CLEAN RINGS \& CLEAN GROUP RINGS

Nicholas A. Immormino

A Dissertation

Submitted to the Graduate College of Bowling Green State University in partial fulfillment of the requirements for the degree of

DOCTOR OF PHILOSOPHY

August 2013

Committee:

Warren Wm. McGovern, Advisor

Rieuwert J. Blok, Advisor

Sheila J. Roberts, Graduate Faculty Representative

Mihai D. Staic ii ABSTRACT

Warren Wm. McGovern, Advisor

Rieuwert J. Blok, Advisor

A ring is said to be clean if each element in the ring can be written as the sum of a unit and an idempotent of the ring. More generally, an element in a ring is said to be clean if it can be written as the sum of a unit and an idempotent of the ring. The notion of a clean ring was introduced by

Nicholson in his 1977 study of lifting idempotents and exchange rings, and these rings have since been studied by many different authors.

In our study of clean rings, we classify the rings that consist entirely of units, idempotents, and quasiregular elements. It is well known that the units, idempotents, and quasiregular elements of any ring are clean. Therefore any ring that consists entirely of these types of elements is clean. We prove that a ring consists entirely of units, idempotents, and quasiregular elements if and only if it is a boolean ring, a local ring, isomorphic to the direct product of two division rings, isomorphic to the full matrix ring M2(D) for some division ring D, or isomorphic to the ring of a Morita context with zero pairings where both of the underlying rings are division rings. We also classify the rings that consist entirely of units, idempotents, and nilpotent elements.

In our study of clean group rings, we show exactly when the group ring \BbbZ( p)Cn is clean, where

\BbbZ( p) is the localization of the integers at p, and Cn is the cyclic group of order n. It is well known that the group ring \BbbZ(7) C3 is not clean even though the group ring \BbbZ( p)C3 is quasiclean, semiclean, and \Sigma -clean for any prime p, and 2-clean for any prime p \not = 2. We prove that \BbbZ( p)C3 is clean if and only if p \not \equiv 1 modulo 3. More generally, we prove that the group ring \BbbZ( p)Cn is clean if and only if p is a primitive root of m, where n = pkm and p does not divide m. We also consider the problems of classifying the groups G whose group rings RG are clean for any clean ring R, and of classifying the rings R such that the group ring RG of any locally finite group G over the ring R is clean. iii ACKNOWLEDGMENTS

I would like to thank all of the wonderful mathematics teachers that I had the opportunity to work with and learn from since I first became enamored with mathematics in the fourth grade. In particular, I would like to thank Thomas W. Hungerford, Curtis D. Bennett, Corneliu G. Hoffman,

Sergey Shpectorov, and Warren Wm. McGovern for teaching me about ring theory. I would like to thank Rieuwert J. Blok, Mihai D. Staic, and Sheila J. Roberts for their service on my dissertation committee, and I would like to thank Warren Wm. McGovern for being my advisor. I would also like to thank Tong Sun and Marcia Seubert for all their help, especially over the past year. iv

LIST OF SYMBOLS

\BbbZ the integers \BbbQ the rational numbers \BbbZ/n \BbbZ the integers modulo n

\BbbZ( n) the rational numbers with denominators relatively prime to n \BbbZp the p-adic integers \phi the Euler phi function ordn(a) the multiplicative order of a modulo n A \times B the direct product of A and B

Eij the ijth matrix unit

Mn(R) the set of n \times n matrices with entries from R

Tn(R) the set of n \times n upper triangular matrices with entries from R R[X] the set of polynomials with coefficients in R

Cn the cyclic group of order n

Sn the symmetric group on n elements

Cp\infty the Pr\"ufer p-group \langleg \rangle the cyclic group generated by g Gk the direct product of k copies of the group G

Gp the p-primary component of the group G \scrU (R) the set of all units of the ring R \scrI \scrD(R) the set of all idempotents of the ring R \scrQ \scrR(R) the set of all quasiregular elements of the ring R J(R) the Jacobson radical of the ring R

\BbbF q the finite field of order q char K the characteristic of the field K

\Phin the nth cyclotomic polynomial over a field T (R,M) the trivial extension of R by M (A, B, M, N, \psi ,\varphi) the Morita context of A, B, M, N, \psi , and \varphi RG the group ring of G over R \Delta the augmentation ideal of a groupring v

TABLE OF CONTENTS

CHAPTER ONE: Clean Rings 1

\S1 Some Properties of Clean Rings...... 1

Basic Properties...... 1

Lifting Idempotents...... 2

Two More Properties...... 4

Some Notes...... 6

\S2 Some Examples of Clean Rings...... 8

Division, Boolean \& Local Rings...... 8

Perfect \& Semiperfect Rings...... 10

Regular \& \pi -Regular Rings...... 14

Some Notes...... 17

\S3 An Interesting Subclass of Clean Rings...... 20

Abelian Rings...... 20

Nonabelian Rings...... 24

A Special Case...... 29

Conclusions...... 32

Some Notes...... 33

CHAPTER TWO: Clean Group Rings 36

\S4 Preliminaries...... 36

Some Results on Semiperfect Group Rings...... 36

Some Results on Clean Group Rings...... 38

Some Notes...... 41

\S5 Clean Group Rings, I...... 43

The Group Ring \BbbZ( p)Cn ...... 43 vi

The Group Ring \BbbZ( p)G ...... 46 Two More Results...... 47

Some Notes...... 48

\S6 Clean Group Rings, II...... 50

Abelian Groups...... 51

Nonabelian Groups...... 55

\S7 Clean Group Rings, III...... 56

Three Examples...... 56

Some Notes...... 58 REFERENCES 60 INDEX 67 1

CHAPTER ONE Clean Rings

A ring is said to be clean if every element in the ring can be written as the sum of a unit and an idempotent of the ring. More generally, an element in a ring is called clean if it can be written as the sum of a unit and an idempotent of the ring. These rings were introduced by Nicholson [52] in his study of lifting idempotents and exchange rings. In this chapter we review some well-known properties and examples of clean rings, and then we classify the rings that belong to an interesting subclass of clean rings. For a more general introduction to clean rings, see the survey of clean rings given by Nicholson and Zhou [55]. For a different perspective on this class of rings, see the history of commutative clean rings given by McGovern [47]. In what follows, all rings are associative with identity 1 \not = 0, all modules are unitary, and we generally use the notation and terminology of [41] and [42], which we list as our main references on noncommutative rings.

§1 Some Properties of Clean Rings

In this section we review some useful properties of clean rings. We begin by showing that every homomorphic image of a clean ring is clean, that every direct product of clean rings is clean, and that every full matrix ring Mn(R) with entries from a clean ring R is clean. Then we consider the notion of lifting idempotents, and we show that a ring R is clean if and only if for any ideal I of R such that I \subseteq J(R) the quotient ring R/I is clean and idempotents lift modulo I. These properties are well known, but proofs are included for the sake of completeness.

Basic Properties

The next two results were proved by Anderson and Camillo [1]. 2 (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 preserved by every ring homomorphism, the result follows.

(1.2) Proposition. Every direct product of clean rings is clean.

Proof. Since multiplication in a direct product of rings is defined componentwise, an element in a 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 rings is also defined componentwise, the result follows from a simple computation.

A matrix with exactly one entry equal to 1 and all other entries equal to 0 is sometimes said to be a matrix unit and is denoted by Eij when the entry in the ith row and jth column is 1. Notice

n that the set of n \times n matrix units \{E ii\}i =1 is a finite set of mutually orthogonal idempotents inthe full matrix ring Mn(R) for any ring R, and notice that the sum E11 + E22 + \cdot \cdot + Enn is equal to the n \times n identity matrix. In general, a finite set of mutually orthogonal idempotents whose sumis equal to the identity 1 is said to be a complete set of orthogonal idempotents. The following result is due to Han and Nicholson [29].

(1.3) Theorem. Every full matrix ring Mn(R) with entries from a clean ring R is clean.

Proof. Han and Nicholson [29, Theorem] showed that if the identity 1 of a ring R can be written as a finite sum 1= e1 + e2 + \cdot \cdot + en of mutually orthogonal idempotents ei such that each corner ring eiRei is clean, then the ring R is clean---the interested reader is encouraged to see their proof.

n Since the set of n \times n matrix units \{ Eii\} i=1 is a complete set of orthogonal idempotents for the full matrix ring Mn(R) with each corner ring EiiMn(R)Eii isomorphic to R, the result follows.

Lifting Idempotents

For an ideal I of a ring R, we say that idempotents lift modulo \bfitI if for each element x \in R such that x - x2 \in I there is some idempotent e of R with e - x \in I. Meanwhile, a ring is said to 3 be exchange if it satisfies the exchange property of Crawley and J\'onsson[21] when regarded as a left module over itself---the interested reader should21 see[ , pp. 797--798] for the formal definition of the exchange property. In his study of lifting idempotents and exchange rings, Nicholson [52,

Theorem 2.1] proved that a ring is exchange if and only if idempotents lift modulo every left ideal of the ring, and he proved that this notion is left-right symmetric. Moreover, he introduced clean rings as an example of a class of exchange rings when he proved the following.

(1.4) Proposition. Every clean ring is exchange.

Proof. Let R be any clean ring, let x be any element in R, and let I be any left ideal of R such that x - x2 \in I. Since a ring is exchange if and only if idempotents lift modulo every left ideal of the ring according to Nicholson [52, Theorem 2.1], it suffices for us to show that there exists some idempotent e of R with e - x \in I. Since R is a clean ring, we can write x = u + f for some unit u and idempotent f of R. Let v denote the inverse of u, and check that e = v(1 - f)u is idempotent with e - x = v(x - x2) \in I. This completes the proof.

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

Proof. This result follows immediately from (1.4) since idempotents lift modulo every left (right) ideal of an exchange ring according to Nicholson [52, Theorem 2.1].

For an ideal I of a ring R, we say that units lift modulo \bfitI if for each element x \in R such that x + I is a unit of the quotient ring R/I there is some unit u of R with u - x \in I. It is well known that for any ring R and any ideal I \subseteq J(R), an element x + I is a unit of R/I if and only if x is a unit of R. In particular, this tells us that units lift modulo any ideal of a ring R that is contained in its Jacobson radical. This fact is used to prove the following characterization of clean rings due to Han and Nicholson [29].

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

Proof. Suppose that the ring R is clean. Then the quotient ring R/I is clean as a result of (1.1), and idempotents lift modulo I as a result of (1.5). This proves the necessity. Conversely, suppose 4 that R/I is clean and that idempotents lift modulo I. We need to show that every element r \in R is clean. Let r be any element in R. Since the quotient ring R/I is clean, we can write r + I as the sum r + I = (x + I) + (e + I) of a unit x + I and an idempotent e + I of R/I. Since idempotents lift modulo I, we can assume that e is an idempotent of R. Since I \subseteq J(R) by hypothesis, we also know that units lift modulo I. Since (r - e) + I = (r + I) - (e + I) = x + I is a unit of R/I, this tells us that r - e is a unit of R. It now follows that r can be written as the sum of a unit and an idempotent of R by writing r = (r - e) + e. This proves the sufficiency. Therefore R is clean if and only if R/I is clean and idempotents lift modulo I.

It is well known that idempotents lift modulo every nil ideal of a ring---the interested reader is encouraged to see Koh [39] for an elementary proof of this result. Since every nil ideal of a ring R is contained in its Jacobson radical, we have the following 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 quotient ring R/N is clean.

Proof. This result follows immediately from (1.6) since every nil ideal of a ring R is contained in its Jacobson radical and idempotents lift modulo every nil ideal.

Two More Properties

A ring R is said to be semipotent (or sometimes an I0-ring) if each left ideal of R that is not contained in its Jacobson radical contains a nonzero idempotent. It is well known that this notion is left-right symmetric. Moreover, it is not difficult to see that a ring R is semipotent if and only if for every element a \in R that is not contained in J(R) there is a nonzero x \in R such that xax = x.

A semipotent ring R is said to be potent (or sometimes an I-ring) if idempotents lift modulo its

Jacobson radical. Nicholson [52, Proposition 1.9] proved that every exchange ring is potent. Since every clean ring is exchange according to (1.4), we have the following.

clean =\Rightarrow exchange =\Rightarrow potent =\Rightarrow semipotent

These implications are known to be irreversible. In particular, Camillo and Yu [11, p. 4746] proved that the ring in a well-known example due to Bergman [30, Example 1] is exchange but not clean. 5 Meanwhile, Nicholson [52, p. 272] showed that a potent ring need not be exchange, and Nicholson and Zhou [56, Example 25] showed that a semipotent ring need not be potent.

An element x in a ring R is said to be quasiregular if there is some element y \in R such that x + y = xy = yx. It is not difficult to see that an element x in a ring R is quasiregular if and only if 1 - x is a unit of R. Meanwhile, it is well known that every left (right) ideal of a ring R that is not contained in its Jacobson radical must contain an element that is not quasiregular. Following

Han and Nicholson [29], we give a direct proof of the fact 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 \nsubseteq J(R). We need to show that I contains a nonzero idempotent. As we mentioned above, it is well known that any left ideal of a ring R that is not contained in its Jacobson radical must contain an element that is not quasiregular. Since I \nsubseteq J(R), this tells us that there must be some a \in I that is not quasiregular. Since R is a clean ring, we can write a = u + e for some unit u and idempotent e of R. Since a is not quasiregular, it follows that e \not = 1. Now let v denote the inverse of u, and check that v(1 - e)u is idempotent. Moreover, notice that v(1 - e)u = v(1 - e)(a - e) = v(1 - e)a \in I. Since u and v are both units, it follows that v(1 - e)u is a nonzero idempotent of I. This completes the proof.

(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 said to be primitive if the corner ring eRe contains no nontrivial idempotents---in other words, no idempotents other than 0and e. It can be shown that a nonzero idempotent e of a ring R is primitive if and only if Rf = Re for any nonzero idempotent f of R such that f \in Re. This was noted by Nicholson [51, p. 364]. Meanwhile, an idempotent e of a ring R is said to be local if the corner ring eRe is local. Since a local ring contains no nontrivial idempotents, every local idempotent is primitive. The following result due to Nicholson [51] shows that the converse is true for any semipotent ring.

