollowsfromcondition(iv)thatatotallyacycliccomplexoffreemodules,inwhichfourconsecutivemoduleshaverank1,hastheform(2.0.1).Thismeans,inparticular,thatoveralocalringR,anytotallyre exivemoduleMwithRi(M)=1for06i63hastheformM=R=(x),wherexisanexactzerodivisor.FormodulesovershortlocalringsweproveastrongerstatementinTheorem(5.3).Thenextlemmaisfolklore;itisprovedforGorensteinringsin[10].(2.3)Lemma.LetRbelocalandletTbeaminimaltotallyacycliccomplexofnitelygeneratedfreeR-modules.Thefollowingconditionsareequivalent:(i)ThemoduleCoker@Tiisindecomposableforsomei2Z.(ii)ThemoduleCoker@Tiisindecomposableforeveryi2Z.Inparticular,everysyzygyofanindecomposabletotallyre exiveR-moduleisin-decomposableandtotallyre exive.Proof.Foreveryi2ZsetMi=Coker@Ti.Bydenition,themodulesMiandHomR(Mi;R)aretotallyre exiveandonehasMi=HomR(HomR(Mi;R);R) 6L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDforeveryi2Z.Assumethatforsomeintegerj,themoduleMjisindecom-posable.Foreveryij,themoduleMjisasyzygyofMi,andeverysum-mandofMihasinniteprojectivedimension,asTisminimalandtotallyacyclic.ThusMiisindecomposable.Further,iftherewereanon-trivialdecompositionMi=KNforsomei&#x-530;j,thenthedualmoduleHomR(Mi;R)woulddecom-poseasHomR(K;R)HomR(N;R),wherebothsummandswouldbenon-zeroR-modulesofinniteprojectivedimension.However,HomR(Mj;R)isasyzygyofHomR(Mi;R),soHomR(Mj;R)wouldthenhaveanon-trivialdecompositionandsowouldMj=HomR(HomR(Mj;R);R);acontradiction.(2.4)Lemma.LetTandFbecomplexesofnitelygeneratedfreeR-modules.IfTistotallyacyclic,andthemodulesFiarezerofori0,thenthecomplexT RFistotallyacyclic.Proof.ThecomplexT RFisacyclicby[4,lem.2.13].AdjointnessofHomandtensoryieldstheisomorphismHomR(T RF;R)=HomR(F;HomR(T;R)).AsthecomplexHomR(T;R)isacyclic,soisHomR(F;HomR(T;R))by[4,lem.2.4].(2.5)Denition.Forn2NandelementsyandzinRwithyz=0,letLn(y;z)bethecomplexdenedasfollowsLn(y;z)i=(Rfor0&#x-530;i&#x-530;�n+10elsewhereand@Ln(y;z)i=(yforieven�zforiodd.IfwandxformanexactpairofzerodivisorsinR,andTisthecomplex(2.0.1),thenthedierentialsofthecomplexT RLn(y;z)haveaparticularlysimpleform;seeRemark(2.7).InSections3and6westudymoduleswhosepresentationma-triceshavethisform.(2.6)Construction.Letn2N,letInbethennidentitymatrix,andletridenoteitsithrow.ConsiderthennmatricesIon,Ien,Jon,andJendenedbyspecifyingtheirrowsasfollows(Ion)i=(riiodd0ievenand(Ien)i=(0ioddriieven(Jon)i=(ri+1iodd0ievenand(Jen)i=(0ioddri+1ievenwiththeconventionrn+1=0.TheequalityIon+Ien=Inisclear,andthematrixJn=Jon+JenisthennnilpotentJordanblockwitheigenvaluezero.Forelementsw,x,y,andzinRletMn(w;x;y;z)betheR-modulewithpresen-tationmatrixn(w;x;y;z)=wIon+xIen+yJon+zJen;itisanuppertriangularnnmatrixwithwandxalternatingonthediagonal,andwithyandzalternatingonthesuperdiagonal:n(w;x;y;z)=0BBBBBBB@wy000:::0xz00:::00wy0:::000xz0000w..................1CCCCCCCA: BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES7Forn=1thematrixhasonlyoneentry,namelyw.Forn=2,thematrixdoesnotdependonz,sowesetM2(w;x;y)=M2(w;x;y;z)foreveryz2R.Notethatif(R;m)islocalandw,x,y,andzarenon-zeroelementsinm,thenMn(w;x;y;z)isminimallygeneratedbynelements.(2.7)Remark.AssumethatwandxformanexactpairofzerodivisorsinR,andletTbethecomplex(2.0.1),positionedsuchthat@T1ismultiplicationbyw.Letn2NandletyandzbeelementsinRthatsatisfyyz=0.ItfollowsfromLemma(2.4)thatthecomplexT RLn(y;z)istotallyacyclic.Itiselementarytoverifythatthedierential@T RLn(y;z)iisgivenbythematrixn(w;x;y;z)forioddandbyn(x;w;�y;�z)forieven,cf.Construction(2.6).Inparticular,themoduleMn(w;x;y;z)istotallyre exive.3.FamiliesofindecomposablemodulesofdifferentsizeWithappropriatelychosenringelementsasinput,Construction(2.6)yieldstheinnitefamiliesofmodulesintheBrauer{ThralltheoremsadvertisedinSection1.Inthissectionwebegintoanalyzetherequirementsontheinput.(3.1)Theorem.Let(R;m)bealocalringandassumethatwandxareelementsinmnm2,thatformanexactpairofzerodivisors.Assumefurtherthatyandzareelementsinmnm2withyz=0andthatoneofthefollowingconditionsholds:(a)Theelementsw,x,andyarelinearlyindependentmodulom2.(b)Onehasw2(x)+m2andy;z62(x)+m2.Foreveryn2N,theR-moduleMn(w;x;y;z)isindecomposable,totallyre exive,andnon-free.Moreover,Mn(w;x;y;z)hasconstantBettinumbers,equalton,anditsminimalfreeresolutionisperiodicofperiodatmost2.Theproofof(3.1)|whichtakesupthebalanceofthesection|employsanaux-iliaryresultofindependentinterest,Proposition(3.2)below;itsproofisdeferredtotheendofthesection.(3.2)Proposition.Let(R;m)bealocalring,letnbeapositiveinteger,andletw,x,y,andzbeelementsinmnm2.(a)Assumethatw,x,andyarelinearlyindependentmodulom2.Ifniseven,thenMn(w;x;y;z)isindecomposable.Ifnisodd,thenMn(w;x;y;z)orMn(x;w;y;z)isindecomposable.(b)Ify=2(w;x)+m2andz=2(x)+m2hold,thenMn(w;x;y;z)isindecomposable.ProofofTheorem(3.1).Letnbeapositiveinteger;inviewofRemark(2.7),allweneedtoshowisthattheR-moduleMn(w;x;y;z)isindecomposable.(a):Assumethatw,x,andyarelinearlyindependentmodulom2.ByRe-mark(2.7)themoduleMn(w;x;y;z)istherstsyzygyofMn(x;w;�y;�z).IfthemoduleMn(x;w;y;z)isindecomposable,thensoistheisomorphicmoduleMn(x;w;�y;�z),anditfollowsfromLemma(2.3)thatMn(w;x;y;z)isinde-composableaswell.ThusbyProposition(3.2)(a)themoduleMn(w;x;y;z)isindecomposable.(b):Undertheassumptionsw2(x)+m2andy;z=2(x)+m2,theconditionsinProposition(3.2)(b)aremet,sotheR-moduleMn(w;x;y;z)isindecomposable. 8L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDProofofProposition(3.2).Letnbeapositiveintegerandletw,x,y,andzbeelementsinmnm2.Itisconvenienttoworkwithapresentationmatrixn(w;x;y;z)forM=Mn(w;x;y;z)thatoneobtainsasfollows.Setp=dn 2eandletbethennmatrixobtainedfromInbypermutingitsrowsaccordingto123456


n1p+12p+23p+3:::p+p;with=0ifnisoddand=1ifniseven.Setn(w;x;y;z)=n(w;x;y;z)�1.Ifniseven,thennistheblockmatrixn(w;x;y;z)=wIpyIpzJpxIp;andifnisodd,thenn(w;x;y;z)isthematrixobtainedfromn+1(w;x;y;z)bydeletingthelastrowandthelastcolumn.ToverifythatMisindecomposable,assumethat"2HomR(M;M)isidempo-tentandnottheidentitymap1M.Thegoalistoshowthat"isthezeromap.TheonlyidempotentautomorphismofMis1M,so"isnotanisomorphism.Thus"isnotsurjective,asMisnoetherian.Set=n(w;x;y;z)andconsiderthecommutativediagramwithexactrows(1)Rn// B Rn// A M// " 0Rn// Rn// M// 0obtainedbylifting".LetA=(aij)and BbethennmatricesobtainedfromAandBbyreducingtheirentriesmodulom.Toprovethat"isthezeromap,itsucestoshowthatthematrixAisnilpotent.Indeed,ifAisnilpotent,thensoisthemap":M=mM!M=mM.As"isalsoidempotent,itisthezeromap.NowitfollowsfromNakayama'slemmathatthemap1M�"issurjectiveandhenceanisomorphism.As1M�"isidempotent,itfollowsthat1M�"istheidentitymap1M;thatis,"=0.Claim.IfthematrixAisuppertriangularwithidenticalentriesonthediagonal,i.e.a11=a22=


=ann,thenitisnilpotent.Proof.Since"isnotsurjective,thematrixAdoesnotrepresentasurjectivemapand,byNakayama'slemma,neitherdoesA.Therefore,thediagonalentriesofAcannotallbenon-zero,whencetheyareallzero,andAisnilpotent.Denoteby~w,~x,~y,and~ztheimagesofw,x,y,andzinm=m2,andletVbethek-subspaceofm=m2spannedby~w,~x,~y,and~z.Considerthefollowingpossibilities:(I)Theelements~w,~x,~y,and~zformabasisforV.(II)Theelements~x,~y,and~zformabasisforV,andk~w=k~xholds.(III)Theelements~w,~x,and~yformabasisforV.(IV)Theelements~xand~yformabasisforV,andonehask~w=k~xand~z=2k~x.Undertheassumptionsonw,x,andyinpart(a),oneoftheconditions(I)or(III)holds.Undertheassumptionsinpart(b),oneoftheconditions(I){(IV)holds.Indeed,ifVhasdimension4,then(I)holds.Ifthatisnotthecase,thenthedimensionofVis2or3.IncasedimkV=2,theelements~xand~yformabasisforV,whereas~wand~xcannotbelinearlyindependent;thus(IV)holds.Incase BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES9dimkV=3,condition(II)or(III)holds,dependingonwhetherornottheequalityk~w=k~xholds.Therestoftheproofissplitintwo,accordingtotheparityofn.ToprovethatMn(w;x;y;z)isindecomposable,itsucestoprovethatthematrixAisnilpotent.Thisishowweproceedundereachoftheconditions(I),(II),and(IV),andundercondition(III)whenniseven.Whennisodd,andcondition(III)holds,weshowthatoneofthemodulesMn(w;x;y;z)andMn(x;w;y;z)isindecomposable.Case1:niseven.Letedenotethematrixobtainedfrombyreducingtheentriesmodulom2.WriteAand BasblockmatricesA=A11A12A21A22and B=B11B12B21B22;whereAijandBijareppmatriceswithentriesfromk.By(1),theequalityA=Bholds;itimpliesanequalityofblockmatrices(2)~wA11+~zA12Jp~yA11+~xA12~wA21+~zA22Jp~yA21+~xA22=~wB11+~yB21~wB12+~yB22~zJpB11+~xB21~zJpB12+~xB22:Assumerstthatcondition(I)or(III)holds,sothattheelements~w,~x,and~yarelinearlyindependent.Thentheequalityoftheupperrightblocks,~yA11+~xA12=~wB12+~yB22,yieldsA11=B22andA12=0=B12:Fromtheblocksonthediagonals,onenowgetsA11=B11;A21=0=B21;andA22=B22:ThusthematrixAhastheform�A1100A11
.Finally,theequalityofthelowerleftblocksyieldsA11Jp=JpA11.SinceJpisnon-derogatory,thisimpliesthatA11belongstothealgebrak[Jp]ofpolynomialsinJp.Thatis,thereareelementsc0;:::;cp�1inksuchthatA11=c0Ip+c1Jp+


