# Some Notes on Semiabelian Rings

Authors: Junchao Wei; Nanjie Li
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

It is proved that if a ringR is semiabelian, then so is
the skew polynomial ring R[x;σ], where σ is an endomorphism
of R satisfying σ(e)=e for all e∈E(R). Some characterizations
and properties of semiabelian rings are studied.


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## Body

## 1. Introduction

Throughout the paper, all rings are associative with identities. We always useN(R) and E(R) to denote the set of all nilpotent elements and the set of all idempotent elements of R.According to [1], a ring R is called semiabelian if every idempotent of R is either right semicentral or left semicentral. Clearly, a ring R is semiabelian if and only if either eR(1-e)=0 or (1-e)Re=0 for every e∈E(R), so, abelian rings (i.e., every idempotent of R is central) are semiabelian. But the converse is not true by [1, Example 2.2].A ringR is called directly  finite if ab=1 implies ba=1 for any a,b∈R. It is well known that abelian rings are directly finite. In Theorem 2.7, we show that semiabelian rings are directly finite.An elemente of a ring R is called a left minimal idempotent if e∈E(R) and Re is a minimal left ideal of R. A ring R is called left min-abel [2] if every left minimal idempotent element of R is left semicentral. Clearly, abelian rings are left min-abel. In Theorem 2.7, we show that semiabelian rings are left min-abel.A ringR is called left idempotent reflexive if for any e∈E(R) and a∈R, aRe=0 implies eRa=0. Theorem 2.5 shows that R is abelian if and only if R is semiabelian and left idempotent reflexive.In [3], Wang called an element e of a ring R an op-idempotent if e2=-e. Clearly, op-idempotent need not be idempotent. For example, let R=Z/3Z. Then 2̅∈R is an op-idempotent, while it is not an idempotent. In [4], Chen called an element e∈Rpotent in case there exists some integer n≥2 such that en=e. We write p(e) for the smallest number n of such. Clearly, idempotent is potent, while there exists a potent element which is not idempotent. For example, (100-1)∈M2(Z) is a potent element, while it is not idempotent. We use Eo(R) and PE(R) to denote the set of all op-idempotent elements and the set of all potent elements of R. In Corollaries 2.2 and 2.3, we observe that every semiabelian ring can be characterized by its op-idempotent and potent elements.IfR is a ring and σ:R→R is a ring endomorphism, let R[x;σ] denote the ring of skew polynomials over R; that is all formal polynomials in x with coefficients from R with multiplication defined by xr=σ(r)x. In [1], Chen showed that R is a semiabelian ring if and only if R[x] is a semiabelian ring. In Theorem 2.13, we show that if R is a semiabelian ring with an endomorphism σ satisfying σ(e)=e for all e∈E(R), then R[x;σ] is semiabelian.


## 2. Main Results

It is well known that an idempotente of a ring R is left semicentral if and only if 1-e is right semicentral. Hence we have the following theorem.Theorem 2.1.
The following conditions are equivalent for a ringR.(1)
R is a semiabelian ring.(2)
For anye∈E(R), eR(1-e)∪(1-e)Re is an ideal of R.(3)
For anye∈E(R), eR(1-e)∪(1-e)Re=eR(1-e)+(1-e)Re.Proof.
(1)⇒(2) assume that e∈E(R). Since R is semiabelian, e is either left semicentral or right semicentral. If e is right semicentral, then eR(1-e)=0 and 1-e is left semicentral. Thus R(1-e)ReR=(1-e)R(1-e)ReRe=(1-e)Re and eR(1-e)∪(1-ve)Re=(1-e)Re=R(1-e)ReR is an ideal of R. Similarly, if e is left semicentral, then eR(1-e)∪(1-e)Re=ReR(1-e)R is also an ideal of R.
(2)⇒(3) is clear.
(3)⇒(1) assume that e∈E(R). If e is neither left semicentral nor right semicentral, there exist a,b∈R such that (1-e)ae≠0 and eb(1-e)≠0. By (3), (1-e)ae+eb(1-e)∈(1-e)Re∪eR(1-e). If (1-e)ae+eb(1-e)∈(1-e)Re, then eb(1-e)=e((1-e)ae+eb(1-e))(1-e)∈e((1-e)Re)(1-e)=0, a contradiction; if (1-e)ae+eb(1-e)∈eR(1-e), then (1-e)ae=0, it is also a contradiction. Hence e is either left semicentral or right semicentral.Evidently,R is semiabelian if and only if either eR(1-e)=0 or (1-e)Re=0 for every e∈E(R). On the other hand, an element e of R is op-idempotent if and only if -e is idempotent. Hence, by Theorem 2.1, we have the following corollary.Corollary 2.2.
