aa r X i v : . [ m a t h . R A ] M a y On completely prime submodules
David SsevviiriDepartment of MathematicsMakerere University, P.O BOX 7062, Kampala UgandaEmail: [email protected], [email protected]
Abstract
The formal study of completely prime modules was initiated by N. J. Groe-newald and the current author in the paper; Completely prime submodules,
Int. Elect. J. Algebra , , (2013), 1–14. In this paper, the study of completelyprime modules is continued. Firstly, the advantage completely prime mod-ules have over prime modules is highlited and different situations that lead tocompletely prime modules given. Later, emphasis is put on fully completelyprime modules, (i.e., modules whose all submodules are completely prime).For a fully completely prime left R -module M , if a, b ∈ R and m ∈ M , then abm = bam , am = a k m for all positive integers k , and either am = abm or bm = abm . In the last section, two different torsion theories induced by thecompletely prime radical are given. Keywords : Domain; Prime module; Completely prime module; Completely primeradical; Torsion theory
MSC 2010
Mathematics Subject Classification:16S90, 16D60, 16D99
Completely prime modules were first formally studied in [8] as a generalization ofprime modules. These modules had earlier appeared informally in most cases asexamples in the works of: Andrunakievich [1], De la Rosa and Veldsman [5, p. 466,Section 5.6], Lomp and Pe˜na [12, Proposition 3.1], and Tuganbaev [18, p. 1840]which were published in the years 1962, 1994, 2000 and 2003 respectively. In [1] and[5] these modules were called modules without zero-divisors, in [12] they were notgiven any special name and [18] they were called completely prime modules. In thispaper just like in [8], we follow the nomenclature of Tuganbaev.
Definition 1.1
Let R be a ring. A left R -module M for which RM = { } is . completely prime if for all a ∈ R and every m ∈ M , am = 0 implies m = 0 or aM = { } ;2. completely semiprime if for all a ∈ R and every m ∈ M , a m = 0 implies aRm = { } ;3. prime if for all a ∈ R and every m ∈ M , aRm = { } implies m = 0 or aM = { } . A submodule P of M is a completely prime (resp. completely semiprime, prime) sub-module if the factor module M/P is a completely prime (resp. completely semiprime,prime) module. A completely prime module is prime but not conversely in gen-eral. Over a commutative ring, completely prime modules are indistinguishablefrom prime modules.
Example 1.1
We know that the ring R = M n ( F ) of all n × n matrices over a field F is prime but not completely prime, i.e., it is not a domain. Since for a unitalring R , R is prime (resp. completely prime) if and only if the module R R is prime(resp. completely prime), see [8, Proposition 2.4], we conclude that the R -module R (where R = M n ( F )) is prime but not completely prime.Example 1.2 below is motivated by Example 3.2 in [6]. Example 1.2
Let M = (cid:26)(cid:18) ¯0 ¯0¯0 ¯0 (cid:19) , (cid:18) ¯0 ¯0¯1 ¯1 (cid:19) , (cid:18) ¯1 ¯1¯0 ¯0 (cid:19) , (cid:18) ¯1 ¯1¯1 ¯1 (cid:19)(cid:27) where entries ofmatrices in M are from the ring Z = { ¯0 , ¯1 } of integers modulo 2 and R = M ( Z )the ring of all 2 × M is a prime R -module which isnot completely prime. Proof:
First, we show that M is simple and hence prime since all simple modulesare prime. Let r = (cid:18) a bc d (cid:19) ∈ R , then rM = (cid:26)(cid:18) ¯0 ¯0¯0 ¯0 (cid:19) , (cid:18) a ac c (cid:19) , (cid:18) b bd d (cid:19) , (cid:18) a + b a + bc + d c + d (cid:19)(cid:27) ⊆ M for any a, b, c, d ∈ Z . The would be non-trivial proper submodules, namely; N = (cid:26)(cid:18) ¯0 ¯0¯0 ¯0 (cid:19) , (cid:18) ¯1 ¯1¯0 ¯0 (cid:19)(cid:27) , N = (cid:26)(cid:18) ¯0 ¯0¯0 ¯0 (cid:19) , (cid:18) ¯0 ¯0¯1 ¯1 (cid:19)(cid:27) , and N = (cid:26)(cid:18) ¯0 ¯0¯0 ¯0 (cid:19) , (cid:18) ¯1 ¯1¯1 ¯1 (cid:19)(cid:27) are not closed under multiplication by R since for a and c odd, rN N , for b and d odd, rN N and for a odd but b, c, d even, rN N . Now, take