ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 Min (Max)-CS Modules I. M. A. Hadi, R . N. Majeed Departme nt of Mathematics, College of Ibn-Al-Haitham , Unive rsity of Baghdad Received in:25 August 2011, Accepte d in:20 Septe mber 2011 Abstract. In this p aper, we give a comp rehensive st udy of min (max)-CS modules such as a closed submodule of min-CS module is min-CS. Amongst other results we show that a direct summand of min (max)-CS module is min (max)-CS module. One of interested theorems in this p aper is, if R is a nonsingular ring then R is a max-CS ring if and only if R is a min-CS ring. Key words: CS-module, min-CS module, max-CS module, uniform-CS module. 1- Introduction Throughout the p aper all rings R are commutative with identity and all R-modules are unitary . We write A  M and A  e M to indicate that A is a submodule of M and A is an essential submodule of M , resp ectively. Recall that anR-module M is called an extending module (or, CS-module) if every submodule is essential in a direct summand of M or M is extending if and only if every closed submodule is a direct summand, [1, p .55]. In this p aper definitions, notations, examp les and fundamental results of min (max)-CS modules are introduced. 1.1 Definition: [2] An R-module M is called min-CS module if every minimal closed submodule of M is a direct summand of M . A ring R is called min-CS if it is min-CS R-module. 1.2 Definition: [2] An R-module M is called max-CS module if every maximal closed submodule of M with nonzero annihilator is a direct summand of M . A ring R is max-CS if it is max-CS R-module. Recall that an R-module M is -injective (quasi-continuous) if and only if M satisfies C1 (M is extending) and C3, where M is said to satisfy the C3 if the sum of any two direct summands of M with zero intersection is a direct summand of M . [3, p .18] 1.3 Remarks and Examp les 1. Every CS-module is min-CS and max-CS. Proof: It follows directly by [1, p .55]. 2. Every semisimple module is max-CS and min-CS. In p articular ℤ2, ℤ3, ℤ6, ℤ10,…, ℤ30 as a ℤ-module is max-CS and min-CS. Proof: By [1, p .55], every semisimple module is CS. Hence the result follows by remark 1. ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 3. Every uniform R-module M is min-CS and max-CS. In p articular each of ℤ-module ℤ, ℤ4, ℤ8, ℤ9, ℤ16 is min-CS and max-CS. 4. If R is a semisimple ring, then every R-module M is injective, by [4, theorem 1.18, p.29]. Hence M is max (min)-CS module since every injective module is CS. By [1, p .16]. 5. The ℤ-module ℤ12 is max-CS and min-CS. The submodules of ℤ12 are 2 , 3 , 4 , 6 , 0          and ℤ12. Since 1 23 4     ¢ . So each of 3  and 4  are direct summands. Hence they are closed submodules. 2   e ℤ12 and e6 3     imp ly hat 2  and 6  are not closed. Thus M is CS and so max-CS and min-CS. 6. It is easy to check that each of the ℤ-modules ℤ18 and ℤ24 are min-CS and max-CS. 7. Let M be a module whose lattice of submodules is the following: 2 1 2 M N N N 0     / \ \ / It is clear that N1 is closed in M , but it is not a direct summand of M . So M is not CS. Also N1 is a minimal closed submodule of M . Hence M is not a min-CS module. Not ice that N1  N2  e M , so it is not closed submodule of M . It follows that N1 is a max-closed submodule of M . Hence M is not a max-CS module. 8. Every -injective is min-CS and max-CS. Proof: It follows by the definition of -injective module and remark 1.3 (1). 