PII. S0161171202109203 http://ijmms.hindawi.com © Hindawi Publishing Corp.
ON MATLIS DUALIZING MODULES
EDGAR E. ENOCHS, J. A. LÓPEZ-RAMOS, and B. TORRECILLAS
Received 30 September 2001
We consider rings admitting a Matlis dualizing moduleE. We argue that ifRadmits two such dualizing modules, then a module is reflexive with respect to one if and only if it is reflexive with respect to the other. Using this fact we argue that the number (whether finite or infinite) of distinct dualizing modules equals the number of distinct invertible (R,R)-bimodules. We show by example that this number can be greater than one.
2000 Mathematics Subject Classification: 16D20, 16D90.
1. Notation and preliminaries. Throughout this paper,Ris always an associative ring,RM(resp.,MR) means thatMis a left (right)R-module, andRMRmeans thatMis a left-right (R,R)-bimodule. We denote byE(M)the injective envelope of the (left or right)R-moduleM.
We recall from [4] the following definition.
Definition 1.1. A ring R has a Matlis dualizing module if there is an (R,R)-bimoduleE such thatRE andER are both injective cogenerators, and such that the canonical mapsR→HomR(RE,RE)andR→HomR(ER,RR)are both bijections.
If suchEexists, it will be called a Matlis dualizing module forR.
We note that, each of the maps
R →HomRRE,RE, R →HomRER,RR (1.1)
is a morphism of(R,R)-bimodules. IfMis a left (right) module, when there is such a Matlis dualizing moduleE, we denote byM∨the right (left)R-module HomR(M,E). If the canonical mapM→M∨∨is an isomorphism,MisE-reflexive. IfRis a ring having a Matlis dualizing module, then it is clear thatR andE are examples of E-reflexive modules (both of them as left and rightR-modules).
Note that, the duality given by such a Matlis dualizing module is what in [8] and its reference is called a Morita self-duality.
Several examples of Matlis dualizing modules are given in [4]. With the following one we give a partial answer to [8, Question 4.16].
Example1.2. LetRbe a left and right Noetherian ring having a Matlis dualizing moduleE. Then, by [5, Theorem 1],E[x−1]is an injective cogenerator as left and right R[[x]]-module. Furthermore, by using [7, Theorem 4.1],
HomR[[x]]Ex−1,Ex−1HomR(E,E)[x]R[x] (1.2)
Lemma1.3. IfRhas a Matlis dualizing moduleEand0→M→M→M→0is an
exact sequence of left or rightR-modules, thenMis E-reflexive if and only if M and MareE-reflexive.
Proof. The lemma follows easily by applying the snake lemma to the commutative
diagram
0 M M M 0
0 M∨∨ M∨∨ M∨∨ 0.
(1.3)
IfRhas a Matlis dualizing moduleE, then recall from [1] that a left (right)R-module Mis finitely cogenerated byEor simply finitely cogenerated ifMmay be embedded in a finite direct sum of copies ofE. An immediate consequence of the foregoing lemma is that every finitely generated and every finitely cogenerated left or rightR-module isE-reflexive.
Osofsky in [6] showed that, ifRES defines a Morita duality, then no infinite direct sum of nonzero modules is reflexive. The following lemma is a particular case of this result. We include here a proof for completeness.
Lemma1.4. LetE be a Matlis dualizing module forRand(Mi)i∈Ia family of left
(right)R-modules. Then⊕i∈IMiisE-reflexive if and only if eachMiisE-reflexive and Mi=0except for a finite number ofi.
Proof. IfMi=0 except for a finite number ofi, and eachMiisE-reflexive, then
HomRHomR⊕i∈IMi,E,E ⊕i∈IHomRHomRMi,E,E ⊕i∈IMi. (1.4)
Conversely, we show first that no infinite direct sum of nonzero modules isE-reflexive. So consider⊕i∈IMiwithMi≠0 for alli∈I, and the canonical morphism
⊕i∈IMi →HomRHomR⊕i∈IMi,E,EHomR
i∈I
HomRMi,E,E
, (1.5)
defined by
xii∈I →f∈HomR
i∈I
HomRMi,E,E
, (1.6)
wheref ((Φi)i∈I)=i∈IΦi(xi). The image of(xi)i∈Iis zero in⊕i∈IHomR(Mi,E)which is contained in i∈IHomR(Mi,E)if and only ifxi=0 for alli∈I, that is,(xi)i∈I= 0. Now since ⊕i∈IHomR(Mi,E)≠ i∈IHomR(Mi,E) and E is a cogenerator, HomR ( i∈IHomR(Mi,E)/⊕i∈IHomR(Mi,E),E)≠ 0, which means that there is a nonzero f: i∈IHomR(Mi,E)→E and such thatf|⊕i∈IHomR(Mi,E)=0. But from the foregoing, fcannot be the image of(xi)i∈I∈ ⊕i∈IMi, and so⊕i∈IMiis notE-reflexive.