(1.10) Proposition. Every primitive idempotent in a semipotent ring is local. 6 Proof. Let R be a semipotent ring, and let e be any primitive idempotent of R. We need to show that e is a local idempotent of R. In particular, we need to show that the corner ring eRe is a local ring. Let a be any element in the ring eRe that is not contained in J(eRe). It is well known that

J(eRe) = J(R) \cap eRe for any idempotent e of a ring R---see Lam [41, Theorem 21.10]. This tells us that a is not contained in J(R), and hence Ra \nsubseteq J(R). Since R is a semipotent ring, it follows that the left ideal Ra contains some nonzero idempotent f. Notice that Rf \subseteq Ra \subseteq Re. Since e is a primitive idempotent of R, this implies that Rf = Re, which forces Ra = Re. In particular, this tells us that e is contained in Ra, and hence there is some element x \in R such that xa = e. Since a is an element of eRe, it follows that (exe)a = e, and hence a has a left inverse in eRe. Since this is the case for every element of eRe that is not contained in J(eRe), it follows that the corner ring eRe is a local ring. This completes the proof.

(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 and

Nicholson [29, Lemma] showed that if the corner rings eRe and fRf are both clean, then R is also clean. The converse is not true in general according to Ster\v [59, 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 are exchange by [52, Proposition 1.10], and the corner rings of any potent ring (resp. semipotent ring) are potent (resp. semipotent) as a result of [51, Corollary 1.7].

(2) An idempotent e of a ring R is said to be a full idempotent if ReR = R. It is not known whether the corner rings of a clean ring that arise from a full idempotent of the ring are clean. It is also not known whether being a clean ring is Morita invariant. In fact, since every full matrix ring over a clean ring is clean according to (1.3), being clean is Morita invariant if and only if the corner rings of a clean ring that arise from a full idempotent of the ring are clean. It is not known to the author whether the underlying ring must be clean in order for a full matrix ring to be clean. 7 (3) A ring is called abelian (or sometimes normal) if every idempotent in the ring is central.

It is easy to see that a ring R is abelian if and only if eRf = fRe = 0 for every idempotent e of R with complement f = 1 - e. Nicholson [52, Proposition 1.8] showed that an abelian ring is clean if and only if it is exchange. It follows from Burgess and Stephenson [8, Theorem 3.4] that an abelian ring is clean if and only if each of its Pierce stalks is local.

(4) A ring R is called quasinormal if eRfRe = 0 for every noncentral idempotent e of R with complement f = 1 - e. It is easy to check that every abelian ring is quasinormal, while the upper triangular matrix ring T2(\BbbZ/ 2\BbbZ) is quasinormal but not abelian. Therefore this notion is a proper generalization of that of an abelian ring. As we noted above, it is well known that an abelian ring is clean if and only if it is exchange. More generally, Wei and Li [67, Proposition 4.1] proved that a quasinormal ring is clean if and only if it is exchange.

(5) A ring is called left quasi-duo if every maximal left ideal of the ring is two-sided. To be right quasi-duo requires instead that every maximal right ideal of the ring is two-sided. It is not known whether this notion is left-right symmetric. Yu [73, Theorem 4.2] proved that a left (right) quasi-duo ring is clean if and only if it is exchange. Since every quasinormal exchange ring is both left and right quasi-duo according to Wei and Li [67, Theorem 3.12], this generalizes the fact that every quasinormal exchange ring is clean.

(6) A ring R is said to be a ring of bounded index (of nilpotency) if there is some positive integer n such that xn = 0 for every nilpotent element x \in R. Chen [12, Corollary 2] showed that every exchange ring of bounded index is clean. We should note that an exchange ring of bounded index need not be left (right) quasi-duo. For example, it is not difficult to see that the full matrix ring M2(\BbbZ/ 2\BbbZ) is an exchange ring of bounded index that is neither left nor right quasi-duo.

(7) A ring is said to be root clean if every element in the ring can be written as the sum of a unit and a square root of 1. Hiremath and Hegde [32, Proposition 2.12] showed that a ring is root clean if and only if it is a clean ring in which 2 is invertible. The sufficiency was proved earlier by

Camillo and Yu [11, Proposition 10] when they showed that a ring in which 2 is invertible is clean if and only if it is root clean. 8 (8) As we mentioned above, Camillo and Yu [11, p. 4746] proved that the ring in a well-known example due to Bergman [30, Example 1] is exchange but not clean. More specifically, the ring in

Bergman's example is an exchange ring in which 2 is invertible that is not root clean. Since a ring in which 2 is invertible is clean if and only if it is root clean according to [11, Proposition 10], the ring in Bergman's example is indeed exchange but not clean. The interested reader is encouraged to see [11], [30], [43], [59], and [60] for more on the ring in Bergman's example. No other example of an exchange ring that is not clean is 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 \pi -regular rings. We review some useful properties of these rings and of some closely related rings. In particular, we are interested in the relationships between these rings and clean rings.

Division, Boolean \& Local Rings

The following lemma shows that the units, idempotents, and quasiregular elements of any ring are clean. In particular, this tells us that any ring that consists entirely of these types of elements is a clean ring. We study this interesting subclass of clean rings in \S 3.

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

Proof. Any unit u can be written as the sum of a unit and an idempotent by writing u = u + 0.

Meanwhile, any idempotent e with complement f = 1 - e can be written as the sum of a unit and an idempotent by writing e = (e - f) + f since the difference e - f is a square root of 1, and hence a unit. Finally, any quasiregular element x can be written as the sum of a unit and an idempotent by writing x = (x - 1) + 1 since an element x is quasiregular if and only if x - 1 a unit. Therefore the units, idempotents, and quasiregular elements of any ring are clean.

(2.2) Corollary. Every division ring, boolean ring, and local ring is clean. 9 Proof. This result follows immediately from (2.1) since all division rings, boolean rings, and local rings consist 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 and only if it is clean (resp. exchange, potent, semipotent) and has no nontrivial idempotents. The first direct proof that a ring is local if and only if it is clean and contains no nontrivial idempotents was given by Anderson and Camillo [1]. It was proved earlier by Warfield [65] that a ring is local if and only if it is exchange and has no nontrivial idempotents, while Nicholson [51] proved that a ring is local if and only if it is potent (or semipotent) and has no nontrivial idempotents.

(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 well know that every local ring has no nontrivial idempotents. Since every local ring is clean according to (2.2), it follows that (1) \Rightarrow (2). Meanwhile, we know that every clean ring is exchange by (1.4), every exchange ring is potent according to Nicholson [52, Proposition 1.9], and every potent ring is semipotent by definition. It follows that (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5). Finally, recall from (1.10) that every primitive idempotent in a semipotent ring is local. Since the identity 1 is a primitive idempotent in any ring that contains no nontrivial idempotents, it follows that (5) \Rightarrow (1).

This completes 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 nontrivial idempotents. 10

Perfect \& Semiperfect Rings

A ring is called left artinian if every descending chain of left ideals in the ring has a minimal element---in other words, if for every descending chain I1 \supseteq I2 \supseteq I3 \supseteq \cdot of left ideals in the ring there is some positive integer n such that In = In+1 = In+2 = \cdot \cdot . To be right artinian requires instead that this property holds for every descending chain of right ideals in the ring. We will say 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, and hence every nil one-sided ideal of 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 turns out, 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 every homomorphic image of a left (resp. right) artinian ring is left (resp. right) artinian, and hence the quotient ring R/J(R) is semisimple for any left (resp. right) artinian ring R. It is well known that a 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 full matrix rings over division rings---this is the Wedderburn-Artin Theorem. Since every division ring is clean according to (2.2), the result follows from (1.3) and (1.2).

(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, and hence is a nil ideal. It follows from (1.7) that R is clean if and only if the quotient ring R/J(R) is clean.

Since R/J(R) is semisimple when R is left (right) artinian, the result now follows from (2.5).

(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 the 11 Jacobson radical of every left (resp. right) artinian ring is nilpotent, it follows that every left (resp. right) artinian ring is semiprimary. It is well known that a commutative ring is semiprimary if and only if it is isomorphic to a finite direct product of (commutative) local rings each of which hasa nilpotent maximal ideal.

(2.8) Proposition. Every semiprimary ring is clean.

Proof. Let R be any semiprimary ring. Then its Jacobson radical is nilpotent, and hence is a nil ideal. It follows from (1.7) that R is clean if and only if the quotient ring R/J(R) is clean. Since

R/J(R) is semisimple by definition when R is semiprimary, the result now follows from (2.5).

A subset of a ring is called left \bfitT -nilpotent if for every sequence of elements \{a 1, a2, a3,... \} in the subset there is some positive integer n such that a1a2 \cdot \cdot an = 0. To be right \bfitT -nilpotent requires instead that an \cdot \cdot a2a1 = 0. We will call a subset \bfitT -nilpotent if it is both left and right T -nilpotent---this notion is not left-right symmetric. It is not difficult to see that the implications below hold for any one-sided ideal of a ring.

nilpotent =\Rightarrow T -nilpotent =\Rightarrow left T -nilpotent =\Rightarrow nil

A ring R is said to be left perfect if R/J(R) is semisimple and J(R) is left T -nilpotent. To be right perfect requires instead that J(R) is right T -nilpotent. We will say that a ring is perfect if it is both left and right perfect---this notion is not left-right symmetric. Since every nilpotent ideal of a ring is (left and right) T -nilpotent, it follows that every semiprimary ring is perfect. It is well known that a commutative ring is perfect if and only if it is isomorphic to a finite direct product of

(commutative) local rings each of which has a T -nilpotent maximal ideal. In general, 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 according to Bass [5, Theorem P]---note the 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, and hence is a nil ideal. It follows from (1.7) that R is clean if and only if the quotient ring R/J(R) is clean. Since R/J(R) is semisimple by definition when R is left (right) perfect, the result now follows from (2.5). 12 A ring R is said to be semiperfect if R/J(R) is semisimple and idempotents lift 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 only if it is isomorphic to a finite direct product of (commutative) local rings. In general,48 Mueller[ , Theorem 1] proved that a ring is semiperfect if and only if its identity 1 can be written as the finite sum of orthogonal local idempotents. Moreover, he also proved that a ring is semiperfect if and only if it contains no infinite set of orthogonal idempotents and every primitive idempotent in the ring is local. Aringis said to be orthogonally finite (or sometimes I-finite) if there exists no infinite set of orthogonal idempotents in the ring. The next result shows the precise relationship between semiperfect rings and clean rings (resp. exchange rings, potent rings, semipotent rings). In particular, it tells us that a ring is semiperfect if and only if it is clean (resp. exchange, potent, semipotent) and orthogonally finite. This characterization of semiperfect rings is due to Camillo andYu[11].

(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 semiperfect ring is clean as noted above, it follows that (1) \Rightarrow (2). Meanwhile, we know that every clean ring is exchange by (1.4), every exchange ring is potent according to Nicholson [52, Proposition 1.9], and every potent ring is semipotent by definition. It follows that (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5). Finally, recall from (1.10) that every primitive idempotent in a semipotent ring is local. Since a ring is semiperfect if and only if it is orthogonally finite and every primitive idempotent in the ring is local according to Mueller [48, Theorem 1], it follows that (5) \Rightarrow (1). This completes the proof.

A ring R is said to be semilocal if R/J(R) is semisimple. It is well known that every semilocal ring is orthogonally finite, and it follows from (2.10) that a semilocal ring is clean if and only if it is semiperfect. Moreover, this tells us that a semilocal ring is clean if and only if idempotents lift 13 modulo its Jacobson radical. It is well known that a commutative ring is semilocal if and only if it has only finitely many maximal ideals. More generally, it is known that every ring R that has only finitely many maximal ideals is semilocal, while the converse is true if the quotient ring R/J(R) is commutative. It is not difficult to see that a semilocal ring need not be clean. For example, notice that the ring \BbbZ(6) of all rational numbers with denominators relatively prime to 6 (when written in lowest terms) is semilocal since it is a commutative ring that has only two maximal ideals, but it is not clean as a result of (2.3) since it is a non-local ring with no nontrivial idempotents.

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

Proof. As mentioned above, it is well known that every semilocal ring is orthogonally finite. The result now follows from (2.10) since every semiperfect ring is semilocal by definition.

Bass [5] introduced the notions of perfect and semiperfect rings as homological generalizations of semiprimary rings. In particular, semiperfect rings will play a central role in our study of clean group rings in the next chapter. The following implications are well known.

left/right artinian =\Rightarrow semiprimary =\Rightarrow left perfect =\Rightarrow semiperfect =\Rightarrow 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 [42, Ex. 20.5].

Meanwhile, since every semiprimary ring is (left and right) perfect, any left perfect ring that is not right perfect is not semiprimary---the interested reader should see Bass5 [ , p. 476] for an example of such a ring. Finally, it is not difficult to see that the ring \BbbZ(2) of all rational numbers with odd denominators (when written in lowest terms) is semiperfect but not perfect, and that the infinite direct product \BbbZ(2) \times \BbbZ(2) \times \cdot is clean but not semiperfect. In addition to the relationships above, we should note that every division ring is artinian, and that every local ring is semiperfect. Of course these implications are also known to be irreversible.

For example, the ring \BbbZ/ 4\BbbZ of integers modulo 4 is artinian but not a division ring, and the direct product \BbbZ/ 4\BbbZ \times \BbbZ/ 4\BbbZ is semiperfect but not local. 14

Regular \& \bfitpi -Regular Rings

A ring R is said to be regular---in the sense of von Neumann62 [ ]---if for every element r \in R there is some element x \in R such that rxr = r. More generally, an element r in a ring R is said to be regular if there is some element x \in R such that rxr = r. It is not difficult to see that a ring R is regular if and only if for every element a \in R there is an idempotent e of R such that Re = Ra.

In other words, a ring is regular if and only if every principal left ideal of the ring is generated by an idempotent. Since the Jacobson radical of any ring contains no nonzero idempotents, it follows that the Jacobson radical of any regular ring must be equal to 0. Warfield [66, Theorem 3] proved that every regular ring is exchange, but a regular ring need not be clean according to Camillo and

Yu [11, p. 4746]. A ring is said to be indecomposable if it cannot be written as a direct product of two (nonzero) rings. It is well known that a ring is indecomposable if and only if it contains no nontrivial central idempotents. The following theorem due to Lee, Yi, and Zhou [43] tells us that a regular ring is clean if and only if its indecomposable images are 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 [43, Corollary 7].

