On g-Semisymmetric Rings

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 R is bounded right g-semisymmetric with boundary 1 from right if and only if R is symmetric, whenever R is regular. It is shown that a ring R is strongly regular if and only if R is regular and bounded right g-semisymmetric with boundary 1 from right. For a right p.p.-ring R it is shown that R is reduced if and only if R is symmetric, if and only if R is bounded right g-semisymmetric ring with boundary 1 from left, if and only if R is IFP, if and only if R is abelian. We prove that there is a special subring of the ring of 3 × 3 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.


Introduction
Throughout this paper, all rings are associated with identity and all modules are unitary. For a subset X of R, the left right annihilator of X in R is denoted by l X r X . If X {a}, we usually abbreviate l a r a . According to Lambic 1 , a ring R is called symmetric if abc 0 then acb 0 for a, b ∈ R. A ring R is called reduced if it has no nonzero nilpotent elements. Reduced rings are symmetric according to 2, Theorem 1.3 . According to Lee  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 R is called right g-semisymmetric if for a, b, c ∈ R with abc 0, there exist two positive integers n n c , l l b such that ac n b l 0. A ring R is called bounded right gsemisymmetric with boundary n from left if for a, b, c ∈ R with abc 0, there exists two positive integers n, l l b such that ac s b l 0, for all s ≥ n. Clearly, symmetric rings are right g-semisymmetric. Examples 2.2 and 2.21 are given to show that there exist right gsemisymmetric 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 R is right p.p.-ring, by Theorem 2.19.

G-Semisymmetric Rings
0, there exist two positive integers n n b , l l a such that mb n a l 0. A ring R is called right g-semisymmetric if for a, b, c ∈ R with abc 0, there exist two positive integers n n c , l l b such that ac n b l 0.
2 A left R-module M is called g-semisymmetric if for m ∈ M and a, b ∈ R with abm 0, there exist two positive integers n n b , l l a such that b n a l m 0. A ring R is called left g-semisymmetric if for a, b, c ∈ R with abc 0, there exist two positive integers n n b , l l a such that b n a l c 0.
3 A right R-module M is called bounded g-semisymmetric with boundary n from left if for m ∈ M and a, b ∈ R with mab 0, there exist two positive integers n, l l a such that mb s a l 0, for all s ≥ n. A ring R is called bounded right g-semisymmetric with boundary n from left if for a, b, c ∈ R with abc 0, there exist two positive integers n, l l b such that ac s b l 0, for all s ≥ n.
4 A right R-module M is called bounded g-semisymmetric with boundary l from right if for m ∈ M and a, b ∈ R with mab 0, there exist two positive integers n n b , l such that mb n a s 0, for all s ≥ l. A ring R is called bounded right g-semisymmetric with boundary l from right if for a, b, c ∈ R with abc 0, there exist two positive integers n n c , l such that ac n b s 0, for all s ≥ l.
5 A left R-module M is called bounded g-semisymmetric with boundary n from left if for m ∈ M and a, b ∈ R with abm 0, there exist two positive integers n, l such that abm 0, then b s a l m 0, for all s ≥ n. A ring R is called bounded left g-semisymmetric with boundary n from left if for a, b, c ∈ R with abc 0, there exist two positive integers n, l l a such that b s a l c 0, for all s ≥ n.
6 A left R-module M is called bounded g-semisymmetric with boundary l from right if for m ∈ M and a, b ∈ R with abm 0, there exist two positive integers n n b , l such that abm 0, then b n a s m 0, for all s ≥ l. A ring R is called bounded left g-semisymmetric with boundary l from right if for a, b, c ∈ R with abc 0, there exist two positive integers n n b , l such that b n a s c 0, for all s ≥ l.
Every symmetric ring is right g-semisymmetric ring, the converse is not true as illustrated by the following example, due originally to Bell  1 M is g-semisymmetric.
2 All cyclic submodules of M are g-semisymmetric.
Proof. 1 ⇒ 2 Let N mR be a cyclic submodules of M, and let m ∈ N. Since M is gsemisymmetric, then for a, b ∈ R with m ab 0, it implies that m b m a n 0, and some positive integers m m b ,n n a . Hence N is g-semisymmetric.
2 ⇒ 1 Let a, b ∈ R, m ∈ M such that mab 0. Since the cyclic R-module mR is semisymmetric, then there exist positive integers m m b , n n a such that mb m a n 0. Therefore M is g-semisymmetric.
Proposition 2.5. The following conditions are equivalent for a ring R.
i R is strongly regular.
ii Every right R-module is flat and g-semisymmetric with boundary 1 from right.
iii Every cyclic right R-module is flat and g-semisymmetric with boundary 1 from right.
iv R is regular and bounded right g-semisymmetric with boundary 1 from right.
Proof. i ⇒ ii Let R be a strongly regular ring, and let M be a right R-module. Then M is flat module. Let m ∈ M and r, s ∈ R with mrs 0, and let I {x ∈ R | mx 0}. Since R is strongly regular, then the right ideal I of R is a two-sided ideal and R has no nilpotent elements. Hence R R/I has no nilpotent elements. Since rs ∈ I, then s m r n 2 0 and hence s m r 0. This shows that ms m r 0. Therefore M is bounded g-semisymmetric with boundary 1 from right.
ii ⇒ iii Clear.
iii ⇒ iv Suppose that every cyclic right R-module is flat and g-semisymmetric with boundary 1 from right. Since every cyclic right R-module is flat, then R is a regular ring 7, Theorem 4.21 . Since every cyclic right R-module is g-semisymmetric with boundary 1 from right, then R R is g-semisymmetric with boundary 1 from right proving that the ring R is bounded right g-semisymmetric with boundary 1 from right. iv ⇒ i Let R be regular and bounded right g-semisymmetric with boundary 1 from right. Suppose that x ∈ R with x 2 0. Since R is regular, then there exists y ∈ R such that x xyx. Since R is bounded right g-semisymmetric ring with boundary 1 from right and y x xy 0, then y xy n x l 0 for all l ≥ 1. Since x xyx xyxyx xyxyxyx · · · xy n x, then yx y xy n x 0. Therefore x xyx 0. Hence R has no nonzero nilpotent element and R is strongly regular ring. Corollary 2.6. If a ring R is regular and bounded right g-semisymmetric with boundary 1 from right, then R is reduced.

