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ADJOINT REGULAR RINGS
HENRY E. HEATHERLY and RALPH P. TUCCI
Received 10 September 2001
LetRbe a ring. Thecircle operationis the operationa◦b=a+b−ab, for alla,b∈R. This operation gives rise to a semigroup called theadjoint semigrouporcircle semigroupofR. We investigate rings in which the adjoint semigroup is regular. Examples are given which illustrate and delimit the theory developed.
2000 Mathematics Subject Classification: 16N20, 16U99, 20M17, 20M18, 20M19.
1. Introduction. This paper continues the authors’ investigation of adjoint semi-groups of rings [13,14]. HereRwill always denote a ring (not necessarily commuta-tive and not necessarily with unity). The Jacobsoncircleoradjoint operation,a◦b= a+b−ab, for eacha,b∈R, yields a monoid(R,◦), theadjoint semigroupofR. Here we primarily consider the situation whereRis a (von Neumann) regular ring or where
(R,◦)is (von Neumann) regular. Previous work along these lines has been done by Du [7] and Clark [3]. Our viewpoint is that of the interplay between semigroup properties of(R,◦)or(R,·)and ring properties ofR.
Forx∈R, we uselR(x)andrR(x)for the left and right annihilator sets ofxinR, respectively. When no ambiguity will arise, we use simplyl(x)andr(x). The Jacobson radical ofRis denoted by(R).
LetSbe a semigroup. We useE(S)for the set of idempotents inSandZ(S)for the center ofS. Frequent use will be made of the fact thatE(R,·)=E(R,◦)andZ(R,·)= Z(R,◦). Consequently we useE(R)andZ(R)for these sets, respectively.
Of particular interest here are the following types of regular semigroups. LetTbe a regular semigroup. If the idempotents ofTcommute among one another,Tis said to beinverse; this is equivalent to the condition that the von Neumann inverse of each element inTis unique, [5, Theorem 4.11]. If the idempotents ofT are central, thenT
is said to beClifford. It is well known that a Clifford semigroup is a union of groups, [5, Theorem 1.17]. If the idempotents form a subsemigroup, thenTis calledorthodox. If each element commutes with one of its von Neumann pseudoinverses, thenT is completely regular. This last condition is equivalent to the condition thatT is a union of groups, [19, Theorem II.1.4].
2. Preliminaries. In this section, we present some preliminary results. LetR1 be the standard Dorroh extension of a ring R to the ring R1, which has unity. Recall that this embedsR as an ideal ofR1, so we can identifyRas the ideal ¯RinR1. We will at times useR1in conjunction withR in order to make use of the simplifying attributes of having an identity for calculation. Recall that the mappingφ:x→1−x
is an isomorphism from(R1,◦)onto(R1,·), andφrestricted toRyields an injective homomorphism. Observe that ifRhas identity, then this same mapping,x→1−x, yields an isomorphism from(R,◦)onto(R,·). This observation makes the next result immediate. (See also [9, Lemma 20].)
Proposition2.1. Leta,b∈R. Then
(a) a◦b∈E(R)if and only if(1−a)(1−b)=1−(a◦b); (b) a◦b◦a=aif and only if(1−a)(1−b)(1−a)=(1−a);
(c) a◦R=b◦Rif and only if(1−a)R1=(1−b)R1;
(d) ais adjoint regular inRif and only if1−ais a regular element inR1. The next result is immediate, but it is useful enough to warrant stating.
Proposition 2.2. An element a∈R is adjoint regular if and only if there exist
b∈R,e=e2∈Rsuch thata+b−ab=eandea=e.
Note that a quasi-regular element satisfies the conditions ofProposition 2.2with
e=0. Hence, the conditions for adjoint regularity are a natural generalization of the condition for quasi-regularity.
Du [7, Theorem 1] has shown that if R is a regular ring, then (R,◦)is a regular monoid. We next give a different proof of that result.
Proposition2.3. IfRis a regular ring, then(R,◦)is a regular monoid.
Proof. There is an injective ring homomorphism,φ:R→R∗, that embedsRas an ideal, ˆR, in the regular ringR∗which has identity [10, Theorem 1]. Consequently,φ: (R,◦)→(R∗,◦)is an isomorphism. SinceR∗has identity we have that(R∗,·)(R∗,◦) and that each ideal of(R∗,·)is regular. So each ideal of(R∗,◦)is regular. Hence(R,ˆ◦) is regular and so is its isomorphic image(R,◦).