9. Let M be the ℤ-module ℤ8  ℤ2. M is not CS-module, since there exists a submodule N = { (2 , 1 ), ( 4, 0), ( 6, 1 ), (0, 0) } which is closed but not a direct summand. M oreover N is minimal closed, so M is not min-CS module. On the other hand M is a max-CS module, since the only maximal closed submodules of M are ℤ8  ( 0 ) and < (3, 1) >, and (ℤ8  ( 0 ))  (( 0 ) ℤ2) = M and < (3, 1) >  (( 0 )  ℤ2) = M . Hence we deduce that M is max-CS. 10. Let M 1 and M 2 be two R-modules such that M 1 is isomorphic to M 2 (M 1≅M 2), then M 1 min (max)-CS if and only if M 2 is min (max)-CS. M ahmoud A.Kamal and Amany M .M enshawy in [5] gave the following: An R-module M is called min-CS if every simple submodule of M is essential in a direct summand. However this definition of min-CS is different from definition 1.1, since the ℤ-module M = ℤ8  ℤ2 is not min-CS in our sense. However it is min-CS (in sense of Kamal and M enshawy ) since ℤ8 is CS, so min-CS (in sense of Kamal and M enshawy ) and ℤ2 is simple, so semisimp le. ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 Hence by app lying [5, lemma 3, p.166], M is min-CS (in sense of Kamal and M enshawy ). 1.4 Prop osition: Let M be an R-module, and let I be an ideal of R such that I ⊆ annM . M is max- CS R-module then M is max-CS (R/I)-module and the converse is true if annM ≠ annN, for all maximal closed submodule N ≨ M . Proof: Let N be a maximal-closed (R/I)-submodule of M and annR/IN ≠ 0R/I = I. It is easy to see that N is a maximal closed submodule of M . Since annR/IN ≠ I = 0R/I, so there exists r + I  R/I with rI such that r + I  annN, hence r ≠ 0 and rN = 0. Thus annRN ≠ 0 and so that N is a direct summand of M . Conversely , let N be a maximal closed R-submodule of M with annN ≠ 0. Hence N is a maximal closed (R/I)-submodule. Now, since ann M  ann N, there exists r  ann N and r  ann M . Thus r  I; that is 0R/I = I ≠ r + I and (r + I)N = rN = 0. Hence ann R/IN ≠ 0R/I. But M is a max-CS (R/ I)-module, so N is a direct summand. 1.5 Prop osition: Let M be an R-module, let I be an ideal of R such that I  ann M . Then M is min-CS R- module if and only if M is min-CS (R/ I)-module. Proof: It is st raight forward, so it is omitted. Recall that an R-module M is called a uniform extending (or uniform-CS) if every uniform submodule is essential in a direct summand. [1, p .55]. Al-Hazmi in [2,p .24], mentioned that min-CS and uniform-CS are equivalent concepts without p roof. We shall p rove this equivalence, but first we need the following lemmas. 1.6 Lemma: Let N be a submodule of an R-module M . N is minimal closed if and only if N is uniform-closed (that is every closed submodule of N is essential in N). Proof: () It is enough to p rove that N is uniform Let V, W be two nonz ero submodules of N. Supp ose V  W = (0). Hence, there exists V'  N such that V' is a relative comp lement of V and hence V' is closed in N. Since N is minimal closed of M thus N is closed in M , so by [4, p rop osition 1.5, p.18], V' is closed in M and 0 ≠V'  N. Thus N is not minimal closed submodule of M , which is a contradiction. Therefore, N is uniform-closed submodule. () Sup p ose that t here exists a closed submodule V of M such that V  N. But N is uniform. So V  e N. Hence V = N, since V is closed. Thus N is minimal closed. 1.7 Lemma: If U is a uniform submodule of M such that U e K  M . Then K is uniform. Proof: ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 Let V and W be two nonzero submodules of K. Since U  e K, U  W ≠ (0) and U  V ≠ (0). But U is uniform submodule of M . So (U  V)  (U  W) ≠ (0). Hence U  (V  W) ≠ (0). Thus V  W ≠ (0); that is K is uniform. 1.8 Lemma: Let U be a submodule of an R-module M . Then U is uniform closed if and only if U is maximal uniform; that is U is maximal in the collection of uniform submodules of M . Proof: () Sup p ose there exists a uniform submodule V of M such that U  V. Since V is uniform so U  e V. But U is closed so U = V. Thus U is a maximal-uniform submodule. () It is enough to show that U is closed. Sup p ose there exists V  M such that U  e V. Hence by lemma1.7, V is uniform. Thus U = V, since U is maximal uniform, so t hat U is closed. By combining lemma 1.6 and lemma 1.8, we have the following: 1.9 Corollary: Let U be a submodule of an R-module. Then the following are equivalent: (1) U is minimal-closed. (2) U is uniform-closed. (3) U is maximal-uniform. Now, we can give the proof of the following result. 1.10 Prop osition: [2, p .24] Let M be an R-module. M is uniform-CS if and only if M is min-CS. Proof: () Let U be a maximal uniform submodule of M . Since M is uniform-CS, so U is essential in a direct summand V. Then by lemma 1.7, V is uniform. Hence U = V since U is a maximal uniform submodule. Thus U is a direct summand of M , it follows that every minimal-closed is a direct summand by lemma 1.6. So that M is a min-CS module. () Let U be a uniform submodule of M . By [4, Exc.13, p.20], there exists a closed submodule V of M such that U  e V. Hence by lemma 1.7, V is uniform. Thus V is a closed-uniform submodule of M , and hence by lemma 1.6, V is minimal-closed. So that V is a direct summand of M , since M is min-CS module. It follows that U is essential in a direct summand. Thus M is uniform-CS. The following result is given in [2, lemma 3.1.1, p .45], we give the details of p roof. 1.11 Prop osition: Let M be an indecomp osable R-module with a uniform submodule. If M is a min-CS module, then M is uniform. Proof: ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 By hy p othesis M has a uniform submodule, say U. By [4, Exc.13, p.20], there exists a closed submodule K of M such that U  e K. Hence by lemma 1.7, K is a uniform submodule of M . Let C = {K: K is a uniform submodule of M and U  K}. Hence C ≠ , and so that by Zorn's lemma C has a maximal element say W. It is clear that W is a maximal-uniform. So it is a minimal-closed submodule of M . Thus W is a direct summand of M , since M is a min-CS module. Then W  V = M for some submodule V  M . But M is indecomp osable, hence V = (0). Thus W = M ; that is M is uniform. 1.12 Corollary: Let M be an indecomp osable R-module with a uniform submodule. Then M is min-CS module if and only if M is a uniform module. 1.13 Prop osition: Let M be an R-module. The following st atements are equivalent for a module M :- (1) M is a min-CS module. (2) For every minimal-closed submodule A of M , there is a decomposition M = M 1  M 2 such that A is a submodule of M 1 and M 2 is a complement of A in M . Proof: (1)  (2) Let A be a minimal-closed submodule in M . Therefore A is a direct summand of M since M is a min-CS module. That is M = A  M ' for some submodule M ' of M . It is clear that A is a submodule of A, and it is easy to check that M ' is a complement of A. (2)  (1) To p rove M is a min-CS module. Let A be a minimal-closed submodule of M . Therefore, there is a decomposition M = M 1  M 2, where M 1 and M 2 are two submodules of M and A is a submodule of M 1 and M 2 is a complement of A in M . Since M 2 is a complement of A in M , then A  M 2  e M . But A is a closed submodule in M and A is a submodule in M 1, therefore A is closed in M 1. Therefore, A  M 2 is closed in M 1  M 2 = M by [4, Exc.15, p.20]. Thus A  M 2 = M . So that A is a direct summand in M . Hence M is a min-CS module. 1.14 Prop osition: Let M be a min-CS R-module. If N is a closed submodule, then N is a min-CS module. Proof: Let U be a minimal closed submodule of N. Since N is a closed submodule in M , then by [4, p rop osition 1.5, p.18], U is closed in M . We claim that U is a minimal closed submodule of M . To p rove our assertion: Sup p ose there exists a closed submodule V of M such that V  U. But V  U  N, implies V is a closed submodule in N by [4, p .18]. But U is a minimal closed submodule of N so U = V. Thus U is a minimal closed submodule of M , and hence U is a direct summand of M , since M is a min-CS module. Hence M = U  W for some W  M . It follows that N = (U  W)  N and by modular law, N = U  (W  N); that is U is a direct summand of N and so N is a min-CS module. ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 1.15 Remark: The converse of the p revious p rop osition is not true in general For examp le: The ℤ-module ℤ8  ℤ2 is not min-CS, but N = ℤ8  (0) ≅ ℤ8 is min-CS. 1.16 Corollary: A direct summand of min-CS module is min-CS. Proof: Let M be a min-CS module and let N be a direct summand of M . Hence by [4, Exc.3, p.19], N is a closed submodule of M . Hence, N is a min-CS module by p rop osition 14. 1.17 Corollary: Let M be an R-module. If M  M is min-CS, then M is a min-CS module. Proof: It follows by corollary 1.16. Corollary 1.16, lead us to give the following examp le:- Let M = ℤ8  ℤ2  ℤ3 as a ℤ-module. Thus M is not min-CS because if it is, then N = ℤ8  ℤ2  M is min-CS by corollary 1.16, which is a contradiction. 1.18 Remark: The condition N is closed can not be drop p ed from p rop osition 1.14 as the following examp le shows:- Let M be the ℤ-module ℤ16  ℤ2. M is min-CS. Let N = ( 2 )  ℤ2. It is clear that N  e M , so N is not closed in M , also N is isomorphic to ℤ8  ℤ2 which is not a min-CS module. 1.19 Prop osition: Let M be a finitely generated or multiplication R-module. Then if every maximal submodule is a direct summand, then M is a max-CS module. Proof: Let A be a maximal closed submodule of M . Since M is a finitely generated or multiplication module. Then by [6, theorem 2.3.11, p .28] and by [7, theorem 2.5 (1)], there exists a maximal submodule B of M such that A  B. But B is a direct summand, by hy p othesis; hence B is a closed submodule by [4, Exc.3, p.19]. It follows that A = B. Thus A is a direct summand of M . Hence M is a max-CS module. The converse of the p revious p rop osition is not true in general, as the following examp le shows: 1.20 Examp le: ℤ as a ℤ-module is a max-CS module; also ℤ is a multip lication ℤ-module. ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 But 2ℤ as a ℤ-module is a maximal submodule of ℤ and it is not a direct summand. Not e that we have an analogous result to corollary 1.16, for max-CS modules, that is a direct summand of max-CS module is max-CS module, but first we p rove the following: 1.21 Prop osition: Let M be an R-module, and let N be a closed submodule of M . If M is a max-CS module, then (M /N) is a max-CS R-module, provided M is not faithful. Proof: Let (B/N) be a maximal closed submodule in (M /N), with annR(B/N)≠ 0. We claim that B is a maximal closed submodule in M . To p rove our assertation: First , assume that B  e L  M . But N is closed in M and N  B  L, so N is closed in L. Hence by [4, p rop osition 1.4, p.18], N  B  e L implies B/N  e L/N  M /N. Hence (B/N) = (L/N), since (B/N) closed in (M /N). Thus B = L and B is closed in M . Now, assume there exists a closed submodule B' of M such that B  B', hence N  B'  M . Then, by [4, Exc.16, p.20], (B'/N) is closed in (M /N) and so that (B/N)  (B'/N). This imp lies (B/N) = (B'/N), since (B/N) is a maximal closed submodule in (M /N). M oreover annRB  annRM ≠ 0, so annB ≠ 0. Since M is max-CS then B  K = M for some K  M , and hence (B/N)((K+ N)/N) = (M /N). It follows that (M /N) is a max-CS module. 