+cp�1Jp�1p;see[13,thm.3.2.4.2].Inparticular,thematricesA11and,therefore,Aareupper-triangularwithidenticalentriesonthediagonal.ByClaim,Aisnilpotentasdesired.Undereithercondition(II)or(IV),theelements~xand~yarelinearlyindependent,~zisnotink~x,andtheequalityk~w=k~xholds.Inparticular,thereisanelementt6=0inksuchthatt~w=~x.Fromtheo-diagonalblocksin(2)oneobtainsthefollowingrelations:A11=B22tA12=B12andA22Jp=JpB11A21=tB21:(3)If(II)holds,thentheblocksonthediagonalsin(2)yieldA11=B11;A22=B22;andA21=0:ThusthematrixAhastheform�A11A120A11
,andtheequalityA11Jp=JpA11holds.Asabove,itfollowsthatthematricesA11and,therefore,Aareupper-triangularwithidenticalentriesonthediagonal.Thatis,AisnilpotentbyClaim.Nowassumethat(IV)holds.Thereexistelementsrands6=0inksuchthat~z=r~x+s~y.Comparisonoftheblocksonthediagonalsin(2)nowyieldsA11+rtA12Jp=B11sA12Jp=B21andA22=rJpB12+B22A21=sJpB12:(4) 10L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDCombinetheseequalitieswiththosefrom(3)togetA=A11A12stJpA12A11+rtJpA12:ItfollowsfromtheequalitiesJpA12=t�1JpB12=(st)�1A21=s�1B21=A12Jp;derivedfrom(3)and(4),thatthematrixA12commuteswithJp;henceitbelongstok[Jp].Similarly,thechainofequalitiesJpA11=JpB11�rtJpA12Jp=JpB11�rJpB12Jp=A22Jp�rJpB12Jp=B22Jp=A11JpshowsthatA11isink[Jp].ThusallfourblocksinAbelongtok[Jp].Fornotationalbliss,identifyk[Jp]withtheringS=k[]=(p),wherecorrespondstotJp.Withthisidentication,Atakestheformofa22matrixwithentriesinS:fgsgf+rg:AsAisnotinvertible,thedeterminantf2+frg�sg2belongstothemaximalideal()ofS.Itfollowsthatfisin(),whenceonehasA2p=0asdesired.Case2:nisodd.Setq=p�1,wherep=dn 2e=n+1 2.Thepresentationmatrixtakestheform=wIpyHzKxIq;whereHandKarethefollowingblockmatricesH=Iq01qandK=�0q1Iq
.Noticethatthereareequalities(5)HX=X01mandX0K=�0m01X0
foreveryqmmatrixXandeverym0qmatrixX0.Furthermore,itisstraight-forwardtoverifytheequalities(6)HK=JpandKH=Jq:AsinCase1,writeA=A11A12A21A22and B=B11B12B21B22;where,now,AijandBijarematricesofsizemimj,form1=pandm2=q.WitheasdenedinCase1,therelationAe=e B,derivedfrom(1),yields:(7)~wA11+~zA12K~yA11H+~xA12~wA21+~zA22K~yA21H+~xA22=~wB11+~yHB21~wB12+~yHB22~zKB11+~xB21~zKB12+~xB22:Assumerstthatcondition(I)or(III)holds,sothattheelements~w,~x,and~yarelinearlyindependent.Fromtheequalityoftheupperrightblocksin(7)onegets(8)A11H=HB22andA12=0=B12:Inviewof(5),comparisonoftheblocksonthediagonalsnowyieldsA11=B11B21=0andA21H=0A22=B22:(9) BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES11Fromtherstequalityin(8)andthelastequalityin(9)onegetsinviewof(5)(10)A11=A2201q;hereandinthefollowingthesymbol`'inamatrixdenotesanunspeciedblockofappropriatesize.Togleaninformationfromtheequalityofthelowerleftblocksin(7),assumerstthat~wand~zarelinearlyindependent.Thenonehas(11)A21=0andA22K=KB11:Combinethiswith(10)andthesecondequalityin(8)toseethatthematrixAhastheform(12)A=0@A2201q 0pq 0qp A221A:TheequalitiesA11Jp=A11HK=HB22K=HA22K=HKB11=JpA11andA22Jq=A22KH=KB11H=KA11H=KHB22=JqA22;derivedfrom(6),(8),(9),and(11),showthatA11isink[Jp]andA22isink[Jq].ItfollowsthatAisuppertriangularwithidenticalentriesonthediagonal.ThusAisnilpotent,andMn(w;x;y;z)isindecomposable.If,ontheotherhand,~zand~warelinearlydependent,then~zand~xarelinearlyindependent,as~wand~xarelinearlyindependentbyassumption.ItfollowsfromwhatwehavejustshownthatMn(x;w;y;z)isindecomposable.Undereithercondition(II)or(IV),theelements~xand~yarelinearlyindependent,~zisnotink~x,andtheequalityk~w=k~xholds.Inparticular,thereisanelementt6=0inksuchthatt~w=~x.Comparetheo-diagonalblocksin(7)togettA12=B12A11H=HB22andA21=tB21A22K=KB11:(13)If(II)holds,thenacomparisonoftheblocksonthediagonalsin(7)combinedwith(5)implies(14)A11=B11;A12=0=B21;andA22=B22:Itfollowsfrom(13)and(14)thatthematrixA21iszero.InviewoftheequalityA11H=HB22from(13),itnowfollowsthatAhastheformgivenin(12).Usingtheequalitiesin(13)and(14),onecanrepeattheargumentsabovetoseethatA11isink[Jp]andA22isink[Jq],andcontinuetoconcludethatAisnilpotent.Finally,assumethat(IV)holds.Thereexistelementsrands6=0inksuchthat~z=r~x+s~y.Comparisonoftheblocksonthediagonalsin(7)yieldsA11+rtA12K=B11sA12K=HB21andA21H=sKB12A22=B22+rKB12:(15)TheequalitysA12K=t�1HA21,obtainedfrom(13)and(15),shows,inviewof(5),thatthematricesA12andA21havethefollowingform:A12=EandA21=�0q1�
; 12L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDwhereEand�areqqmatrices,andthelastrowofEiszero.Fromtheequalitiesin(6),(13),and(15)onegetsA21Jp=sKB12K=stKA12K=tKHB21=JqA21:Fromhereitisstraightforwardtoverifythat�commuteswithJq;i.e.�belongstok[Jq].Similarly,fromtheequalitiesA12Jq=s�1HB21H=(st)�1HA21H=t�1HKB12=JpA12;itfollowsthatEbelongstok[Jq].SincethelastrowinEiszero,allentriesonthediagonalofEarezero,andthematrixisnilpotent.Therstequalityin(15)cannowbewrittenasA11=B11�00q1rtE:Combinethiswiththelastequalityin(13)togetA=0@a11 0q1A22�rtE E 0q1� A221A:TheequalitiesA11Jp=HB22K=H(A22�rKB12)K=HA22K�rHKB12K=HKB11�rtHKA12K=JpA11showthatA11isink[Jp],andasimilarchainofequalitiesshowsthatA22isink[Jq].ItfollowsthatallentriesonthediagonalofAareidentical.Let
bethematrixobtainedbydeletingtherstrowandrstcolumninAandwriteitinblockform
=A22�rtEE�A22:AsAisnotinvertible,onehas0=detA=a11(det
).Ifa11isnon-zero,then
hasdeterminant0;inparticular,itisnotinvertible.Consideredasa22matrixovertheartinianlocalringk[Jq],itsdeterminant(A22)2�rtEA22��Ebelongstothemaximalideal(Jq).AsEisnilpotent,itbelongsto(Jq)andhencesodoesA22;thiscontradictstheassumptionthata11isnon-zero.Ifthediagonalentrya11is0,thenthematrixA22isnilpotent,whichimpliesthat
isnilpotentand,nally,thatAisnilpotent.4.Brauer{ThrallIovershortlocalringswithexactzerodivisorsLet(R;m;k)bealocalring.TheembeddingdimensionofR,denotedemb:dimR,istheminimalnumberofgeneratorsofm,i.e.thedimensionofthek-vectorspacem=m2.TheHilbertseriesofRisthepowerseriesHR()=P1i=0dimk(mi=mi+1)i.Intherestofthissection(R;m)isalocalringwithm3=0.Themainresult,Theorem(4.4),togetherwith(4.1.1)andTheorem(3.1),establishesTheorem(1.1).TowardstheproofofTheorem(4.4),werstrecapitulateafewfactsabouttotallyre exivemodulesandexactzerodivisors.IfRisGorenstein,theneveryR-moduleistotallyre exive;see[1,thm.(4.13)and(4.20)].IfRisnotGorenstein,thenexistenceofanon-freetotallyre exiveR-moduleforcescertainrelationsamonginvariantsofR.Thefactsin(4.1)areprovedbyYoshino[19];seealso[6]forthenon-gradedcase. BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES13(4.1)Totallyre exivemodules.AssumethatRisnotGorensteinandsete=emb:dimR.IfMisatotallyre exiveR-modulewithoutfreesummandsandminimallygeneratedbynelements,thentheequalities(4.1.1)lengthRM=neandRi(M)=nforalli�0hold.Moreover,m2isnon-zeroandthefollowinghold:(4.1.2)(0:m)=m2;dimkm2=e�1;andlengthR=2e:Inparticular,eisatleast3,andtheHilbertseriesofRis1+e+(e�1)2.Letkbeaeld.FortheGorensteinringR=k[x]=(x3),twooftherelationsin(4.1.2)fail.Thisringalsohasanexactzerodivisor,x2,inthesquareofthemaximalideal;forringswithembeddingdimension2orhigherthiscannothappen.(4.2)Exactzerodivisors.Sete=emb:dimRandassumee�2.SupposethatwandxformanexactpairofzerodivisorsinR.Asm2iscontainedintheannihilatorofm,thereisaninclusionm2(x),whichhastobestrict,as(w)=(0:x)isstrictlycontainedinm.Thusxisaminimalgeneratorofmwithxm=m2:Bysymmetry,wisaminimalgeneratorofmwithwm=m2.Letfv1;:::;ve�1;wgbeaminimalsetofgeneratorsform,thentheelementsxv1;:::;xve�1generatem2,anditiselementarytoverifythattheyformabasisform2asak-vectorspace.Itfollowsthattherelationsin(4.1.2)holdand,inaddition,thereisanequalitylengthR(x)=e:Notethatthesocle(0:m)ofRhasdimensione�1overk,soRisGorensteinifandonlyife=2.(4.3)Lemma.Assumethat(R;m)hasHilbertseries1+e+(e�1)2withe�2.Foreveryelementx2mnm2thefollowinghold:(a)Theideal(x)inRhaslengthatmoste.(b)Thereexistsanelementw2mnm2thatannihilatesx,andifwgenerates(0:x),thenwandxformanexactpairofzerodivisorsinR.(c)Iftheequalitieswx=0andwm=m2=xmhold,thenwandxformanexactpairofzerodivisorsinR.Proof.(a):Byassumption,thelengthofmis2e�1.Asxande�1otherelementsformaminimalsetofgeneratorsform,theinequality2e�1�lengthR(x)+e�1holds;whencelengthR(x)isatmoste.(b):AdditivityoflengthonshortexactsequencesyieldslengthR(0:x)=lengthR�lengthR(x)�e;som2isproperlycontainedin(0:x);chooseanelementwin(0:x)nm2.Thereisaninclusion(x)(0:w),andiftheequality(w)=(0:x)holds,thentwolengthcountsyieldlengthR(0:w)=lengthR�lengthR(w)=lengthR(x):Thus(x)=(0:w)holds;hencewandxformanexactpairofzerodivisorsinR.(c):Asbothideals(w)and(x)strictlycontainm2,theequalitieslengthR(x)=e=lengthR(w)holdinviewofpart(a).Aswandxannihilateeachother,simplelengthcountsshowthattheyformanexactpairofzerodivisorsinR. 14L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGAND(4.4)Theorem.Let(R;m)bealocalringwithm3=0andemb:dimR�3.AssumethatwandxformanexactpairofzerodivisorsinR.Foreveryelementyinmn(w;x)thereexistsanelementz2mnm2suchthattheR-modulesMn(w;x;y;z)areindecomposableandtotallyre exiveforalln2N.Proof.By(4.2)theequalitiesin(4.1.2)holdforR.Letybeanelementinmn(w;x).ByLemma(4.3)(b)theideal(0:y)containsanelementzofmnm2.Sinceyisnotcontainedintheideal(w;x)=(w;x)+m2,theelementzisnotin(x)=(x)+m2.ThedesiredconclusionnowfollowsfromTheorem(3.1).ThekeytothetheoremisthatexistenceofanexactzerodivisorinRimpliestheexistenceofadditionalelementssuchthattheconditionsinTheorem(3.1)aresatised.Thisphenomenondoesnotextendtoringswithm4=0.(4.5)Example.LetFbeaeldandsetS=F[x;y;z]=(x2;y2z;yz2;y3;z3);itisastandardgradedF-algebrawithHilbertseries1+3+52+33,andxisanexactzerodivisorinS.Setn=(x;y;z)Sandletvbeanelementinnn((x)+n2).Astraightforwardcalculationshowsthattheannihilator(0:v)iscontainedinn2.5.ExactzerodivisorsfromtotallyreflexivemodulesLet(R;m)bealocalringwithm3=0.IfRisnotGorenstein,thenacyclictotallyre exiveR-moduleiseitherfreeorgeneratedbyanexactzerodivisor.Indeed,ifitisnotfree,thenby(4.1.1)ithasconstantBettinumbers,equalto1,sobyRemark(2.2)itisgeneratedbyanexactzerodivisorinR.Thenextresultsimproveonthiselementaryobservation;inparticular,Corollary(5.4)shouldbecomparedtoRemark(2.2).(5.1)Lemma.Let(R;m)bealocalringwithm3=0andletF:F2�!F1 ��!F0'��!F�1beanexactsequenceofnitelygeneratedfreeR-modules,wherethehomomor-phismsarerepresentedbymatriceswithentriesinm.Let beanymatrixthatrepresents .Foreveryrow rof thefollowinghold:(a)Theidealr,generatedbytheentriesof r,containsm2.(b)Ifdimkm2isatleast2andHomR(F;R)isexact,then rhasanentryfrommnm2,theentriesin rfrommnm2generater,andmr=m2holds.Proof.Let andbethematricesfor andwithrespecttobasesB1,B0,andB�1forF1,F0,andF�1.Foreveryp�1,letfe1;:::;epgbethestandardbasisforRp.Thematrixisofsizelm,and isofsizemn,wherel,mandndenotetheranksofF�1,F0,andF1,respectively.WemaketheidenticationsF�1=Rl,F0=Rm,andF1=Rn,bylettingB�1,B0,andB1correspondtothestandardbases.Themap isnowleftmultiplicationbythematrix ,andisleftmultiplicationby.Foreveryx2m2andeverybasiselementeiinRmonehas'(xei)=0;indeed,hasentriesinm,theentriesofxeiareinm2,andm3=0holdsbyassumption.ByexactnessofF,theelementxeiisintheimageof ,and(a)follows.(b):Fixq2f1;:::;mgandlet qbetheqthrowof =(xij);westartbyprovingthefollowing: BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES15Claim.Everyentryfromm2in qiscontainedintheidealgeneratedbytheotherentriesin q.Proof.Assume,towardsacontradiction,thatsomeentryfromm2in qisnotintheidealgeneratedbytheotherentries.Afterapermutationofthecolumnsof ,onecanassumethattheentryxq1isinm2butnotintheideal(xq2;:::;xqn).Sincetheelementxq1eqbelongstoKer'=Im ,thereexistelementsaiinRsuchthat (Pni=1aiei)=xq1eq.Inparticular,onehasPni=1aixqi=xq1,whenceitfollowsthata1isinvertible.ThennmatrixA=0BBB@a10


