R- Annihilator Small Submodules المقاسات الجزئية الصغيرة من النمط تالف R

Transcription

1 ISSN: R- Annihilator Small Submodules Hala K. Al-Hurmuzy*, Bahar H. Al-Bahrany Department of Mathematics, College of Science, Baghdad University, Baghdad, Iraq Abstract Let R be an associative ring with identity and let M be a unitary left R- module.we call a submodule K of M, R-annihilator small if K+T=M,where T is a submodule of M, implies that ann(t)=0, where ann(t) indicates annihilator of T in R.The sum of all such submodules of M contains the Jacobson radical J(M) and the singular submodule Z(M).When M is finitly generated and faithful, we study and in this paper.conditions when is R-annihilator-small and =, J(M) and are given. Keywords: Small submodules, annihilators, R-annihilator- small submodules. الخالصة المقاسات الجزئية الصغيرة من النمط تالف R هالة قحطان الهرمزي*, بهار حمد البح ارني قسم الرياضيات, كلية العلوم, جامعة بغداد, بغداد, الع ارق لتكن R حلقة تجميعية ذات عنصر محايد وليكن M مقاسا ايس ار على الحلقة R نقول لمقاس جزئي N من M=K+T وT هو مقاس جزئي من M يقود الى ann = 0 حيث ان لكل الموديوالت الجزئية الصغيرة من النمط تالف R يحوي على و واعطينا شروط ليكون M بأنه صغير من النمط تالف R اذا ann هو تالف T في R. المجموع Rad(M) و Z(M) في حالة كون M منتهي التولد ومخلص. درسنا. موديول جزئي صغير من النمط تالف R و = 1-Introduction Throughout this paper all rings are associative ring with identity and modules are unitary left modules. In [1], Nicholson and Zhou defined annihilator small right(left) ideals as follows : a left ideal A of a ring R is called annihilator-small if A+T=R,where T is a left ideal, implies that r(t)=0, where r(t) indicates the right annihilator. Kalati and.keskin consider this problem for modules in [2] as follows:- let M be an R- module and S=End(M). A submodule K of M is called annihilator-small if K+T =M, T a submodule of M,implies that (T) =0, where indicates the right annihilator of T over S= End(M), where (T)={f t }. These observation lead us to introduce the following concept. A submodul N of an R- module M is called R-annihilator- small if N+T=M, T a submodule of M, implies that (T) =0, where (T) ={r }. In fact, the set of all elements k such that Rk is semisubmodule and annihilator-small.and contains both the Jacobson radical and the singular submodule when M is finitely generated and faithful. The submodule generated by is a submodule of M analogue of the Jacobson radical that contains every R-annihilator small submodules. In this work we give some basic properties of R- annihilator-small submodules and various. * hala_hamdi73@yahoo.com 129

