Generalized periodic and generalized Boolean rings

© Hindawi Publishing Corp.

GENERALIZED PERIODIC AND GENERALIZED BOOLEAN RINGS

HOWARD E. BELL and ADIL YAQUB

(Received 21 August 2000)

Abstract.We prove that a generalized periodic, as well as a generalized Boolean, ring is either commutative or periodic. We also prove that a generalized Boolean ring with central idempotents must be nil or commutative. We further consider conditions which imply the commutativity of a generalized periodic, or a generalized Boolean, ring.

2000 Mathematics Subject Classification. 16D70, 16U80.

Throughout,Rdenotes a ring,Nthe set of nilpotents,Cthe center, andEthe set of idempotents ofR. A ringRis calledperiodicif for everyxinR, there exist distinct positive integersm,nsuch thatxm_=xn_{. We now formally state the definitions of a}

generalized periodicring and a generalized Boolean ring.

Definition1. A ringRis calledgeneralized periodicif for everyxinRsuch that x∉(N∪C), we havexn_−xm_{∈(N∩C), for some positive integersm,nof opposite} parity.

Definition2. A ringRis calledgeneralized Booleanif for everyxinRsuch that x∉(N∪C), there exists anevenpositive integernsuch thatx−xn_∈(N∩C).

Theorem 3. IfR is a generalized periodic ring, thenR is either commutative or periodic.

Proof. LetNandCdenote the set of nilpotents and the center ofR, respectively. We distinguish three cases.

Case1 (N⊆C). Thenx∉C impliesx∉(N∪C), and hence there exist distinct positive integersm,nsuch thatxm_−xn_{∈N, withn > m. Suppose(x}m_−xn₎k_=0.

Then, as is readily verified,

x−xn−m+1k_xk(m−1)_{=0, (1)}

which, in turn, implies that

x−xn−m+1km_=x−xn−m+1k_xk(m−1)_g(x)

=0, (2)

where

g(λ)∈Z[λ]. (3)

We have thus shown that

Recall that, in our present case, we assumed thatN⊆C, and hence by (4),

x−xn−m+1_{∈C, ∀x∉C, (n−m+1>1). (5)}

Since (5) is trivially satisfied ifx∈C, we see that

x−xn(x)_{∈C, for somen(x) >1,wherex∈R(arbitrary). (6)}

Therefore,Ris commutative, by a well-known theorem of Herstein [3].

Case2(C ⊆N). Thenx∉N impliesx∉(N∪C), and hence there exist distinct positive integersm,nsuch thatxn_−xm_{∈N, withn > m. Repeating the argument}

used to prove (4), we see that

x−xn−m+1_{∈N, ∀x∉N, (n−m+1>1). (7)}

Since (7) is trivially satisfied for allx∈N, we conclude that

x−xk(x)_{∈N, for somek(x) >1,wherex∈R(arbitrary). (8)}

By a well-known theorem of Chacron [2], equation (8) implies thatRis periodic.

Case3(C ⊆NandN ⊆C). In this case, let

z∈C\N, u∈N\C. (9)

Equation (9) readily implies thatz+u∉Candz+u∉N, and hence (seeDefinition 1)

(z+u)n_−(z+u)m_{∈N, for some integersn > m≥1. (10)}

Sincezcommutes with the nilpotent elementu, (10) implies that

zn_−zm_{+u∈N, whereu∈N, ucommutes withz. (11)}

Hencezn_−zm_{∈N, withn > m≥1. Now, a repetition of the argument used in the}

proof of (4) shows that

z−zn−m+1_{∈N, ∀z∈C\N, (n−m+1>1). (12)}

Trivially,

x−xk_{∈N, ∀x∈N,∀k∈Z}+_{. (13)}

Finally, ifx∉(N∪C), then

xn_−xm_{∈N, for some integersn > m≥1. (14)}

Again, repeating the argument used in the proof of (4), we see that

x−xn−m+1_{∈N, ∀x∉(N∪C), (n−m+1>1). (15)}

Combining (12), (13), and (15), we conclude that

x−xk(x)_{∈N, for somek(x) >1,wherex∈R(arbitrary). (16)}

Corollary4. IfRis a generalized Boolean ring, thenR is either commutative or periodic.

This follows at once, since a generalized Boolean ring is necessarily a generalized periodicring (see Definitions1and2).

