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A COMBINATORIAL COMMUTATIVITY PROPERTY FOR RINGS
HOWARD E. BELL and ABRAHAM A. KLEIN
Received 24 August 2001
We study commutativity in ringsRwith the property that for a fixed positive integern, xS=Sxfor allx∈Rand alln-subsetsSofR.
2000 Mathematics Subject Classification: 16U80.
1. Introduction. In [2], we discussedP∞-ringsR, which were defined by the prop-erty that
XY=Y X (1.1)
for all infinite subsetsX,YofR; and in an earlier paper [1], the first author discussed
Pn-rings, defined by the property that (1.1) holds for alln-subsetsX, Y ofR. For a fixed positive integern, we now define aQn-ring to be a ringRwith the property that
xS=Sx ∀x∈R, ∀n-subsetsSofR. (1.2)
Clearly, every commutative ring is aQn-ring for arbitraryn; moreover, there exist badly noncommutativeQn-rings, since every ring with fewer thannelements is aQn -ring. Our purpose is to identify conditions which forceQn-rings to be commutative or nearly commutative.
It is obvious that everyQn-ring is aPn-ring and everyPn-ring is aP∞-ring. We make no use of the results onPn-rings in [1], and most of our results are of a different sort than those in [1]. However, a special case of the theorem onP∞-rings in [2] plays a crucial role in our study.
2. Preliminaries. We begin with some notation. LetRbe an arbitrary ring, not nec-essarily with 1. The symbolsD,N,Z, andC(R)denote the set of zero divisors, the set of nilpotent elements, the center, and the commutator ideal, respectively; and|R|
denotes the cardinal number ofR. ForY being an element or subset ofR, the symbols
CR(Y ),A(Y ),Ar(Y ), andA(Y )denote the centralizer ofYand the left, right, and two-sided annihilators ofY. Forx,y∈R, the setLx,yis defined to be{w∈R|xy=wx}. We give three lemmas, the first of which is rather trivial and the other two of which are not.
Lemma2.1. LetRbe aQn-ring with|R| ≥n. Then
(iii) Nis an ideal;
(iv) ifRis not commutative andx∈Z, thenR\(A(x)∪CR(x))andR\(Ar(x)∪ CR(x))are nonempty.
Proof. (i) is obvious; and ife is idempotent, the fact thateR=Reyieldsex=
exe=xefor allx∈R, soe∈Z. Moreover, (i) enables us to prove (iii) by adapting the standard proof thatNis an ideal in commutative rings. Finally, ifx∈Z thenCR(x) is a proper subgroup of(R,+); and (i) implies thatA(x)andAr(x)are also proper subgroups of(R,+). Since a group cannot be the union of two proper subgroups, (iv) is immediate.
Lemma2.2. IfRis an infiniteQn-ring, thenRis commutative.
Proof. Since everyQn-ring is aP∞-ring, we could simply invoke the theorem of
[2], which states that everyP∞-ring is either finite or commutative. However, the proof in [2] is long and involved, so we prefer to give a more elementary proof.
LetRbe a noncommutativeQn-ring. We may assume thatR is not aQm-ring for anym < n. Since allQ1-rings are commutative,n >1, and there existx∈Rand an
(n−1)-subsetHofRsuch thatxH=Hx; and we may assume thatxHis not a subset ofHx. We may also assume thatR\H= ∅, since otherwiseRis finite.
For anya∈R\H,x(H∪{a})=(H∪{a})x, so if we takeh∈Hfor whichxh∈Hx, we have
xh=ax. (2.1)
Since (2.1) holds for alla∈R\H, it follows that for fixedb∈R\H,R\H⊆b+A(x). Moreover, ifc ∈A(x), thenxh=bx=(b+c)x, so b+c ∈H. ThereforeR\H= b+A(x), hence|R\H| = |A(x)|and|R\A(x)| = |H|. But sinceA(x)is a proper subgroup of R, |R\A(x)| ≥ |A(x)|, that is,|H| ≥ |R\H|; and the finiteness ofH yields the finiteness ofR.
Lemma2.3(see [4]). IfRis a finite ring withN⊆Z, thenRis commutative.
In view ofLemma 2.2, we assume henceforth thatRis finite.
3. Commutativity ofQn-rings with1
Theorem3.1. IfRis anyQn-ring with1such that|R|> n, thenRis commutative.
Proof. ByLemma 2.3, we need only to show thatN⊆Z; and sinceu∈Nimplies
1+uis invertible, it suffices to prove that invertible elements are central.
Suppose, then, thatxis a noncentral invertible element andy∈CR(x). IfHis any (n−1)-subset ofRwhich excludesy, the conditionx({y}∪H)=({y}∪H)xyields
z∈Hsuch that
xy=zx. (3.1)
The bound on|R|inTheorem 3.1is best possible, as the following example shows. The rings of this example were introduced by Corbas in [3].