2 Characterizations We abbreviate the Jacobson radical as Rad(M) and the singular submodule as Z(M) for any R- module M. The notations N and N mean respectively that a submodule N of M is essential and maximal in the module M. See [1] / [2]. 2. R- annihilator small submodules. In this section we give various characterizations of R-annihilator-small submodules.we start this section by the following definition: Definition (2.1) We say that a submodule N of a module M is R- annihilator small in M(R-a-small) if N+X=M, X a submodule of M,implies that annx=0, we write N M in this case. Examples (2.2.):- 1- R-a-small submodule need not be small, for example, consider Z as Z- module, every proper submodule of Z is Z-a-small.But is the only small submodule of Z. 2-A small submodule need not be R- a- small, for example, consider as Z- module, has no Z-asmall submodule. But and {, } are small in. Now, we give some basic properties of R-a-small submodules Proposition(2-3):- Let A and B be submodules a module M such that A B. if A B,then A M. Proof:- Let M=A+X, X M. By Modular Law, B=A+( X B) Since A B,then ann(x B)=0. One can easily show that annx ann(x B), therefor annx=0. Thus A M. Proposition(2-4):- Let A and B be sub modules of a module M such that A B. If B M, Then A M. Proof:- Let M=A+X,Then M =B+X. Since B M, then annx=0. Proposition(2-5):- let M and N be two R- modules and f:m N be an epimorphism. If H N, then (H) M. Proof:- Let M = (H)+X, X M.Then f( M) = (H)+X))= (H))+f(X).Since f is an epimorphism, then N=H +f(x).but H N, therefor annf(x)=0. Clearly that annx annf(x) and hence annx=0. Thus (H) M. Note. Let f:m N be an epimorphism. Then the image of R-a-small submodule of M need not be R- a-small in N as the following example shows:- Consider Z and as Z modules and let be the natural epimorphism. Z. But = is not Z-a-small in, where = + and ann =4Z. Note. The sum of two R-a-small submodules of a module M need not be R-a-small submodule. For example, In Z as Z- module. each of 2Z and 3Z are Z-a-small submodule of Z. But Z=3Z+2Z is not Z-a-small in Z. We prove the following :- Proposition(2-6):- Let and be a R- modules. if, then. Proof:- Let, i=1,2 be the projection maps. Since, then by prop(2.5) = and =. By prop(2.4) ( )=. Remark (2.7):- Let R be an integeral domain and let M be a torsion free module. Then every proper submodule of M is R-a-small in M. Proof:- Clear Remark (2.8):- Let M be a faithful R-module, then every small submodule is a-small. Proof:- Clear. Proposition(2-9):- Let M be a faithful R- module and N be a submodule of M such that, then N M. Proof:- Let M=N+K, Then 0= annm =ann(n+k). One can easily show that ann(n+k)= annn annk.then 0= annn annk. But,therefor annk=0 Thus N M. Proposition(2-10):- Let R be an integral domain and M be a faithful R-module. Then every submodule N of M such that annn is R-a-small. Proof:- Let M=N+K, Then 0= annm =ann(n+k)= annn annk, 130

3 Since annn and R is integral domain, then by [4]. Therefor annk=0. Thus N is R-a-small submodule of M. Proposition(2-11):- Let R be an integral domain and M be a faithful and torsion module. Then every finitely generated submodule N of M is R-a-small. Proof:- Let N= + + +R be a submodule of M and M=N+K. Then 0=annM=ann(N+K) = annn annk =(ann( + + +R )) annk=( ) annk Since M is torsion, then annr, i=1,2,,n. But R is an integral domain, therefor annr is essential in R, i=1,2,,n and Hence is essential in R, by [3] Hence annk=0. Thus N M. 3-Characterizations of R-annihilator-small submodules In this section we give various characterizations of R-annihilator-small submodules. Compare the following with prop 2 in [1] Proposition (3-1):- Let M be a finitely generated module and K M,then K+RadM)+Z(M) M. Proof:- Let M= + + +R, M, i=1,2,,n and Let M = K+Rad(M)+Z(M)+X. Since M is finitely generated, then Rad(M) M, by [3] And hence M=. So = + +,,, i=1,2,,n. It s easy to show that M= + + +R +X. But k M, therefor ann ( + + +R +X)=0. Hence ( )) annx=0. Since (M), i=1,2,,n Then ann( ) R, i=1,2,,n and hence ) R, by [4]. So ann X=0. Thus K+Rad(M)+Z(M) M. By the same way one can prove the following Proposition Proposition(3-2):- Let M be a module and K M. If Rad(M) M and Z(M) is finitely generated, then K+Rad(M)+Z(M) M. The following theorem gives a characterizations of cyclic R-a-small submodules Theorem( 3.3):- Let M= be a module and k M. Then the following statements are equivalent:- 1. R M 2. =0, Proof:- (1) (2) Let For each. Then +, Then M= ) + R. Since, then 0= ann( )) =. (2) (1) Let M = R + B. Then for each, +, and Now let t annb. +. Since =0, then t ( ) =0,. So t ann( ) =0, Hence t ) =0 Now, compare it with lemma 4 in [1]. Theorem ( 3.4):- Let R be a commutative ring, M= be a module and k M. Then the following statements are equivalent:- 1. R M 2. =0, 3. There exists s.t b Rbk, Proof:- (1) (2) By Th (3.3) (2) (3) Let. Assume that b Rbk,. Then b = b k = b k,. Therefor b ), And hence =0 which is a contradiction (3) (2) Let and hence b ) 131

