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Krull Dimension and Classical Krull Dimension ofModulesJaime 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 , MéxicoPublished online: 13 Mar 2014.

To cite this article: Jaime Castro Pérez & José Ríos Montes (2014) Krull Dimension and Classical Krull Dimension of Modules,Communications in Algebra, 42:7, 3183-3204, DOI: 10.1080/00927872.2013.781611

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Communications in Algebra®, 42: 3183–3204, 2014Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927872.2013.781611

KRULL DIMENSION AND CLASSICAL KRULLDIMENSION OF MODULES

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, México

Using the concept of prime submodule defined by Raggi et al. in [16], for M ∈ R-Modwe define the concept of classical Krull dimension relative to a hereditary torsion theory� ∈ M-tors. We prove that if M is progenerator in ��M�, � ∈ M-tors such that M has�-Krull dimension then cl�K� dim�M� ≤ k��M�. Also we show that if M is noetherian,�-fully bounded, progenerator of ��M�, and M ∈ ��, then cl · K� dim�M� = k��M�.

Key Words: Classical Krull dimension; Krull dimension; Prime submodules; Semiprime submodules;Torsion theory.

2010 Mathematics Subject Classification: 16S90; 16D50; 16P50; 16P70.

INTRODUCTION

The classical Krull-dimension and Krull-dimension for a ring have beenstudied by several authors. For commutative noetherian rings the Krull dimensionas introduced by Gabriel and Rentschler in [7] (cf. also Michler [14]) coincideswith the classical Krull-dimension. Krause in [11] extended the concepts of bothdimensions to arbitrary ordinals and he investigated the relationship between thesedimensions. He showed that the classical Krull dimension of a left noetherian ringR does not exceed the left Krull dimension of R. Later, Krause in [12] showed thatthese dimensions are equal for every fully left bounded left noetherian ring.

Later, a more general situation was considered by Heakyung Lee andPeter L. Vachuska in [13]. For a hereditary torsion theory � in R-Mod theyconsidered the �-Krull dimension of R (K� − dim�R�) and they defined the classical�-Krull dimension of R�cl�K� dim�R��. In that paper they showed that, if R has�-Krull dimension, then the classical �-Krull dimension exists and cl�K� − dim�R� ≤K� − dim�R�. Moreover they showed that if R is �-fully bounded, �-noetherian withKrull dimension which is not �-torsion, then K� − dim�R� = cl�K� − dim�R�.

Received July 5, 2012; Revised February 1, 2013. Communicated by T. Albu.Dedicated to the memory of Professor Francisco Raggi.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,México; E-mail: jcastrop@itesm.mx

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3184 CASTRO PÉREZ AND RÍOS MONTES

In 2001 Toma Albu, Günter Krause, and Mark L. Teply in [2] investigatedthe relationship between �-Krull dimension and the classical �-Krull dimension of aring R with �-Krull dimension and bijective �-Gabriel correspondence. They showedthat the two dimensions differ by at most 1 and are equal whenever R is either rightfully �-bounded with Ass�M� �= ∅ for every �-torsion-free right R-module M �= 0 orR is right �-noetherian with bijective �-Gabriel correspondence.

In this paper, using the concept of prime submodule defined by Raggi et al.in [16], for M ∈ R-Mod we define the classical Krull dimension relative to a torsiontheory � ∈ M-tors denoted as cl�K� − dim�M�. The classical �-Krull dimension ofa module is a special instance of the more general concept of the classical Krulldimension of a poset �X�≤� introduced in [1]. We obtain results about the classical�-Krull dimension of a module M ∈ R-mod and � ∈ M-tors. We prove that if Mis progenerator of the category ��M� such that M has �-Krull dimension, thencl�K� − dim�M� ≤ K� − dim�M�. Also for an R-module M and � ∈ M-tors we usethe concept of �-fully bounded module defined in [6] and we prove that if M isnoetherian, progenerator of ��M�, �-fully bounded and M ∈ ��, then cl�K� dim�M� =k��M�.

In order to do this, we organized the paper in tree sections. In Section 1we give some results about the classical �-Krull dimension of M for M ∈ R-modand � ∈ M-tors. In particular we show (Proposition 1.8) that if each subset of fullyinvariant submodules of M has maximal elements, then the following conditions areequivalent:

i) M is a prime module;ii) cl�K� dim�M/P� < cl�K� dim�M� for all submodules 0 �= P prime in M with

M/P ∈ ��;iii) cl�K� dim�M/N� < cl�K� dim�M� for all fully invariant submodules 0 �= N of M

with M/N ∈ ��.

For for M ∈ R-mod and � ∈ M-tors Jategaonkar introduced in [10] the relativeKrull dimension.

In Section 2, for M ∈ R-Mod and � ∈ M-tors, N ∈ ��M�, we define the �-Krulldimension of N as the natural extension of the relative Krull dimension givenby Jategaonkar in [10]. In particular we prove (Proposition 2.10) that if M ∈ R-Mod, � ∈ M-tors, N ∈ ��M� such that N is finitely generated and M has �-Krulldimension, then N has �-Krull dimension and k��N� ≤ k��M�. Furthermore we alsoshow (Corollary 2.11) that if N ∈ ��M� such that M and N have �-Krull dimension,then k��N� ≤ k��M�.

In Section 3, we investigate the relationship between the �-Krull dimensionand the classical �-Krull dimension of an R-module M . In particular we prove thatif M is progenerator in ��M�, � ∈ M-tors such that M has �-Krull dimension theni) (Proposition 3.18) Spec��M� is a noetherian poset, ii) (Corollary 3.19) M hasclassical �-Krull dimension, and iii) (Theorem 3.21) cl�K� dim�M� ≤ k��M�. Finallywe prove in (Theorem 3.27) that if M is noetherian, �-fully bounded, progeneratorof ��M�, and M ∈ ��, then cl�K� dim�M� = k��M�. This last result generalizes theresult given in [12] for rings.

In what follows, R will denote an associative ring with unity and R-Mod willdenote the category of unitary left R-modules. Let M and X be R-modules. X is said

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KRULL DIMENSION AND CLASSICAL 3185

to be M-generated if there exists an R-epimorphism from a direct sum of copies ofM onto X. The category ��M� is defined as the full subcategory of R-Mod containingall R-modules X which are isomorphic to a submodule of an M-generated module.

Let M-tors be the frame of all hereditary torsion theories on ��M�. For a familyM� of left R-modules in ��M�, let ��M�� be the greatest element of M-tors forwhich all the M are torsion free, and let �M�� denote the least element of M-torsfor which all the M are torsion. ��M�� is called the torsion theory cogenerated bythe family M�� and �M�� is the torsion theory generated by the family M�� Inparticular, the greatest element of M-tors is denoted by �, and the least element ofM-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 free class,and the torsion functor associated to �, respectively. For N ∈ ��M�, N is called �-cocritical if N ∈ �� and for all 0 �= N ′ ⊆ N , N/N ′ ∈ ��. We say that N is cocritical ifN is �-cocritical for some � ∈ M-tors. For N ∈ ��M�, let N denote the injective hullof N in ��M�. If N is an essential submodule of M , we write N ⊆ess M . If N is afully invariant submodule of M , we write N ⊆FI M . For � ∈ M-tors and M ′ ∈ ��M�,a submodule N of M ′ is �-pure in M ′ if M ′/N ∈ ��.

A module N ∈ ��M� is called singular in ��M�, or M-singular, if there is anexact sequence in ��M�0 → K → L → N → 0, with K essential in L. The class � ofall M-singular modules in ��M� is closed by taking submodules, factor modules anddirect sums. Therefore, any L ∈ ��M� has a largest M-singular submodule ��L� =∑f�N� �N ∈ � and f ∈ Hom�N�L��. L is called non-M-singular if ��L� = 0.

Let M ∈ R-Mod. In [3, Definition 1.1] the annihilator in M of a class � of R-modules is defined as AnnM��� = ⋂

K∈� K, where

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

Also in Beachy [3, Definition 1.5] a product is defined in the following way. Let N ;be a submodule of M . For each module X ∈ R-Mod, N · X = AnnX���, where � isthe class of modules W , such that f�N� = 0 for all f ∈ Hom�M�W�. For M ∈ R-Modand K, L submodules of M , in Bican et al. in [4] is defined the product KML asKML = ∑

f�K� � f ∈ Hom�M�L��. Moreover, Beachy showed in [3, Proposition 5.5]that, if M is projective in ��M� and N is a any submodule of M , then N · X = NMX,for any R-module X ∈ ��M�.

Let M ∈ R-Mod and N �= M a fully invariant submodule of M . N is prime inM if for any K, L fully invariant submodules of M we have that KML ⊆ N impliesthat K ⊆ N or L ⊆ N . We say that M is a prime module if 0 is prime in M see [16,Definitions 13 and 16].

