Prime Submodules and Local Gabriel Correspondence in σ[ M ]

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Prime Submodules and Local Gabriel Correspondencein σ[M]Jaime Castro Pérez a & José Ríos Montes ba Departamento de Matemáticas, Instituto Tecnológico y de Estudios Superiores deMonterrey, Tlalpan, Méxicob Instituto de Matemáticas, Universidad Nacional Autónoma de México, Area de laInvestigación Científica, Circuito Exterior, C.U., MéxicoVersion of record first published: 17 Jan 2012.

To cite this article: Jaime Castro Pérez & José Ríos Montes (2012): Prime Submodules and Local Gabriel Correspondence inσ[M], Communications in Algebra, 40:1, 213-232

To link to this article: http://dx.doi.org/10.1080/00927872.2010.529095

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Communications in Algebra®, 40: 213–232, 2012Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927872.2010.529095

PRIME SUBMODULES AND LOCAL GABRIELCORRESPONDENCE IN ��M�

Jaime Castro Pérez1 and José Ríos Montes21Departamento de Matemáticas, Instituto Tecnológico y de EstudiosSuperiores de Monterrey, Tlalpan, México2Instituto de Matemáticas, Universidad Nacional Autónoma de México,Area de la Investigación Científica, Circuito Exterior, C.U., México

We consider the concept of prime submodule defined by Raggi et al. [7]. Wefind equivalent conditions for a module M progenerator in ��M�, with �M -Gabrieldimension, to have a one-to-one correspondence between the set of isomorphism classesof indecomposable �-torsion free injective modules in ��M� and the set of �-puresubmodules prime in M , where � is a hereditary torsion theory in ��M�. Also we givea relation between the concept of prime M-ideal given by Beachy and the conceptof prime submodule in M . We obtain that if M is progenerator in ��M�, then theseconcepts are equivalent.

Key Words: Gabriel correspondence; Preradicals; Prime submodules; Torsion theory.

2000 Mathematics Subject Classification: Primary 16S90; Secondary 16D50, 16P50, 16P70.

INTRODUCTION

A classical result due to Matlis [12] says that, if R is a commutative Noetherianring, then there is a one to one correspondence between the indecomposableinjective R-modules and the prime ideals of R. Gabriel [13] showed that thiscorrespondence remains true for any left Noetherian ring R which satisfies thecondition known as condition H . In Krause [14], it is showed that left fully boundedNoetherian rings are characterized as those having bijective Gabriel correspondence.

Recently, some results have appeared characterizing rings with bijectiveGabriel correspondence which are not neccesarily Noetherian nor commutative. Let� be a hereditary torsion theory in R-Mod. Albu et al. [15] obtained conditions for Rwith �-Krull dimension equivalent to having local Gabriel correspondence relativeto �.

Let M be an R-module, and let ��M� denote the full category whoseobjects are R-modules isomorphic to submodules of M-generated modules. Beachy

Received January 18, 2010; Revised September 14, 2010. Communicated by T. Albu.Dedicated to the memory of Lucía Pérez Maya (1938–2007) and BerthaMontes Flores (1927–2010).Address correspondence to Prof. Jaime Castro Pérez, Departamento de Matemáticas, Instituto

Tecnológico y de Estudios Superiores de Monterrey, Calle del Puente 222, Tlalpan, 14380 México,D.F. México; E-mail: [email protected]

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214 CASTRO PÉREZ AND RÍOS MONTES

in Beachy [1] defines the concept of prime M-ideal and extends the Gabrielcorrespondence to M-injective modules by showing that if M is a left Noetherianmodule satisfying condition H and such that Hom�M�X� �= 0 for all modulesX ∈ ��M�, then there is a bijective correspondence between the isomorphism classesof indecomposable M-injective modules in ��M� and prime M-ideals.

For � ∈ M-tors, we consider a filtration �j in gen���, which is called the �M -Gabriel filtration of � ∈ ��M�. M has �M -Gabriel dimension if there exists an ordinali such that M ∈ �i

.In this article, using the concept of prime submodule defined in Raggi et al. [7,

Definition 13], we find necessary and sufficient conditions for a module M projectivein ��M�, generator of the category ��M� and with �M -Gabriel dimension, to havea one-to-one correspondence between set of representatives of isomorphism classesof indecomposable �-torsion free injective modules in ��M� and the set of �-puresubmodules prime in M . In order to do this, we organized the article in foursections. Section 1 is devoted to defining and studying the concept of primenessin M . Sections 2 and 3 are devoted to defining and studying the concepts of �M -module, �M -ass, and �M -dimension in ��M�. With these tools in hand, in Section 4we give necessary and sufficient conditions for a module M , projective in ��M�,generator of the category ��M�, and with �M -Gabriel dimension, to have localGabriel correspondence with respect to �, where � ∈ M-tors. Examples are given toillustrate the theory.

Let R be an associative ring with unity and R-Mod be the category of unitaryleft R-modules. Let M-tors be the frame of all hereditary torsion theories in ��M�.For a family �M� of left R-modules in ��M�, let ���M�� be the greatest elementof M-tors for which all the M� are torsion free, and let ��M�� denote the leastelement of M-tors for which all the M� are torsion. ���M�� is called the torsiontheory cogenerated by the family �M�, and ��M�� is the torsion theory generatedby the family �M�. In particular, the maximal element of M-tors is denoted by �

and the minimal element of M-tors is denoted by . If � is an element of M-tors,gen��� denotes the interval ��� ��.

Let � ∈ M-tors. By ������ t�, we denote the torsion class, the torsion freeclass, and the torsion functor associated to �, respectively. For N ∈ ��M�, N iscalled �-cocritical if N ∈ �� and for all 0 �= N ′ ⊆ N , we have that N/N ′ ∈ ��. Wesay that N is cocritical if N is �-cocritical for some � ∈ M-tors. A torsion theory� ∈ M-tors is irreducible if, for �′� �′′ ∈ M-tors with �′ ∧ �′′ = �, we have that �′ =� or �′′ = �. For N ∈ ��M�, let N denote the injective hull of N in ��M�. If N isa fully invariant submodule of M , we write N ⊆FI M . For all other concepts andterminology concerning torsion theories in ��M�, see Wisbauer [10, 11]. For torsiontheoretic dimensions, the reader is referred to Golan [6], Castro et al. [3–5].

1. PRIME SUBMODULES

The following definition was given in Bican et al. [2].

Definition 1.1. Let M ∈ R-Mod and K, L be two submodules of M . Put KML =∑�f�K� � f ∈ Hom�M�L�.

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 215

Notice that, if K and L are fully invariant submodules of M , then this productagrees with the product defined in Raggi et al. [7, Definition 11].

On the other hand, we know that, given a submodule N of M , there exists asubmodule N ⊂ M such that N is the least fully invariant submodule of M whichcontains N . In fact, N = ∑

�f�N� � f ∈ Hom�M�M�. Notice that N = NMM .Also notice that if K and L are submodules of M , then∑

�f�K� � f ∈ Hom�M�L� = ∑�f�K� � f ∈ Hom�M�L��

Therefore, KML = KML. Besides, it is not difficult to prove that, if K is asubmodule of M , then KM�_� is a preradical over the ring R.

Remark 1.2.

i) Let M ∈ R-Mod and K , L be submodules of M ; then KML ⊂ K ∩ L;ii) Let I , J be left ideals of R; then the product IRJ as in Definition 1.1 coincides

with the usual product of ideals.

