# On g-Semisymmetric Rings

Authors: Farahat S. Aly; Mohammed O. Al Mestady
Journal: ISRN Algebra
 (2012)

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## 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 thatR 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.


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## Body

## 1. Introduction

Throughout this paper, all rings are associated with identity and all modules are unitary. For a subsetX 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 and Zhou [3], a left R-module M is reduced if a2m=0 implies aRm=0, for all a∈R,m∈M. Abelian rings are rings in which each idempotent is central. According to Buhphang and Rege [4], a left R-module M is semicommutative, if am=0 implies aRm=0, for all a∈R, m∈M. 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 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 acnbl=0. A ring R is called bounded right g-semisymmetric 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 acsbl=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 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 R is right p.p.-ring, by Theorem 2.19.


## 2. G-Semisymmetric Rings

Definition 2.1.
(1) A right R-module M is called g-semisymmetric if for m∈M and a,b∈R with mab=0, there exist two positive integers n=n(b),l=l(a) such that mbnal=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 acnbl=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 bnalm=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 bnalc=0.
(3) A rightR-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 mbsal=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 acsbl=0, for all s≥n.
(4) A rightR-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 mbnas=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 acnbs=0, for all s≥l.
(5) A leftR-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 bsalm=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 bsalc=0, for all s≥n.
(6) A leftR-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 bnasm=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 bnasc=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 [6, Example  9] with changes in its multiplications.Example 2.2.
LetS={a,b}be the semigroup with multiplicationa2=ba=a,
b2=ab=b. PutT=F2S,
which is a four-element semigroup ring without identity. This ringTis bounded right g-semisymmetric ring with boundary 2 from right and with boundary 1 from left, butTis neither symmetric ring nor reversible.Remark 2.3.
(1) A ringR is left g-semisymmetric if and only if the module RR is g-semisymmetric. (2) A ring R is right g-semisymmetric if and only if the module RR is g-semisymmetric.Proposition 2.4.
The following conditions are equivalent for a rightR-module M.(1)
M is g-semisymmetric.(2)
All cyclic submodules ofM are g-semisymmetric.Proof.
(1)⇒(2) Let N=mR be a cyclic submodules of M, and let m′∈N. Since M is g-semisymmetric, then for a,b∈R with m′ab=0, it implies that m′bman=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 mbman=0. Therefore M is g-semisymmetric.Proposition 2.5.
The following conditions are equivalent for a ringR.(i)
R is strongly regular.(ii)
Every rightR-module is flat and g-semisymmetric with boundary 1 from right.(iii)
Every cyclic rightR-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. Sincers∈I, then ((s¯)m(r¯)n)2=0¯ and hence (s¯)mr¯=0¯. This shows that msmr=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 RR 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 x2=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)nxl=0 for all l≥1. Since x=xyx=xyxyx=xyxyxyx=⋯=(xy)nx, then yx=y(xy)nx=0. Therefore x=xyx=0. Hence R has no nonzero nilpotent element and R is strongly regular ring.Corollary 2.6.
If a ringR is regular and bounded right g-semisymmetric with boundary 1 from right, then R is reduced.A one-sided idealI 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.
LetR be a von Neumann regular ring. Then the following conditions are equivalent:(1)
R is right (left) duo,(2)
R is reduced,(3)
R is bounded right g-semisymmetric with boundary 1 from right,(4)
R is symmetric,(5)
R is IFP,(6)
R 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 ifR is a finite direct sum of local bounded right g-semisymmetric rings from left.
(4) A ringR 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 thatR is a direct product of bounded right g-semisymmetric rings Ri,i∈I with boundaries 1 from left and 1 from right. Let xi,yi,zi∈Ri, i∈I with (x1,x2,…)(y1,y2,…)(z1,z2,…)=(0,0,…). Then xiyizi=0, i=1,2,…. Since Ri,i∈I are bounded right g-semisymmetric rings with boundaries 1 from left and 1 from right, then xizisiyini=0, for all si≥1,andni≥1,i=1,2…. Therefore (x1,x2,…)(z1,z2,…)s(y1,y2,…)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 thatR 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=∑i=1neiRsuch that each eiRei 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 ei is central and eiR is an ideal of R,i=1,2,…,n. Thus eiR=eiRei, for all i=1,2,…,n. It follows that each eiR is bounded right g-semisymmetric ring with boundary 1 from left, by (1).
Conversely, suppose thatR is a finite direct sum of local bounded right g-semisymmetric 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 Lemma2.16, 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.
Letθ: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)
IfAM is g-semisymmetric, then so is RM,(2)
Ifθ is onto andRM is g-semisymmetric, then so isAM.Proof.
(1) SupposeAM is g-semisymmetric, and let a,b∈R, m∈M such that abm=0. Then 0=abm=θ(ab)m=θ(a)θ(b)m. SinceAM is g-semisymmetric, then there exist positive integers s=s(b),t=t(a) such that θ(b)sθ(a)tm=0. Hence bsatm=θ(bsat)m=θ(bs)θ(at)m=θ(b)s(θ(a))tm=0. Therefore RM is g-semisymmetric.