Since every ideal of a regular semigroup is regular, we immediately have that ifR
is a regular ring andXis an ideal of(R,◦), then(X,◦)is regular.
By using the powerful Fuchs-Halperin result, this proof completely bypasses the calculations used in Du’s proof [7]. Also, using similar methods, we can obtain analo-gous results by replacing the termregular inProposition 2.3byorthodoxorinverse. Finally, note that the converse ofProposition 2.3does not hold, as any example of a Jacobson radical ring will illustrate.
3. Equivalent conditions under regularity. In this section, we discuss various equivalent conditions toRstrongly regular and toRadjoint completely regular. Exam-ples are given which illustrate these results and show limitations to extending them.
(2) for eacha∈Rthere existsc∈Rsuch thata=c2a; (3) Ris regular andE(R)⊆Z(R).
A ringRsatisfying any of (and hence all of) these conditions is called astrongly regular ring, [6]. Note that these conditions are not equivalent for semigroups, [19, Theorem II.1.4]. The next proposition ties strongly regular with several other conditions on the multiplicative semigroup of a ring. The proposition is a compilation of results from various sources. First some terminology is needed.
Following [17] we say that a semigroupS isE-solidif, whenevere,f ,g∈E(S)such thateᏸfg, there existsh∈E(S)such thatehᏸg. (Hereᏸandare the standard Green’s relations, [5, page 47].) Recall that the core of S, denoted byC(S), is the subsemigroup ofS generated byE(S), [19, page 89]. SoS is orthodox if and only ifS
is regular andE(S)=C(S). Letᏼbe a semigroup property. ThenS islocally-ᏼifeSe has propertyᏼfor everye∈E(S).
Proposition3.1. LetRbe a regular ring. The following are equivalent: (a) Ris strongly regular;
(b) Ris orthodox;
(c) Ris completely regular; (d) Ris inverse;
(e) Ris Clifford; (f) Ris locally inverse; (g) RisE-solid; (h) Ris locallyE-solid;
(i) C(R)is completely regular.
Proof. As mentioned above, (a)(e). So (e)⇒(b) is trivial, while (b)⇒(e) follows from [22, Remark 15]. It is known that (a)(c), even for semigroups [19, page 58]. Next, (d)(e) comes from [13], while (d)(f)(g)(h) comes from [17], and (g)(i) comes from [12, Theorem 3]. This completes the logical circuit.
This by no means exhausts the vast number of known equivalent conditions toR
strongly regular, but it suffices for our purposes and gives a good sample of what is known. For more equivalent conditions toRstrongly regular see [5,6,11,16,18,19,
21,22].
We next consider various equivalent conditions on the adjoint semigroup of a reg-ular ring.
Proposition3.2. LetRbe regular. The following are equivalent: (a) Ris adjoint completely regular;
(b) Ris adjoint Clifford; (c) Ris strongly regular.
Proof. Assume (a). Du [8, Lemma 11] has shown that(R,◦)completely regular im-plies thatE(R)is closed under the ring multiplication. SinceRis regular, this yields thatRis orthodox. ThusRis Clifford and byProposition 3.1Ris strongly regular. Con-sequently,Ris adjoint regular. Conversely,Rregular and adjoint Clifford implies that
a Clifford semigroup is a union of groups,(R,◦)Clifford implies(R,◦) completely regular, yielding (b)⇒(a) and finishing the logical chain.
It is known [13] thatRis adjoint inverse if and only ifRis adjoint Clifford. Recently Du [9] has shown that adjoint orthodox implies adjoint completely regular. However, an adjoint completely regular ring need not be adjoint Clifford, as the next example shows.
Example3.3. LetS be a right zero semigroup with more than one element and letR be the semigroup ringZ2[S]. So every nonzero element ofRhas the formx= e1+ ··· +en, for some distincte1,...,en∈S. Lety=f1+ ··· +fm, wheref1,...,fm are distinct terms from S. Then xy=ny. So xy =0, if n is even, andxy =y, if n is odd. Thenx is an idempotent if and only if n is odd, and if y is also an idempotent, then xy=y. Consequently E(R) is closed under ring multiplication. Observe thatR=N(R)∪E(R); henceR is adjoint orthodox. Since elements inE(R)
are completely regular in(R,◦), and since elements inN(R)are quasi-regular inR, and hence are completely regular inR, we have thatRis adjoint completely regular. However, since(e1+e2)y(e1+e2)=0, for eachy∈R, we see thatRis not regular. Also,Ris not a Jacobson radical ring. Sincee1e2≠e2e1, for distincte1,e2∈S, we see thate1◦e2≠e2◦e1, and henceRis not adjoint inverse.