ISRN Algebra
A one-sided ideal I of a ring R is said to have the insertion-of-factors principle or simply IFP if ab ∈ I implies aRb ⊆ I for a, b ∈ R. Hence the ring R is called IFP ring if the zero ideal of R 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 R be a von Neumann regular ring. Then the following conditions are equivalent:

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 R is a finite direct sum of local bounded right g-semisymmetric rings from left. (4) A ring R is strongly regular if and only if R is regular and bounded right g-semisymmetric ring with boundary 1 from left if and only if R is regular and bounded right g-semisymmetric ring with boundary 1 from right.
Proof. 1 Trivial.
2 Assume that R is a direct product of bounded right g-semisymmetric rings R i , i ∈ I with boundaries 1 from left and 1 from right. Let x i , y i , z i ∈ R i , i ∈ I with x 1 , x 2 , . . . y 1 , y 2 , . . . z 1 , z 2 , . . . 0, 0, . . . . Then x i y i z i 0, i 1, 2, . . . . Since R i , i ∈ I are bounded right g-semisymmetric rings with boundaries 1 from left and 1 from right, then x i z s i i y n i i 0, for all s i ≥ 1, and n i ≥ 1, i 1, 2 . . . . Therefore x 1 , x 2 , . . . z 1 , z 2 , . . . s y 1 , y 2 , . . . n 0, for all s ≥ 1 and n ≥ 1. Hence R is bounded right g-semisymmetric ring with boundaries 1 from left and 1 from right.
3 Assume that R is semiperfect bounded right g-semisymmetric ring with boundary 1 from left. Since R is semiperfect, R has a finite orthogonal set of local idempotents whose sum is 1 1, Proposition 3.7.2 . Hence we consider R n i 1 e i R such that each e i Re i is a local ring. Since R is bounded right g-semisymmetric rings with boundary 1 from left, then R is abelian by Lemma 2.16, whence every e i is central and e i R is an ideal of R, i 1, 2, . . . , n. Thus e i R e i Re i , for all i 1, 2, . . . , n. It follows that each e i R is bounded right g-semisymmetric ring with boundary 1 from left, by 1 .
Conversely, suppose that R is a finite direct sum of local bounded right gsemisymmetric rings with boundary 1 from left. Then, by 2 , and the fact that local rings are semiperfect, R 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 R is regular, then this is equivalent to R be strongly regular by 12, Theorem 3.7 which is equivalent to the condition R is regular and bounded right g-semisymmetric ring with boundary 1 from right, by Proposition 2.5.