1.22 Corollary: A direct summand of a max-CS R-module M is a max-CS module, p rovided M is not faithful. Proof: Since N is a direct summand of M , so M = N  W for some W  M . Hence (M /W) isomorphic to N by second isomorp hism theorem. But (M /W) is a max-CS by p rop osition 1.21. So N is a max-CS module, by remark 1.3 (10). 1.23 Corollary: Let M be an R-module. If M  M is a max-CS module. Then M is max-CS module p rovided M is not faithful. Proof: It follows by corollary 1.22. Recall that a p rop er submodule N of an R-module M is called p rime submodule if whenever r  R, x  M , r x  N implies x  N or r  [N:M ]. An R-module M is called a p rime module if annRM =annRN for each submodule N of M . Equivalently , M is called a p rime module if (0) is a p rime submodule of M , [8]. 1.24 Corollary: Let M be a not faithful p rime max-CS R-module. If f : M  M ' is an epimorp hism, then M ' is a max-CS module. Proof: M is a max-CS module. Then (M /ker f) ≅ M ', by the 1 st fundamental theorem. We claim that ker f is clsed submodule in M . To show this, let ker f ≨ e L  M . Assume there exists 0 ≠ x  L such that x  ker f. ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 Then there exists 0 ≠ r  R such that 0 ≠ r x  ker f. Hence rf(x) = 0. Since M is a prime module, so either f(x) = 0 or r  annM . But f(x) ≠ 0 since x  ker f, hence r  annM and this implies r x = 0 which is a contradiction. Thus ker f = L and so that ker f is a closed submodule of M . Since annM ≠ 0 by hy p othesis. Then by p rop osition 1.21, (M /ker f) is a max-CS module and hence by remark 1.3 (10) M ' is a max-CS module. 1.25 Corollary: Let M be a max-CS not faithful R-module. If N is a p rime submodule of M and [N R : M ] = annRM . Then (M /N) is a max-CS R-module. Proof: Since N is a p rime R-submodule of M , then (M /N) is a torsion free (R/[N:M ])-module, [8, p .61-69]. Hence (M /N) is a torsion free (R/annM )-module, and so t hat N is closed (R/annM )-module of M by [9, remark 3.3,p.48]. It follows that N is closed R-submodule of M . Then by p rop osition 1.15, (M /N) is a max-CS R-module. 1.26 Corollary: If M is a max-CS not faithful R-module and N is a submodule of M , such that (M /N) is torsion free, then (M /N) is max-CS. Proof: Since (M /N) is torsion free R-module. Then N is closed submodule in M by [9, remark 3.3, p .48]. Hence the result follows by p rop osition 1.21. Now we have the following note for min-CS modules. 1.27 Remark: The homomorp hic image of min-CS need not be a min-CS, as the following examp les show. 1.28 Examp les: (1) Let M be the ℤ-module ℤℤ. It is clear that M is a min-CS module. Let N = 8ℤ2ℤ, and let : M  (M /N) be the natural projection. Since (M /N) is isomorphic to ℤ8ℤ2, so (M /N) = (M ) is not min-CS module. (2) Let M be the ℤ-module ℤℤ2. M is min-CS. Let N = 8ℤ( 0 ), N  e ℤ( 0 ), but (M /N) = ℤ8ℤ2 is not min-CS. (3) Let M = ℤ8ℤ be a ℤ-module, M is a CS-module. So M is a min-CS module. Let N = (0) ( 2) . (M /N) is isomorphic to ℤ8ℤ2 which is not min-CS module. 1.29 Theorem: Let M be an R-module. If M is a faithful, finitely generating and multiplication R-module, then M is a max-CS module if and only if R is a max-CS ring. Proof: () If M is a max-CS R-module. Let I be a maximal closed ideal of R such that ann I≠0. We claim that N = IM is a maximal closed R-submodule of M . First of all N = [IM :M ]M since M is multiplication and by [10, p rop osition 3.31] [IM :M ] is a closed ideal of R. But M is finitely generating faithful multip lication, then by [7, theorem 3.1] I = [IM :M ]. ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 Thus by [10, p rop osition 3.31, ch.3], N = [IM :M ]M is a closed submodule of M . Now, to p rove that N is a maximal submodule of M . Sup p ose there exists a closed submodule W of M such that N  W. Hence N = IM = [N:M ]M  [W:M ]M = W and by [7, theorem 3.1], I = [N:M ]  [W:M ]. On the other hand by [10, p rop osition 3.31, chap ter three], I = [N:M ] and [W:M ] are closed ideals of R. Hence I = [W:M ], since I is a maximal closed ideal of R. It follows N = W and N is maximal closed submodule of M . But M is faithful multip lication, so annRN = annRI ≠ 0. Thus N is a direct summand of M . So that N  L = M for some L  M . But N = IM , L = [L:M ]M , so t hat IM  [L:M ]M = M and hence (I  [L:M ])M = M . But by [7, theorem 1.6] (I  [L:M ])M = IM  [L:M ]M = (0), and so (I  [L:M ])M = 0; that is (I  [L:M ])  annM = (0). T hus I  [L:M ] = (0), so that M = RM = (I  [L:M ])M and hence R = I  [W:M ] by [7, theorem 6.1]. () Let N be a maximal closed submodule of M with annN ≠ 0. Since N is closed, then by [10, p rop osition 3.1] [N:M ] is closed in R. We claim that [N:M ] is a maximal closed ideal in R. To show this, assume J is a closed ideal in R such that [N:M ]  J, hence N = [N:M ]M  JM . But JM = [JM :M ]M and by [7, theorem 6.1] [JM :M ] = J and by [10, p rop osition 3.1] JM is a closed submodule of M . It follows that N = [N:M ]M = JM , since N is a maximal closed submodule of M . Then by [7, theorem 3.1], [N:M ] = J; that [N:M ] is a maximal closed. But annN = ann[N:M ] since M is faithful multip lication. Thus [N:M ] is a maximal closed ideal of R, with annR[N:M ] ≠ 0. On t he other hand R is a max-CS ring, so [N:M ] is a direct summand of R. Thus R = [N:M ]  T where T is an ideal of R, and hence M = RM = ([N:M ]  T)M = [N:M ]M  TM . But by [7, theorem 1.6], [N:M ]M  TM = ([N:M ]  T)M = 0M = 0 So M = [N:M ]M  TM ; that is M = N  TM . By the same argument of p roof of theorem 1.29, we have the following: 1.30 Prop osition: Let M be a faithful, multiplication and finitely generated R-module, then M is a min-CS ring if and only if R is min-CS ring. Next we have for nonsingular rings, the concepts min-CS ring and max-CS ring are equivalent. But first we need the following which is given in [2, lemma 2.1.1, p .31]. We give the details of p roof. 1.31 Lemma: For a ring R, a comp lement of minimal (maximal) closed ideal of R is a maximal (minimal) closed ideal of R. Proof: () Let I be a minimal closed ideal of R. Let J be the relative comp lement of I. Then by [4, p rop osition 1.3, p.17], (I  J)  e R, and J is closed in R, by [4, p rop osition 1.4, p.18]. Now, we shall show that J is a maximal closed ideal in R. Assume that there exists a closed ideal J* in R, such that J  J*   R. It follows that J*  I ≠ (0) because J is t he largest ideal in R such that I  J = (0). ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 On the other hand I is a minimal closed ideal in R, so it is uniform closed by lemma 1.6. Hence J* I  e I. But J  e I, hence (J* I)  J  e I  J  e R and so that (J* I)  J  e R, by [4, p rop osition 1.1(a), p .16]. But J* I  J* and J < J*, hence (J* I)  J  J*. Thus J*  e R, which is a contradiction. () If T is a maximal closed ideal of R and V its comp lement. To p rove V is minimal closed in R, it is enough to show that V is uniform by lemma 1.6. Assume A  B = (0) for some two nonz ero ideals A and B of V. Now, there exists a closed ideal I in R such that A  T  e I, by [4, Exc.13, p.20]. Hence T   I which is a contradiction, since I is closed in R and T is a maximal closed ideal in R. Thus V is uniform closed, that is V is minimal closed. 