Original Paper
SOME NOTES ON REFLEXIVE MODULES
MasahisaSATO ZhaoyongHUANG Received June 3, 1998 Abstract In commutative ring theory， G−Matlis reflexive modules over a commutative complete semilocal noetherian ring have been studied． We attempt to show that these can be explained naturally through general theory we give in this paper on reflexive modules in non−commutative ring theory． In our observation， it is essential to study relations between chain conditions and reflexive conditions．
1，INTRODUCTION
Throughout this paper， R and S are associative rings with identity， all modules are unital． We always assume that a bimodule REs is a cogenerator both as a left R−module and as a right S−module． Let M be a left R−module（resp． a right 8−module）， we call HomR（RM， REs）（resp． Homs（Ms， REs））the dual module ofルI with respect to E， and denote it by M”（resp． Mti）． A module RM（resp． Ms）is said to be E−re且exive ifthe canonical homomorphismσM：M→ル1’ti（resp．σM：M→ル1‖＊）， de丘ned by
σM（x）（f）＝f（x）fbr any x∈Mand f∈ル1＊（resp． f∈Mu）is an isomorphism． It
is easy to see thatσM is always a monomorphism． Reca皿that a moduleルI is丘nitely cogenerated if for every set Z【 of submodules of M，∩Mi∈m Mi＝Oimplies∩Mi∈3 Mi＝O for some finite subset｛ζof 2t． By l 1， Proposition 10．10】， every quotient module ofルI is finitely cogenerated if and only if M is artinian． If a module E is a finitely cogenerated cogenerator， by l 1， Proposition 10．2］， a moduleルf is finitely cogenerated if and only if there is a monomorphism O→M→En fbr some positive integer n． In section 2， we investigate chain conditions under E−duality． First we prove that if R and S are E−reflexive and RE and Es are injective and・Es （resp． RE）is finitely cogen− erated， then RM is a noetherian（resp． artinian）if and only if HomR（M， E）is artinian （resp． noetherian）in Theorem 3 and Theorem 4 i Moreover， if R is a left noetherian ring and S is a right noetherian ring and RE and Es are injective and finitely cogeperated， then RM is noetherian（resp． artinian）if and only if HomR（M， N）is an artinian（resp． a noetherian）right S−module for any bimodule RNs which has an E−coresolution（Theorem 7and Theorem 7’）． ’ In section 4， we apply the above results to G−Matlis re且exive modules（refer to section 4fbr the definition）in commutative ring theory and we shbw that results on G−Matlis ＊The s㏄ond author’s research was supported by the Administration of Education of Japan as Schol− arship in Yamanashi University． 1991Mathematics Subject Classification．16E10，16D50，16P40． Key words． reflexive modules， noetherian modules， artinian modules．SOME NOTES ON REFLEXIVE MODULES reflexive modules over a commutative complete semilocal noetherian ring are treated as the special case of our duality theory stated in section 3．
2． PRELIMARY LEMMAs
Lemma l A’雄R−module RM is E・・reflexive if and o鳩ザα吻んt S−mαdule Mg isE一耐e励e．
Proof・We have that（σM）＊σM・＝IM・by l3，23．5 Proposition｜． Since RE is a cogenerator， σMis an isomorphism if and only ifσM． is an isomorphism． So we conclude that RM is E−reflexive if and only ifルfξis E−reflexive． 口 Lemma 2τんeμ・ωing stαtements are equivαlent． ル10reover， equivαlent to tんe above properties： （4） Evengノ．fiηitely cogenerated right S−module is E一γefle vive． Pro（’f， It is easy to see the equivalence of（1）and（2）from Lemma 1． （1）⇔（3）．（3）implies（1）is trivial． Now suppose RR is an E−reflexive left R−moduleand RM is丘nitely generated． Then we have an epimorphism RRn→RM→Ofbr some