A ring R is said to be unit regular if for every element r \in R there is a unit u of R such that rur = r. More generally, an element r in a ring R is said to be unit regular if there is some unit u \in R such that rur = r. It is not difficult to see that a ring R is unit regular if and only if every element r \in R can be written as the product r = ue of a unit u and an idempotent e of R. Camillo and Khurana [9, Theorem 1] proved that every unit regular ring is clean, but a single unit regular element in a ring need not be clean according to Khurana and Lam [38, Example 4.5].

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

Proof. This proof is omitted---see Camillo and Khurana [9, Theorem 1].

A ring R is said to be strongly regular if for every element r \in R there is some element x \in R such that r2x = r. It is well known that this notion is left-right symmetric, and that every strongly regular ring is regular. Moreover, Azumaya [4, Lemma 1] proved that a ring R is strongly regular 15 if and only if for every element r \in R there is a unique z \in R such that rzr = r and zrz = z with rz = zr. It is not difficult to see that every abelian regular ring is strongly regular. Meanwhile,a ring is said to be reduced if it contains no nonzero nilpotent elements. Forsythe and McCoy [26,

Lemmas 1 \& 3] proved that every reduced ring is abelian, and that a ring is strongly regularifand only if it is regular and reduced. Since every abelian regular ring is strongly regular, it follows that a ring is strongly regular if and only if it is an abelian regular ring. The following proposition tells us that every strongly regular ring is unit regular and clean. These results are well known.

(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 \in R such that rzr = r with rz = zr according to Azumaya [4, Lemma 1]. Notice that e = rz is idempotent, and that u = r - (1 - e) is a unit of R with inverse v = ze - (1 - e). It now follows that r can be written as the product of a unit and an idempotent of R by writing r = ue, and it can be written as the sum of a unit and an idempotent of R by writing r = u + (1 - e). Therefore every strongly regular ring is unit regular and clean.

A ring R is called \bfitpi -regular if for every element r \in R there is some element x \in R such that rnxrn = rn for some positive integer n. Notice that the equation rnxrn = rn implies that xrn is an idempotent, and that xrn = 0 if and only if rn = 0. This tells us that every non-nil left ideal of a

\pi -regular ring contains a nonzero idempotent. Since the Jacobson radical of any ring contains no nonzero idempotents, it follows that every \pi -regular ring is semipotent with a nil Jacobson radical.

A semipotent ring with a nil Jacobson radical is sometimes known as a Zorn ring, and it follows immediately from (1.7) that a Zorn ring R is clean if and only if the quotient ring R/J(R) is clean.

A ring R is called strongly \bfitpi -regular if for every element r \in R there is some element x \in R such that rn+1x = rn for some positive integer n. These rings were studied for nearly thirty years before Dischinger [22, Th\'eor\`eme1] proved that this notion is left-right symmetric. It follows from

Azumaya [4, Theorem 3] that a ring R is strongly \pi -regular if and only if for every element r \in R there is some z \in R such that rnzrn = rn with rz = zr for some positive integer n. The following result was first proved by Burgess and Menal [7, Proposition 2.6].

(2.15) Theorem. Every strongly \pi -regular ring is clean. 16 Proof. This proof is omitted---see Nicholson [53, Theorem 1] for an elementary proof of this result whose original proof involves sheaf-theoretic representations.

A commutative ring R is said to be zero-dimensional---in regard to Krull dimension---if each prime ideal of the ring is maximal. It is well known that a commutative ring is zero-dimensional if and only if it is strongly \pi -regular, and hence every zero-dimensional commutative ring is clean as a result of (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. In particular, 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 a ring is strongly regular if and only if it is strongly \pi -regular and reduced.

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

Proof. As mentioned above, it is well known that a commutative ring is zero-dimensional if and only if it is strongly \pi -regular. The result now follows from (2.15).

Von Neumann [62] introduced regular rings as a proper generalization of semisimple rings. The interested reader is encouraged to see Goodearl [27] and Tuganbaev [61] for more on regular rings and rings close to regular. The following implications are well known.

commutative regular =\Rightarrow strongly regular =\Rightarrow unit regular =\Rightarrow regular = = = = \Rightarrow \Rightarrow \Rightarrow \Rightarrow

zero-dimensional comm. =\Rightarrow strongly \pi -regular =\Rightarrow clean =\Rightarrow exchange

These implications are known to be irreversible. For example, every noncommutative division ring is strongly regular but not a zero-dimensional commutative ring, and hence both of the horizontal implications on the left are irreversible. Meanwhile, Chen [17, Remark 4.3] showed that the direct \prod \infty product n=1 Mn(F ) of full matrix ring over any field F is unit regular but not strongly \pi -regular, which tells us that both of the horizontal implications in the middle are irreversible, while Camillo and Yu [11, p. 4746] proved that the ring in an example due to Bergman [30, Example 1] is regular but not clean, which tells us that both of the horizontal implications on the right are irreversible.

Finally, notice that the ring \BbbZ/ 4\BbbZ of integers modulo 4 is a zero-dimensional commutative ring but is not regular, and hence all four of the vertical implications above are also irreversible. 17 In addition to the relationships above, we should mention that every left (right) perfect ring is strongly \pi -regular. In particular, it is easy to see that a ring R is strongly \pi -regular if and only if for every element a \in R the descending chain aR \supseteq a2R \supseteq a3R \supseteq \cdot of principal right ideals in R has a minimal element. Since a ring is left perfect if and only if every descending chain of principal right ideals in the ring has a minimal element according to Bass [5, Theorem P], this tells us that every left perfect ring is strongly \pi -regular. Moreover, since the notion of a strongly \pi -regular ring is left-right symmetric according to Dischinger [22, Th\'eor\`eme1], it follows that every left or right perfect ring is strongly \pi -regular. This implication is known to be irreversible. For example, notice that any infinite direct product of fields is strongly \pi -regular but neither left nor right perfect.

Some Notes

(1) A ring is said to be left noetherian if every ascending chain of left ideals in the ring has a maximal element---in other words, if for every ascending chain I1 \subseteq I2 \subseteq I3 \subseteq \cdot of left ideals in the ring there is a positive integer n such that In = In+1 = In+2 = \cdot \cdot . To be right noetherian requires instead that this property holds for every ascending chain of right ideals. We will say that a 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 that the converse is true for all semiprimary rings.

(2) It is not difficult to see that a noetherian ring need not be clean. For example, theringof integers \BbbZ is noetherian but not clean, and the ring \BbbZ(6) of all rational numbers with denominators relatively 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 orthogonally finite ring is clean (resp. exchange, potent, semipotent) if and only if it is semiperfect by (2.10), it follows that a left (right) noetherian ring is clean (resp. exchange, potent, semipotent) if and only if it is semiperfect. Moreover, since a ring is semiperfect if and only if it is orthogonally finite and every primitive idempotent in the ring is local according to Mueller [48, Theorem 1], it follows that a left (right) noetherian ring is semiperfect (resp. clean, exchange, potent, semipotent) if and only if every primitive idempotent in the ring is local. 18 (3) It is well known that the notions of regular, unit regular, and strongly regular coincide in the commutative case. Examples of commutative regular rings include all boolean rings and fields.

Meanwhile, Jacobson [33, Theorem 11] showed that if for every element r in a ring R there is some integer n > 1 such that rn = r, then the ring R is commutative. It is not difficult to see that every such ring is regular (hence commutative regular), and therefore is clean as a result of (2.14).

(4) As we mentioned above, it is not difficult to see that the ring \BbbZ/ 4\BbbZ of integers modulo 4 is a zero-dimensional commutative ring but is not regular. More generally, it can be shown that the ring \BbbZ/n \BbbZ of integers modulo n is zero-dimensional for any positive integer n, but is regular if and only if n is square-free. This result was first proved by Ehrlich24 [ , Theorem 5].

(5) 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-dimensional commutative rings, and it is well known that every semiregular ring is exchange. The relationship between semiperfect rings, semiregular rings, clean rings, and exchange rings is shown below.

semiperfect =\Rightarrow clean = = \Rightarrow \Rightarrow

semiregular =\Rightarrow exchange

These implications are known to be irreversible. For example, Nicholson [52, p. 271] showed that a commutative exchange ring need not be semiregular. Since a commutative ring is clean if and only if it is exchange by (1.4), this tells us that a clean ring need not be semiregular, and hence both of the horizontal implications above are irreversible. Meanwhile, we know that the ring in Bergman's example [30, Example 1] is regular but not clean according to Camillo and Yu [11, p. 4746]. Since every regular ring is semiregular, this tells us that a semiregular ring need not be clean, and hence both of the vertical implications above are also irreversible. No example of an exchange ring that is neither semiregular nor clean is known to the author.

(6) As we mentioned above, a semipotent ring with a nil Jacobson radical is sometimes known as a Zorn ring. It is not difficult to see that R is a Zorn ring if and only if for every non-nilpotent a \in R there is some b \in R such that ab is a nonzero idempotent of R. These rings were introduced 19 by Kaplansky [35] in his study of semisimple alternative rings. The relationship between \pi -regular rings, exchange rings, Zorn rings, and potent rings is shown below.

\pi -regular =\Rightarrow Zorn = = \Rightarrow \Rightarrow

exchange =\Rightarrow potent

These implications are known to be irreversible. For example, McGovern [45, Example 20] showed that a commutative Zorn ring need not be clean. Since a commutative ring is clean if and only if it is exchange by (1.4), this tells us that a Zorn ring need not be exchange, and hence both of the horizontal implications above are irreversible. Meanwhile, it is easy to see that the ring \BbbZ(2) of all rational numbers with odd denominators (when written in lowest terms) is exchange but not Zorn, and hence both of the vertical implications above are also irreversible.

(7) Azumaya [4, Theorem 5] proved that a ring of bounded index (of nilpotency) is \pi -regular if and only if it is strongly \pi -regular. The general relationship between regular rings, \pi -regular rings, strongly \pi -regular rings, and Zorn rings is shown below.

regular =\Rightarrow \pi -regular =\Rightarrow Zorn

strongly =\Rightarrow \pi -regular

These implications are known to be irreversible. In particular, it is not difficult to check that the upper triangular matrix ring T2(\BbbZ/ 2\BbbZ) is strongly \pi -regular but not regular, and hence a \pi -regular ring need not be regular. Moreover, Chen [17, Remark 4.3] gave an example of a unit regular ring that is not strongly \pi -regular, and hence a \pi -regular ring need not be strongly \pi -regular. Finally, we noted above that a Zorn ring need not be \pi -regular as a result of McGovern [45, Example 20].

(8) A ring R is said to have stable range one if for any elements a, b \in R with Ra + Rb = R, there is some element y \in R such that a + yb is a unit of R---this notion is left-right symmetric. It is well known that every semiperfect ring, unit regular ring, and strongly \pi -regular ring has stable range one. Meanwhile, every left (right) quasi-duo exchange ring has stable range one according to 20 Chen and Li [15, Corollary 3], and every exchange ring of bounded index (of nilpotency) has stable range one according to Yu [72, Corollary 4]. It is not known to the author whether every exchange ring with stable range one is clean, but every semiregular ring with stable range one is clean as a result of Aydo\u gdu,Lee, and Ozcan\" [2, Theorem 2.1]. In general, a clean ring need not have stable range one according to Nicholson and Varadarajan [54, Corollary].

(9) A ring R is said to have idempotent stable range one if for any elements a, b \in R with

Ra + Rb = R, there exists some idempotent e of R such that a + eb is a unit of R---this notion is left-right symmetric. It is not difficult to see that every ring that has idempotent stable rangeone is clean. In particular, this follows almost immediately from the fact that Ra + R( 1)- = R for any element a in a ring R. Chen [13, Theorem 12] showed that an abelian ring is clean if and only if it has idempotent stable range one. Meanwhile, Wang et al. [64] proved that all semiperfect rings 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 quasiregular elements. We know from (2.1) that the units, idempotents, and quasiregular elements of any ring are clean, and thus every ring that consists entirely of these types of elements is a clean ring. But exactly which rings belong to this interesting subclass of clean rings? We give a complete solution to 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 the direct product of two division rings, while any nonabelian ring that consists entirely of these types of elements is isomorphic to either the full matrix ring M2(D) for some division ring D, or to the ring 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 elements by 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 21 define the following subsets of aring R.

\scrU (R) = \{ r \in R | r is a unit of R\}

\scrI \scrD(R) = \{ r \in R | r is an idempotent of R\}

\scrQ( \scrR R) = \{ r \in R | r is a quasiregular element of R\}

The first part of the following theorem was proved independently by Chen18 andCui[ ] and by the author (unpublished), and was proved earlier for commutative rings by Anderson and Camillo [1].

We give a new proof of this result. Recall that every reduced ring is abelian according to Forsythe and McCoy [26, Lemma 1].

(3.1) Theorem. Let R be any ring.

(1) R = \scrU (R) \cup \scrI\scrD(R) if and only if R is a division ring or a boolean ring.

(2) R = \scrU (R) \cup( \scrQ\scrR R) if and only if R is a local ring.

(3) R = \scrI \scrD(R) \cup( \scrQ\scrR 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 difficult to see that any such ring is reduced, and therefore is abelian. In particular, the ring R either has no 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 \times B be any direct product decomposition of R. If either of the rings A or B contained a non-idempotent x, then there would be an element (x, 0) or (0, x) in A \times B that was neither a unit nor an idempotent of A \times B. Since the ring R consists entirely of units and idempotents, this tells us that A and B are both boolean rings. Since the direct product of any two boolean rings is itself a boolean ring, it follows that the ring 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 the element 1 - r \in R must be either idempotent or quasiregular for any r \in R. It is not difficult to see that r is an 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, and hence it must be either a division ring or a boolean ring by (1). Therefore every ring that consists 22 entirely of idempotents and quasiregular elements is either a division ring or a boolean ring. The converse is easy to check.

It is not difficult to see that division rings, boolean rings, and local rings are not theonlyrings that consist entirely of units, idempotents, and quasiregular elements. For example, the interested reader is encouraged to check that the direct product \BbbZ/ 2\BbbZ \times \BbbZ/ 3\BbbZ consists entirely of these types of elements even though it is not a division ring, a boolean ring, or a local ring. More generally, we show that the direct product of any two division rings consists entirely of units, idempotents, and quasiregular 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 in any such ring is a unit (resp. idempotent, quasiregular element) of that ring if and only if the entry in each of its components is a unit (resp. idempotent, quasiregular element) of its ring. Since the ring 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 \in R must satisfy at least one of the following conditions, where each of the entries denoted by \ast can be any nonzero, nonidentity element of its ring.

r = (1, 1), (1, \ast), (\ast , 1), or (\ast, \ast) \in \scrU(R)

r = (0, 0), (0, 1), (1, 0), or (1, 1) \in \scrI\scrD(R)

r = (0, 0), (0, \ast ), (\ast , 0), or (\ast, \ast) \in \scrQ\scrR(R)

Therefore the direct product of any two division rings consists entirely of units, idempotents, and quasiregular 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 two division rings. Incidentally, this result tells us that every decomposable ring that consists entirely of these types of elements is necessarily abelian. 23 (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.