The following result characterizesE-reflexive modules in terms of some submodule. This result is shown by using Enochs’s argument in [3, Proposition 1.3] to characterize Matlis reflexive modules over a complete local Noetherian ring.
Proposition1.5. LetRbe a ring having a Matlis dualizing moduleE. Then a left
(right)R-moduleMisE-reflexive if and only if it contains a finitely generated submodule
N, such thatM/Nis finitely cogenerated.
Proof. If M contains such a submodule N, then M is E-reflexive by using
Lemma 1.3.
Conversely, ifM=0 there is nothing to prove. IfM≠0, then it contains a finitely generated submodule N1 such that the socle of M/N1, Soc(M/N1), is not zero. If Soc(M/N1) is not essential inM/S1, let L/N1∩Soc(M/N1)=0 with N1⊂ L, then there is a finitely generated R-module N2 such that N1 ⊂N2 ⊂ L, and such that Soc(L/N2)≠0. Then, Soc(M/N1)→Soc(M/N2) is injective but not surjective. Re-peating the process we observe that we must stop, since otherwise ifT= ∪Nnthen Soc(M/T )is an infinite direct sum, which is not possible sinceM/T isE-reflexive and then Soc(M/T )would also beE-reflexive, in contradiction withLemma 1.4. The result follows from [1, Proposition 10.7].
Remark1.6. Proposition 1.5shows that the reflexive property of a module is
some-thing that depends only on the internal structure of such a module.
As a result of the preceding remark, we get the following immediate but important consequence.
Corollary1.7. LetRbe a ring admitting several Matlis dualizing modules. Suppose
thatE1andE2are two of these modules and letMbe a left (right)R-module. ThenM
isE1-reflexive if and only if it isE2-reflexive.
Proposition1.8. Let(R,ᏹ)be a local commutative ring having a Matlis dualizing
moduleE(not necessarilyE(R/ᏹ)). ThenRis complete.
Proof. We show that the dual of any Artinian left (right)R-moduleMis complete.
IfMis Artinian, then it is embedded in a finite direct product of copies ofE. Then, we have an epimorphism
HomR(E,E)n →HomR(M,E) (1.7)
for some natural numbern, that is,M∨if finitely generated. ButM∨has also a struc-ture as ˆR-module sinceEhas such a structure. Then it follows immediately thatM∨is finitely generated as ˆR-module and so it is complete. The result follows now from the fact that there is a correspondence between submodules ofMand quotient modules ofM∨ and therefore, ifMis Noetherian, thenM∨is Artinian and this is the case for RandR∨E.
EE(R/ᏹ), both as a left and as a right module, but in generalEis not isomorphic toE(R/ᏹ)as an(R,R)-bimodule.
2. Matlis dualizing modules. InSection 1we mentioned the possibility that a ring admits different and not necessarily isomorphic Matlis dualizing modules. We start this section by giving an example of a ring having more than one Matlis dualizing module.
Example2.1. LetAbe a finite-dimensional algebra over a fieldk. If we consider AM, we have the dual M∗=Homk(M,k). If dimk(M) <∞then M is reflexive with respect to this duality (cf. [2]). From this it may be deduced thatAEA=Homk(A,k)is a Matlis dualizing module forA.
Now, letσ :A→Abe an automorphism and letAE
Abe such thatAE=AE, but if x∈Eanda∈A, we definexa=xσ (a)(inE). Iff:AEA→AEAis an isomorphism (as bimodules) thenf:AE→AEis an isomorphism, and we know that forf (x)=xcfor somec∈A. Sincefis an isomorphism then such acis invertible. Butf (xa)=f (x)a which means that(xa)c=xcσ (a)for everyx∈E. This gives us thatσ (a)=c−1ac for alla∈A. Now ifσ is not of this form, we get thatEandEare not isomorphic as (A,A)-bimodules.