a = (cid:18) (cid:19) ∈ R and m = (cid:18) ¯1 ¯1¯1 ¯1 (cid:19) ∈ M . It follows that am = 0 but aM = { } since a = (cid:18) (cid:19) (cid:18) ¯1 ¯1¯0 ¯0 (cid:19) = (cid:18) ¯1 ¯1¯0 ¯0 (cid:19) = 0. Thus, M is not completely prime.2 .1 Notation All modules considered are left unital modules defined over rings. The rings areunital and associative. Let M be an R -module. If S is a subset of M and m ∈ M \ S ,by ( S : m ) we denote the set { r ∈ R : rm ∈ S } . If N is a submodule of a module M , we write N ≤ M . If N ≤ M , ( N : M ) is the ideal { r ∈ R : rM ⊆ N } which isthe annihilator of the R -module M/N . For an R -module M , End R ( M ) denotes thering of all R -endomorphisms of M . This paper contains five sections. In Section 1, we give an introduction, define someof the notation used and describe how the paper is organized. In Section 2, we statethe advantage completely prime modules have over prime modules. They behave asthough they are defined over a commutative ring, a behaviour prime modules do nothave in general. The aim of Section 3 is two fold; we provide situations under whicha module becomes completely prime and furnish concrete examples for completelyprime modules. In Section 4, we define completely co-prime modules by drawingmotivation from how prime, completely prime and co-prime modules are defined. Achart of implications is established between completely co-prime modules, co-primemodules, completely prime modules and prime modules, see Proposition 4.2. InProposition 4.2 it is established that the notion of completely co-prime modules is thesame as that for fully completely prime modules, i.e., modules whose all submodulesare completely prime. Many other equivalent formulations for completely co-primemodules are given. It is shown that if M is a fully completely prime R -module, thenfor all a, b ∈ R and every m ∈ M , abm = bam , am = a k m for all positive integers k , and either am = abm or bm = abm . In Section 5, which is the last section, wegive two torsion theories induced by the completely prime radical of a module. Onthe class of IFP modules (i.e., modules with the insertion-of-factor property), thefaithful completely prime radical is hereditary and hence leads to a torsion theory,see Theorem 5.1. Lastly, we show in Theorem 5.2 that the completely prime radicalis also hereditary on the class of semisimple R -modules and therefore it inducesanother torsion theory. 3 Advantage of completely prime modules overprime modules
Where as prime modules form a much bigger class than that of completely primemodules, completely prime modules possess nice properties which prime moduleslack in general. Completely prime modules over noncommutative rings behave likemodules over commutative rings. In particular, they lead to the following propertieson an R -module M :P1. for all a, b ∈ R and m ∈ M , abm = 0 implies bam = 0;P2. for all subsets S of M and m ∈ M \ S , ( S : m ) is a two sided ideal of R ;P3. for all a ∈ R and m ∈ M , am = 0 implies arm = 0 for all r ∈ R ;P4. the prime radical of M coincides with its completely prime radical, i.e., theintersection of all prime submodules of M coincides with the intersection ofall its completely prime submodules.A module which satisfies property P1, P3 and P4 is respectively called symmetric,IFP (i.e., has insertion-of-factor property) and 2-primal. Properties P2 and P3 areequivalent. To prove the claims made in this section, one only needs to prove thefollowing implications for a module:completely prime ⇒ completely semiprime ⇒ symmetric ⇒ IFP ⇒ P of an R -module M is said to be symmetric (resp. IFP) if the module M/P is symmetric (resp. IFP).A comparison with what happens for rings indicates that these results on modulesare what one would expect. Every domain (completely prime ring) is reduced (i.e.,completely semiprime) so it is symmetric, IFP and 2-primal, see [13]. Note that theIFP condition is called SI in [13]. The notions of IFP and symmetry first existedfor rings before they were extended to modules. An R -module M is completely prime if and only if for all nonzero m ∈ M , (0 : m ) =(0 : M ). This characterisation is used in the proof of Proposition 3.1, in Example3.2, in Proposition 3.3 and in Section 4. Proposition 3.1