positive integer n． Because Es is injective， we obtain the fbllowing commutative diagram with exact rows： （1） 1砒isαn E−reflexzve left R−mo（tule（i． e．， R＝End（E8））． （2） E8 乞3 αηE−reゾ7exil／e rigん￡5一η・］Lo〔iule． ∬E8 iS iηjeCtive， tんen￡んe∫靴）ZIOωing Pγ⇔per老y iS eO磁t／alent tO tんeαbOVeργりpertie8： （3） 」Ever3t finitely geneTated left・R−module is E−r匂eeStve． ザES iS injeCtiveαη∂疏彦吻C・generated， tんen tんeμ・ωing C・nditi・n iSRRn
⊥・・n 一一一ｨ RM −一→0
⊥・M（RRn）＊tt−→RM“ti−→o
SinceσRπis an isomorphism，σM is an epimorphism． ButσM is always a monomorphism， σMis an isomorphism and RM is E−reflexive． （2）⇔（4）That（4）implies（2）is trivial since Es is finitely cogenerated． Now suppose E is an E−reflexive right S−module and suppose Ns is丘nitely cogenerated． Then we have an exact sequence O→Ns→E§→E§／Ns−一〉 O for some positive integer n， Because Es is injective， we obtain the fbllowing commutative diagram with exact rows： 0−一→ ノVs −一→ ⊥・No−→N』＊一一・
E§ ⊥aEn （E§）u＊一
一
E§／Ns
！・E・・N （E§／Ns）‖＊一一
｡0
SinceσEn is an isomorphism， Cok（σN） Ker（σEn／N）tO． ButσN is a monomorphism， henceσN is an isomorphism and／Vs is E−reflexive．口 ＼Lemma 1’A吻ん￡S−m（嚇e的is・E−reflexive if and・吻ザα雄R−m・dule RNti i8
E−reflexive． Lemma・2’7Weμ・ωing stαtements aTe equivalent． （1） Ss ts an E−refleStveγ吻尼S−module k． e．， S＝＝End（RE）ノ． （2）RE ts an Eイeflexive left R−module．∬RE is卿iec物e， tんen・tんeμ0ωm9ρπ）pe吻ts equivαlent t・tんe above PT℃）perties：
（3）Euε卿輌W gene耐ed吻尼8−m・dule is・E−rej7exive．
M・reOVer」if RE iS卿ieCtiVe and finitely COgenerated， tんeπeαCん（ザεんeαbOUe卿ρ酷e3 法Sequivalent to tんe following condition： （4） Ever3t∫initely cogenβrated Iφ」R−module is Eイeゾ1ピ涜ue． Theorem 3 Suppose RR and Ss are E−Teflexiveαnd Es is injective and finitely co− genemte〔i． Then／bγ’αIfiプ1．R−module Rル1， tんe／bI↓oωづ7Lg statementsんold．（1）RM Z3疏鞠geηe耐・nt if and・晦ガM静輌τely c・generated
（2） ］ア、RE is inワiective，診んen児ルI is noetんerian勾Fαnd only if M三｝isαrtiniαn． Proof．（1）We prove that there is an ．epimorphism RRn→RM→Oif and only if there is a monomorphism O→．Mg−→E§， where n is a positive integer． Suppose there isamonomorphism O→Mき→瑠． Then we have an epimorphism（REn）tt→RM＊ロ→
0．But RM＊is E−reflexive， so M is E−re且exive by Lemma 1． Since RR is E−reflexive， （REn）‖i≧i RRrn． Hence R1㍗→RM→Ois exact． The converse is clear． So we have RM is finitely generated if and only ifルf』is finitely cogenerated． （2）Suppose RM is noetherian and Ks is any right S−submodule of Mき． From（1）， Lemma 2 and Lemma l we know that RM andルfきare E−reflexive and also any quotient module of Mきis E−reflexive since RE and Es are injective． An epimorphism M蓋一→Mき／K→Oinduces a monomorphism O・・→R（M＊／K）‖→RM＊tt…≡RM． But RM is
noetherian， so R（M＊／K）ロis finitely generated． Then it fbllows from（1）that M＊／K竺 （M＊／K）tt＊is a丘nitely cogenerated right S−module and Mきis artinian by Anderson and Fuller［1， Proposition 10．10］．Conversely， suppose M蓋is artinian， and suppose O→RK→、RM is exact． Then
ルlg→Kき→Ois exact． But M苔is artinian， so Kきis finitely cogenerated． Therefbre from（1）we know that RK is finitely generated and RM is noetherian．□ We can give the similar result of the fbrmer theorem fbr an S−module as fbllows．Theorem 3’SupP・se R．R and Ss are E−reflexive and RE is物ヒc励eα閲輌吻c・−