In any case,A→HomA(AE,AE)=HomA(AE,AE)Ais an isomorphism andA→ HomA(EA,EA)is a monomorphism. But, sinceEAis an injective cogenerator,EA is also an injective cogenerator and if Soc(E
A)=Soc(EA), we get thatEAEA. Then
dimkHomAE A,EA
=dimkHomAEA,EA=dimk(A), (2.1)
soA→HomA(EA,EA)is also an isomorphism and thereforeAEAis a Matlis dualizing module forA.
Now, ifp is a prime,k=Z/(p),Gis a commutative group such that|G| =pn for some positive integern, and we consider group algebraA=kG,Ais local, so there is only one simple module overA. Then Soc(E
A)=Soc(EA)is simple sinceAis commu-tative and we only have to find an automorphism ofAwhich is not a conjugation by any unit ofA.
So takep=2,k=Z/(2)and let the groupG=H×HwhereHis the multiplicative group with two elementsH= {1,h}. Takeσ:kG→kGinduced by the automorphism inGgiven by
h1,h2
→h2,h1
. (2.2)
If{(1,1),(1,h),(h,1),(h,h)}is a basis forkG, we have that, for any element inkG (in matrix notation),σ ((x,y,z,t))=(x,z,y,t), and sincekGis commutative, we get thatσ cannot be a conjugation for some unit ofA.
Theorem2.2. LetRbe a ring having Matlis dualizing modules, and letE1andE2
be two of such modules. Denote byP andQthe(R,R)-bimodulesHomR(E1R,E2R)and HomR(E2R,E1R). Then
(i) P⊗RQRandQ⊗RPR;
(iii) PandQare finitely generated projective (as left and rightR-modules);
(iv) PHomR(QR,RR)andQHomR(PR,RR).
Proof. Denote by HomR(Ei,Ej)the(R,R)-bimodule HomR(EiR,EjR).
(i) What we want to show is that the morphisms
HomRE1,E2
⊗RHomRE2,E1 f
→HomRE2,E2
R,
HomRE2,E1⊗RHomRE1,E2 g
→HomRE1,E1R,
(2.3)
defined byf (σ⊗τ)=σ τ and g(τ⊗σ )=τσ for every σ ∈P and everyτ ∈Q, respectively, are isomorphisms.
Since E1 is an injective cogenerator, f is an isomorphism if and only if when HomR(−,E1)is applied, the resultant morphism
HomRR,E1
E1 →HomRHomRE1,E2
⊗RHomRE2,E1
,E1
(2.4)
is also an isomorphism. But sinceE1is E1-reflexive, byCorollary 1.7, E1 is alsoE2 -reflexive and vice versa, and therefore we get that
HomRHomRE1,E2⊗RHomRE2,E1,E1 HomRHomRE1,E2
,HomRHomRE2,E1
,E1
HomRHomRE1,E2
,E2
E1
(2.5)
and it follows thatf is an isomorphism. Forg, the reasoning is analogous. (ii) Sincegis an isomorphism, in particular,gis an epimorphism, and so
1=
i∈I
τiσi (2.6)
withσi∈P andτi∈Qfor alli. Now lethi:P→Rdefined byhi(σ )=τiσ for each i. Then, we haveh:P(I)→Rand by (2.6) it follows thatPis a generator. Analogous reasoning is used forQ.
(iii) Let h : P →R(I) HomR(E
1,E1)(I) and let h :R(I) →P defined by h(σ )= (σ τi)i∈I and h((ri)i∈I)=i∈Iriσi. It is clear now that (hh)(σ )=h((σ τi)i∈I)=
i∈I(σ τi)σi=σi∈Iσiτi=σ using (2.6). It follows thatPR(I), and soPis finitely generated projective. Similarly, we prove thatQis finitely generated projective.
