ISRN AlgebraVolume 2012 (2012), Article ID 415207, 9 pagesdoi:10.5402/2012/415207

Research Article

On g-Semisymmetric Rings

Farahat S. Aly1,2 and Mohammed O. Al Mestady¹

¹Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia²Department of Mathematics, Faculty of Science, AL-Azhar University, Nasr City, P.O. Box 11884, Cairo, Egypt

Received 22 February 2012; Accepted 15 March 2012

Academic Editors: H. Chen and F. U. Coelho

Copyright © 2012 Farahat S. Aly and Mohammed O. Al Mestady. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We introduce right (left) g-semisymmetric ring as a new concept to generalize the well-known concept: symmetric ring. Examples are given to show that these classes of rings are distinct. They coincide under some conditions. It is shown that is bounded right g-semisymmetric with boundary 1 from right if and only if is symmetric, whenever is regular. It is shown that a ring is strongly regular if and only if is regular and bounded right g-semisymmetric with boundary 1 from right. For a right -ring it is shown that is reduced if and only if is symmetric, if and only if is bounded right g-semisymmetric ring with boundary 1 from left, if and only if is IFP, if and only if is abelian. We prove that there is a special subring of the ring of matrices over a ring without zero divisors which is bounded right g-semisymmetric with boundary 2 from left and boundary 2 from right. Also we show that flat left modules over bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right are bounded left g-semisymmetric with boundaries 1 from left and 1 from right.

1. Introduction

Throughout this paper, all rings are associated with identity and all modules are unitary. For a subset of , the left (right) annihilator of in is denoted by (). If , we usually abbreviate (). According to Lambic [1], a ring is called if then for . A ring is called reduced if it has no nonzero nilpotent elements. Reduced rings are symmetric according to [2, Theorem  1.3]. According to Lee and Zhou [3], a left R-module is reduced if implies , for all . Abelian rings are rings in which each idempotent is central. According to Buhphang and Rege [4], a left R-module is semicommutative, if implies , for all , . Reduced rings are symmetric [2, Theorem  1.3]. Commutative rings are symmetric. Semicommutative rings are abelian [5, Lemma  2.7]. Several examples in the indicated references were given to show that the converse of these implications is not necessary to be true, for example, [2, Example II.5] is an example of noncommutative nonreduced symmetric ring. g-semisymmetric rings are defined and studied herein. A ring is called right g-semisymmetric if for with exist two positive integers such that . A ring is called bounded right g-semisymmetric with boundary from left if for with exists two positive integers such that , for all . Clearly, symmetric rings are right g-semisymmetric. Examples 2.2 and 2.21 are given to show that there exist right g-semisymmetric rings which are not symmetric. Bounded right g-semisymmetric ring with boundary 1 from left is abelian. This is false for rings without identity, by Example 2.2. Also its converse is not necessary true as shown from Example 2.17. The converse holds if is right ring, by Theorem 2.19.

2. G-Semisymmetric Rings

Definition 2.1. A right -module is called g-semisymmetric if for and with exist two positive integers such that . A ring is called right g-semisymmetric if for with exist two positive integers such that . A left -module is called g-semisymmetric if for and with exist two positive integers such that . A ring is called left g-semisymmetric if for with exist two positive integers such that .(3) A right -module is called bounded g-semisymmetric with boundary from left if for and with exist two positive integers such that , for all . A ring is called bounded right g-semisymmetric with boundary from left if for with exist two positive integers such that , for all .(4) A right -module is called bounded g-semisymmetric with boundary from right if for and with exist two positive integers such that , for all . A ring is called bounded right g-semisymmetric with boundary from right if for with exist two positive integers such that , for all .(5) A left -module is called bounded g-semisymmetric with boundary from left if for and with exist two positive integers such that , then , for all . A ring is called bounded left g-semisymmetric with boundary from left if for with exist two positive integers such that , for all .(6) A left -module is called bounded g-semisymmetric with boundary from right if for and with exist two positive integers such that , then , for all . A ring is called bounded left g-semisymmetric with boundary from right if for with exist two positive integers such that , for all .