A nonzero R-module M is monoform if M has the property that, for eachsubmodule N every homomorphism f � N → M is either zero or a monomorphism.An R-module M is said to have enough monoforms if every nonzero submodule ofM contains a monoform submodule. If M ∈ R-Mod, we denote by Sat��M� = of all�-pure submodules of M .

For details about concepts and terminology concerning torsion theories in��M�, see [20, 21]. For torsion theoretic dimensions, the reader is referred toGolan [8] and [19].

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1. PRELIMINARIES

The Classical � -Krull Dimension of a Module M

In this section we will suppose that M is projective in ��M�. For M ∈ R-Modand � ∈ M-tors, we denote

Spec��M� = P ⊂ M �P is �-pure and P is prime in M��

Let � ∈ M-tors. The classical Krull dimension relative to a torsion theory �, isdefined as follows: Set Spec−1

� �M� = ∅, and for an ordinal > −1, define

Spec� �M� ={P ∈ Spec��M� �P ≤ Q ∈ Spec��M� ⇒ Q ∈ ⋃

�<

Spec�� �M�

}�

If an ordinal with Spec� �M� = Spec��M� exists, then the smallest suchordinals is called the classical �-Krull dimension of M ; it is denoted by cl�K� dim�M�.

The classical Krull dimension of a module is a special instance of the moregeneral concept of the classical Krull dimension of a poset �X�≤� introduced in [1],we briefly describe below.

For ordinals ≥ −1, define subsets X ⊆ X recursively as follows. Start bysetting X−1 = ∅. Let ≥ 0, and suppose that X� ⊆ X has been defined for each� < . Set

X ={x ∈ X � ∀y ∈ X� x < y ⇒ y ∈ ⋃

�<

X�

}

This construction leads to an ascending filtration

X−1 ⊆ X0 ⊆ X1 ⊆ · · · ⊆ X ⊆ · · · �

called the classical Krull filtration of X. Since X is a set, there exists a leastordinal � such that X� = X�+1. Obviously, X� �= ∅ ⇐⇒ X0 �= ∅ ⇐⇒ X has at leastone maximal element.

The poset X is said to have classical Krull dimension cl�K dim�X� = � ifX = X�.

Lemma 1.1 ([1, Proposition 1.4]). The following conditions are equivalent for aposet X:

i) X has classical Krull dimension;ii) X is a noetherian poset.

Notice that the filtration Spec� �M��≥−1 of Spec��M� defined at the beginning ofthis section is nothing else than the classical Krull filtration of Spec��M�.

Remark 1.2. Let M be an R-module, and let � be a hereditary torsion theory on��M�. Then M has classical �-Krull dimension if and only if the poset �Spec��M��⊆�is noetherian. In fact it follows of Lemma 1.1.

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KRULL DIMENSION AND CLASSICAL 3187

Notice that if M is �-noetherian, then the poset �Spec��M��⊆� is noetherian.Therefore, M has classical �-Krull dimension.

Proposition 1.3. Let M be an R-module. If N , K, L, P are fully invariant submodulesof M such that N ⊆ K; N ⊆ L; N ⊆ P, then �K/N�M/N �L/N� ⊆ P/N if and only ifKML ⊆ P.

Proof. Suppose that �K/N�M/N �L/N� ⊆ P/N . Let f � M → L be a morphism. Nowwe consider the morphism � � f � M → L/N where �: L → L/N is the naturalprojection. As N is fully invariant submodule of M , then �� � f��N� = 0. Thus wecan consider the morphism F � M/N → L/N such that F�x + N� = f�x�+ N . Byhypothesis, F�K/N� ⊆ P/N . Therefore, F�x + N� = f�x�+ N ∈ P/N for all x ∈ K. Sof�x� ∈ P. Hence f�K� ⊆ P. Therefore, KML ⊆ P.

Now suppose that KML ⊆ P. Let F � M/N → L/N be a morphism. As M isprojective in ��M�, then there exists a morphism f � M → L such that the followingdiagram commutes:

,

where �1 and �2 are the natural projections.Thus �F � �1��K� = F�K/N� = ��2 � f��K� = �2�f�K��. As f � M → L and

KML ⊆ P, then f�K� ⊆ P. Hence �2�f�K�� ⊆ P/N . Therefore, �K/N�M/N �L/N� ⊆ P/N .

Corollary 1.4. Let M ∈ R-Mod, and let P, N be fully invariant submodules of M suchthat N ⊆ P ⊆ M . Then P is prime in M if and only if P/N is prime in M/N .

Proof. Let P be prime in M , and let K/N , L/N be fully invariant submodules ofM/N such that �K/N�M/N �L/N� ⊆ P/N . By Proposition 1.3, KML ⊆ P. As P is primein M , then by [5, Proposition 1.11] K ⊆ P or L ⊆ P. Hence K/N ⊆ P/N or L/N ⊆P/N . Therefore, P/N is prime in M/N .

Now we suppose that P/N is prime in M/N . By [16, Proposition 18]�M/N�/�P/N� is prime module. So M/P is prime module. As M is projective in ��M�,then P is prime in M . �

Remark 1.5. If M ∈R-Mod, �∈M-tors, and N is a fully invariant submodule of M ,then �N denotes the direct image of � in the category ��M/N�, where ��N = L ∈��M/N� �L ∈ ��� see [21, Definition 9.1]. It is clear that �N ∈ �M/N�-tors. Moreover,if L ∈ ��M/N�, then K ⊆ L �L/K ∈ ��N � = K ⊆ L �L/K ∈ ���. In fact let K ⊆ Lsuch that L/K ∈ ��N . If L/K � ��, then there exists 0 �= L′/K ⊆ L/K such thatL′/K ∈ ��. As L′/K ∈ ��M/N�, then L′/K ∈ ��N . So L/K � ��N a contradiction.Therefore, L/K ∈ �� and K ⊆ L �L/K ∈ ��N � ⊆ K ⊆ L �L/K ∈ ���. Analogouslywe prove that K ⊆ L �L/K ∈ ��� ⊆ K ⊆ L �L/K ∈ ��N �.

As a consequence of Remark 1.5, we have that Spec�N �M/N�=P/N ⊆M/N �P/N is prime inM/N and �M/N�/�P/N�∈��N �=P/N ⊆M/N �P/N is primeinM/N and �M/N�/�P/N�∈���=Spec��M/N�. Hence Spec�N �M/N�=Spec��M/N�.Then if � ∈ M-tors and N is a fully invariant submodule of M we can write

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cl�K� dim�M/N� instead of cl�K�N dim�M/N�, which will be used frequently, withoutfurther mention.

Proposition 1.6. Let M ∈ R-Mod, � ∈ M-tors, and ≥ 0, then the followingconditions hold:

i) If N and P are fully invariant submodules of M such that N ⊆ P and P is primein M , then P ∈ Spec� �M� if and only if P/N ∈ Spec� �M/N�;

ii) If cl�K� dim�M� = , then cl�K� dim�M/N� ≤ for all fully invariant submodules Nof M;

iii) If M is a prime module such that cl�K� dim�M� = and 0 �= P is a prime submoduleof M , then cl�K� dim�M/P� < .

Proof. i) Let = 0 and P ∈ Spec0� �M�. Then P is maximal prime in M suchthat M/P ∈ ��. By Corollary 1.4, P/N is prime in M/N . Moreover, P/N is �-purein M/N . Now we claim that P/N is maximal prime in M/N . In fact, let P ′/Nsubmodule prime in M/N such that P/N ⊆ P ′/N . By Corollary 1.4, P ′ is prime inM . Since P is maximal prime in M , then P = P ′. Thus P/N ∈ Spec0� �M/N�. Inversely,let P/N ∈ Spec0� �M/N�. Then P/N is �-pure in M/N and maximal prime in M/N . ByCorollary 1.4, P is prime in M . We claim that P is �-pure in M and maximal primein M . In fact, it is clear M/P ∈ ��. Now let P ′ be a prime submodule of M such thatP ⊆ P ′ ⊆ M and M/P ′ ∈ ��. Then by Corollary 1.4 we have that P ′/N is prime inM/N . Moreover, �M/N�/�P ′/N� ∈ ��. Since P ⊆ P ′ ⊆ M , then P/N ⊆ P ′/N . As P/Nis maximal prime in M/N , then P/N = P ′/N . Therefore, P = P ′. So P ∈ Spec0� �M�.