1.32 Not e: By a similar argument of p roof of lemma 1.31, we get analogous results for submodules. Recall that a ring R is semip rime if for each x  R, x 2 = 0, then x = 0, [4, p .2]. Now, we can give the following theorem: 1.33 Theorem: Let R be a nonsingular ring. Then R is max-CS if and only if R is min-CS. Proof: () If R is a max-CS ring. Let I be a minimal closed ideal in R. By [4, theorem 2.38, p .65], I = ann annI, hence annI ≠ 0, but R is nonsingular, so R is semip rime by [4, p rop osition 1.27(b), p.35], which implies I  annI = 0. Let J be the relative comp lement of I, so by lemma 1.31, J is a maximal closed ideal in R. Also by [4, theorem 2.38, p.65], J = ann annJ, so annJ ≠ 0. Therefore J is a direct summand of R since R is max-CS. It follows that J = for some idempotent e in R. On t he other hand, IJ  I  J = (0), imp lies that J  annI. But I  annI = (0), so J = annI since J is t he largest ideal in R such that I  J = (0). Thus annJ = ann annI = I and so < 1 – e > = I. It follows that I is a direct summand of R. () If R is a min-CS ring. Let I be a maximal closed ideal in R, with annI ≠ 0. By [4, theorem 2.38, p .65], I = ann annI. Let J be a relative comp lement of I. So J is a minimal closed ideal of R, by Lemma 1.31. Hence J is a direct summand of R, since R is min-CS. Thus J = < e > for some idempotent e  R. But IJ  I  J = (0), so I  ann J = ann = <1 – e >. But <1 – e > is closed ideal of R, since it is a direct summand of R. It follows that I = <1 – e > because I is a maximal closed ideal of R. Thus I is a direct summand of R. Recall that, a ring R is called semihereditary if every finitely generating ideal of R is p rojective, [4, p.10]. 1.34 Corollary: Let R be a semiheriditary ring. Then R is a min-CS ring if and only if R is a max-CS ring. Proof: Since R is a semiheriditary ring. Then R is a nonsingular ring, [4, p .36]. So we get the result by Theorem 1.33. ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 Recall that a ring R is called regular (Von Neumann) if for every aR there is an x  R such that a x a = a, that is a = a 2 x, [4, p .10]. 1.35 Corollary: If R is a regular ring. Then R is min-CS if and only if R is a max-CS. The following lemma is needed for the following corollary . 1.36 Lemma: [4, Exc.5, p.36] For a commutative ring, R/Z(R) is a nonsingular ring. Proof: Sup p ose there exists x = x + Z(R)  Z(R/Z(R)) and x  Z(R), so x ≠ 0 and ann(x) is not essential in R, also annR/Z(R)( x + Z(R))  e R/Z(R). Then for each I/Z(R) with I  Z(R), (I/Z(R)) annR/Z(R)( x+Z(R))≠O R/Z(R)=Z(R). So there exists a + Z(R)  ann R/Z(R)( x + Z(R)), a  I such that a  Z(R). Then a ≠ 0. Thus ax + Z(R) = Z (R), annR(a) is not essential in R and a  I. Hence ax  Z(R) and annR(a) is not essential in R. So that ann(ax)  e R and ann(a) is not essential in R. Now, since ann(ax)  e R, then for each nonzero ideal J of R, J  ann(ax) ≠ 0. Hence there exists y  ann(ax) and y ≠ 0. T hus yax = 0. (1) If ya = 0 then 0 ≠ y  ann(a) thus 0 ≠ y  Jann(a) so that ann(a)  e R which is a contradiction. (2) If ya ≠ 0, t hen 0≠yaann(x)J. Hence ann(x)  e R which is a contradiction. (3) If yx = 0 then 0 ≠ y  ann(x), so that 0 ≠ y  ann(x)  J. Thus ann(x)  e R which is a contradiction. (4) If yx ≠ 0, t hen 0≠ yxann(a)J, so that ann(a) e R which is a contradiction. (5) If ax = 0 then 0 ≠ a ann(x)  I, that is ann(x)  e R which is a contradiction. Thus our assumption is false and so Z (R/Z(R)) = Z(R) = OR/Z(R). 1.37 Corollary Let R be a ring. Then R/Z(R) is min-CS if and only if R/Z(R) is max-CS. Proof: By lemma 1.36 R/Z(R) is nonsingular, so t he result follows by theorem 1.33. Before we give the following corollary , we need the following lemma. 