Let M be an R -module. If every nonzero endomorphism f ∈ End R ( M ) is a monomorphism, then M is a completely prime module. roof: Let r ∈ R such that r (0 : M ) and let 0 = m ∈ M . Then there exists n ∈ M such that rn = 0. The endomorphism g : M → M given by g ( x ) = rx isnonzero since g ( n ) = rn = 0. By hypothesis, g is a monomorphism. Thus, g ( m ) = rm = 0 since by assumption m = 0. So, r (0 : m ). Hence, (0 : m ) ⊆ (0 : M )which shows that (0 : m ) = (0 : M ) for all 0 = m ∈ M since the reverse inclusionalways holds.According to Reyes [15, Definition 2.1], a left ideal P of a ring R is completely prime if for any a, b ∈ R such that P a ⊆ P , ab ∈ P implies that either a ∈ P or b ∈ P . Example 3.1 If P is a left ideal of a ring R which is completely prime in the senseof Reyes, then R/P is a completely prime module and S = End R ( R/P ) is a domain.This is because, according to [15, Proposition 2.5], P is a completely prime left idealof R if and only if every nonzero f ∈ S := End R ( R/P ) is injective if and only if S isa domain and the right S -module R/P is torsion-free. Now apply Proposition 3.1.
Example 3.2
A torsion-free module is completely prime and faithful. If M istorsion-free, (0 : m ) = { } for all 0 = m ∈ M . So, (0 : M ) ⊆ (0 : m ) = { } andhence (0 : M ) = (0 : m ) = { } for all 0 = m ∈ M .Let N be a submodule of an R -module M , the zero divisor set of the R -module M/N is the setZd R ( M/N ) := { r ∈ R : there exists m ∈ M \ N with rm ∈ N } . In Proposition 3.2, we characterise completely prime submodules in terms of zerodivisor sets of their factor modules.
Proposition 3.2
A submodule N of an R -module M is a completely prime submod-ule if and only if ( N : M ) = Zd R ( M/N ) . In particular, Zd R ( M/N ) is a completelyprime ideal of R whenever N is a completely prime submodule of M . Proof:
Suppose ( N : M ) = Zd R ( M/N ), i.e., T m ∈ M \ N ( N : m ) = S m ∈ M \ N ( N : m ).Then this equality is possible if and only if the set { ( N : m ) : m ∈ M \ N } is asingleton. Thus, N is a completely prime submodule by [8, Proposition 2.5]. For theconverse, if N is a completely prime submodule, it follows by [8, Proposition 2.5] thatthe set { ( N : m ) : m ∈ M \ N } is a singleton. So, T m ∈ M \ N ( N : m ) = S m ∈ M \ N ( N : m )and ( N : M ) = Zd R ( M/N ). The last statement follows from the fact that, if N isa completely prime submodule of an R -module M , then ( N : M ) is a completelyprime ideal of R . 5 orollary 3.1 An R -module M is completely prime if and only if (0 : M ) = Zd R ( M ) . In particular, Zd R ( M ) is a completely prime ideal of R whenever M is acompletely prime module. Corollary 3.2 If M is a faithful completely prime module, then the set Zd R ( M ) isa domain. Proposition 3.3 If M is a uniform module, then M is completely prime if andonly if every cyclic submodule of M is a completely prime module. Proof:
The if part is clear. For the converse, we prove by contradiction. Supposethere exists 0 = m ∈ M such that (0 : M ) = (0 : m ), i.e., (0 : M ) ( (0 : m ).Then, there exists a ∈ R and 0 = x ∈ M such that am = 0 and ax = 0. Since M is uniform, there exists a nonzero element z such that z ∈ Rm ∩ Rx . z, m ∈ Rm and z, x ∈ Rx . Since by hypothesis, Rm and Rx are completely prime modules, itfollows that (0 : z ) = (0 : m ) = (0 : Rm ) and (0 : z ) = (0 : x ) = (0 : Rx ). Hence,(0 : m ) = (0 : x ) which contradicts the fact that am = 0 and ax = 0.Completely prime modules are generalizations of torsion-free modules. Torsion-free modules form the module analogue of domains. If M is a faithful completelyprime R -module, then R is a domain. We show in Propositions 3.5 and 3.6 (resp.Proposition 3.4) that under “suitable conditions” the R -module M is completelyprime whenever R (resp. End R ( M )) is a domain. We define a retractable module anda torsionless module first. A module M is retractable if for any nonzero submodule N of M , Hom R ( M, N ) = { } . An R -module M is torsionless if for each 0 = m ∈ M there exists f ∈ Hom R ( M, R ) such that f ( m ) = 0. Free modules, generators andsemisimple modules are retractable. Torsionless modules over semiprime rings arealso retractable, see [17, Sec. 2, p.685]. Proposition 3.4
