Some Notes Clean r Ngs

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Some Notes On Clean Rings

Nicholas A. Immormino


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Clean Rings

A ring is said to be clean if every element in the ring can be written as the sum of a unit andan idempotent of the ring. More generally, an element in a ring is called clean if it can be writtenas the sum of a unit and an idempotent of the ring. These rings were introduced by Nicholson [ 34]in his study of lifting idempotents and exchange rings. In these notes we review some well-knownproperties and examples of clean rings, and then we classify the rings that belong to an interestingsubclass of clean rings. For a more general introduction to clean rings, see the survey of clean ringsgiven by Nicholson and Zhou [37]. For a different perspective on this class of rings, see the historyof commutative clean rings given by McGovern [29].

§1 Some Properties of Clean Rings

In this section we review some useful properties of clean rings. We begin by showing that everyhomomorphic image of a clean ring is clean, that a direct product of rings

Ri is clean if and only

if each of the rings Ri is clean, and that a full matrix ring M n(R) is clean if the underlying ring Ris clean. Then we consider the notion of lifting idempotents modulo an ideal, and we prove that aring R is clean if and only if for any ideal I of R such that I ⊆ J (R) the quotient ring R/I is cleanand idempotents lift modulo I . These results are well-known, but proofs are included for the sakeof completeness.

Basic Properties

The next two results were proved by Anderson and Camillo [1].

(1.1) Proposition. Every homomorphic image of a clean ring is clean.

Proof . Since multiplication is preserved by every ring homomorphism, the homomorphic image of a unit (resp. idempotent) is a unit (resp. idempotent) of its ring. Since addition is also preservedby every ring homomorphism, the result follows.

(1.2) Proposition. A direct product of rings

Ri is clean if and only if each ring Ri is clean.

Proof . Since multiplication in a direct product of rings is defined componentwise, an element ina direct product of rings is a unit (resp. idempotent) of that ring if and only if the entry in each of its components is a unit (resp. idempotent) of its ring. Since addition in a direct product of ringsis also defined componentwise, the result follows from a simple computation.

The next result is due to Han and Nicholson [18].

1


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(1.3) Theorem. A full matrix ring M n(R) is clean if the underlying ring R is clean.

Proof . Han and Nicholson [18, Theorem] showed that if the identity 1 of a ring R can be writtenas a finite sum 1 = e1 + e2 + · · · + en of mutually orthogonal idempotents ei such that each cornerring eiRei is clean, then the ring R is clean—the interested reader is encouraged to see their proof.Since the set of matrix units {E ii}

n

i=1 is a complete set of orthogonal idempotents for M n(R) with

each corner ring E iiM n(R)E ii isomorphic to R, the result follows.

Lifting Idempotents

For an ideal I of a ring R, we say that idempotents lift modulo I if for each element x ∈ Rsuch that x − x2 ∈ I there is some idempotent e of R with e − x ∈ I . Meanwhile, a ring is said tobe exchange if it satisfies the exchange property of Crawley and Jónsson [13] when regarded as aleft module over itself. Exchange rings were introduced by Warfield [46], and he proved that theirdefinition is left-right symmetric. The interested reader is encouraged to see either [13] or [46] forthe formal definition of the exchange property. In his study of lifting idempotents and exchangerings, Nicholson [34] called a ring suitable if idempotents lift modulo every left ideal of the ring,and he proved that this definition is left-right symmetric. Moreover, he showed that these suitablerings coincide with the exchange rings of Warfield. In other words, a ring is exchange if and only if idempotents lift modulo every left (equivalently, right) ideal of the ring. Nicholson [34] introducedclean rings as an example of a class of exchange rings when he proved the following.

(1.4) Proposition. (1) Every clean ring is exchange. (2) An abelian ring is clean if and only if it is exchange.

Proof . (1) Let R be a clean ring, let x be any element in R, and let I be any left ideal of R suchthat x − x2 ∈ I . As we mentioned above, a ring is exchange if and only if idempotents lift modulo

every left ideal of the ring according to Nicholson [34]. Thus it suffices to show that there is someidempotent e of R with e − x ∈ I . Since R is a clean ring, we can write x = u + f for some unit uand idempotent f of R. Let v denote the inverse of u, and check that e = v(1 − f )u is idempotentwith e − x = v(x − x2) ∈ I as desired. Therefore every clean ring is exchange.

(2) This proof is omitted—see Nicholson [34, Proposition 1.8].

(1.5) Corollary. Idempotents lift modulo every left ideal of a clean ring.

Proof . This result follows from (1.4) since idempotents lift modulo every left (equivalently, right)ideal of an exchange ring according to Nicholson [34].

For an ideal I of a ring R, we say that units lift modulo I if for each element x ∈ R such that

x + I is a unit of the quotient ring R/I there is some unit u of R with u − x ∈ I . It can be shownthat for any ring R and any ideal I ⊆ J (R), an element x + I is a unit of R/I if and only if x is aunit of R. In particular, this tells us that units lift modulo any ideal of a ring R that is containedin its Jacobson radical. We use this fact to prove the following characterization of clean rings dueto Han and Nicholson [18].

(1.6) Theorem. Let I be any ideal of a ring R such that I ⊆ J (R). Then R is clean if and only if the quotient ring R/I is clean and idempotents lift modulo I .


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Proof . Suppose that R is a clean ring. Then R/I is clean as a result of (1.1), while idempotentslift modulo I by (1.5). This proves the necessity. Conversely, suppose that the quotient ring R/I isclean and that idempotents lift modulo I , and let r be any element in R. Since R/I is a clean ring,we can write r + I = x + e + I for some unit x + I and idempotent e + I of R/I . Moreover, sinceidempotents lift modulo I , we can assume that e is an idempotent of R. Since I ⊆ J (R), we also

know that units lift modulo I . In particular, since r − e + I = x + I is a unit of R/I , this tells usthat r − e is a unit of R. It follows that r can be written as the sum of a unit and an idempotentof R by writing r = (r − e) + e. This proves the sufficiency. Therefore R is clean if and only if thequotient ring R/I is clean and idempotents lift modulo I .

It is well known that idempotents lift modulo every nil ideal of a ring—for an elementary proof of this important classical result, the interested reader should see Koh [24]. Since every nil ideal of a ring R is contained in its Jacobson radical, the following result is a corollary to (1.6).

(1.7) Corollary. Let N be any nil ideal of a ring R. Then R is clean if and only if the quotientring R/N is clean.

Proof .

As noted above, this result follows from (1.6) since every nil ideal of a ring R is containedin its Jacobson radical, and idempotents lift modulo every nil ideal.

Two More Properties

A ring R is said to be semipotent if each left ideal of R that is not contained in its Jacobsonradical contains a nonzero idempotent—this notion is left-right symmetric. It is not difficult to seethat a ring R is semipotent if and only if for each a ∈ R that is not contained in J (R) there existssome nonzero x ∈ R such that xax = x. A semipotent ring R is said to be potent if idempotentslift modulo its Jacobson radical—these rings are sometimes called I -rings, while semipotent rings

are sometimes known as I 0-rings. Nicholson [34, Proposition 1.9] proved that every exchange ringis potent. Since every clean ring is exchange by (1.4), we have the following.

clean =⇒ exchange =⇒ potent =⇒ semipotent

These implications are known to be irreversible. In particular, Camillo and Yu [ 7, p. 4746] provedthat the ring in a well-known example due to Bergman [19, Example 1] is exchange but not clean.Meanwhile, Nicholson [34, p. 272] showed that a potent ring need not be exchange, and Nicholsonand Zhou [38, Example 25] showed that a semipotent ring need not be potent. Following the leadof Han and Nicholson [18], we give a direct proof that every clean ring is semipotent.

(1.8) Proposition. Every clean ring is semipotent.

Proof . Let R be any clean ring, and let I be any left ideal of R such that I \ J (R). We need toshow that I contains a nonzero idempotent. It is well known that every quasiregular left ideal of aring R is contained in its Jacobson radical. Since I \ J (R), this tells us that there is some elementa ∈ I that is not quasiregular. Since R is a clean ring, we can write a = u + e for some unit u andidempotent e of R. Let v denote the inverse of u, and check that v(1 − e)u is idempotent and thatv(1 − e)u = v(1 − e)(a − e) = v(1 − e)a is an element of I . Meanwhile, notice that e − a = −u is aunit of R. Since a is not quasiregular, this tells us that e \= 1. Since u and v are both units of R, it


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follows that v(1 − e)u \= 0, and hence v(1 − e)u is a nonzero idempotent of I . Therefore every cleanring is semipotent.

(1.9) Corollary. Every clean ring is potent.

Proof . This result follows immediately from (1.8) and (1.5).

A nonzero idempotent e of a ring R is called a primitive idempotent if the corner ring eRecontains no nontrivial idempotents—in other words, if the ring eRe contains no idempotents otherthan 0 and e. It can be shown that a nonzero idempotent e of a ring R is a primitive idempotentif and only if for any nonzero idempotent f of R the inclusion f ∈ Re implies that Rf = Re. Thisfact was pointed out by Nicholson [33, p. 346]. Meanwhile, an idempotent e of a ring R is called alocal idempotent if the corner ring eRe is a local ring. Since a local ring contains no nontrivialidempotents, every local idempotent is primitive. The following result due to Nicholson [33] showsthat the converse is true for any semipotent ring.

(1.10) Proposition. Every primitive idempotent in a semipotent ring is local.

Proof .

Let R be a semipotent ring, and let e be any primitive idempotent of R. We need to showthat e is a local idempotent of R. In particular, we need to show that the corner ring eRe is a localring. Let a be any element in the ring eRe that is not contained in J (eRe). It is well known thatJ (eRe) = J (R) ∩ eRe for any idempotent e of a ring R—see Lam [25, Theorem 21.10]. It followsthat a is not contained in J (R), and hence Ra \ J (R). Since R is a semipotent ring, this tells usthat the left ideal Ra contains some nonzero idempotent f . Notice that Rf ⊆ Ra ⊆ Re. Since e isa primitive idempotent of R, this implies that Rf = Re, which forces Ra = Re. In particular, thistells us that e is contained in Ra, and hence there is some element x ∈ R such that xa = e. Sincea is an element of eRe, it follows that (exe)a = e, and hence a has a left inverse in eRe. Since thisis the case for every element of eRe that is not contained in J (eRe), it follows that the corner ringeRe is a local ring. Therefore every primitive idempotent of R is local.

(1.11) Corollary. Every primitive idempotent in a clean ring is local.

Proof . This result follows immediately from (1.8) and (1.10).

Some Notes

(1) Let R be any ring, and let e be any idempotent of R with complement f = 1 − e. Han andNicholson [18, Lemma] showed that if the corner rings eRe and f Rf are both clean, then R is alsoclean. The converse is not true in general according to Šter [40, Example 3.4]. In other words, the

corner rings of a clean ring need not be clean. However, the corner rings of any exchange ring areexchange by [34, Proposition 1.10], and the corner rings of any potent ring (resp. semipotent ring)are potent (resp. semipotent) by [33, Corollary 1.7].

(2) An idempotent e of a ring R is said to be a full idempotent if ReR = R. It is not knownif the corner rings of a clean ring that arise from a full idempotent of the ring are clean. It is alsonot known if the converse of (1.3) is true—in other words, it is not known if the underlying ring Rmust be clean in order for a full matrix ring M n(R) to be clean. Moreover, it is not known if beingclean is Morita invariant.


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(3) A ring R is called quasinormal if eRfRe = 0 for every noncentral idempotent e of R withcomplement f = 1 − e. It is easy to check that every abelian ring is quasinormal, while the uppertriangular matrix ring T 2(Z/2Z) is quasinormal but not abelian. Therefore this notion is a propergeneralization of that of an abelian ring. We know from (1.4) that an abelian ring is clean if andonly if it is exchange. More generally, Wei and Li [ 47, Proposition 4.1] showed that a quasinormal

ring is clean if and only if it is exchange.

(4) A ring is called quasi-duo if every one-sided maximal ideal of the ring is in fact two-sided.Yu [49, Theorem 4.2] showed that a quasi-duo ring is clean if and only if it is exchange. As it turnsout, this result generalizes the above result of Wei and Li since every quasinormal exchange ring isquasi-duo according to [47, Theorem 3.12].

(5) A ring R is said to have artinian primitive factors if the quotient ring R/U is artinianfor every primitive ideal U of R—the interested reader should see Lam [25, Definition 11.3] for thedefinition of a primitive ideal. Chen [8, Theorem 1] showed that a ring with artinian primitivefactors is clean if and only if it is exchange.

(6) A ring is said to be root clean if each element in the ring can be written as the sum of aunit and a square root of 1. Hiremath and Hegde [20, Proposition 2.12] showed that a ring is rootclean if and only if it is a clean ring in which 2 is invertible. The sufficiency was proved earlier byCamillo and Yu [7]. More specifically, they showed that a ring in which 2 is invertible is clean if and only if it is root clean—see [7, Proposition 10].

(7) As we mentioned above, Camillo and Yu [7, p. 4746] proved that the ring in a well-knownexample due to Bergman [19, Example 1] is exchange but not clean. More specifically, the ring inBergman’s example is an exchange ring in which 2 is invertible that is not root clean. Since a ringin which 2 is invertible is clean if and only if it is root clean according to [ 7, Proposition 10], thering in Bergman’s example is indeed exchange but not clean. The interested reader is encouragedto see [7], [19], [27], and [40] for more on this example. No other examples of exchange rings that

are not clean are known to the author.

§2 Some Examples of Clean Rings

In this section we look at several important classes of rings whose rings are known to be clean.These include semiperfect rings, unit regular rings, and strongly π-regular rings. We review someuseful properties of these rings and of some closely related rings. In particular, we are interestedin the relationships between these rings and clean rings. All rings of interest are defined below.

Division, Boolean & Local Rings

The following lemma due to Anderson and Camillo [1] shows that the units, idempotents, andquasiregular elements of any ring are clean. The immediate result of this lemma is that every ringthat consists entirely of these types of elements is a clean ring. We study this interesting subclassof clean rings in §3.


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(2.1) Lemma. The units, idempotents, and quasiregular elements of any ring are clean.

Proof . Since 0 is idempotent, any unit u can be written as the sum of a unit and an idempotentby writing u = u + 0. Meanwhile, notice that 2e − 1 is a square root of 1 (hence a unit) when e isidempotent. Since 1 − e is idempotent when e is an idempotent, any idempotent e can be writtenas the sum of a unit and an idempotent by writing e = (2e − 1) + (1 − e). Finally, since 1 − x is a

unit by definition when x is quasiregular, any quasiregular element x can be written as the sum of a unit and an idempotent by writing x = −(1 − x) + 1. This completes the proof.

(2.2) Corollary. Every division ring, boolean ring, and local ring is clean.

Proof . This follows immediately from (2.1) since every division ring, boolean ring, and local ringconsists entirely of units, idempotents, and quasiregular elements.

The following theorem shows the precise relationship between local rings and clean rings (resp.exchange rings, potent rings, semipotent rings). In particular, it tells us that a ring is local if andonly if it is clean (resp. exchange, potent, semipotent) and has no nontrivial idempotents. The firstdirect proof that (1) ⇔ (2) below was given by Anderson and Camillo [1]. The fact that (1) ⇔ (3)was proved earlier by Warfield [45], and the equivalences (1) ⇔ (4) and (1) ⇔ (5) were essentiallyproved by Nicholson [33].

(2.3) Theorem. The following are equivalent for any ring R.

(1) R is local;

(2) R is clean and has no nontrivial idempotents;

(3) R is exchange and has no nontrivial idempotents;

(4) R is potent and has no nontrivial idempotents;

(5) R is semipotent and has no nontrivial idempotents.

Proof . It is not difficult to see that a local ring has no nontrivial idempotents. Since every local

ring is clean by (2.2), it follows that (1) ⇒ (2). Meanwhile, recall that every clean ring is exchangeby (1.4), that every exchange ring is potent according to Nicholson [34, Proposition 1.9], and thatevery potent ring is semipotent by definition. This tells us that (2) ⇒ (3) ⇒ (4) ⇒ (5). It remainsonly for us to show that every semipotent ring with no nontrivial idempotents is local. Recall thatevery primitive idempotent in a semipotent ring is local according to (1.10). Since the identity 1 isa primitive idempotent in any ring with no nontrivial idempotents, it follows that (5) ⇒ (1). Thiscompletes the proof.

(2.4) Corollary. An integral domain is clean if and only if it is local.

Proof . This result follows immediately from (2.3) since an integral domain contains no nontrivialidempotents.

Perfect & Semiperfect Rings

A ring R is said to be left artinian if every descending chain of left ideals in R has a minimalelement—in other words, if for every descending chain I 1 ⊇ I 2 ⊇ I 3 ⊇ · · · of left ideals in the ringthere is some positive integer n such that I n = I n+1 = I n+2 = · · · . To be right artinian requiresinstead that this property holds for every descending chain of right ideals in the ring. We will say


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that a ring is artinian if it is both left and right artinian—this notion is not left-right symmetric.It is well known that the Jacobson radical of every left (resp. right) artinian ring is nilpotent, andhence every nil one-sided ideal in a left (resp. right) artinian ring is nilpotent.

A left artinian ring is said to be semisimple if its Jacobson radical is equal to 0. As it turnsout, this notion is left-right symmetric. In particular, it follows that any left or right artinian ring

whose Jacobson radical is equal to 0 is (left and right) artinian. It is not difficult to see that everyhomomorphic image of a left (resp. right) artinian ring is left (resp. right) artinian, and hence thequotient ring R/J (R) is semisimple for any left (resp. right) artinian ring R. It is well known thata commutative ring is semisimple if and only if it is isomorphic to a finite direct product of fields.

(2.5) Proposition. Every semisimple ring is clean.

Proof . It is well known that every semisimple ring is isomorphic to a finite direct product of fullmatrix rings over division rings—this is the Wedderburn-Artin Theorem. Since every division ringis clean by (2.2), every full matrix ring with entries from a division ring is clean as a result of (1.3).Since every direct product of clean rings is clean by (1.2), the result now follows. Therefore everysemisimple ring is clean.

(2.6) Proposition. Every left (right) artinian ring is clean.

Proof . Let R be any left (right) artinian ring. Then its Jacobson radical is nilpotent (hence nil),and it follows from (1.7) that the ring R is clean if and only if R/J (R) is clean. Since the quotientring R/J (R) is semisimple, as mentioned above, for any left (right) artinian ring R, the result nowfollows from (2.5). Therefore every left (right) artinian ring is clean.

(2.7) Corollary. Every finite ring is clean.

Proof . This result follows immediately from (2.6) since every finite ring is artinian.

A ring R is said to be semiprimary if R/J (R) is semisimple and J (R) is nilpotent. Since theJacobson radical of every left (resp. right) artinian ring is nilpotent, it follows that every left (resp.right) artinian ring is semiprimary.

(2.8) Proposition. Every semiprimary ring is clean.

Proof . Let R be any semiprimary ring. Then its Jacobson radical is nilpotent (hence nil), and itfollows from (1.7) that the ring R is clean if and only if R/J (R) is clean. Since the quotient ringR/J (R) is semisimple by definition for any semiprimary ring R, the result now follows from (2.5).Therefore every semiprimary ring is clean.

A subset of a ring R is called left T -nilpotent if for any sequence {a1, a2, a3, . . . } of elements

in the subset there is some positive integer n such that a1a2 · · · an

= 0. To be right T

-nilpotentrequires instead that an · · · a2a1 = 0. We will call a subset T -nilpotent if it is both left and rightT -nilpotent—this notion is not left-right symmetric. It is not difficult to see that the implicationsbelow hold for any one-sided ideal of a ring.

nilpotent =⇒ T -nilpotent =⇒ left T -nilpotent =⇒ nil

A ring R is said to be left perfect if the quotient ring R/J (R) is semisimple and J (R) is leftT -nilpotent. To be right perfect requires instead that J (R) is right T -nilpotent. We will say that


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a ring is perfect if it is both left and right perfect—this notion is not left-right symmetric. Sinceevery nilpotent ideal of a ring is both left and right T -nilpotent, it follows that every semiprimaryring is perfect. It is well known that a commutative ring is perfect if and only if it is isomorphic toa finite direct product of (commutative) local rings each of which has a T -nilpotent maximal ideal.In general, Bass [4, Theorem P] showed that a ring is left (resp. right) perfect if and only if every

descending chain of principal right (resp. left) ideals in the ring has a minimal element—notice theunusual switch from left to right.

(2.9) Proposition. Every left (right) perfect ring is clean.

Proof . Let R be any left (resp. right) perfect ring. Then its Jacobson radical is left (resp. right)T -nilpotent (hence nil), and it follows from (1.7) that the ring R is clean if and only if R/J (R) isclean. Since R/J (R) is semisimple by definition for any left (right) perfect ring R, the result nowfollows from (2.5). Therefore every left (right) perfect ring is clean.

A ring R is said to be semiperfect if the quotient ring R/J (R) is semisimple and idempotentslift modulo J (R). Since every semisimple ring is clean according to (2.5), it follows from (1.6) that

every semiperfect ring is clean. It is well known that a commutative ring is semiperfect if and onlyif it is isomorphic to a finite direct product of (commutative) local rings. In general, it was shownby Mueller [31, Theorem 1] that a ring is semiperfect if and only if its identity 1 can be written asa finite sum of orthogonal local idempotents. Mueller also showed that a ring is semiperfect if andonly if every primitive idempotent is local and there is no infinite set of orthogonal idempotents inthe ring. A ring is sometimes said to be orthogonally finite if it has no infinite set of orthogonalidempotents. The following characterization of semiperfect rings is due to Camillo and Yu [7]. Ittells us that a ring is semiperfect if and only if it is clean (resp. exchange, potent, semipotent) andorthogonally finite.

(2.10) Theorem. The following are equivalent for any ring R.

(1) R is semiperfect;

(2) R is clean and orthogonally finite;

(3) R is exchange and orthogonally finite;

(4) R is potent and orthogonally finite;

(5) R is semipotent and orthogonally finite.

Proof . It is well known that every semiperfect ring is orthogonally finite. Since every semiperfectring is clean, as noted above, this tells us that (1) ⇒ (2). Meanwhile, recall that every clean ring isexchange by (1.4), that every exchange ring is potent according to Nicholson [34, Proposition 1.9],and that every potent ring is semipotent by definition. It follows that (2) ⇒ (3) ⇒ (4) ⇒ (5). Weonly need to show that every orthogonally finite semipotent ring is semiperfect. Recall that everyprimitive idempotent in a semipotent ring is local according to (1.10). Since a ring is semiperfect

if and only if it is orthogonally finite and every primitive idempotent in the ring is local accordingto Mueller [31, Theorem 1], it follows that (5) ⇒ (1). This completes the proof.

A ring R is called semilocal if the quotient ring R/J (R) is semisimple. It is known that everysemilocal ring is orthogonally finite, and it follows immediately from (2.10) that a semilocal ring isclean if and only if it is semiperfect. In particular, this means that a semilocal ring is clean if andonly if idempotents lift modulo its Jacobson radical. It is well known that a commutative ring issemilocal if and only if it has only a finite number of maximal ideals. More generally, it is knownthat any ring R that has only finitely many maximal ideals is semilocal, and that the converse is


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true when the quotient ring R/J (R) is commutative. It is not difficult to see that a semilocal ringneed not be clean. For example, the ring Z(6) of all rational numbers with denominators relativelyprime to 6 (when written in lowest terms) is semilocal since it is a commutative ring with only twomaximal ideals, but it is not clean according to (2.3) since it is a non-local ring with no nontrivialidempotents.

(2.11) Proposition. A semilocal ring is clean if and only if it is semiperfect.

Proof . As we mentioned above, it is known that every semilocal ring is orthogonally finite. Sinceevery semiperfect ring is semilocal by definition, the result follows from (2.10).

Bass [4] introduced the notions of perfect and semiperfect rings as homological generalizationsof semiprimary rings. Semiperfect rings in particular will play a central role in our study of cleangroup rings in the next chapter. The implications below should now be clear.

left/right artinian =⇒ semiprimary =⇒ left perfect =⇒ semiperfect =⇒ clean

These implications are known to be irreversible. For an example of a noncommutative ring that is

semiprimary but neither left nor right artinian, the interested reader should see Lam [26, Ex. 20.5].Meanwhile, since every semiprimary ring is (left and right) perfect, any left perfect ring that is notright perfect is not semiprimary—the interested reader should see Bass [4, p. 476] for an exampleof such a ring. Finally, it is not difficult to see that the ring Z(2) of all rational numbers with odddenominators (when written in lowest terms) is semiperfect but not perfect, and that the infinitedirect product Z(2) × Z(2) × · · · is clean but not semiperfect.

In addition to the relationships above, we should note that every division ring is artinian, andthat every local ring is semiperfect. Of course these implications are also known to be irreversible.For example, the ring Z/4Z of integers modulo 4 is artinian but not a division ring, and the directproduct Z/4Z × Z/4Z is semiperfect but not local.

Regular & π-Regular Rings

A ring R is said to be regular—in the sense of von Neumann [42]—if for every element r ∈ Rthere exists an element x ∈ R such that rxr = r. More generally, an element r in a ring R is calledregular if there exists an element x ∈ R such that rxr = r. It is not difficult to see that a ring Ris regular if and only if every principal left (equivalently, right) ideal of the ring is generated by anidempotent—in other words, if for every element a ∈ R there exists some idempotent e of R suchthat Re = Ra. Moreover, since the Jacobson radical of any ring contains no nonzero idempotents,it follows that the Jacobson radical of a regular ring must be equal to 0. Warfield [46, p. 34] provedthat every regular ring is exchange, but a regular ring need not be clean according to Camillo andYu [7, p. 4746]. The following result due to Lee, Yi, and Zhou [27] gives a necessary and sufficientcondition for a regular ring to be clean.

(2.12) Theorem. A regular ring is clean if and only if its indecomposable images are clean.

Proof . This proof is omitted—see Lee, Yi, and Zhou [27, Corollary 7].

A ring R is called unit regular if for every element r ∈ R there exists a unit u of R such thatrur = r. More generally, an element r in a ring R is called unit regular if there exists some unitu ∈ R such that rur = r. It is not difficult to see that a ring R is unit regular if and only if every


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element r ∈ R can be written as the product r = ue of a unit u and an idempotent e of R. Camilloand Khurana [6, Theorem 1] proved that every unit regular ring is clean, but a single element in aring can be unit regular without being clean according to Khurana and Lam [ 23, Example 4.5].

(2.13) Theorem. Every unit regular ring is clean.

Proof .

This proof is omitted—see Camillo and Khurana [6, Theorem 1].

A ring R is called strongly regular if for every element r ∈ R there exists an element x ∈ Rsuch that r2x = r. This definition is left-right symmetric. In fact, Azumaya [3, Lemma 1] showedthat a ring R is strongly regular if and only if for each element r ∈ R there exists a unique elementz ∈ R such that rzr = r and z rz = z with rz = zr. It can be shown that a ring is strongly regularif and only if it is both regular and reduced, and that a ring is strongly regular if and only if it isan abelian regular ring. It is also well known that every strongly regular ring is unit regular, andtherefore every strongly regular ring is clean by (2.13). We give a direct proof of this result.

(2.14) Proposition. Every strongly regular ring is unit regular and clean.

Proof .

Let r be any element in a strongly regular ring R. Then there is some element z ∈ R suchthat rzr = r with rz = zr. It is not difficult to check that e = rz is idempotent, while the elementu = r − (1 − e) is a unit of R with the inverse v = ze − (1 − e). Thus we can write r as the productof a unit and an idempotent of R by writing r = ue, and as the sum of a unit and an idempotentof R by writing r = u + (1 − e). Therefore every strongly regular ring is unit regular and clean.

It is well known that the notions of regular, unit regular, and strongly regular are equivalent inthe commutative case. Examples of commutative regular rings include all boolean rings and fields.Meanwhile, Jacobson [21, Theorem 11] showed that if for every element r in a ring R there exists apositive integer n such that rn = r, then every element in R has finite additive order and the ringR is commutative. This notion is a proper generalization of that of a boolean ring. Moreover, it is

easy to see that every such ring is strongly regular, and therefore is a commutative regular ring.A ring R is called π-regular if for every element r ∈ R there exists an element x ∈ R such that

rnxrn = rn for some positive integer n. Notice that the equation rnxrn = rn implies that xrn is anidempotent, and that xrn is equal to 0 if and only if rn = 0. In particular, this tells us that everynon-nil left ideal of a π-regular ring contains a nonzero idempotent—this result was pointed out byKaplansky [22, p. 63]. It follows that every π-regular ring is semipotent and its Jacobson radical isa nil ideal. A semipotent ring with a nil Jacobson radical is sometimes called a Zorn ring. Sincethe Jacobson radical of every π-regular ring is a nil ideal, it now follows from (1.7) that a π-regularring R is clean if and only if the quotient ring R/J (R) is clean.

A ring R is called strongly π-regular if for every element r ∈ R there exists an element x ∈ Rsuch that rn+1x = rn for some positive integer n. These rings were studied for nearly thirty years

before Dischinger [14] proved that their definition is left-right symmetric. Azumaya [3, Theorem 3]showed that a ring R is strongly π-regular if and only if for every element r ∈ R there exists someelement z ∈ R such that rnzrn = rn with rz = zr for some positive integer n. The following resultwas first proved by Burgess and Menal [5, Proposition 2.6].

(2.15) Theorem. Every strongly π-regular ring is clean.

Proof . This proof is omitted—see Nicholson [35, Theorem 1] for an elementary proof of this resultwhose original proof involves sheaf-theoretic representations.


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A commutative ring R is called zero-dimensional—in regard to Krull dimension—if its primeideals are all maximal. It is well known that a commutative ring is zero-dimensional if and only if it is strongly π-regular, and therefore every zero-dimensional commutative ring is clean as a resultof (2.15). Meanwhile, it is also known that a commutative ring R is zero-dimensional if and only if the quotient ring R/J (R) is regular and J (R) is a nil ideal. This tells us that a commutative ring

is regular if and only if it is zero-dimensional and reduced. More generally, it can be shown that aring is strongly regular if and only if it is strongly π-regular and reduced.

(2.16) Corollary. Every zero-dimensional commutative ring is clean.

Proof . As mentioned above, every zero-dimensional commutative ring is strongly π-regular, andtherefore is clean by (2.15).

Von Neumann [42] introduced regular rings as a proper generalization of semisimple rings. Theinterested reader is encouraged to see Goodearl [17] and Tuganbaev [41] for more on regular ringsand rings close to regular. The implications below are well known.

commutative regular =⇒ strongly regular =⇒ unit regular =⇒ regular= ⇒

= ⇒

= ⇒

= ⇒

zero-dimensional comm. =⇒ strongly π-regular =⇒ clean =⇒ exchange

These implications are known to be irreversible. In particular, since any noncommutative divisionring (e.g. Hamilton’s quaternions) is strongly regular but not commutative, the pair of horizontalimplications on the left is irreversible. Meanwhile, Chen [11, Remark 4.3] constructed an exampleof a ring that is unit regular but not strongly π-regular, which tells us that the pair of horizontalimplications in the middle is irreversible, and Camillo and Yu [7, p. 4746] proved that the ring ina well-known example due to Bergman [19, Example 1] is regular but not clean, which shows thatthe pair of horizontal implications on the right is irreversible. Finally, notice that the ring Z/4Z of

integers modulo 4 is a zero-dimensional commutative ring but is not regular, and therefore all fourvertical implications above are irreversible—more generally, the ring Z/nZ of integers modulo n iszero-dimensional for any positive integer n, but is regular if and only if n is square-free accordingto Ehrlich [16, Theorem 5].

In addition to the relationships above, we should mention that every left (right) perfect ring isstrongly π-regular. It is an easy exercise to check that a ring R is strongly π-regular if and only if for any element a ∈ R the descending chain Ra ⊇ Ra2 ⊇ Ra3 ⊇ · · · of principal left ideals in R hasa minimal element. Since a ring R is right perfect if and only if every descending chain of principalleft ideals in R has a minimal element according to Bass [4, Theorem P], it follows that every rightperfect ring is strongly π-regular. In fact, since the notion of a strongly π-regular ring is left-rightsymmetric, this tells us that every left or right perfect ring is strongly π-regular. This implication

is known to be irreversible. For example, any infinite direct product of fields is strongly π-regular,but not perfect.

Some Notes

(1) A ring is said to be left noetherian if every ascending chain of left ideals in the ring hasa maximal element—in other words, if for every ascending chain I 1 ⊆ I 2 ⊆ I 3 ⊆ · · · of left ideals inthe ring there is a positive integer n such that I n = I n+1 = I n+2 = · · · . To be right noetherian


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requires instead that this property holds for every ascending chain of right ideals. We will say thata ring is noetherian if it is both left and right noetherian—this notion is not left-right symmetric.It is well known that every left (resp. right) artinian ring is left (resp. right) noetherian, and thatthe converse is true for all semiprimary rings—this is the Hopkins-Levitzki Theorem.

(2) It is not difficult to see that a noetherian ring need not be clean. For example, the ring of integers Z is noetherian but not clean, and the ring Z(6) of all rational numbers with denominatorsrelatively prime to 6 (when written in lowest terms) is both semilocal and noetherian but not clean.It is well known that every left (right) noetherian ring is orthogonally finite. Since an orthogonallyfinite ring is clean (resp. exchange, potent, semipotent) if and only if it is semiperfect by (2.10), itfollows that a left (right) noetherian ring is clean (resp. exchange, potent, semipotent) if and onlyif it is semiperfect. Moreover, since a ring is semiperfect if and only if it is orthogonally finite andevery primitive idempotent in the ring is local according to Mueller [31, Theorem 1], it follows thata left (right) noetherian ring is semiperfect (resp. clean, exchange, potent, semipotent) if and onlyif every primitive idempotent in the ring is local.

(3) A ring R is called semiregular if the quotient ring R/J (R) is regular and idempotents lift

modulo J (R). This class of rings includes all regular rings, semiperfect rings, and zero-dimensionalcommutative rings, and it is well known that every semiregular ring is exchange. The relationshipbetween semiperfect rings, semiregular rings, and exchange rings is shown below.

semiperfect =⇒ clean= ⇒

= ⇒

semiregular =⇒ exchangeThese implications are known to be irreversible. In particular, we gave an example above of a ringthat is clean but not semiperfect, while Monk [30, Theorem 2] showed that an exchange ring neednot be semiregular—more specifically, he showed that a commutative exchange (hence clean) ringwhose Jacobson radical is equal to 0 need not be regular. Meanwhile, we also mentioned above thatthe ring in a well-known example due to Bergman [19, Example 1] is regular but not clean. Sinceevery regular ring is semiregular, it follows that the ring in Bergman’s example is semiregular butnot clean, and therefore both of the vertical implications above are irreversible. No example of anexchange ring that is neither regular nor clean is known to the author.

(4) A ring R is said to have stable range one if for any elements a, b ∈ R with Ra + Rb = R,there is some element y ∈ R such that a + yb is a unit of R—this notion is left-right symmetric. Itis well known that every semiperfect ring, unit regular ring, and strongly π-regular ring has stablerange one. Moreover, Chen and Li [10, Corollary 3] proved that every quasi-duo exchange ring hasstable range one, and Yu [48, Theorem 1] proved that every exchange ring with artinian primitivefactors has stable range one. However, it follows immediately from [36, Corollary] that a clean ring

need not have stable range one.

(5) A semiregular ring is called unit semiregular if it has stable range one. It is well knownthat a regular ring has stable range one if and only if it is unit regular. More generally, Aydo ğdu,Lee, and Özcan [2, Theorem 2.1] showed that a semiregular ring R has stable range one if and onlyif R/J (R) is unit regular and idempotents lift modulo J (R). Since every unit regular ring is cleanby (2.13), it follows from (1.6) that every semiregular ring with stable range one is clean—in otherwords, every unit semiregular ring is clean. It is not known to the author whether every exchangering with stable range one is clean.


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(6) A ring R is said to have idempotent stable range one if for any elements a, b ∈ R withRa + Rb = R, there exists some idempotent e of R such that a + eb is a unit of R—this notion isleft-right symmetric. Chen [9, Theorem 12] proved that every ring with idempotent stable rangeone is clean—this follows easily from the fact that Ra + R(−1) = R for any element a in a ring R.He also proved that the converse is true for abelian rings, and hence an abelian ring is clean if and

only if it has idempotent stable range one. Meanwhile, Wang et al. [ 44] proved that all semiperfectrings and unit regular rings have idempotent stable range one.

§3 An Interesting Subclass of Clean Rings

In this section, we study the rings that consist entirely of units, idempotents, and quasiregularelements. We know from (2.1) that the units, idempotents, and quasiregular elements of any ringare clean, and thus every ring that consists entirely of these types of elements is a clean ring. Butexactly which rings belong to this interesting subclass of clean rings? We give a complete solutionto this problem below. In particular, we prove that any abelian ring that consists entirely of units,idempotents, and quasiregular elements is either a boolean ring, a local ring, or isomorphic to thedirect product of two division rings, while any nonabelian ring that consists entirely of these typesof elements is isomorphic to either the full matrix ring M 2(D) for some division ring D, or to thering of a Morita context with zero pairings where both of the underlying rings are division rings.

Abelian Rings

We begin our study of rings consisting entirely of units, idempotents, and quasiregular elementsby classifying those rings that consist entirely of any two of these types of elements. In particular,we prove that any such ring must be either a division ring, a boolean ring, or a local ring. Let us

define the following subsets of a ring R.

U (R) = {r ∈ R | r is a unit of R}

ID(R) = {r ∈ R | r is an idempotent of R}

QR(R) = {r ∈ R | r is a quasiregular element of R}

The first part of the following theorem was proved independently by Chen and Cui [12] and by theauthor (unpublished). This was proved earlier for commutative rings by Anderson and Camillo [1].We give a new proof of this result.

(3.1) Theorem. Let R be any ring.

(1) R = U (R) ∪ ID(R) if and only if R is a division ring or a boolean ring.

(2) R = U (R) ∪ QR(R) if and only if R is a local ring.

(3) R = ID(R) ∪ QR(R) if and only if R is a division ring or a boolean ring.

Proof . (1) Suppose that the ring R consists entirely of units and idempotents. It is not difficultto see that any such ring is reduced, and therefore is abelian. In particular, the ring R either hasno nontrivial idempotents, and hence is a division ring, or it has a nontrivial central idempotent,and hence is decomposable. Suppose that R is decomposable, and let A × B be any direct product


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decomposition of R. If either of the rings A or B contained a non-idempotent x, then there wouldbe an element (x, 0) or (0, x) in A × B that was neither a unit nor an idempotent of A × B. Sincethe ring R consists entirely of units and idempotents, this tells us that A and B are both booleanrings. Since the direct product of any two boolean rings is itself a boolean ring, it follows that thering R is boolean in this case. Therefore every ring that consists entirely of units and idempotents

is either a division ring or a boolean ring. The converse is easy to check.(2) This result is well known, and its proof is omitted.(3) Suppose that the ring R consists entirely of idempotents and quasiregular elements. Then

for any element r ∈ R, the element 1 − r is either idempotent or quasiregular. It is not difficult tocheck that r is idempotent if (and only if) 1 − r is idempotent, and that r is a unit if (and only if)1 − r is quasiregular. This tells us that the ring R consists entirely of units and idempotents, andhence it must be either a division ring or a boolean ring by (1). Therefore every ring that consistsentirely of idempotents and quasiregular elements is either a division ring or a boolean ring. Theconverse is easy to check.

It is not difficult to see that division rings, boolean rings, and local rings are not the only ringsthat consist entirely of units, idempotents, and quasiregular elements. For example, the interested

reader is encouraged to check that the direct product Z/2Z×Z/3Z consists entirely of these typesof elements even though it is not a division ring, a boolean ring, or a local ring. More generally, weshow that the direct product of any two division rings consists entirely of units, idempotents, andquasiregular elements.

(3.2) Proposition. Let R be the direct product of two division rings. Then R consists entirely of units, idempotents, and quasiregular elements.

Proof . Since multiplication in a direct product of rings is defined componentwise, an element inany such ring is a unit (resp. idempotent, quasiregular element) of that ring if and only if the entryin each of its components is a unit (resp. idempotent, quasiregular element) of its ring. Since thering R is a direct product of two division rings, this tells us that an element in R is a unit (resp.quasiregular element) of the ring R if and only if the entry in each of its components is a nonzero(resp. nonidentity) element of its ring. It is not difficult now to see that each element r ∈ R mustsatisfy at least one of the following conditions, where each of the entries denoted by ∗ can be anynonzero, nonidentity element of its ring.

r = (1, 1), (1, ∗), (∗, 1), or (∗, ∗) ∈ U (R)

r = (0, 0), (0, 1), (1, 0), or (1, 1) ∈ ID(R)

r = (0, 0), (0, ∗), (∗, 0), or (∗, ∗) ∈ QR(R)

Therefore the direct product of any two division rings consists entirely of units, idempotents, andquasiregular elements.

The next result shows that any decomposable ring that consists entirely of units, idempotents,and quasiregular elements must be either a boolean ring or isomorphic to a direct product of twodivision rings. Incidentally, this result tells us that every decomposable ring that consists entirelyof these types of elements is necessarily abelian.

(3.3) Theorem. Let R be a decomposable ring that consists entirely of units, idempotents, and quasiregular elements. Then R is a boolean ring or a direct product of two division rings.


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Proof . Let A × B be any direct product decomposition of the ring R. Since multiplication in adirect product of rings is defined componentwise, an element in A × B is a unit (resp. idempotent,quasiregular element) of the ring A × B if and only if the entry in each of its components is a unit(resp. idempotent, quasiregular element) of its ring. If either of the rings A or B contained someelement x that was neither a unit nor an idempotent of its ring, then there would be an element

(x, 1) or (1, x) in A × B that was not a unit, an idempotent, or a quasiregular element of A × B.Since the ring R consists entirely of units, idempotents, and quasiregular elements, it follows thatthe rings A and B must both consist entirely of units and idempotents. We know from (3.1) thatany such ring is either a division ring or boolean ring. If either of the rings A or B contained someunit u other than 1, while the other ring contained some nontrivial idempotent e, then there wouldbe an element (u, e) or (e, u) in A × B that is not a unit, an idempotent, or a quasiregular elementof A × B. Since the ring R consists entirely of units, idempotents, and quasiregular elements, thistells us that A and B are either both boolean rings or both division rings. Since the direct productof any two boolean rings is itself a boolean ring, it follows that the ring R is either a boolean ringor a direct product of two division rings. Therefore every decomposable ring that consists entirelyof units, idempotents, and quasiregular elements is either a boolean ring or a direct product of twodivision rings.

We are now ready to classify the abelian rings that consist entirely of units, idempotents, andquasiregular elements. In particular, we prove that any abelian ring that consists entirely of thesetypes of elements must be either a boolean ring, a local ring, or isomorphic to a direct product of two division rings.

(3.4) Theorem. Let R be an abelian ring that consists entirely of units, idempotents, and quasi-regular elements. Then R is a boolean ring, a local ring, or a direct product of two division rings.

Proof . Since the ring R is abelian, it either has no nontrivial idempotents or a nontrivial centralidempotent. If R has no nontrivial idempotents, then it consists entirely of units and quasiregularelements, and hence is a local ring by (3.1). Meanwhile, if R has a nontrivial central idempotent,then it is decomposable, and hence is either a boolean ring or a direct product of two division ringsby (3.3). Therefore every abelian ring that consists entirely of units, idempotents, and quasiregularelements is either a boolean ring, a local ring, or a direct product of two division rings.

(3.5) Corollary. Let R be a commutative ring that consists entirely of units, idempotents, and quasiregular elements. Then R is a boolean ring, a commutative local ring, or a direct product of two fields.

Proof . This follows immediately from (3.4).

Nonabelian Rings

It still remains to classify the nonabelian rings that consist entirely of units, idempotents, andquasiregular elements. We give a complete solution to this problem below. In particular, we provethat any nonabelian ring that consists entirely of units, idempotents, and quasiregular elements isisomorphic to either the full matrix ring M 2(D) for some division ring D, or to the ring of a Moritacontext with zero pairings where both of the underlying rings are division rings. We begin with thefollowing lemma.


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(3.6) Lemma. Let R be any ring that consists entirely of units, idempotents, and quasiregular elements. Then eRe is a division ring for every noncentral idempotent e of R.

Proof . Let e be a nontrivial idempotent of R with complement f = 1 − e, and let us identify thering R with its Peirce decomposition with respect to e.

R ∼=eRe eRf

f Re f Rf

It is not difficult to check that the inverse of a diagonal matrix (if one exists) is a diagonal matrix.Since the ring R consists entirely of units, idempotents, and quasiregular elements, it follows thatthe subring of all diagonal matrices in this decomposition of R also consists entirely of these typesof elements. Moreover, this subring is isomorphic to the direct product eRe × f Rf , and since e isa nontrivial idempotent of R, we know that eRe and f Rf are both (nonzero) rings. It now followsfrom the proof of (3.3) that the corner ring eRe is either a division ring or a boolean ring. We needto show that eRe must be a division ring when e is noncentral. We prove the contrapositive.

Suppose that the corner ring eRe is not a division ring. Since it must be either a division ringor a boolean ring, this tells us that eRe must contain a nontrivial idempotent—in other words, it

must contain some idempotent other than 0 or e. Let a be any nontrivial idempotent of eRe withcomplement b = e − a, and consider the following matrices, where each entry denoted by x can beany element in eRf , while each entry denoted by y can be any element in f Re.

X 1 =

a x0 0

, X 2 =

b x0 0

, X 3 =

a 0y 0

, X 4 =

b 0y 0

Since X 1 and X 2 both have a row in which each entry is 0, while X 3 and X 4 both have a column inwhich each entry is 0, none of these matrices is a unit of R. Now consider the following matrix.

I − X 1 =

e 00 f

−

a x0 0

=

b −x0 f

In particular, notice that b would need to have a left inverse in eRe in order for this matrix I − X 1to have a left inverse in R. However, this element b cannot have a left inverse in eRe since it is anontrivial idempotent of eRe. This tells us that I − X 1 is not invertible in R, and hence X 1 is notquasiregular in R. Similar arguments show that the matrices X 2, X 3, and X 4 are not quasiregularin R. Since the ring R consists entirely of units, idempotents, and quasiregular elements, it followsthat each of these matrices must be idempotent. A simple computation shows that the matrix X 1is idempotent if and only if ax = x, while X 2 is idempotent if and only if bx = x. Moreover, sincex is an element of eRf , we also know that ex = x. These last three equalities tell us that x = 0, asfollows, while a similar argument shows that y = 0.

x = bx = (e − a)x = ex − ax = x − x = 0

Since this is the case for any elements x ∈ eRf and y ∈ f Re, it follows that eRf = 0 = f Re, andhence e is a central idempotent of R. More specifically, this tells us that e is a central idempotentof R when the ring eRe is a not division ring. Therefore eRe is a division ring for every noncentralidempotent e of R by contrapositive.

The next result classifies the full matrix rings M n(R) that consist entirely of units, idempotents,and quasiregular elements for some integer n > 1. In particular, it shows that the ring of all n × nmatrices with entries from some arbitrary ring consists entirely of these types of elements for some


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integer n > 1 if and only if it is equal to the full matrix ring M 2(D) for some division ring D . Thesufficiency is essentially due to Li [28]. Recall that two matrices X and Y are said to be similar if there is an invertible matrix P such that P −1XP = Y .

(3.7) Proposition. Let R be the ring of all n × n matrices with entries from some arbitrary ring.If n > 1, then R consists entirely of units, idempotents, and quasiregular elements if and only if itis equal to M 2(D) for some division ring D.

Proof . Suppose that the ring R consists entirely of units, idempotents, and quasiregular elementsfor some integer n > 1, and let us write R = M n(S ). Since n > 1, the matrix unit E = E 11 and itscomplement F = I − E 11 are both noncentral idempotents of R. Since the ring R consists entirelyof units, idempotents, and quasiregular elements, it follows from (3.6) that the ring F RF must bea division ring. Moreover, notice that we have the following.

F RF ∼= S × · · · × S n−1 times

Since the direct product on the right-hand side of the isomorphism above must be a division ring,it follows that n = 2 and that S must be a division ring. Therefore R is equal to M 2(D) for somedivision ring D . This proves the necessity.

Conversely, suppose that the ring R is equal to M 2(D) for some division ring D . Then for anymatrix X ∈ R, it follows immediately from a technical lemma due to Li [28, Lemma 2.4] that if X is neither a unit nor a quasiregular element of R, then it must be similar to the following matrix.

Y =

1 10 0

Notice that this matrix Y is idempotent. It is not difficult to check that any matrix similar to anidempotent is idempotent. In particular, this tells us that if X is neither a unit nor a quasiregular

element of the ring R, then it must be an idempotent of R. Therefore R consists entirely of units,idempotents, and quasiregular elements. This proves the sufficiency.

A ring is called semiprime if it has no nonzero nilpotent ideals. Du [15] proved that if eRe isa division ring for every noncentral idempotent e of a nonabelian semiprime ring R, then that ringR must be isomorphic to the full matrix ring M 2(D) for some division ring D. We prove that everynonabelian semiprime ring that consists entirely of units, idempotents, and quasiregular elementsis isomorphic to M 2(D) for some division ring D.

(3.8) Theorem. Let R be any nonabelian ring that consists entirely of units, idempotents, and quasiregular elements. If R is semiprime, then it is isomorphic to M 2(D) for some division ring D .

Proof .

Suppose that R is semiprime. Since the ring R consists entirely of units, idempotents, andquasiregular elements, it follows immediately from (3.6) that the ring eRe must be a division ringfor every noncentral idempotent e of R. As we noted above, Du [15, Lemma 21] proved that everynonabelian semiprime ring that has this property is isomorphic to the full matrix ring M 2(D) forsome division ring D . Therefore R is isomorphic to M 2(D) for some division ring D.

Let A, B be some pair of rings, let M be any (A, B)-bimodule, let N be any (B, A)-bimodule,and let ψ : M × N → A and ϕ : N × M → B be some binary functions, written multiplicatively asψ(x, y) = xy and ϕ(y, x) = yx. The 6-tuple (A ,B,M ,N,ψ ,ϕ) is called a Morita context if these


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binary functions ψ and ϕ preserve addition in each variable and satisfy the following associativityconditions for every element a ∈ A, b ∈ B, x ∈ M , and y ∈ N .

a(xy) = (ax)y, (xb)y = x(by), (xy)a = x(ya),

b(yx) = (by)x, (ya)x = y(ax), (yx)b = y(xb).

In this case, the binary functions ψ and ϕ are called pairings, and if ψ and ϕ are both equal to 0,then the Morita context (A ,B,M ,N,ψ ,ϕ) is said to be a Morita context with zero pairings.

Let (A ,B,M ,N,ψ ,ϕ) be any Morita context, and notice that the following set of matrices is aring with respect to the usual operations for addition and multiplication of matrices when ψ and ϕare used to define multiplication between elements of the bimodules M and N .

A M N B

=

a xy b

\\\\ a ∈ A, b ∈ B, x ∈ M, y ∈ N

Any such ring is known as the the ring of a Morita context. The next result shows that if eitherof the bimodules M or N is nonzero, then the ring of any Morita context ( A ,B,M ,N,ψ ,ϕ) with

zeropairings consists entirely of units, idempotents, and quasiregular elements if and only if the ringsA and B are both division rings.

(3.9) Proposition. Let R be the ring of a Morita context (A ,B,M ,N,ψ ,ϕ) with zero pairings.If either of the bimodules M or N is nonzero, then R consists entirely of units, idempotents, and quasiregular elements if and only if A and B are both division rings.

Proof . Suppose that the ring R consists entirely of units, idempotents, and quasiregular elementsfor some bimodules M and N , at least one of which is nonzero. Since at least one of the bimodulesM or N is nonzero, the matrix units E = E 11 and F = E 22 are both noncentral idempotents of R.Since the ring R consists entirely of units, idempotents, and quasiregular elements, it now followsfrom (3.6) that E RE and F RF must both be division rings. Moreover, notice that E RE ∼= A andthat F RF ∼= B. Therefore the rings A and B must both be division rings when either of M or N is nonzero. This proves the necessity.

Conversely, suppose that A and B are both division rings. Then any nonzero element in eitherof these rings is a unit of its ring. Let a be any nonzero element in A with inverse u, and let b beany nonzero element in B with inverse v . Then for any elements x ∈ M and y ∈ N , since R is thering of a Morita context with zero pairings, we have the following.

a xy b

−1

=

u −uxv

−vyu v

This shows that a matrix in R is a unit (resp. quasiregular element) of the ring R when each entryon its main diagonal is a nonzero (resp. nonidentity) element of its ring. It is not difficult now tosee that each matrix X ∈ R must satisfy at least one of the following conditions, where each of theentries denoted by ∗ can be any nonzero, nonidentity element of its ring, while each entry denoted


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by x can be any element in M , and each entry denoted by y can be any element in N .

X =

1 xy 1

,

1 xy ∗

,

∗ xy 1

, or

∗ xy ∗

∈ U (R)

X = 0 00 0, 0 x

y 1, 1 x

y 0, or 1 0

0 1 ∈ ID(R)X =

0 xy 0

,

0 xy ∗

,

∗ xy 0

, or

∗ xy ∗

∈ QR(R)

Therefore R consists entirely of units, idempotents, and quasiregular elements when A and B areboth division rings. This proves the sufficiency.

Nicholson [32] called a ring an NJ-ring if the ring consists entirely of regular and quasiregularelements. Examples of NJ-rings include all regular rings and local rings. Moreover, since the unitsand idempotents of any ring are regular, every ring that consists entirely of units, idempotents, andquasiregular elements is an NJ-ring. The converse is not true in general, since a regular ring need

not be clean. Nicholson [32] showed that an NJ-ring must be either a regular ring, a local ring, orisomorphic to the ring of a Morita context with zero pairings where the underlying rings are bothdivision rings. We prove that every nonabelian, nonsemiprime ring that consists entirely of units,idempotents, and quasiregular elements is isomorphic to the ring of some Morita context with zeropairings where the underlying rings are both division rings.

(3.10) Theorem. Let R be any nonabelian ring that consists entirely of units, idempotents, and quasiregular elements. If R is not semiprime, then it is isomorphic to the ring of a Morita contextwith zero pairings where the underlying rings are both division rings.

Proof . Suppose that R is not semiprime. Then R is a nonabelian, nonsemiprime ring that consistsentirely of units, idempotents, and quasiregular elements. Since every ring that consists entirely of

these types of elements is an NJ-ring, it follows that R is a nonabelian, nonsemiprime NJ-ring. Aswe noted above, Nicholson [32, Theorem 2] showed that an NJ-ring must be either a regular ring,a local ring, or isomorphic to the ring of a Morita context with zero pairings where the underlyingrings are both division rings. In particular, since every regular ring is semiprime, while every localring is abelian, it follows that every nonabelian, nonsemiprime NJ-ring must be isomorphic to thering of some Morita context with zero pairings where the underlying rings are both division rings.Therefore R is isomorphic to the ring of a Morita context with zero pairings where the underlyingrings are both division rings.

A Special Case

It is well known that every nilpotent element is quasiregular. In particular, if xn = 0 for someinteger n > 1, then an easy computation shows that 1 + x + x2 + · · · + xn−1 is an inverse of 1 − x.We classified the rings that consist entirely of units idempotents, and quasiregular elements above.But when do these rings consist entirely of units, idempotents, and nilpotent elements? We give acomplete solution to this problem below. Specifically, we prove that any abelian ring that consistsentirely of units, idempotents, and nilpotent elements is either a boolean ring or a local ring witha nil Jacobson radical, while any nonabelian ring that consists entirely of these types of elements


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is isomorphic to either the full matrix ring M 2(Z/2Z), or to the ring of a Morita context with zeropairings where the underlying rings are both Z/2Z.

We begin our study of rings consisting entirely of units, idempotents, and nilpotent elements byclassifying the abelian rings that consist entirely of these types of elements. It is easy to check thatall boolean rings and local rings with a nil Jacobson radical consist entirely of units, idempotents,

and nilpotent elements. The next result shows that any abelian ring that consists entirely of thesetypes of elements must be either a boolean ring or a local ring with a nil Jacobson radical.

(3.11) Proposition. Let R be any abelian ring that consists entirely of units, idempotents, and nilpotent elements. Then R is a boolean ring or a local ring with a nil Jacobson radical.