Every symmetric ring is right g-semisymmetric ring, the converse is not true as illustrated by the following example, due originally to Bell [6, Example  9] with changes in its multiplications.

Example 2.2. Let be the semigroup with multiplication , . Put , which is a four-element semigroup ring without identity. This ring is bounded right g-semisymmetric ring with boundary 2 from right and with boundary 1 from left, but is neither symmetric ring nor reversible.

Remark 2.3. (1) A ring is left g-semisymmetric if and only if the module is g-semisymmetric. (2) A ring is right g-semisymmetric if and only if the module is g-semisymmetric.

Proposition 2.4. The following conditions are equivalent for a right -module . (1) is g-semisymmetric.(2)All cyclic submodules of are g-semisymmetric.

Proof. Let be a cyclic submodules of , and let . Since is g-semisymmetric, then for with , it implies that , and some positive integers ,. Hence is g-semisymmetric. Let , such that . Since the cyclic -module is semisymmetric, then there exist positive integers such that . Therefore is g-semisymmetric.

Proposition 2.5. The following conditions are equivalent for a ring . (i) is strongly regular.(ii)Every right -module is flat and g-semisymmetric with boundary 1 from right.(iii)Every cyclic right -module is flat and g-semisymmetric with boundary 1 from right.(iv) is regular and bounded right g-semisymmetric with boundary 1 from right.

Proof. Let be a strongly regular ring, and let be a right -module. Then is flat module. Let and with , and let . Since is strongly regular, then the right ideal of is a two-sided ideal and has no nilpotent elements. Hence has no nilpotent elements. Since, then and hence . This shows that . Therefore is bounded g-semisymmetric with boundary 1 from right. Clear. Suppose that every cyclic right -module is flat and g-semisymmetric with boundary 1 from right. Since every cyclic right -module is flat, then is a regular ring [7, Theorem  4.21]. Since every cyclic right -module is g-semisymmetric with boundary 1 from right, then is g-semisymmetric with boundary 1 from right proving that the ring is bounded right g-semisymmetric with boundary 1 from right. Let be regular and bounded right g-semisymmetric with boundary 1 from right. Suppose that with . Since is regular, then there exists such that . Since is bounded right g-semisymmetric ring with boundary 1 from right and , then for all . Since , then . Therefore . Hence has no nonzero nilpotent element and is strongly regular ring.

Corollary 2.6. If a ring is regular and bounded right g-semisymmetric with boundary 1 from right, then is reduced.

A one-sided ideal of a ring is said to have the insertion-of-factors principle (or simply IFP) if implies for . Hence the ring is called IFP ring if the zero ideal of has the IFP. Such rings are also known as semicommutative rings or rings satisfying SI condition or ZI rings, see [6, 8–10]. The equivalences of (1), (2), (4), (5), and (6) in the following proposition are in [11, Proposition  2.7 (7)]. By Corollary 2.6 and the fact that every symmetric ring is bounded right g-semisymmetric with boundary 1 from right, we state without proof the following proposition.

Proposition 2.7. Let be a von Neumann regular ring. Then the following conditions are equivalent: (1) is right (left) duo,(2) is reduced,(3) is bounded right g-semisymmetric with boundary 1 from right,(4) is symmetric,(5) is IFP,(6) is abelian.

Proposition 2.8. (1) The class of right g-semisymmetric rings is closed under subrings. (2) The class of bounded right g-semisymmetric rings with boundaries 1 from left and 1 from right is closed under direct products. (3) A ring is semiperfect and bounded right g-semisymmetric with boundary 1 from left if and only if is a finite direct sum of local bounded right g-semisymmetric rings from left. (4) A ring is strongly regular if and only if is regular and bounded right g-semisymmetric ring with boundary 1 from left if and only if is regular and bounded right g-semisymmetric ring with boundary 1 from right.