Proposition 1.3. Let M ∈ R-Mod and K, K′ be submodules of M . Then:

1) If K ⊂ K′, then KMX ⊂ K′MX for every X ∈ R-Mod;

2) If X ∈ R-Mod and Y ⊆ X, then KMY ⊆ KMX;3) MMX = trM�X�, for every X ∈ R-Mod;4) 0MX = 0, for every X ∈ R-Mod;5) KMX = 0, if and only if f�K� = 0 for all f ∈ Hom�M�X�;6) If X� Y are submodules for any module N ∈ R-Mod, then KMX + KMY ⊆

KM�X+ Y�;7) If �Kii∈I is a family of submodules of M , then �

∑i∈I Ki�MN = ∑

i∈I KiMN ;8) If �Xii∈I is a family of R-modules, then KM�

⊕i∈I Xi� =

⊕i∈I �KMXi�.

Proof. 1) As K ⊂ K′ and as KMX = ∑�f�K� � f ∈ Hom�M�X�, then for every f ∈

Hom�M�X�, we have that f�K� ⊆ f�K′�. Therefore, KMX ⊆ K′MX.

2) As Y ⊆ X, then for f ∈ Hom�M� Y� we have that f ∈ Hom�M�X�;therefore, KMY ⊆ KMX.

3) MMX = ∑�f�M� � f ∈ Hom�M�X� = trM�X�.

4) This is clear.

5) KMX = ∑�f�K� � f ∈ Hom�M�X� = 0 ⇐⇒ f�K� = 0 for every f ∈

Hom�M�X�.

6) Let X and Y be submodules of N ; then KMX ⊆ KM�X + Y� and KMY ⊆KM�X + Y�. Therefore, KMX + KMY ⊆ KM�X + Y�.

7) �∑

i∈I Ki�MN = ∑�f�

∑i∈I Ki� � f ∈ Hom�M�N� = ∑

�∑

i∈I f�Ki� � f ∈Hom�M�N� = ∑

i∈I �∑

f�Ki� � f ∈ Hom�M�N� = ∑i∈I KiMN .

8) As KM�_� is a preradical, then we get the result.

Notice that in (6) the equality is not true in general. Consider the followingexample.

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216 CASTRO PÉREZ AND RÍOS MONTES

Example 1.4. Let R = � and p ∈ R be a prime number. If M = �, K = �, X =��p� = � a

b∈ � �p � b, Y = � a

pn∈ � � n ∈ �, then X + Y = �, KMX = �M��p� = 0,

KMY = �MY = 0. On the other hand, KM�X + Y� = KM� = �M� = �. Therefore,KM�X + Y� �⊆ KMX + KMY .

Let M ∈ R-Mod. In Beachy [1, Definition 1.1] is defined the annihilator in Mof a class � of R-modules as AnnM

��� = ∩K∈�K, where

� = �K ⊆ M � there exists W ∈ � and f ∈ Hom�M�W� with K = ker f�

If the class � consists of a single module X, we use the notationAnnM

�X� and AnnM�X� = ∩�ker f � f ∈ Hom�M�X�. Also notice that if N ⊆ M , then

AnnM�M/N�⊆N .Note that if � and �′ are subclasses of R-Mod, such that �′ ⊆ �, then

AnnM��� ⊆ AnnM

��′�. In particular, if X ∈ R-Mod and Y ⊆ X, then AnnM�X� ⊆

AnnM�Y�. Moreover, AnnM

�X� = M if and only if Hom�M�X� = 0. In fact, ifAnnM

�X� = M , then MMX = 0. So by 1.3 (5), we have that f�M� = 0 for all f ∈Hom�M�X�. Therefore, Hom�M�X� = 0.

In Beachy [1, Definition 1.2] a submodule N of M is called an M-ideal ifthere is a class � of modules in ��M� such that N = AnnM

���. Also in Beachy [1,Definition 1.5] a product is defined in the following way. Let N be a submodule ofM . For each module X ∈ R-Mod, N · X = AnnX

���, where � is the class of modulesW , such that f�N� = 0 for all f ∈ Hom�M�W�.

So we have that NMX ⊆ N · X. Notice that the product defined by Beachy isdifferent from the product defined here. In order to see this, consider the followingexample.

Example 1.5. Let R = � and p ∈ R be a prime number. M = �p� , N = �p, andX = �p� . As �p is a fully invariant submodule of �p� , then NMX = �pM�p� =�p. On the other hand, by Beachy [1, Definition 1.9(c)] we know that N · X is thesmallest M-ideal that contains N . Moreover, M does not have nontrivial M-idealssee Beachy [1, Example 2.1]. So the smallest M-ideal containing N is �p� . Therefore,N · X = �p� . Hence we have that NMX �= N · X.

We know that NMX = NMX, where N is the smallest fully invariant submoduleof M which contains N . Therefore, we can suppose without loss of generality thatN is a fully invariant submodule of M .

For a fully invariant submodule N of M , the preradical �MN �_� is defined asfollows: For a K ∈ R-Mod, �MN �K� =

∑�f�N� � f ∈ Hom�M�K�. �MN �_� is defined in

Raggi et al. [16, Definition 4].Following Stenstrom [9, VI Proposition 1.5], we denote by r the smallest

radical larger than r. Using this concept and the product defined by Beachy in [1,Definition 1.5], we have the following proposition.

Proposition 1.6. Let M ∈ R-Mod and N be a fully invariant submodule of M , thenthe following conditions hold:

i) The preradicals �MN �_� and N · �_� have the same pretorsion-free class;

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 217

ii) �MN �_� = N · �_�;iii) �MN �_� is a radical if and only if �MN �_� = N · �_�.

Proof. i) By Proposition 1.2, NMX = 0 if and only if f�N� = 0 for all f ∈Hom�M�X�. So by Beachy [1, Definition 1.5], we have that NMX = 0 if and onlyif N · X = 0. On the other hand, we know that �MN �X� = NMX. Therefore, thepreradicals �MN �_� and N · �_� have the same pretorsion-free class.

ii) Since N · �_� is a radical, then this follows from Stenstrom [9, VIProposition 1.4].

iii) It is clear.

Remark 1.7. For a fully invariant submodule K of M , KM�M/K� = 0 is in generalfalse. In order to see this, we consider M = �p� and K = �p. It is clear that K is afully invariant submodule of M and M

K� M . So we have that KM�

MK� � KM�M� =

K �= 0.

Proposition 1.8. Let M ∈ R-Mod be a quasi-projective module and K a fullyinvariant submodule of M . Then KM�

MK� = 0.

Proof. By Proposition 1.3(5), we have that KM�M/K� = 0 if and only if f�K� =0 for all f ∈ Hom�M�M/K�. Let f ∈ Hom�M�M/K�. As M is a quasi-projectivemodule, then there exists f ∈ End�M� such that the following diagram commutes:

So f�K� = ��f ��K� = ��f �K��, where � � M → M/K is the natural projection.Since K is a fully invariant submodule of M , then ��f �K�� ⊆ ��K� = 0. So we havethat f�K� = 0. Therefore, KM�M/K� = 0.

In the next proposition, we give a description of AnnM��� in terms of the

product−M−.