Proof. Let A \times B be any direct product decomposition of the ring R. Since multiplication in a direct product of rings is defined componentwise, an element in A \times B is a unit (resp. idempotent, quasiregular element) of the ring A \times 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 some element 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 \times B that was not a unit, an idempotent, or a quasiregular element of A \times B.

Since the ring R consists entirely of units, idempotents, and quasiregular elements, it follows that the rings A and B must both consist entirely of units and idempotents. We know from (3.1) that any such ring is either a division ring or boolean ring. If either of the rings A or B contained some unit u other than 1, while the other ring contained some nontrivial idempotent e, then there would be an element (u, e) or (e, u) in A \times B that is not a unit, an idempotent, or a quasiregular element of A \times B. Since the ring R consists entirely of units, idempotents, and quasiregular elements, this tells us that A and B are either both boolean rings or both division rings. Since the direct product of any two boolean rings is itself a boolean ring, it follows that the ring R is either a boolean ring or a direct product of two division rings. Therefore every decomposable ring that consists entirely of units, idempotents, and quasiregular elements is either a boolean ring or a direct product of two division rings.

We are now ready to classify the abelian rings that consist entirely of units, idempotents, and quasiregular elements. In particular, we prove that any abelian ring that consists entirely of these types 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 central idempotent. If R has no nontrivial idempotents, then it consists entirely of units and quasiregular elements, and hence is a local ring by (3.1). Meanwhile, if R has a nontrivial central idempotent, 24 then it is decomposable, and hence is either a boolean ring or a direct product of two division rings by (3.3). Therefore every abelian ring that consists entirely of units, idempotents, and quasiregular elements 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, and quasiregular elements. We give a complete solution to this problem below. In particular, we prove that any nonabelian ring that consists entirely of units, idempotents, and quasiregular elements is isomorphic to either the full matrix ring M2(D) for some division ring D, or to the ring of a Morita context with zero pairings where both of the underlying rings are division rings. We begin with the following lemma.

(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 the ring R with its Peirce decomposition with respect to e.

\Biggl( \Biggr) eRe eRf R =\sim fRe fRf

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 that the subring of all diagonal matrices in this decomposition of R also consists entirely of these types of elements. Moreover, this subring is isomorphic to the direct product eRe \times fRf, and since e is a nontrivial idempotent of R, we know that eRe and fRf are both (nonzero) rings. It now follows 25 from the proof of (3.3) that the corner ring eRe is either a division ring or a boolean ring. We need to 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 ring or 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 with complement b = e - a, and consider the following matrices, where each entry denoted by x can be any element in eRf, while each entry denoted by y can be any element in fRe. \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) a x b x a 0 b 0 X1 = ,X2 = ,X3 = ,X4 = 0 0 0 0 y 0 y 0

Since X1 and X2 both have a row in which each entry is 0, while X3 and X4 both have a column in which each entry is 0, none of these matrices is a unit of R. Now consider the following matrix. \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) e 0 a x b x- I - X1 = - = 0 f 0 0 0 f

In particular, notice that b would need to have a left inverse in eRe in order for this matrix I - X1 to have a left inverse in R. However, this element b cannot have a left inverse in eRe since it is a nontrivial idempotent of eRe. This tells us that I - X1 is not invertible in R, and hence X1 is not quasiregular in R. Similar arguments show that the matrices X2, X3, and X4 are not quasiregular in R. Since the ring R consists entirely of units, idempotents, and quasiregular elements, it follows that each of these matrices must be idempotent. A simple computation shows that the matrix X1 is idempotent if and only if ax = x, while X2 is idempotent if and only if bx = x. Moreover, since x is an element of eRf, we also know that ex = x. These last three equalities tell us that x = 0, as follows, 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 \in eRf and y \in fRe, it follows that eRf = 0 = fRe, and hence e is a central idempotent of R. More specifically, this tells us that e is a central idempotent of R when the ring eRe is a not division ring. Therefore eRe is a division ring for every noncentral idempotent e of R by contrapositive.

The next result classifies the full matrix rings Mn(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 \times n 26 matrices with entries from some arbitrary ring consists entirely of these types of elements for some integer n > 1 if and only if it is equal to the full matrix ring M2(D) for some division ring D. The sufficiency is essentially due to44 Li[ ]. 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 \times 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 it is equal to M2(D) for some division ring D.

Proof. Suppose that the ring R consists entirely of units, idempotents, and quasiregular elements for some integer n > 1, and let us write R = Mn(S). Since n > 1, the matrix unit E = E11 and its complement F = I - E11 are both noncentral idempotents of R. Since the ring R consists entirely of units, idempotents, and quasiregular elements, it follows from (3.6) that the ring FRF must be a division ring. Moreover, notice that we have the following.

FRF =\sim S \times \cdot\times S \underbrace{} \underbrace{} n 1- \mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{s}

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 M2(D) for some division ring D. This proves the necessity.

Conversely, suppose that the ring R is equal to M2(D) for some division ring D. Then for any matrix X \in R, it follows immediately from a technical lemma due to Li [44, 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. \Biggl( \Biggr) 1 1 Y = 0 0

Notice that this matrix Y is idempotent. It is not difficult to check that any matrix similar toan idempotent 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 [23] proved that if eRe is a division ring for every noncentral idempotent e of a nonabelian semiprime ring R, then that ring 27

R must be isomorphic to the full matrix ring M2(D) for some division ring D. We prove that every nonabelian semiprime ring that consists entirely of units, idempotents, and quasiregular elements is isomorphic to M2(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 M2(D) for some division ring D.

Proof. Suppose that R is semiprime. Since the ring R consists entirely of units, idempotents, and quasiregular elements, it follows immediately from (3.6) that the ring eRe must be a division ring for every noncentral idempotent e of R. As we noted above, Du [23, Lemma 21] proved that every nonabelian semiprime ring that has this property is isomorphic to the full matrix ring M2(D) for some division ring D. Therefore R is isomorphic to M2(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 \psi : M \times N \rightarrow A and \varphi : N \times M \rightarrow B be some binary functions, written multiplicatively as

\psi (x, y) = xy and \varphi (y, x) = yx. The 6-tuple (A, B, M, N, \psi ,\varphi) is called a Morita context if these binary functions \psi and \varphi preserve addition in each variable and satisfy the following associativity conditions for every element a \in A, b \in B, x \in M, and y \in 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 \psi and \varphi are called pairings, and if \psi and \varphi are both equal to 0, then the Morita context (A, B, M, N, \psi ,\varphi) is said to be a Morita context with zero pairings.

Let (A, B, M, N, \psi ,\varphi) be any Morita context, and notice that the following set of matrices is a ring with respect to the usual operations for addition and multiplication of matrices when \psi and \varphi are used to define multiplication between elements of the bimodules M and N. \Biggl( \Biggr) \Biggl\{ \Biggl( \Biggr) \bigm| \Biggr\} AM a x \bigm| = \bigm| a \in A, b \in B, x \in M, y \in N NB y b \bigm|

Any such ring is known as the ring of a Morita context. The next result shows that if either of the bimodules M or N is nonzero, then the ring of any Morita context (A, B, M, N, \psi ,\varphi) with zero pairings consists entirely of units, idempotents, and quasiregular elements if and only if the rings

A and B are both division rings. 28 (3.9) Proposition. Let R be the ring of a Morita context (A, B, M, N, \psi ,\varphi) 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 elements for some bimodules M and N, at least one of which is nonzero. Since at least one of the bimodules

M or N is nonzero, the matrix units E = E11 and F = E22 are both noncentral idempotents of R. Since the ring R consists entirely of units, idempotents, and quasiregular elements, it now follows from (3.6) that ERE and FRF must both be division rings. Moreover, notice that ERE =\sim A and that FRF =\sim 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 either of these rings is a unit of its ring. Let a be any nonzero element in A with inverse u, and let b be any nonzero element in B with inverse v. Then for any elements x \in M and y \in N, since R is the ring of a Morita context with zero pairings, we have the following.

\Biggl( \Biggr) 1- \Biggl( \Biggr) a x u uxv- = y b vyu- v

This shows that a matrix in R is a unit (resp. quasiregular element) of the ring R when each entry on its main diagonal is a nonzero (resp. nonidentity) element of its ring. It is not difficult now to see that each matrix X \in R must satisfy at least one of the following conditions, where each of the entries denoted by \ast can be any nonzero, nonidentity element of its ring, while each entry denoted by x can be any element in M, and each entry denoted by y can be any element in N. \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) 1 x 1 x \ast x \ast x X = , , , or \in \scrU(R) y 1 y \ast y 1 y \ast \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) 0 0 0 x 1 x 1 0 X = , , , or \in \scrI\scrD(R) 0 0 y 1 y 0 0 1 \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) \Biggl( \Biggr) 0 x 0 x \ast x \ast x X = , , , or \in( \scrQ\scrR R) y 0 y \ast y 0 y \ast

Therefore R consists entirely of units, idempotents, and quasiregular elements when A and B are both division rings. This proves the sufficiency. 29 Nicholson [50] called a ring an NJ-ring if the ring consists entirely of regular and quasiregular elements. Examples of NJ-rings include all regular rings and local rings. Moreover, since the units and idempotents of any ring are regular, every ring that consists entirely of units, idempotents, and quasiregular elements is an NJ-ring. The converse is not true in general since a regular ring need not be clean. Nicholson [50] 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 underlying rings are both division 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 zero pairings 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 context with 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 consists entirely 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. As we noted above, Nicholson [50, 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 underlying rings are both division rings. In particular, since every regular ring is semiprime, while every local ring is abelian, it follows that every nonabelian, nonsemiprime NJ-ring must be isomorphic to the ring 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 underlying rings are both division rings.

A Special Case

It is well known that every nilpotent element is quasiregular. In particular, if xn = 0 for some integer n > 1, then an easy computation shows that 1 + x + x2 + \cdot \cdot + 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 a 30 complete solution to this problem below. Specifically, we prove that any abelian ring that consists entirely of units, idempotents, and nilpotent elements is either a boolean ring or a local ring with a nil Jacobson radical, while any nonabelian ring that consists entirely of these types of elements is isomorphic to either the full matrix ring M2(\BbbZ/ 2\BbbZ), or to the ring of a Morita context with zero pairings where the underlying rings are both \BbbZ/ 2\BbbZ. We begin our study of rings consisting entirely of units, idempotents, and nilpotent elements by classifying the abelian rings that consist entirely of these types of elements. It is easy to check that all 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 these types 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 central idempotent. Suppose that R has no nontrivial idempotents. Then it consists entirely of units and nilpotent (hence quasiregular) elements, and thus it is a local ring by (3.1). Moreover, since every nonunit of R is nilpotent, it is a local ring with a nil Jacobson radical. Suppose instead that R has a nontrivial central idempotent. Then R is a decomposable ring. Let A \times B be any direct product decomposition 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 rings or both division rings. If either of A or B contained a unit u other than 1, then there would be an element (u, 0) or (0, u) in A \times B that is not a unit, an idempotent, or a nilpotent element of A \times B.

Since the ring R consists entirely of units, idempotents, and nilpotent elements, this tells us that

A and B are both boolean rings. Since the direct product of two boolean rings is itself a boolean ring, it follows that the ring R is boolean in this case. Therefore every abelian ring that consists entirely of units, idempotents, and nilpotent elements is either a boolean ring or a local ring with a nil Jacobson radical.

It still remains to classify the nonabelian rings that consist entirely of units, idempotents, and nilpotent elements. We begin with the following corollary to (3.6). 31 (3.12) Corollary. Let R be any ring that consists entirely of units, idempotents, and nilpotent elements. Then eRe is isomorphic to \BbbZ/ 2\BbbZ 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 e of R. However, if the ring eRe contained some unit u \not = e, then the following matrix would not be a unit, an idempotent, or a nilpotent element of the Peirce decomposition of R with respect to e.

\Biggl( \Biggr) u 0 X = 0 0

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

It is easy to see that the full matrix ring M2(\BbbZ/ 2\BbbZ) consists entirely of units, idempotents, and nilpotent elements. The following result shows that every nonabelian semiprime ring that consists entirely of these types of elements must be isomorphic to M2(\BbbZ/ 2\BbbZ).

(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 M2(\BbbZ/ 2\BbbZ) .

Proof. Suppose that the ring R is semiprime. Since it consists entirely of units, idempotents, and nilpotent (hence quasiregular) elements, we know from (3.8) that R must be isomorphic to M2(D) for some division ring D. Let us identify R with the full matrix ring M2(D), and notice that the \sim matrix unit E = E11 is a noncentral idempotent of R with ERE = D. Moreover, since the ring R consists entirely of units, idempotents, and nilpotent elements, it follows from (3.12) that the ring

ERE is isomorphic to \BbbZ/ 2\BbbZ. Therefore R is isomorphic to M2(\BbbZ/ 2\BbbZ).

It is easy to check that the ring of any Morita context with zero pairings where the underlying rings are both isomorphic to \BbbZ/ 2\BbbZ consists entirely of units, idempotents, and nilpotent elements. The following result shows that every nonabelian, nonsemiprime ring that consists entirely of these types of elements must be isomorphic to the ring of a Morita context with zero pairings where the underlying rings are both \BbbZ/ 2\BbbZ. In this case, it can be shown that the underlying bimodules must both be elementary abelian 2-groups. 32 (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 context with zero pairings where the underlying rings are both \BbbZ/ 2\BbbZ.

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 the ring 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, \psi ,\varphi) with zero pairings where the rings A and B are both division rings, and notice that the matrix units E = E11 and F = E22 are noncentral idempotents of R with ERE =\sim A and FRF =\sim B. Since the ring R consists entirely of units, idempotents, and nilpotent elements, it now follows from (3.12) that the rings ERE and

FRF must both be isomorphic to \BbbZ/ 2\BbbZ. Therefore R is isomorphic to the ring of a Morita context with zero pairings where the underlying rings are both \BbbZ/ 2\BbbZ.