Proof. (1) Trivial.(2) Assume that is a direct product of bounded right g-semisymmetric rings with boundaries 1 from left and 1 from right. Let , with . Then , . Since are bounded right g-semisymmetric rings with boundaries 1 from left and 1 from right, then , for all . Therefore , for all and . Hence is bounded right g-semisymmetric ring with boundaries 1 from left and 1 from right.(3) Assume that is semiperfect bounded right g-semisymmetric ring with boundary 1 from left. Since is semiperfect, has a finite orthogonal set of local idempotents whose sum is 1 [1, Proposition  3.7.2]. Hence we consider such that each is a local ring. Since is bounded right g-semisymmetric rings with boundary 1 from left, then is abelian by Lemma 2.16, whence every is central and is an ideal of . Thus , for all . It follows that each is bounded right g-semisymmetric ring with boundary 1 from left, by (1).Conversely, suppose that is a finite direct sum of local bounded right g-semisymmetric rings with boundary 1 from left. Then, by , and the fact that local rings are semiperfect, is bounded right g-semisymmetric ring with boundary 1 from left.(4) By Lemma 2.16, every bounded right g-semisymmetric ring with boundary 1 from left with identity is abelian. Moreover, as is regular, then this is equivalent to be strongly regular by [12, Theorem  3.7] which is equivalent to the condition is regular and bounded right g-semisymmetric ring with boundary 1 from right, by Proposition 2.5.

Proposition 2.9. Let be a ring homomorphism and a left -module; then is a left -module via . Moreover, (1)If is g-semisymmetric, then so is ,(2)If is onto and is g-semisymmetric, then so is.

Proof. (1) Suppose is g-semisymmetric, and let , such that . Then . Since is g-semisymmetric, then there exist positive integers such that . Hence . Therefore is g-semisymmetric.(2) Let , such that . Since is onto, there exists such that , . Now . Since is g-semisymmetric, then there exist positive integers such that and . Hence is g-semisymmetric.

Lemma 2.10 (see [10, Proposition  2.6]). Suppose that is a flat left -module. Then for every exact sequence where is -free, one has for each right ideal of ; in particular, one has for each element of .

Lemma 2.11. Let be a bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right, then every free left -module is bounded g-semisymmetric with boundaries 1 from left and 1 from right.

Proof. Since is free module, then is isomorphic to a (possibly infinite) direct sum of copies of , see [7]. Since is bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right, then is bounded g-semisymmetric with boundaries 1 from left and 1 from right, by Proposition 2.8.

Now we are ready to prove the following proposition.

Proposition 2.12. Flat left modules over bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right are bounded left g-semisymmetric with boundaries 1 from left and 1 from right.

Proof. Let be a flat module over bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right. Let and be such that . Suppose that for the epimorphism the sequence is exact. Now there exists such that . This implies that . Hence . Therefore , by Lemma 2.10. Hence for some , , yielding . Since is free -module over bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right, then is bounded g-semisymmetric with boundaries 1 from left and 1 from right, by Lemma 2.11. Therefore , for all . Hence and so gives , for all . Since , then implies , for all . Hence , for all . Thus is bounded g-semisymmetric with boundaries 1 from left and 1 from right.

In the following propositions denotes the -endomorphism ring of . The associativity is deduced from the generalized associativity situation in the standard Morita context without explicit mention, where is the left -,rightbimodule .

A torsionless module is an module such that is a direct product of copies of , or, equivalently, if , then there exists such that . If is faithful module, then is a submodule of a direct product of copies of . The following proposition is an application of Remark 2.3 and Proposition 2.8.

Proposition 2.13. The following conditions are equivalent. (1) is a bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right.(2)Every torsionless left -module is bounded g-semisymmetric with boundaries 1 from left and 1 from right.(3)Every submodule of a free left -module is bounded g-semisymmetric with boundaries 1 from left and 1 from right.(4)There exists a faithful, bounded g-semisymmetric left -module with boundaries 1 from left and 1 from right.

An application of Propositions 2.13 and 2.9 yields the following proposition.