Proposition 1.9. Let M ∈ R-Mod and � be a class of left R-modules. Then:

AnnM��� = ∑

�N ⊂ M �NMX = 0 for all X ∈ ��

Proof. Let N ⊂ M such that NMX = 0, then by Proposition 1.3, we have thatf�N� = 0 for all f ∈ Hom�M�X�. So by Beachy [1, Definition 1.5] N · X = 0. ThusN ⊆ AnnM

�X� by Beachy [1, Definition 1.6]. So∑�N ⊂ M �NMX = 0 for all X ∈

� ⊆ AnnM���. On the other hand, we know that AnnM

��� ⊂ AnnM�X� for all

X ∈ �. Then by Beachy [1, Definition 1.6] AnnM��� · X = 0 for all X ∈ �. Thus

AnnM���MX = 0 for all X ∈ �. So AnnM

��� ⊆ ∑�N ⊂ M �NMX = 0 for all X ∈ �.

Therefore, AnnM��� = ∑

�N ⊂ M �NMX = 0 for all X ∈ �.

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218 CASTRO PÉREZ AND RÍOS MONTES

Notice that NMX = NMX for all X ∈ R-Mod, where N is the smallest fullyinvariant submodule of M that contains N . Then we have that

AnnM��� = ∑

�N ⊆FI M �NMX = 0 for all X ∈ ��

We give the definition of prime submodule in M wich was given in Raggiet al. [7, Definition 13].

Definition 1.10. Let M ∈ R-Mod and N �= M a fully invariant submodule of M .N is prime in M if for any K, L fully invariant submodules of M , we have that�MK �L� = KML ⊆ N implies that K ⊆ N or L ⊆ N .

We note that, if I ⊆ R is an ideal, then I is prime in R in the sense ofDefinition 1.10 if and only if I is a prime ideal. We also notice that, if N ⊂ M isprime in M and K is a fully invariant submodule of M such that MMK = 0, thenK ⊆ N . In fact, as MMK = 0 ⊂ N and N is prime in M , then M ⊂ N or K ⊂ N . ThusK ⊂ N . In particular, every submodule prime in M contains all the fully invariantsubmodules K of M such that HomR�M�K� = 0.

In Beachy [1, Definition 2.1], the concept of M-prime module is defined asfollows: the R-module X is called M-prime module if Hom�M�X� �= 0 and AnnM

�Y� =AnnM

�X� for all submodules Y ⊆ X such that Hom�M� Y� �= 0. And in Beachy [1,Definition 3.1], Beachy defines the M-ideal P be prime M-ideal, if there exists anM-prime module X such that P = AnnM

�X�. In general, the concept of prime M-ideal is different of the concept of submodule prime in M as in Definition 1.10. Inorder to see this, we consider the �-module M = �p� , where p is a prime number.Since HomR�M�K� = 0 for all K � M , then M is M-prime module by Beachy [1,Definition 2.1]. So by Beachy [1, Definition 3.1], we have that AnnM

�M� = 0 isprime M-ideal. On the other hand, 0 is not prime in M since �p M�p2 = 0 with�p �= 0, �p2 �= 0. Moreover, M = �p� does not have prime submodules in M since�pn M�pm = 0 for all n �= m and �pn �= 0, �pm �= 0.

We know that an ideal P of R is prime ideal if and only if for any I and J leftideals of R such that IJ ⊂ P, then I ⊂ P o J ⊂ P. If M is projective in ��M�, thenthis result is true with the concept of prime in M in the sense of Definition 1.10.

Proposition 1.11. Let M projective in ��M� and let P be a fully invariant submoduleof M , then the following conditions are equivalent:

i) P is prime in M;ii) For any submodules L and K of M , LMK ⊆ P implies that L ⊆ P or K ⊆ P.

Proof. i) ⇒ ii) Let L and K be submodules of M . Inasmuch as LMK = LMK,where L is the minimal fully invariant submodule of M which contains N , we cansuppose without loss of generality that L is a fully invariant submodule of M . Nowif LMK ⊆ P, then �LMK�MM ⊆ PMM = P. Since M is projective in ��M�, then wehave that �_�M�_� = �_� · �_� by Beachy [1, Proposition 5.5a)]. Now by Beachy [1,Proposition 5.6], we have that �L · K� ·M = L · �K ·M�. So �LMK�MM = LM�KMM�,Thus LM�KMM� ⊆ P. As KMM is a fully invariant submodule of M , then either L ⊆P or KMM ⊆ P. Since K ⊆ KMM , then either L ⊆ P or K ⊆ P.

ii) ⇒ i) It is clear.

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 219

In general, it is not true that, if N is a maximal fully invariant submodule ofM , N is a prime in M . In order to see this, we consider the following example.

Example 1.12. Let R = �2 � ��2 ⊕ �2�. The ring R is called the trivial extensionof �2 by �2 ⊕ �2. This ring can be described as R = {(

a �x�y�0 a

) � a ∈ �2� �x� y� ∈ �2 ⊕�2

}. R has only one maximal ideal I = {(

0 �x�y�0 0

) � �x� y� ∈ �2 ⊕ �2

}, and it has three

simple ideals: J1, J2, J3, which are isomorphic, where J1 ={(

0 �0�0�0 0

)�(0 �1�0�0 0

)}, J2 ={(

0 �0�0�0 0

)�(0 �0�1�0 0

)}, and J3 =

{(0 �0�0�0 0

)�(0 �1�1�0 0

)}. Then the lattice of ideals of R has

the form

R is artinian and R-Mod has only one simple module. Let S be the simplemodule. By Raggi et al. [8, Theorem 2.13], we know that there is a lattice anti-isomorphism between the lattice of ideals of R and the lattice of fully invariantsubmodules of E�S�. Thus the lattice of fully invariant submodules of E�S� has treemaximal submodules K, L, and N . Moreover, E�S� contains only one simple moduleS. Therefore, the lattice of fully invariant submodules of E�S� has the form

Let M = E�S�. As K ∩ L = S and KML ⊆ K ∩ L, then KML ⊆ S. On the otherhand, we consider the morphism

f � M�→ M

N� S

i↪→ L�

where � is the natural projection and i is the inclusion. So f�K� = S. Therefore, S ⊆KML. So we have that KML = S. Thus KML ⊆ N but K � N and L � N . Therefore,N is not prime in M . Analogously, we prove that neither K nor L are prime in M .Also we note that KMK = S and SMK = 0. Thus AnnM

�K� = S.We claim that KMS = S. In fact, if � is the natural projection from M onto

M/N , then ��K� = S since MN� S. Thus KMS = S. Analogously, we prove that

LMS = S and NMS = S. Moreover, it is not difficult to prove that AnnM�S� = K ∩

L ∩ N = S and S is not prime in M . Notice that Soc�R� = J1 + J2 + J3 = J1 ⊕ J2and Soc�R� ⊂es R. As S is the only simple module, then Soc�R� � S ⊕ S. So E�S ⊕S� = E�R�. Thus E�R� = E�S�⊕ E�S� = M ⊕M . Therefore, ��M� = ��R� = R-Mod.Moreover, Hom�M�K� �= 0 for all K ∈ ��M�, but M is not a generator of ��M�.

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220 CASTRO PÉREZ AND RÍOS MONTES

From the previous arguments we can draw the following conclusions:

1) M does not have submodules prime in M ;2) S is the only one M-ideal. Moreover, S is the only one prime M-ideal;3) M is the only one indecomposable injective module in ��M�.

The following proposition provides a sufficient condition for a maximal fullyinvariant submodule of M to be prime in M .

Proposition 1.13. Let M ∈ R-Mod. Suppose M generates all its fully invariantsubmodules. If N is a maximal fully invariant submodule of M , then N is prime in M .