1.38 Lemma: [4, Exc.13, p.37] If R is a nonsingular ring, then R[x1, x2, …, xn] is a nonsingular ring. Proof: First we shall p rove that R[x] is a nonsingular. Let f(x)  Z(R[x]), then annR[x]f(x)  e R[x]. Assume f(x) = a0 + a1x + … + anx n , where ai  R, for each i = 0, 1, …, n. annR[x]f(x) = annR[x]a0  annR[x]a1x  …  annR[x]anx n  e R[x]. So, annR[x]a0 e R[x], annR[x]a1x  e R[x], …, annR[x]anx n  e R[x]. We claim that annR(ai)  e R, for all i = 0, 1, …, n. To p rove this. Supp ose there exists J  R and J ≠ 0 such that annRa0  J = 0. J ≠ 0 so J[x] is an ideal in R[x] and J[x] ≠ 0. Hence J[x]  annR[x]a0 ≠ 0. Let 0 ≠ g(x)  annR[x]a0 and g(x)  J[x], such that g(x) = b0 + b1x + … + bmx m , where bi  J, for all i = 0, 1, …, m and there exists k  {0,1,…,m} with bk ≠ 0. Since g(x)  annR[x](a), then (b0 + b1x + …+ bk x k +… + bmx m )a0 = 0. ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 Thus b0 a0 + b1 a0 x + …+ bk a0 x k + … + bm a0 x m = 0 which imp lies b0 a0 = 0, b1 a0 = 0, …, bk a0 = 0 , …, bm a0 = 0. But bk a0 = 0 and 0 ≠ bk  J imp lies that J  annRa0 ≠ 0, which is a contradiction with our assump tion. Hence annRa0  eR, so that a0  Z(R). But R is nonsingular, hence a0 = 0. By the same way we can p rove that each of a1,…, am belongs to Z(R)=(0). Thus f(x) = 0 and Z(R[x]) = (0); that R[x] is nonsingular. Then by induction R[x1, x2,…, xn] is a nonsingular ring. 1.39 Corollary: Let R be a nonsingular ring. Then R[x1, x2,…, xn] is min-CS if and only if R[x1, x2,…, xn] is a max-CS ring. Proof: By lemma 38 if R is a nonsingular ring, then R[x1, x2,…, xn] is a nonsingular ring. Hence the result follows by theorem 33. 1.40 Not e: A direct sum of min (max)-CS modules will be discussed in another p aper. Re ferences 1. Dung,N.V.; Huy nh, D.V.; Smith, P.F. and Wisbauer, R., (1994), Extending M odules, John Wiley and Sons, Inc. New York. 2. Husain Suleman.S.Al-Hazmi, (2005), A Study of CS and -CS Rings and M odules, Ph.D.Thesis, College of Arts and Sciences of Ohio University . 3. Saad.H.M ohamed, Bruno J. M uller, (1990), Continuous and Discrete M odules, Combridge Univ. Press, New York, Port Chester M elbourne Sy dney. 4. Goodearl, K.R. (1976), Ring Theory, Nonsingular Rings and M odules, M arcel Dekker, Inc. NewYork and Basel. 5. M ahmoud A. Kamal and Amany M . M enshawy , (2007), J. Egy p t. M ath.Soc., 15(2), . 157-168. 6. Kasch,F., (1982), M odules and Rings, Academic Press, Inc. London. 7. Z.A.EL-Bast and P.F.Smith, (1988), M ultip lication M odules, Communication in Algebra, 10(4),.755-779. 8. C.P.Lu, (1984), Prime Submodules of M odules, Comment.M ath. Univ.St.Paul, 33, . 61-69. 9. Zeinab Talib Salman Al-Zubaidey, (2005), On Purely Extending M odules, M .Sc. Thesis, College of Science, University of Baghdad. 10. Ahmed, A.A. (1992), On Submodules of M ultip lication M odules, M .Sc. Thesis, University of Baghdad. مقاسات التوسع) أعظم(أصغر ة مجلة إبن الھیثم للعلوم الصرفة و التطبیقی 2012 السنة 25 المجلد 1 العدد Ibn A l-Hai tham Journal f or Pure and Applied Science No. 1 Vol. 25 Year 2012 ، رنا نوري مجید ھاديانعام محمد علي جامعة بغداد- ابن الھیثم -كلیة التربیة -قسم الریاضیات 2011 ایلول20:قبل البحث في ، 2011 أب 25:استلم البحث في الخالصة ن أصـغر مقـاس ) أعظـم( في ھذا البحث نعطي دراسة واسعة ألصغر ي المغلـق ـم اس الجزـئ ل المـق مقاسـات التوسـع مـث مقـاس توسـع ) أعظم(ومن بین النتائج االخرى نستعرض أنھ مركبة المجموع المباشر ألصغر . توسع ھو أصغر مقاس توسع ت الحلقـة ) أعظم(ھو أصغر ط إذا كانـت Rیـر مفـرده فـان الحلقـة غRمقاس توسـع وكـذلك اذا كاـن ـع إذا وفـق أعظـم حلقـة توس . أصغر حلقة توسع والتي ھي واحدة من المبرھنات المھمة في ھذا البحثRالحلقة . مقاس توسع، أصغر مقاس توسع، أعظم مقاس توسع، مقاس توسع منتظم:الكلمات المفتاحیة