Let M be a retractable R -module and S = End R ( M ) . If S is adomain, then M is a completely prime module. Proof:
By [19, Proposition 1.7], S is a domain if and only if any nonzero endo-morphism of M is a monomorphism. By Proposition 3.1, M is a completely primemodule. Proposition 3.5
Let M be a torsionless R -module, if R is a domain, then M is acompletely prime module. Proof:
Suppose am = 0 for some a ∈ R and m ∈ M but m = 0 and aM = { } . M torsionless implies f ( m ) = 0 for some f ∈ Hom R ( M, R ). Now, a = 0 and f ( m ) = 0imply af ( m ) = 0 since R is a domain. Thus, f ( am ) = 0 and am = 0 which is acontradiction. 6 xample 3.3 By [2, p. 477], a submodule of a projective module is a torsionlessmodule. Thus, if R is a domain, a submodule of a projective module is a completelyprime module by Proposition 3.5. Proposition 3.6
A free module M over a domain R is completely prime. Proof:
Suppose am = 0 for some a ∈ R and m ∈ M . If m = 0, M is a completelyprime module. Suppose m = 0. Then am = a P ni =1 r i m i = P ni =1 ( ar i ) m i = 0 forsome r i ∈ R and m i ∈ M with i ∈ { , , · · · , n } . M being free implies ar i = 0. m = 0 implies there exists j ∈ { , , · · · , n } such that r j = 0. ar j = 0 implies a = 0since R is a domain and r j = 0. Hence, aM = { } and M is completely prime. Recall that an R -module M for which RM = { } is:1. prime , if for all nonzero submodules N of M , (0 : N ) = (0 : M );2. completely prime , if for all nonzero elements m of M , (0 : m ) = (0 : M );3. co-prime [19], if for all nonzero submodules N of M , ( N : M ) = (0 : M ).These definitions motivate us to define completely co-prime modules. Definition 4.1 An R -module M for which RM = { } is completely co-prime if forall submodules N of M and all elements m ∈ M \ N , ( N : m ) = (0 : M ) . Proposition 4.1
For any R -module M , we have the following implications:completely co-prime ⇒ completely prime ⇒ prime. ⇓ co-prime Proof:
For { } 6 = N ≤ M and m ∈ M \ N , we have (0 : M ) ⊆ ( N : M ) ⊆ ( N : m )and (0 : M ) ⊆ (0 : m ) ⊆ ( N : m ). If M is completely co-prime, (0 : M ) = ( N : m )so that we respectively obtain ( N : M ) = (0 : M ) and (0 : M ) = (0 : m ) forall 0 = m ∈ M . Thus, M is respectively co-prime and completely prime. Toprove completely prime implies prime, let { } 6 = N ≤ M and 0 = m ∈ N . Then(0 : M ) ⊆ (0 : N ) ⊆ (0 : m ). If M is completely prime, (0 : M ) = (0 : m ) so that(0 : M ) = (0 : N ). This is true for all { } 6 = N ≤ M . Thus, M is prime.7 roposition 4.2 The following statements are equivalent for any R -module M with RM = { } :1. M is completely co-prime,2. the set { ( N : m ) } is a singleton for all N ≤ M and m ∈ M \ N ;3. M is fully completely prime, i.e., every submodule of M is a completely primesubmodule;4. M is completely prime and (0 : m ) = ( N : m ) for all N ≤ M and m ∈ M \ N ;5. M is co-prime and ( N : M ) = ( N : m ) for all N ≤ M and m ∈ M \ N ;6. M is completely prime and for all a ∈ R , N ≤ M and m ∈ M \ N , am ∈ N implies am = 0 ;7. for all N ≤ M and m ∈ M \ N , am ∈ N implies aM = { } ;8. the set { Zd R ( M/N ) : N ≤ M } is a singleton;9. (0 : M ) = Zd R ( M/N ) for all N ≤ M . Proof:
Elementary.From Proposition 4.2(1) and Proposition 4.2(3) we see that the notion of completelyco-prime modules coincides with that of fully completely prime modules. From nowonwards we use the two interchangeably.A module is fully prime if all its submodules are prime submodules.