Proof . Since the ring R is abelian, it either has no nontrivial idempotents or a nontrivial centralidempotent. Suppose that R has no nontrivial idempotents. Then it consists entirely of units andnilpotent (hence quasiregular) elements, and thus it is a local ring by (3.1). Moreover, since everynonunit of R is nilpotent, it is a local ring with a nil Jacobson radical. Suppose instead that R hasa nontrivial central idempotent. Then R is a decomposable ring. Let A × B be any direct productdecomposition of R. Since the ring R consists entirely of units, idempotents, and nilpotent (hence

quasiregular) elements, it follows from (3.3) that the rings A and B are either both boolean ringsor both division rings. If either of A or B contained a unit u other than 1, then there would be anelement (u, 0) or (0, u) in A × B that is not a unit, an idempotent, or a nilpotent element of A × B.Since the ring R consists entirely of units, idempotents, and nilpotent elements, this tells us thatA and B are both boolean rings. Since the direct product of two boolean rings is itself a booleanring, it follows that the ring R is boolean in this case. Therefore every abelian ring that consistsentirely of units, idempotents, and nilpotent elements is either a boolean ring or a local ring witha nil Jacobson radical.

It still remains to classify the nonabelian rings that consist entirely of units, idempotents, andnilpotent elements. We begin with the following corollary to (3.6).

(3.12) Corollary. Let R be any ring that consists entirely of units, idempotents, and nilpotentelements. Then eRe is isomorphic to Z/2Z for every noncentral idempotent e of R.

Proof . Since the ring R consists entirely of units, idempotents, and nilpotent (hence quasiregular)elements, we know from (3.6) that eRe must be a division ring for every noncentral idempotent eof R. However, if the ring eRe contained some unit u \= e, then the following matrix would not bea unit, an idempotent, or a nilpotent element of the Peirce decomposition of R with respect to e.

X =

u 00 0

Therefore eRe must be isomorphic to Z/2Z for every noncentral idempotent e of R.

It is easy to see that the full matrix ring M 2(Z/2Z) consists entirely of units, idempotents, andnilpotent elements. The following result shows that every nonabelian semiprime ring that consistsentirely of these types of elements must be isomorphic to M 2(Z/2Z).

(3.13) Proposition. Let R be any nonabelian ring that consists entirely of units, idempotents,and nilpotent elements. If R is semiprime, then it is isomorphic to M 2(Z/2Z).


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Proof . Suppose that the ring R is semiprime. Since it consists entirely of units, idempotents, andnilpotent (hence quasiregular) elements, we know from (3.8) that R must be isomorphic to M 2(D)for some division ring D. Let us identify R with the full matrix ring M 2(D), and notice that thematrix unit E = E 11 is a noncentral idempotent of R with E RE ∼= D. Moreover, since the ring Rconsists entirely of units, idempotents, and nilpotent elements, it follows from (3.12) that the ring

ERE is isomorphic to Z/2Z. Therefore R is isomorphic to M 2(Z/2Z).It is easy to check that the ring of any Morita context with zero pairings where the underlying

rings are both isomorphic to Z/2Z consists entirely of units, idempotents, and nilpotent elements.The following result shows that every nonabelian, nonsemiprime ring that consists entirely of thesetypes of elements must be isomorphic to the ring of a Morita context with zero pairings where theunderlying rings are both Z/2Z. In this case, it can be shown that the underlying bimodules mustboth be elementary (abelian) 2-groups.

(3.14) Proposition. Let R be any nonabelian ring that consists entirely of units, idempotents,and nilpotent elements. If R is not semiprime, then it is isomorphic to the ring of a Morita contextwith zero pairings where the underlying rings are both Z/2Z.

Proof . Suppose that the ring R is not semiprime. Since it consists entirely of units, idempotents,and nilpotent (hence quasiregular) elements, we know from (3.10) that R must be isomorphic to thering of a Morita context with zero pairings where the underlying rings and are both division rings.Let us identify R with the ring of some Morita context (A ,B,M ,N,ψ ,ϕ) with zero pairings wherethe rings A and B are both division rings, and notice that the matrix units E = E 11 and F = E 22are noncentral idempotents of R with E RE ∼= A and F RF ∼= B . Since the ring R consists entirelyof units, idempotents, and nilpotent elements, it now follows from (3.12) that the rings ERE andF RF must both be isomorphic to Z/2Z. Therefore R is isomorphic to the ring of a Morita contextwith zero pairings where the underlying rings are both Z/2Z.

Some Notes

(1) A ring is said to be strongly clean if each element in the ring can be written as the sumof a unit and an idempotent that commute. It follows easily from the proof of (2.1) that every ringthat consists entirely of units, idempotents, and quasiregular elements is strongly clean. It can beshown that every ring that consists entirely of these types of elements is also optimally clean inthe sense of Shifflet [39]. We should note that optimally clean implies strongly clean by definition.

(2) As we mentioned in §1, it is known that the corner rings of a clean ring need not be clean.However, the corner rings of any strongly clean ring must be strongly clean according to S ánchez

Campos (unpublished), while the corner rings of any optimally clean ring must be optimally cleanaccording to Shifflet [39, Proposition 3.1.5]. Similarly, it can be shown that the corner rings of anyring that consists entirely of units, idempotents, and quasiregular elements also consists entirely of units, idempotents, and quasiregular elements.

(3) As we mentioned in §1, it is not known whether being clean is Morita invariant. However,we know from (3.7) that consisting entirely of units, idempotents, and quasiregular elements is notMorita invariant. Moreover, being strongly clean (or optimally clean) is also not Morita invariantaccording to Wang and Chen [43, Example 1].


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(4) A ring is called Dedekind finite (or sometimes, von Neumann finite) if ab = 1 impliesthat ba = 1 for any pair of elements a, b in the ring. It can be shown that every ring that consistsentirely of units, idempotents, and quasiregular elements is Dedekind finite. It not known whethera strongly clean ring or an optimally clean ring must be Dedekind finite, but a clean ring need notbe Dedekind finite according to Nicholson and Varadarajan [36, Corollary].

(5) Recall that a ring R is called quasinormal if eRfRe = 0 for every noncentral idempotente of R with complement f = 1 − e. It can be shown that every nonabelian, nonsemiprime ring thatconsists entirely of units, idempotents, and quasiregular elements is quasinormal. More generally,Nicholson [32] showed that every nonsemiprime NJ-ring is quasinormal. Unfortunately, this resultis hidden in the proof of [32, Theorem 2].

(6) We know from (3.6) that eRe must be a division ring for every noncentral idempotent e of any ring R that consists entirely of units, idempotents, and quasiregular elements. It can be shownthat the converse is true for any nonabelian quasinormal ring R, but it is not known whether theconverse is true for every nonabelian ring R.

(7) Let R be any ring, and let M be any (R, R)-bimodule. Then the following set of matricesis a ring with respect to the usual operations for addition and multiplication of matrices.

T (R, M ) =

r x0 r

\\\\ r ∈ R, x ∈ M

Any such ring is known as a trivial extension. It can be shown that a trivial extension T (R, M )consists entirely of units, idempotents, and quasiregular elements if and only if the ring R consistsentirely of these types of elements and the equation ex = xf holds for every nontrivial idempotente of R and every element x ∈ M , where f = 1 − e. It follows easily from this result that a trivialextension of the form T (R, R) consists entirely of units, idempotents, and quasiregular elements if and only if R is local. In this case, the trivial extension T (R, R) is also local.

(8) Let A, B be a pair of rings, let X be any (A, B)-bimodule, let Y be any (B, A)-bimodule,and consider the following sets of matrices.

R =

a 00 b

\\\\ a ∈ A, b ∈ B

, M =

0 xy 0

\\\\ x ∈ X, y ∈ Y

Notice that the set R is a ring with respect to the usual operations for addition and multiplicationof matrices, while the set M is an (R, R)-bimodule. Now consider the trivial extension T (R, M ).

T (R, M ) =

a 0 0 x0 b y 0

0 0 a 00 0 0 b

\\\\\\\\ a ∈ A, b ∈ B, x ∈ X, y ∈ Y

It is not difficult to see that the trivial extension T (R, M ) is isomorphic to the ring of the Moritacontext (A ,B,X,Y ,ψ ,ϕ) where ψ and ϕ are zero pairings. It follows immediately from (3.9) thatT (R, M ) consists entirely of units, idempotents, and quasiregular elements for some M \= 0 if andonly if A and B are both division rings. Meanwhile, it follows from (3.10) that every nonabelian,nonsemiprime ring that consists entirely of units, idempotents, and quasiregular elements must beisomorphic to T (R, M ) for some division rings A and B.


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