Conclusions

The next two theorems summarize the main results from this section. In particular, they tell us exactly when a ring consists entirely of units, idempotents, and quasiregular elements, and exactly when a ring consists entirely of units, idempotents, and nilpotent elements.

(3.15) Theorem. A ring R consists entirely of units, idempotents, and quasiregular elements if and only if it satisfies one (or more) of the following.

(1) R is a boolean ring;

(2) R is a local ring;

(3) R is isomorphic to the direct product of two division rings;

(4) R is isomorphic to the full matrix ring M2(D) for some division ring D; (5) R is isomorphic to the ring of a Morita context with zero pairings where both of the

underlying rings are division rings.

Proof. The necessity follows immediately from (3.4), (3.8), and (3.10), and the sufficiency follows from (3.1), (3.2), (3.7), and (3.9). 33 (3.16) Theorem. A ring R consists entirely of units, idempotents, and nilpotent elements if and only if it satisfies one (or more) of the following.

(1) R is a boolean ring;

(2) R is a local ring with a nil Jacobson radical;

(3) R is isomorphic to the full matrix ring M2(\BbbZ/ 2\BbbZ) ; (4) R is isomorphic to the ring of a Morita context with zero pairings where both of the

underlying rings are \BbbZ/ 2\BbbZ.

Proof. The necessity follows immediately from (3.11), (3.13), and (3.14), and the sufficiency can be directly verified and is left for the interested reader---we should note that the sufficiency of(4) follows easily from the proof of (3.10).

An element r in a ring is said to be potent if there is some positive integer n such that rn = r.

Since every potent element is strongly regular, and hence clean by (2.14), it would be interesting to know exactly when a ring consists entirely of potent and quasiregular (or nilpotent) elements.

Some Notes

(1) A ring is said to be strongly clean if each element in the ring can be written as the sum of a unit and an idempotent that commute. It follows easily from the proof of (2.1) that every ring that consists entirely of units, idempotents, and quasiregular elements is strongly clean. It can be shown that every ring that consists entirely of these types of elements is also optimally clean in the sense of Shifflet [58]. We should note that optimally clean implies strongly clean by definition.

(2) As we mentioned in \S 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\'anchez

Campos (unpublished), while the corner rings of any optimally clean ring must be optimally clean according to Shifflet58 [ , Proposition 3.1.5]. Similarly, it can be shown that the corner rings of any ring that consists entirely of units, idempotents, and quasiregular elements also consists entirely of units, idempotents, and quasiregular elements.

(3) As we mentioned in \S1 , it is not known whether being clean is Morita invariant. However, 34 we know from (3.7) that consisting entirely of units, idempotents, and quasiregular elements is not

Morita invariant. Meanwhile, being strongly clean (or optimally clean) is also not Morita invariant according to Wang and Chen [63, Example 1].

(4) A ring is called Dedekind finite (or sometimes von Neumann finite) if ab = 1 implies that ba = 1 for any pair of elements a, b in the ring. It can be shown that every ring that consists entirely of units, idempotents, and quasiregular elements is Dedekind finite. It not known whether a strongly clean ring or an optimally clean ring must be Dedekind finite, but a clean ring need not be Dedekind finite according to Nicholson and Varadarajan [54, Corollary].

(5) Recall that a ring R is said to have stable range one if for any pair of elements a, b \in R with Ra + Rb = R, there exists an element y \in R such that a + yb is a unit of R. It is not difficult to see that every ring with stable range one is Dedekind finite. As we noted above, it can be shown that every ring that consists entirely of units, idempotents, and quasiregular elements is Dedekind finite. More specifically, it can be shown that every such ring has stable range one. In fact,itcan be shown that every such ring has idempotent stable range one.

(6) Recall that a ring R is called quasinormal if eRfRe = 0 for every noncentral idempotent e of R with complement f = 1 - e. It can be shown that every nonabelian, nonsemiprime ring that consists entirely of units, idempotents, and quasiregular elements is quasinormal. More generally,

Nicholson [50] showed that every nonsemiprime NJ-ring is quasinormal. Unfortunately, this result is hidden in the proof of [50, Theorem 2].

(7) An idempotent e of a ring R is sometimes said to be a division idempotent if the corner ring eRe is a division ring. We know from (3.6) that every noncentral idempotent of any ring that consists entirely of units, idempotents, and quasiregular elements is a division idempotent. It can be shown that the converse is true for any nonabelian quasinormal ring. In other words, it can be shown that a nonabelian quasinormal ring consists entirely of units, idempotents, and quasiregular elements if and only if every noncentral idempotent of the ring is a division idempotent. However, it is not known to the author whether this is true for every nonabelian ring that consists entirely of units, idempotents, and quasiregular elements. 35 (8) Let R be any ring, and let M be any (R,R)-bimodule. Then the following set of matrices is a ring with respect to the usual operations for addition and multiplication of matrices.

\Biggl\{ \Biggl( \Biggr) \bigm| \Biggr\} r x \bigm| T (R,M) = \bigm| r \in R, x \in M 0 r \bigm|

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 R consists entirely of these types of elements and the equation ex = xf holds for every x \in M and for every nontrivial idempotent e of R with complement f = 1 - e. Meanwhile, it can be shown that a trivial extension of the form T (R,R) consists entirely of units, idempotents, and quasiregular elements if and only if the underlying ring R is local. We should note that T (R,R) is a local ring when R is local.

(9) 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.

\Biggl\{ \Biggl( \Biggr) \bigm| \Biggr\} \Biggl\{ \Biggl( \Biggr) \bigm| \Biggr\} a 0 \bigm| 0 x \bigm| R = \bigm| a \in A, b \in B ,M = \bigm| x \in X, y \in Y 0 b \bigm| y 0 \bigm|

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

\left\{\left( \right) \bigm| \right\} a 0 0 x \bigm| \bigm| 0 b y 0 \bigm| T (R,M) = \bigm| a \in A, b \in B, x \in X, y \in Y 0 0 a 0 \bigm| \bigm| 0 0 0 b \bigm|

It is not difficult to see that the trivial extension T (R,M) is isomorphic to the ring of the Morita context (A, B, X, Y, \psi) ,\varphi where \psi and \varphi are zero pairings. It follows immediately from (3.9) that

T (R,M) consists entirely of units, idempotents, and quasiregular elements for some M \not = 0 if and only 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 be isomorphic to the trivial extension T (R,M) for some division rings A and B. 36

CHAPTER TWO Clean Group Rings

§4 Preliminaries

In this section we review some known results on clean group rings and semiperfect group rings that will be useful in what follows. In particular, we are interested in several results on semiperfect group rings due to Woods [69].

Some Results on Semiperfect Group Rings

We begin our study of clean group rings by looking at some results on semiperfect group rings.

The following result due to Woods [69] gives necessary conditions on the coefficient ring R and the underlying group G in order for a group ring RG to be semiperfect. In particular, it tells us that R must be a semiperfect ring and G must be a torsion group in order for RG to be semiperfect.

(4.1) Proposition. The coefficient ring R must be semiperfect and the underlying group G must be a torsion group in order for a group ring RG to be semiperfect.

Proof. It is well known that the quotient ring RG/\Delta is isomorphic to R for any group ring RG, where \Delta is the augmentation idealof RG. Since every homomorphic image of a semiperfect ring is semiperfect, it follows that the coefficient ring R must be semiperfect in order for a group ring RG to be semiperfect. Meanwhile, Woods [69, Theorem 3.2] proved that the underlying group G must be a torsion group in order for a group ring RG to be semiperfect. This completes the proof.

The following result due to Woods [69] gives two examples of groups G whose group rings RG are semiperfect for every semiperfect ring R. In particular, it tells us that the group rings RC2 and

RS3 are semiperfect for every semiperfect ring R. Since the coefficient ring R must be semiperfect 37 in order for a group ring RG to be semiperfect according to (4.1), this tells us that RC2 and RS3 are semiperfect if and only if R is semiperfect. Moreover, it can be shown that no other nontrivial cyclic group has this property, but it is not known whether any other nontrivial symmetric group has this property.

(4.2) Proposition. The group rings RC2 and RS3 are semiperfect for every semiperfect ring R.

Proof. This proof is omitted---see Woods69 [ , pp. 128--129].

The exponent of a group G is the smallest positive integer n (if one exists) such that gn = 1 for every element g \in G. It is not difficult to see that every finite group has an exponent, whilean infinite group need not have an exponent in general. The following result due toWoods[69] tells us that the group ring RG of any finite abelian group G of exponent n over any ring R is semiperfect if and only if RCn is semiperfect. In particular, since the group ring RC2 is semiperfect for every semiperfect ring R according to (4.2), this tells us that the group ring RG of any finite elementary abelian 2-group G is semiperfect for every semiperfect ring R.

(4.3) Proposition. Let R be a ring, and let G be any finite abelian group of exponent n. Then the group ring RG is semiperfect if and only if RCn is semiperfect.

Proof. This proof is omitted---see Woods69 [ , Proposition 5.7].

The next result is essentially due to Burgess [6]. It gives necessary and sufficient conditions for the group ring RG of an abelian group G over any local ring R to be semiperfect. The p-primary component of an abelian group G will mean the trivial subgroup when p = 0.

(4.4) Theorem. Let R be a local ring with char R/J(R) = p \geq 0, and let G be an abelian group with p-primary component Gp. Then the group ring RG is semiperfect if and only if the quotient group G/Gp is finite and the group ring R(G/Gp) is semiperfect.

Proof. This proof is omitted---see Woods69 [ , Lemma 5.1].

(4.5) Corollary. Let R be a local ring with char R/J(R) = p \geq 0, and let G be an abelian group with p-primary component Gp. Then the group ring RG is semiperfect if and only if the quotient 38 group G/Gp is a finite group of exponent n and the group ring RCn is semiperfect.

Proof. This result follows from (4.4) and (4.3) since every finite group has an exponent.

The following theorem due to Woods [69] gives necessary and sufficient conditions for the group ring RG of an abelian group G over a commutative local ring R to be semiperfect in terms of the polynomial rings R[X] and R[X], where R denotes the quotient ring R/J(R).

(4.6) Theorem. Let R be a commutative local ring with char R = p \geq 0, and let G be an abelian group with p-primary component Gp. Then the group ring RG is semiperfect if and only if G/Gp is a finite group of exponent n and every monic factor of Xn - 1 in R[X] can be lifted to a monic factor of Xn - 1 in R[X].

Proof. This proof is omitted---see Woods69 [ , Theorem 5.8].

The following example due to Woods [69] tells us that the group ring \BbbZ(7) C3 is not semiperfect. More importantly, this example shows how we can use the result in (4.6) to determine whether the group ring RG of an abelian group G over a commutative local ring R is semiperfect.

(4.7) Example. \BbbZ(7) C3 is not semiperfect.

Proof. We know from (4.6) that the group ring \BbbZ(7) C3 is semiperfect if and only if every monic 3 3 factor of X - 1 in the polynomial ring \BbbF 7[X] can be lifted to a monic factor of X - 1 in \BbbZ(7) [X]. 3 2 Since X - 1 factors as (X - 1)(X - 2)(X - 4) over \BbbF 7, while it factors as (X - 1)(X + X + 1) 2 over \BbbZ(7) with X + X + 1 irreducible, it follows that the group ring \BbbZ(7) C3 is not semiperfect.

Some Results on Clean Group Rings

The following result due to McGovern [46] gives necessary conditions on the coefficient ring R and the underlying group G in order for a commutative group ring RG to be clean. In particular, it tells us that R must be a clean ring and G must be a torsion group in order for a commutative group ring RG to be clean. It is not known if the underlying group G must be a torsion group in order for a noncommutative group ring RG to be clean, but it is known that the coefficient ring R must be clean in order for any group ring RG to be clean. 39 (4.8) Proposition. The coefficient ring R must be clean and the underlying group G must be a torsion group in order for a commutative group ring RG to be clean.

Proof. It is well known that the quotient ring RG/\Delta is isomorphic to R for any group ring RG, where \Delta is the augmentation idealof RG. Since every homomorphic image of a clean ring is clean by (1.1), it follows that the coefficient ring R must be clean in order for any group ring RG to be clean. Meanwhile, McGovern [46, Proposition 2.7] showed that the underlying group G must be a torsion group in order for a commutative group ring RG to be clean. This completes the proof.

We know from (4.2) that the group rings RC2 and RS3 are semiperfect (hence clean) for every semiperfect ring R. The following result due to Khurana and Kumar [37] tells us that these group rings are also clean for every abelian clean ring R. It is not known if the group rings RC2 and RS3 are clean for every clean ring R, but we will see in \S5 that the group ring RCn does not have this property for any cyclic group of order n > 2.

(4.9) Proposition. The group rings RC2 and RS3 are clean for every abelian clean ring R.

Proof. This proof is omitted---see Khurana and Kumar [37, Theorem 4].

The next result shows that the group ring RG of any finite group G over a semilocal ring R is clean if and only if it is semiperfect. The proof uses the well-known fact that the group ring RG of any finite group G over a semilocal ring R is semilocal.

(4.10) Proposition. Let R be a semilocal ring, and let G be a finite group. Then the group ring

RG is clean if and only if it is semiperfect.

Proof. As we mentioned above, it is well known that the group ring RG of a finite group G over a semilocal ring R is semilocal---see Chin19 [ , Proposition 2.3]. Since a semilocal ring is clean if and only if it is semiperfect according to (2.11), the result follows.

(4.11) Corollary. Let R be a semilocal ring, and let G be a finite abelian group of exponent n.

Then the group ring RG is clean if and only if RCn is clean.

Proof. This follows easily from (4.3) and (4.10). 40

The following example due to Han and Nicholson [29] tells us that the group ring \BbbZ(7) C3 is not clean. We will see in \S 5 that a group ring of the form \BbbZ( p)C3 is clean if and only if p \not \equiv 1 modulo 3.

More generally, we will show that the group ring \BbbZ( p)Cn is clean if and only if p is a primitive root of m, where n = pkm and p does not divide m.

(4.12) Example. \BbbZ(7) C3 is not clean.

Proof. This result follows immediately from (4.10) and (4.7).

The following proposition due to Han and Nicholson [29] tells us that the group ring RG of any locally finite group G over a boolean ring R is clean. This result was extended to a larger class of rings by Chin and Chen [20]. We will look at their result and some other examples of rings R such that the group ring RG of any locally finite group G over the ring R is clean in \S7 .

(4.13) Proposition. Let R be a boolean ring, and let G be a locally finite group. Then the group ring RG is clean.

Proof. This proof is omitted---see Han and Nicholson [29, Proposition 4].

(4.14) Corollary. Let R be a boolean ring, and let G be an abelian group. Then the group ring

RG is clean if and only if G is locally finite.

Proof. The necessity follows immediately from (4.8) since every boolean ring is commutative, and the sufficiency follows from (4.13).

Nicholson [49, Theorem] proved that the group ring RG of a locally finite p-group G is local for any local ring R with p \in J(R). The following result due to Zhou [75] tells us that the group ring

RG of a locally finite p-group G is clean for any clean ring R with p \in J(R).

(4.15) Theorem. Let G be a locally finite p-group, and let R be any ring with p \in J(R). Then the group ring RG is clean if and only if R is clean.

Proof. This proof is omitted---see Zhou [75, Theorem 4]. 41 (4.16) Corollary. Let G be an abelian p-group, and let R be any ring with p \in J(R). Then the group ring RG is clean if and only if R is clean.

Proof. This result follows from (4.15) since every abelian p-group is locally finite.

We will say that a group ring RG satisfies a ring-theoretic property ``locally"" if the group ring

RH satisfies that property for every finitely generated subgroup H of G. For instance, we will say that a group ring RG is clean locally if RH is clean for every finitely generated subgroup H of G.

The following proposition shows that every group ring that is clean locally is clean---this fact was noted by Chen and Zhou [16].

(4.17) Proposition. Every group ring that is clean locally is clean.

Proof. Let RG be any group ring that is clean locally, let \lambda be any element in RG, and let H be the subgroup of G generated by the support of \lambda . Since the support of any element in a group ring is finite by definition, H is a finitely generated subgroup of G. Since RG is clean locally, it follows that the group ring RH is clean. Meanwhile, it is not difficult to see that any element that is clean in some (unitary) subring S of a ring R is also clean in that ring R. Since RH is a clean (unitary) subring of RG containing \lambda , it follows that \lambda is clean in RG. Since this is the case for any element

\lambda \in RG, the group ring RG is clean. Therefore every group ring that is clean locally is clean.

The following example due to Chen and Zhou [16] shows that the converse of (4.17) is not true in general. More specifically, it tells us that the group ring \BbbZ(7) S3 is clean but not clean locally. It is not known to the author whether the converse of (4.17) is true in the commutative case.

(4.18) Example. \BbbZ(7) S3 is clean but not clean locally.

Proof. This follows easily from (4.9) and (4.12) since every local ring is clean by (2.3).

Some Notes

(1) Woods [68, Theorem] proved that a group ring RG is left (resp. right) perfect if and only if the coefficient ring R is left (resp. right) perfect and the underlying group G is finite. In particular, it follows that a group ring RG is left (resp. right) perfect locally if and only if the coefficient ring 42 R is left (resp. right) perfect and the underlying group G is locally finite. Notice that every group ring that is left (resp. right) perfect is left (resp. right) perfect locally, and that the converse is not true in general since a locally finite group need not be finite. For example, the Pr\"ufer p-groups are locally finite but not finite, and hence the group ring RG of any Pr\"ufer p-group G over a left (resp. right) perfect ring R is left (resp. right) perfect locally but not left (resp. right) perfect.

(2) It is not difficult to see that a semiperfect group ring need not be semiperfect locally, and that a group ring that is semiperfect locally need not be semiperfect. For example, the group ring

\BbbZ(7) S3 is semiperfect according to (4.2) but not semiperfect locally as a result of (4.7). Meanwhile, let k be any field of characteristic 0, and let G be a Pr\"ufer p-group. It follows from (4.6) that the group ring kG is semiperfect locally but not semiperfect. We should note that this group ring kG is also perfect locally but not perfect according to Woods [68, Theorem].

(3) We know from \S2 that every left (right) perfect ring is semiperfect, while every semiperfect ring is clean. In particular, this tells us that every group ring that is left (right) perfect locally is semiperfect locally, and that every group ring that is semiperfect locally is clean locally. Moreover, since every left (resp. right) perfect group ring is left (resp. right) perfect locally (as noted above), while every group ring that is clean locally is clean by (4.17), we have the following.

left perfect =\Rightarrow semiperfect =\Rightarrow clean = \Rightarrow \Rightarrow = left perfect locally =\Rightarrow semiperfect locally =\Rightarrow clean locally

It is not difficult to show that these implications are irreversible. For example, we mentioned in \S2 that the ring \BbbZ(2) of all rational numbers with odd denominators (when written in lowest terms) is semiperfect but not perfect, and that the infinite direct product \BbbZ(2) \times \BbbZ(2) \times \cdot is clean but not semiperfect. Moreover, since the coefficient ring R must be perfect in order for a group ring RG to be perfect, it follows that the group ring \BbbZ(2) C2 is not perfect locally even though it is semiperfect locally by (4.2), and since the coefficient ring R must be semiperfect in order for a group ring RG to be semiperfect, the group ring (\BbbZ(2) \times \BbbZ(2) \times \cdot )C2 is not semiperfect locally even though it is clean locally by (4.9). Finally, we noted above that the group ring RG of a Pr\"ufer p-group G over a left (resp. right) perfect ring R is left (resp. right) perfect locally but not left (resp. right) perfect, 43 and we know from (4.18) that the group ring \BbbZ(7) S3 is clean but not clean locally. Therefore each of the implications above is indeed irreversible.

§5 Clean Group Rings, I

In this section we show exactly when the commutative group ring \BbbZ( p)Cn is clean, where p is a prime and n is a positive integer. In particular, we prove that the group ring \BbbZ( p)Cn is clean if and k only if p is a primitive root of m, where n = p m and p \nmid m.

The Group Ring Z(\bfitp )\bfitC \bfitn

We begin by restating the result in (4.6) for the group ring \BbbZ( p)Cn, where p is a prime and n is a positive integer. This result gives a necessary and sufficient condition for the group ring \BbbZ( p)Cn to be clean in terms of the polynomial rings \BbbZ( p)[X] and \BbbFp [X].

k (5.1) Theorem. Let p be a prime, let n be any positive integer, and write n = p m where p \nmid m. m Then \BbbZ( p)Cn is clean if and only if each monic factor of X - 1 in \BbbF p[X] can be lifted to a monic m factor of X - 1 in \BbbZ( p)[X].

Proof. This result follows immediately from (4.6) and (4.10).

We can improve upon the result in (5.1) by looking at the irreducible monic factors of Xn - 1 in the polynomial rings \BbbZ( p)[X] and \BbbF p[X], where p is any prime, and n is any positive integer such that p \nmid n. But first we need some definitions and some well-known results. A polynomial f over a field K is said to be squarefree in K[X] if there exists no nonconstant polynomial g \in K[X] such that g2 divides f. It is well known that a polynomial f is squarefree if f and its derivative f \prime have no nonconstant factor in common. In particular, it follows easily from this fact that Xn - 1 is squarefree over any field whose characteristic does not divide n.

Let K be a field, let n be any positive integer, and consider the polynomials \Phi1 , \Phi2 , \cdot \cdot\in K[X] that satisfy the following relation, where the product is taken over all positive divisors d of n.

n \prod X - 1 = \Phid (X) d| n 44

These polynomials \Phi1 , \Phi2 ,... are called cyclotomic polynomials, and for any positive integer n,

n the nth cyclotomic polynomial \Phin over K can be computed recursively by dividing X - 1 by the product of all \Phid (X) such that d is less than n and is a positive divisor of n. It is well known that each cyclotomic polynomial over a field of characteristic 0 is an irreducible monic polynomial with integer coefficients. Meanwhile, each cyclotomic polynomial over a field of characteristic p > 0 is a monic polynomial with coefficients in \BbbZ/p \BbbZ, but need not be irreducible in general. Let a and n be any relatively prime positive integers. The smallest positive integer \ell such that

\ell a \equiv 1 modulo n is called the multiplicative order of \bfita modulo \bfitn, and is denoted ordn(a). It is well known that a\phi (n) \equiv 1 modulo n for any relatively prime positive integers a and n, where \phi is the Euler phi function. In particular, this tells us that ordn(a) \leq \phi (n). A positive integer a is said to be a primitive root modulo \bfitn if ordn(a) = \phi (n). It is not difficult to see that a is a primitive root of n if and only if for every integer x relatively prime to n there is some integer k \leq \phi (n) such that ak \equiv x modulo n. Moreover, it can be shown that a is a primitive root of n if and only if it is also a primitive root of d for each positive divisor d of n.

(5.2) Lemma. Let p be any prime, and let m be any positive integer such that p \nmid m. Then each m m monic factor of X - 1 in \BbbF p[X] can be lifted to a monic factor of X - 1 in \BbbZ( p)[X] if and only if p is a primitive root of m.

Proof. To avoid any possible confusion, let \Phin denote the nth cyclotomic polynomial over \BbbQ, and let \Psi n denote the nth cyclotomic polynomial over \BbbF p. As we mentioned above, it is well known that each cyclotomic polynomial over a field of characteristic 0 is an irreducible monic polynomial with integer coefficients. In particular, this means that \Phid is an irreducible monic polynomial in \BbbZ( p)[X] m for each positive divisor d of m, and since X - 1 is equal to the product of all \Phid (X) such that d

m divides m, these are precisely the irreducible monic factors of X - 1 in \BbbZ( p)[X]. Meanwhile, it is not difficult to see that \Phin \equiv \Psi n modulo p for any prime p and any positive integer n. Since \Phid is

m an irreducible monic factor of X - 1 in \BbbZ( p)[X] for each positive divisor d of m, this tells us that m \Psi d can be lifted to a monic factor of X - 1 in \BbbZ( p)[X] for each positive divisor d of m. m Suppose that \Psi d is irreducible over \BbbF p for each positive divisor d of m. Since X - 1 is equal to the product of all \Psi d(X) such that d divides m, these are precisely the irreducible monic factors of

m m X - 1 in \BbbF p[X]. Since \Psi d can be lifted to a monic factor of X - 1 in \BbbZ( p)[X] for each positive 45

m divisor d of m, this tells us that each irreducible monic factor of X - 1 over \BbbF p can be lifted to m m a monic factor of X - 1 in \BbbZ( p)[X]. Since each monic factor of X - 1 in \BbbFp [X] is a product of m m irreducible monic factors of X - 1 in \BbbFp [X], it follows that each monic factor of X - 1 in \BbbF p[X] m can be lifted to a monic factor of X - 1 in \BbbZ( p)[X]. Suppose instead that \Psi d is reducible for some m positive divisor d of m. Then X - 1 has more irreducible monic factors over \BbbF p than over \BbbZ( p). As we noted above, it is well known that Xm - 1 is squarefree over any field whose characteristic

m does not divide m. In particular, since p \nmid m by hypothesis, this tells us that X - 1 is squarefree m over \BbbFp , and since X - 1 has more irreducible monic factors over \BbbFp than over \BbbZ( p), it follows m m that not all monic factors of X - 1 in \BbbF p[X] can be lifted to monic factors of X - 1 in \BbbZ( p)[X]. m m Therefore each monic factor of X - 1 in \BbbF p[X] can be lifted to monic factors of X - 1 in \BbbZ( p)[X] if and only if \Psid is irreducible over \BbbFp for each positive divisor d of m.

It now suffices to show that \Psi d is irreducible over \BbbFp for each positive divisor d of m if and only if p is a primitive root of m. Since p \nmid m by hypothesis, \Psi d is irreducible over \BbbF p for each positive divisor d of m if and only if \Psi m is irreducible over \BbbF p according to [57, Theorem 11.2.6], and \Psim is irreducible over \BbbF p if and only if p is a primitive root of m according to [57, Theorem 11.2.10]. This completes the proof.

k (5.3) Theorem. Let p be a prime, let n be any positive integer, and write n = p m where p \nmid m.

Then \BbbZ( p)Cn is clean if and only if p is a primitive root of m.

Proof. This result follows immediately from (5.1) and (5.2).

It is known that \BbbZ( p)C3 is quasiclean [14], semiclean [71], and \Sigma -clean [70] for any prime p, and is 2-clean [70] for any prime p \not = 2. The following result tells us exactly when \BbbZ( p)C3 is clean.

In particular, it shows that the group ring \BbbZ( p)C3 is clean if and only if p \not \equiv 1 modulo 3.

(5.4) Example. Let p be any prime. Then \BbbZ( p)C3 is clean if and only if p \not \equiv 1 modulo 3.

Proof. Since every positive integer is a primitive root of 1, it follows immediately from (5.3) that the group ring \BbbZ( p)C3 is clean if and only if p is equal to 3 or is a primitive root of 3. In particular, this means that when p \not = 3, the group ring \BbbZ( p)C3 is clean if and only if ord3(p) = \phi (3) = 2. Since ord3(p) = 1 when p \equiv 1 modulo 3, and ord3(p) = 2 when p \equiv 2 modulo 3, the result follows. 46

The Group Ring Z(\bfitp )\bfitG

We can now show exactly when a commutative group ring \BbbZ( p)G is semiperfect.

(5.5) Theorem. Let p be a prime, and let G be an abelian group with p-primary component Gp.

Then the group ring \BbbZ( p)G is semiperfect if and only if the quotient group G/Gp is a finite group of exponent n and p is a primitive root of n.

Proof. Since G is an abelian group, we know from (4.5) that the group ring \BbbZ( p)G is semiperfect if and only if the quotient group G/Gp is a finite group of exponent n and the group ring \BbbZ( p)Cn is semiperfect. Since the group ring \BbbZ( p)Cn is semiperfect if and only if it is clean according to (4.10), the result now follows from (5.3).

(5.6) Example. Let p and q be any primes. Then \BbbZ( p)Cq\infty is semiperfect if and only if p = q.

Proof. This result follows immediately from (5.5) since Cq\infty is an infinite abelian q-group.

The next result tells us that when the quotient group G/Gp is a finite group of exponent n, the commutative group ring \BbbZ( p)G is clean if and only if p is a primitive root of n. The proof uses the fact due to Chin [19, Theorem 1] that the group ring RG of an abelian group G over any semilocal ring R with char R/J(R) = p > 0 is semilocal when the quotient group G/Gp is finite.

(5.7) Corollary. Let p be a prime, and let G be an abelian group with p-primary component Gp.

If the quotient group G/Gp is a finite group of exponent n, then \BbbZ( p)G is clean if and only if p is a primitive root of n.

Proof. Suppose that G/Gp is a finite group of exponent n. As we mentioned above, it is known that the group ring RG of an abelian group G over a semilocal ring R with char R/J(R) = p > 0 is semilocal when the quotient group G/Gp is finite---see Chin19 [ , Theorem 1]. Since \BbbZ( p) is local, this tells us that the group ring \BbbZ( p)G is semilocal. Since a semilocal ring is clean if and only if it is semiperfect according to (2.11), the result now follows from (5.5).

We know from (4.4) that the quotient group G/Gp must be finite in order for the commutative group ring \BbbZ( p)G to be semiperfect. Since every group ring that is clean locally is clean by (4.17), 47 the following example tell us that G/Gp need not be finite in order for \BbbZ( p)G to be clean.

(5.8) Example. Let p be any prime, and let q be an odd prime.

(1) If p = q, then \BbbZ( p)Cq\infty is clean. 2 (2) If p \not = q, then \BbbZ( p)Cq\infty is clean locally if and only if p is a primitive root of q .

Proof. (1) This result follows immediately from (4.15).

(2) Suppose that the group ring \BbbZ( p)Cq\infty is clean locally for some p \not = q. Then \BbbZ( p)H must be clean for every finitely generated subgroup H of Cq\infty . Since the Pr\"ufergroup Cq\infty has a subgroup

2 isomorphic to Cq2 , it follows from (5.3) that p is a primitive root of q . This proves the necessity. Conversely, suppose that p is a primitive root of q2. It is well known that for any odd prime q, every primitive root of q2 is a primitive root of qk for every positive integer k. In particular, since q is an odd prime by hypothesis, and since each subgroup of the Pr\"ufergroup Cq\infty is isomorphic to

Cqk for some positive integer k, it follows from (5.3) that group ring \BbbZ( p)Cq\infty is clean locally. This proves the sufficiency.

(5.9) Example. Let p be a prime. Then \BbbZ( p)C2\infty is clean locally if and only if p = 2.

Proof. Suppose that the group ring \BbbZ( p)C2\infty is clean locally. Then \BbbZ( p)H must be clean for every finitely generated subgroup H of C2\infty . Since the Pr\"ufergroup C2\infty has a subgroup isomorphic to

C2k for every positive integer k, it follows from (5.3) that p is either equal to 2 or a primitive root of 2k for every positive integer k. Since the integer 2k has no primitive roots for any k > 2 by the

Primitive Root Theorem, it follows that p = 2. This proves the necessity.

Conversely, suppose that p = 2. Since each subgroup of the Pr\"ufer group C2\infty is isomorphic to

C2k for some positive integer k, it follows from (5.3) that group ring \BbbZ( p)C2\infty is clean locally. This proves the sufficiency.

Two More Results

Since every odd prime is a primitive root of 2, it follows immediately from (5.3) that the group ring \BbbZ( p)C2 is clean for every prime p. The next result shows that no other nontrivial cyclic group has this property. This result was proved independently by Chen and Zhou [16, Example 3.6] and by the author (unpublished). 48 (5.10) Proposition. Let n be an integer. If n > 2, then there are infinitely many primes p such that the group ring \BbbZ( p)Cn is not clean.

Proof. A classical result in number theory due to Dirichlet (c. 1837) tells us that for any relatively prime positive integers a and b, the arithmetic progression a, a + b, a + 2b, . . . contains infinitely many prime numbers. In other words, for any relatively prime positive integers a and b, there are infinitely many primes p that are congruent to a modulo b. In particular, for any positive integer n, there are infinitely many primes p that are congruent to 1 modulo n. Since any such prime p is not a primitive root of n for any integer n > 2, the result now follows from (5.3).

It is well known that a positive integer has a primitive root if and only if it is equal to 1, 2, 4, q\ell, or 2q\ell for some odd prime q and some positive integer \ell . This result is due to Gauss (c. 1801) and is known as the Primitive Root Theorem---the interested reader is encouraged to see Guichard28 [ ] for an algebraic proof of this number-theoretic result. The following proposition tells us that when n is equal to 1, 2, 4, q\ell, or 2q\ell for some odd prime q and some positive integer \ell , there are infinitely many primes p such that the group ring \BbbZ( p)Cn is clean. It can be shown that for any other positive integer n, there are at most two primes p such that the group ring \BbbZ( p)Cn is clean.

(5.11) Proposition. Let n be an integer. If n = 1, 2, 4, q\ell, or 2q\ell for some odd prime q and some positive integer \ell , then there are infinitely many primes p such that the group ring \BbbZ( p)Cn is clean.

Proof. Suppose that n = 1, 2, 4, q\ell, or 2q\ell for some odd prime q and some positive integer \ell. Then n has a primitive root according to the Primitive Root Theorem. Since any primitive root r of n is relatively prime to n, it follows from Dirichlet's theorem on primes in arithmetic progressions that there are infinitely many primes p that are congruent to r modulo n. Since any such prime is also a primitive root of n, the result now follows from (5.3).

Some Notes

k (1) Let p be any prime, let n be any positive integer, and write n = p m where p \nmid m. We can use the free computational knowledge engine Wolfram| Alpha to determine whether p is a primitive root of m. The following query will return True if p is a primitive root of m, and False otherwise. 49 MultiplicativeOrder[p,m] = EulerPhi[m]

In particular, since the group ring \BbbZ( p)Cn is clean if and only if p is a primitive root of m by (5.3), a result of True implies that \BbbZ( p)Cn is clean, and a result of False implies that \BbbZ( p)Cn is not clean. We should note that these conclusions are only valid when p is a prime, n is a positive integer, and

k n = p m where p \nmid m, as described above.

(2) Let p be any prime, and let n be a positive integer. The following query in Wolfram|Alpha will return True if the group ring \BbbZ( p)Cn is clean, and False otherwise.

Select[Divisors[n], MultiplicativeOrder[p, \#] = EulerPhi[\#] \&] = Select[Divisors[n], !Divisible[\#, p] \&]

More specifically, this query will return True if p is a primitive root of each positive divisor d of n such that p \nmid d, and False otherwise. It is not difficult to see that the group ring \BbbZ( p)Cn is clean if and only if this condition is satisfied. This query gives us a way to determine whether the group

k ring \BbbZ( p)Cn is clean without factoring n into the form n = p m where p \nmid m. We should note that the conclusions drawn from this query are only valid if p is a prime, as described above.

k (3) Let p be a prime, let n be any positive integer, and write n = p m where p \nmid m. Then the group ring \BbbZ( p)Cn is clean if and only if p is a primitive root of m according to (5.3). Meanwhile, we know from the Primitive Root Theorem that m has a primitive root if and only if it is equal to

\ell \ell 1, 2, 4, q , or 2q for some odd prime q and positive integer \ell . This tells us that in order for \BbbZ( p)Cn to be clean, n must be equal to 1, 2, 4, pk, q\ell, 2pk, 2q\ell, 4pk, pkq\ell, or 2pkq\ell for some odd prime q and some positive integers k and \ell .

k (4) Let p be a prime, let n be any positive integer, and write n = p m where p \nmid m. It follows immediately from the previous note that if n is divisible by three odd primes or is divisible by two odd primes and 4, then the group ring \BbbZ( p)Cn is not clean for any prime p. The following query in Wolfram| Alpha will return a list of the positive integers n < 1000 such that n is divisible by three odd primes or is divisible by two odd primes and 4.

Select[Range[1000], PrimeNu[\#] Greater 3 | | (PrimeNu[\#] = 3 \&\& (OddQ[\#] | | Divisible[\#,4])) \&]

As it turns out, these are not the only positive integers n such that group ring \BbbZ( p)Cn is not clean for any prime p. However, these are the only positive integers that have this property based solely 50 on the structure of their prime decomposition. We should note that this sequence is not currently listed in the On-Line Encyclopedia of Integer Sequences (OEIS).

(5) As noted above, the integers that are divisible by three odd primes or that are divisible by two odd primes and 4 are not the only integers n such that the group ring \BbbZ( p)Cn is not clean for any prime p. For example, it can be shown that the group ring \BbbZ( p)Cn is not clean for any prime p when n = 39, 55, 56, 68, 78, 95, 110, etc., even though none of these integers n is divisible by three odd primes or is divisible by two odd primes and 4. In particular, this result follows from (5.3) and the Primitive Root Theorem.

(6) Let p be any prime, and let q be an odd prime. It is well known that for any odd prime q, any primitive root of q2 is a primitive root of q\ell for every positive integer \ell, and any odd primitive root of q2 is also a primitive root of 2q\ell for every positive integer \ell. It now follows from (5.3) that

2 k \ell k \ell when p is a primitive root of q , the group ring \BbbZ( p)Cn is clean for any n equal to p q or 2p q for some nonnegative integers k and \ell . For example, since 3 is a primitive root of 25, this tells us that the group ring \BbbZ(3) Cn is clean for n = 2, 3, 5, 6, 9, 10, 15, 18, 25, 27, 30, 45, 50, 54, 75, etc. Of course these are not the only positive integers n such that the group ring \BbbZ(3) Cn is clean.

§6 Clean Group Rings, II

In this section we consider the problem of classifying the groups G whose group rings RG are clean if and only if the coefficient ring R is clean. We know from (4.8) that the coefficient ring R must be clean in order for the group ring RG to be clean. The problem is to determine the groups

G whose group rings RG are clean for every clean ring R. This seems to be a difficult problem in general. In fact, it is not even known if there is a nontrivial group that has this property. However, several partial solutions to this problem are known. For example, McGovern [46] proved that the group ring RC2 is clean for every commutative clean ring R, and we know from (5.10) that no other nontrivial cyclic group has this property. Together these results classify the cyclic groups G whose group rings RG are clean for every commutative clean ring R. More generally, we can classify the reduced abelian groups G that have this property. 51

Abelian Groups

We begin with the problem of determining the groups G whose group rings RG are semiperfect for every semiperfect ring R. In particular, the following result classifies the abelian groups G that have this property. The first part of this result is essentially due to Woods69 [ ].

(6.1) Proposition. Let R be any ring, and let G be a finite elementary abelian 2-group.

(1) The group ring RG is semiperfect if and only if R is semiperfect.

(2) No other abelian group has this property.

Proof. (1) Let k be an arbitrary positive integer. Since every finite elementary abelian 2-group is

k isomorphic to a finite direct product of copies of C2, it suffices to show that the group ring RC2 k \sim k 1- is semiperfect if and only if R is semiperfect. Since RC2 = (RC2 )C2, it follows from (4.2) and

k induction that the group ring RC2 is semiperfect if and only if R is semiperfect. Therefore RG is semiperfect if and only if R is semiperfect.

(2) Let G be any nontrivial abelian group that has the property in (1). Since the group ring

RG is semiperfect whenever the coefficient ring R is semiperfect, it follows from (4.4) that G/Gp must be finite for every prime p. This forces G to be finite, which means that G is a finite abelian group of exponent n for some positive integer n. It now follows from (4.3) that the group ring RG is semiperfect if and only if RCn is semiperfect. Since no cyclic group of order greater than 2 has the property in (1) as a result of (5.10), this tells us that G must have exponent 2. Therefore G is a finite elementary abelian 2-group.

(6.2) Corollary. Let R be any ring, and let G be an elementary abelian 2-group.

(1) The group ring RG is semiperfect locally if and only if R is semiperfect.

(2) No other group has this property.

Proof. (1) This follows immediately from (6.1) since every finitely generated subgroup of G is a finite elementary abelian 2-group.

(2) Let G be any group that has the property in (1). Then for any element g \in G, the group ring R\langleg \rangle is semiperfect whenever R is semiperfect. Since no cyclic group of order greater than 2 52 has this property according to (5.10), every nontrivial element g \in G must be of order 2. Therefore

G is an elementary abelian 2-group.

The following lemma generalizes the result in (5.10).

(6.3) Lemma. Let G be any group with a cyclic direct summand of order greater than 2. Then there are infinitely many primes p such that the group ring \BbbZ( p)G is not clean.

Proof. Let H be any cyclic direct summand of G that has order greater than 2. Then the group ring \BbbZ( p)H is a homomorphic image of \BbbZ( p)G for any prime p. Since every homomorphic image of a clean ring is clean by (1.1), it suffices to show that there are infinitely many primes p such that the group ring \BbbZ( p)H is not clean. In particular, the result now follows from (5.10).

An abelian group G is called divisible if for every positive integer n and every element g \in G, there is some x \in G such that xn = g. An abelian group G is called reduced if it has no nontrivial divisible subgroups. The following technical lemma shows that any reduced torsion abelian group that has no cyclic direct summands of order greater than 2 is an elementary abelian 2-group. The proof uses several well-known facts about abelian groups that can be found in [36].

(6.4) Lemma. Let G be any reduced torsion abelian group that has no cyclic direct summands of order greater than 2. Then G is an elementary abelian 2-group.

Proof. It is well known that every torsion abelian group is a direct sum of primary groups---for a proof, see [36, Theorem 1]. Since G is a torsion abelian group, it is a direct sum of primary groups.

Moreover, since G is a reduced torsion abelian group, each p-primary component of G is a reduced torsion abelian group. It is known that every reduced torsion abelian group has a nontrivial cyclic direct summand---see36 [ , Theorem 9]. Since the group G has no cyclic direct summands of order greater than 2 by hypothesis, this tells us that G must be a 2-group. It can be shown that every reduced primary abelian group of infinite exponent has cyclic direct summands of arbitrarily high orders---see36 [ , Exercise 27]. Since G is a primary abelian group with no cyclic direct summands of order greater than 2, this tells us that G must have finite exponent. It is well known that every abelian group of finite exponent is a direct sum of cyclic groups---see36 [ , Theorem 6]. Since G is 53 a 2-group with no cyclic direct summands of order greater than 2, this tells us that G is a direct sum of copies of C2. Therefore G is an elementary abelian 2-group.

The following result classifies the reduced abelian groups G whose group rings RG are clean for every commutative clean ring R. The first part of this result is due to McGovern46 [ ].

(6.5) Proposition. Let R be any ring, and let G be an elementary abelian 2-group.

(1) If R is commutative, then the group ring RG is clean if and only if R is clean.

(2) No other reduced abelian group has this property.

Proof. (1) This proof is omitted---see McGovern46 [ , Corollary 3.14].

(2) Let G be any reduced abelian group that has the property in (1). Then the group ring RG is clean whenever the coefficient ring R is a commutative clean ring. It follows from (4.8) that the group G must be torsion, and it follows from (6.3) that G has no cyclic direct summands of order greater than 2. Therefore G is a reduced torsion abelian group with no cyclic direct summands of order greater than 2, and the result now follows from (6.4).

(6.6) Corollary. Let R be any ring, and let G be an elementary abelian 2-group.

(1) If R is commutative, then the group ring RG is clean locally if and only if R is clean.

(2) No other group has this property.

Proof. This proof is similar to the proof of (6.2) and should be omitted.

A ring R is called semiabelian if its identity 1 can be written as a finite sum 1= e1 + \cdot \cdot + en of mutually orthogonal idempotents ei such that each corner ring eiRei is abelian---this definition is due to Zhou [75]. Examples of semiabelian rings include all abelian rings and semiperfect rings.

The next result classifies the reduced abelian groups G whose group rings RG are clean for every semiabelian clean ring R. The first part of this result is due to Zhou[75].

(6.7) Proposition. Let R be any ring, and let G be an elementary abelian 2-group.

(1) If R is semiabelian, then the group ring RG is clean if and only if R is clean.

(2) No other reduced abelian group has this property. 54 Proof. (1) This proof is omitted---see Zhou75 [ , Corollary 22].

(2) Let G be any reduced abelian group that has the property in (1). Then the group ring RG is clean whenever the coefficient ring R is a semiabelian clean ring. In particular, this tells us that

RG is clean for every commutative clean ring R. The result now follows from (6.5)(2).

(6.8) Corollary. Let R be any ring, and let G be an elementary abelian 2-group.

(1) If R is semiabelian, then the group ring RG is clean locally if and only if R is clean.

(2) No other group has this property.

Proof. This proof is similar to the proof of (6.2) and should be omitted.

The following result classifies the reduced abelian groups G whose group rings RG are clean for every clean ring R in which 2 is \pi -regular. Examples of clean rings in which 2 is \pi -regular include all unit regular rings and strongly \pi -regular rings. The first part of this result is due to Zhou[75].

(6.9) Proposition. Let R be any ring, and let G be an elementary abelian 2-group.

(1) If 2 is \pi -regular in R, then the group ring RG is clean if and only if R is clean.

(2) No other reduced abelian group has this property.

Proof. (1) This proof is omitted---see Zhou75 [ , Theorem 8].

(2) Let G be any reduced abelian group that has the property in (1). Then the group ring RG is clean whenever the coefficient ring R is a clean ring in which 2 is \pi -regular. Since 2 is \pi -regular in \BbbZ( p) for any prime p > 2, this tells us that the group ring \BbbZ( p)G is clean for every prime p > 2. It follows from (4.8) that G must be torsion, and it follows from (6.3) that G has no cyclic direct summands of order greater than 2. Therefore G is a reduced torsion abelian group with no cyclic direct summands of order greater than 2, and the result now follows from (6.4).

(6.10) Corollary. Let R be any ring, and let G be an elementary abelian 2-group.

(1) If 2 is \pi -regular in R, then RG is clean locally if and only if R is clean.

(2) No other group has this property.

Proof. This proof is similar to the proof of (6.2) and should be omitted. 55

Nonabelian Groups

The following result gives us some examples of nonabelian groups G whose group rings RG are semiperfect for every semiperfect ring R. No other nonabelian groups that have this property are known to the author. This result is essentially due to Woods [69].

(6.11) Proposition. Let R be any ring and let G = H \times K where H is a finite direct product of copies of S3 and K is a finite elementary abelian 2-group. Then the group ring RG is semiperfect if and only if R is semiperfect.

Proof. Since G = H \times K, we know that RG =\sim (RH)K. Moreover, since K is a finite elementary abelian 2-group, it now follows from (6.1) that the group ring RG is semiperfect if and only if the ring RH is semiperfect. Let k be an arbitrary positive integer. Since the group H is a finite direct

k product of copies of S3, it suffices to show that the group ring RS3 is semiperfect if and only if R k \sim k - 1 is semiperfect. Since RS3 = (RS3 )S3, it follows from (4.2) and induction that the group ring

k RS3 is semiperfect if and only if R is semiperfect. Therefore the group ring RG is semiperfect if and only if R is semiperfect.

The next two results give us some examples of nonabelian groups G whose group rings RG are clean for every clean ring R that is semiabelian or in which 2 and 3 are both \pi -regular. No other nonabelian groups that have either of these properties are known to the author. These two results are due to Zhou [75].

(6.12) Proposition. Let R be any ring and let G = H \times K where H is a finite direct product of copies of S3 and K is an elementary abelian 2-group. If R is semiabelian, then the group ring RG is clean if and only if R is clean.

Proof. This proof is omitted---see Zhou [75, Corollary 22].

(6.13) Proposition. Let R be any ring and let G = H \times K where H is a finite direct product of copies of S3 and K is an elementary abelian 2-group. If 2 and 3 are both \pi -regular in R, then the group ring RG is clean if and only if R is clean. 56 Proof. This proof is omitted---see Zhou [75, Theorem 10].

§7 Clean Group Rings, III

In this section we look at some examples of rings R such that the group ring RG of any locally finite group G over the ring R is clean. Every boolean ring has this property as a result of (4.13).

More generally, Chin and Chen [20, Theorem 3.3] showed that every ring that has artinian prime factors has this property. We prove that every henselian ring also has this property. In particular, we prove that the group ring RG of any locally finite group G over a henselian ring R is clean. It is not known in general whether the underlying group G must be locally finite in order for a group ring RG to be clean, but G must be torsion abelian in order for a commutative group ring RG to be clean according to (4.8).

Three Examples

A ring R is said to have artinian prime factors if the quotient ring R/P is artinian for every prime ideal P of R. This class of rings includes all left (right) perfect rings, strongly regular rings, and zero-dimensional commutative rings. Meanwhile, Fisher and Snider [25, Theorem 2.1] showed that a ring R is strongly \pi -regular if and only if R/P is strongly \pi -regular for every prime ideal P of R. Since every artinian ring is strongly \pi -regular, it follows that every ring with artinian prime factors is strongly \pi -regular. The converse is true for commutative rings since every commutative strongly \pi -regular ring is zero-dimensional. In particular, this means that a commutative ring has artinian prime factors if and only if it is zero-dimensional. Meanwhile, the following result due to

Chin and Chen [20] tells us that the group ring RG of any locally finite group G over a ring R that has artinian prime factors is strongly \pi -regular, and therefore is clean as a result of (2.15).

(7.1) Proposition. Let R be a ring that has artinian prime factors, and let G be a locally finite group. Then the group ring RG is strongly \pi -regular.

Proof. This proof is omitted---see Chin and Chen20 [ , Theorem 3.3].

(7.2) Corollary. Let R be a zero-dimensional commutative ring, and let G be an abelian group. 57 Then the group ring RG is clean if and only if G is locally finite.

Proof. The necessity follows immediately from (4.8), and the sufficiency follows from (7.1).

For any ring R, and for any polynomial f in R[X], let f\= denote the image of f in R[X], where

R denotes the quotient ring R/J(R). Following Azumaya [3], we will say that the Hensel lemma

holds for \bfitf if for any given relatively prime polynomials g0, h0 \in R[X] such that g0h0 = f\= with g0

monic, there exist g, h \in R[X] such thatg \= = g0, h\= = h0, and gh = f with g monic. A commutative local ring R is said to be henselian if the Hensel lemma holds for each monic polynomial in R[X].

This class of rings includes all fields and complete local rings. Meanwhile, we prove that the group

ring RG of any locally finite group G over a henselian ring R is clean.

(7.3) Proposition. Let R be a henselian ring, and let G be a locally finite group. Then the group

ring RG is clean.

Proof. As we mentioned in \S 4, it is well known that the group ring RG of a finite group G over a

semilocal ring R is semilocal---see Chin19 [ , Proposition 2.3]. Meanwhile, idempotents lift modulo

the Jacobson radical of the group ring RG of a finite group G over a henselian ring R according to

Azumaya [3, Theorem 24]. Since every henselian ring is semilocal, this tells us that the group ring

RG of any locally finite group G over a henselian ring R is semiperfect locally. Since every group

ring that is semiperfect locally is clean according to (2.10) and (4.17), the result follows.

(7.4) Corollary. Let R be a henselian ring, and let G be an abelian group. Then the group ring

RG is clean if and only if G is locally finite.

Proof. The necessity follows immediately from (4.8), and the sufficiency follows from (7.3).

The following result due to Camillo et al. [10] tells us that the group ring RG of a locally finite

group G over a left (right) pure-injective ring R is clean. This result is included for completeness,

and the definition of pure-injective need not concern us here. However, we should note that such

rings are sometimes said to be algebraically compact.

(7.5) Proposition. Let R be a left (right) pure-injective ring, and let G be a locally finite group.

Then the group ring RG is clean. 58 Proof. It is well known that a group ring RG is left (resp. right) pure-injective if and only if the coefficient ring R is left (resp. right) pure-injective and the underlying group G is finite---this result is due to Zimmermann [76, Theorem 1]. In particular, since every left (right) pure-injective ring is clean according to Camillo et al. [10, Corollary 3.13], the result now follows from (4.17).

(7.6) Corollary. Let R be any commutative pure-injective ring, and let G be an abelian group.

Then the group ring RG is clean if and only if G is locally finite.

Proof. The necessity follows immediately from (4.8), and the sufficiency follows from (7.5).

In addition to these three examples of rings R such that the group ring RG of any locally finite group G over the ring R is clean, we should mentioned the following result due to Zhou [75], which gives a necessary and sufficient condition for a ring R to have this property. More specifically, this result tells us that the group ring RG of a locally finite group G over a ring R is clean if and only if SG is clean for every indecomposable image S of R. This result does not lead us to any further examples of rings R such that the group ring RG is clean for every locally finite group G.

(7.7) Proposition. Let R be any ring, and let G be a locally finite group. Then the group ring

RG is clean if and only if SG is clean for every indecomposable image S of R.

Proof. This proof is omitted---see Zhou [75, Proposition 17].

Some Notes

(1) As we mentioned above, it is not known if the underlying group G must be locally finite in order for a group ring RG to be clean. Moreover, it is not known if the underlying group must be locally finite in order for a group ring to be semiperfect. Burgess [6, p. 648] claimed that the group

G need not be locally finite in order for a group ring RG to be semiperfect. Since every semiperfect ring is clean according to (2.10), this would mean that the underlying group G need not be locally finite in order for a group ring RG to be clean. Burgess gave an example to support his claim, but his proof is known to be flawed---see Josephy34 [ , pp. 22--23].

(2) Azumaya [4, Theorem 5] proved that a ring of bounded index (of nilpotency) is \pi -regular if 59 and only if it is strongly \pi -regular. It follows from Hirano [31, Theorem 2] that a ring of bounded index is strongly \pi -regular if and only if it has artinian prime factors. Since every zero-dimensional commutative ring has artinian prime factors, and every ring with artinian prime factors is strongly

\pi -regular, we have the following relationships between these classes of rings.

zero-dimensional commutative =\Rightarrow artinian prime factors =\Rightarrow strongly \pi -regular

strongly \pi -regular =\Rightarrow with bounded index

Yu [74, Example 2.3] gave an example of a noncommutative ring that is not of bounded index but has artinian prime factors. In particular, this tells us that both of the diagonal implications above are irreversible. It is not known to the author whether the other implication above is irreversible, but it can be shown that a zero-dimensional commutative ring need not be of bounded index, and that a strongly \pi -regular ring of bounded index need not be commutative.

(3) It is known that the ring of p-adic integers \BbbZp is henselian but not zero-dimensional for any prime p. Since a commutative ring has artinian prime factors if and only if it is zero-dimensional, it follows that a henselian ring need not have artinian prime factors. Since every henselian ring is a commutative local ring by definition, we also know that a ring with artinian prime factors need not be henselian. For example, notice that the direct product of any two noncommutative division rings is a noncommutative, nonlocal ring with artinian prime factors.

(4) It is not difficult to see that the commutative local ring \BbbZ( p) is neither zero-dimensional nor henselian for any prime p. In particular, this follows immediately from (7.2) and (7.4) since there are infinitely many positive integers n such that the group ring \BbbZ( p)Cn is not clean for any prime p as a result of (5.3) and the Primitive Root Theorem. For a more direct proof of the fact that \BbbZ( p) is neither zero-dimensional nor henselian for any prime p, notice that it is an integral domain but not a field for any prime p, and notice that the Hensel lemma does not hold for X2 + X + p in the polynomial ring \BbbZ( p)[X] for any prime p. 60

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INDEX

\pi -regular ring, 15 perfect locally, 41 perfect ring, 11 abelian ring, 7 Pierce stalk, 7 artinian prime factors, 56 potent element, 33 artinian ring, 10 potent ring, 4 primitive idempotent, 5 bounded index (of nilpotency), 7 primitive root modulo n, 44 pure-injective ring, 57 clean element, 1 clean locally, 41 quasi-duo ring, 7 clean ring, 1 quasinormal ring, 7, 34 complete set of orthogonal idempotents, 2 quasiregular element, 5 cyclotomic polynomial, 44 reduced abelian group, 52 Dedekind finite, 34 reduced ring, 15 divisible abelian group, 52 regular element, 14 division idempotent, 34 regular ring, 14 root clean ring, 7 exchange property, 3 exchange ring, 3 semiabelian ring, 53 exponent of a group, 37 semilocal ring, 12 semiperfect locally, 41 full idempotent, 6 semiperfect ring, 12 semipotent ring, 4 Hensel lemma holds for f, 57 semiprimary ring, 10 henselian ring, 57 semiprime ring, 26 idempotent stable range one, 20, 34 semiregular ring, 18 idempotents lift modulo I, 2 semisimple ring, 10 indecomposable ring, 14 similar matrices, 26 squarefree polynomial, 43 local idempotent, 5 stable range one, 19, 34 strongly \pi -regular ring, 15 matrix units, 2 strongly clean ring, 33 Morita context, 27 strongly regular ring, 14 Morita invariant, 6 multiplicative order modulo n, 44 T -nilpotent, 11 trivial extension, 35 NJ-ring, 29 noetherian ring, 17 unit regular element, 14 unit regular ring, 14 optimally clean ring, 33 units lift modulo I, 3 orthogonally finite, 12 zero-dimensional ring, 16 pairing, 27 Zorn ring, 15, 18