Microsoft Word - 235-244 235 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 2-Regular Modules II Nuhad S.AL-Mothafar Dept. of Mathematic/ College of Science/ University of Baghdad Ghaleb A. Humod Dept. of Mathematic/ College of Education for Pure Science (Ibn Al-Haitham)/ University of Baghdad Received in:28/March/2015,Accepted in:7/June/2015 Abstract An R-module M is called a 2-regular module if every submodule N of M is 2-pure submodule, where a submodule N of M is 2-pure in M if for every ideal I of R, I2MN = I2N, [1]. This paper is a continuation of [1]. We give some conditions to characterize this class of modules, also many relationships with other related concepts are introduced. Key Words: 2-pure submodules, 2-regular modules, pure submodule, regular modules. 236 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 0- Introduction Throughout this paper, R is a commutative ring with identity and all R-modules are unitary. A submodule N of an R-module M is called 2-pure submodule if for every ideal I of R, I2MN = I2N. If every submodule of M is 2-pure, then M is said to be 2-regular module. This work consists of two sections. In the first section we give some properties of 2-regular rings. Next we present a characterization of 2-regular modules. In the second section we illustrate some relationships between the concept 2-regular modules and other modules such as semiprime divisible, projective and multiplication modules. 1- 2-Regular Modules In this section, we first define 2-regular rings and study some of its properties. Next we consider some conditions to characterize 2-regular modules. Definition (1.1): [1] An ideal I of a ring R is called 2-pure ideal of R if for each ideal J of R, J2  I = J2I. If every ideal of a ring R is 2-pure ideal, then we say R is 2-regular ring. Remarks and Examples (1.2): (1) It is clear every (von Neumman) regular ring is 2-regular ring, but the converse is not true, for example: the ring Z4 is 2-regular ring, since every ideal of Z4 is 2-pure. But Z4 is not regular since the ideal {0, 2} is not pure because {0, 2} {0, 2} {0, 2}  , on the other hand {0, 2} {0, 2} {0}  implies {0, 2} {0, 2} {0, 2} {0, 2}   . (2) It is clear that {0} and R are always 2-pure ideals of any ring R. (3) Every field is 2-regular ring. (4) Let R be an integral domain. If R is 2-regular ring, then R is a field. Proof: Let I be an ideal of R. Since R is 2-regular ring then J2  I = J2I for every ideal J of R. If we take J = I implies I2 = I3. Thus for each element 0  a  R, < a >2 = < a >3, hence a2  < a >3. Let a2 = r a3 for some r  R, then a2(1 – ra) = 0 but R is domain and a  0 implies 1 – ra = 0, thus 1 = ra. Therefore a is an invertible element of R. Thus R is a field (5) If R is a 2-regular ring then every prime ideal of R is a maximal ideal. Proof: Let I be a prime ideal of R. Since R is a 2-regular ring then R I is 2-regular by [1,Cor.3.2]. But R I is a domain since I is a prime ideal. Thus R I is a field by the above remark. Therefore I is a maximal ideal. (6) Every 2-regular ring is nearly regular, where a ring R is called nearly regular if R J(R) is regular ring, see [2], where J(R) = the intersection of all maximal ideals of R. Proof: 237 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 Let R be a 2-regular ring. Then R J(R) is 2-regular by corollary (1.2.3). So by above remark (5), every prime ideal of R J(R) is a maximal ideal and since R J 0 J(R)       , therefore by [3], R J(R) is regular. Proposition (1.3): Let M be 2-regular R-module then for every element x of Mand every element r  R, r2x = r2tr2x for some t  R. Proof: Let x be an element of M and r be an element of R. Since r2x  r2M and r2x  < r2x > implies r2x  r2M  < r2x >. But M is 2-regular, then r2M< r2x >=r2 < r2x >. Thus, r2xr2 < r2x > implies r2x=r2t r2x for some tR. Proposition (1.4): Let M be a module over principal ideal ring R. If for every element x of M and every element r  R, r2x = r2tr2x for some t  R implies M is a 2-regular module. Proof: Let N be a submodule of M and I is an ideal of R. First, to prove r2M  N = r2N for every element r  R. Let x  r2M  N implies x  r2M, x  N. Thus x = r2m for some m  M. Then x = r2tr2m for some t  R by hypothesis. Hence x  r2N. But R is a principal ideal ring. Therefore I2M  N = I2N. Proposition (1.5): Let M be a cyclic R-module. If for every element x of M and every element r of R, r2x = r2tr2x for some t  R, implies M is a 2-regular module. Proof: Let M = Rm be a cyclic module for some m  M. Let N be a submodule of M and I is an ideal of R. Let y  I2M  N then y  I2M and y  N. Thus y = r2m = r2tr2m  r2N for some t  R and r  I. Therefore y  I2N implies M is 2-regular. The proof of the following result is similar to that of propositions (1.3) and (1.4). Corollary (1.6): Let R be a 2-regular ring then for every element a  R, a2 = a2t a2 for some t  R, and the converse is true if R is a principal ideal ring. Proposition (1.7): Let R be a principal ideal ring and M be an R-module. The following statements are equivalent: (1) M is 2-regular module. (2) R R ann(x) is 2-regular for every element x of M. (3) For every element x of M and every element r of R, r2x = r2tr2x for some t  R. 238 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 Proof: (1)  (3) It follows by Proposition (1.3). (3)  (1) By Proposition (1.4). (1)  (2) Let r + R ann(x)  R R ann(x) where x  M and r  R. Since M is 2-regular, then r2x = r2tr2x for some t  R. Thus r2 – r2tr2  R ann(x) implies R R ann(x) is 2-regular. (2)  (1) Let x  M and r  R. Since R R ann(x) is 2-regular, then r2 + R ann(x) = (r2 + R ann(x) (t + R ann(x) )(r2 + R ann(x) ) for some t  R. Thus r2x = r2tr2x implies M is 2-regular. We have the following results: Corollary (1.8): Let R be a principal ideal ring. Then R is 2-regular if and only if all R-modules are 2-regular. Proof: () Let R be 2-regular ring and M is an R-module. Then R R ann(x) is 2-regular for every element x  M by [1,Cor.(3.3)]. Therefore M is 2-regular by proposition (1.7). () Assume all R-modules are 2-regular. Thus R is 2-regular R-module. By Proposition (1.7), R R ann(x) is 2-regular for some every element x  R, so if take x = 1  R implies R R R R ann(x) 0     , therefore R is 2-regular. Corollary (1.9): Let R be a principal ideal ring. Then R is a 2-regular if and only if R is 2-regular R-module. Proof: By the same argument of Corollary (1.8). Corollary (1. 10): Let R be a principal ideal ring. If R R ann(M) is 2-regular then M is 2-regular R-module. Proof: Let x be a non-zero element of M. Since R R ann(M) ann(x) , there exists an epimorphism f: R R R R ann(M) ann(x)  defined by f (r + R ann(M) ) = r + R ann(x) . Therefore R R ann(x) is 2-regular by [1,Cor.(3.3)]. Then M is 2-regular by Proposition (1.7). 239 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 2- Regular Modules and Other Related Modules In this section, we study the relationships between 2-regular modules and other modules such as semiprime, divisible, projective and multiplication modules. Recall that a proper submodule N of an R-module M is called a semiprime submodule if for every r  R, x  M, k  Z+ such that rkx  N implies rx  N implies rx  N, see [4]. Equivalently, a proper submodule N of M is semiprime if for every r  R, x  M such that r2x  N implies rx  N, see [5]. An R-module M is called semiprime if <0> is a semiprime submodule of M. The proof of the following result follows by [5]. Proposition (2.1): Let R be a principal ideal ring and M is an R-module. If every proper submodule of M is semiprime then M is a 2-regular module. The converse is not true, for example: The module Z4 as Z-module is 2-regular but <0> is not semiprime. The following proposition gives a partial converse of proposition (2.1). Proposition (2.2): Let M be 2-regular and semiprime R-module then every proper submodule of M is semiprime. Proof: Let N be a proper submodule of M and r2x  N where r  R, x  M implies r2x  r2M  N = r2N since M is 2-regular. Then r2x = r2n for some n  N, thus r2(x – n)  <0>. But <0> is semiprime, hence rx = rn  N. Therefore N is semiprime submodule of M. Before we give a consequence of Proposition (2.2), we need the following lemma: Lemma (2.3): Let M be 2-regular and semiprime R-module then J(R)M = <0>. Proof: Let r  J(R) and x  M then r2x=r2tr2x for some t  R since