The following conditions are equivalent for a ringR.(1)
R is a semiabelian ring.(2)
For anye∈Eo(R), eR(1+e)=0 or (1+e)Re=0.(3)
For anye∈Eo(R), eR(1+e)∪(1+e)Re is an ideal of R.(4)
For anye∈Eo(R), eR(1+e)∪(1+e)Re=eR(1+e)+(1+e)Re.Clearly, for anye∈PE(R), ep(e)-1∈E(R), Re=Rep(e)-1, and eR=ep(e)-1R. Hence, by Theorem 2.1, we have the following corollary.Corollary 2.3.
The following conditions are equivalent for a ringR.(1)
R is a semiabelian ring.(2)
For anye∈PE(R), eR(1-ep(e)-1)=0 or (1-ep(e)-1)Re=0.(3)
For anye∈PE(R), eR(1-ep(e)-1)∪(1-ep(e)-1)Re is an ideal of R.(4)
For anye∈PE(R), eR(1-ep(e)-1)∪(1-ep(e)-1)Re=eR(1-ep(e)-1)+(1-ep(e)-1)Re.Using Theorem2.1, Corollaries 2.2 and 2.3, we have the following corollary.Corollary 2.4.
LetR be a semiabelian ring. If e∈E(R), g∈Eo(R) and h∈PE(R), then:(1)
ifReR=R, then e=1,(2)
ifRgR=R, then g=-1,(3)
ifRhR=R, then hp(h)-1=1.Call a ringR  idempotent reversible if gRe=0 implies eRg=0 for e,g∈E(R). Clearly, abelian rings are left idempotent reflexive, and left idempotent reflexive rings are idempotent reversible. But we do not know that whether idempotent reversible rings must be left idempotent reflexive. It is easy to see that a ring R is left idempotent reflexive if and only if for any a∈N(R), aRe=0 implies eRa=0. (In fact, it is only to show the sufficiency: Let a∈R and e∈E(R) satisfy aRe=0. If eRa=0, then eba≠0 for some b∈R. Since eba∈N(R) and (eba)Re=0, by hypothesis, eR(eba)=0, this implies eba=ee(eba)=0, which is a contradiction. Hence eRa=0, R is a left idempotent reflexive ring.)LetD be a division ring. Then the 2-by-2 upper triangular matrix ring UT2(D)=(DD0D) is not idempotent reversible. In fact, (1000),(0101)∈E(UT2(D)) and (0101)(DD0D)(1000)=0, but (1000)(DD0D)(0101)=(0D00)≠0. On the other hand, by [1, Example 2.2], UT2(D) is a semiabelian ring.We have the following theorem.Theorem 2.5.
The following conditions are equivalent for a ringR.(1)
R is an abelian ring.(2)
R is a semiabelian ring and idempotent reversible ring.(3)
R is a semiabelian ring and left idempotent reflexive ring.(4)
R is a semiabelian ring and for any a∈J(R), aRe=0 implies eRa=0.Proof.
(1)⇒(3)⇒(2)⇒(1) and (3)⇒(4) are trivial.
Now lete∈E(R). If R is semiabelian, then e is either right semicentral or left semicentral. If e is right semicentral, then (1-e)ReR(1-e)=0. Since R(1-e)Re⊆N(R)∩J(R), (4) implies (1-e)R(1-e)Re=0. Hence (1-e)Re=0. This shows that e is central; if e is left semicentral, then 1-e is right semicentral. Hence 1-e and so e is also central. Thus (4)⇒(1) holds.Since semiprime rings are left idempotent reflexive, we have the following corollary by Theorem2.5.Corollary 2.6.
Semiprime semiabelian rings are abelian.Theorem 2.7.
LetR be a semiabelian ring and e∈E(R). Then,(1)
eR(1-e)Re=(1-e)ReR(1-e)=0,(2)
Ifa∈R and ae=0, then Rera⊆N(R) for all r∈R,(3)
eR(1-e)⊆J(R).Proof.
