XIANG Yue-ming

(College of Mathematics and Computer Science, Hunan Normal University, Changsha 410006, China)

1 Introduction

ArightR-moduleTiscalledtorsionlessiftheevaluationmapσ:T→T**isinjection.AringRissaidtoberight∏-coherent[5]ifeveryfinitelygeneratedtorsionlessrightR-moduleisfinitelypresented.TherightFGT-injectivedimensionofarightR-moduleM[6],denotedbyFGT-id(M),isdefinedastheleastnonnegativeintegernsuchthatExtn+1(T,M)=0foranyfinitelygeneratedtorsionlessrightR-moduleT.SetFGT-I.dim(R) =sup{FGT-id(M):M∈MR}andcallFGT-I.dim(R)therightFGT-injectivedimensionofR.TheleftFGT-flatdimensionofaleftR-moduleF,denotedbyFGT-fd(F),isdefinedastheleastnonnegativeintegernsuchthatTorn+1(T,F)=0foranyfinitelygeneratedtorsionlessrightR-moduleT.ArightR-moduleMiscalledFGT-injectiveifExt1(T,M)=0foranyfinitelygeneratedtorsionlessrightR-moduleT.AleftR-moduleFiscalledFGT-flatifTor1(T,F)=0foranyfinitelygeneratedtorsionlessrightR-moduleT(see[6]).AleftR-moduleMisFGT-flatifandonlyifM+isFGT-injectivebythestandardisomorphismExt1(T,M+)?Tor1(M,T)+foranyfinitelygeneratedtorsionlessrightR-moduleT.WewriteTIfortheclassofallFGT-injectiveR-modules.Following[4],TIisclosedunderextensions,directsummandsanddirectproducts.IfRisa∏-coherentring,thenTIisclosedunderdirectsums,directlimitsandpuresubmodules.

Thetheoryof(pre)coversand(pre)envelopestakesanimportantpartintheoryofringsandmodules,homologicalalgebra,representationtheoryofalgebraandsoon. (Pre)envelopesand(pre)coversofmoduleshavebeenstudiedbymanyauthors[2,4,7-9].Inparticular,thecokernelof(pre)envelopeandthekernelof(pre)coverarealsoconsidered[10-12].InSection2ofthisarticle,weintroducetheconceptsofTI-injectiveandTI-flatmodulesandshowsomeoftheirgeneralproperties.Section3isdevotedtoTI-dimensionsofmodulesandrings.

2 TI-injective modules and TI-flat modules

Definition1ArightR-moduleMissaidtobeTI-injectiveifExt1(N,M)=0foranyFGT-injectiverightR-moduleN.AleftR-moduleFiscalledTI-flatifTor1(N,F)=0foranyFGT-injectiverightR-moduleN.

Remark1(1)Bywakamutsu’sLemma,anykernelofanFGT-injectivecoverisTI-injective.

(2)AleftR-moduleFisTI-flatifandonlyifF+isTI-injectivebythestandardisomorphismExt1(N,F+) ?Tor1(N,F)+foranyFGT-injectiverightR-moduleN.

Proposition1ThefollowingareequivalentforarightR-moduleM:

(1)MisTI-injective.(2)Foreveryexactsequence0→M→A→B→0withanFGT-injective,A→BisanFGT-injectiveprecoverofB.(3)MisthekernelofanFGT-injectiveprecoverf:A→BwithAinjective.(4)Misinjectivewithrespecttoeveryexactsequence0→K→A→C→0,whereCisFGT-injective.

ProofTheproofissimilartothatof[10,Proposition2.4].

Itisclearthateveryinjective(flat)moduleisTI-injective(TI-flat)byDefinition1.Theconverseisnottrueingeneral.However,ifRisaright∏-coherentring,wehavethefollowingproposition.

