On Semiprime Goldie Modules 6 1 Jaime Castro P´erez∗ 0 2 Escuela de Ingenier´ıa y Ciencias, Instituto Tecnol´ologico y de n Estudios Superiores de Monterrey a Calle del Puente 222, Tlalpan, 14380, M´exico D.F., M´exico. J Mauricio Medina B´arcenas† Jos´e R´ıos Montes‡ 3 1 Angel Zald´ıvar§ Instituto de Matem´aticas, Universidad Nacional ] A Aut´onoma de M´exico R Area de la Investigaci´on Cient´ıfica, Circuito Exterior, C.U., . 04510, M´exico D.F., M´exico. h t a January 15, 2016 m [ 1 v Abstract 6 ForanR-moduleM,projectiveinσ[M]andsatisfyingascendingchain 3 4 condition (ACC) on left annihilators, we introduce theconcept of Goldie 3 module. We also use the concept of semiprime module defined by Raggi 0 et. al. in [16]to givenecessary and sufficient conditions for an R-module . M, to be a semiprime Goldie module. This theorem is a generalization 1 of Goldie’s theorem for semiprime left Goldie rings. Moreover, we prove 0 6 that M is a semiprime (prime) Goldie module if and only if the ring 1 S = EndR(M) is a semiprime (prime) right Goldie ring. Also, we study : the case when M is a duomodule. v i Keywords: Prime module, Semiprime module, Goldie module, Essentially X compressible module, Duo module. r a 2010 Mathematics Subject Classification: 16D50, 16D80, 16P50,16P70. Introduction Goldie’s Theorem states that a ring R has a semisimple artinian classical left quotient ring if and only if R is a semiprime ring with finite uniform dimension andsatisfiesACC onleft annihilators. Wisbauer provesin([20],Theorem11.6) ∗[email protected],correspondingauthor †[email protected] ‡[email protected] §[email protected] 1 a versionofGoldie’s Theoreminterms ofmodules. For a retractableR-module M with S = End (M) the following conditions are equivalent: 1. M is non R M-singular with finite uniform dimension and S is semiprime, 2. M is non M-singular with finite uniform dimension and for every N ≤ M there exists e a monomorphism M → N, 3. End (M) is semisimple left artinian and it is R c the classical left quotient ring of S, here M denotes the M-injective hull of M. Also, in [8] the authors study when the encdomorphism ring of a semiprojective module is a semiprime Goldie ring. In this paper we give another generalization of Goldie’s Theorem. For this, we use the product of submodules of a module M defined in [3] to say when a module is a semiprime module. This product extends the product of left ideals of a ring R, so R is a semiprime module (over itself) if and only if R is a semiprime ring in the usual sense. In orderto havea definition of Goldie Module such thatit extends the clas- sicaldefinition of left Goldie ring,we introduce whatascending chaincondition onleft annihilatorsmeans ona module. A leftannihilatorin M is a submodule of the form A = Ker(f) for some X ⊆End (M). This definition with X Tf∈X R R=M is the usual concept of left annihilator. The mainconceptofthis workis thatanR-module M is a Goldiemodule if M satisfiesACConleftannihilatorsandhasfiniteuniformdimension. Weprove some characterizations of semiprime Goldie modules (Theorem 2.8, Theorem 2.22 and Corollary 2.23) which generalize the Goldie’s Theorem and extends the Theorem 11.6 of [20] and corollary 2.7 of [8]. We organize this paper in three