Before proving the next theorem, we prove the following lemma.

Lemma5. LetRbe a generalized periodic ring. Ifeis any nonzero central idempotent inRanda∈N, thenea∈C.

Proof. The proof is by contradiction. Suppose the lemma is false, and let

η0∈N, eη0∉C. (17)

Sincee∈Candη0∈N, thereforeeη0is nilpotent. Let

eη0α∈C, ∀α≥α0, α0minimal. (18)

Sinceeη0∉C(see (17)), thereforeα0>1. Letη=(eη0)α0−1. Then, η=eη0α0−1∈N, η∉C(by the minimality ofα0),

ηk_{∈C, ∀k≥2, e∈C, e}2_{=e =0, e∉N.} (19) Equation (19) implies thate+η∉Cande+η∉N, and hence (seeDefinition 1)

e+ηm−e+ηn∈C, (20)

wherem_{,nare ofoppositeparity. Combining (20) and (19), we see that (keep in mind} thateη=η; see (19))

m_{−neη∈C, (21)}

wherem_{−n is an odd integer. Equation (19) also implies that (−e+η)is not in}

(N∪C), so

−e+ηm−−e+ηn∈N, (22)

wherem_{,nare ofoppositeparity. Combining (19) and (22), we see that}

(−e)m

−(−e)n

∈N, (23)

and hence 2e∈N, sincem_{andnare ofoppositeparity. Therefore,(2e)}γ_=0,γ∈Z+_, and thus 2γ_{e=0, which implies that}

2γ_{eη∈C; γ∈Z}+_{. (24)}

Now, combining (21) and (24), keeping in mind that(2γ_{,m−n)=1, we see that} eη∈C, and hence, by (19),η=eη∈C, which contradicts (19). This contradiction proves the lemma.

Theorem6. SupposeRis a generalized periodic ring, and suppose that there exists an elementcinC, withc =0, such that

cx,y=0 impliesx,y=0,∀x,y∈R. (25)

ThenRis commutative.

Proof. We distinguish two cases.

Case1(c∈N). In this case,ck_{=0 for some positive integerk, and hence}

ck_{x,y=0, ∀x,y∈R. (26)}

Combining (25) and (26), we see that

ck_{x,y=0 ⇒cc}k−1_{x,y=0 ⇒c}k−1_x,y=0⇒ck−1_x,y=0

⇒ ··· ⇒cx,y=0 ⇒x,y=0. (27)

Thus,ck_{[x,y]=0 implies[x,y]=0, and henceRis commutative.}

Case 2 (c∉N). In view of Theorem 3, we may assume that R is periodic. This implies, in particular, thatcm_{is idempotent for some positive integerm. Furthermore,} cm _{=0 (sincec∉Nin our present case). The net result is (sincec∈Calso)}

cm_{=e is a nonzero central idempotent inR. (28)}

Leta∈N. ByLemma 5and equation (28), we haveea∈C, and hence[ea,x]=0 for allx∈R, which implies

cm_a,x=cm_{[a,x]=0, ∀x∈R. (29)}

The argument used inCase 1ofTheorem 6shows that

cm_{[a,x]=0 implies[a,x]=0, (30)}

and hence (see (29))

[a,x]=0 ∀x∈R, ∀a∈N. (31)

Thus,Ris a periodicring with the property thatN⊆C. By a well-known theorem of Herstein [4], it follows thatRis commutative, and the theorem is proved.

Corollary7. Suppose thatRis a generalized periodic ring with identity1. Then, Ris commutative.

Corollary 7follows at once by takingc=1 inTheorem 6.

Since a generalized Boolean ring is also a generalized periodic ring, therefore we have the following corollary.

Corollary9. Suppose thatRis a generalized periodic ring containing a central element which is not a zero divisor. ThenRis commutative.

This follows at once, since the hypotheses of this corollary imply the hypotheses ofTheorem 6.

Theorem10. SupposeRis a generalized periodic ring. Suppose, further, that there exists a nonzero central elementcsuch that

ca=0 impliesa=0,∀a∈N. (32)

ThenRis commutative.

Proof. In [1], the authors proved the following result:

IfRis a generalized periodicring, then the nilpotents

Nform an ideal andR/Nis commutative. (33) Letx,y ∈R. By (33), for all ¯x,y¯ inR/N, ¯xy¯=y¯x¯, and hence[x,y]∈N. Taking a=[x,y]∈Nin (32), we see that (32) yields

cx,y=0 impliesx,y=0, ∀x,y∈R. (34)

The theorem now follows at once fromTheorem 6.