ArightR-moduleMiscalledreduced[2]ifMhasnononzeroinjectivesubmodules.Similartotheproofsof[10,Proposition2.3],[10,Proposition2.5]，[10,Theorem2.6]respectively,wehave

Proposition2LetRbearight∏-coherentring.Thenthefollowingstatementshold:

(1)ArightR-moduleMisinjectiveifandonlyifMisTI-injectiveandFGT-id(M) ≤1.

(2)AleftR-moduleFisflatifandonlyifFisTI-flatandFGT-fd(F)≤1.

Proposition3LetMbearightR-moduleoveraright∏-coherentringR.Thenthefollowingareequivalent:

(1)MisreducedTI-injective. (2)MisthekernelofanFGT-injectivecoverf:A→BwithAinjective.

Theorem1LetMbearightR-moduleoveraright∏-coherentringR.Thenthefollowingareequivalent:

(1)MisTI-injective.

(2)MisadirectsumofaninjectiverightR-moduleandareducedTI-injectiverightR-module.

Proposition4LetSbeasimpleR-moduleoveracommutativeringR.Thenthefollowingareequivalent:

(1)SisTI-injective. (2)SisTI-flat.

ProofSupposethat{Si}i∈IisanirredundantsetofrepresentativesofthesimpleR-modules.LetE=E(⊕i∈ISi),theinjectivehullof⊕i∈ISi,thenEisaninjectivecogenerator.ForanyFGT-injectiveR-moduleN,thereexistsanisomorphismExt1(N,Hom(S,E))?Hom(Tor1(N,S),E).NotethatHom(S,E)?S.ThusSisTI-injectiveifandonlyifExt1(N,Hom(S,E)) =0ifandonlyifHom(Tor1(N,S),E)=0ifandonlyifTor1(N,S)=0ifandonlyifSisTI-flat.

Proposition5LetRbeacommutative∏-coherentringandFbeaflatR-module.Thenthefollowingstatementshold:

(1)MisTI-injectiveifandonlyifHom(F,M)isTI-injective.

(2)NisTI-flatifandonlyifF?NisTIV-flat.

Proof(1)(?)holdsbylettingF=R.(?).ForanyFGT-injectiveR-moduleEandflatR-moduleF,weclaimthatE?FisFGT-injective.Infact,anyfinitelygeneratedtorsionlessR-moduleTisfinitelypresentedsinceRis∏-coherent,thenthereisanexactsequence0→K→P→T→0withFandKfinitelygeneratedandPfree,soPandKarefinitelypresented.Ontheotherhand,thesequenceHom(P,E)?F→Hom(K,E)?F→0isexactsinceEisFGT-injective.Furthermore,wehavethefollowingcommutativediagram:

Hom(P,E)?F→Hom(K,E)?F→0

α↓β↓

Hom(P,E?F) →Hom(K,E?F).

SincePandKarefinitelypresented,by[2,Theorem3.2.14],αandβareisomorphic.ThenHom(P,E?F)→Hom(K,E?F)→0isexact.ThusExt1(T,E?F)=0,andhenceE?FisFGT-injective.

NowweprovethatHom(F,M)isTI-injective.Thereexistsanexactsequence0→K1→P1→E→0withP1projective.ThenwehaveaninducedexactsequenceHom(P1?F,M)→Hom(K1?F,M)→Ext1(E?F,M)=0.

SothesequenceHom(P1,Hom(F,M))→Hom(K1,Hom(F,M))→0isexact.ThusExt1(E,Hom(F,M))=0,asdesired.

(2)NisTI-flatifandonlyifN+isTI-injectiveifandonlyifHom(F,N+)isTI-injectiveby(1)ifandonlyif(F?N)+isTI-injectivebythestandardisomorphism(F?N)+?Hom(F,N+)ifandonlyifF?NisTI-flat.

RecallthatRisaQF-ring[1]ifRisleft(right)noetherianandRR(RR)isinjective.HerewehaveanewcharacterizationofQF-rings.

Theorem2RisaQF-ringifandonlyifeveryrightR-moduleisTI-injective.