Now let > 0, and assume �i� holds for all � < . Let P ∈ Spec� �M�. Fromthe definition of Spec� �M�, we know that P ∈ Spec� �M� if and only if for eachQ ∈ Spec��M� such that P � Q we have Q ∈ Spec�� �M� for a some � < . Since P ∈Spec��M�, then by Corollary 1.4 P/N ∈ Spec��M/N�. Now let Q/N ∈ Spec��M/N�such that �P/N� � �Q/N�. Since Q/N ∈ Spec��M/N�, then by the Corollary 1.4 Q ∈Spec��M�. Moreover, as P � Q, then Q ∈ Spec�� �M� for some � < . By inductionhypothesis, we get Q ∈ Spec�� �M� if and only if Q/N ∈ Spec�� �M/N�. Hence weobtain that P ∈ Spec� �M� if and only if P/N ∈ Spec� �M/N�.

ii) We know that Spec� �M� = Spec��M�. Now let P/N ∈ Spec��M/N�, then byCorollary 1.4 P ∈ Spec��M� = Spec� �M�. By �i�, we have that P/N ∈ Spec� �M/N�.Thus Spec��M/N� ⊆ Spec� �M/N�. Hence cl�K� dim�M/N� ≤ .

iii) As M is prime module such that cl�K� dim�M� = and 0 � P is prime inM , then P ∈ ⋃

�< Spec�� �M�. So there exits � < such that P ∈ Spec�� �M�. Now let

Q/P ∈ Spec��M/P�. By Corollary 1.4, Q is prime in M . Since P ⊆ Q, there exits anordinal � such that � ≤ � and Q ∈ Spec���M�. Hence by (i) Q/P ∈ Spec���M/P�. So wehave that Spec��M/P� ⊆ ⋃

�≤� Spec���M� = Spec�� �M�. Therefore, cl�K� dim�M/P� ≤

� <

Proposition 1.7. Let M ∈ R-Mod, and let be an ordinal such that cl�K� dim�M� ≥ ≥ 0. If cl�K� dim�M/N� < for all fully invariant submodules N of M , then M is aprime module and cl�K� dim�M� = .

Proof. Let P ∈ Spec��M�. If P is maximal prime in M , then P ∈ Spec� �M�.Let us suppose that P is not maximal prime in M and P � Q ∈ Spec��M�.

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KRULL DIMENSION AND CLASSICAL 3189

Since cl�K� dim�M/P� ≤ � < , then Q/P ∈ Spec�� �M/P�. By Proposition 1.6 (i),Q ∈ Spec�� �M�. So P ∈ Spec� �M�. Therefore, we have that Spec��M� ⊆ Spec� �M�.Hence cl�K� dim�M� ≤ . So cl�K� dim�M� = . Now suppose that M is nota prime module. Then by [16, Definition 13 and 16] there exist 0 �= Land 0 �= N fully invariant submodules of M such that LMN = 0. Let � =maxcl�K� dim�M/N�� cl�K� dim�M/L��, and let P be a submodule prime inM . So LMN ⊆ P. Thus L ⊆ P or N ⊆ P. If L ⊆ P, then by �ii� we havethat cl�K� dim�M/P� = cl�K� dim��M/L�/�P/L�� ≤ cl�K� dim�M/L� ≤ � < . SinceP ∈ Spec��M�, then by Proposition 1.6 (i), P/L ∈ Spec��M/L�. Therefore, P/L ∈Spec�� �M/L�. By Proposition 1.6 (i), we have that P ∈ Spec�� �M�. Hence Spec��M� ⊆Spec�� �M�. So cl�K� dim�M� ≤ � < a contradiction. Thus M is prime module.

Proposition 1.8. Let M ∈ R-Mod. Suppose that each subset of fully invariantsubmodules of M has maximal elements, then the following conditions are equivalent:

i) M is prime module;ii) cl�K� dim�M/P� < cl�K� dim�M� for all submodules 0 �= P prime in M with M/P ∈

��;iii) cl�K� dim�M/N� < cl�K� dim�M� for all fully invariant submodules 0 �= N of M

with M/N ∈ ��.

Proof. i� ⇒ ii� By [16, Definition 13] we know that each submodule P prime in Mis a fully invariant submodule of M . Since each subset of fully invariant submodulesof M has maximal elements, then the poset �Spec��M��⊆� is noetherian. So byCorollary 1.4 and Remark 1.5, the poset �Spec��M/P��⊆� is noetherian for eachP prime in M . So by Lemma 1.1, cl�K� dim�M/P� exists. Now if cl�K� dim�M� = ,then by Proposition 1.6, we have that cl�K� dim�M/P� < for all 0 �= P prime in Mand M/P ∈ ��.

ii� ⇒ iii� Suppose there exists a fully invariant submodule 0 �= N of M withM/N ∈ �� such that cl�K� dim�M/N� = cl�K� dim�M� = . By hypothesis, we cansuppose N is maximal with respect to cl�K� dim�M/N� = cl�K� dim�M�. Let 0 �=L/N be a fully invariant submodule of M/N such that �M/N�/�L/N� ∈ ��. Hencecl�K� dim��M/N�/�L/N�� = cl�K� dim�M/L�. As N is a fully invariant submoduleof M and N � L, then by maximality of N , we have that cl�K� dim�M/L� < =cl�K� dim�M/N�.

Therefore, cl�K� dim��M/N�/�L/N�� < cl�K� dim�M/N�. So by Proposition 1.7,M/N is primemodule.Hence by [16, Proposition 18],N is prime inM . So by ii),N = 0, acontradiction.

iii� ⇒ i� As cl�K� dim�M/N� < cl�K� dim�M� for all fully invariantsubmodules 0 �= N of M with M/N ∈ ��, then by Proposition 1.7 M is a primemodule.

2. THE � -KRULL DIMENSION OF A MODULE M

Let M ∈ R-Mod and � ∈ M-tors. It is clear that Sat��M� is an upper continuousmodular lattice see [20, Proposition 4.1, p. 207]. A module N ∈ ��M� is �-artinian(�-noetherian) if Sat��N� is artinian (noetherian). The �-Krull dimension k��N� of a

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3190 CASTRO PÉREZ AND RÍOS MONTES

module N ∈ ��M� is the Krull dimension (or deviation) K dim�Sat��N�� of the posetSat��N�. Thus k��N� = −1 if and only if N ∈ ��, and k��N� = for an ordinal ≥ 0if k��N� ≮ and, given any descending chain

N ⊃ N1 ⊃ N2 ⊃ · · · ⊃ Ni ⊃ Ni+1 ⊃ · · ·

of �-pure submodules, k��Ni/Ni+1� < for all but finitely many i. Note that if M =R and � = , then k��N� = K dim�N� = K dim���N�� is the Krull dimension of theR-module N in the sense of Gordon and Robson [9]. A module 0 �= N ∈ ��M� iscalled �-critical if it is �-torsion free with �-Krull dimension and k��N/L� < k��N�for every 0 �= L ⊆ N . A �-critical module N with k��N� = is called -�-critical.

Note that from Remark 1.5 we have that if P is a fully invariant submoduleof M and � ∈ M-tors, then �P ∈ �M/P�-tors where ��P = L ∈ ��M/P� �L ∈ ���.Moreover, if N ∈ ��M/P�, then Sat�P �N� = Sat��N�. Thus k�P �N� = k��N�, which willbe used frequently, without further mention.

Definition 2.1. Let M ∈ R-Mod and � ∈ M-tors. If N ∈ ��M� and L is a submoduleof N , the �-purification L of L in N is

�L = ∩N ′ ⊆ N �N ′ ∈ Sat��N� and L ⊆ N ′��

Notice that �L ∈ Sat��N�. Also note that, if L ∈ Sat��N�, then �L = L.

Lemma 2.2. Let M ∈ R-Mod, � ∈ M-tors and N ∈ ��M�. If L′ and L are submodulesof N , then the following conditions hold:

i) L ⊆ L;ii) t��N/L� = L/L;iii) If L ⊆ L′ then L ⊆ L′;iv) L = L;v) L+ L′ = L+ L′;vi) If L ⊆ L′, then �L′/L� = L′/L;vii) L ∩ L′ = L ∩ L′.

Proof. It is easy.

Lemma 2.3. Let N ∈ ��M� and � ∈ M-tors. Then the following conditions hold:

i) Let L be a submodule of N . Then N has �-Krull dimension if and only if L and N/Lhave �-Krull dimension. Moreover, k��N� = supk��L�� k��N/L��;

ii) Let Ni�i∈I be a finite family of modules in ��M�, then N = ⊕i∈I Ni has �-

Krull dimension if and only if each Ni has �-Krull dimension. Moreover k��N� =supk��Ni��.

Proof. i) It follows from [15, Proposition 3.2.1].

ii) It follows from i).

The following results are not difficult to show, so we will omit their proofs.