9eneratαd． Then∫bγ’αrigん孟S−mαduleハls，￡んe／b”oωing 8tatemen’tsんold． （1） 」∼Sis．tinitely generated乞『and only乞if RNti X3 JFinitelyαogenerated． （2）lf Es is inject醜tんen」Vs i8 n・e￡んeriαn if and・nly if RNtt ts artinian．Theorem 4 SupPo8e RR and Ss are E−refle蜘eαnd RE is吻セc伽eαηd輌τ吻c・−
generated．7Weη∫brα峨1∼覗・dule Rルf， tんeμ0ωτη98鋤emeηε8ん・ld．（1）RMτ8伽吻c・generated if and・晦σ鳩τ3疏診ely geηe耐e∂．
（2）lf Es i8物ie・tive， then RM is artiniαn if and・吻σMξ加・etんerian． Pro（’f，（1）Suppose RM is finitely cogenerated， Then there is an exact sequence O→ Rハ4→REn・So 3§！；lil（En）5−→Mg→Ois also exact， and henceル1きis finitely generated． Conversely， suppose Mξis finitely generated， that is， there is an exact sequence S§→Mξ→0，then O→RM＊‖→REn is exact． SinceσM：RM→Rハ4＊‖is monomorphism，
there is a monomorphism RM→REn． Hence RM is丘nitely cogenerated．
（2）Since an artinian left R−module is finitely cogenerated， it is E−reflexive by Lemma 2’．So RM is artinian if and only if RM＊‖is artinian． Then it fbllows from Theorem 3’ that RルI is artinian if and only if Mξis noetherian． □SOME NOTES ON REFLEXIVE MODULES In the fbllowing we give the corresponding result with respect to an 5−module． Theorem 4’SupP・se RR and Ss are E−reflentve Es is卿；ective andfinitely c・generated． Then／brα吻ん診5−mo醐e」VS，諺んeμoω仇98￡α￡emeπ君8んold．
（1）∧伝飴伽Wc・genemted if and・nly if RNU i3輌ε吻geπe耐ed．
（2）砺E乞8卿iective， then∧ls is’artinian if and・nly if RNti is n・etんertαn．3．MAIN REsuLTs
In this section， we study chain conditions under the E−duality and give our main results． Proposition 5 Assume RE is injective． if Es is artiniαn（resp． noetんetiaη，），君んen R τ3α雄πoe舌んerian（resp．αrtinian）痂9． Proof． Let 11⊂12 be two different left ideals of R． Then rE（∫1）⊃rE（12）， where rEO is the right annihilator． Since 11≠12， there is a non−zero element x in I2 but not in∫1．Because RE is an injective cogenerator， there is an R−homomorphism∫：R／11→RE
such that∫＠十∫1）≠0． So xf（1十11）＝f（x十Il）≠Oand hence∫（1十11）￠rE（12）． But it is clear that∫（1十∫1）∈rE（11）． Soγ’E（11）≠rE（12）． SupPose R is not a left noetherian（resp． artinian）ring， then there is an ascending（resp． descending）chain of left ideals of R which does not terminate． From the above argument it fbllows that there is a descending（resp． ascending）chain of submodules of E8 which does not terminate， which is a contradication since E8 is artinian（resp． noetherian）．□ Proposition 5’As8ume Es is injective． if RE isαrtiniαn（resp． noetんerian）， then 5 乞3α吻ん彦noetherian（πe8ρ．αrtinian）吻9． Corollary 6 SUpρ08e RR and 58 are E−reflexive． （1）lfRE i8 injective and Es ts物iective and finitely cogenerated， tんen RE is n・etんeπαη 勾Fand onl3ノザ8isαrigん￠αrtinian ring． （2） if RE is injectiveαnd finitely cogenerated and E8 i8 injective， then RE isαrtinianザαπdo叫ザSi8α吻んt・noetんemπ噺9．
Proof．（1）It fbllows from Theorem 3 and Proposition 5． （2）It follows from Theorem 4 and Proposition 5’．□ Abimodule RNs is said to have an E−coresolution if there is an exact sequence O→R／Vs−→Eo→E1→…→Ei→＿such that each Ei is a direct sums of finitely many
copies of REs．Theorem 7 Supρose RR and Ssαre E−neflexive and REαnd Es are卿iective and
finiteZy C・genertzted． Then／b閲雄R−module RM，τんeμ・ωing statements are equivαlent． （1）RM is n・etheWiαn． （2） HomR（RM， RNs）’is anαrtinian吻んt S−module for any bimodule R／Vsωんicんんα8 an E−core801％tion． （3） Ext6（Rル1， R∧ls）i8αnαrtiniαn right S−module ichんα3αn E＿COT℃301ution，for any integer n≧0αndαny
bimo掘e RNsωん
Pπ）（’f，（2）⇒（1）It is clear that REs has an E−coresolution． So our claim fbllows from Theorem 3． （1）⇒（2）Suppose RNs is a bimodule which has an E−coresolution． Then RNs can be embedded into REき for some positive integer k．So HomR（RM， R／V8）⊂HomR（RM， RE§）2㊥ltl HomR（RM， REs）＝㊥㌧1 M＊． If RM is noetherian， then Mg is artinian by Theorem 3．So㊥峯l M＊and HomR（RM， RNs）are artinian right S−modules． （3）⇒（2）It is trivial． （2）⇒（3）There is nothing to do fbr the case n＝0． Now suppose n≧1and RNs is a bimodule which has an E−coresolution as fbllows： 0→RNs→Eo−d！2， Ei」5…→」Ei一曳．．．
The exact sequence O→RNs→Eo→Im〔4）→Oinduces an epimorphism HomR（RM，
Imdo）→Extk（RM， RNs）→Oand isomorphisms Ext監＋1（RM， R／Vs）2Ext隻（RM， Imdo）
for any「n≧1． Then HomR（RM， Imalo）is an artinian right S−module by（2）． So Extk（R．M， R／Vs）is also an artinian right S−module． Since lmdo has also an E−coresolution， Ext｝t（Rル1， Imdo）is an artinian right S−module from the above argument． Hence Ext三（Rル1， RIVs）is an artinian right S−module． We have the desired result by repeating this procedure．口 Theorem 7’Sμppose lf R Rαnd SsαTe E−reflexive and R Eαnd Es are iηj’eCtive and finitely C・generated．τWeπ∫brα嫌R−m（ntule RM， tんe∫foll・wing statements are equivalent． （1） RM友5 artinian． （2） HomR（RM， R／Vs）isα noetherian right S−module∫for any bimodute RNsωhich hαs απE−COηe801μtion． （3）bim
Ext2（R
αdule R∧lsωんM，Rハls）ts a n・etんerian吻んt s−m・醐e∫brα仰煽切erη≧O
icんんα8αn E＿COηe301Ution．αndαny
4．APPLIcATIoNs
If」R is a commutative semilocal ring and J is the Jacobson radical of R， then E（R／J）， the injective envelope of R／」， is a finitely cogenerated injective cogenerator． ルlv ＝＝ HomR（M，E（」R／」））is called the G−Matlis dual module of M． If the homomorphismτM： M→Mvv＝HomR（HomR（M， E（R／J）），E（R／」））， defined by 7M（x）（∫）＝∫（x）fbr any x∈Mand∫∈ルlv， is an isomorphism， then〃is said to be G−Matlis reHexive l2】． In particular， if R is a commutive semilocal noetherian ring， it fbllows from l2， Proposition 31and【4， Lemma 2】that R is complete、if and only if R is G−Matlis reflexive． In this case， S＝End（RE（R／J））＝R， particularlyτM coincides withσM． The notion of G−Matlis reflexive module is described as a special case of the E−reHexive module discussed f（）r non−commutative ring in section 3． In this section we prove the fbllowing results in l21 and l4］by applying our theorems． With these facts and by applying Theorem 3， we have the fbllowing results． Corollary 8（【4， Theorem ID 1アRCαα）Mmutαtive complete semitocα1 ring， then ノ’or an R−module・Mωe・hαve（1）M28疏W geηe耐ed（re8P． finitely c・generateのif and・nly if MV i3輌εely
C・generated（resp．輌蜘geηe耐eの．
（2）Mis n・eth・riαn（resρ． artinian）if and・nly if MV is artinian（re・P． n・eth・παη）． Corollary 9（【2， Proposition 41）∬Risαcommutαtive complete semilocαt noethe一 磁η噺9，tんen・an・R−m・dule・M・is・n・e￡んerian（resp． artiniαn）ifαnd・nly if MV is artiniαn （resp． n・etんerian）． If R is commutative complete semilocal noetherian ring， then R is G−Matlis re且exive （12，Proposition 31）． So by applying Corollary 6， we have the fbllowing conclusion．SOME NOTES ON REFLEXIVE MODULES