Proposition 2.9. Let θ : R → A be a ring homomorphism and M a left A-module; then M is a left R-module via r · m θ r · m. Moreover, 1 If A M is g-semisymmetric, then so is R M,
2 If θ is onto and R M is g-semisymmetric, then so is A M.  s s b , t t a such that θ b s θ a t m 0. Hence b s a t m θ b s a t m  θ b s θ a t m θ b s θ a t m 0. Therefore R M is g-semisymmetric.

Proof. 1 Suppose
2 Let a, b ∈ A, m ∈ M such that abm 0. Since θ is onto, there exists r, s ∈ R such that θ r a, θ s b. Now 0 abm θ r θ s m rsm. Since R M is gsemisymmetric, then there exist positive integers t t s , n n r such that s t r n m 0 and b t a n m θ s t θ r n m θ s t θ r n m s t r n m 0. Hence A M is g-semisymmetric.
Lemma 2.10 see 10, Proposition 2.6 . Suppose that M is a flat left R-module. Then for every exact sequence 0 → K → F → M → 0 where F is R-free, one has IF ∩ K IK for each right ideal I of R; in particular, one has xF ∩ K xK for each element x of R.

Lemma 2.11. Let R be a bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right, then every free left R-module M is bounded g-semisymmetric with boundaries 1 from left and 1 from right.
Proof. Since M is free module, then M is isomorphic to a possibly infinite direct sum of copies of R, see 7 . Since R is bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right, then R M 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. Proof. Let R M be a flat module over bounded left g-semisymmetric ring R with boundaries 1 from left and 1 from right. Let m ∈ M and a ∈ R be such that abm 0. Suppose that for the epimorphism β : F → M the sequence 0 → K → F → M → 0 is exact. Now there exists y ∈ F such that β y m. This implies that β aby abm 0. Hence aby ∈ ker β Im K K. Therefore aby ∈ abF ∩ K abK, by Lemma 2.10. Hence for some k ∈ K, aby abk, yielding ab y − k 0. Since F is free R-module over bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right, then R F is bounded g-semisymmetric with boundaries 1 from left and 1 from right, by Lemma 2.11. Therefore b n a s y − k 0, for all n ≥ 1 and s ≥ 1. Hence b n a s y b n a s k and so β b n a s y β b n a s k gives b n a s β y b n a s β k , for all n ≥ 1 and s ≥ 1. Since k ∈ ker β, then b n a s β k 0 implies b n a s β y 0, for all n ≥ 1 and s ≥ 1. Hence b n a s m 0, for all n ≥ 1 and s ≥ 1. Thus R M is bounded g-semisymmetric with boundaries 1 from left and 1 from right.

ISRN Algebra
A torsionless R-module M is an R-module M such that M is a direct product of copies of R, or, equivalently, if 0 / m ∈ M, then there exists q ∈ M * such that mq / 0. If M is faithful R-module, then R is a submodule of a direct product of copies of M. The following proposition is an application of Remark 2.3 and Proposition 2.8. An application of Propositions 2.13 and 2.9 yields the following proposition.
Proposition 2.14. For an R-module M, let R denote the ring R/ann M . Then one has the following.

The left R-module M is g-semisymmetric if and only if the left R-module M is g-symmetric.
2 If the left R-module M is bounded g-semisymmetric with boundaries 1 from left and 1 from right, then R is bounded left g-semisymmetric with boundaries 1 from left and 1 from right.  Proof. Assume that R is bounded right g-semisymmetric ring with boundary 1 from left and e is an idempotent. Then e − e 2 0 gives e 1 − e 0. Hence for all x ∈ R there exists a positive integer n such that ex s 1 − e n 0 for all s ≥ 1. Therefore ex exe. And since 1 − e e 0, then xe exe. Therefore e is central.
The previous lemma is false for rings without identity. Indeed, the ring T 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. Proof. Let R be abelian right p.p.-ring. Let a 2 0. Since R is right p.p.-ring, then r a eR, for some idempotent e of R. Since R is abelian and a ∈ r a , then a ea ae 0 and hence R 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.   a 1 a 2 d 3 a 3 b 1 d 2 a 2 a 3 d 1 0,   a 1 a 2 c 3 a 1 b 2 d 3 a 1 a 3 c 2 a 2 b 1 d 3 a 3 b 1 d 2 a 2 a 3 c 1 0.
Therefore we have the following cases: 1 if a 1 0, a 2 / 0, a 3 / 0, then A 0, impossible, 2 if a 1 / 0, a 2 0, a 3 / 0, then B 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. Since Z is a ring without zero divisors, then R 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 A ACB / 0, and hence R is not symmetric ring. Also we notice that AB 2 0 and ACB / 0 and therefore R R is not reduced.