sections. Section 1 proves several results for semiprime modules. We also generalize Theorem 10.24 of [10] to semiprime artinian modules. Insection2weintroducetheconceptofGoldiemodules. Weprovethemain Theorem of this paper and a characterization of semiprime Goldie modules. We also obtain some examples of Goldie modules. We also prove that if M has finitely many minimal prime submodules P ,...,P in M such that M/P 1 t i (1 ≤ i≤ t) has finite uniform dimension, then M is Goldie module if and only if eachM/P is Goldie module for(1≤i≤t). We also givea descriptionof the i submodule Z(N) with N ∈σ[M]. Inthelastsectionweapplythepreviousresultstoduomoduleswhichextend results for commutative rings. In [13] the authors say that they do not know a duo module with a quotient not duo, in this section we show an example. Throughout this paper R will be an associative ring with unit and R-Mod will denote the category of unitary left R-modules. A submodule N of an R- moduleM isdenotedbyN ≤M. IfN isapropersubmodulewewriteN <M. We use N ≤ M for an essential submodule. Let M and X be R-modules. X e is said to be M-generated if there exists an epimorphism from a direct sum of copies of M onto X. Every R-module X has a largestM-generatedsubmodule called the trace of M in X, defined by trM(X)= {f(M)|f : M → X}. The P categoryσ[M]isdefinedasthesmallestfullsubcategoryofR-Modcontainingall R-modulesX whichareisomorphictoasubmodule ofanM-generatedmodule. 2 A module N ∈σ[M] is called singular in σ[M] or M-singular, if there is an exactsequenceinσ[M], 0→K →L→N →0withK ≤ L. TheclassS ofall e M-singular modules in σ[M] is closed under submodules, quotients and direct sums. Therefore, any L∈σ[M] has a largest M-singular submodule Z(L)=X{f(N)|N ∈S andf ∈HomR(N,L)} L is called non M-singular if Z(L)=0. Let M be an R-module. In [2] the annihilator in M of a class C of modules is defined as Ann (C)= K, where M TK∈Ω Ω={K ≤M|thereexistsW∈C andf ∈Hom (M,W)withK=Ker(f)} R Also in [2], the author defines a product in the following way: Let N ≤M. For eachmodule X,N·X =Ann (C)whereC isthe classofmodulesW suchthat M f(N)=0 for all f ∈Hom (M,W). R For an R-module M and K,L submodules of M, in [3] the product K L is M defined by K L = {f(K)|f ∈ Hom (M,L)}. Moreover, in [2] it is showed M P R that if M is projective in σ[M], and N ≤ M, then N ·X = N X for every M module X. A nonzero R-module M is called monoform if for each submodule N of M and each morphism f : N → M, f is either zero or a monomorphism. M has enough monoforms if each nonzero submodule of M contains a monoform submodule. Let M-tors be the frame of all hereditary torsion theories on σ[M]. For a family {M } of modules in σ[M], let χ({M }) the greatest element of M-tors α α for which all M are torsion free. Let ξ({M }) be the least element of M-tors α α for which all M are torsion. ξ({M }) and χ({M }) are called the hereditary α α α torsiontheory generatedby the family {M } and the hereditary torsiontheory α cogenerated by the same family. In particular, the greatest and least elements in M-tors are denoted by χ and ξ respectively. If τ ∈M −tors, let Tτ, Fτ and t denotethetorsionclass,thetorsionfreeclassandthepreradicalassociatedto τ τ, respectively. For details about concepts and terminology concerning torsion theories in σ[M], see [19] and [20]. 1 Semiprime Modules Definition 1.1. Let M ∈R−Mod and K, L submodules of M. Put K L= M {f(K)|f ∈ Hom (M,L)}. For the properties of this product see [4] Propo- P R sition 1.3. Definition 1.2. Let M ∈R−Mod. We say a fully invariant submodule N ≤ M is a prime submodule in M if for any fully invariant submodules K,L ≤ M such that K L ≤ N, then K ≤ N or L ≤ N. We say M is a prime module if M 0 is a prime submodule. 3 Proposition 1.3. Let M be projective in σ[M] and P a fully invariant sub- module of M. The following conditions are equivalent: 1. P is prime in M. 2. For any submodules K, L of M containing P and such that K L ≤ P, M then K =P or L=P. Proof. 1⇒2: By Proposition 1.11 of [4]. 2⇒1: Suppose that K, L are submodules of M such that K L≤P. M We claimthat K (L+P)≤P. Since K L≤L∩P, by Proposition5.5of M M [2] K (L/L∩P)=0 so K (L+P/P)=0. Thus K (L+P)≤P. M M M On the other hand, (K+P) (L+P)=K (L+P)+P (L+P)≤P M M M because P is fully invariant in M. Then, by hypothesisK+P =P orL+P =P, hence K ≤P orL≤P. Definition1.4. WesayafullyinvariantsubmoduleN ≤M isasemiprimesub- module inM ifforanyfullyinvariantsubmoduleK ≤M suchthatK K ≤N, M then K ≤N. We saidM is a semiprime module if 0 is a semiprime submodule. Lemma 1.5. Let M be projective in σ[M] and N a fully invariant submodule of M. The following conditions are equivalent: 1. N is semiprime in M. 2. For any submodule K of M, K K ≤N implies K ≤M. M 3. For any submodule K ≤ M containing N such that K K ≤ N, then M K =N. Proof. 1 ⇒ 2 : Let K ≤ M such that K K ≤ N. Consider the submodule M K M ofM. ThisistheminimalfullyinvariantsubmoduleofM whichcontains M K and K X =(K M) X for every module X. Hence by Proposition 1.3 of M M M [4] we have that K K =(K M) K ≤((K M) K) M)≤N M M M M M M M M Since N isfully invariantsubmodule ofM thenN M =N andby Proposition M 5.5 of [2] (K M) (K M)=((K M) K) M)≤N. Since N is semiprime M M M M M M in M, K M ≤N. Hence K ≤N. M 2⇒1: By definition. 1⇔3: Similar to the proof of Proposition 1.3. In Remark 1.26 below, we give an example where the associativity of the product (·) (·) is not true in general. M Definition 1.6. Let M ∈ R−Mod and N a fully invariant submodule of M. We define the powers of N as: 4 1. N0 =0 2. N1 =N 3. Nm =N Nm−1 M Lemma 1.7. Let M be projective in σ[M] and N semiprime in M. Let J be a fully invariant submodule of M such that Jn ≤N then J ≤N. Proof. By induction on n. If n=1 the result is clear. Supposen>1andthePropositionisvalidforn−1. Wehavethat2n−2≥n then J2n−2 ≤N so (Jn−1)2 =Jn−1 Jn−1 ≤N M since N is semiprime Jn−1 ≤N then J ≤N. Proposition 1.8. Let S :=End (M) and assume M generates all its submod- R ules. If N is a fully invariant submodule of M such that Hom (M,N) is a R prime (semiprime) ideal of S, then N is prime (semiprime) in M. Proof. Let K and L be fully invariant submodules of M such that K L≤N. M Put I = Hom (M,L) and J = Hom (M,K). Let m ∈ M and f g ∈ R R P i i IJ. Since g ∈ J and g (m) ∈ K then f (g (m)) ∈ K L ≤ N. Hence i i P i i M IJ ≤ Hom (M,N). Since Hom (M,N) is prime (semiprime) in S, then I ≤ R R Hom (M,N) or J ≤ Hom (M,N). Hence trM(L) := Hom(M,L)M ≤ N or R R trM(K)≤N and since M generates all its submodules then L≤N or K ≤N. Thus N is a prime (semiprime) submodule. Next definition aper in [9] Definition 1.9. A module M is retractable if Hom (M,N) 6= 0 for all 0 6= R N ≤M Corollary 1.10. Let S := End (M) with M retractable. If S is a prime R (semiprime) ring then M is prime (semiprime). Proof. Let K and L be fully invariant submodules of M such that K L = 0. M Since Hom (M,0) is a prime (semiprime) ideal of S then by the proof of 1.8, R trM(K)=0 o trM(L)=0. Since M is retractable, K =0 or L=0. Hence 0 is prime (semiprime) in M. Thus M is prime (semiprime). Proposition 1.11. Let M be projective in σ[M] and N a proper fully invariant submodule of M. The following conditions are equivalent: 1. N is semiprime in M. 2. If m∈M is such that Rm Rm≤N, then m∈N. M 3. N is an intersection of prime submodules. 5 Proof. 1⇒2: By Lemma 1.5. 2 ⇒3 : Since N is proper in M, let 0 6= m ∈M \N. Then Rm Rm (cid:2) 0 0M 0 N. Now, let 0 6= m ∈ Rm Rm but m ∈/ N Then Rm Rm (cid:2) N and 1 0M 0 1 1M 1 Rm Rm ≤ Rm Rm . We obtain a sequence of non-zero elements of M, 1M 1 0M 0 {m ,m ,...} such that m ∈/ N for all i and Rm Rm ≤Rm Rm . 0 1 i i+1M i+1 iM i By Zorn’s Lemma there exists a fully invariant submodule P of M with N ≤P, maximal with the property that m ∈/ P for all i . i WeclaimP isaprimesubmodule. LetK andLsubmodulesofM containing P. Since P ≤ K and P ≤ L, then there exists m and m such that m ∈ K i j i and m ∈ L. Suppose i ≤ j, then Rm Rm ≤ K and by construction m ∈ j iM i j Rm Rm and thus m ∈ K. If we put k = max{i,j}, then m ∈ K and iM i j k mk ∈ L. Hence, RmkMRmk ≤ KML, and so KML (cid:2) P. By Proposition 1.3, P is prime in M. 3⇒1: It is clear. Proposition 1.12. Let 06=M be a semiprime module and projective in σ[M]. Then M has minimal prime submodules in M. Proof. By the proofPropositionof1.11, M has prime submodules. Let P ≤M be a prime submodule. Consider Γ = {Q ≤ P|Q is prime}. This family is not empty because P ∈ Γ. Let C = {Q } be a descending chain in Γ. Let i N,K ≤M befullyinvariantsubmodulesofM suchthatN K ≤ C. Suppose M T that N (cid:2) TC. Then there exists Qj such that N (cid:2) Qj and N (cid:2) Ql for all Q ≤ Q . Therefore K ≤ Q for all Q ≤ Q , and since C is a chain then l j l l j K ≤ C. Therefore C ∈Γ. By Zorn’s Lemma Γ has minimal elements. T T Remark 1.13. Notice that if M is projective in σ[M] and M has prime sub- modules in M, then M has minimal prime submodules. Corollary 1.14. Let 0 6= M be a semiprime module and projective in σ[M]. Then 0=\{P ≤M|P isaminimalprimeinM}. Proof. Letx∈ {P ≤M|P isaminimalprimeinM}andQ≤M beaprime T submodule in M. By Proposition1.12 there exists a minimal prime submodule P such that P ≤Q then x∈Q and x is in the intersection of all primes in M. By Proposition 1.11, x=0. Lemma 1.15. Let M ∈ R−Mod and N a minimal submodule of M. Then N2 =0 or N is a direct summand of M. Proof. Suppose that N N 6= 0. Then there exists f : M → N such that M f(N) 6= 0. Since 0 6= f(M) ≤ N and N is a minimal submodule, f(M) = N. On the other hand, Ker(f)∩N ≤ N, since f(N) 6= 0 then Ker(f)∩N = 0. We have that (M/Ker(f)) ∼= N and since N is a minimal submodule Ker(f), then is a maximal submodule of M. Thus Ker(f)⊕N =M. Corollary 1.16. Let M be a retractable module. If N is a minimal submodule in a semiprime module M, then N is a direct summand. 