M is 2-regular, r2x(1 – r2t)=0 implies 1 – r2t is invertible in R. Then r2x = 0, but M is semiprime thus rx = 0. Therefore J(R)M = <0>. Recall that an R-module M is called semisimple if every submodule of M is a summand. The sum of all simple submodules of a module M is called the socle of M is denoted by Soc(M), moreover if Soc(M) = 0 , then M has no simple submodule and if Soc(M) = M then M is semisimple module, see [6]. A commutative ring is a local ring in case it has a unique maximal ideal, see [7]. Corollary (2.4): Let R be a local ring and M is 2-regular and semiprime R-module then M is a semisimple and hence is regular. 240 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 Proof: Since R is a local ring, then R J(R) is a simple ring and hence is semisimple. By [6], Soc(M) = M ann(J(R)) = {m  M; mJ(R) = 0}. But J(R)M = <0> by lemma (2.3), thus Soc(M) = M. Therefore M is semisimple. Now, we have the following: Proposition (2.5): Let N be a semiprime submodule of an R-module M and K is a 2-pure submodule of M containing N, then K N is semiprime submodule in M N . Proof: Let r2(x + N)  K N for some r  R and x + N  M N . Then r2x  K, imples r2x  r2M  K = r2K since K is 2-pure in M. Let r2x = r2m for some m  K. Thus r2(x – m) = 0  N implies r(x – m)  N since N is semiprime submodule in M, hence r(x + N) = rm + N  K N . Therefore K N is semiprime submodule in M N . Corollary (2.6): Let N be a semiprime submodule of an R-module M and K is a 2-pure in M with N  K then K is semiprime submodule in M. Proof: Let r2x  K for some r  R and x  M. Thus r2(x + N)  K N , but K N is semiprime in M K by Proposition (2.5) therefore r(x + N)  K N . Hence rx  K, that is K is semiprime in M. Let R be an integral domain, an R-module M is said to be divisible if and only if rM = M for every non-zero element r of R, see [8]. An R-module M is said to be a prime module if R R ann(M) ann(N) for every non-zero submodule N of M, see [9]. Proposition (2.7): Let M be a module over a principal ideal domain R and N is a divisible R-submodule of M then N is a 2-pure submodule in M. Proof: Since N is divisible then for each rR, r2N=N. Therefore N  r2M = r2N. 241 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 Remark (2.8): The converse of proposition (2.7) is not true, for example: the submodule {0, 2} of the module Z4 as Z-module where {0, 2} is 2-pure in Z4, but is not divisible since there exists 2  Z and 2{0, 2} = {0} . That is 2{0, 2}  {0, 2} . The following proposition gives a condition under which the converse of proposition (2.7) is true. Proposition (2.9): Let M be divisible module over a principal ideal domain R and N is a 2-pure in M then N is divisible. Proof: Assume N is 2-pure in M, let m  N and r  R. Since M is divisible implies m = r2x for some x  M. But m = r2x  r2M  N = r2N  rN. Therefore N = rN. As an immediate consequence we have the following: Corollary (2.10): Let R be a principal ideal domain and every proper submodule of an R-module M is divisible then M is 2-regular. The converse is true if M is divisible. Proof: Follows by Propositions (2.7) and (2.9). Corollary (2.11): Let R be a principal ideal domain and M is 2-regular and divisible R-module then M is prime module. Proof: By above corollary (2.10), every submodule N of M is divisible. Thus rN = N for every r  R. Therefore R R ann(N) ann(M) 0   . Hence M is prime module. Corollary (2.12): Let R be a principal ideal domain and M is 2-regular injective R-module then M is prime module. Proof: Clear We give the following theorem. Theorem (2.13): Let R be any ring. The following statements are equivalent: (1)   R is 2-regular R-module for any index set . (2) Every projective R-module is 2-regulaar module. Proof: 242 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 (1)  (2) Let M be projective R-module then there exists a free R-module F and an R- epimorphism f : F  M, and F    R where  is an index set. We have the following short exact sequence i0 ker R M 0ff      Where i is the inclusion mapping. Since M is projective, the sequence is split implies that   R  ker f  M. But   R is 2-regular R-module. Therefore by [1,Cor.