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KRULL DIMENSION AND CLASSICAL 3191

Lemma 2.4. Let M ∈ R-Mod, � ∈ M-tors and N ∈ ��M�. If L is a submodule of N ,then following conditions hold:

i) L has �-Krull dimension if and only if L has �-Krull dimension. Moreover, k��L� =k��L�;

ii) N/L has �-Krull dimension if and only if N/L has �-Krull dimension. Moreover,k��N/L� = k��N/L�;

iii) If N is a submodule of M and M has �-Krull dimension, then k��N/L� = k��N/L�.

Proposition 2.5. Let M ∈ R-Mod, � ∈ M-tors and N ∈ ��M�. If N ∈ �� and L ∈Sat��N�, then L is a uniform element in the lattice Sat��N� if and only if L is uniform.

Proposition 2.6. Let M ∈ R-Mod, � ∈ M-tors and N ∈ ��M�. Suppose N is − �-critical. Then following conditions hold:

i) L is -�-critical for all 0 �= L ⊆ N ;ii) N is a uniform module.

Proposition 2.7. Let M ∈ R-Mod, � ∈ M-tors, and N ∈ ��M� such that N ∈ ��. If Nhas �-Krull dimension, then

k��N� ≤ supk��N/E�+ 1 �E is essential in N and N/E ∈ ���

= supk��N/E�+ 1 �E is essential in N��

Proposition 2.8. Let M ∈ R-Mod, � ∈ M-tors, and N ∈ ��M�. If N has �-Krulldimension. Then N contains a submodule L such that L is -�-critical for an ordinal .

Proposition 2.9. Let M ∈ R-Mod, � ∈ M-tors, and N ∈ ��M�. Suppose that N ∈ ��

and N has �-Krull dimension. Then N has finite uniform dimension.

Proposition 2.10. Let M ∈ R-Mod, � ∈ M-tors, and N ∈ ��M�. Suppose that N isfinitely generated and M has �-Krull dimension. Then N has �-Krull dimension andk��N� ≤ k��M�.

Proof. Since N ∈ ��M� there exists an M-generated module X such that N ↪→ X.As X is M-generated there exist a set A and an epimorphism f � M�A� → X. LetL = ker f . Then X � �M�A��/L. So N ↪→ �M�A��/L. Since N is finitely generated, thenthere exists a finite subset B of A such that N ↪→ �M�B� + L�/L. On the other hand,we know ��M�B� + L�/L� � M�B�/�L ∩M�B��. Hence N ↪→ M�B�/�L ∩M�B��.

Thus k��N� ≤ k��M�B�/�L ∩M�B��� ≤ k��M

�B��. Since B is finite, then byLemma 2.3 ii� we have that k��M

�B�� = k��M�. Therefore, k��N� ≤ k��M�.

Corollary 2.11. Let M ∈ R-Mod, � ∈ M-tors, and N ∈ ��M�. Suppose that M and Nhave �-Krull dimension. Then k��N� ≤ k��M�.

Proof. We know that N = ∑x∈N Rx. As N has �-Krull dimension, then by [15,

Corollary 3.2.10] k��N� = supk��Rx� � x ∈ N�. Since Rx is finitely generated andRx ∈ ��M�, then by Proposition 2.10. k��Rx� ≤ k��M�. Therefore, k��N� ≤ k��M�.

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Proposition 2.12. Let M ∈ R-Mod, � ∈ M-tors, and N ∈ ��M�. Suppose that N is-�-critical. Then N is monoform.

Proof. Let 0 �= D be a submodule of N and f � D → N be a morphism. As N is-�-critical, then by Proposition 2.6, D is -�-critical. Since �D/ ker f� � Im f ⊆ N ,then k��Im f� = k��D/ ker f�. If ker f �= 0, then k��D/ ker f� < k��D� = k��N� = , acontradiction. Thus ker f = 0.

Corollary 2.13. Let M ∈ R-Mod, � ∈ M-tors, and N ∈ ��M�. Suppose that N has�-Krull dimension. Then L contains a monoform submodule for all 0 �= L ⊆ N .

Proof. Let 0 �= L ⊆ N . By Lemma 2.3 and Proposition 2.8, L contains a submoduleL′ such that L′ is -�-critical, for an ordinal . Hence L′ is monoform.

3. � -KRULL DIMENSION AND CLASSICAL � -KRULL DIMENSION OF AMODULE M

Let M ∈ R-Mod and � ∈ M-tors.For X ⊆ HomR�M�M�, let �X�

= ⋂f∈Xker f � ker f is �-pure in M�. Now we

consider the set �− �M = �X��X ⊆ HomR�M�M��.

Definition 3.1. Let M ∈ R-Mod and � ∈ M-tors. We say M satisfies ascendingchain condition (ACC) on �-pure annihilators, if �− �M satisfies ACC.

If � = , then we simply will say M satisfies ACC on annihilators.It is clear, that if � ≤ �′ ∈ M-tors and M satisfies ACC on �-pure annihilators,

then M satisfies ACC on �′-pure annihilators.Notice that, if N is a submodule of M and AnnM�N� is �-pure in M , then

by [3, Definition 1.1] we have that AnnM�N� ∈ �− �M . Also note that if M = R andX ⊆ Hom�R�R�, then

⋂f∈X ker f is the left annihilator of the set S = f�1� � f ∈ X�,

namely AnnR�S� =⋂

f∈X ker f . Therefore, if � = ∈ R-tors, then − �R is the set ofleft annihilators of the subsets of R.

The following definition was given in [18, Definition 10].

Definition 3.2. Let M ∈ R-Mod and N �FI M . We say that N is semiprime in Mif for any K ⊆FI M such that KMK ⊆ N , then K ⊆ N . We say M is semiprime if 0 issemiprime in M .

If we define N 2 = NMN . Then by induction, for any integer n > 2, we defineNn = NM�N

n−1�. Note that if M is semiprime, then M does not have nonzerosubmodules N such that Nn = 0.

Remark 3.3. If M is projective in ��M� and L, K are submodules of M , then by [5,Definition 1.1] LMK = LMK where L = ∑

f�L� � f ∈ HomR�M�M�� is the minimalfully invariant submodule of M which contains L. Moreover, notice that L = LMM .We claim that, if M is projective in ��M� and N is a fully invariant submodule of M ,then N is semiprime submodule of M if and only if for any submodule K of M , K2 ⊆N implies that K ⊆ N . In fact, suppose that N is semiprime module, and let K be a

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KRULL DIMENSION AND CLASSICAL 3193

submodule of M such that K2 ⊆ N . As K2 = KMK = KMK, then KMK ⊆ N . Hence�KMK�MM ⊆ NMM = N , since N ⊆FI M . On the other hand, since M is projective in��M�, then by [3, Proposition 5.5] �KMK�MM = KM�KMM� = KMK. Thus KMK ⊆ N .As K is fully invariant submodule of M , then K ⊆ N . The converse is obvious.

The next result improves the Proposition 1.3 of [6].

Proposition 3.4. Let � ∈ M-tors. Suppose that M is �-torsion-free, projective in ��M�,and M satisfies ACC on �-pure annihilators. If M is a semiprime module, then M is nonM-singular.

Proof. Let ��M� = Z the M-singular submodule of M in ��M�. Suppose that Z �=0. As M is semiprime module then by Remark 3.3. ZMZ �= 0. Hence there is anonzero morphism j � M → Z. Thus ZMIm j �= 0. Now we consider the family

� = ker h �h ∈ Hom�M�Z� and ZMIm h �= 0�� Then ker j ∈ � . So � �= ∅�

On the other hand, since Z ⊆ M ∈ ��, then ker h is a �-pure submodule of M forall h ∈ Hom�M�Z�. As M satisfies ACC on �-pure annihilators, then the family �has maximal elements. Let f ∈ Hom�M�Z� such that ker f is a maximal elementof � . Now let g ∈ Hom�M�Z�, and consider the composition �g � f� � M → Z. LetK = ker f . Then K ⊆ ker�g � f�. By [6, Proposition 1.2], we have that ker g ⊆ess

M . As 0 �= Im f ⊆ Z ⊆ M , then ker g ∩ Im f �= 0. So there exists m ∈ M such that0 �= f�m� ∈ ker g. Hence m � ker f . Thus K � ker�g � f�. As K is maximal in �and �g � f� ∈ Hom�M�Z�, then ZMIm �g � f� = 0. As Im �g � f� ⊆ Z, then by [5,Proposition 1.3] Im �g � f�MIm �g � f� ⊆ ZMIm �g � f� = 0. So Im �g � f� = 0 sinceM is semiprime. Hence g�f�M�� = 0. So we have shown that f�M�MZ = 0. Again,as f�M� ⊆ Z, then by [5, Proposition 1.3] f�M�Mf�M� ⊆ f�M�MZ = 0. Since M issemiprime, then f�M� = 0 a contradiction. Therefore, M is non-M-singular. �

Lemma 3.5. Let M ∈ R-Mod. Suppose that M is non M-singular, then the followingconditions hold:

i) For all ∅ �= X ⊆ HomR�M�M�, N = ⋂f∈X ker f does not have essential extensions

in M;ii) If M is a projective in ��M� and f , g ∈ HomR�M�M�, such that f�M� ⊆ess M ,

g�M� ⊆ess M , then f�M�Mg�M� ⊆ess M .