Theorem11. A generalized Boolean ringRwith central idempotents is necessarily nil(R=N)or commutative(R=C).

Proof. SinceR is also a generalized periodicring, therefore byTheorem 3,R is commutative or periodic. IfRis commutative, there is nothing to prove. So we may assume thatRis periodic. We now distinguish two cases.

Case1(C⊆N). Recall that, by hypothesis, the setEof idempotents is central, and henceE⊆C⊆N(in the present case). Thus,E⊆N, and henceE= {0}. Therefore,

zero is the only idempotent ofR. (35)

Letx∈R. SinceRis periodic, thereforexk_{is idempotent for some positive integerk,}

and hence by (35),xk_{=0, which proves thatRis nil.} Case2(C ⊆N). Then, for somec∈R, we have

c∈C, c ∈N. (36)

Again, sinceRis periodic,cm_{is idempotent for some positive integerm. Moreover,} cm _{=0 (sincec ∈N). The net result is (see (36))}

e=cm_{is a nonzero central idempotent ofR. (37)}

Now, supposea∈N. Since 0 =e∈Canda∈N, thereforee+a ∈N. Suppose, for the moment, thata ∈C. Thene+a ∈C(sincee∈C), and hencee+a ∈(N∪C). Therefore, byDefinition 2,

SinceRis also a generalized periodicring, therefore byLemma 5(see (37))

eai_{∈C, ∀i∈ {1,...,n−1},0 =e=e}2_{, e∈C, a∈N. (39)} Combining (38) and (39), we see that

a−an_{∈C, ∀a∈N\C. (40)}

Since (40) is trivially satisfied fora∈(N∩C), therefore

a−an_{∈C, ∀a∈N, n≥2. (41)}

We claim that

N⊆C. (42)

The proof is by contradiction. Suppose (42) is false. Then, for somea∈R, we have

a∈N, a ∈C. (43)

Sincea∈N, there exists a positive integerσ0such that aσ_{∈C, ∀σ≥σ}

0, σ0minimal. (44)

Moreover, sincea∉C(see (43)), thereforeσ0>1. Now, applying (41) to the nilpotent elementaσ0−1_{, we see that}

aσ0−1_−aσ0−1n_{∈C, for somen=na}σ0−1_{≥2. (45)}

Furthermore, since(σ0−1)n≥(σ0−1)2≥σ0(sinceσ0≥2), (44) implies that

aσ0−1n_=a(σ0−1)n_{∈C. (46)}

Combining (45) and (46), we conclude thataσ0−1_{∈C, which contradicts the minimality} ofσ0in (44). This contradiction proves (42). SinceRis a periodicring satisfying (42), therefore, by a well-known theorem of Herstein [4],Ris commutative. This completes the proof.

Corollary12. A generalized Boolean ring with central idempotents and commut-ing nilpotents is commutative.

This corollary recovers a result proved by the authors in [1].

Corollary13. IfRis a generalized Boolean ring, and ifRis2-torsion-free, thenR is nil or commutative.

Proof. We claim that all idempotents ofRare central. Suppose not, and suppose eis a noncentral idempotent inR. Then−e∉(N∪C), and hence (seeDefinition 2)

(−e)−(−e)n_{∈C, neven. (47)}

Theorem14. LetR be a generalized Boolean ring in which every finite subring is either commutative or nil. ThenRis either commutative or nil.

Proof. By contradiction. Thus, supposeRis a generalized Boolean ring such that every finite subring ofR is either commutative or nil. Suppose, further, that R is not commutative and not nil either. By Theorem 11, there must exist a noncentral idempotent elementeinR, and hencee∉(C∪N). Thus (seeDefinition 2), since−e∉

(C∪N),

(−e)−(−e)n_{∈(N∩C), neven. (48)}

This implies that 2e∈(N∩C), and hence (2e)k₌₂k_{e=0, for somek∈Z}+_{. Since} e ∈C, we must have the following:

Eitherex−exe =0 for somex∈R,orx_{e−exe =0 for somex∈R. (49)}

Supposeu=ex−exe =0. Then,

eu=u =0=ue=u2_{, (u=ex−exe =0). (50)} Moreover,

2u=[2e,ex]=0 (since 2e∈C). (51) Furthermore, the subring generated byeanduis

e,u =r e+su|r , s∈Z. (52)

Since 2k_{e=0 and 2u=0, the subringe,uisfinite. Indeed,}

e,u =r e+su|1≤r≤2k_{,1≤s≤2. (53)}

On the other hand, ifx_{e−exe =0 for somex∈R(the only other possibility), then} the subring,e,v, generated byeandv=x_{e−exeis (as is readily verified)}

e,v =r e+sv|1≤r≤2k_{,1≤s≤2. (54)}

Again,e,vis afinitesubring ofR. Hence, in either case, we found afinitesubring of R, which is neither commutative (sincee∉C), nor nil (sincee∉N), contradicting our hypothesis. This contradiction proves the theorem.

Remark15. A careful examination of the proof ofTheorem 14shows that we only need to assume that “every subringS, with|S| =2m_{for some positive integerm, is}

commutative or nil” in order for the ground generalized Boolean ringRto be com-mutative or nil. Indeed,|e,u| =2k_·2=2k+1_{, since the representation of anyxin} this subring in the formx=r e+su; r ,s∈Z, isunique. For, suppose x=r e+su andx=r_{e+su. Then,(r−r)e=(s−s)u. Recall that 2u=0, andue=0. Thus,} ifs_{−sis even, then(r−r)e=0, and hencer e=re, su=su. On the other hand,} ifs_{−s is odd, then(r−r)e=u, and hence(r−r)ee=ue=0. Again, we obtain} r e=r_{e, su=su.}

Example16. Let R=           

a b c

0 a2 ₀ 0 0 a   

:a,b,c∈GF(4)      

. (55)

It is readily verified that the idempotents ofRare central and

x−x7_{=0, ∀x∈R\(N∪C), (56)}

butRis neither nil nor commutative. Hence,Theorem 11is not true if we drop the hypothesis that “niseven” in the definition of a generalized Boolean ring.

Example17. Let

R=           

0 a b 0 0 c 0 0 0   

:a,b,c∈GF(3)      

. (57)

This example shows that we cannot drop the hypothesis that “Nis commutative” in Corollary 12. (Note thatRis not commutative.)

Example18. Let

R= 0 0 0 0 , 1 0 1 0 , 0 1 0 1 , 1 1 1 1

: 0,1∈GF(2)

. (58)

This example shows that we cannot drop the hypothesis that “the idempotents are central” inCorollary 12. (Note thatR is not commutative.) This example also shows that we cannot drop the hypothesis that “R is 2-torsion-free” inCorollary 13. Note that, in this ringR,x−x2_{=0 for allx∈R\(N∪C). Even more is true. This ringR} also shows that we cannot drop the hypothesis that “1∈R” inCorollary 7, nor the hypothesis that “1∈R” inCorollary 8.

Returning to the ringRinExample 16, we see that this ring further shows that we cannot drop the hypothesis that “mandnare ofoppositeparity” in the definition of a generalized periodic ring in connection withCorollary 7, or the hypothesis that “nis even” in the definition of a generalized Boolean ring as far asCorollary 8is concerned. (Recall thatx−x7_{=0 for allx∈R\(N∪C).)}

Acknowledgement. This work was supported by the Natural Sciences and Engi-neering Research Council of Canada, Grant No. 3961.

References

[1] H. E. Bell and A. Yaqub,Generalized periodic rings, Int. J. Math. Math. Sci.19(1996), no. 1, 87–92.MR 96h:16036. Zbl 842.16019.

[2] M. Chacron,On a theorem of Herstein, Canad. J. Math.21(1969), 1348–1353.MR 41 #6905. Zbl 213.04302.

[3] I. N. Herstein,A generalization of a theorem of Jacobson. III, Amer. J. Math.75(1953), 105–111.MR 14,613e. Zbl 050.02901.

[4] ,A note on rings with central nilpotent elements, Proc. Amer. Math. Soc.5(1954), 620.MR 16,5c. Zbl 055.26003.

Howard E. Bell: Department of Mathematics, Brock University, Street Catharines, Ontario, CanadaL2S 3A1

E-mail address:[email protected]

Adil Yaqub: Department of Mathematics, University of California, Santa Barbara, CA93106, USA