Proof. Let L and K be fully invariant submodules of M such that LMK ⊆ N .Suppose L �⊆ N . As N is maximal fully invariant submodule of M , then L+ N =M . So we have that �L+ N�MK = MMK. By Proposition 1.3 point 7), �L+ N�MK =LMK + NMK. Moreover, as K is generated by M , then MMK = K. Thus LMK +NMK = K. On the other hand, we know that LMK ⊆ N and NMK ⊆ N ∩ K ⊆ N .Therefore, K ⊆ N .

In general it is false that, for P prime in M , P is a prime M-ideal. In order tosee this, we consider the following example,

Example 1.14. Let F be a field, and let R be the ring of lower triangular 3× 3matrices over F . The maximal left ideals of R can be described as

N1 =0 0 0F F 0F F F

� N2 =

F 0 0F 0 0F F F

� N3 =

F 0 0F F 0F F 0

�

Therefore, Jacobson radical is J =(

0 0 0F 0 0F F 0

). There are three nonisomorphic simple

left R-modules: S1 = R/N1, S2 = R/N2, and S3 = R/N3. We denote

a =0 0 00 1 00 0 0

� b =

0 0 00 0 00 0 1

�

Therefore, RaJa

=(

0 0 00 F 00 F 0

)(

0 0 00 0 00 F 0

) � S2 and Rb =(

0 0 00 0 00 0 F

)� S3. Let A1 =

(0 0 0F 0 0F 0 0

), and let M be

the module M = A1

⊕RaJa

⊕Rb. Then Soc�M� =

(0 0 00 0 0F 0 0

)⊕RaJa

⊕Rb.

On the other hand, we know that Soc�M� is a fully invariant submodule of M .Moreover, we note that Soc�M� is a maximal submodule of M .

We claim Soc�M� is prime in M and Soc�M� is not a prime M-ideal. In fact,A1

⊕RaJa

is the only fully invariant submodule of M which is not contained inSoc�M�. As

[A1

⊕RaJa

]is a direct summand of M , then

[A1

⊕ Ra

Ja

]M

[A1

⊕ Ra

Ja

]=

[A1

⊕ Ra

Ja

]�⊂ Soc�M��

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 221

Thus Soc�M� is prime in M . Now we will see that Soc�M� is not prime M-ideal.We have that M

Soc�M�� S2, then

RaJa

and MSoc�M�

are isomorphic. So we can define amorphism f � M → M

Soc�M�such that f�Soc�M�� �= 0. Thus Soc�M� �⊂ AnnM

(M

Soc�M�

). So

by Beachy [1, Proposition 1.3(4)], Soc�M� is not M-ideal. Therefore, it cannot be aprime M-ideal.

In the following result, we give a relation between the concept defined inBeachy [1, Definition 3.1] and the concept of being prime in M , as in Definition 1.10.

Proposition 1.15. Let M be a projective module in ��M�. If P ⊂ M is a submoduleprime in M , then P is a prime M-ideal.

Proof. As P is a prime submodule of M , then P is a fully invariant submodule ofM . By Beachy [1, Proposition 5.1], P is an M-ideal. Now let N and K be M-ideals,such that K is M-generated. Suppose N · K ⊆ P. Since M is projective in ��M�, thenby Beachy [1, Proposition 5.5] we have that N · K = NMK. So NMK ⊆ P. Since Nand K are M-ideales, using Beachy [1, Proposition 5.1], we have that N and K arefully invariant submodules of M . As P is a prime submodule of M , then N ⊆ P orK ⊆ P. So by Beachy [1, Theorem 5.7(2)], P is a prime M-ideal.

We claim that if M is a generator of the category ��M� and M is a projectivemodule in ��M�, then the concepts of prime M-ideal and submodule prime of M areequivalent. In order to see this, we will prove first the following lemma.

Lemma 1.16. Let M projective in ��M� and X ∈ ��M� such that AnnM�X� = P � M .

Suppose that for all 0 �= Y ⊆ X, we have that AnnM�Y� = P. Then P is prime in M .

Proof. Let N , K be fully invariant submodules of M , such that NMK ⊆ P. Then byProposition 1.3(1) �NMK�MX ⊆ PMX = 0. By Beachy [1, Proposition 5.5(a)] NMK =N · K, since K ⊆ M and X ∈ ��M�. Moreover, �NMK�MX = �N · K� · X. So we havethat �N · K� · X = 0. Now by Beachy [1, Proposition 5.6], �N · K� · X = N · �K · X� =0. So N · �K · X� = N · �KMX� = 0. As KMX ⊆ X ∈ ��M�, then we have that N ·�KMX� = NM�KMX� = 0. If KMX = 0, then K ⊆ AnnM

�X� = P. If KMX �= 0, then N ⊆AnnM

�KMX� = P, since NM�KMX� = 0. Therefore, P is prime in M .

Proposition 1.17. Suppose M is projective in ��M� and Hom�M�X� �= 0 for all X ∈��M�. If P ⊆ M is a prime M-ideal, then P is prime in M .

Proof. Let P ⊆ M be a prime M-ideal, then by Beachy [1, Proposition 3.2],P = AnnM

�X� for an M-generated M-prime module X. Hence, by Beachy [1,Definition 2.1], P �= M . Since M is generator of the category ��M�, thenHom�M� Y� �= 0 for all 0 �= Y ⊆ X. As X is an M-prime module, we have thatAnnM

�Y� = AnnM�X� = P, see Beachy [1, Definition 2.1]. Therefore, P is prime in M

by Lemma 1.16.

Notice that if M is projective in ��M� and Hom�M�X� �= 0 for all X ∈ ��M�,then M is generator of the category ��M�. Hence M contains prime submodules.In fact, if S ∈ ��M� is a simple module, then Hom�M� S� �= 0. So by Lemma 1.15,AnnM

�S� is prime in M . Example 1.12 shows that there exist modules that do nothave prime submodules.

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222 CASTRO PÉREZ AND RÍOS MONTES

Proposition 1.18. Let M ∈ R-Mod. Suppose M is projective in ��M�. Then �MK �ML =

�MKML for any K and L fully invariant submodule S of M .

Proof. It follows from Beachy [1, Proposition 5.6].

The condition of M being projective in ��M� in the previous proposition isnot superfluous. In order to see this, we consider the ring of the Example 1.12.In that ring, SML = 0. In fact, if f ∈ Hom�M�L�, then ker f is a submodule ofM . As the submodules of M are M , K, L, N , and S, then ker f = S, since L is amaximal submodule of M . So SML = 0. Analogously, we prove SMK = 0, SMN = 0.

Now we claim LMN = S. In fact, if we consider the morphism M�→ M

T� S

i↪→ N ,

where � is the natural projection, then we can consider the morphism f = i � �.So we have that ker f = T . Thus f�L� = S. Hence LMN = S. Therefore, �MK �

ML �N� =

�MK �LMN� = KM�LMN� = KMS = S. On the other hand, we have that �MKML�N� =�KML�MN = SMN = 0. So �MK �

ML �N� �= �MKML�N�. Moreover, we know that M = E�S�

is not projective module in ��M� = R-Mod.

Definition 1.19. Let � ∈ R-pr. � is prime in R-pr if � �= 1 and for any �� � ∈ R-pr�� � � implies that � � � or � � �.

Corollary 1.20. Let M ∈ R-Mod. Suppose M is projective in ��M� and � is prime inR-pr and such that ��M� �= M . Then ��M� is prime in M .

Proof. It follows from Proposition 1.18 and from Raggi et al. [7, Theorem 27].