Example 4.1
A fully prime module over a left-duo ring is fully completely prime.For if am ∈ P for some a ∈ R , m ∈ M and P ≤ M , we get aR ⊆ ( P : m ) since( P : m ) is a two sided ideal as R is left-duo. So, aRm ⊆ P . By hypothesis, P is a prime submodule of M , hence m ∈ P or aM ⊆ P which proves that P is acompletely prime submodule.If R is a commutative ring, then fully prime R -modules are indistinguishable fromfully completely prime modules. Fully prime modules over commutative rings werestudied in [4]. A ring is said to be left-duo if every left ideal of that ring is a two sided ideal. xample 4.2 If M is a module such that every factor module of M is torsion-free, then M is completely co-prime and faithful. Observe that a factor module M/N is torsion-free if ( N : m ) = { } for all m ∈ M \ N . Take for instance M := Z = { ¯0 , ¯1 , ¯2 , ¯3 } the group of integers modulo 4 and R := Z = { ¯0 , ¯1 } thering of integers modulo 2. M is an R -module with only one nonzero submodule N := 2 Z = { ¯0 , ¯2 } . For any m ∈ M \ N and a ∈ R , am ∈ N implies a = 0, i.e.,( N : m ) = { } for all m ∈ M \ N . Now, for the zero submodule, if am = 0 with a ∈ R and m ∈ M \ { } , we still get a = 0. So that (0 : m ) = { } . Hence, M isfully (completely) prime. Example 4.3
Fully completely prime rings were studied by Hirano in [11]. If R is a fully completely prime ring such that R has no one sided left ideals, then themodule R R is a fully completely prime module.A module is fully IFP if all its submodules are IFP submodules. Proposition 4.3
A cyclic module over a fully completely prime ring is fully com-pletely prime.
Proof:
We use the fact that a fully completely prime ring is fully IFP. Let M = Rm , N ≤ M and am ∈ N for some a ∈ R and m ∈ M . Then arm ∈ N forsome r ∈ R where m = rm . ar ∈ ( N : m ). Since R is fully IFP, ( N : m ) is a twosided ideal. Thus, a ∈ ( N : m ) or r ∈ ( N : m ) by hypothesis so that aRm ⊆ N or rm ∈ N . From which we obtain aM ⊆ N or m ∈ N . Proposition 4.4
Let R be a left-duo ring such that for every submodule P of an R -module M , ( P : M ) is a maximal ideal of R , then M is a fully completely primemodule. Proof: If P ≤ M and m ∈ M \ P , then ( P : M ) ⊆ ( P : m ). Since R is left-duo,( P : m ) is a two sided ideal. ( P : M ) maximal implies ( P : M ) = ( P : m ), i.e., P isa completely prime submodule of M . Since P was arbitrary, M is a fully completelyprime module. Proposition 4.5
Each of the following statements implies that the R -module M iscompletely co-prime:1. (0 : m ) is a maximal left ideal of R for all = m ∈ M ,2. ( N : m ) is a minimal left ideal of R for all N ≤ M and every m ∈ M \ N . roof: We know that (0 : m ) ⊆ ( N : m ) for any N ≤ M and m ∈ M \ N . If(0 : m ) is maximal as a left ideal of R for all 0 = m ∈ M , then (0 : m ) = ( N : m )for all m ∈ M \ N . On the other hand, if ( N : m ) is minimal as a left ideal of R forall N ≤ M and m ∈ M \ N , then (0 : m ) = ( N : m ) for all m ∈ M \ N . Thus, bothcases imply that M is a completely co-prime module.A ring is said to be a chain ring if its ideals are linearly ordered by inclusion. Achain ring is sometimes called a uniserial ring . Theorem 4.1 [11, Theorem]
The following statements are equivalent:1. R is a fully completely prime ring;2. R is a chain ring satisfying ( a ) = ( a ) for all elements a ∈ R . For modules, we get Theorem 4.2 and Corollary 4.1. A module is fully symmetric ifall its submodules are symmetric submodules.