Example3.2. Letn=p2k, wherepis prime andk >1. Letφbe a nonidentity
auto-morphism ofGF(pk). LetR=GF(pk)×GF(pk), with addition being componentwise and multiplication given by(a,b)(c,d)=(ac,ad+bφ(c)). It is easily shown thatR
is a ring with|R| =nandD= {(0,b)|b∈GF(pk)}; hence ifa≠0,(a,b)is invertible. Thus, ifa≠0,(a,b)R=R(a,b)=R; and ifb≠0,(0,b)R= {(0,bφ(c))|c∈GF(pk)} andR(0,b)= {(0,bc)|c∈GF(pk)}, so that(0,b)R=R(0,b)=D. Thus,Ris aQ
n -ring. Obviously,Ris noncommutative and(1,0)is a multiplicative identity element.
4. Commutativity ofQn-rings: the general case. We begin this section with a near-commutativity theorem, which is reminiscent of [1, Theorem 6].
Theorem4.1. Ifn≤16andRis anyQn-ring, thenC(R)is nil.
Proof. Since everyQk-ring is aQk+1-ring, we may assume thatn=16. If|R| ≥16,
thenNis an ideal byLemma 2.1(iii); andR/Nis a finite ring with no nonzero nilpotent elements, hence is commutative. If|R|<16, it follows easily from the Wedderburn-Artin structure theory thatC(R)is nil.
We proceed to our major commutativity theorems.
Theorem4.2. Letn≥4, and letRbe aQn-ring. If|R|>2n−2, or ifnis even and
|R|>2n−4, thenRis commutative.
Proof. LetRbe aQn-ring which is not commutative, and letx∈Z. Our aim is to
show that|R| ≤2n−2 or|R| ≤2n−4; and sincen−1<2n−4, we may suppose that
|R| ≥n. ByLemma 2.1(iv), there existsy∈R\(Ar(x)∪CR(x)). IfH is any(n−1) -subset which does not containy, we havex({y}∪H)=({y}∪H)x; and sincexy≠ yx, there existsz∈H such thatxy=zx—that is,H∩Lx,y≠∅. We have argued that any(n−1)-subset ofRmust either containyor intersectLx,y—a condition that cannot hold if|R\Lx,y| ≥n; thus,
|R| ≤Lx,y+n−1. (4.1)
We now investigate|Lx,y|. Ifw∈Lx,y, thenLx,y=w+A(x), hence|Lx,y| = |A(x)|. ByLemma 2.1(iv),A(x)≠R, so|Lx,y| = |R|/kfor somek≥2. Substituting into (4.1) gives
|R| ≤ k
k−1(n−1)≤2n−2. (4.2)
Suppose now thatn is even. If A(x) has index at least 3 in(R,+), (4.2) yields |R| ≤ 3(n−1)/2 ≤2n−4. Thus, we may assume that|A(x)| = |R|/2 and show that |R|≠2n−2.
y∈Ar(x),A(x)≠Ar(x), soAr(x)x≠{0}. Nowx(y∪Ar(x))=(y∪Ar(x))xand thereforeAr(x)x⊆ {xy,0}; henceAr(x)x= {0,xy}is an additive subgroup of order 2. Therefore the mapφ:Ar(x)→Ar(x)xgiven bywwxhas kernel of index 2 in Ar(x). But|Ar(x)|is odd, so we have a contradiction; hence|R| ≤2n−4.
As we will see later, the bounds on|R|inTheorem 4.2are best possible; however, under various restrictions, a smaller bound holds.
Theorem4.3. Letn≥4and letRbe aQn-ring with|R|> (3/2)(n−1). ThenRis
commutative if one of the following is satisfied: (i) |R|is odd;
(ii) (R,+)is not the union of three proper subgroups; (iii) Nis commutative;
(iv) R3≠{0}.
Proof. Again we suppose that R is not commutative and x ∈ Z. Since |R|>
(3/2)(n−1) > n, the arguments in the proof ofTheorem 4.2show that |A(x)| = |Ar(x)| = |R|/2—a fact which proves that (i) implies commutativity ofR.
Applying the first isomorphism theorem for groups shows that|xR| = |Rx| =2; hence for anyu∈R\Ar(x)andv∈R\A(x),xR= {0,xu}andRx= {0,vx}. Since xR=RxbyLemma 2.1(i), it follows that ify∈R\(A(x)∪Ar(x)), then{0,xy} = xR=Rx= {0,yx}and therefore y∈CR(x). ThusR=A(x)∪Ar(x)∪CR(x), and we have proved that (ii) implies commutativity ofR.