2. PM-MODULES

Definition 2.1. Let 0 �= N ∈ ��M� and � ∈ M-tors, N is said to be �-�M -module ifN ∈ �� and � ∨ �N� is atom in gen���.

The �-�-modules have been studied in Castro et al. [3, 4]. It is not difficult toprove that the results obtained in these articles about of the �-�-modules are truefor �-�M -modules.

Proposition 2.2. Let � ∈ M-tors with � �= � and N ∈ ��M�.

1) If N is a �-�M -module, then N�X� is a �-�M -module.2) If N is a �-�M -module, then N ′ is �-�M -module, for all 0 �= N ′ ⊆ N . Moreover, we

have that:

i) � ∨ �N� = � ∨ �N ′� ii) ��N ′� = ��N�.

3) If N is a �-�M -module and L is a proper submodule of N which is �-pure, then N/Lis a �-�M -module. Moreover, we have that:

i) � ∨ �N� = � ∨ �N/L� ii) ��N� = ��N/L�.

4) If N is a �-�M -module and � ∈ M-tors, such that � ≥ �, then N ∈ �� or N ∈ ��.

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 223

5) Let � ∈ gen��� and N be a �-�M -module if N ∈ ��, then N is a �-�M -module.6) Let � ∈ M-tors and N be a �-�M -module. If N ∈ ��, then N is �� ∧ ��-�M -module.

Notice that 5) implies that if N is a �-�M -module, then N is ��N�-�M -module.Hence 2� implies that ��N� is an irreducible element in M-tors.

Definition 2.3. A module N ∈ ��M� is called �M -module if there exists � ∈ M-torssuch that N is �-�M -module.

Notice that N is �-�M -module if and only if N is ��N�-�M -module.We denote �M = ����N� ∈ M-tors �N is an �M -module.

Proposition 2.4. Let � ∈ �M , and let N be an �M -module such that M is a �-�M -module. Then ���N� = �.

Proof. It follows from Castro et al. [3, Proposition 1.6].

Remark 2.5. If M is projective in ��M� and H is a submodule of M such thatHMN = 0 for a module N ∈ ��M�, then HM�N/K� = 0 for all submodule K of N . Infact, by Beachy [1, Proposition 5.5(a)], we have that HMN = H · N and HM�N/K� =H · �N/K�. As HMN = 0, then H · N = 0 ⊂ K. Now by Beachy [1, Proposition5.5(b)], H · �N/K� = 0. Thus HM�N/K� = 0.

In the subsequent part of this article, we will suppose that M is projective in��M� and that Hom�M�X� �= 0 for all X ∈ ��M�.

Lemma 2.6. Suppose P ⊆ M is prime in M . Then P contains all fully invariantsubmodules which are not ���M/P�-dense in M .

Proof. Let P prime in M , and let H be a fully invariant submodule of M such thatH is not ���M/P�-dense in M . So M/H � ���M/P�. Hence

Hom�M/H� M/P� �= 0�

Let f � M/H −→ M/P be a nonzero morphism. We claim HM�Im f� = 0. In fact, asH is a fully invariant submodule of M and as M is projective, then by Beachy [1,Proposition 5.1], we have that H is an M-ideal. So by Beachy [1, Proposition 1.3], H ·�M/H� = 0. Now by Beachy [1, Proposition 5.5(a)], HM�M/H� = H · �M/H�. ThusHM�M/H� = 0. Moreover, we know that Im f � �M/H�/�L/H�, where L/H = ker f .As M/H ∈ ��M�, then HM��M/H�/�L/H�� = 0 by Remark 2.5. Hence HM�Im f� = 0.Since P is prime in M , then by Proposition 1.15, P is prime M-ideal. So by Beachy[1, Theorem 5.7(3)], M/P is an M-prime module. On the other hand, we knowHom�M�X� �= 0 for all X ∈ ��M�, then AnnM

�N/P� = AnnM�M/P� for all 0 �= N/P ⊆

M/P since M/P is M-prime module. Moreover, as 0 �= Im f ∩ �M/P� = K/P, thenHM�K/P� ⊂ HM�Im f� = 0. Hence H ⊆ AnnM

�K/P� = P. Therefore, H ⊆ P.

Proposition 2.7. Suppose P ⊆ M is a prime in M . Then M/P is ��M/P�-�M -module.

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224 CASTRO PÉREZ AND RÍOS MONTES

Proof. In order to show that M/P is ��M/P�-�M -module, it is enough to prove that��M/P� ∨ �N/P� = ��M/P� ∨ �M/P� for all 0 �= N/P fully invariant submoduleof M/P. Let 0 �= N/P be a fully invariant submodule of M/P and � = ��M/P� ∨ �N/P�. Let t��M/P� = J/P. Suppose J/P �= M/P. As t� is a radical in ��M�, thenJ/P is a fully invariant submodule of M/P. Moreover, we know that �M/P�/�J/P� ∈��. So M/J ∈ ���M/P�∨ �N/P�. Hence M/J ∈ ���M/P�. Thus J is not ��M/P�-dense inM . We claim that J is a fully invariant submodule of M . In fact, let f � M →M be a morphism, and let � � M → M/P denote the natural projection. Now weconsider the morphism � � f � M → M/P. As P is a fully invariant submodule ofM , then f�P� ⊂ P. Thus �� � f��P� = ��f�P�� ⊂ ��P� = 0. So we can consider themorphism � � f � M/P → M/P such that � � f�x + P� = �� � f��x� = ��f�x��. Onthe other hand, we know that J/P is a fully invariant submodule of M/P. Then� � f� + P� ∈ J/P for all j ∈ J . Hence f��+ P ∈ J/P for all j ∈ J . Thus f�J� ⊂ J .Therefore, J is a fully invariant submodule of M . Now by Lemma 2.6 J ⊂ P. SoJ/P = 0, a contradiction since 0 �= N/P ⊂ t��M/P� = J/P = 0. So M/P is ��M/P�-�M -module.

We denote �M = ���M/P� ∈ M-tors �P is prime in M.

Definition 2.8. Let � ∈ M-tors and N ∈ ��M�. N is called �-�M -module, if N is a�-�M -module and ���N� ∈ �M .

Remark 2.9. For � ∈ R-tors, the �-�-modules have been defined in Castro et al.[3, Definition 1.8]. Definition 2.8 is the same as the given in Castro et al. [3,Definition 1.8]. Moreover, it is not difficult to prove that the results given in Castroet al. [3, Proposition 1.10] for �-�-modules are true for �-�M -modules.

Proposition 2.10. Let � ∈ M-tors and N ∈ ��M� be a �-�M -module. Then thefollowing conditions hold:

1) Every nonzero submodule L of N is a �-�M -module;2) If � ∈ M-tors such that � ≥ � and M ∈ ��, then N is a �-�M -module;3) If L is a proper �-pure submodule of N , then N/L is a �-�M -module;4) If � ∈ M-tors and N ∈ ��, then N is a �� ∧ ��-�M -module.

Definition 2.11. For N ∈ ��M�, N is said to be is �M -module if there exists � ∈ M-tors such that N is �-�M -module.

Remark 2.12. Notice that a module N ∈ ��M� is �M -module if and only if N is���N�-�M -module. In fact, it follows from Castro et al. [3, Proposition 1.12].

Now we give an example of a �M -module.

Example 2.13. Let K be a field, let R be the ring of lower triangular 2× 2 matricesover K. R has two maximal ideals I1 =

(K 0K 0

)and I2 =

(0 0K K

). Therefore, there are

two nonisomorphic simple modules S1 = RI1and S2 = R

I2. Now let M = I1

⊕S2

⊕S1.