(1) Sincee is right semicentral if and only if eR(1-e)=0 and e is left semicentral if and only if (1-e)Re=0, (1) is evident by hypothesis.(2)
Sincea=a(1-e), (ReRa)2=ReRaReRa=ReRa(1-e)ReRa=0 by (1). Hence ReRa⊆N(R), so for any r∈R, Rera⊆N(R).(3)
Since(1-e)e=0, by (2), Rer(1-e)⊆N(R) for all r∈R. This implies Rer(1-e)⊆J(R) for all r∈R. Hence eR(1-e)⊆J(R).Theorem 2.8.
LetR be a semiabelian ring. Then,(1)
R is directly finite,(2)
R is left min-abel.Proof.
(1) Assume thatab=1. Let e=ba. Then e∈E(R) and eb=b. By Theorem 2.7(3), b(1-e)=eb(1-e)∈J(R). Hence 1-e=ab(1-e)∈J(R), which implies 1=e=ba.
(2) Let0≠e∈E(R) and Re be a minimal left ideal of R. Then (1-e)Re≠0 and R(1-e)Re=Re. Since R is a semiabelian ring, by Theorem 2.7(3), (1-e)Re⊆J(R). This implies e∈J(R), that is, e=0 which is a contradiction. Hence (1-e)Re=0, so e is left semicentral. Hence R is a left min-abel ring.For a ringR, a proper left ideal P of R is prime if aRb⊆P implies that a∈P or b∈P. Let Specl(R) be the set of all prime left ideals of R. In [5], it has been shown that if R is not a left quasiduo ring, then Specl(R) is a space with the weakly Zariski topology but not with the Zariski topology.LetR be a ring. Then the set Maxl(R) of all maximal left ideals of R is a compact T1-space by [6, Lemma 2.1]. Recall that a topological space is said to be zero dimensional if it has a base consisting of clopen sets. Where a clopen set in a topological space is a set which is both open and closed.Now, for a left idealI of a ring R, let α(I)={P∈Specl(R)∣I⫅P} and β(I)=Specl(R)∖α(I). If I=Ra for some a∈R, then we write αl(a) and βl(a) for α(Ra) and β(Ra).For any left idealI of R, we let Ul(I)=Maxl(R)∩α(I), Vl(I)=Maxl(R)∩β(I) and let ξ={Ul(∑1≤i≤n∑1≤j1<j2<⋯<ji≤n(-1)i-1ej1ej1⋯eji)∣eji∈E(R),i=1,2,…,n,n∈Z+}.A ringR is called left topologically boolean, or a left tb-ring [7] for short, if for every pair of distinct maximal left ideals of R there is an idempotent in exactly one of them.A ringR is called clean [8] if every element of R is the sum of a unit and an idempotent.The following theorems generalize [6, Lemmas 2.2 and 2.3].Lemma 2.9.
LetR be a semiabelian ring and ei,e,f∈E(R),i=1,2,…,n. Then,(1)
ifN is a maximal left ideal of R and e∉N, then 1-e∈N,(2)
Ul(e)∩Ul(f)=Ul(fe),(3)
Ul(e)∪Ul(f)=Ul(e+f-ef)=Ul(ef),(4)
Ul(e)=Vl(1-e),(5)
⋂i=1nUl(ei)=Ul(e1e2⋯en),(6)
⋃i=1nUl(ei)=Ul(∑1≤i≤n∑1≤j1<j2<⋯<ji≤n(-1)i-1ej1ej1…eji),(7)
Ul(∑1≤i≤n∑1≤j1<j2<⋯<ji≤n(-1)i-1ej1ej1⋯eji)=Vl(1-∑1≤i≤n∑1≤j1<j2<⋯<ji≤n(-1)i-1ej1ej1…eji).In particular, every set in ξ is clopen.Proof.
(1) Sincee∉N, Re+N=R. Let 1=be+n for some b∈R and n∈N. Since eR(1-e)Re=0 by Theorem 2.7(1), eR(1-e)=eR(1-e)n⊆N. Since N is a prime left ideal and e∉N, 1-e∈N.
(2) LetP∈Ul(e)∩Ul(f). Then e∉P and f∉P. By (1), we have 1-e,1-f∈P. Hence 1-e-f+ef=(1-e)(1-f)∈P. Clearly, ef∉P, so P∈Ul(ef). This shows Ul(e)∩Ul(f)⊆Ul(ef). Conversely, if Q∈Ul(ef), then ef∉Q. Since Q is a left ideal, f∉Q. Hence 1-f∈Q by (1). If e∈Q, then 1-e-f+ef=(1-e)(1-f)∈Q implies ef∈Q, which is a contradiction. Hence e∉Q, so Q∈Ul(e)∩Ul(f). Therefore Ul(ef)⊆Ul(e)∩Ul(f). Thus Ul(e)∩Ul(f)=Ul(ef). Similarly, we have Ul(e)∩Ul(f)=Ul(fe).