ProofIfRisaQF-ring,inviewof[4,Remark2.3 (2)],everyFGT-injectiverightR-moduleisinjective,andhenceeveryFGT-injectiverightR-moduleisprojectiveby[1,Theorem31.9].ThuseveryrightR-moduleisTI-injective.Conversely,notethatanyinjectiverightR-moduleMisFGT-injective,byassumption,anyrightR-moduleNisTI-injective,soExt1(M,N)=0,henceMisprojective.Therefore,RisaQF-ringby[1,Theorem31.9]again.

Inthefollowingproposition,weconsiderthecokernelofanFGT-flatpreenvelope.

Proposition6LetRbearight∏-coherentring.Thenthefollowingstatementshold:

(1)IfCisthecokernelofanFGT-flatpreenvelopef:M→FofaleftR-moduleMwithFflat,thenCisTI-flat.

(2)IfLisafinitelypresentedTI-flatleftR-module,thenListhecokernelofanFGT-flatpreenvelopeg:K→PwithPflat.

ProofTheproofissimilartothatof[10,Proposition2.7].

LetKbeasubmoduleofleftR-moduleM.Kiscalledaclosedsubmodule[13]ifM/Kistorsionless.

Proposition7LetRbealeftandright∏-coherentring.Thenthefollowingareequivalent:

(1)FGT-id(RR) ≤1. (2)EveryclosedsubmoduleofafinitelygeneratedTI-flatleftR-moduleisTI-flat.

Proof(1)?(2).LetKbeaclosedsubmoduleofafinitelygeneratedTI-flatleftR-moduleM.ForanyFGT-injectiverightR-moduleN,thereisanexactsequenceTor2(N,M/K)→Tor1(N,K)→Tor1(N,M) = 0.InviewofTheorem5.6.16andTheorem5.6.17of[6],FGT-fd(NR) ≤1.NotethatM/Kisfinitelygeneratedtorsionless,soTor2(N,M/K)=0.ThusTor1(N,K)=0,andhenceKisTI-flat.

(2)?(1).ForanyfinitelygeneratedtorsionlessleftR-moduleM,thereisanexactsequence0→K→F→M→0,whereKisclosedsubmoduleoffinitelygeneratedfreeR-moduleF.Sothereisaninducedexactsequence0=Tor2((RR)+,F)→Tor2((RR)+,M)→Tor1((RR)+,K) →…Byassumption,KisTI-flat.ThenTor1((RR)+,K)=0,andhenceTor2((RR)+,M)=0.SoFGT-fd((RR)+) ≤1.By[6,Theorem5.6.11],FGT-id(RR)≤1.

3 TI-dimensions

LetRbearight∏-coherentring.By[4,Theorem3.4],everyrightR-modulehasaTI-precover.TheneveryrightR-modulehasaleftTI-resolution.Accordingto[9,Theorem3.5]and[4,Lemma3.2],itiseasytoverifythateveryrightR-modulehasaTI-preenvelope.ThuseveryrightR-modulehasarightTI-resolution.SoHom(-,-)isleftbalanced[14]onMR×MRbyTI×TI.LetExtn(-,-)bethenthleftderivedfunctorofHom(-,-)withrespecttothepairTI×TI.Then,fortworightR-modulesMandN,Extn(M,N)canbecomputedbyusingarightTI-resolutionofMoraleftTI-resolutionofN.

Proposition8LetRbearight∏-coherentring.ThefollowingareequivalentforarightR-moduleM:

(1)MisFGT-injective. (2)Thecanonicalmapα:Ext0(M,N)→Hom(M,N)isanepimorphismforanyrightR-moduleN. (3)Thecanonicalmapα:Ext0(M,M)→Hom(M,M)isanepimorphism.

ProofTheproofissimilartothatof[10,Proposition3.1].