If P = Soc�M�, then P = (0 0K 0

)⊕S2

⊕S1. It is clear that P is a maximal submodule

of M and that P is a fully invariant submodule of M . Moreover, N = I1⊕

S2 is theunique fully invariant submodule of M not contained in P. On the other hand, as

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 225

N is a direct sumand of M , we have that NMN = N . Now as N � P, then NMN � P.Thus P is prime in M . Since S2 is isomorphic to M

P, then S2 is ��M/P�-�M -module.

Also notice that R ↪→ M . Hence ��M� = ��R� = R-Mod. Moreover, as S2 is asingular R-module, then S2 es singular in ��M�. Thus M is not projective in ��M�.

3. PM-ASS AND PM-DIMENSION

We associate to each module N ∈ ��M� a set of torsion theories in M-tors; theelements in this set are called �M -associated to N .

Definition 3.1. Let N ∈ ��M�. We denote �M -ass�N� = �� ∈ �M � there is asubmodule L of N such that L is a �-�M -module.

Notice that if N is �M -module, then we have that �M -ass�N� = ����N�. Infact, as N is a �M -module, then ���N� ∈ �M -ass�N�. Now � ∈ �M -ass�N�, and thusthere exists a submodule L of M such that L is a �-�M -module. Hence, ��L� ∈ �M .As � ∈ �M , then by Proposition 2.4 ��L� = �. Moreover, by Proposition 2.2, 2),���N� = ���L� = �. Therefore, �M -ass�N� = ����N�.

In the following proposition, we collect some properties of the �M -ass, thatare straightforward to prove.

Proposition 3.2. Let N ∈ ��M�. Then:

1. For every submodule L of N , �M -ass�L� ⊂ �M -ass�N� ⊂ �M -ass�L� ∪�M -ass�N/L�;

2. If N = ⊕Ni, then �M -ass�N� = ∪�M -ass�Ni�;

3. If L is an essential submodule of N , then �M -ass�L� = �M -ass�N�.

For each � ∈ M-tors, we are going to use the �M -modules in order to define afiltration in gen���.

The �M -filtration of � in M-tors is defined to be a chain of hereditary torsiontheories �0 ≤ �1 ≤ · · · ≤ �i ≤ · · · satisfying the following conditions:

1. �0 = �;2. If i is not limit ordinal, then

�i = �i−1 ∨ ���N ∈ ��M� �N is a �i−1-�M -module��

3. If i is limit ordinal, then

�i = ∨j<i�j�

Since M-tors is a set, there exists a minimal ordinal k such that �k = �k+� forall ordinal �.

Definition 3.3. Let N be a nonzero module in ��M�. N is said to have �-�M -dimension equal to an ordinal h if and only if N is �h-torsion but not �i-torsion forany i < h. In particular, M has �-�M -dimension if and only if N has �-�M -dimensionfor all N ∈ ��M�.

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226 CASTRO PÉREZ AND RÍOS MONTES

Notice that M has �-�M -dimension if and only if �k = � for some ordinal k.The �-�M -dimension of N is denoted by �-�M -dim�N�.

As a direct consequence of the definition, we have the following result.

Proposition 3.4. Let 0−→N ′ → N −→ N ′′ −→ 0 be an exact sequence in��M�. Then�-�M -dim�N� = Sup��-�M -dim�N

′�� �-�M -dim�N′′�, provided that either side exists.

Proposition 3.5. Let ��i be the �M -filtration of � in M-tors, and let k be the ordinalsuch that �k = �k+� for all ordinals �. If �k = �, then the following conditions hold:

1. For all �, �′ ∈ M-tors with � ≤ � < �′ ≤ �, there exists a �-�M -module N such thatN ∈ ��′ ;

2. For all � ∈ M-tors with � ≤ � < �,

� = �∧���N� �N is �-�M -module� ∧ ��

3. For all � ∈ M-tors such that � ≤ � < �, there exists a �-�M -module.

Proof. 1) Let �, �′ ∈ M-tors with � ≤ � < �′ ≤ �. Then there exists an minimalordinal i such that �′ ∧ �i � �. Notice that i is not a limit ordinal and that, �′ ∧ �0 =� ≤ �, so i ≥ 1. As �′ ∧ �i � �, then there exists a nonzero module L such that L ∈��′∧�i ∩��. The choice of i implies �′ ∧ �i−1 ≤ �. Hence L ∈ ��′∧�i−1

. As L ∈ ��′ ,then L ∈ ��i−1

. Therefore, L ∈ ��i−1∩ ��i

. On the other hand, we know that �i =�i−1 ∨ ��N ∈ ��M� �N is �i−1-�M -module. Since L ∈ ��i−1

∩ ��i, then there exists a

module N ∈ ��M� such that N is �i−1-�M -module and Hom�N� L� �= 0. Therefore,there are submodules N ′′ and N ′ of N with N ′′ � N ′ ⊂ N , satisfying that 0 �= N ′

N ′′ ↪→L. As L ∈ ��i−1

, by Proposition 2.10, point 3), then N ′/N ′′ is a �i−1-�M -module sinceN is �i−1-�M -module. As L ∈ ��′ , then by Proposition 2.10 point 4), N ′/N ′′ is a�′ ∧ �i−1-�M -module. Now, as �′ ∧ �i−1 ≤ � and L ∈ ��, then N ′/N ′′ ∈ ��. Hence,by Proposition 2.10 point 2), N ′/N ′′ is �-�M -module. Moreover, as L ∈ ��′ , thenN ′/N ′′ ∈ ��′ .

2) Let � ∈ M-tors be such that � ≤ � < �.If �′ = �∧���N� �N is �-�M -module� ∧ �. Then � ≤ � ≤ �′ ≤ �. Suppose

� < �′. Then by 1) there exists a �-�M -module N such that N ∈ T�′ which is acontradiction. Therefore, we have that � = �′

3) Let � ∈ M-tors be such that � ≤ � < �. By 2), we have that � =�∧���N� �N is �-�M -module� ∧ �. Since � < �, then there exists a �-�M -module.

Corollary 3.6. Let � ∈ M-tors. The following conditions are equivalent:

i) M has �-�M -dimension;ii) For all �, �′ ∈ M-tors with � ≤ � < �′ , there exists a �-�M -module N such that

N ∈ ��′ ;iii) For all � ∈ M-tors with � ≤ � < �

� = �∧���N� �N es �-�M -module��

iv) For all � ∈ M-tors such that � ≤ � < �, there exists a �-�M -module.

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 227

Proof. Let ��i be the �M -filtration of � in M-tors, and let k be the smallestordinal such that �k = �k+� for all ordinals �. Then �k = �. So, by the proof ofProposition 3.5, we have that i) ⇒ ii) ⇒ iii) ⇒ iv). Now, we will prove iv) ⇒ i).Suppose �k < �. Then there exists a �k-�M -module N . Hence we have that N ∈�

�k+1= ��k

, a contradiction.

Definition 3.7. The -�M -dimension of a module N ∈ ��M� is simply called the�M -dimensión of the module N .

Corollary 3.8. Let M ∈ R-Mod. The following conditions are equivalent:

i) M has �M -dimension;ii) For every � ∈ M-tors with � �= �, there exists a �-�M -module N ∈ ��M�;iii) For every � ∈ M-tors with � �= �, � = ∧���N� �N is �-�M -module;iv) For all � ∈ M-tors such that � < �, there exists a �-�M -module.

Notice that, if M has �M -dimension, then every proper torsion theory in ��M�is a meet of irreducible elements of M-tors.

4. GABRIEL CORRESPONDENCE

In this section we will use the concept of Gabriel dimension relative to ahereditary torsion theory � ∈ M-tors.

For each � ∈ M-tors, we define a filtration in gen���. The �M -Gabriel filtrationof � ∈ ��M� is defined to be a chain of torsion theories 0 ≤ 1 ≤ · · · ≤ i ≤ · · · ,satisfying the following conditions:

1. 0 = �;2. If i is not a limit ordinal, then

i = i−1 ∨ �N ∈ ��M� �N es i−1-cocritical�

3. If i is a limit ordinal, then

i = ∨j<ij�

Since M-tors is a set, there exists a smallest ordinal h such that h = h+r forall ordinals r .

A nonzero module N ∈ ��M� is said to have �M -Gabriel dimension equal to anordinal i if N is �i-torsion, but not �j-torsion for any j < i. If N is not �i-torsionfor any i, then its �M -Gabriel dimension is not defined. The �M -Gabriel dimensionof N is denoted by �M -G dim�N�. The M -Gabriel dimension of N is simply calledthe M-Gabriel dimension of N . Notice that the results for �-Gabriel dimension inR-Mod are true for �M -Gabriel dimension in ��M�.

Proposition 4.1. Let � ∈ M-tors. The following conditions are equivalent:

i) M has �M -Gabriel dimension;

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228 CASTRO PÉREZ AND RÍOS MONTES

ii) For all �, �′ ∈ M-tors with � ≤ � < �′, there exists a module N ∈ ��M� such thatN is �-cocritical and N ∈ ��′ ;

iii) For all � ∈ M-tors with � ≤ � < �, there exists a module N ∈ ��M� such that N is�-cocritical.

Notice that if M is a Noetherian module, then M has �M -Gabriel dimension for all� ∈ M-tors with � �= �.

Lemma 4.2. Let � ∈ M-tors and � ∈ gen���. Suppose M has �M -Gabriel dimension.Then for each N ∈ ��M� such that 0 �= N ∈ ��, we have that N contains a ��N�-cocritical submodule.

Proof. Let N ∈ ��M� with 0 �= N ∈ ��. As M has �M -Gabriel dimension, then thereexists a module C ∈ ��M� such that C is a ��N�-cocritical module. So there exists anonzero morphism f � C −→ N . As C is a ��N�-cocritical, then f is monomorphism.Therefore, Im f ∩ N is ��N�-cocritical.

Now we associate to each module N ∈ ��M� a set of prime submodules in M .

Definition 4.3. Let N ∈ ��M�. A fully invariant proper submodule K of M issaid to be associated to N if there exists a nonzero submodule L of N such thatAnnM

�L′� = K for all nonzero submodule L′ of L. Notice that by Proposition 1.16,we have that K is prime in M .

We denote by AssM�N� the set of all submodules prime in M associatedto N . Notice that our AssR�N� coincides with Ass�N� as defined in Stenstrom [9,Chapter VII, §1].

In the following proposition, we collect some straightforward propertiesof AssM .

Proposition 4.4. Let N ∈ ��M�. Then:

1. If 0 −→ N ′ −→ N −→ N ′′ −→ 0 is an exact sequence of modules in ��M�, thenAssM�N

′� ⊆ AssM�N� ⊆ AssM�N′� ∪ AssM�N

′′�;2. If N ′ is an essential submodule of N , then AssM�N

′� = AssM�N�;3. If N = ⊕

i∈I Ni, AssM�N� = ∪i∈IAssM�Ni�;4. If N is a uniform module, then AssM�N� has at most one element. If N is a uniform

module and AssM�N� �= ∅, we denote assM�N� the prime associated to N .

Proposition 4.5. Let P be a prime in M , then AssM�M/P� = �P.

Proof. As M is projective in ��M�, then by Proposition 1.15, P is a prime M-ideal. Hence by Beachy [1, Theorem 5.7], we have that M/P is an M-prime module.Moreover, by Beachy [1, Proposition 1.3, (4)], AnnM

�M/P� = P. On the other hand,we know that M is a generator of the category ��M�. Then Hom�M�L/P� �= 0 forall 0 �= L

P⊆ M

P. Thus AnnM

�L/P� = AnnM�M/P� = P since M/P is M-prime module.

Therefore, AssM�M/P� = �P.

Remark 4.6. Suppose each submodule of M is fully invariant. Let N ∈ ��M� and� ∈ AssM�N�. Then ��M/P� ∈ �M -ass�N�. In fact, as P ∈ AssM�N�, then there exists

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 229

a submodule L of N such that P = AnnM�L′� for all 0 �= L′ ⊂ L. As M is a generator

of the category ��M�, then there exists a nonzero morphism f � M −→ L. Let L′ =Im f . Then we can consider the morphism f � M −→ L′. Hence M/ ker f � L′. ThusAnnM

�M/ ker f� = P. Moreover, as ker f is a fully invariant submodule of M , then byBeachy [1, Proposition 5.1], ker f is a M-ideal since M is projective in ��M�. Now byBeachy [1, Proposition 1.3(4)], we have that AnnM

�M/ ker f� = ker f . So ker f = P.Thus L′ � M/P and ��M/P� ∈ �M -ass�L� ⊂ �M -ass�N�.

Moreover, if N satisfies AssM�L� �= ∅ for all 0 �= L ⊂ N , we claim that���M/P� �P ∈ AssM�N� = �M -ass�N�. In fact, let � ∈ �M -ass�N�. Then there exists asubmodule L of N such that L is �-�M -module. As � ∈ �M , then by Proposition 2.4,� = ��L�. On the other hand, we know that there exists P prime in M such that� = ��M/P�. Now let H ∈ AssM�L�, then there exists a submodule L′ of L suchthat AnnM

�L′′� = H for all 0 �= L′′ ⊂ L′. As M is a generator of the category ��M�,there exists a nonzero morphism g � M −→ L′. If K = ker g, then M

K� Im g ⊂ L′.

Thus AnnM�MK� = H . Moreover, as K is a fully invariant submodule of M , then by

Beachy [1, Proposition 1.3] K = AnnM�M/K�. Thus K = H . So M

H� Im g. As L is �-

�M -module, then ��M/H� = ��Im g� = ��L� = � = ��M/P�. Hence H is not ��M/P�-dense in M . Thus by Lemma 2.6, H ⊆ P. Also notice that H is prime in M andP is not ��M/H�-dense. So by Lemma 2.6, P ⊆ H . Thus P = H . Therefore, P ∈AssM�L� ⊆ AssM�N�.

Notice that if R is a ring such that every left ideal is two-side ideal, then foreach 0 �= N ∈ R-Mod, we have that ���R/P� ∈ R-tors �P ∈ Ass�N� = �R-ass�N� =� − ass�N�; see Castro et al. [3, Remark 2.9].

Lemma 4.7. Let � ∈ M-tors. Suppose M has �-M -Gabriel dimension. If M has �-�M -dimension, then AssM�N� �= ∅ for all N ∈ ��M� such that N ∈ ��.