Having R regular implies thatR is adjoint regular, as we have seen. However, R
being regular does not imply thatRis adjoint inverse, adjoint completely regular, nor adjoint orthodox, as the next example illustrates.
Example3.4. LetAbe a regular ring with identity and letR=M2(A), the ring of 2×2 matrices over A. SoR is regular and henceR is adjoint regular. But there are noncommuting idempotents inR; soRis not adjoint inverse, and hence not adjoint completely regular byProposition 3.2. By [9, Theorem 14],Ris not adjoint orthodox.
4. Decomposition. LetR be a ring, and letS1,...,Sn be subrings ofR. IfR=S1+ ··· +Sn and wheneversi∈Si, i=1,...,n such that s1+ ··· +sn=0, then si=0,
i=1,...,n, then we say that R is a supplementary sum of theSi, i=1,...,n, [1]. This is equivalent to R+ =Σni=1⊕S+i, as a direct sum of abelian groups. We write
R=S1 ··· Sn for such a supplementary sum. We next state as a lemma the well-known two-sided Peirce decomposition, given here without using an identity in the ring. (See [1,15].)
Lemma4.1. Lete∈E(R). ThenR=eRee·lR(e)rR(e)·erR(e)∩lR(e).
Recall [2] thate∈E(R)is said to be aleft semicentral idempotentinRifeRe=Re, (equivalently,er e=r e, for eachr∈R). It is well known that in this caselR(e)is an ideal ofRandR/lR(e)eRe.
Proposition4.2. Letebe a left semicentral idempotent inR.
(b) R=Re⊕llR(Re)as a direct sum of left ideals ofR andR/lR(e)eRe, a ring with unitye;
(c) eR=Ree·lR(Re).
Proof. By definition,Re=eRe. The rest is routine.
In a strictly analogous fashion, we define right semicentral idempotent and obtain a dual result. For a central idempotent we naturally obtain a stronger result.
Corollary4.3. Ifeis a central idempotent inR, thenR=eRe⊕AnnR(eRe), as a direct sum of ideals ofR.
Following Clark and Lewin [4],e∈E(R)is said to be aprincipal idempotentinRif the homomorphic image ofein ¯R=R/J(R)is the identity in ¯R. This implies that if
u∈E(R)such thateu=0=ue, thenu=0. (The latter condition, together withe≠0, is what Albert used in defining theprincipal idempotent[1, page 25].)
Principal idempotents play a key role in our next decomposition, whose proof makes use of the following result due to Du [7, Corollary 2].
Lemma4.4. IfR is adjoint regular andeis a principal idempotent inR, thenR=
eReJ(R), as a supplementary sum of subrings.
We are now ready to give a much shorter proof of the main result in [3, Theorem B].
Proposition4.5. Letebe an idempotent in a ringR. The following are equivalent: (a) Ris adjoint completely regular andeis a principal idempotent inR;
(b) R=eReJ(R)andeReis a strongly regular ring.
Proof. Assume (a). ByLemma 4.4,R=eReJ(R); soR/J(R)eRe. Since the ring
eReinherits the adjoint completely regular condition fromRandeReis regular, by
Proposition 3.2we have that ringeReis strongly regular.
Assume (b). TheneReis adjoint regular, so by [7] the ringRis adjoint regular. Since all the idempotents ofR are central,R is adjoint completely regular. Clearlyeis a principal idempotent ofR.
In view of Proposition 4.5 and the results of Section 3, we immediately have a plethora of conditions equivalent to part (a) ofProposition 4.5.
Proposition4.6. LetR=A⊕Bas a direct sum of left (right) ideals. IfRis adjoint regular, thenAandBare adjoint regular rings.
Proof. Leta∈A. Then there exista1∈A,b1∈Bsuch thata=a◦(a1+b1)◦a= a◦(a1+b1)+a−[a◦(a1+b1)]a. Soa◦(a1+b1)=[a◦(a1+b1)]a∈A. Thena◦(a1+ b1)=a+a1+b1−a(a1+b1), orb1−ab1=a◦(a1+b1)−a−a1∈A. Butb1−ab1∈B; so b1−ab1= 0. Thena= a◦(a1+b1)◦a=[a+a(a1+b1)−a(a1+b1)]◦a, or a=(a+a1−aa1)◦a=a◦a1◦a. Proceed similarly for right ideals.