0a210.........an0


11CCCAisinvertible,asithasdeterminanta1.Therstcolumnofthematrix A,whichistherstrowofthetransposedmatrix( A)T,hasonlyonenon-zeroentry,namelyxq1.As Arepresents ,thematrix( A)TrepresentsthedualhomomorphismHomR( ;R).ByassumptionthesequenceHomR(F;R)isexact,soitfollowsfrompart(a)thattheelementxq1spansm2.Thiscontradictstheassumptionthatm2isak-vectorspaceofdimensionatleast2.ThisnishestheproofofClaim.Suppose,forthemoment,thateveryentryof risinm2.Performingcolumnoperationson resultsinamatrixthatalsorepresents ,sobyClaimonecanassumethat risthezerorow,whichcontradictspart(a).Thus rhasanentryfrommnm2.Afterapermutationofthecolumnsof ,onemayassumethattheentriesxr1;:::;xrtareinmnm2whilexr(t+1);:::;xrnareinm2,wheretisinf1;:::;ng.Claimshowsthat|aftercolumnoperationsthatdonotalterthersttcolumns|onecanassumethattheentriesxr(t+1);:::;xrnarezero.Thustheentriesxr1;:::;xrtfrommnm2generatetheidealr.Finally,afteranotherpermutationofthecolumnsof ,onecanassumethatfxr1;:::;xrsgismaximalamongthesubsetsoffxr1;:::;xrtgwithrespecttothepropertythatitselementsarelinearlyindependentmodulom2.Nowusecolumnoperationstoensurethattheelementsxr(s+1);:::;xrnareinm2.Asabove,itfollowsthatxr1;:::;xrsgenerater.Toverifythelastequalityin(b),notersttheobviousinclusionmrm2.Forthereverseinclusion,letx2m2andwritex=xr1b1+