Proof. Let 0 �= N ∈ ��M�, N ∈ ��, and ��i the �M -filtration of � in M-tors. As Mhas �-�M -dimension, then there exists an ordinal j minimal such that 0 �= t�j �N�. PutN ′ = t�j �N�. Notice that j is not a limit ordinal and N ′ ∈ ��j−1

. Hence N ′ ∈ ��j∩

��j−1. Inasmuch as �j = �j−1 ∨ ��L �L is �j−1-�M -module�, there exists a �j−1-�M -

module L such that Hom�L� N ′�� �= 0. Hence, there exist submodules L′′ � L′ ⊆ Lsuch that 0 �= C = L′/L′′ ↪→ N ′ with N ′ ∈ F�j−1

. Thus by Proposition 2.10, C is a�j−1-�M -module, and by Proposition 2.2, we have that ��C� = ��L�. By Lemma 4.2and Proposition 2.10, we can assume C is cocritical. Since is a C is �j−1-�M -module,then there exists a submodule P prime in M such that ��C� = ��M/P�. Because C iscocritical, we have that there exists a nonzero submodule C ′ of C that is isomorphicto a submodule of M/P. By Proposition 4.5, AssM�C

′� = �P. Hence 0 �= AssM�C′� ⊂

AssM�N′� ⊆ AssM�N�.

We denote by M-sp = ���C� ∈ M-tors �C ∈ ��M� and C is cocritical and��M� a complete set of representatives of isomorphism classes of indecomposable �-torsion free injective modules in ��M�. Notice that if M has �-M-Gabriel dimension,then ��M� �= ∅.

Let Spec��M� denote the set of �-pure submodules prime in M , and letus denote �M�

= ���M/P� ∈ M-tors �P is �-pure submodule prime in M. We want

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230 CASTRO PÉREZ AND RÍOS MONTES

to determine when Spec��M� is large enough for the assignment �� � ��M� −→Spec��M� defined by ���E� = assM�E� to be a bijection. We will say that M has localGabriel correspondence with respect to � if �� is a bijective function.

Lemma 4.8. Let P be prime in M and N ∈ ��M� such that AnnM�K� = AnnM

�N� =P, for all 0 �= K ⊆ N . Then for every nonzero submodule L of N , there exists amonomorphism F � M

P−→ LHom�M�L�.

Proof. The family of morphisms �f � f � M −→ L induces a morphism F � M −→LHom�M�L�, such that �F�x���f� = f�x�. On the other hand, we know that AnnM

�L� =∩�ker f � f ∈ Hom�M�L�. Thus, x ∈ ker F if and only if �F�x���f� = 0 for all f ∈Hom�M�L�, if and only if f�x� = 0 for all f ∈ Hom�M�L�, if and only if x ∈ AnnM

�L�.Therefore, ker F = AnnM

�L� = P. Hence there exists a monomorphism F � MP−→

LHom�M�L�.

Theorem 4.9. Let � ∈ M-tors. Suppose M has �-GM -dimension. The followingconditions are equivalent:

i) �� is bijective;ii) For every N ∈ ��M�, N ∈ ��

�M -ass�N� = ���M/P� �P ∈ AssM�N� �= ∅�

iii) �M -ass�N� �= ∅ for every nonzero �-torsion free module N ∈ ��M�;iv) �M�

= gen��� ∩M-sp;v) M has �-�M -dimension.

Proof. i) ⇒ ii) If N ∈ ��M� with N ∈ ��, then by Lemma 4.2, there exists asubmodule N ′of N such that N ′ is ��N�-cocritical. So, N ′ ∈ ��M�. As �� is bijective,then ���N

′� = assM�N′� �= ∅. Thus AssM�N� �= ∅.

Now let P ∈ AssM�N�. Then there exists a submodule C of N such thatAnnM

�L� = P for all 0 �= L ⊂ C. As M has �M -Gabriel dimension, we can supposethat C is cocritical module. By Lemma 4.8, we know that there exists amonomorphism F � M

P−→ LHom�M�L�. As L ∈ ��, then P is �-pure submodule prime

in M . So by hypothesis, there exists a cocritical submodule J/P of M/P. Hence wehave that J/P and C are indecomposable �-torsion injective modules in ��M� suchthat ���J/P� = assM�J/P� = �P and ���C� = assM�C� = �P.

As �� is an injective map, then J/P � C. Therefore, there exists a submoduleC ′ of C such that C ′ ↪→ J/P. So we have that ��M/P� ∈ �M -ass�N�.

Let N ∈ �� and ��M/P� ∈ PM -ass�N�. Then there is a submodule L of Nsuch that L is ��M/P�-�M -module. By Lemma 4.2, we can assume that L is a��N�-cocritical module. By Proposition 2.4, we have ���L� = ���M/P�. Therefore,Hom�L� M/P� �= 0, and there exists a submodule C of L such that C ↪→ M/P. Thus�P = assM�C� ⊂ assM�L� ⊂ assM�N�.

ii) ⇒ iii) It is clear.

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PRIME SUBMODULES AND LOCAL GABRIEL CORRESPONDENCE 231

iii) ⇒ iv) If ��M/P� ∈ �M�, then M/P ∈ ��. By Lemma 4.2, there is acocritical module C in ��M� such that C ↪→ M/P. By Proposition 2.7, M/P is a �M -module, then we get ���C� = ���M/P�. Therefore, ���M/P� ∈ gen��� ∩M-sp.

Conversely, let us take � ∈ gen��� ∩M-sp, and let C ∈ ��M� be a cocriticalmodule such that � = ���C�. As C ∈ ��, by iii), �M -ass�C� �= ∅. If ���M/P� ∈PM -ass�C�, then there exists C ′ ⊂ C such that C ′ is ���M/P�-�M -module.By Proposition 2.4, ���C ′� = ���M/P�. Therefore, � = ���C� = ���C ′� =���M/P� ∈ �M�.

iv) ⇒ v) Let � ∈ gen���, � �= �. Since M has �M -Gabriel dimension, there is a�-cocritical module C. Then ���C� ∈ gen��� ∩M-sp = �M�. So there is a submoduleP prime in M such that ���C� = ���M/P�. As C is �-cocritical, then C is �-�M -module. So C is �-�M -module. From Corollary 3.6, we get the result.

v) ⇒ i) As M has �-�M -dimension and by Lemma 4.7, we see that �� isa well-defined function. If P ∈ Spec��M�, then M/P ∈ ��. From Lemma 4.2, M/P

contains a nonzero ���M/P�-cocritical module C. Hence, C is an indecomposable�-torsion free injective module in ��M� such that ���C� = assM�C� = assM�C� ⊂assM�M/P� = P. So �� is epic. Now, let E and E′ be �-torsion free indecomposableinjective modules in ��M� such that ���E� = ���E

′� = �P, by Lemma 4.2 there arecocritical modules C and C ′ such that C ↪→ E and C ′ ↪→ E′. Then ���C� = ���E�and ���C ′� = ���E′�. From v) and Corollary 3.6, there exists a module N ∈ ��M�such that N is ���C�-�M -module. Proposition 2.6 implies ���N� = ���C�. On theother hand, since N is a ���C�-�M -module, there is a submodule H , prime inM , such that ���N� = ���M/H�. Hence ���C� = ���M/H�. Inasmuch as C isa cocritical module, there is a nonzero submodule K of C that is isomorphic toa submodule of M/H . Thus assM�K� = H . Since K ⊆ C, we have that assM�K� =assM�C� = assM�E� = P. So H = P and ���C� = ���M/P�.

Analogously, we show that ���C ′� = ���M/P�. From this ���C� = ���C ′�.Since C and C ′ are cocritical modules in ��M�, then we have that E = C = C ′ = E′.Therefore, �� is a bijection, and the proof is complete.

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