6 Proof. Since M is semiprime, N N 6=0. M Theorem 1.17. The following conditions are equivalent for a retractable R- module M: 1. M is semisimple and left artinian. 2. M is semiprime and left artinian. 3. M is semiprime and satisfies DCC on cyclic submodules and direct sum- mands. Proof. 1⇒2: If M is semisimple then it is semiprime. 2⇒3: Since M is left artinian, then it satisfies DCC on cyclic submodules and direct summands. 3 ⇒ 1 : Since M satisfies DCC on cyclic submodules, there exists K a 1 minimal submodule of M. By Corollary 1.16, M =K ⊕L . Now there exists 1 1 K a minimal submodule of L and L = K ⊕ L . With this process we 2 1 1 2 2 obtain a descending chain of direct summands, which by hypothesis it is finite L ⊇L ⊇L ⊇...⊇L . SinceL issimpleandM =K ⊕K ⊕...⊕K ⊕L , 1 2 3 m m 1 2 m m then M is semisimple. Now, if M is semisimple and satisfies DCC on direct summands then M is artinian. Definition 1.18. Let M ∈R−Mod and N ≤M. We say N is an annihilator submodule if N =Ann (K) for some 06=K ≤M. M Lemma 1.19. Let M be semiprime and projective in σ[M]. Let N,L≤M. If L N =0, then N L=0 and L∩N =0. M M Proof. Since L N =0, then M 0=N (L N) L=(N L) (N L). M M M M M M Hence N L=0 . M Now, since L∩N ≤L and L∩N ≤N, then (L∩N) (L∩N)≤L N =0. M M Thus L∩N =0 Corollary 1.20. Let M be semiprime and projective in σ[M]. If N ≤M, then N Ann (N)=0. M M Proposition 1.21. Let M be semiprime and projective in σ[M] and N ≤ M. Then N is an annihilator submodule if and only if N =Ann (Ann (N)) M M 7 Proof. ⇒:By Lemma 1.20N ≤Ann (Ann (N)). Thereis K ≤M suchthat M M N =Ann (K), hence M K N =K Ann (K)=0 M M M and thus K ≤Ann (N). Therefore, M Ann (Ann (N))≤Ann (K)=N M M M It follows that N =Ann (Ann (N)). M M ⇐: By definition of annihilator submodule. Proposition 1.22. Let M be semiprime and N ≤ M. Then, Ann (N) is M the unique pseudocomplement fully invariant of N. Moreover, N Ann (N) L M intersects all fully invariant submodules of M. Proof. Let L≤M be a fully invariant pseudocomplement of N in M. Then L N ≤L∩N =0 M Thus L≤Ann (N). Observe that M (Ann (N)∩N) (Ann (N)∩N)≤(Ann (N)∩N) N =0 M M M M M Since M is semiprime, Ann (N)∩N =0. Thus L=Ann (N). M M Lemma 1.23. Let M be a semiprime module and N ≤ M. Let S be the set of all minimal prime submodules of M which do not contain N. Then Ann (N)= {P|P ∈S}. M T Proof. Put K = {P|P ∈ S}. Any element in K ∩N is in the intersection of T all minimal prime submodules of M which is zero. Then K ∩N = 0. Since K is fully invariant in M, K N ≤ K∩N = 0. Thus, K ≤ Ann (N). Now, let M M P ∈S. Since AnnM(N)MN =0≤P and N (cid:2)P, then AnnM(N)≤K. Lemma 1.24. Let M be projective in σ[M]. If M is semiprime then M is retractable. Proof. Let N ≤ M and suppose Hom (M,N) = 0. Then Ann (N) = M. R M So M N = 0 but N N ⊆ M N = 0. Since M is semprime then N = 0 by M M M Lemma 1.5. Proposition 1.25. Let M be projective in σ[M] and semiprime. The following conditions are equivalent for N ≤M: 1. N is a maximal annihilator submodule. 2. N is an annihilator submodule and is a minimal prime submodule. 3. N is prime in M and N is an annihilator submodule. 8 Proof. 