Lemma4.7. LetRbe adjoint regular.
(a)EitherJ(R)=RorRcontains a nonzero idempotent.
(c)If the modulesRR and RR are indecomposable, then eitherR=J(R)or R is a division ring.
Proof. (a) Letr∈R,r≠0. Then there exists ¯r∈Rsuch thatr◦r¯◦r=r, andr◦r¯ and ¯r◦r are nonzero idempotents in(R,◦). Sor◦r¯and ¯r◦r are also idempotents inR. IfRhas no nonzero idempotent, thenr◦r¯=r¯◦r. So in this case each element ofRis quasi-regular, that is,R=J(R).
(b) IfR≠J(R), then there exists a nonzeror∈Rsuch thate=r◦r¯is a nonzero idempotent inR. ThenR=Re⊕l(e), as a direct sum of left ideals. SinceReandl(e)are submodules ofRR, and sinceRe≠0, we havel(e)=0, and henceeis a right identity of
R. SoR=eR⊕r(e), as a direct sum of right ideals ofR. However,R·r(e)=(Re)r(e)=0, sor(e)is an ideal ofRandr(e)⊆r(R). SinceRis adjoint regular, we have thateReis a regular ring [7, Proposition 2].
(c) Continuing from the proof of (b), sinceRRis indecomposable and sinceeRe≠0, we have that r(e)=0, and hence eis also a left identity. ThusR =eRe, a regular ring with identity. Since the ring is indecomposable, eitherR=J(R)orRis a division ring.
Note that fromLemma 4.7, we have that ifRis indecomposable in terms of both left and right ideal decompositions, then eitherR is a Jacobson radical ring and the regular radical (see [20, Chapter VI]) is zero, orRis equal to its regular radical and the Jacobson radical is zero.
Proposition4.8. LetRbe adjoint regular. IfRhas a right (left) nonzero semicentral idempotent, then there exist submonoidsAandBof(R,◦)such that
(a) R=A◦BandA∩B=0;
(b) R=A⊕Bas a direct sum of adjoint regular right (left) ideals ofR; (c) Ais a regular ring with identity andBis a two-sided ideal ofR.
Proof. Letebe a nonzero right semicentral idempotent inR. ThenR=eR⊕r (e)
as a direct sum of right ideals, whereeR=eReis a ring with unity andr (e)is a two-sided ideal ofR. LetA=eR and B=r (e). Observe thatR=A◦B becauseAB=0. ByProposition 4.6,(A,◦)is regular. SinceAis a ring with unity,Ais regular. Proceed similarly foreleft semicentral.
5. Radicals for adjoint Clifford rings. We show the equivalence of several standard radicals for rings which are adjoint Clifford and obtain a characterization for the Jacobson radical of such rings.
Proposition5.1. IfRis adjoint Clifford, then the Brown-McCoy radical,Ᏻ(R), and (R)are equal.
Example5.2. Proposition 5.1cannot be extended to all adjoint regular rings, even if the ring is also adjoint simple, as the following example illustrates. LetVbe a vector space over a fieldF with dimFV= ℵω, and letωbe the first infinite limit ordinal. Let
R= {φ∈EndFV |rankφ <ℵω}. It is well known that R is a regular ring with no maximal ideal. SoJ(R)=0 andG(R)=R. It is also known that(R,◦)is simple, [4, Example 3B].
In the next proposition, whenRdoes not have unity we use the unity element in the Dorroh extension for convenience of expression.
Proposition5.3. IfRis adjoint Clifford, then(R)= ∩e∈E(1−e)R.
Proof. LetR=eR⊕(1−e)R as ideals. TheneR is adjoint Clifford because the mapφ:R→eR is a ring homomorphism and soφ:(R,◦)→(eR,◦)is a semigroup homomorphism. Therefore(R)⊆(1−e)R. Sinceeis arbitrary, we have that(R)⊆ ∩e∈E(1−e)R.
Conversely, letI= ∩e∈E(1−e)R. Then I is an ideal, hence adjoint Clifford. ButI contains no nonzero idempotent. Therefore(I,◦)is a group, soI⊆(R).
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Henry E. Heatherly: Department of Mathematics, University of Louisiana at Lafayette, Lafayette, LA70504-1010, USA
Ralph P. Tucci: Department of Mathematics and Computer Science, Loyola University, New Orleans, LA70118, USA