(3) and (4) They are also straightforward to prove.
By induction onn, we can show (5), (6) and (7).
Thus every set inξ is clopen.Theorem 2.10.
LetR be a semiabelian clean ring. Then R is a left tb-ring.Proof.
Suppose thatM and N are distinct maximal left ideals of R. Let a∈M∖N. Then Ra+N=R and 1-xa∈N for some x∈R. Clearly, xa∈M∖N. Since R is clean, there exist an idempotent e∈E(R) and a unit u in R such that xa=e+u. If e∈M, then u=xa-e∈M from which it follows that R=M, a contradiction. Thus e∉M. If e∉N, then 1-e∈N by Lemma 2.9 (1) and hence u=(1-e)+(xa-1)∈N. It follows that N=R which is also not possible. We thus have that e is an idempotent belonging to N only.Theorem 2.11.
LetR be a semiabelian ring. If R is a left tb-ring, then ξ forms a base for the weakly Zariski topology on Maxl(R). In particular, Maxl(R) is a compact, zero-dimensional Hausdorff space.Proof.
Similar to the proof of [6, Proposition 2.5], we can complete the proof.A ringR is called von Neumann regular if a∈aRa for all a∈R and R is said to be unit-regular if for any a∈R, a=aua for some u∈U(R). A ring R is called strongly regular if a∈a2R for all a∈R. Clearly, strongly regular ⇒ unit-regular ⇒von Neumann regular. Since von Neumann regular rings are semiprime, it follows that von Neumann regular rings are left idempotent reflexive. And it is well known that R is strongly regular if and only if R is von Neumann regular and abelian. In view of Theorem 2.5, we have the following corollary.Corollary 2.12.
The following conditions are equivalent for a ringR.(1)
R is strongly regular.(2)
R is unit-regular and semiabelian.(3)
R is von Neumann regular and semiabelian.Following [9], a ring R is called left NPP if for any a∈N(R), Ra is projective left R-module, and R is said to be n-regular if for any a∈N(R), a∈aRa. A ring R is said to be reduced if a2=0 implies a=0 for each a∈R, or equivalently, N(R)=0. Obviously, reduced rings are n-regular and abelian, and n-regular rings are left NPP and semiprime. Using Theorem 2.5, the following theorem gives some new characterization of reduced rings in terms of semiabelian rings.Theorem 2.13.
The following conditions are equivalent for a ringR.(1)
R is reduced.(2)
R is n-regular and semiabelian.(3)
R is left NPP, semiprime, and semiabelian.Proof.
(1)⇒(2)⇒(3) are trivial.
(3)⇒(1) let a∈R such that a2=0. Since R is left NPP, l(a)=Re,e∈E(R). Hence ea=0 and a=ae because a∈l(a). Since R is semiabelian and aRa=(1-e)aeR(1-e)a⊆(1-e)ReR(1-e)a, aRa=0 by Theorem 2.7. Since R is semiprime, a=0, which shows that R is reduced.IfR is a ring and σ:R→R is a ring endomorphism, let R[x;σ] denote the ring of skew polynomials over R; that is all formal polynomials in x with coefficients from R with multiplication defined by xr=σ(r)x. Note that if R(σ) is the (R,R)-bimodule defined by RR(σ)=RR and m∘r=mσ(r), for all m∈R(σ) and r∈R, then R[x;σ]/(x2)≅R∝R(σ).Theorem 2.14.
LetR be a semiabelian ring. If σ is a ring endomorphism of R satisfying σ(e)=e for all e∈E(R). Then R[x;σ] is semiabelian.Proof.
Letf(x)=e0+e1x+⋯+enxn∈E(R[x;σ]). Then
(2.1)e02=e0,e1=e0e1+e1σ(e0),e2=e0e2+e1σ(e1)+e2σ2(e0),⋮en=e0en+e1σ(en-1)+e2σ2(en-2)+⋯+en-1σn-1(e1)+enσn(e0).
Sincee0∈E(R), σ(e0)=e0 by hypothesis. Hence we have the following equations:
(2.2)e1=e0e1+e1e0,e2=e0e2+e1σ(e1)+e2e0,⋮en=e0en+e1σ(en-1)+e2σ2(en-2)+⋯+en-1σn-1(e1)+ene0.