(iv) Leth:P→HomR(QR,RR)defined byh(σ )(τ)=σ τ for allσ ∈P andτ∈Q. If h(σ )=0 then we get that σ τ =0 for allτ ∈Q. But σ =σ1=σ (i∈Iτiσi)=
i∈Iσ (τiσi)=i∈I(σ τi)σi=0, sohis injective. Let nowϕ∈HomR(Q,R). Then for allτ∈Q,
ϕ(τ)=ϕ(1τ)=ϕ
i∈I τiσi τ =ϕ
i∈I
τiσiτ
=ϕ
i∈I τiσiτ
=
i∈I
ϕτiσiτ
=
i∈I
ϕτiσiτ=h(σ )(τ),
whereσ =i∈I(ϕ(τi)σi). It follows that PHomR(Q,R). Similarly, we can prove thatQHomR(PR,RR).
Now by analogy with the commutative case, we give the following definition.
Definition2.3. LetRbe a ring and letPbe an(R,R)-bimodule. We will say thatPis invertible if it is left and right finitely generated projective and ifP⊗RHomR(P,R)R and HomR(P,R)⊗RPR.
Theorem2.4. LetRbe a ring admitting Matlis duality. Then there exists a bijective
correspondence between Matlis dualizing modules and invertible modules.
Proof. LetMbe a left (right)R-module and denoteM∗=HomR(M,R). Note first
that, ifF is a finitely generated free module, then the natural morphism
F∗⊗RN →f HomR(F,N) (2.8)
defined byf (g⊗n)=h∈HomR(F,N)whereh(x)=g(x)nis clearly an isomorphism. It follows then thatP∗⊗RNHomR(P,N)forPfinitely generated projective.
We show now that ifP is invertible andEis Matlis dualizing, thenP∗⊗E is also Matlis dualizing. First, since P is finitely generated projective, so isP∗. Therefore P∗⊗RE is a direct summand ofRn⊗
REEn for some positive integernwhich is injective, and soP∗⊗REis also injective.
Now ifNis any (left or right)R-module,
HomRN,P∗⊗REHomRN,HomR(P,E)HomRN⊗RP,E (2.9)
which is not zero sinceEis a cogenerator, this shows thatP∗⊗REis also a cogenerator. Moreover,
HomRP∗⊗RE,P∗⊗REHomRE,HomRP∗,P∗⊗E
HomRE,HomRP∗,HomR(P,E) HomRE,HomRP∗⊗P,E
HomRE,HomR(R,E)HomR(E,E)R
(2.10)
which shows thatP∗⊗REis a Matlis dualizing module.
On the other hand, if ¯Eis a Matlis dualizing module, then byTheorem 2.2we have that HomR(E,E)¯ is invertible.
If we denote byᏰandᏵ, respectively the classes of Matlis dualizing and invertible modules, the correspondence may be then represented by the following diagram:
Ᏸ
HomR(−,E) Ᏽ (−)∗⊗RE
If ¯Eis Matlis dualizing, then (
−)∗⊗
RE◦Hom(−,E)E¯=HomRHomRE,E¯ ,R⊗RE
HomRHomRE,E¯ ,EE¯ (2.12)
since HomR(E,E)¯ is finitely generated projective and ¯EisE-reflexive byCorollary 1.7. Now, ifPis invertible,
HomR(−,E)◦(−)∗⊗RE(P)=HomRP∗⊗RE,EHomRHomR(P,E),EP
(2.13)
sincePis finitely generated projective and also isE-reflexive.
Remark2.5. Note that all elements inᏵare finitely generated projective, and so, by the proof ofTheorem 2.4(−)∗⊗RE=HomR(−,E), so in other words, ifRhas a Matlis dualizing moduleE, the theorem shows that, in this case, there is a duality between Matlis dualizing modules and invertible modules, which determines the Picard group ofR.
Acknowledgments. This paper was written while the second author was visiting
University of Lexington, Kentucky, and was supported by Ministerio de Educacion y Cultura grant EX 00 34845300. Second and third authors are partially supported by DGES grant PB98-1005 and Junta de Andalucía FQM 0211.
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Edgar E. Enochs: Department of Mathematics, University of Kentucky, Lexington,
KY40506, USA
E-mail address:[email protected]
J. A. López-Ramos: Departamento de Algebra y Análisis Matemático, Universidad de
Almería,04120Almería, Spain
E-mail address:[email protected]
B. Torrecillas: Departamento de Algebra y Análisis Matemático, Universidad de
Almería,04120Almería, Spain