+xrsbswithbi2R.Ifsomebiwereaunit,thenthelinearindependenceoftheelementsxrimodulom2wouldbecontradicted.Thuseachbiisinm,andtheproofiscomplete.Theconditiondimkm2�2inpart(b)ofthelemmacannotberelaxed:(5.2)Example.Letkbeaeld;thelocalringR=k[x;y]=(x2;xy;y3)hasHilbertseries1+2+2.ThesequenceR2(xy)����!Rx�!R(xy)���!R2isexactandremainsexactafterdualization,buttheproduct(x;y)xiszero. 16L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGAND(5.3)Theorem.Let(R;m)bealocalringwithm3=0ande=emb:dimR�3.Letxbeanelementofmnm2;thefollowingconditionsareequivalent.(i)TheelementxisanexactzerodivisorinR.(ii)TheHilbertseriesofRis1+e+(e�1)2,andthereexistsanexactsequenceofnitelygeneratedfreeR-modulesF:F3�!F2�!F1 ��!F0�!F�1suchthatHomR(F;R)isexact,thehomomorphismsarerepresentedbyma-triceswithentriesinm,and isrepresentedbyamatrixinwhichsomerowhasxasanentryandnootherentryfrommnm2.Proof.IfxisanexactzerodivisorinR,thenthecomplex(2.0.1)suppliesthedesiredexactsequence,andRhasHilbertseries1+e+(e�1)2;see(4.2).Toprovetheconverse,let =(xij)beamatrixofsizemnthatrepresents andassume,withoutlossofgenerality,thatthelastrowof hasexactlyoneentryx=xmqfrommnm2.ByLemma(5.1)(b)thereisanequalityxm=m2and,therefore,thelengthof(x)isebyLemma(4.3)(a).AsRhaslength2e,additivityoflengthonshortexactsequencesyieldslengthR(0:x)=e.Let(wij)beannpmatrixthatrepresentsthehomomorphismF2!F1.Thematrixequality(xij)(wij)=0yieldsxwqj=0forj2f1;:::;pg;itfollowsthattheidealr=(wq1;:::;wqp)iscontainedin(0:x).ByLemma(5.1)(b)someentryw=wqlisinmnm2,andthereareinclusions(1)(w)+m2r(0:x):Nowtheinequalitiese6lengthR((w)+m2)6lengthR(0:x)=eimplythatequalitiesholdthroughout(1);inparticular,(w)+m2=rholds.ThisequalityandLemma(5.1)(b)yieldwm=mr=m2;hencewandxformanexactpairofzerodivisorsbyLemma(4.3)(c).(5.4)Corollary.Let(R;m)bealocalringwithm3=0.IfRisnotGorenstein,thenthefollowingconditionsareequivalent:(i)ThereisanexactzerodivisorinR.(ii)Foreveryn2Nthereisanindecomposabletotallyre exiveR-modulethatispresentedbyanuppertriangularnnmatrixwithentriesinm.(iii)Thereisatotallyre exiveR-modulewithoutfreesummandsthatispresentedbyamatrixwithentriesinmandarow/columnwithonlyoneentryinmnm2.Proof.Sete=emb:dimR;itisatleast2asRisnotGorenstein.If(i)holds,theneisatleast3,see(4.2),so(ii)followsfromTheorem(4.4).ItisclearfromLemma(5.1)that(iii)followsfrom(ii).Toprovethat(iii)implies(i),let beapresentationmatrixforatotallyre exiveR-modulewithoutfreesummandsandassume|possiblyafterreplacing withitstranspose,whichalsopresentsatotallyre exivemodule|thatsomerowof hasonlyoneentryinmnm2.By(4.1)theHilbertseriesofRis1+e+(e�1)2,andeisatleast3,soitfollowsfromTheorem(5.3)thatthereisanexactzerodivisorinR. BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES17Thecorollarymanifestsastrongrelationbetweentheexistenceofexactzerodi-visorsinRandexistenceoftotallyre exiveR-modulesofanysize.Aqualitativelydierentrelationisstudiedlater;seeRemark(8.7).Theserelationsnotwithstand-ing,totallyre exivemodulesmayexistintheabsenceofexactzerodivisors.InSection9weexhibitalocalring(R;m),whichhasnoexactzerodivisors,andatotallyre exiveR-modulethatispresentedbya22matrixwithallfourentriesfrommnm2.Thustheconditionontheentriesofthematrixin(5.4)(iii)issharp.6.Familiesofnon-isomorphicmodulesofthesamesizeInthissectionwecontinuetheanalysisofConstruction(2.6).(6.1)Denition.Let(R;m;k)bealocalring.GivenasubsetKk,asubsetofRthatcontainsexactlyoneliftofeveryelementinKiscalledaliftofKinR.(6.2)Theorem.Let(R;m;k)bealocalringandletLbealiftofkinR.Letw,x,y,y0,andzbeelementsinmnm2andletnbeaninteger.AssumethatwandxformanexactpairofzerodivisorsinRandthatoneofthefollowingholds.(a)n=2andtheelementsw,x,y,andy0arelinearlyindependentmodulom2;(b)n=2,theelementsx,y,andy0arelinearlyindependentmodulom2,andtheelementwbelongsto(x)+m2;(c)n�3,theelementsw,x,y,andy0arelinearlyindependentmodulom2,andthefollowinghold:z62(w)+m2,z62(x)+m2,and(y;y0)(0:z);or(d)n�3,theelementsx,y,andy0arelinearlyindependentmodulom2,andthefollowinghold:w2(x)+m2,z62(x)+m2,and(y;y0)(0:z).ThenthemodulesinthefamilyfMn(w;x;y+y0;z)g2Lareindecomposable,totallyre exive,andpairwisenon-isomorphic.Theproofof(6.2)takesupthebalanceofthissection;hereisthecornerstone:(6.3)Proposition.Let(R;m)bealocalringandletw,x,y,y0,andzbeelementsinmnm2.Assumethatthefollowinghold:z62(w)+m2;z62(x)+m2;y62(w;x)+m2;andy0=2(w;y)+m2:IfandareelementsinRwith�62m,andn�3isaninteger,thentheR-modulesMn(w;x;y+y0;z)andMn(w;x;y+y0;z)arenon-isomorphic.Thenextexampleshowsthattheconditionn�3in(6.3)cannotberelaxed.(6.4)Example.Let(R;m)bealocalringwithemb:dimR�3andassumethat2isaunitinR.Letw,x,andybelinearlyindependentmodulom2,andsety0=y�2�1x.Theequality110�1wy00x=wy0�2y0x100�1showsthattheR-modulesM2(w;x;0y+y0)andM2(w;x;�2y+y0)areisomorphic.TogetastatementsimilartoProposition(6.3)for2-generatedmodules,itsucestoassumethaty0isoutsidethespanofw,x,andymodulom2. 18L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGAND(6.5)Proposition.Let(R;m)bealocalringandletw,x,y,andy0beelementsinmnm2.Assumethatoneofthefollowingconditionsholds:(a)Theelementsw,x,y,andy0arelinearlyindependentmodulom2.(b)Theelementsx,y,andy0arelinearlyindependentmodulom2,andtheele-mentwbelongsto(x)+m2.IfandareelementsinRwith�62m,thentheR-modulesM2(w;x;y+y0)andM2(w;x;y+y0)arenon-isomorphic.Proof.LetandbeinR.Assumethat(a)holdsandthattheR-modulesM2(w;x;y+y0)andM2(w;x;y+y0)areisomorphic.ItfollowsthatthereexistmatricesAandBinGL2(R)suchthattheequalityA(wIo+xIe+(y+y0)Jo)=(wIo+xIe+(y+y0)Jo)Bholds;herethematricesIo=Io2,Ie=Ie2,andJo=Jo2areasdenedinConstruc-tion(2.6).Thegoalistoprovethat�isinm.Afterreductionmodulom,theequalityaboveyields,inparticular,AJo�Jo B=0=AJo�Jo B;and,therefore,(�)Jo B=0.Asthematrix BisinvertibleandJoisnon-zero,thisimpliesthat�isinmasdesired.If(b)holds,thedesiredconclusionisprovedunderCase1inthenextproof.Proofof(6.3).LetandbeelementsinRandassumethatMn(w;x;y+y0;z)andMn(w;x;y+y0;z)areisomorphicasR-modules.ItfollowsthatthereexistmatricesAandBinGLn(R)suchthattheequality(1)A(wIo+xIe+(y+y0)Jo+zJe)=(wIo+xIe+(y+y0)Jo+zJe)Bholds;herethematricesIo=Ion,Ie=Ien,Jo=Jon,andJe=JenareasdenedinConstruction(2.6).Thegoalistoprovethat�isinm.Case1:wisin(x)+m2.Underthisassumption,onecanwritew=rx+,where2m2,andrewrite(1)asx(r(AIo�IoB)+AIe�IeB)+y(AJo�JoB)+y0(AJo�JoB)+z(AJe�JeB)=
;(2)where
isamatrixwithentriesinm2.Theassumptionsonw,x,y,andy0implyy=2(x)+m2andy0=2(x;y)+m2,sotheelementsx,y,andy0arelinearlyindependentmodulom2.Thereexistelementsvisuchthatv1;:::;ve�3;x;y;y0formaminimalsetofgeneratorsform.Writez=sx+ty+uy0+e�3Xi=1divi;substitutethisexpressioninto(2),andreducemodulomtoget0=r(AIo�Io B)+AIe�Ie B+s(AJe�Je B);(3)0=AJo�Jo B+t(AJe�Je B);(4)0=AJo�Jo B+u(AJe�Je B);and(5)0=di(AJe�Je B);fori2f1;:::;e�3g.(6) BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES19Theargumentsthatfollowusetherelationsthat(3){(6)inducebetweentheentriesofthematricesA=(aij)and B=(bij).Withtheconventionahl=0=bhlforh;l2f0;n+1g,itiselementarytoverifythatthefollowingsystemsofequalitiesholdforiandjinf1;:::;ngandelementsfandginR=m:(AIo�Io B)ij=8���&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:aij�bijiodd,jodd�bijiodd,jevenaijieven,jodd0ieven,jeven(7)(AIe�Ie B)ij=8���&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:0iodd,joddaijiodd,jeven�bijieven,joddaij�bijieven,jeven(8)(fAJo�gJo B)ij=8���&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:gb(i+1)jiodd,joddfai(j�1)�gb(i+1)jiodd,jeven0ieven,joddfai(j�1)ieven,jeven(9)(AJe�Je B)ij=8���&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:ai(j�1)iodd,jodd0iodd,jevenai(j�1)�b(i+1)jieven,jodd�b(i+1)jieven,jeven(10)Eachoftheequalities(3){(6)inducesfoursubsystems,whicharereferredtobysubscripts`oo',`oe',`eo',and`ee',where`oo'standsfor'ioddandjodd'etc.Forexample,(4)oereferstotheequalities0=ai(j�1)�b(i+1)j,forioddandjeven.Letn&#x]TJ ;� -1;.93; Td;&#x [00;2;thegoalistoprovetheequality=,asthatimplies�2m.Firstassumedi6=0forsomei2f1;:::;e�3g,then(6)yieldsAJe�Je B=0.From(4)and(5)onethengetsAJo�Jo B=0=AJo�Jo Bandthus(�)Jo B=0.As BisinvertibleandJoisnon-zero,thisyieldsthedesiredequality=.Henceforthweassumedi=0foralli2f1:::;e�3g.Byassumption,zisnotin(x)+m2,souortisnon-zero.Incaseuisnon-zero,(5)ooand(5)eoyieldbh1=0for1h6n.Iftisnon-zero,then(4)ooand(4)eoyieldthesameconclusion.Thematrix Bisinvertible,soeachofitscolumnscontainsanon-zeroelement.Itfollowsthatb11isnon-zero,andby(3)ooonehasa11=b11.From(4)oeand(5)oeonegetsa11=b22and(�)a11=0,whencetheequality=holds.Forn=2theargumentsaboveestablishtheassertioninProposition(6.5)underassumption(b)ibid.Case2:wisnotin(x)+m2.Itfollowsfromtheassumptiony=2(w;x)+m2thatw,x,andyarelinearlyindependentmodulom2,sothereexistelementsvisuchthatv1;:::;ve�3;w;x;yformaminimalsetofgeneratorsform.Writey0=pw+qx+ry+e�3Xi=1civiandz=sw+tx+uy+e�3Xi=1divi: 20L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDSubstitutetheseexpressionsinto(1)andreducemodulomtoget0=AIo�Io B+p(AJo�Jo B)+s(AJe�Je B);(11)0=AIe�Ie B+q(AJo�Jo B)+t(AJe�Je B);(12)0=AJo�Jo B+r(AJo�Jo B)+u(AJe�Je B);and(13)0=ci(AJo�Jo B)+di(AJe�Je B);fori2f1;:::;e�3g:(14)AsinCase1,setA=(aij)and B=(bij)sothattheequalities(7){(10)hold.Thesubscripts`oo',`oe',`eo',and`ee'areused,asinCase1,todenotethesubsystemsinducedby(11){(14).Letn�3;thegoalis,again,toprovetheequality=.Inthefollowing,handlareintegersinf1;:::;ng.First,noticethatifcm6=0forsomem2f1;:::;e�3g,then(14)oeimpliesa11=b22and,inturn,(13)oeyields(�)a11=0.Toconclude=,itmustbeveriedthata11isnon-zero.Tothisend,assumerstdm=0;from(14)eeand(14)oooneimmediatelygets(15)ah1=0=bh1forheven.Aszisinmnm2,oneofthecoecientsd1;:::;de�3;s;t;uisnon-zero.Ifuoroneofd1;:::;de�3isnon-zero,then(13)eoor(14)eoyieldsbh1=0forh�1odd.Ifsortisnon-zero,thenthesameconclusionfollowsfrom(15)combinedwith(11)eoorwith(12)eo.Nowthatbh1=0holdsforallh�2,itfollowsthatb11isnon-zero.Finally,(11)ooyieldsa11=b11,sotheentrya11isnotzero,asdesired.Nowassumedm6=0,then(14)ooand(14)eoimmediatelygivebh1=0forh�1.Asabove,weconcludethatb11isnon-zero,whencea116=0by(11)oo.Thisconcludestheargumentundertheassumptionthatoneofthecoecientsciisnon-zero.Henceforthweassumeci=0foralli2f1;:::;e�3g;itfollowsthatqisnon-zero,asy0=2(w;y)+m2byassumption.Ifdi6=0forsomei2f1;:::;e�3g,then(14)ooyieldsa12=0,asnisatleast3.From(12)oeonegetsa11=b22,andthen(13)oeimplies(�)a11=0.Toseethata11isnon-zero,noticethat(14)eoyieldsbh1=0forh�1odd,while(12)ooyieldsbh1=0forheven.Itfollowsthatb11isnon-zero,andthena116=0,by(11)oo.Thusthedesiredequality=holds.Thisconcludestheargumentundertheassumptionthatoneofthecoecientsdiisnon-zero.Henceforthweassumedi=0foralli2f1;:::;e�3g.Tonishtheargument,wedealseparatelywiththecasesu6=0andu=0.Assumerstu6=0.Byassumptiononehasn�3,so(13)eoyieldsa22=b33,andthenitfollowsfrom(12)eothatb23iszero.From(13)ooonegetsa12=0,andthentheequalitya11=b22followsfrom(12)oe.Inturn,(13)oeimplies(�)a11=0.Toseethata11isnon-zero,noticethatthereareequalitiesbh1=0forh�1oddby(13)eoandbh1=0forhevenby(12)oo.Itfollowsthatb11isnon-zero,andtheequalitya11=b11holdsby(11)oo.Asaboveweconclude=.Finally,assumeu=0;theassumptionsz=2(w)+m2andz=2(x)+m2implythatsandtarebothnon-zero.From(13)eeand(13)ooonegets:(16)(+r)ah1=0=(+r)bh1forheven.Firstassumethat+riszero.If+riszero,thenthedesiredequality=holds.If+risnon-zero,then(16)givesah1=0forheven.Moreover,(13)oeimplies(+r)ah1=(+r)b(h+1)2=0,soah1=0forhoddaswell,whichisabsurdasAisinvertible.Nowassumethat+risnon-zero.From(16)onegets BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES21bh1=0forhevenand,inturn,(12)eoyieldsbh1=0forh�1odd.Itfollowsthatb11isnon-zero,andthenonehasa116=0by(11)oo.Byassumption,nisatleast3,so(13)oogives(+r)b23=0,whichimpliesb23=0.Now(12)ooyieldsa12=0,andthen(12)oeimpliesa11=b22.From(13)oeonegets(�)a11,whencetheequality=holds.Proofof(6.2).Undertheassumptionsinpart(a)or(b),itisimmediatefromTheorem(3.1)thatthemoduleM2(w;x;y+y0)isindecomposableandtotallyre exiveforevery2R.ItfollowsfromProposition(6.5)thatthemodulesinthefamilyfM2(w;x;y+y0)g2Larepairwisenon-isomorphic.Fixn�3.Proposition(6.3)showsthat,undertheassumptionsinpart(c)or(d),themodulesinthefamilyfMn(w;x;y+y0;z)g2Larepairwisenon-isomorphic.Moreover,forevery2R,onehas(y+y0)z=0,soitfollowsfromTheorem(3.1)themoduleMn(w;z;y+y0;z)istotallyre exiveandindecomposable.7.Brauer{ThrallIIovershortlocalringswithexactzerodivisorsInthissection(R;m;k)isalocalringwithm3=0.Together,(4.1.1)andthetheorems(3.1),(7.4),(7.6),and(7.8)establishTheorem(1.4).(7.1)Remark.AssumethatRhasembeddingdimension2.IfthereisanexactzerodivisorinR,thentheHilbertseriesofRis1+2+2,andtheequality(0:m)=m2holds;see(4.2).Therefore,RisGorenstein,andtheequalityxm=m2holdsforallx2mnm2;inparticular,everyelementinmnm2isanexactzerodivisorbyLemma(4.3)(c).Ontheotherhand,ifRisGorenstein,thenonehasHR()=1+2+2and(0:m)=m2.ItisnowelementarytoverifythattheequalitieslengthR(x)=2=lengthR(0:x)holdforeveryx2mnm2,soeverysuchelementisanexactzerodivisorinRbyLemma(4.3)(c).ItfollowsfromworkofSerre[18,prop.5]thatRisGorensteinifandonlyifitiscompleteintersection;thusthefollowingconditionsareequivalent:(i)ThereisanexactzerodivisorinR.(ii)Everyelementinmnm2isanexactzerodivisor.(iii)Riscompleteintersection.Incontrast,ifRhasembeddingdimensionatleast3,andkisalgebraicallyclosed,thenthereexistelementsinmnm2thatarenotexactzerodivisors.ThisfactfollowsfromLemma(7.3),anditisessentialforourproofofTheorem(7.4).(7.2)Remark.AssumethatRhasHilbertseries1+e+f2,andletxbeanelementinm.Theequalityxm=m2holdsifandonlyifthek-linearmapfromm=m2tom2givenbymultiplicationbyxissurjective.Letxbeamatrixthatrepresentsthismap;itisanfematrixwithentriesink,sotheequalityxm=m2holdsifandonlyifxhasrankf.Assumethatthe(in)equalitiesf=e�1�1hold.IfwandxareelementsinRwithwx=0,thentheyformanexactpairofzerodivisorsifandonlyifbothmatricesxandwhaveanon-zeromaximalminor;cf.Lemma(4.3)(c).(7.3)Lemma.Assumethat(R;m;k)hasHilbertseries1+e+f2andthatkisalgebraicallyclosed;setn=e�f+1.Ifonehas26f6e�1andv0;:::;vn2marelinearlyindependentmodulom2,thenthereexistr0;:::;rn2R,atleastoneofwhichisinvertible,suchthattheideal(Pnh=0rhvh)misproperlycontainedinm2. 22L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDProof.Letfx1;:::;xegbeaminimalsetofgeneratorsformandletfu1;:::;ufgbeabasisforthek-vectorspacem2.Forh2f0;:::;ngandj2f1;:::;egwritevhxj=fXi=1hijui;wheretheelementshijareink.Inthefollowing,rdenotestheimageinkoftheelementr2R.Forr0;:::;rninRandv=Pnh=0rhvhtheequalityvm=m2holdsifandonlyifthefematrixv= nXh=0rhhij!ijhasrankf;seeRemark(7.2).LetXbethematrixobtainedfromvbyreplacingr0;:::;rnwithindeterminates0;:::;n.Thenon-zeroentriesinXarethenho-mogeneouslinearformsin0;:::;n,andinthepolynomialalgebrak[0;:::;n]theidealIf(X),generatedbythemaximalminorsofX,hasheightatmostn;see[2,thm.(2.1)].AsIf(X)isgeneratedbyhomogeneouspolynomials,itfollowsfromHilbert'sNullstellensatzthatthereexistsapoint(0;:::;n)inkn+1nf0gsuchthatallmaximalminorsofthematrix(Pnh=0hhij)ijvanish.Letr0;:::;rnbeliftsof0;:::;ninR,thenatleastoneofthemisnotinm,andtheideal(Pnh=0rhvh)misproperlycontainedinm2.(7.4)Theorem.Let(R;m;k)bealocalringwithm3=0,emb:dimR�3,andkalgebraicallyclosed.IfwandxformanexactpairofzerodivisorsinR,thenthereexistelementsy,y0,andzinmnm2,suchthatforeveryliftLofknf0ginRandforeveryintegern�3themodulesinthefamilyfMn(w;x;y+y0;z)g2Lareindecomposable,totallyre exive,andpairwisenon-isomorphic.Proof.AssumethatwandxformanexactpairofzerodivisorsinR.By(4.2)theHilbertseriesofRis1+e+(e�1)2,soitfollowsfromLemma(7.3)thatthereexistsanelementz2mnm2suchthatzmisproperlycontainedinm2.Inparticular,onehaslengthR(z)eand,therefore,lengthR(0:z)&#x-297;ebyadditivityoflengthonshortexactsequences.Itfollowsthatthereexisttwoelements,callthemyandy0,in(0:z)thatarelinearlyindependentmodulom2.TheinequalitylengthR(z)eimpliesthatzisnotin(w)=(w)+m2andnotin(x)=(x)+m2.Ifyandy0werebothin(w;x)=(w;x)+m2,thenxwouldbein(y;y0),whichisimpossibleasz=2(w).Withoutlossofgenerality,assumey=2(w;x)+m2.Ify0werein(w;y)=(w;y)+m2,thenwwouldbein(y;y0),whichisalsoimpossible.NowletLbealiftofknf0ginRandletn&#x-282;3beaninteger.ItfollowsfromProposition(6.3)thatthemodulesinthefamilyfMn(w;x;y+y0;z)g2Larepairwisenon-isomorphic.Ifwisin(x)=(x)+m2,thenitfollowsfromTheorem(3.1)thatthemodulesinthefamilyfMn(w;x;y+y0;z)g2Lareindecomposableandtotallyre exive.Indeed,forevery2Rtheelementy+y0annihilatesz,whenceitisnotin(x)=(x)+m2.Ifwisnotin(x)=(x)+m2,thenitfollowsfromtheassumptiony=2(w;x)+m2thatw,x,andyarelinearlyindependentmodulom2.Thereexistelementsvisuch BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES23thatv1;:::;ve�3;w;x;yformaminimalsetofgeneratorsform.Writey0=pw+qx+ry+e�3Xi=1civi:Asyz=0holdsandtheelementsyandy0arelinearlyindependentmodulom2,wemayassumer=0.For2L,theelementsw,x,andy+y0arelinearlyindependent.Indeed,ifthereisarelationsw+tx+u(y+y0)=(s+up)w+(t+uq)x+uy+ue�3Xi=1civi2m2;thenuisinmas=2m,andthensandtareinmaswandxarelinearlyindependentmodulom2.NowitfollowsfromTheorem(3.1)thatthemodulesinthefamilyfMn(w;x;y+y0;z)g2Lareindecomposableandtotallyre exive.(7.5)Remark.AssumethatRhasembeddingdimension3andthatmisminimallygeneratedbyelementsv,w,andx,wherewandxformanexactpairofzerodivisors.Letyandy0beelementsinmandletLbeasubsetofR.ItiselementarytoverifythateachmoduleinthefamilyfM2(w;x;y+y0)g2LisisomorphictoeitherR=(w)R=(x)ortotheindecomposableR-moduleM2(w;x;v).Thustherequirementn�3inTheorem(7.4)cannotberelaxed.Wenowproceedtodealwith1-and2-generatedmodules.(7.6)Theorem.Let(R;m;k)bealocalringwithm3=0,emb:dimR�2,andkinnite.IfthereisanexactzerodivisorinR,thenthesetN1=fR=(x)jxisanexactzerodivisorinRgoftotallyre exiveR-modulescontainsasubsetM1ofcardinalitycard(k),suchthatthemodulesinM1arepairwisenon-isomorphic.Proof.AssumethatthereisanexactzerodivisorinR;thenRhasHilbertseries1+e+(e�1)2;see(4.2).Letfx1;:::;xegbeaminimalsetofgeneratorsformandletfu1;:::;ue�1gbeabasisform2.Forhandjinf1;:::;egwritexhxj=e�1Xi=1hijui;wheretheelementshijareink.Forr1;:::;reinRletrhdenotetheimageofrhink,andsetx=r1x1+