Lemma1LetRbearight∏-coherentringandMarightR-module.ThenFGT-id(M)=rightTI-dimM.Moreover,FGT-I.dim(R)=gl.rightTI-dimMR.

ProofNotethatrightTI-resolutionisexactbecauseinjectivemoduleisFGT-injective.ItiseasytoseethatFGT-id(M)≤rightTI-dimM.Conversely,assumethatFGT-id(M)=n<∞.Let0→M→F0→F1→…→Fn→…bearightTI-resolutionofM.Thenwehaveanexactsequence0→M→F0→F1→…→Fn-1→L→0,whereL=Coker(Fn-2→Fn-1).By[6,Proposition5.5.4],LisFGT-injective,andhencerightTI-dimM≤n.

Proposition9LetRbearight∏-coherentring.ThefollowingareequivalentforarightR-moduleM:

(1)FGT-id(M)≤1.

(2)Thecanonicalmapα:Ext0(M,N)→Hom(M,N)isamonomorphismforanyrightR-moduleN.

ProofTheproofissimilartothatof[10,Proposition3.3].

AringRiscalledaD-ring[13]ifI=rl(I)foreveryrightidealIandL=lr(L)foreveryleftidealLofR,wherel(I)(resp.r(L))denotestheleft(resp.right)annihilatorofI(resp.L).ItisshownthatFGT-injectiveR-modulescoincidewithinjectiveR-modulesoveraD-ring(see[6,Proposition5.5.1]).

Proposition10LetRbearight∏-coherentring.If

(1)FGT-id(RR)≤1.(2)Thecanonicalmapα:Ext0(R,N) →Hom(R,N)isamonomorphismforanyrightR-moduleN.(3)EveryTI-injectiverightR-modulehasanepimorphicTI-precover.Then(1)?(2)and(1)?(3).IfRisaD-ring,then(3)?(1).

Proof(1)?(2)followsfromProposition9.

(1)?(3).LetMbeaTI-injectiverightR-module.ByProposition1,MiskernelofaTI-precoverofarightR-moduleN.ThenwehavealeftTI-resolution…→F1→F0→N→0,whereM=Ker(F0→N)andFiisFGT-injectiveforalli.Ontheotherhand,thereisanexactsequence0→R→F0→F1→0,whereFiisFGT-injectivefori=0,1.So0→Hom(F1,N)→Hom(F0,N)→Hom(R,N)isexact.ThusExtk(R,N) = 0fork≥ 1andExt0(R,N) →Hom(R,N)ismonomorphicbyProposition9.ButcomputingExt0(R,N)usingaleftTI-resolutionofN,weseethatF1→F0→NisexactatF0,soMadmitsanepimorphicTI-precover.

(3)?(1).IfRisaD-ring,forleftTI-resolution…→F1→F0→N→0ofarightR-moduleNwithK1=Ker(F0→N),byProposition1,K1isTI-injective.Byassumption,K1hasanepimorphicTI-precoverF1→K1,henceHom(R,F1) →Hom(R,K1)isepimorphic.By[15,Lemma2.2],FGT-id(RR) ≤ 1.

Proposition11LetRbearight∏-coherentringandanintegern≥ 2.ThefollowingareequivalentforarightR-moduleM. (1)rightTI-dimM≤n.(2)Extn+k(M,N)=0forallrightR-modulesNandk≥-1.(3)Extn-1(M,N)=0forallrightR-modulesN.

Proof(1)?(2).Let0→M→E0→E1→…→En→0bearightTI-resolutionofM.Then0→Hom(En,N)→Hom(En-1,N)→Hom(En-2,N)isexactandsoExtn-1(M,N)=Extn(M,N)=0.Inaddition,itisclearthatExtn+k(M,N)=0forallk≥ 1.Hence(2)follows.

(2)?(3)istrivial. (3)?(1).Let0→M→E0→E1→…bearightTI-resolutionofMandletC=im(En-2→En-1).ThenExtn-1(M,En-1/C)=0byassumption.ButthenEn-1/C→Enhasaretract.HenceEn-1/CisFGT-injective.So0→M→E0→E1→…→En-1→En-1/C→0mustbearightTI-resolutionofM.