Proof. i) Let X ⊆ HomR�M�M� and M ′ ⊆ M such that⋂

f∈X ker f ⊆ess M′. Let

g ∈ X. As⋂

f∈X ker f ⊆ess M′. Then ker g ∩M ′ ⊆ess M

′. We claim ker g ∩M ′ = M ′.In fact, suppose that ker g ∩M ′ �= M ′. Let g � M ′ → M be the restriction of themorphism g. Then ker g = ker g ∩M ′. Thus �M ′/ ker g� ↪→ M , a contradiction sinceM ′/ ker g is M-singular and M is non-M-singular. So ker g ∩M ′ = M ′. Hence wehave that

⋂f∈X ker f = �

⋂f∈X ker f� ∩M ′ = ⋂

f∈X�ker f ∩M ′� = M ′.

ii) Let 0 �= N ⊆ g�M�. By [3, Proposition 5.5], f�M�Mg�M� =∑h∈Hom�M�g�M�� h�f�M��. Since g � M → M , then g ∈ Hom�M� g�M��. As 0 �= N ⊆

g�M�, then there exists 0 �= L ⊆ M , such that g�L� = N . Since f�M� ⊆ess M , then

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f�M� ∩ L �= 0. We claim that f�M� ∩ L � ker g. In fact, suppose that f�M� ∩ L ⊆ker g. As 0 �= N = g�L�, then ker g � L. On the other hand, we know that f�M� ⊆ess

M . Hence f�M� ∩ L ⊆ess M ∩ L = L. Thus ker g ⊆ess L. Now let g � L → M bethe restriction of the morphism g. Hence we have that ker g = ker g ∩ L = ker g.Therefore, �L/ ker g� ↪→ M , a contradiction since M is non-M-singular. Thusf�M� ∩ L � ker g. So 0 �= g�f�M� ∩ L� ⊆ g�f�M�� ∩ g�L� = g�f�M�� ∩ N . Thusf�M�Mg�M� ⊆ess g�M�.

Proposition 3.6. Let M ∈ R-Mod. Suppose that ��M� = 0 and M has finite uniformdimension. Then M satisfies the ascending chain condition on annihilators.

Proof. Suppose that there is an infinite chain of annihilators⋂f∈X1

ker f �⋂f∈X2

ker f � · · · � ⋂f∈Xn

ker f � · · ·

where Xi ⊆ HomR�M�M� for all i. As M is non-M-singular, then by Lemma 3.5,⋂f∈X1

ker f is not essential in⋂

f∈X2ker f . So there exists 0 �= N1 ⊆

⋂f∈X2

ker f suchthat

⋂f∈X1

ker f ⊕ N1 ⊆⋂

f∈X2ker f . Again since

⋂f∈X2

ker f �⋂

f∈X3ker f , then

there exist 0 �= N2 ⊆⋂

f∈X3ker f such that

⋂f∈X2

ker f ⊕ N2 ⊆⋂

f∈X3ker f . Hence we

have that⋂

f∈X1ker f ⊕ N1 ⊕ N2 ⊆

⋂f∈X3

ker f . In this way we can give the module⋂f∈X1

ker f ⊕ N1 ⊕ N2 ⊕ N3 ⊕ · · · ⊕ Nn ⊕ · · · ⊆ M , a contradiction since M has finiteuniform dimension.

Notice that, if ��M� = 0 and M has finite uniform dimension, then M satisfiesthe ascending chain condition on �-pure annihilators for all � ∈ M-tors.

Proposition 3.7. Let M ∈ R-Mod such that M is a semiprime module. If 0 �= N is amonoform submodule of M , then there exists f ∈ Hom�M�N� such that ker f ∩ N = 0.

Proof. As M is semiprime, then NMN �= 0. Thus there exists a morphism f � M →N such that f�N� �= 0. Let f � N → N be the restriction of the morphism f . Thenf �N� = f�N� �= 0. Since N is monoform, then f is a monomorphism. On the otherhand, we have that ker f = ker f ∩ N . Hence ker f ∩ N = 0.

An R-module M is said to have enough monoforms if every nonzerosubmodule of M contains a monoform submodule.

Notice that, if N ∈ ��M� has �-Krull dimension, then by Corollary 2.13 wehave that N has enough monoforms.

Proposition 3.8. Let M ∈ R-Mod such that M is a semiprime module and projectivein ��M�. Suppose that M has enough monoforms. Then M is non-M-singular.

Proof. Suppose that ��M� �= 0. Then there exists 0 �= N ∈ ��M�, such that N isM-singular and f ∈ Hom�N�M� such that 0 �= f�N�. As f�N� is M-singular andf�N� ⊆ M , then we can suppose that there exists 0 �= M ′ ⊆ M such that M ′ isM-singular. Moreover, since M has enough monoforms, then we can suppose thatM ′ is monoform. By Proposition 3.6, there is F ∈ HomR�M�M� such that ker F ∩M ′ = 0. Since M ′ is M-singular, there exists an exact sequence 0 → K → L →

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KRULL DIMENSION AND CLASSICAL 3195

M ′ → 0 in ��M� such that K ⊆ess L. On the other hand, we know that M is projectivein ��M�. Hence there exists g � M → L such that the following diagram commutes:

As K ⊆ess L, then g−1�K� ⊆ess M . Since 0 �= M ′ ⊆ M , then g−1�K� ∩M ′ �= 0. Asg�g−1�K� ∩M ′� ⊆ K, then

0 = h�g�g−1�K� ∩M ′�� = �h � g��g−1�K� ∩M ′� = F�g−1�K� ∩M ′��

Thus g−1�K� ∩M ′ ⊆ ker F . Hence ker F ∩M ′ �= 0, a contradiction. Therefore,��M� = 0.

Corollary 3.9. Let M be projective in ��M�, � ∈ M-tors, and P prime in M such thatM/P ∈ ��. Suppose M has �-Krull dimension. Then the following conditions hold:

i) M/P has finite uniform dimension;ii) M/P has ACC on annihilators.

Proof. i) As M has �-Krull dimension, then by Lemma 2.3, M/P has �-Krulldimension. By Proposition 2.9, M/P has finite uniform dimension.

ii) As M/P has �-Krull dimension, then by Corollary 2.13, M/P hasenough monoforms. By [16, Proposition 18], M/P is prime module. Hence byProposition 3.8, M/P is non-M/P-singular. By i�, we have that M/P has finiteuniform dimension. Hence by Proposition 3.6, M/P satisfies ACC on annihilators.

Proposition 3.10. Let M ∈ R-Mod, � ∈ M-tors. If k��M� ≥ 0, then k��M/f�M�� <k��M� for all monomorphisms f � M → M

Proof. If f�M� = M , then the result is clear. Suppose that f�M� � M . We claimthat f�f�M�� � f�M�. In fact as f�M� � M , then there exists m ∈ M such that m �f�M�. Now if f�m� ∈ f�f�M��, then there exists x ∈ f�M� such that f�x� = f�m�.Inasmuch as f is monomorphism, then x = m. Hence m ∈ f�M�, a contradiction.Therefore, f�f�M�� � f�M�. Analogously, we may show that f�f�f�M��� � f�f�M��.

If we denote f �n� =n︷ ︸︸ ︷

f � f � · · · � f , then f �n+1��M� � f �n��M�. Thus we have the chain

M � f�M� � f�f�M�� � · · · � f �n��M� � f �n+1��M� � · · · �

On the other hand, we know that Mf→ f�M�

f→ f�f�M��f→ · · · f→ f �n��M�

f→f �n+1��M�

f→ · · · are monomorphisms.

We claim that M/f�M� � f�M�/f�f�M��. In fact, we consider Mf→ f�M�

�→f�M�/f�f�M�� which is monomorphism, where � is the natural projection. So�� � f��f�M�� = ��f�f�M��� = 0. Thus f�M� ⊆ ker � � f . Now let m ∈ ker�� � f�.

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Then 0 = �� � f��m� = ��f�m��. So f�m�+ f�f�M�� = 0. Hence f�m� ∈ f�f�M��.Thus there exists x ∈ f�M� such that f�m� = f�x�. As f is monomorphism,then m = x. Hence m ∈ f�M�. Thus ker�� � f� = f�M�. So we have shown thatM/f�M� � f�M�/f�f�M��. Analogously, we can prove that f �n−1��M�/f �n��M� �f �n��M�/f �n+1��M�.