Ife0 is right semicentral, then e0e1=e0e1+e0e1e0=e0e1+e0e1, which implies e0e1=0. Hence e1=e1e0.
Assume thate0ei=0 and ei=eie0 for i=1,2,…,n-1. Then
(2.3)e0en=e0en+e0ene0=e0en+e0en,
so
(2.4)e0en=0,en=e1σ(en-1)+e2σ2(en-2)+⋯+en-1σn-1(e1)+ene0=e1e0σ(en-1)+e2e0σ2(en-2)+⋯+en-1e0σn-1(e1)+ene0=e1σ(e0en-1)+e2σ2(e0en-2)+⋯+en-1σn-1(e0e1)+ene0=ene0.
Hencef(x)e0=e0+e1σ(e0)x+⋯+enσn(e0)xn=e0+e1x+⋯+enxn=f(x) and e0f(x)=e0.
For anyg(x)=b0+b1x+⋯+bmxm∈R[x;σ], we have e0g(x)e0=∑0≤i≤me0biσi(e0)xi=∑0≤i≤me0bie0xi=∑0≤i≤me0bixi=e0g(x). Thus f(x)g(x)f(x)=f(x)e0g(x)f(x)=f(x)e0g(x)e0f(x)=f(x)e0g(x)e0=f(x)e0g(x)=f(x)g(x), which implies f(x) is right semicentral in R[x;σ]. Similarly, if e0 is left semicentral in R, then we can show that f(x) is left semicentral in R[x;σ]. Hence R[x;σ] is a semiabelian ring.Corollary 2.15.
LetR be a semiabelian ring. If σ is a ring endomorphism of R satisfying σ(e)=e for all e∈E(R). Then R[x;σ]/(x2) is semiabelian.Proof.
Since every elementf(x) of R[x;σ]/(x2) can be written f(x)=a0+a1x with x2=0, by the same proof as Theorem 2.14, we can complete the proof.Corollary 2.16.
LetR be a semiabelian ring. If σ is a ring endomorphism of R satisfying σ(e)=e for all e∈E(R). Then R∝R(σ) is semiabelian.Corollary 2.17.
LetD be a division ring with an endomorphism σ, then D[x;σ]/(x2) is semiabelian.A ringR is called left MC2 [2] if for any a∈R and e∈MEl(R), aRe=0 implies eRa=0. Clearly, left idempotent reflexive rings are left MC2. We do not know whether idempotent reversible rings are left MC2. But we know that there exists a left MC2 ring which is not idempotent reversible. In fact, there exists a semiabelian ring R which is not abelian (see the example above Theorem 2.5), by [1, Corollary 2.4], R[x] is a semiabelian ring which is not abelian. Hence, by Theorem 2.5, R[x] is not idempotent reversible. But R[x] is a left MC2 ring.The authors in [10, Theorem 4.1] showed that if R is a left MC2 ring containing an injective maximal left ideal, then R is a left self-injective ring. And [11, Proposition 5] showed that if R is a left idempotent reflexive ring containing an injective maximal left ideal, then R is a left self-injective ring.Proposition 2.18.
LetR be an idempotent reversible ring. If R contains an injective maximal left ideal, then R is a left self-injective ring.Proof.
LetM be an injective maximal left ideal of R. Then R=M⊕N for some minimal left ideal N of R. Hence we have M=Re and N=R(1-e) for some e2=e∈R. If MN=0, then we have eR(1-e)=0. Since R is idempotent reversible, (1-e)Re=0. So e is central. Now let L be any proper essential left ideal of R and f:L→N any non-zero left R-homomorphism. Then L/U≅N, where U=kerf is a maximal submodule of L. Now L=U⊕V, where V≅N=R(1-e) is a minimal left ideal of R. Since e is central, V=R(1-e). For any z∈L, let z=x+y, where x∈U,y∈V. Then f(z)=f(x)+f(y)=f(y). Since y=y(1-e)=(1-e)y, f(z)=f(y)=f(y(1-e))=yf(1-e). Since x(1-e)=(1-e)x∈V∩U=0, xf(1-e)=f(x(1-e)=f(0)=0. Thus f(z)=yf(1-e)=yf(1-e)+xf(1-e)=(y+x)f(1-e)=zf(1-e). Hence RN is injective. If MN≠0, by the proof of [10, Proposition 5], we have that RN is injective. Hence R=M⊕N is left self-injective.Recall that a ringR is left pp if every principal left ideal of R is projective. As an application of Proposition 2.18, we have the following result.Corollary 2.19.