Proposition 2.14. For an -module M, let denote the ring . Then one has the following. (1)The left -module is g-semisymmetric if and only if the left -module is g-symmetric.(2)If the left -module is bounded g-semisymmetric with boundaries 1 from left and 1 from right, then is bounded left g-semisymmetric with boundaries 1 from left and 1 from right.(3)If the right -module is bounded g-semisymmetric from left, then the ring is bounded right g-semisymmetric with boundaries 1 from left and 1 from right.

An application of Proposition 2.9 yields (1); since the left -, right -bimodule is faithful as a left -module and is also faithful as a right -module, applying of Proposition 2.13 we get (2) and (3).

Let be a right -module. Then as in [10] is called(1)reduced if , then , ;(2)ZI (zero-insertive ring) if , then , .

Proposition 2.15. Let be a right -module . Then, (1)if is reduced, then is symmetric [10, Proposition  2.2],(2)if is symmetric, then is [10, Proposition  2.2].

Lemma 2.16. If is bounded right g-semisymmetric ring with boundary 1 from left, then is abelian.

Proof. Assume that is bounded right g-semisymmetric ring with boundary 1 from left and is an idempotent. Then gives . Hence for all there exists a positive integer such that for all . Therefore . And since , then . Therefore is central.

The previous lemma is false for rings without identity. Indeed, the ring in Example 2.2 is a ring without identity and it is a bounded right g-semisymmetric ring with boundary 2 from right and 1 from left which is nonabelian ring. Also its converse is not necessary true as shown from the following example.

Example 2.17. We use [1, Example  2.10], as a counter example. Let , where is the full matrix ring over the ring of integers. Since the zero and the identity matrices are only the idempotent elements in , then R is abelian ring. Since and for any positive integers m and n, then R is not right g-semisymmetric ring.

A ring be a right ring if for any , for some idempotent of .

Proposition 2.18. Let be a right ring. If is abelian, then is reduced.

Proof. Let be abelian right ring. Let . Since is right ring, then , for some idempotent of . Since is abelian and , then and hence is reduced.

Since every reduced ring is symmetric, bounded right g-semisymmetric ring with boundary 1 from left and IFP, since every bounded right g-semisymmetric ring with boundary 1 from left is abelian, by Lemma 2.16 and since every reduced ring, symmetric ring, and IFP ring are abelian, then we deduce the following theorem from the above proposition.

Theorem 2.19. Let be right ring. Then the following are equivalent. (1) is reduced.(2) is symmetric.(3) is bounded right g-semisymmetric ring with boundary 1 from left.(4) is IFP.(5) is abelian.

Theorem 2.20. Let be a ring without zero divisors and . Then is bounded right g-semisymmetric with boundary 2 from left and boundary 2 from right.

Proof. Suppose that ,, such that . Then,, , , . Therefore we have the following cases: (1) if , , , then , impossible,(2) if , , then and ; in this case, ,(3) if , , , then , impossible, if , , , then , , and ;hence ,(5) if , , , then ,, and which implies that ,(6) if , , then , and which implies that . These cases prove that R is bounded right g-semisymmetric ring with boundary 2 from left and right.

The following example gives a bounded right g-semisymmetric ring with boundary 2 from left and right which is not symmetric.

Example 2.21. Let . Then is a bounded right g-semisymmetric ring with boundary 2 from left and boundary 2 from right which is not symmetric. Since is a ring without zero divisors, then is bounded right g-semisymmetric ring with boundary 2 from left and boundary 2 from right, by the above theorem. This ring is not symmetric, indeed; suppose , , , then and , and hence is not symmetric ring. Also we notice that and and therefore is not reduced.

Acknowledgments

This Project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under Grant no. 03-41/429. The authors, therefore, acknowledge with thanks DSR technical and financial support. The authors would like to express their great appreciation to the professor Mohammed Hessain Fahmy for helpful discussions, They had with him, during the preparation of this work. The authors thank the referee for his or her comments, which considerably improved the presentation.

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