Now we denote by f �n��M� the �-purification of f �n��M� in M for all n ∈ . Sowe obtain the chain

M ⊇ f�M� ⊇ f�f�M�� ⊇ · · · ⊇ f �n��M� ⊇ f �n+1��M� � · · · �

Since M has �-Krull dimension, there exists n0 ∈ such that

k��f�i��M�/f �i+1��M�� < k��M� for all i � n0�

On the other hand, by Lemma 2.4, we know that k��f�i��M�/f �i+1��M�� =

k��f�i��M�/f �i+1��M��. Thus k��f

�i��M�/f �i+1��M�� < k��M� for all i � n0. SinceM/f�M� � f�M�/f�f�M�� � · · · � f �i��M�/f �i+1��M�, then k��M/f�M�� < k��M�.

Lemma 3.11. Let M ∈ R-Mod such that M is non-M-singular. Let f , g ∈HomR�M�M� such that f�M� ⊆ess M and g�M� ⊆ess M . Then f�g�M�� ⊆ess M .

Proof. Let 0 �= N ⊆ f�M� and L = f−1�N�. As g�M� ⊆ess M , then g�M� ∩ L �= 0.Moreover, g�M� ∩ L ⊆ess L. We claim that g�M� ∩ L � ker f . In fact, supposethat g�M� ∩ L ⊆ ker f . Since ker f ⊆ L, then ker f ⊆ess L. Now let f : L → Nbe the restriction of f . So ker f = ker f ∩ L = ker f . Thus ker f ⊆ess L. Hence�L/ ker f � ↪→ N ⊆ M , a contradiction since M is non-M-singular. So we haveshown that g�M� ∩ L � ker f . Therefore, 0 �= f�g�M� ∩ L� ⊆ f�g�M�� ∩ f�L� ⊆f�g�M�� ∩ N . Hence f�g�M�� ⊆ess f�M�.

Proposition 3.12. Let M ∈ R-Mod be a semiprime module such that M is projectivein ��M�. Let � ∈ M-tors, and suppose that M ∈ �� and M satisfies ACC on �-pureannihilators. If f ∈ HomR�M�M� and f�M� ⊆ess M , then f is monomorphism

Proof. By Proposition 3.4, M is non-M-singular. Now we consider the chain

ker f ⊆ ker�f � f� ⊆ ker�f � f � f� ⊆ · · · ⊆ ker f �n� ⊆ · · · �

As M ∈ ��, then ker f �n� is �-pure in M for all n ∈ . So there exists n0 ∈ such that ker f �n0� = ker f �n0+1�. By Lemma 3.11, we know that f �n0��M� ⊆ess M . Iff is not monomorphism, then ker f �= 0. Thus f �n0��M� ∩ ker f �= 0. Let 0 �= K =f �n0��M� ∩ ker f . Then there exists a submodule T of M such that K = f �n0��T�.Hence 0 = f�K� = f�f �n0��T�� = f �n0+1��T�. Thus T ⊆ ker f �n0+1� = ker f �n0�. So K =f �n0��T� = 0, a contradiction. Therefore, f is monomorphism.

Proposition 3.13. Let M ∈ R-Mod be a semiprime module such that M is aprogenerator in ��M�. Let � ∈ M-tors, and suppose that M ∈ �� such that M satisfiesACC on �-pure annihilators and M has finite uniform dimension. Then for everyN ⊆ess M , there exists f � M → N such that f is monomorphism.

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KRULL DIMENSION AND CLASSICAL 3197

Proof. Since N ⊆ess M , then by Proposition 3.12, it is enough to show thatthere exists f ∈ Hom�M�N� such that f�M� ⊆ess N . Let 0 �= f1 ∈ Hom�M�N�. Iff1�M� ⊆ess N , then we have finished. Suppose that f1�M� is not essential in N .Then there exists 0 �= N1 ⊆ N such that f1�M� ∩ N1 = 0. Moreover, we can supposethat N1 is a pseudocomplement of f1�M� in N . Thus f1�M�⊕ N1 ⊆ess N . Nowlet 0 �= f2 ∈ Hom�M�N1�. If f2�M� ⊆ess N1, then f1�M�⊕ f2�M� ⊆ess f1�M�⊕ N1.As f1�M�⊕ N1 ⊆ess N , then f1�M�⊕ f2�M� ⊆ess N . Since �f1 + f2��M� = f1�M�⊕f2�M�, then �f1 + f2��M� ⊆ess N . As �f1 + f2� ∈ Hom�M�N�, then we have finished.Now if f2�M� �ess N1, then there exists 0 �= N2 ⊆ N1 such that f2�M� ∩ N2 = 0.Moreover, we can suppose that f2�M�⊕ N2 ⊆ess N1. Thus f1�M�⊕ f2�M�⊕ N2 ⊆ess

f1�M�⊕ N1 ⊆ess N . As M has finite uniform dimension, then this process isfinite. Therefore, there exists n ∈ such that f1�M�⊕ f2�M�⊕ · · · ⊕ fn�M� ⊆ess N .Since �f1 + f2 + · · · + fn��M� = f1�M�⊕ f2�M�⊕ · · · ⊕ fn�M�, then �f1 + f2 + · · · +fn��M� ⊆ess N .

As �f1 + f2 + · · · + fn� ∈ Hom�M�N�, then we have finished.

Theorem 3.14. Let M ∈ R-Mod be a semiprime module such that M is aprogenerator in ��M�. Let � ∈ M-tors, and suppose that M has �-Krull dimension andM ∈ ��. Then k��M� = supk��M/N�+ 1 �N ⊆ess M and M/N ∈ ���.

Proof. By Proposition 2.7, we have that

k��M� ≤ supk��M/N�+ 1 �N ⊆ess M and M/N ∈ ����

Now let N ⊆ess M such that 0 �= M/N ∈ ��. As M has �-Krull dimension, then byProposition 2.9 we know that M has finite uniform dimension. Since M has �-Krulldimension, then by Corollary 2.13, M has enough monoforms. By Proposition 3.8,we have that ��M� = 0. Moreover, since M has finite uniform dimension, byProposition 3.6, M satisfies the ascending chain condition on annihilators; inparticular, M satisfies ACC on �-pure annihilators.

On the other hand, since N ⊆ess M and M ∈ ��, then by Proposition 3.13there exists f � M → N such that f is monomorphism. As f�M� ⊆ N � M ,then by Proposition 3.10 we have that k��M� > k��M/f�M��. Moreover, byLemma 2.3, we have that k��M/f�M�� ≥ k��M/N�. Thus k��M� > k��M/N�. Hencewe obtain that k��M� ≥ k��M/N�+ 1. Therefore, k��M� ≥ supk��M/N�+ 1 �N ⊆ess

M and M/N ∈ ���. So we have shown that k��M� = supk��M/N�+ 1 �N ⊆ess

M and M/N ∈ ���.Notice that if M is a prime module, then N ⊆ess M for all 0 �= N fully invariant

submodules of M . In fact let 0 �= L ⊆ M . As N is a fully invariant submodule of M ,then NML ⊆ N ∩ L. So NML �= 0, since M is prime. Therefore, N ∩ L �= 0.

The following example shows that the equality in Theorem 3.14 is not true ingeneral.

Example 3.15. Let R = 2 � �2 ⊕ 2�. The trivial extension of 2 by 2 ⊕ 2.This ring can be described as

R ={(

a �x� y�0 a

)∣∣∣∣ a ∈ 2� �x� y� ∈ 2 ⊕ 2

}�

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3198 CASTRO PÉREZ AND RÍOS MONTES

R has only one maximal ideal I = {(0 �x�y�0 0

) � �x� y� ∈ 2 ⊕ 2

}and 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 following form:

R is artinian and R-Mod has only one simple module. Let S be the simplemodule. By [17, Theorem 2.13], we know that there is a lattice anti-isomorphismbetween the lattice of ideals or R and the lattice of fully invariant submodulesof E�S�. Thus the lattice of fully invariant submodules of E�S� has tree maximalelements, K, L, and N . Moreover, E�S� contains only one simple module S.Therefore, the lattice of fully invariant submodules of E�S� has the following 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 inM . Also we note that KMK = S and SMK = 0. Thus AnnM�K� = S. Moreover, sinceSMK = 0, then SMS = 0. So by Definition 3.2, M is not semiprime.

We claim that KMS = S. In fact, if � is the natural projection from M ontoM/N , then ��K� = S since M/N � S. Thus KMS = S. Analogously, we prove thatLMS = S and NMS = S. Moreover, it is not difficult to prove that AnnM�S� = K ∩L ∩ N = S. So S is not prime in M . Notice that Soc�R� = J1 + J2 + J3 = J1 ⊕ J2and Soc�R� ⊂ess 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�H� �= 0 for all H ∈ ��M�, but M is not a generator of ��M�.

From the previous arguments, we can conclude as follows:

1) M is not semiprime;2) M is finite. Therefore, M/N is artinian module for all N ⊆ M . Hence k �M/N� =

0 for all N � M ;

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KRULL DIMENSION AND CLASSICAL 3199

3) M is a uniform module.So 0 = k �M� �= 1 = supk �M/N�+ 1 � 0 �= N ⊆ M�= supk �M/N�+ 1 �N ⊆ess M and M/N ∈ � �.

Proposition 3.16. Let M ∈ R-Mod be a prime module such that M is a progeneratorin ��M� and � ∈ M-tors. Suppose that M ∈ �� and M has �-Krull dimension. Thenk��M/N� < k��M� for all 0 �= N fully invariant submodules of M .

Proof. By Theorem 3.14, we have that

k��M� = supk��M/N�+ 1 �N ⊆ess M and M/N ∈ ����

Now let 0 �= N be a fully invariant submodule of M . As M is prime, then N ⊆ess

M . By Lemma 2.4, we know that k��M/N� = k��M/N�. Thus k��M/N�+ 1 ≤ k��M�.Hence k��M/N� < k��M�.

The following example shows that the inequality in Proposition 3.16 is not truein general.

Example 3.17. We consider the -module M = p� , where p is a prime number.We claim that 0 is not prime in M . In fact, as Hom�M�K� = 0 for all K � M ,then pnMpm = 0, but pn �= 0 and pm �= 0. So M is not prime. Now let N ⊆ M .We know that N is a fully invariant submodule of M . Moreover M � M/N for allN � M . Since M is artinian then k �M� = 0. Therefore k �M/N� = 0 for all N � M .Hence k �M/N� ≮ k �M�. Notice that in this example M = p� is not a generator ofthe category ��M�.

Proposition 3.18. Let M ∈ R-Mod such that M is a progenerator in ��M� and � ∈M-tors. If M has �-Krull dimension, then Spec��M� is a noetherian poset.

Proof. If Spec��M� = ∅, then we have the result. Suppose that Spec��M� �= ∅.Let P1, P2 ∈ Spec��M� such that P1 � P2. Since P1 is prime in M , then by [16,Proposition 18] M/P1 is a prime module. Moreover, by [6, Proposition 1.5],we have that M/P1 is progenerator of the category ��M/P1�. Since P1 and P2

are fully invariant submodules of M , then P2/P1 is a fully invariant submoduleof M/P1. On the other hand, we know M has �-Krull dimension. Then M/P1

has �-Krull dimension. As M/P1 ∈ ��, then M/P1 ∈ ��P1 . Since M/P1 is primemodule and P2/P1 is fully invariant submodule of M/P1, then by Proposition 3.4,we have that k�P1 ��M/P1�/�P2/P1�� < k�P1 �M/P1�. Hence k���M/P1�/�P2/P1�� <k��M/P1�. Since M/P2 � �M/P1�/�P2/P1�, then k��M/P2� < k��M/P1�. Consequently,a strictly ascending chain of �-pure submodules prime in M leads to a strictlydecreasing sequence of ordinals. Since the latter must be finite, the assertion follows.

Notice that in the Example 3.15, Spec��M� = ∅ for all � ∈ M-tors.

Corollary 3.19. Let M ∈ R-Mod such that M is a progenerator in ��M� and � ∈ M-tors. If M has �-Krull dimension, then M has classical �-Krull dimension.

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3200 CASTRO PÉREZ AND RÍOS MONTES

Proof. By Proposition 3.18, we have Spec��M� is noetherian. Then by Lemma 1.1,M has classical �-Krull dimension.

The following example shows that the inverse of Corollary 3.19 is false.

Example 3.20. Let R = and � = p ∈ �p is prime �. We consider the R-module M = ⊕

p∈� p. It is clear that if 0 �= N is a submodule of M , then there exists�′ ⊆ � such that N = ⊕

p∈�′ p. Moreover, N is a fully invariant submodule of Mfor all N ⊆ M .

Now for q ∈ �, let �q = �− q� and Nq =⊕

p∈�qp. So Nq is a maximal

fully invariant submodule of M for all q ∈ �. On the other hand, it is clear M is aprogenerator of the category ��M�. Hence by [5, Proposition 1.13], Nq is prime in M

for all q ∈ �. We claim that the modules Nq are the only submodules prime in M .In fact, let N be a submodule of M such that N �= Nq for any q ∈ �. Thus there areq, q′ distinct prime numbers and �′ ⊆ � such that N = ⊕

p∈�′ p and q, q′ are notelements of �′. So �q�M�q′� = 0. Hence �q�M�q′� ⊆ N but q � N and q′ � N .Therefore, N is not prime in M . So Spec �M� = Nq � q ∈ ��. Since every Nq is amaximal fully invariant submodule of M , then Nq is maximal submodule prime inM for all q ∈ �. Hence cl�K� dim�M� = 0. On the other hand, we know that M hasno finite uniform dimension. Hence M has no �-Krull dimension for all � ∈ M-tors,in particular for � = .

Theorem 3.21. Let M ∈ R-Mod such that M is a progenerator in ��M� and � ∈ M-tors. If M has �-Krull dimension, then cl�K� dim�M� ≤ k��M�.

Proof. By Corollary 3.19, we know that cl�K� dim�M� exists. Let us supposethat k��M� = . If = −1, then M ∈ ��. So Spec��M� = ∅. Hence cl�K� dim�M� =−1. Now suppose that M � ��. Thus k��M� = ≥ 0. If Spec��M� = ∅, thencl�K� dim�M� = −1. So we have the result. Now we will show by inductionon that cl�K� dim�M� ≤ . Let = 0. So cl�K� dim�M� < k��M�. Now supposethat Spec��M� �= ∅. We claim that P is maximal �-pure prime in M for all P ∈Spec��M�. In fact let P, Q ∈ Spec��M� such that P � Q. So Q/P is an invariantsubmodule of M/P. By [16, Proposition 18], M/P is a prime module. Hence Q/P ⊆ess

M/P. As M/P ∈ ��, then by Proposition 3.16, k���M/P�/�Q/P�� < k��M/P�. Hencek��M/Q� < k��M/P� ≤ k��M� = = 0. Thus k��M/Q� = −1. So M/Q ∈ ��. ButM/Q ∈ ��. Since Q is prime in M , Q � M , a contradiction. Hence P is a maximal�-pure submodule prime in M for all P ∈ Spec��M�. Therefore, cl�K� dim�M� = 0.So we have the result. Suppose that > 0 and that the conclusion holds for all� < . Let P, Q ∈ Spec��M� such that P � Q. Newly by Proposition 3.16 we havethat k���M/P�/�Q/P�� < k��M/P�. Hence k��M/Q� < k��M/P� ≤ k��M� = . If � =k��M/Q�, then by induction hypothesis cl�K� dim�M/Q� ≤ � = k��M/Q� < . By thedefinition of classical �-Krull dimension, we have that Spec�� �M/Q� = Spec��M/Q�.As M/Q is prime, then Q/Q = 0 ∈ Spec��M/Q�. Hence Q/Q = 0 ∈ Spec�� �M/Q�. ByProposition 1.6, Q ∈ Spec�� �M�. Therefore, P ∈ Spec� �M�. So Spec��M� ⊆ Spec� �M�.Thus Spec��M� = Spec� �M�. Hence cl�K� dim�M� ≤ .

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KRULL DIMENSION AND CLASSICAL 3201

Remark 3.22. By [16, Proposition 18], we know that if M is projective in ��M� andP is a proper fully invariant submodule of M , then P is prime in M if and only ifM/P is a prime module. We claim that a proper fully invariant submodule P of M isprime in M if and only if for every L, K fully invariant submodules of M such thatP � L and P � K, then KML � P. In fact, let L, K be fully invariant submodules ofM such that P � L and P � K, then by [16, Definition 13] KML � P. Inversely, letK/P and L/P be fully invariant submodules of M/P such that �K/P�M�L/P� = 0 =P/P, then by Proposition 1.3, KML ⊆ P. If K/P �= 0 and L/P �= 0, then P � K andP � L . By hypothesis, we have that KML � P, a contradiction. Therefore, K/P = 0or L/P = 0. So M/P is a prime module.

Lemma 3.23. Let M ∈ R-Mod such that M is noetherian and projective in ��M�. Thenfor all fully invariant submodules N of M there exist P1� P2� � � � � Pn such that N ⊆ Pi

for 1 ≤ i ≤ n and �Pn�M�Pn−1�M � � � �P2�M�P1� ⊆ N .

Proof. If N is a submodule prime in M , then we have the result. If N is notprime in M , then by Remark 3.22, there exist L and K fully invariant submodulesof M such that N � L, N � K, and LMK ⊆ N . If L and K are prime in M ,then we have the result. If L is not prime in M , there exist L1 and L2 fullyinvariant submodules in M such that L � L1, L � L2, but �L1�M�L2� ⊆ L. By [5,Proposition 1.3], we have that ��L1�M�L2��MK ⊆ LMK. So ��L1�M�L2��MK ⊆ N .Moreover, N � L � L1. As M is noetherian the chain N � L � L1 � · · · must befinite. So there exists a fully invariant submodule Ln of M such that Ln is primein M and �Ln�M�Ln−1�M � � � M�K� ⊆ N . Therefore, there exist P1� P2� � � � � Pn such thatN ⊆ Pi for 1 ≤ i ≤ n and �Pn�M�Pn−1�M � � � �P2�M�P1� ⊆ N .

Lemma 3.24. Let M ∈ R-Mod such that M is a progenerator in ��M�. Let � ∈ M-tors.Suppose that P is a submodule prime in M such that M/P satisfies ACC on annihilatorsand has finite uniform dimension. Then P is either �-pure or �-dense.

Proof. Suppose that P is not �-pure in M . Then 0 �= t��M/P� = N/P. We claim thatN/P ⊆ess M/P. In fact, if N/P is not essential in M/P, then there exists 0 �= L/P ⊆M/P such that �N/P� ∩ �L/P� = 0. As N/P is a fully invariant submodule of M/P,then by [5, Remark 1.2] �N/P�M/P�L/P� ⊆ �N/P� ∩ �L/P� = 0. Hence we have that�N/P�M/P�L/P� = 0. As P is prime in M , then By [16, Proposition 18] M/P is a primemodule. Thus N/P = 0 or L/P = 0, a contradiction. Therefore, N/P ⊆ess M/P. Onthe other hand by [6, Proposition 1.5], we have that M/P is progenerator in ��M/P�.Since M/P satisfies ACC on annihilators and has finite uniform dimension, then byProposition 3.13 there exists a monomorphism f � M/P → N/P. So M/P ↪→ N/P.Thus M/P ∈ ��. Hence P is �-dense in M .

Notice that if M is noetherian, then M is �-noetherian for all � ∈ M-tors. ThusM has �-Krull dimension for all � ∈ M-tors.

Proposition 3.25. Let M be an R-module projective in ��M�, � ∈ M-tors. If M isnoetherian, then there exists a submodule P prime in M such that k��M� = k��M/P�.Further, if M is not �-torsion, then P is �-pure in M .

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3202 CASTRO PÉREZ AND RÍOS MONTES

Proof. By Lemma 3.23, there exist submodules Pn, Pn−1� � � � � P1, prime in Msuch that �Pn�M�Pn−1�M � � � �P2�M�P1� = 0. Now we consider the chain M ⊇ P1 ⊇�P2�M�P1� ⊇ �P3�M�P2�M�P1� ⊇ · · · ⊇ �Pn�M�Pn−1�M � � � �P2�M�P1� = 0. By propertiesof the �-Krull dimension, we have that

k��M� = supk��M/P1�� k��P1��

= supk��M/P1�� k��P1/��P2�M�P1���� k���P2�M�P1���

= supk��M/P1�� k��P1/��P2�M�P1����

k����P2�M�P1��/��P3�M�P2�M�P1���� � � � � k���Pn−1�M�Pn−2�M � � � �P2�M�P1����

On the other hand, by [5, Proposition 1.5] and [3, Proposition 5.5], we havethat P1/��P2�M�P1�� ∈ ��M/P2�. Since P1 and P2 are fully invariant submodulesof M , then by Corollary 2.11 a submodule of M/P2. Therefore, we have thatk��P1/��P2�M�P1��� ≤ k��M/P2�. Analogously, we have that

��Pn−1�M�Pn−2�M � � � � � � �P2�M�P1��/��Pn�M�Pn−1�M � � � �P2�M�P1�� ∈ ��M/Pn�

and

k����Pn−1�M�Pn−2�M � � � � � � �P2�M�P1��/��Pn�M�Pn−1�M � � � �P2�M�P1��� ≤ k��M/Pn��

So we have shown that k��M� ≤ supk��M/P1�� k��M/P2�� � � � � k��M/Pn�� ≤ k��M�.Thus there exists Pi such that k��M� = k��M/Pi�. Since M is noetherian, then Mhas �-Krull dimension and M/P ∈ ��, so by Corollary 3.9, M/Pi has finite uniformdimension and satisfies ACC on annihilators. Moreover, since M is not �-torsion,we have k��M� > −1. Thus k��M/Pi� > −1. So M/Pi is not �-torsion. Thus byLemma 3.24, we have that Pi is �-pure in M .

The following definition was given in [6, Definition 2.1], and we include it herefor the convenience of the reader.

Definition 3.26. Let M ∈ R-Mod and � ∈ M-tors. The module M is �-bounded ifevery �-pure essential submodule of M contains a nonzero fully invariant submoduleof M . M is fully �-bounded if, for every submodule P prime in M , the module M/Pis �-bounded.

Notice that if M is projective in ��M�, fully �-bounded, and N is a fullyinvariant submodule of M , then M/N is fully �-bounded. In fact. Let P/Nsubmodule prime in M/N . Then by Corollary 1.4, P is prime in M . Hence�M/N�/�P/N� � M/P is �-bounded.

Theorem 3.27. Let M ∈ R-Mod and � ∈ M-tors. Suppose that M is a noetherian,progenerator of ��M�, �-fully bounded, and M ∈ ��. Then

cl�K� dim�M� = k��M��

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KRULL DIMENSION AND CLASSICAL 3203

Proof. Let M ∈ ��. We proceed by induction on k��M�. Let k��M� = , and weassume first that M is a prime module. If = 0, then k��M� = 0. By Theorem 3.21,we have that cl�K� dim�M� = 0. Now let > 0, and suppose that for any ′ < ,every noetherian �-torsionfree, �-fully bounded, prime module M ′, and progeneratorof ��M ′� with k��M

′� = ′ satisfies cl�K� dim�M ′� = ′. By Theorem 3.21, we havethat cl�K� dim�M� ≤ k��M�. Suppose that cl�K� dim�M� = � < = k��M�. FromTheorem 3.14, we have that

k��M� = supk��M/N�+ 1 �N ⊆ess M and M/N ∈ ����

Hence there exists N ⊆ess M such that M/N ∈ �� and � < k��M/N�+ 1 ≤ . So � ≤k��M/N� ≤ . On the other hand, since M is �-fully bounded and N is �-pure in M ,there exists I ⊆ N such that I is a fully invariant submodule of M . As M is prime,then I ⊆ess M . Thus � ≤ k��M/N� ≤ k��M/I� = k��M/I�, where I is the �-purificationof I in M . Since I ⊆ess M , then I ⊆ess M . Hence k��M/I�+ 1 ≤ . So k��M/I� < .As M is noetherian, then M/I is noetherian. So by Proposition 3.25, there exists a �-pure submodule P/I prime in M/I such that k��M/I� = k���M/I�/�P/I�� = k��M/P�.As I is a fully invariant submodule of M and P/I is a fully invariant submoduleof M/I , then by [3, Proposition 5.1 and 5.2] P is a fully invariant submodule of M .By [16, Proposition 18], we have that P is prime in M with M/P ∈ ��. Moreover,by [6, Proposition 1.5], M/P is progenerator of ��M/P�. Therefore, we have shownthat � ≤ k��M/P� ≤ . By induction hypothesis, cl�K� dim�M/P� = k��M/P�. But byProposition 1.6, � = cl�K� dim�M� > cl�K� dim�M/P� ≥ �, a contradiction.

Now if M is not prime, then by Proposition 3.25, there exists a submoduleP prime in M with M/P ∈ �� such that k��M� = k��M/P�. By [16, Proposition 18]and [6, Proposition 1.5], M/P is a prime module and a progenerator of��M/P�. Moreover, as M is noetherian, �-fully bounded, then M/P is noetherianand �-fully bounded. Therefore, we have that cl�K� dim�M/P� = k��M/P�. ByProposition 1.6, cl�K� dim�M/P� ≤ cl�K� dim�M�. Thus k��M� ≤ cl�K� dim�M�. Nowby Theorem 3.21, we have that k��M� = cl�K� dim�M�.

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3204 CASTRO PÉREZ AND RÍOS MONTES

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