(2) Leta,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 RM is g-semisymmetric, then there exist positive integers t=t(s),n=n(r) such that strnm=0 and btanm=θ(st)(θ(r))nm=θ(st)θ(rn)m=strnm=0. HenceAM is g-semisymmetric.Lemma 2.10 (see [10, Proposition  2.6]).
Suppose thatM 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.
LetR 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.
SinceM 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 RM 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.
LetRM 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)=ImK=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, thenRF is bounded g-semisymmetric with boundaries 1 from left and 1 from right, by Lemma 2.11. Therefore bnas(y-k)=0, for all n≥1ands≥1. Hence bnasy=bnask and so β(bnasy)=β(bnask) gives bnasβ(y)=bnasβ(k), for all n≥1ands≥1. Since k∈kerβ, then bnasβ(k)=0 implies bnasβ(y)=0, for all n≥1ands≥1. Hence bnasm=0, for all n≥1ands≥1. ThusRM is bounded g-semisymmetric with boundaries 1 from left and 1 from right.In the following propositionsE(M) denotes the R-endomorphism ring of M. The associativity is deduced from the generalized associativity situation in the standard Morita context (R,M,M*,E(M)) without explicit mention, where M* is the left E(M)-,rightR-bimodule HomR(M,R).A torsionlessR-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.Proposition 2.13.
The following conditions are equivalent.(1)
R is a bounded left g-semisymmetric ring with boundaries 1 from left and 1 from right.(2)
Every torsionless leftR-module is bounded g-semisymmetric with boundaries 1 from left and 1 from right.(3)
Every submodule of a free leftR-module is bounded g-semisymmetric with boundaries 1 from left and 1 from right.(4)
There exists a faithful, bounded g-semisymmetric leftR-module with boundaries 1 from left and 1 from right.An application of Propositions2.13 and 2.9 yields the following proposition.Proposition 2.14.
For anR-module M, let R¯ denote the ring R/ann(M). Then one has the following.(1)
The leftR-module M is g-semisymmetric if and only if the left R¯-module M is g-symmetric.(2)
If the leftR¯-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.(3)
If the rightE(M)-module M is bounded g-semisymmetric from left, then the ring E(M) is bounded right g-semisymmetric with boundaries 1 from left and 1 from right.An application of Proposition2.9 yields (1); since the left R-, right E(M)-bimodule M is faithful as a left R¯-module and is also faithful as a right E(M)-module, applying (4)⇒(1) of Proposition 2.13 we get (2) and (3).LetM be a right R-module. Then as in [10] M is called(1)
reduced ifma2=0, then mRa=0, a∈R,m∈M;(2)
ZI (zero-insertive ring) ifma=0, then mRa=0, a∈R,m∈M.Proposition 2.15.
LetR be a right R-module M. Then,(1)
ifM is reduced, then M is symmetric [10, Proposition  2.2],(2)
ifM is symmetric, then M is ZI [10, Proposition  2.2].Lemma 2.16.
IfR is bounded right g-semisymmetric ring with boundary 1 from left, then R is abelian.Proof.
Assume thatR is bounded right g-semisymmetric ring with boundary 1 from left and e is an idempotent. Then e-e2=0 gives e(1-e)=0. Hence for all x∈R there exists a positive integer n such that exs(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 ringT 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. LetR={(abcd)∈M2×2(ℤ):a≡d(mod2),b=c≡0(mod2)},
whereM2×2(ℤ)is the full matrix ring over the ring of integers. Since the zero and the identity matrices are only the idempotent elements inR,
then R is abelian ring. Since(00-22)(0202)(2402)=0and(00-22)(2402)m(0202)n≠0for any positive integers m and n, then R is not right g-semisymmetric ring.A ringR be a right p.p.-ring if for any a∈R, r(a)=eR for some idempotent e of R.Proposition 2.18.
LetR be a right p.p.-ring. If R is abelian, then R is reduced.Proof.
LetR be abelian right p.p.-ring. Let a2=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 Lemma2.16 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.
LetR be right p.p.-ring. Then the following are equivalent.(1)
R is reduced.(2)
R is symmetric.(3)
R is bounded right g-semisymmetric ring with boundary 1 from left.(4)
R is IFP.(5)
R is abelian.Theorem 2.20.
LetS be a ring without zero divisors and R={(abc0ad00a)∣a,b,c,d∈S}. Then R is bounded right g-semisymmetric with boundary 2 from left and boundary 2 from right.Proof.
Suppose that0≠A=(a1b1c10a1d100a1),0≠B=(a2b2c20a2d200a2),0≠C=(a3b3c30a3d300a3)∈R such that ABC=0. Then,a1a2a3=0,a1a2b3+a1a3c3+a2a3b1=0,a1a2d3+a3b1d2+a2a3d1=0,a1a2c3+a1b2d3+a1a3c2+a2b1d3+a3b1d2+a2a3c1=0. Therefore we have the following cases:
(1) ifa1=0, a2≠0, a3≠0, then A=0, impossible,
(2) ifa1≠0, a2=0, a3≠0,then B=(0b20000000),and C=(a3b3c30a3000a3); in this case, ACB2=0,
(3) ifa1≠0, a2≠0, a3=0, then C=0, impossible,
(4) if a1=0, a2=0, a3≠0, then A=(00c100d1000), B=(0b2c2000000), and C=(a3b3c30a3d300a3);
henceACB2=0,
(5) ifa1=0, a2≠0, a3=0, then A=(00c100d1000),B=(a2b2c20a2d200a2), and C=(0b3c3000000) which implies that AC2B=0,
(6) ifa1≠0, a2=0, a3=0,then A=(a1b1c10a1d100a1),B=(00c200d200a), and C=(0b3c3000000) which implies that ACB2=0.
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.
LetR={(abc0ad00a)∣a,b,c,d∈ℤ}. 						ThenRis a bounded right g-semisymmetric ring with boundary 2 from left and boundary 2 from right which is not symmetric.
Sinceℤ 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=(121010001), B=(003001000), C=(011001000), then ABC=0 and ACB≠0, and hence R is not symmetric ring. Also we notice that AB2=0 and ACB≠0 and therefore RR is not reduced.



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