+rexe.Theequalityxm=m2holdsifandonlyifthe(e�1)ematrixx= eXh=1rhhij!ijhasranke�1;seeRemark(7.2).Forj2f1;:::;egletj(x)bethemaximalminorofxobtainedbyomittingthejthcolumn.Noticethateachminorj(x)isahomogeneouspolynomialexpressionintheelementsr1;:::;re.Thecolumnvector(1(x)�2(x)


(�1)e�1e(x))Tisinthenull-spaceofthematrixx,sotheelementw=1(x)x1�2(x)x2+


+(�1)e�1e(x)xeannihilatesx.Let1(w);:::;e(w)bethemaximalminorsofthematrixw;theyarehomogeneouspolynomialexpressionsin1(x);:::;e(x)and,therefore, 24L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDinr1;:::;re.ByRemark(7.2)theelementsxandwformanexactpairofzerodivisorsifandonlyifbothmatricesxandwhaveranke�1.Forj2f1;:::;egdenejandjtobethehomogeneouspolynomialsink[1;:::;e]obtainedfromj(x)andj(w)byreplacingtheelementsrhbythein-determinatesh,forh2f1;:::;eg.IntheprojectivespacePe�1k,thecomplementEoftheintersectionofvanishingsetsZ(1)\


\Z(e)isanopenset.NopointinEisintheintersectionZ(1)\


\Z(e),aseachpolynomialjisapolynomialin1;:::;e.Therefore,eachpoint(1:


:e)inEcorrespondstoanexactzerodivisorasfollows.Letr1;:::;rebeliftsof1;:::;einR;thentheelementx=r1x1+


+rexeisanexactzerodivisor.ItisclearthattwodistinctpointsinEcorrespondtonon-isomorphicmodulesinN1.TakeasM1anysubsetofN1suchthattheelementsofM1areinone-to-onecorrespondencewiththepointsinE.Byassumption,thesetEisnon-empty,and,ifkisinnite,thenEhasthesamecardinalityask.(7.7)Remark.ThemodulesinM1areinone-to-onecorrespondencewiththepointsofanon-emptyZariskiopensetinPe�1k.SeealsoRemark(8.8).(7.8)Theorem.Let(R;m;k)bealocalringwithm3=0,emb:dimR�3,andkinnite.IfthereisanexactzerodivisorinR,thenthesetN2=M2(w;x;y)w,x,andyareelementsinR,suchthatwandxformanexactpairofzerodivisorsoftotallyre exiveR-modulescontainsasubsetM2ofcardinalitycard(k),suchthatthemodulesinM2areindecomposableandpairwisenon-isomorphic.Proof.AssumethatthereisanexactzerodivisorinRandletM1bethesetofcyclictotallyre exiveR-modulesaordedbyTheorem(7.6);thecardinalityofM1iscard(k).FromM1onecanconstructanothersetofthesamecardinality,whoseelementsareexactpairsofzerodivisors,suchthatforanytwoofthem,sayw;xandw0;x0,onehas(x)6=(x0)and(w)6=(x0).Giventwosuchpairs,chooseelementsyandy0inmnm2suchthaty=2(w;x)andy0=2(w0;x0).ByTheorem(4.4),themodulesM2(w;x;y)andM2(w0;x0;y0)areindecomposableandtotallyre exive.SupposethattheR-modulesM2(w;x;y)andM2(w0;x0;y0)areisomorphic;thenthereexistmatricesA=(aij)andB=(bij)inGL2(R)suchthattheequalityAwy0x=w0y00x0Bholds.Inparticular,thereareequalitiesa21w=b21x0anda21y+a22x=b22x0:Therstoneshowsthattheentriesa21andb21areelementsinm,andthenitfollowsfromthesecondonethata22andb22areinm.ThusAandBeachhavearowwithentriesinm,whichcontradictstheassumptionthattheyareinvertible.8.ExistenceofexactzerodivisorsInprevioussectionsweconstructedfamiliesoftotallyre exivemodulesstartingfromanexactpairofzerodivisors.Nowweaddressthequestionofexistenceofexactzerodivisors;inparticular,weproveTheorem(1.3);seeRemark(8.7). BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES25Alocalring(R;m)withm3=0andembeddingdimension1is,byCohen'sStructureTheorem,isomorphictoD=(d2)orD=(d3),where(D;(d))isadiscretevaluationdomain.Ineithercase,disanexactzerodivisor.Inthefollowingwefocusonringsofembeddingdimensionatleast2.(8.1)Remark.Let(R;m)bealocalringwithm3=0.Itiselementarytoverifythatelementsinmannihilateeachotherifandonlyiftheirimagesintheassociatedgradedringgrm(R)annihilateeachother.Thusanelementx2misanexactzerodivisorinRifandonlyifx2m=m2isanexactzerodivisoringrm(R).Let(R;m;k)beastandardgradedk-algebrawithm3=0andembeddingdi-mensione�2.AssumethatthereisanexactzerodivisorinR;by(4.2)onehasHR()=1+e+(e�1)2.Ifeis2,thenitfollowsfrom[17,Hilfssatz7]thatRiscompleteintersectionwithPoincareseries1Xi=0Ri(k)i=1 1�2+2=1 HR(�);henceRisKoszulbyaresultofLofwall[15,thm.1.2].Ifeisatleast3,thenRisnotGorensteinand,therefore,RisKoszulby[6,thm.A].ItisknownthatKoszulalgebrasarequadratic,andthegoalofthissectionistoprovethatifkisinnite,thenagenericquadraticstandardgradedk-algebrawithHilbertseries1+e+(e�1)2hasanexactzerodivisor.RecallthatRbeingquadraticmeansitisisomorphictok[x1;:::;xe]=q,whereqisanidealgeneratedbyhomogeneousquadraticforms.TheidealqcorrespondstoasubspaceVofthek-vectorspaceWspannedbyfxixjj16i6j6eg.ThedimensionofWism=e(e+1) 2,andsinceRhasHilbertseries1+e+(e�1)2,theidealqisminimallygeneratedbyn=e2�e+2 2quadraticforms;thatis,dimkV=n.Inthisway,RcorrespondstoapointintheGrassmannianGrassk(n;m).(8.2)Denition.Lete�2beanintegerandsetGk(e)=Grassk(n;m),wheren=e2�e+2 2andm=e(e+1) 2.PointsinGk(e)areinbijectivecorrespondencewithk-algebrasofembeddingdimensionewhosedeningidealisminimallygeneratedbynhomogeneousquadraticforms.Forapoint2Gk(e)letRdenotethecorrespondingk-algebraandletMdenotetheirrelevantmaximalidealofR.NoticethatHR()hastheform1+e+(e�1)2+P1i=3hiiforevery2Gk(e).ConsiderthesetsEk(e)=f2Gk(e)jthereisanexactzerodivisorinRgandHk(e)=f2Gk(e)jHR()=1+e+(e�1)2gandrecallthatasubsetofGk(e)iscalledopen,ifitmapstoaZariskiopensetunderthePluckerembeddingGk(e),!PNk,whereN=�mn
�1.ThesetsEk(e)andHk(e)arenon-empty:(8.3)Example.Letkbeaeldandlete�2beaninteger.Thek-algebraR=k[x1;:::;xe] (x21)+(xixjj26i6j6e)islocalwithHilbertseries1+e+(e�1)2.Onehas(0:x1)=(x1);inparticular,x1isanexactzerodivisorinR. 26L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGAND(8.4)Theorem.Foreveryeldkandeveryintegere�2thesetsEk(e)andHk(e)arenon-emptyopensubsetsoftheGrassmannianGk(e).ThefactthatHk(e)isopenandnon-emptyisaspecialcaseof[11,thm.1];forconvenienceaproofisincludedbelow.Recallthatapropertyissaidtoholdforagenericalgebraoveraninniteeldifthereisanon-emptyopensubsetofanappropriateGrassmanniansuchthateverypointinthatsubsetcorrespondstoanalgebrathathastheproperty.(8.5)Corollary.Letkbeaninniteeldandlete�2beaninteger.Agenericstandardgradedk-algebrawithHilbertseries1+e+(e�1)2hasanexactzerodivisor.Proof.TheassertionfollowsasthesetEk(e)\Hk(e)isopenbyTheorem(8.4)andnon-emptybyExample(8.3).(8.6)Remark.Let(R;m)bealocalring.FollowingAvramov,Iyengar,andSega(2008)wecallanelementxinRwithx2=0andxm=m2aConcageneratorofm.Concaprovesin[7,sec.4]thatifkisalgebraicallyclosed,thenthesetCk(e)=f2Gk(e)jthereisaConcageneratorofMgisopenandnon-emptyinGk(e).Ifm3iszero,thenitfollowsfromLemma(4.3)(c)thatanelementxisaConcageneratorofmifandonlyiftheequality(0:x)=(x)holds.Inparticular,thereisaninclusionCk(e)\Hk(e)Ek(e)\Hk(e):Ifkisalgebraicallyclosed,thenitfollowsfromConca'sresultcombinedwithThe-orem(8.4)andExample(8.3)thatbothsetsarenon-emptyandopeninGk(e);inthenextsectionweshowthattheinclusionmaybestrict.(8.7)Remark.Let(R;m;k)bealocalk-algebrawithm3=0andassumethatitisnotGorenstein.IfRadmitsanon-freetotallyre exivemodule,thenithasembeddingdimensione�3andHilbertseries1+e+(e�1)2;see(4.1).SetTk(e)=f2Hk(e)jRadmitsanon-freetotallyre exivemoduleg:WedonotknowifthisisanopensubsetofGk(e),butitcontainsthenon-emptyopensetEk(e)\Hk(e);hencetheassertioninTheorem(1.3).Proofof(8.4).Letkbeaeld,lete�2beaninteger,andletSdenotethestan-dardgradedpolynomialalgebrak[x1;:::;xe].SetN=�mn
�1,wherem=e(e+1) 2andn=e2�e+2 2.ThePluckerembeddingmapsapointintheGrassman-nianGk(e)tothepoint(0():


:N())intheprojectivespacePNk,where0();:::;N()arethemaximalminorsofanymnmatrixcorrespondingto.Thecolumnsofsuchamatrixgivethecoordinates,inthelexicographicallyorderedbasisB=fxixjj16i6j6egforS2,ofhomogeneousquadraticformsq1;:::;qn.ThealgebraRisthequotientringS=(q1;:::;qn).ThesetsEk(e)andHk(e)arenon-emptybyExample(8.3).LetZbeanmnmatrixofindeterminatesandlet0;:::;NdenotethemaximalminorsofZ.WeproveopennessofeachsetEk(e)andHk(e)inGk(e)byprovingthatthePluckerem-beddingmapsittothecomplementinPNkofthevanishingsetforanitecollectionofhomogeneouspolynomialsink[0;:::;N]. BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES27OpennessofEk(e).LetbeapointinGk(e),letbeacorrespondingmatrix,andletqbethedeningidealforR=R.Foralinearform`=a1x1+


+aexeinS,letldenotetheimageof`inR.Fori2f1;:::;eglet[xi`]denotethecolumnthatgivesthecoordinatesofxi`inthebasisB.Multiplicationbyldenesak-linearmapfromR1toR2.Byassumption,onehasdimkR2=dimkR1�1,sotheequalitylR1=R2holdsifandonlyiflisannihilatedbyaunique,uptoscalarmultiplication,homogeneouslinearforminR.Set`=�[x1`]j[x2`]j


j[xe`]j
;itisanm(m+1)matrixwithentriesink.TheequalitylR1=R2holdsifandonlyiftheequality`S1+q2=S2holds,andthelatterholdsifandonlyifthematrix`hasmaximalrank.Fori2f1;:::;m+1gleti(`)denotethemaximalminorobtainedbyomittingtheithcolumnof`.Thecolumnsofarelinearlyinde-pendent,so`hasmaximalrankifandonlyifoneoftheminors1(`);:::;e(`)isnon-zero.Noticethateachoftheminors1(`);:::;e(`)isapolynomialexpressionofdegreee�1inthecoecientsa1;:::;aeof`andlinearinthePluckercoordinates1();:::;N().Thecolumnvector(1(`)�2(`)


(�1)mm+1(`))Tisinthenull-spaceofthematrix`,sotheelementeXi=1(�1)i�1i(`)xi`=eXi=1(�1)i�1i(`)[xi`]isinthecolumnspaceofthematrix.Set`0=Pei=1(�1)i�1i(`)xi;itfollowsthat`0`belongstotheidealq,soonehasl0l=0inR.Bythediscussionabove,listheunique,uptoscalarmultiplication,annihilatorinR1ofl0ifandonlyifoneofthemaximalminors1(`0);:::;e(`0)of`0isnon-zero.Moreover,ifoneof1(`0);:::;e(`0)isnon-zero,thenalsooneoftheminors1(`);:::;e(`)isnon-zerobythedenitionof`0.ThuslisanexactzerodivisorinRifandonlyifoneof1(`0);:::;e(`0)isnon-zero.Noticethateachoftheseminorsisapolynomialexpressionofdegreeein1();:::;N()anddegree(e�1)2ina1;:::;ae.Let1;:::;ebeindeterminatesandsetL=1x1+


+exe.Fori2f1;:::;egletidenotethemaximalminorofthematrix�[x1L]j[x2L]j


j[xeL]jZ
;obtainedbyomittingtheithcolumn.Eachminoriisapolynomialofdegreee�1intheindeterminates1;:::;eandlinearin0;:::;N.Fori2f1;:::;egsetFi=i(1;�2;:::;(�1)e�1e);eachFiisapolynomialofdegree(e�1)2in1;:::;eandofdegreeein0;:::;N.ConsiderF1;:::;Feaspolynomialsin1;:::;ewithcoecientsink[0;:::;N],andletPdenotethecollectionofthesecoecients.ThealgebraRhasanexactzerodivisor,i.e.thepointbelongstoEk(e),ifandonlyifoneofthepolynomialsFiinthealgebra(k[0;:::;N])[1;:::;e]isnon-zero,thatis,ifandonlyifthePluckerembeddingmapstoapointinthecomplementofthealgebraicvariety\P2PZ(P)PNk:OpennessofHk(e).LetbeapointinGk(e),letbeacorrespondingmatrix,andletq=(q1;:::;qn)bethedeningidealforR.Clearly,belongstothesubsetHk(e)ifandonlyiftheequalityS1q2=S3holds.Setc=�e+23
andtakeas 28L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDk-basisforS3thechomogeneouscubicmonomialsorderedlexicographically.Forahomogeneouscubicformf2S3,let[f]denotethecolumnthatgivesitscoordinatesinthisbasis.TheequalityS1q2=S3holdsifandonlyifthecnematrix=�[x1q1]j[x1q2]j


j[x1qn]j[x2q1]j


j[xeqn]
hasmaximalrank,i.e.rankcasonehasc6ne.SetE=�[x1(x21)]j[x1(x1x2)]j


j[x1(x2e)]j[x2(x21)]j


j[xe(x2e)]
;itisacmematrix,andeachcolumnofEisidenticaltoacolumnintheccidentitymatrixIc.Inparticular,Ehasentriesfromthesetf0;1g.Let
bethematrixe,thatis,theblockmatrixwithecopiesofonthediagonaland0elsewhere;itisamatrixofsizemene.Itisstraightforwardtoverifytheequality=E
.Setg=ne�candletCdenotethecollectionofallsubsetsoff1;:::;megthathavecardinalityg.Fori2f1;:::;megletridenotetheithrowoftheidentitymatrixIme.Foreach =ft1;:::;tgginCsetE =0BBB@rt1...rtg E1CCCA;itisannemematrixwithentriesfromthesetf0;1g.Claim.Thematrixhasmaximalrankifandonlyifthereexistsa 2CsuchthatE
hasnon-zerodeterminant.Proof.AssumethatthenenematrixE
hasnon-zerodeterminant.Onecanwritethedeterminantasalinearcombinationwithcoecientsinf�1;0;1gofthec-minorsofthesubmatrixE
,so=E
hasmaximalrank.Toprovetheconverse,assumethatE
hasmaximalrank,i.e.rankc.Thematrix
hasmaximalrank,ne,sotherowsof
spankne.Therefore,onecanchoose =ft1;:::;tgginCsuchthatrowsnumbert1;:::;tgin
togetherwiththerowsofE
spankne.Thatis,therowsofE
spankne,sothedeterminantofE
isnon-zero.ThedeterminantsP =det(E Ze),for 2C,yield�meg
homogeneouspoly-nomialsofdegreeeink[0;:::;N].UnderthePluckerembedding,ismappedtoapointinthecomplementofthealgebraicvariety\ 2CZ(P )PNkifandonlyifdet(E e)isnon-zeroforsome 2C.ByClaimsucha existsifandonlyifhasmaximalrank,thatis,ifandonlyifbelongstoHk(e).(8.8)Remark.Let(R;m;k)bealocalk-algebrawithm3=0.TheargumentthatshowstheopennessofEk(e)yieldsadditionalinformation.Namely,ifthereisanexactzerodivisorinR,thenoneofthepolynomialsFiinthevariables1;:::;eisnon-zero,andeverypointinthecomplementofitsvanishingsetZ(Fi)Pe�1kcor-respondstoanexactzerodivisor.Thusifkisinnite,thenagenerichomogeneouslinearforminRisanexactzerodivisor. BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES299.Shortlocalringswithoutexactzerodivisors|AnexampleLetkbeaeld;setR=k[s;t;u;v]=(s2;sv;t2;tv;u2;uv;v2�st�su)andm=(s;t;u;v)R:Concamentionsin[7,Example12]thatalthoughRisastandardgradedk-algebrawithHilbertseries1+4+32and(0:m)=m2,empiricalevidencesuggeststhatthereisnoelementx2Rwith(0:x)=(x);thatis,thereisnoConcageneratorofm.Proposition(9.1)conrmsthis,andtogetherwithProposition(9.2)itexhibitspropertiesofRthatframetheresultsintheprevioussections.Inparticular,thesepropositionsshowthatnon-freetotallyre exivemodulesmayexistevenintheabsenceofexactzerodivisors,andthatexactzerodivisorsmayexistalsointheabsenceofConcagenerators.(9.1)Proposition.Thefollowingholdforthek-algebraR.(a)ThereisnoelementxinRwith(0:x)=(x).(b)Ifkdoesnothavecharacteristic2or3,thentheelementss+t+2u�vand3s+t�2u+4vformanexactpairofzerodivisorsinR.(c)Assumethatkhascharacteristic3.If#2kisnotanelementoftheprimesubeldF3,thentheelement(1�#)s+#t+u+visanexactzerodivisorinR.IfkisF3,thentherearenoexactzerodivisorsinR.(d)Ifkhascharacteristic2,thentherearenoexactzerodivisorsinR.(9.2)Proposition.TheR-modulepresentedbythematrix=t�t+u�vt+u�vs+uisindecomposableandtotallyre exive,itsrstsyzygyispresentedby =�t+v2s+t�u+2vt+us�u+v;anditsminimalfreeresolutionisperiodicofperiod2.Proofof(9.1).Foranelementx=s+t+ u+vinmwedenotetheimagesof,, ,andinR=m=kbya,b,c,andd.Forxandx0=0s+0t+ 0u+0vtheproductxx0canbewrittenintermsofthebasisfst;su;tugform2asfollows:(1)xx0=(ab0+ba0+dd0)st+(ac0+ca0+dd0)su+(bc0+cb0)tu:Theproductxmisgeneratedbytheelementsxs=bst+csu,xt=ast+ctu,xu=asu+btu,andxv=dst+dsu.ByRemark(7.2)theequalityxm=m2holdsifandonlyifthematrixx=0@ba0dc0ad0cb01Ahasanon-zero3-minor;thatis,ifandonlyifoneof1(x)=�2abc;2(x)=cd(c�b);3(x)=bd(c�b);and4(x)=�ad(c+b)(2)isnon-zero.Thuselementsxandx0withxx0=0formanexactpairofzerodivisorsifandonlyifoneoftheminors1(x);:::;4(x)isnon-zero,andoneoftheminors1(x0);:::;4(x0)ofthematrixx0isnon-zero. 30L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGAND(a):Fortheequality(0:x)=(x)tohold,xmustbeanelementinmnm2.Ifx=s+t+ u+vsatisesx2=0,then(1)yields2ab+d2=0;2ac+d2=0;and2bc=0:Itfollowsthatborciszeroandthenthatdiszero.Thuseachoftheminors1(x);:::;4(x)iszero,whencexdoesnotgenerate(0:x).(b):Fortheelementsx=s+t+2u�vandx0=3s+t�2u+4v,itisimmediatefrom(1)thattheproductxx0iszero,while(2)yields3(x)=�1and1(x0)=12.Ifkdoesnothavecharacteristic2or3,then1(x0)isnon-zero,soxandx0formanexactpairofzerodivisorsinR.(c):Assumethatkhascharacteristic3.IfkproperlycontainsF3,thenchooseanelement#2knF3andsetx=(1�#)s+#t+u+v:Observethat2(x)=1�#isnon-zero.If#isnota4throotofunity,setx0=(1+#)s+t�#u�(1+#2)v;andnotethattheminor2(x0)=#(1+#)(1+#2)isnon-zero.From(1)onereadilygetsxx0=0,soxandx0formanexactpairofzerodivisors.If#isa4throotofunity,thenonehas#2=�1.Setx00=(1�#)s+t+#u�v;thentheminor2(x00)=#(1�#)isnon-zero,anditisagainstraightforwardtoverifytheequalityxx00=0.Therefore,xandx00formanexactpairofzerodivisors.Assumenowk=F3andassumethattheelementsx=s+t+ u+vandx0=0s+0t+ 0u+0vformanexactpairofzerodivisorsinR.Oneoftheminors1(x);:::;4(x)isnon-zero,andoneoftheminors1(x0);:::;4(x0)isnon-zero,soitfollowsfrom(2)thatabcordisnon-zeroandthata0b0c0ord0isnon-zero.Firstassumedd06=0;wewillshowthatthesixelementsa;a0;b;b0;c;c0arenon-zeroandderiveacontradiction.Supposeb=0,then(1)yieldscb0=0andab06=0,whichforcesc=0.This,however,contradictstheassumptionthatoneoftheminors1(x);:::;4(x)isnon-zero,cf.(2).Therefore,bisnon-zero.Aparallelargumentsshowthatcisnon-zero,andbysymmetrytheelementsb0andc0arenon-zero.Supposea=0,thenitfollowsfrom(1)thata0isnon-zero,asdd06=0byassumption.However,(1)alsoyieldsa(b0�c0)=a0(c�b);sotheassumptiona=0forcesc=b,whichcontradictstheassumptionthatoneoftheminors1(x);:::;4(x)isnon-zero.Thusaisnon-zero,andbysymmetryalsoa0isnon-zero.Withoutlossofgenerality,assumea=1=a0,then(1)yields(3)b0+b=c0+candbc0+cb0=0:Eliminateb0betweenthesetwoequalitiestoget(4)b(c0�c)+c(c0+c)=0:Ascandc0arenon-zeroelementsinF3,theelementsc0�candc0+caredistinct,andtheirproductis0.Thusoneandonlyoneofthemis0,whichcontradicts(4)asbothbandcarenon-zero.Nowassumedd0=0.Withoutlossofgenerality,assumethatdiszero,thenabcisnon-zero.Itfollowsfrom(2)thattheelementsa0,b0,andc0cannotallbezero, BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES31andthen(1)showsthattheyareallnon-zero.Thusallsixelementsa;a0;b;b0;c;c0arenon-zero,andasabovethisleadstoacontradiction.(d):Assumethatkhascharacteristic2andthattheelementsx=s+t+ u+vandx0=0s+0t+ 0u+0vformanexactpairofzerodivisorsinR.From(2)onegetsd6=0,d06=0,b6=c,andb06=c0.Arguingasinpart(c),itisstraightforwardtoverifythatthesixelementsa;a0;b;b0;c;c0arenon-zero.Withoutlossofgenerality,assumea=1=a0.From(1)thefollowingequalitiesemerge:b0+b+dd0=0;c0+c+dd0=0;andbc0+cb0=0:Thersttwoequalitiesyieldb0+b6=0andc0+c6=0.Further,eliminationofdd0yieldsb0=b+c+c0.Substitutethisintothethirdequalitytogetb(c0+c)+c(c0+c)=0:Asc0+cisnon-zerothisimpliesb=c,whichisacontradiction.Proofof(9.2).Itiseasytoverifythattheproducts and arezero,whenceF:


�!R2��!R2 ��!R2��!R2�!


;isacomplex.WeshallrstprovethatFistotallyacyclic.Toseethatitisacyclic,onemustverifytheequalitiesIm =KerandIm=Ker .Assumethattheelement�xx0
isinKer.Itisstraightforwardtoverifythatm2R2iscontainedintheimageof ,sowemayassumethatxandx0havetheformx=as+bt+cu+dvandx0=a0s+b0t+c0u+d0v,wherea;b;c;danda0;b0;c0;d0areelementsink.Usingthatfst;su;tugisabasisform2,theassumption�xx0
=0canbetranslatedintothefollowingsystemofequations0=8��������&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;&#x]TJ ;� -1;.93; Td;&#x [00;:a�a0�d0a0�d0c+b0�c0a�d+b0a�d+a0+c0b+c+b0Fromhereonederives,inorder,thefollowingidentitiesa0=d0;a=2a0;d=2a0+b0;c0=�a0+b0;c=�a0;andb=a0�b0;whichimmediatelyyield�xx0
= �b0a0
.ThisprovestheequalityIm =Ker.Similarly,itiseasytocheckthatm2R2iscontainedinIm,andforanelement(1)xx0=as+bt+cu+dva0s+b0t+c0u+d0v;wherea;a0;:::;d;d0areelementsink,onendsthat �xx0
=0implies�xx0
=�b0a0
.ThisprovestheequalityIm=Ker ,soFisacyclic.ThedierentialsinthedualcomplexHomR(F;R)arerepresentedbythematricesTand T.Oneeasilycheckstheinclusionsm2R2ImTandm2R2Im T.Moreover,foranelementoftheform(1)onendsT�xx0
=0implies�xx0
= T�b0a0�2b0
and T�xx0
=0implies�xx0
=T��b0a0
: 32L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGANDThisprovesthatalsoHomR(F;R)isacyclic,sothemodule,M,presentedbyistotallyre exive.Moreover,therstsyzygyofMispresentedby ,andtheminimalfreeresolutionofMisperiodicofperiod2.ToprovethatMisindecomposable,assumethatthereexistmatricesA=(aij)andB=(bij)inGL2(R),suchthatABisadiagonalmatrix;i.e.theequalities0=(a12b22)s+((a11+a12)b12�a11b22)t+(a12b12+(a11+a12)b22)u�(a12b12+a11b22)v(2)and0=(a22b21)s+((a21+a22)b11�a21b21)t+(a22b11+(a21+a22)b21)u�(a22b11+a21b21)v(3)hold.AsthematricesAandBareinvertible,neitherhasaroworacolumnwithbothentriesinm.Sincetheelementss,t,u,andvarelinearlyindependentmodulom2,itfollowsfrom(2)thata12b22isinm.Assumethata12isinm,thena11anda22arenotinm.Italsofollowsfrom(2)thata12b12+a11b22isinm,whichforcestheconclusionb222m.However,thisimpliesthatb21isnotinm,soa22b21isnotinmwhichcontradicts(3).Aparallelargumentshowsthatalsotheassumptionb222mleadstoacontradiction.ThusMisindecomposable.10.Familiesofnon-isomorphicmodulesofinfinitelengthMostavailableproofsoftheexistenceofinnitefamiliesoftotallyre exivemodulesarenon-constructive.Intheprevioussectionswehavepresentedconstructionsthatapplytolocalringswithexactzerodivisors.In[12]Holmgivesadierentconstruction;itappliestoringsofpositivedimensionwhichhaveaspecialkindofexactzerodivisors.Hereweprovideonethatdoesnotdependonexactzerodivisors.(10.1)Construction.Let(R;m)bealocalringandlet{=fx1;:::;xegbeaminimalsetofgeneratorsform.LetNbeanitelygeneratedR-moduleandletN1beitsrstsyzygy.LetFNbeaprojectivecover,andconsidertheelement:0�!N1�!F�!N�!0inExt1R(N;N1).Fori2f1;:::;egandj2Nrecallthatxjiisthesecondrowinthediagram:0// N1// xji F//  N// 0xji:0// N1!(i;j)// P(i;j)// N// 0;wheretheleft-handsquareisthepushoutofalongthemultiplicationmapxji.ThediagramdenesP(i;j)uniquelyuptoisomorphismofR-modules.SetP({;N)=fP(i;j)j16i6e;j2Ng;notethateverymoduleinP({;N)canbegeneratedbyR0(N)+R1(N)elements.(10.2)Lemma.Let(R;m)bealocalringandletNbeanitelygeneratedR-module.If,forsomeminimalset{=fx1;:::;xegofgeneratorsform,thesetP({;N)containsonlynitelymanypairwisenon-isomorphicmodules,thentheR-moduleExt1R(N;N1)hasnitelength. BRAUER{THRALLFORTOTALLYREFLEXIVEMODULES33Proof.Let{=fx1;:::;xegbeaminimalsetofgeneratorsformandassumethatP({;N)containsonlynitelymanypairwisenon-isomorphicmodules.Givenanindexi2f1;:::;eg,thereexistpositiveintegersmandnsuchthatthemod-ulesP(i;m)andP(i;n+m)areisomorphic.Sincexn+miequalsxni(xmi),themo-duleP(i;n+m)comesfromthepushoutof!(i;m)alongthemultiplicationmapxni;cf.Construction(10.1).Thusthereisanexactsequence0�!N1��!N1P(i;m)�!P(i;n+m)�!0;where=�xni�!(i;m)
.ItfollowsfromtheisomorphismP(i;m)=P(i;n+m)andMiyata'stheorem[16]thatthissequencesplits.Hence,itinducesasplitmonomorphismExt1R(N;N1)Ext1R(N;)��������!Ext1R(N;N1)Ext1R(N;P(i;m)):Letbealeft-inverseofExt1R(N;),setmi=mandnoticethattheelementxmii=Ext1R(N;)(xmii)=�xn+mii0
=xn+mii�0
belongstomn+miExt1R(N;N1).Foreveryindexi2f1;:::;egletmibethepositiveintegerobtainedabove.Withh=m1+


+methereisaninclusionmhExt1R(N;N1)mh+1Ext1R(N;N1),soNakayama'slemmayieldsmhExt1R(N;N1)=0.(10.3)Theorem.Let(R;m)bealocalringandlet{=fx1;:::;xegbeaminimalsetofgeneratorsform.Ifthereexistsatotallyre exiveR-moduleNandaprimeidealp6=msuchthatNpisnotfreeoverRp,thenthesetP({;N)containsinnitelymanyindecomposableandpairwisenon-isomorphictotallyre exiveR-modules.Proof.ItfollowsfromtheassumptionsonNthateverymoduleinthesetP({;N)istotallyre exiveandthattheR-moduleExt1R(N;N1)hasinnitelength,asitssupportcontainstheprimeidealp6=m.By(10.2)thesetP({;N)containsinnitelymanypairwisenon-isomorphicmodules.EverymoduleinP({;N)isminimallygeneratedbyatmostR0(N)+R1(N)elements;seeConstruction(10.1).Therefore,everyinnitecollectionofpairwisenon-isomorphicmodulesinP({;N)containsinnitelymanyindecomposablemodules.AcknowledgmentsItisourpleasuretothankManojKumminiandChristopherMonicoforhelpfulconversationsrelatedtosomeofthematerialinthispaper.References1.MauriceAuslanderandMarkBridger,Stablemoduletheory,MemoirsoftheAmericanMath-ematicalSociety,No.94,AmericanMathematicalSociety,Providence,R.I.,1969.MR02696852.WinfriedBrunsandUdoVetter,Determinantalrings,LectureNotesinMathematics,vol.1327,Springer-Verlag,Berlin,1988.MR9539633.Ragnar-OlafBuchweitz,Gert-MartinGreuel,andFrank-OlafSchreyer,Cohen-Macaulaymod-ulesonhypersurfacesingularitiesII,Invent.Math.88(1987),no.1,165{182.MR8770114.LarsWintherChristensen,AndersFrankild,andHenrikHolm,OnGorensteinprojective,injectiveand atdimensions|Afunctorialdescriptionwithapplications,J.Algebra302(2006),no.1,231{279.MR22366025.LarsWintherChristensen,GregPiepmeyer,JanetStriuli,andRyoTakahashi,FiniteGoren-steinrepresentationtypeimpliessimplesingularity,Adv.Math.218(2008),no.4,1012{1026.MR2419377 34L.W.CHRISTENSEN,D.A.JORGENSEN,H.RAHMATI,J.STRIULI,ANDR.WIEGAND6.LarsWintherChristensenandOanaVeliche,Acyclicityoverlocalringswithradicalcubezero,IllinoisJ.Math.51(2007),no.4,1439{1454.MR24174367.AldoConca,Grobnerbasesforspacesofquadricsoflowcodimension,Adv.inAppl.Math.24(2000),no.2,111{124.MR17489658.VesselinN.GasharovandIrenaV.Peeva,Boundednessversusperiodicityovercommutativelocalrings,Trans.Amer.Math.Soc.320(1990),no.2,569{580.MR9673119.In^esB.HenriquesandLianaM.Sega,FreeresolutionsovershortGorensteinlocalrings,Math.Z.,onlineNovember2009,19pp.PreprintarXiv:0904.3510v2[math.AC].10.JurgenHerzog,RingemitnurendlichvielenIsomorphieklassenvonmaximalen,unzerlegbarenCohen-Macaulay-Moduln,Math.Ann.233(1978),no.1,21{34.MR046315511.MelvinHochsterandDanLaksov,Thelinearsyzygiesofgenericforms,Comm.Algebra15(1987),no.1-2,227{239.MR87697912.HenrikHolm,Constructionoftotallyre exivemodulesfromanexactpairofzerodivisors,preprint,2010,arXiv:1002.0419v1[math.AC].13.RogerA.HornandCharlesR.Johnson,Matrixanalysis,CambridgeUniversityPress,Cam-bridge,1985.MR83218314.DavidA.JorgensenandLianaM.Sega,Independenceofthetotalre exivityconditionsformodules,Algebr.Represent.Theory9(2006),no.2,217{226.MR223836715.ClasLofwall,Onthesubalgebrageneratedbytheone-dimensionalelementsintheYonedaExt-algebra,Algebra,algebraictopologyandtheirinteractions(Stockholm,1983),LectureNotesinMath.,vol.1183,Springer,Berlin,1986,pp.291{338.MR84645716.TakehikoMiyata,Noteondirectsummandsofmodules,J.Math.KyotoUniv.7(1967),65{69.021458517.GunterScheja,UberdieBettizahlenlokalerRinge,Math.Ann.155(1964),155{172.MR016281918.Jean-PierreSerre,Surlesmodulesprojectifs,SeminareDubreil.Algebraettheoriedesnom-bres,tome14,no.1,exp.no.2(Paris),Secretariatmathematique,1960/61,pp.1{16.19.YujiYoshino,ModulesofG-dimensionzerooverlocalringswiththecubeofmaximalidealbeingzero,Commutativealgebra,singularitiesandcomputeralgebra(Sinaia,2002),NATOSci.Ser.IIMath.Phys.Chem.,vol.115,KluwerAcad.Publ.,Dordrecht,2003,pp.255{273.MR2030276L.W.C.DepartmentofMath.andStat.,TexasTechUniversity,Lubbock,TX79409,U.S.A.E-mailaddress:lars.w.christensen@ttu.eduURL:http://www.math.ttu.edu/~lchristeD.A.J.DepartmentofMathematics,UniversityofTexas,Arlington,TX76019,U.S.A.E-mailaddress:djorgens@uta.eduURL:http://dreadnought.uta.edu/~daveH.R.DepartmentofMath.andStat.,TexasTechUniversity,Lubbock,TX79409,U.S.A.Currentaddress:MathematicsDepartment,SyracuseUniversity,Syracuse,NY13244,U.S.A.E-mailaddress:hrahmati@syr.eduJ.S.DepartmentofMath.andC.S.,FairfieldUniversity,Fairfield,CT06824,U.S.A.E-mailaddress:jstriuli@fairfield.eduURL:http://www.faculty.fairfield.edu/jstriuliR.W.DepartmentofMathematics,UniversityofNebraska,Lincoln,NE68588,U.S.A.E-mailaddress:rwiegand1@math.unl.eduURL:http://www.math.unl.edu/~rwiegand1