The following conditions are equivalent for a ringR.(1)
R is a von Neumann regular left self-injective ring with Soc(RR)≠0.(2)
R is an idempotent reversible left pp ring containing an injective maximal left ideal.Proof.
(1)⇒(2) is trivial.
(2)⇒(1) by Proposition 2.18, R is a left self-injective ring. Hence, by [12, Theorem 1.2], R is left C2, so, R is von Neumann regular because R is left pp. Also we have Soc(RR)≠0 since there is an injective maximal left ideal.By [13], a ring R is said to be left HI if R is left hereditary containing an injective maximal left ideal. Osofsky [14] proved that left self-injective left hereditary ring is semisimple Artinian. We can generalize the result as follows.Corollary 2.20.
The following conditions are equivalent for a ringR.(1)
R is a semisimple Artinian ring.(2)
R is an idempotent reversible left HI ring.According to [8], an element x∈R is called exchange if there exists e∈E(R) such that e∈xR and 1-e∈(1-x)R, and x is said to be clean if x=e+u where e∈E(R) and u∈U(R). By [8], clean elements are exchange and the converse holds when R is an abelian ring. A ring R is called exchange (clean) ring if every element of R is an exchange (clean) element.Proposition 2.21.
LetR be a semiabelian ring. If x∈R is an exchange element, then x is a clean element.Proof.
Sincex is an exchange element, there exists e∈E(R) such that e∈xR and 1-e∈(1-x)R. Let e=xy and 1-e=(1-x)z where y=ye,z=z(1-e)∈R. Then (x-(1-e))(y-z)=xy-xz-(1-e)y+(1-e)z=xy+(1-x)z-(1-e)y-ez=e+1-e-(1-e)y-ez=1-(1-e)y-ez. Since R is a semiabelian ring, e is either left semicentral or right semicentral. If e is left semicentral, then (1-e)y=(1-e)ye=0 and (ezR)2=ezRezR=ez(1-e)RezR⊆eR(1-e)ReR=0 by Theorem 2.7(1). Hence ez∈J(R). Similarly, if e is right semicentral, then ez=ez(1-e)=0 and (1-e)y∈J(R). This implies 1-(1-e)y-ez∈U(R), so (x-(1-e))(y-z)∈U(R). Since R is a semiabelian ring, by Theorem 2.8, R is a directly finite ring. Hence x-(1-e)∈U(R), which implies x is a clean element.Corollary 2.22.
IfR is a semiabelian exchange ring, then R is a clean ring.Theorem 2.23.
LetR be a semiabelian ring and a,b∈R. If ab=0, then aE(R)b⊆J(R).Proof.
Letab=0 and e∈E(R). Since R is a semiabelian ring, either e is left semicentral or e is right semicentral. If e is left semicentral, then (Raeb)2=RabRaeb=0. If e is right semicentral, then (Raeb)2=RaebRab=0. Hence Raeb⊆J(R) for each e∈E(R), which implies aE(R)b⊆J(R).Corollary 2.24.
LetR be an abelian ring and a,b∈R. If ab=0, then aE(R)b⊆J(R).The converse of Corollary2.24 is not true, in general.Example 2.25.
letF be a field, and R=(FF0F). Evidently,E(R)=⋃x∈F{(0000),(1001),(1x00),(0x01)}, J(R)=(0F00). Let A=(a1b10c1),B=(a2b20c2)∈R and AB=0. Then a1a2=c1c2=0. Since A(1x00)B=(0a1b2+a1xc200)∈J(R) and A(0x01)B=(0a1xc2+b1c200)∈J(R). Hence AE(R)B⊆J(R), but R is not an abelian ring.A ringR is called EIFP if a,b∈R, ab=0 implies aE(R)b⊆J(R). Clearly, semiabelian rings are EIFP by Theorem 2.23. But the converse of Theorem 2.23 is not true, in general.Example 2.26.
Take the ringR in Example 2.25, and let S=R⊕R. Then S is EIFP, but not semiabelian. Indeed, take e1=E11+E12 and e2=E12+E22 in R, where Eij are matrix units. Then (e1;e2) is an idempotent. By a direct computation, (e1;e2) is neither left nor right semicentral. Hence R⊕R is not semiabelian while R⊕R is EIFP.



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*Source: 154636-2011-08-04.xml*