1 ⇒ 2 : Let K ≤ M such that N = Ann (K). Let L,H ≤ M be fully M invariant submodules of M such that LMH ≤ N. Assume H (cid:2) N. Then 0 6= H K. Hence Ann (K) ≤ Ann (H K), but since Ann (K) is a maximal M M M M M annihilator submodule, then Ann (K)=Ann (H K). M M M As M is projective in σ[M], by Proposition 5.5 of [2], we have that L (H (H K))=(L H) (H K)≤N (H K)=0 M M M M M M M M Now, since H (H K)≤H K, then M M M Ann (K)=Ann (H K)≤Ann (H (H K)) M M M M M M Therefore Ann (H K)=Ann (H (H K)). Thus L≤Ann (K)=N. M M M M M M Now, let P ≤ M be a prime submodule of M such that P < N. We have that N K = 0 ≤ P. So K ≤ P < N. Hence K K = 0. Thus K = 0, but M M M is semiprime, a contradiction. It follows that N is a minimal prime submodule of M. 2⇒3: By hypothesis. 3⇒1: Suppose N <K with K an annihilator submodule. Then Ann (K) K =0≤N M M Since N is prime in M, then Ann (K) ≤ N < K. By Proposition 1.22 M Ann (K)∩K =0,henceAnn (K)=0. SinceK isanannihilatorsubmodule, M M by Proposition 1.21, K =Ann (Ann (K))=Ann (0)=M. M M M Remark 1.26. Following the notation of Example 1.12 of [4] Let R = Z ⋊ 2 (Z ⊕Z ). ThisringhasonlyonemaximalidealI andithasthreesimpleideals: 2 2 J , J , J , which are isomorphic. Then, the lattice of ideals of R has the form 1 2 3 R • I • ❁ ✂ ❁ ✂ ❁ ✂ ❁ ✂ ❁ ✂ ❁ ✂ ❁ ✂ ❁ ✂ J•1 J•2 J•3 ❁ ❁❁❁❁❁❁ ✂✂✂✂✂✂ ❁ ✂ ❁ ✂ 0 ✂ • Moreover, R is artinian and R-Mod has only one simple module up to iso- morphism. Let S be a simple module. By Theorem 2.13 of [15], the lattice of fully invariant submodules of E(S) has tree maximal submodules N, L and K, and it has the form 9 E(S) • ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ K L N • • • ❄ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ S ⑧ • 0 • Put M =E(S). Since K∩L=S and K L≤K∩L, then K L≤S. On M M the other hand consider the composition M π // M/N ∼= //S i //L f = i◦π where π is the natural projection and i is the inclusion. Then, f(K)=S andS ≤KML. Thus,KML=S. NoticethatKML≤N butK (cid:2)N and L (cid:2) N. Hence N is not prime in M. Analogously, we prove that neither K nor L are prime in M. We also note that K K =S. Moreover, π(K)=S, M so K S =S. In the same way L S =S and N S =S M M M Let g : M → K be a non zero morphism. If Ker(g)∩S = 0 then g is a monomorphism, a contradiction. So Ker(g)∩S = S. Thus S K = 0 and M Ann (K) = S. Analogously Ann (L) = S = Ann (N) = Ann (S). Since M M M M S S ≤ S K, S S = 0. Thus M is not semiprime. Hence, S is a maximal M M M annihilator submodule of M which is not prime because K K =S. With this M wecanseethatassociativityisnottrueingeneral,becauseL (K S)=L S = M M M S and (L K) S =S S =0. Notice that, in this example Hom (M,H)6=0 M M M R forallH ∈σ[M]inparticularM isretractable,butM isnotprojectiveinσ[M]. Proposition 1.27. Let M be projective in σ[M] and semiprime. For N ≤M, if N = Ann (U) with U ≤ M a uniform submodule, then N is a maximal M annihilator in M. Proof. Suppose that N < K with K an annihilator submodule in M. Since N = Ann (U) by Proposition 1.22, K ∩U 6= 0. By hypothesis U is uniform M and thus K∩U ≤ U. Then e (K∩U)⊕Ann (U)≤ U ⊕Ann (U) M e M Now,notice that if L≤ M, by Proposition1.22 (U⊕Ann (U))∩L6=0. So FI M ((K∩U)⊕Ann (U))∩L6=0. Therefore, K∩L6=0 andK intersects all fully M invariant submodules of M . Since K∩Ann (K)=0 and Ann (K)≤ M, M M FI then Ann (K)=0. Thus, K =Ann (Ann (K))=Ann (0)=M. M M M M 10