(3.4)] M is 2-regular module. (2)  (1) Assume that every projective R-module is 2-regular module. Since R is projective R-module, then   R is projective because the direct sum of projective modules is projective. Therefore   R is 2-regular R-module for any index set . Recall that an R-module M is called multiplication module if for every submodule N of M there exists an ideal I of R such that N = IM, see [10] We have the following: Proposition (2.14): If M is a finitely generated faithful multiplication R-module. The following statements are equivalent: (1) R is 2-regular ring. (2) M is 2-regular R-module. Proof: (1)  (2) Let N be a submodule of M and I is an ideal of R. Since I2M  N = I2M  JM for some ideal J of R = (I2  J)M since M is faithful multiplication, see [10] = (I2J)M since R is 2-regular = I2(JM) = I2N Therefore M is 2-regular. (2)  (1) Let I and J be ideals of R. Since (I2  J)M = I2M  JM because M is faithful multiplication = I2(JM) since M is 2-regular = (I2J)M Thus I2  J = I2J since M is finitely generated faithful multiplication, see [10]. Therefore R is 2-regular ring. Recall that an R-module M is said to be I-multiplication module if each submodule N of M of the form JM for some idempotent ideal J of R, see [11]. It is clear that every I-miltiplication module is multiplication but not the converse. Clearly the two concepts multiplication and I-multiplication modules are equivalent over regular rings. However we have the following: Proposition (2.15): If M is I-multiplication and 2-regular R-module then M is regular module. Proof: Let N be a submodule of M and I is an ideal of R. Since 243 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 IM  N = IM  JM = IM  J2M for some idempotent J = J2 = J2(IM) since M is 2-regular = (I2J)M since R is 2-regular = I(J2M) = I(JM) = IN Therefore M is regular module. Proposition (2.16): If M is I-multiplication and 2-regular R-module then every submodule N of M is I-multiplication as R-module. Proof: Let N be a submodule of M and K is any submodule in N, then K is a submodule of M and K = IM = I2M for some idempotent ideal I of R. Since K = N  K = N  I2M = I2N because M is 2-regular = IN Thus N is I-multiplication R-module. References 1. Nuhad, S.AL-Mothafar and Ghaleb, A.Humod, 2-Regular Modules, to appear. 2. Naoum, A.G. and Nuhad, S.AL-Mothafar, (1994),Nearly Regular Ring and Nearly Regular Modules, Mutah. Journal For Research and Studies, 9, 6. 3. Aziz, B.B., (1975), Regular rings and Bear rings, M.Sc.Thesis, University of Baghdad. 4. .Duns, (1988), Modules and One-Sided Ideal in Ring Theory and Algebra, 19, 755-779. 5. Athab, E.A., (1996), Prime Submodules and Semiprime Submodules, M.Sc. Thesis, University of Baghdad, Iraq. 6. Kasch, F., (1982), Modules and Rings, Academic Press, New York. 7. Anderson, F.W. and Fuller K.R., (1992), Rings and Categories of Modules, Springer Verlag, New York. 8. Sharp, D.W. and Vamos, P., (1972), Injective Modules, Combrdige University, Press. 9. Desale, G.and Nicholson, W.K., (1981), Endoprimitive Rings, J.Algebra, 70, 548-560. 10. EL-Bast, Z.A., and Smith, P.F., (1988), Multiplication Modules, Commun. Algebra, 16, 4, 755-779. 11. Abbas, M.S., (1990), On Fully Stable Modules, Ph.D. Thesis, University of Baghdad, Iraq. 244 | Mathematics 2015) عام 3العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 28 (3) 2015 II 2-المقاسات المنتظمة من النمط نھاد سالم عبد الكريم جامعة بغداد /كلية العلوم /قسم الرياضيات غالب أحمد حمود جامعة بغداد / )ابن الھيثم(كلية التربية للعلوم الصرفة /قسم الرياضيات 2015/حزيران/7،قبل البحث في:2015/نيسان/ 28استلم البحث في: خالصةال اذا كان كل مقاس 2 –بأنه منتظم من النمط Mحلقة إبدالية ذات محايد. يقال ان المقاس Rإذ Rمقاسا ً على Mليكن اذا حقق Mفي 2-بأنه نقي من النمط Nإذ يقال عن المقاس الجزئي 2-ھو مقاس جزئي نقي من النمط Mجزئي في N2N = IM2I لكل مثاليI فيR ،[1] . تمييزاً . في القسم االول من ھذا البحث أعطينا [1] 2-في ھذا البحث نستمر بدراسة مفھوم االنتظام من النمط وانواع اخرى من 2-. في القسم الثاني درسنا العالقة بين المقاسات المنتظمة من النمط2-للمقاسات المنتظمة من النمط المقاسات. ، المقاسات الجزئية النقية، 2 –، المقاسات النتظمة من النمط 2 –المقاسات الجزئية النقية من النمط : الكلمات المفتاحية المنتظمة.المقاسات