AclassCofrightR-modulesissaidtobecoresolving[8]ifE∈CforallinjectiveR-modulesE,ifCisclosedunderextensions,andifgivenanexactsequenceofrightR-modules0→A′→A→A→ 0,A″∈CwheneverA′,A∈C.By[4,Lemma3.2],TIiscoresolvingoveraright∏-coherentring.Thenwehavethefollowingproposition.

Proposition12LetRbearight∏-coherentringandanintegern≥2.Considerthefollowingstatements.

(1)leftTI-dimN≤n-2. (2)Extn+k(M,N)=0forallrightR-modulesMandk≥-1. (3)Extn-1(M,N)=0forallrightR-modulesM.

Then(1)?(2)?(3).IfRisaD-ring,then(3)?(1).

Corollary1LetRbearight∏-coherentringandanintegern≥2.Considerthefollowingstatements.

(1)gl.leftTI-dimMR≤n-2. (2)gl.rightTI-dimMR≤n. (3)Extn+k(M,N) =0forallrightR-modulesM,Nandk≥-1. (4)Extn-1(M,N)=0forallrightR-modulesM,N.

Then(1)?(3)?(4)?(2).IfRisaD-ring,then(2)?(1).

ProofItfollowsfromProposition11andProposition12.

Ahomomorphismg:M→CwithC∈CissaidtobeaC-envelopewiththeuniquemappingproperty[7]ifforanyhomomorphismg′:M→C′withC′∈C,thereisauniquehomomorphismf:C→C′suchthatfg=g′.Dually,wehavethedefinitionofaC-coverwiththeuniquemappingproperty.

Intheendofthissection,westudytheright∏-coherentringswithFGT-I.dim(R) ≤ 2.

Theorem3Thefollowingareequivalentforaright∏-coherentringR:

(1)FGT-I.dim(R) ≤ 2. (2)gl.rightTI-dimMR≤2. (3)EveryrightR-modulehasaTI-coverwiththeuniquemappingproperty. (4)EveryTI-injectiverightR-modulehasaTI-coverwhichismonomorphic.

(3)?(2).ForanyrightR-moduleM,thereexistsaTI-coverf:F→Mwiththeuniquemappingproperty,hence0→F→M→0isaleftTI-resolution.Thusgl.leftTI-dimMR=0,sogl.rightTI-dimMR≤2byCorollary3.7.

(1)?(4).LetMbeaTI-injectiverightR-module.ThenMisakernelofaTI-precoverF0→NofarightR-moduleN.ThuswehavealeftTI-resolution…→F1→F0→N→0,whereM=Ker(F0→N)andK2=Ker(F1→F0).Byassumption,FGT-id(K2) ≤2,soHom(K2,F2)→Hom(K2,K2)isepimorphicby[15,Lemma2.2].ThusK2isFGT-injective.Ontheotherhand,f:F1→MisaTI-precoverofM,thenwehavetheexactsequence0→K2→F1→im(f)→0.Thusim(f)isFGT-injectivesinceTIiscoresolving.Thereforetheinclusionim(f) →MisaTI-coverwhichismonomorphic.

(4)?(1).LetMbearightR-moduleandlet…→F1→F0→N→0beanyleftTI-resolutionofarightR-moduleNandK2=Ker(F1→F0),K1=Ker(F0→N).K1hasamonomorphicTI-coverF→K1byassumption.ThusK2⊕F?F1intermsof[2,Lemma8.6.3].SoK2isFGT-injective.ThusF2→K2isasplitepimorphism.ThenHom(M,F2) →Hom(M,K2)isepimorphic,andhenceFGT-id(M)≤2by[15,Lemma2.2]again.ThereforeFGT-I.dim(R) ≤ 2.

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