Theorem 4.2 If M is a fully completely prime R -module such that a, b ∈ R and m ∈ M , then:1. abm = bam ,2. am = a k m for all positive integers k ,3. either am = abm or bm = abm . Proof:
Note first that a fully completely prime module is both fully symmetricand fully completely semiprime.1. Since abm ∈ Rabm , M fully symmetric implies bam ∈ Rabm such that
Rbam ⊆ Rabm . Similarly,
Rabm ⊆ Rbam . Thus,
Rabm = Rbam . So, abm − bam ∈ R ( abm − bam ) = Rabm − Rbam = { } such that abm = bam .2. am ∈ Ram . So, a k m ∈ Ram and Ra k m ⊆ Ram for any positive integer k .For the reverse inclusion, we know that a k m ∈ Ra k m . M is fully completelysemiprime, therefore am ∈ Ra k m such that Ram ⊆ Ra k m . Then, Ram = Ra k m for all positive integers k . It follows that am − a k m ∈ Ram − Ra k m = { } . Hence, am = a k m as required.10. From abm ∈ Rbm , we get
Rabm ⊆ Rbm . Similarly, we obtain
Rbam ⊆ Ram .Since by 1,
Rabm = Rbam we have
Rabm ⊆ Ram . We now seek to get reverseinclusions. abm ∈ Rabm , if m ∈ Rabm , Ram ⊆ Rabm and
Rbm ⊆ Rabm andwe are through. Suppose m Rabm . Since M is fully completely prime abm ∈ Rabm implies aM ⊆ Rabm or bm ∈ Rabm such that
Ram ⊆ Rabm or Rbm ⊆ Rabm which are the required inclusions. Hence, either
Ram = Rabm or Rbm = Rabm so that either am = abm or bm = abm . Corollary 4.1
Suppose an R -module M is torsion-free and fully completely prime,then1. R is potent and hence it is commutative and fully completely prime,2. for all a, b ∈ R , a = ab or b = ab . Proof:
Elementary.
Remark 4.1
Corollary 4.1 generalizes Example 4.2. A torsion theory in the category R -mod of R -modules is a pair ( T , F ) of classes ofmodules in R -mod such that:1. Hom( T, F ) = { } for all T ∈ T , F ∈ F ;2. if Hom( C, F ) = { } for all F ∈ F , then C ∈ T ;3. if Hom( T, C ) = { } for all T ∈ T , then C ∈ F .A functor γ : R -mod → R -mod is called a preradical if γ ( M ) is a submodule of M and f ( γ ( M )) ⊆ γ ( N ) for each homomorphism f : M → N in R -mod. A radical γ is apreradical for which γ ( M/γ ( M )) = { } for all M ∈ R -mod. A radical γ is hereditary if N ∩ γ ( M ) = γ ( N ) for all submodules N of M . In general, γ ( N ) ⊆ N ∩ γ ( M ).So, to check for hereditariness of γ , it is enough to show that the reverse inclusion, N ∩ γ ( M ) ⊆ γ ( N ) holds. Proposition 5.1 provides a criterion for γ to be a radicaland for γ to be a hereditary radical. 11 emma 5.1 [14, Proposition 1] Let M be any non-empty class of modules closedunder isomorphisms, i.e., if A ∈ M and A ∼ = B , then B ∈ M . For any M ∈ M define γ ( M ) = ∩{ K : K ≤ M, M/K ∈ M} . It is assumed that γ ( M ) = M if M/K
6∈ M for all K ≤ M . Then1. γ ( M/γ ( M )) = { } for all modules M ;2. if M is closed under taking non-zero submodules, γ is a radical;3. if M is closed under taking essential extensions, then γ ( M ) ∩ N ⊆ γ ( N ) forall N ≤ M , i.e., γ is hereditary. It was shown in [8, Examples 3.5 and 3.6] that the completely prime radical β co onthe category R -mod is in general not hereditary. We define a faithful completelyprime radical β fco as β fco ( M ) := ∩{ N ≤ M : M/N is a faithful completely prime module } and show that on the class of IFP modules, this faithful completely prime radical ishereditary. Later, in Theorem 5.2, we show that on a class of semisimple modules β co is also hereditary. We write β co ( M ) = M (resp. β fco ( M ) = M ) if M hasno completely prime submodules (resp. if there are no faithful completely primemodules M/N for all submodules N of M ). Theorem 5.1
The following statements hold for a class of IFP modules:1. faithful completely prime modules are closed under taking essential extension,2. the faithful completely prime radical β fco is hereditary, i.e., β fco ( N ) = N ∩ β fco ( M ) for any submodule N of M ;3. τ β fco = {T β fco , F β fco } where T β fco = { M : M is an IFP module and β fco ( M ) = M } and F β fco = { M : M is an IFP module and β fco ( M ) = { }} is a torsion theory; . the faithful completely prime radical is idempotent, i.e., ( β fco ) = β fco . Proof:
1. Suppose N is an essential submodule of an R -module M such that N is a faith-ful completely prime module. We show that M is also faithful and completelyprime. Let a ∈ R and m ∈ M such that am = 0. If m = 0, M is completelyprime. Suppose m = 0. Since N is an essential submodule of M , there exists r ∈ R such that 0 = rm ∈ N . am = 0 implies arm = 0 since by hypothesis wehave a class of IFP modules. N completely prime together with the fact that0 = rm ∈ N lead to a ∈ (0 : rm ) = (0 : N ). In general, (0 : M ) ⊆ (0 : N ). N faithful implies (0 : M ) = (0 : N ) = { } so that a ∈ (0 : M ) = { } . Then a = 0 such that aM = { } and M is faithful.2. Since faithful completely prime modules are closed under taking essential ex-tension, β fco ( N ) = N ∩ β fco ( M ) by Lemma 5.1(3) since in general β fco ( N ) ⊆ N ∩ β fco ( M ).3. Follows from [16, Proposition 3.1] and paragraph between Propositions 2.2and 2.3 of [16].4. Follows from [16, Proposition 2.3]. Theorem 5.2
For a class of semisimple R -modules, the following statements hold:1. the completely prime radical is hereditary, i.e., β co ( N ) = N ∩ β co ( M ) for anysubmodule N of M ;2. τ β co = {T β co , F β co } where T β co = { M : M is semisimple and β co ( M ) = M } and F β co = { M : M is semisimple and β co ( M ) = { }} is a torsion theory;3. the completely prime radical is idempotent, i.e., β co = β co . Proof:
13. If M is a semisimple R -module, then every submodule N of M is a directsummand. From [8, Corollary 3.8], β co ( N ) = N ∩ β co ( M ) for every directsummand N of M .2. Follows from [16, Proposition 3.1] and paragraph between Propositions 2.2and 2.3 of [16].3. Follows from [16, Proposition 2.3]. Corollary 5.1 If R is a semisimple Artinian ring, then each of the statements inTheorem 5.2 holds. Proof:
A module over a semisimple Artinian ring is semisimple. The rest followsfrom Theorem 5.2.
Remark 5.1
Theorem 5.2 and Corollary 5.1 still hold when “completely primeradical” is replaced with any one of the following radicals: prime radical, s -primeradical, l -prime radical, weakly prime radical and classical completely prime radical.The module radicals: s -prime radical (also called K¨othe upper nil radical), l -primeradical (also called Levitzki radical), weakly prime radical (also called classical primeradical) and classical completely prime radical were respectively defined and studiedin [10], [7], [3] and [6]. References [1] V. A. Andrunakievich and Ju. M. Rjabuhin,
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