We now show thatx∈N. SinceRis not commutative, it follows fromTheorem 3.1 thatRdoes not have 1, henceR=D; and ifx∈N, some power ofxis an idempotent zero divisore≠0. SinceA(x)⊆A(e)andA(e)≠R, we must haveA(x)=A(e) and similarlyAr(x)=Ar(e). Buteis central byLemma 2.1(ii), henceA(x)=Ar(x)= A(x)⊆CR(x). Thus, ify∈A(x),{0,xy} =xR=Rx= {0,yx}andyis also inCR(x), contrary to the assumption thatx∈Z. Butx was an arbitrary noncentral element; hence, if there exist two noncommuting elements, both must be nilpotent. Thus (iii) forces commutativity ofR.
To complete our proof, we show that our assumption thatRis not commutative forcesR3= {0}. Forx∈Z, the fact thatx∈NyieldsA
r(x2)⊋Ar(x), soAr(x2)=R; hencex2R=Rx2= {0}. If we choosey∈R\(A
r(x)∪CR(x))andw∈R\(A(x)∪ CR(x)), theny2R=Ry2= {0}; moreover,{0,xy} =xR=Rx= {0,wx}, soxy=wx. Thus,xR2=xyR=wxR= {wxy,0} = {xy2,0} = {0}. Ifz∈Z, thenx+z∈Zso
(x+z)R2= {0}; thereforeR3= {0}as required.
We now give examples showing that the bounds on|R|in Theorems4.2and4.3are best possible.
Example4.4. LetRbe the algebra overGF(2)with basisx,y,x2and multiplication
defined byxy=x2=y2, 0=yx=x2y=yx2=xx2=x2x=x2x2. Then{0,x2} =
A(R). It is easily verified that for any u∈A(R), the setsA(u), Ar(u), {w∈R | uw=x2}and{w∈R|wu=x2}all have 4 elements; hence for any 5-subsetSofR,
Example4.5. LetRbe the algebra overGF(3)with basisx,y,x2and multiplication
defined as in the previous example. An argument similar to the one above shows that
R is aQ19-ring with|R| =27, so the bound(3/2)(n−1)inTheorem 4.3cannot be reduced.
5. Further results for smalln. By definition allQ1-rings are commutative, and it is easy to see that allQ2-rings are commutative; and since there exist rings of order 4 which are not commutative, not allQ5-rings are commutative. It is natural to ask: what is the largestnsuch that allQn-rings are commutative? Our next theorem gives the answer.
Theorem5.1. Ifn≤4, allQn-rings are commutative.
Proof. Since everyQk-ring is aQk+1-ring, we may assumen=4. ByTheorem 4.2
any counterexampleRwould have|R| ≤4; and since all rings of order less than 4 are commutative, we need only to consider rings of order 4.
Suppose, then, thatRis a counterexample andxandyare noncommuting elements withxy=0. ThenR= {0,x,y,x+y}. Since idempotents are central, any ofx2=x,
x2=y,x2=x+y would forcex andyto commute; hencex2=0. It is now easily checked that the conditionxR=Rxcannot hold.
Not surprisingly, a better result holds for rings with 1.
Theorem5.2. Ifn≤8, then everyQn-ring with1is commutative.
Proof. We may assume thatn=8. Suppose thatRis a counterexample. ByTheorem
3.1,|R| ≤8; and since all rings with 1 having fewer than 8 elements are commutative,
|R| =8 andR is indecomposable. Since idempotents are central, we therefore have no idempotents except 0 and 1; hence every element is either nilpotent or invertible. Sinceu∈Nimplies 1+u is invertible, it follows fromLemma 2.3that there exists a pairx,yof noncommuting invertible elements. The group of units is not commu-tative and has at most 7 elements, hence is isomorphic toS3. Thus, there exists a unique nonzero nilpotent elementu, which byLemma 2.3is not central; and there is therefore an invertible elementwsuch thatuw=wu. But in view ofLemma 2.1(iii),
wuanduware nonzero nilpotents, so we have a contradiction.
Theorem 5.2is best possible; the ring of upper-triangular 2×2 matrices overGF(2)
is aQ9-ring with 1 which is not commutative.
Acknowledgments. The authors are grateful to Professor B. H. Neumann for
suggesting that we study Qn-rings. The first author was supported by the Natural Sciences and Engineering Research Council of Canada, Grant No. 3961.
References
[1] H. E. Bell,A setwise commutativity property for rings, Comm. Algebra25(1997), no. 3, 989–998.
[4] I. N. Herstein,A note on rings with central nilpotent elements, Proc. Amer. Math. Soc.5 (1954), 620.
Howard E. Bell: Department of Mathematics, Brock University, St. Catharines, Ontario, CanadaL2S 3A1
E-mail address:[email protected]
Abraham A. Klein: Sackler Faculty of Exact Sciences, School of Mathematical Sciences, Tel Aviv University, Tel Aviv69978, Israel