4 Therefor b = k,. So b Rbk. By our a assumption, b=0. The following theorem is analogous to prop7 in [1]. Theorem ( 3.5):- Let R be a commutative ring, let M= be a module and K M. Then the following statements are equivalent:- 1) K 2) =0, Proof:- (1) (2) Let,. Then =, And hence M= (. But K, therefor 0= ann( ( )=. (2) (1) Let M=K+A. Then for each, =,,. Hence, for each. So M= (. Now let t ann A. Therefor t(,. So t =0. Thus anna = 0 and K. Definition (3.6):- Let M be a module and, we say that k is R-a-small in M if Rk. Let = {k }. Note that Z(M) and Rad(M), when M is finitely generated and faithful, by prop (3.1) and prop(2.4). Note. is not closed under addition in general. For example Consider Z as Z-module.The sum of R-a-small need not be R-a-small. Clearly that 3Z Z and 2Z Z. But Z=3Z+2Z is not R-a-small in Z. Remark (3.7):- Let M be a module and.then R. Proof:- Let r. Clearly that Rrk. By prop(2.4), Rrk and hence rk. Thus. Remark (3.8):- Let M be a module and, then A. Proof:- Let x, then Rx and hence Rx, by prop(2.4). Thus. As we have seen, the sum of R-a-small submodules need not be R-a-small (consider 3Z+2Z in Z) With this in mind we define. Definition (3.9):- Let M be a module. Let R-a-small submodule of M be the sum of R-a-small submodules of M. If M has no R-a-small submodule, we write =M. It is clear that in every module, but this may not be equality (consider Z as Z-module). Proposition (3-10):- Let M be a module such that.then 1) - is a submodule of M, contains every R- a- small submodule of M. 2) - = { for each i, n }. 3) - is generated by. 4) - if M finitely generated, then Rad and. Proof:-. Proposition (3-11):- Let M be a module such that.then the following statements are equivalent:- 1) is closed under addition, that is a finite sum of R-a - small elements is R-a -small. 2) =. Proof:- (1) (2) Let + + +, and, i=1,,n Then, by prop (2.4). Hence, i=1,,n. By our assumption, Thus =. (2) (1) Assume that = and let x, y. Since, then x, y. But is a submodule of M, by prop (3.10), therefor x+y =. Thus is closed under addition. Proposition(3-12):- Let M be a module such that.consider the following statements:- (1) - (2) - if K and L then K+L (3) - is closed under addition, that is the sum of R- a small elements is R- a small. (4) - = Then (1) (2) (3) (4).If M is finitely generated, then (1) 132

5 Proof:- (1) (2) Assume that and let K and L be R- a small submodules of M. Then K+L. But, therefor K+L, by prop (2.4). (3) Clear (3) (4) By prop (3.11) To show that (2) (1) let M= + +.+R. And let M= +X. Then = +, And, i=1,,n. Therefor M = +X. Since, i=1,,n Then Hence, i=1,,n. By our assumption, So annx=0. Proposition(3-13):- Let M be a finitely generated module such that. Then :- (1) - is the unique largest R- a -small submodule of M. (2) - = { w maximal submodule of M with }. Proof:- ( 1) - Clear (2) - Let a { w maximal submodule of M with }. Claim that Ra. assume not. Then M=Ra+X, X and annx 0. Since, then M + X. But M is finitely generated,then there exist, a maximal submodule such that + X B, by [3]. Now, if a B we get B=M which is a contradiction. So a. But a { w maximal submodule of M with }. which is a contradiction. Thus Ra and hence a. Reference 1. Nicholson, W.K. and Zhou, Y Annihilator small right ideals, Algebra Colloq., 18(1), pp: Amouzegar Kalati, T. and Keskin-Tutuncu, D Annihilator-Small Sub Modules, Bulletine of the Iranian Mathematical Society, 39(6), pp: Goodearl, K.R Rings Theory, Non-singular Rings and Modules, Mercel Dekker, New York. 4. Kasch F Modules and Rings, Academic press, London. 5. Anderson, D. and Camillo, V.P Armendariz rings and Gaussian rings, Comm. In Algebra. 26, pp: