37? A 8 / J

jNO. 1703

IT-REGULAR RINGS

DISSERTATION

Presented to the Graduate Council of the

University of North Texas in Partial

Fulfillment of the Requirements

For the Degree of

DOCTOR OF PHILOSOPHY

By

Ayman R. Badawi, B.S., M.S.

Denton, Texas

May, 1993

37? A 8 / J

jNO. 1703

IT-REGULAR RINGS

DISSERTATION

Presented to the Graduate Council of the

University of North Texas in Partial

Fulfillment of the Requirements

For the Degree of

DOCTOR OF PHILOSOPHY

By

Ayman R. Badawi, B.S., M.S.

Denton, Texas

May, 1993

Badawi, Ayman R., IT-Regular Rings. Doctor of Philosophy

(Mathematics), May, 1993, 47 pp., bibliography, 29 titles.

All rings studied in this dissertation are associative

with 1*0. A ring R is called semicommutative if for every

a,beR, there exist h,feR such that ab=ha and ba=af. A ring R

is called regular if for every xeR there exists yeR such

that xyx=x. Where R is called 7r-regular if for every xeR

there exists n>l such that xn is regular.

The dissertation focuses on the structure of w-regular

(regular) rings. In chapter two we show that a semicommutat-

ive ring R is ir-regular if and only if R/Nil (R) is regular,

where Nil(R) is the two-sided ideal of all nilpotent

elements in R. In chapter three we give a necessary and

sufficient condition on a ring R so that it is 7 r - r e g u l a r . In

chapter four we state a necessary and sufficient condition

on a 7r-regular ring R so that it is semicommutative. In

chapter five we show that R-{0} is a union of disjoint

multiplicative groups if and only if R is semicommutative

regular. In chapter six under the assumption that 2 is a

unit in a ring R, we discover that the idempotents of R are

in the center of R if and only if the set of units

{ xeR : x2=l } is a multiplicative subgroup in R and

consequently we show that a regular ring R is commutative if

and only if the set of units of R is commutative.

In the last chapter we give a short proof of a well-known

result on matrices and we state its relation with 7r-regular

rings.

ACKNOWLEDGMENTS

I wish to thank my unselfish advisor Prof. Nick Vaughan,

my mother Fouziya, my brothers Basheer, Latif, Kalid, Hasan,

my sister Iman, for their support throughout my education,

and last but not least my Father Rateb who died when I was

ten years old.

111

TABLE OF CONTENTS

Chapter Page

I. INTRODUCTION 1

II. SEMICOMMUTATIVE 7T-REGULAR RINGS 4

III. CONDITIONS EQUIVALENT TO SEMICOMMUTATIVE

TT-REGULAR ELEMENTS 12

IV. WHEN A IT-REGULAR RING IS SEMICOMMUTATIVE 19

V. PARTITION RINGS 24

VI. A RELATION BETWEEN THE IDEMPOTENTS AND

A CLASS OF UNITS IN A RING 30

VII. Rn CONTAINS A DIVISION RING IF AND ONLY IF

R DOES

BIBLIOGRAPHY

xv

CHAPTER I

INTRODUCTION.

The subject of Von Neumann regular rings is a portion of

noncommutative ring theory which was originally introduced

by Von Neumann [4] in order to clarify certain aspects of

operator algebras. Much of the impetus behind the develop-

ment of regular rings is due to this and a number of other

connections with functional analysis. As would be expected

with any good concept, regular rings have also been exten-

sively studied for their own sake, and most ring theorists

are at least aware of the connections between regular rings

and the rings they happen to be interested in. For further

reading and references on the subject see [1].

Throughout this chapter let R be an associative ring

with 1. An element xeR is called regular if for some yeR, we

have xyx=x. If every element in R is regular, then R is

called regular. McCoy [3] generalized the concept of regular

rings to w-regular rings. An element xeR is called 7r-regular

if for some n>l we have xn is regular. If every element in R

is 7r-regular, then R is called 7r-regular. On the other hand,

Kaplansky and Arnes [2] introduced the concept, strongly

regular ring. A ring R is called strongly regular if for

every xeR, there exists yeR such that x2y=x. In the second

chapter we introduce the concept, semicommutative ring. In

1

2

the second chapter we give a characterization of all

semicommutative 7r-regular rings. In the third chapter, we

concentrate on the elements of a ring R. We state eguivalent

conditions to 7r-regular ( regular ) elements in R. In the

fourth chapter we state a necessary and sufficient condition

such that a 7r-regular ring R be semicommutative. In the

fifth chapter we introduce the concept, partition ring: a

ring R is called a partition ring if R-{0} is a union of

disjoint multiplicative groups. We show that a ring R is a

partition ring if and only if it is semicommutative regular

if and only if it is strongly regular. In the sixth chapter

we study a relation between the idempotents of a ring R and

the set { xeR : x2=l }, and consequently we show that a

regular ring R is commutative if and only if the set of unit

in R is commutative. In the last chapter, we give an

alternative proof of a well-known result on matrices and we

state a connection between this result and 7r-regular rings.

REFERENCES

1. Goodearl, K. R., Von Neumann Regular Rings. London,

Pitman publishing, Inc. 1979.

2. Kaplansky, I. and Arnes, R.F., " Topological Representa

tion of Algebra," Trans. American Math. Soc.. 63

(1948), 457-481.

3. McCoy, N., H., " Generalized Regular Rings," Bull. Amer.

Math. Soc.. 45 (1939), 175-178.

4. Von Neumann, J., " On Regular Rings," Proc. Nat. Acad.

SCI. U.S.A.. 22 ( 1936), 16-22.

CHAPTER II

SEMI COMMUTATIVE 7T-REGULAR RINGS

Let K be a ring with 1 and aeK. If for every beK there

exist r,heK such that ab=ra and ba=ah, then a is called

semicommutative. If every element of K is semicommutative,

then K is called a semicommutative ring. Observe that every

commutative ring is semicommutative. Let S be a direct sum

of two noncommutative division rings. Then S is an example

of a semicommutative ring which is not commutative. The ring

K is called jr-regular [1] if and only if for every xeK there

exist n>l and yeK such that xnyxn=xn and unit 7r-regular ring

if for every xeK there exist m>l and a unit u of K such that

xmuxm=xm.

Let R be a semicommutative ring. In this chapter, we

show that all ^-regular rings are unit 7r-regular rings and

the set Nil(R) of all nilpotent elements of R is a two-sided

ideal of R. We show that R is a 7r-regular ring if and only

if R/Nil(R) is regular. Moreover, we show that if 2, that is

1+1, is a unit in a 7r-regular ring R, then every element of

R is a sum of two units in R.

Notations. Let S be a ring with 1. Then

1. Id(S) denotes the set of all idempotents of S.

2. Nil(S) denotes the set of all nilpotent elements

of S.

5

3. U(R) denotes the set of all units of S.

4. C(S) denotes the center of S.

5. J(S) denotes the Jacobson radical of S.

6. If a,beS, then (a,b) denotes the two-sided ideal

of S generated by a,b, and (a,b)S denotes the right ideal of

S generated by a,b, and S(a,b) denotes the left ideal of S

generated by a,b.

7. Let I be an ideal of S and xeS. Then [x] denotes

the element x+l in R/I.

8. Let zeS. Suppose for some x,yeS, z is written as

a product of powers of x and y. Then we define E(z,x) to be

the sum of exponents of x as they appeared in z, and E(x,y)

to be the sum of exponents of y as they appeared in z. For

example, if z=x3yx2y3, then E(z,x)=5 and E(z,y)=4.

Throughout this chapter, the capital letter R denotes a

semicommutative ring with 1.

Lemma 2.1. The set Nil(R) is a two-sided ideal of R.

Proof. Let x,yeNil(R). Then for some n,m>l, x^y^o.

Consider the expansion of (x+y)"*". Let z be a term in the

expansion. Then E(z,x)+E(z,y)=n+m. suppose E(z,x)=k>n. Since

R is semicommutative, for some deR we have z=xkd=0. Suppose

E(z,x)=h<n. Then E(z,y)=w>m. Since R is semicommutative, for

some feR we have z=ywf=0. Thus (x+yj^^o. Let qeNil(R), and

a,beR. Then for some r>l, qr=0. Since R is semicommutative,

(agb) r=qre=0 for some eeR. Thus Nil(R) is a two-sided ideal

of R. •

6

Lemma 2.2. The set ld(R) is a subset of C(R).

Proof. Let eeld(R) and xeR. Then for some y,zeR, ex=ye

and xe=ez. Hence exe=ye=ex and exe=ez=xe and eeC(R). "

The following well-known Lemma is needed.

Lemma 2.3. Let K be a ring with 1 and I be a two-sided

nil ideal of R. If [c]eId(K/I)f then there exists eeld(K)

such that [e]=[c] in K/I.

Proof, see [9].

Theorem 2.1. The ring R is 7r-regular if and only if

R/Nil(R) is regular.

Proof. Suppose R is 7r-regular. Let [x] eR/Nil (R) . Then

for some yeR and n>l, xnyxn=xn. Thus e=xnyeId(R) and

therefore l-eeld(R). Since l-eeC(R), ((l-e)x)n=(l-e)xn=

(l-xny)xn=0. Thus (l-e)x =(l-x"y)x 6 Nil(R). Thus

[x][xn'1y][x] = [xny][x] = [x].

Suppose K=R/Nil(R) is regular. Let xeR. Then for some

[y]eK, [x][y][x]=[x] in K. Thus [xy]eId(K). By Lemma 2.3,

there exists eeld(R) such that [e]=[xy] in K. Thus [e][x]=

[x] in K. Hence [e][x][y]=[xy]=[e] in K. Thus for some

weNil(R), exy-e=w and therefore exy=e+w in R. Since

Nil(R)cJ(R), v=l+weU(R). Hence exy+l-e=v. Let u be the

multiplicative inverse of v. Then exyu+(l-e)u=l. Thus

e(exyu+(l-e)u)=e. Hence exyu=e. Since [e][x]=[x] in K,

(1-e)xeNil (R) . Thus for some m>l, ((l-e)x),n=(l-e)xm=0. Hence

xm=exm. Since exyu=e, (exyu)m=em=e. Since R is

semicommutative, for some deR we have (exyu)m=(ex)md=e.

7

Since eeC(R) , (ex)md=exmd=e. Since exm=xm, we have xmdxm=xm.

Thus R is 7r-regular. m

Theorem 2.2. Let K be a ring with 1 such that Id(K)cC(k)

and xeK is regular, then x is unit regular.

Proof. If xyx=x, then xy,yxeId(K)cC(K). Hence, xy=

x(yx)y=(xy)(yx)=y(xy)x=yx. Let u=x+xy-l and v=xy+xy2-l.

Since xy=yx and xyx=x, we have uv=vu=

x2y+x2y2-x+ (xy) 2+xyxy2-xy-xy-xy2+l=

x+xy-x+xy+xy2-xy-xy-xy2+l=l. Moreover, xvx=x2yx+x2y2x-x2=

x2+x-x2=x. m

Corollary 2.1. Let K be a 7r-regular ring with 1 such

that Id (K) cc (K) . Then K is unit 7r-regular. In particular, if

R is jr-regular, then R is unit 7r-regular.

Proof. Let xeK. Then for some n>l, xn is regular. Hence,

by Theorem 2.2, xn is unit regular. If R is 7r-regular, then

by Lemma 2.2 we have Id(R)cC(R) and therefore the claim is

evident. •

Lemma 2.4. Let K=R/Nil(R) and ueR. Then [u]eU(K) if and

only if ueU(R).

Proof. Let a,b,xeR. Since R is semicommutative, axb=

xd=fx for some d,feR. Thus (a,b)=(a,b)R=R(a,b). In

particular, if (a)=R, then aeU(R). Also, observe that if

[u]eU(R/Nil(R)), then for some [v]eR/Nil(R), [u][v]=[l] and

hence uv=l+z for some zeNil(R). Since Nil(R)cJ(R), l+zeU(R)

and therefore (u)=R. Thus ueU(R). Hence [u]eU(R/Nil(R)) if

and only if ueU(R).

8

Lemma 2.5. Let K be a unit regular ring ( i.e. For every

xeK there exists ueU(R) such that xux=x) and xeK. Then x=ev

for some eeld(R) and veU(R).

Proof. Let xeK. Then for some ueU(K), xux=x. Thus

xueld(K). Let v be the multiplicative inverse of u in K.

Then x=xuv a product of an idempotent and a unit of K. m

Theorem 2.3. The ring R is 7r-regular if and only if for

every xeR there exist eeld(R), ueU(R), and weNil(R) such

that x=eu+w.

Proof. Suppose R is 7r-regular. Let xeR. By Theorem 2.1,

K=R/Nil(R) is regular. Thus K is unit regular by Theorem

2.2. By Lemma 2.4, [x]=[c][u] for some [c]eId(K) and

[u]eU(K). By Lemma 2.3, there exists eeld(R) such that

[e]=[c] in K and by Lemma 2.4 we have ueU(R). Thus [x]=

[e][u] implying x-eu=w for some weNil(R). Hence x=eu+w.

Let xeR such that x=eu+w for some eeld(R), ueU(R), and

weNil(R). Then for some n>l, w"=0. Consider the expansion of

xn=(eu+w)n. Observe that (eu)n=eun since eeC(R) and that the

sum of the other terms are in Nil(R) and of the form ed,

deR, by semicommutativity. But ed=e(ed), so replacing d by

ed we can assume deNil(R). Hence xn= (eu+w) n=eun+ed for some

deNil(R). But un+d=veU(R) . Hence xn=ev. Let r be the

multiplicative inverse of v in R. Then xnrx"=xn. •

Example 1. Let K be a Principal Ideal Domain with 1, and

m be a nonzero nonunit element of K. Then m= p i 1 p " 2 . . .p"" •

9

Where the a^s^l and the p^s are distinct primes in K. Let

S=K/(m) and I=(p1p2.. .pn) an ideal of K. Then Nil(S) =1/(m) .

Thus S/Nil(S)«K/I is isomorphic to a finite direct sum of

fields by the Chinese remainder Theorem. Hence S/Nil(S) is

regular. Thus S is 7r-regular.

Example 2. Let R be a semicommutative Artinian ring.

Since J(R) is nilpotent and Nil(R) is an ideal of R,

Nil(R)=J(R). We know that R/Nil(R)=R/J(R) is isomorphic to a

finite direct sum of simple rings. Furthermore semicommutat-

ive simple ring is a division ring. Thus R/Nil(R) is isomor-

phic to a finite direct sum of division rings. Thus R/Nil(R)

is regular and therefore R is jr-regular.

Example 3. Let R be a finite semicommutative ring. Then

R is Artinian and hence it is jr-regular. In particular, all

finite commutative rings are jr-regular ring.

Related Results.

A ring K is called (s,2)-ring [8] if every element in K

is a sum of two units of K.

The following Theorem gives a characterization of all

semicommutative (s,2)-rings.

Theorem 2.4. A jr-regular ring R is (s,2)-ring if and

only if every element in Id(R) is a sum of two units of R.

Proof. Suppose every element in Id(R) is a sum of two

units of R. Let xeR. Since R is 7r-regular, x=eu+w for some

eeld(R), ueU(R), and weNil(R) by Theorem 2.3. Since e=v+r

for some v,reU(R), x=vu+ru+w. Since weJ(R) and rueU(R),

10

ru+weU(R). Thus x is a sum of two units of R. The second

direction is obvious. "

In [4], it has been shown that if K is a 7r-regular ring

whose primitive factor rings are Artinian, and which does

not have Z/2Z as a homomorphic image, then every element of

K is a sum of two units in K. However, for a semicommutative

7r-regular ring R, we give an alternative proof of this fact.

Corollary 2.2. Let R be a 7r-regular ring such that

2=(l+l)eU(R). Then every element of R is a sum of two units

in R.

Proof. Let eeld(R). Since (l-2e)(l-2e)=l, l-2eeU(R).

Since 2eU(R), e is a sum of two units. Hence by Theorem 2.4

the claim is clear. •

REFERENCES

1. Chacron, M. "On Algebraic Rings," Bull. Austral. Math.

Soc.. 1 (1969), 385-389.

2. Ehrlich, Gertrude, " Unit Regular Rings," Portugal.

Math.. 27 (1968), 209-212.

3. Ehrlich, Gertrude," Units and One-sided Units in Regular

Rings," Trans. Amer. Math. Soc.. 216 (1976), 81-90.

4. Fisher, Joe W. and Snider, Robert L., " Rings Generated

by Their Units," J. Algebra. 42(1976), 363-368.

5. Fisher, Joe W. and Snider, Robert L., " On The Von

Neumann Regularity of Rings with Regular Prime Factor

Rings," Pac. J. Math.. 54 (1974), 135-144.

6. Goodearl K R, Von Neumann Regular Rings. London, Pitman

Publishing, Inc., 1979).

7. Henriksen, Melvin, " two Classes of Rings That are

Elementary Divisor Rings," Arch. Math. (Basel). 24

(1973), 133-141.

8. Henriksen, Melvin, " Two Classes of Rings that Are

Generated By Their Units," J. Algebra. 31 (1974),

182-193.

9. Lam T. Y., A first Course in Noncommutative Rings.

New York, Springer-Verlag, Inc., 1991.

11

CHAPTER III

CONDITIONS EQUIVALENT TO SEMICOMMUTATIVE

7T-REGULAR ELEMENTS

A regular element x of R is called completely regular if

whenever for some ycR we have xyx=x, then xy=yx and x is

called strongly regular if whenever for some zeK we have

xzx=x, then zxx=x.

Throughout this paper, the capital letter R denotes an

associative ring with 1. The following Lemmas are needed.

Lemma 3.1. Let xeR and suppose x is regular. Then there

exist e,ceId(R) and heR such that ex=x, xc=x, xh=e, and

hx=c.

Proof. Since x is regular, for some yeR we have xyx=x.

Let e=xy, c=yx, and h=yxy. Then e,c,h satisfy the claim of

the Lemma. m

Lemma 3.2. Suppose every element in Id(R) is

semicommutative. Then Id(R)<=c(R).

Proof. Let eeld(R) and xeR. Then for some h,deR we have

ex=he and xe=ed. Hence, exe=he=ex and exe=ed=xe. Thus,

ex=xe. m

Lemma 3.3. Suppose every element in Id(R) is strongly

regular. Then Id(R)cC(R).

Proof. Let eeld(R) and xeR. Then z=e-(1-e)xeeld(R).

Moreover, zez=z. Since z is strongly regular, we have

12

13

zez=ezz=z. But ezz=e. Hence, e=z=e-(l-e)xe. Thus, (1-e)xe=0

and therefore exe=xe. Furrther, w=(1-e)-ex(1-e)eld(R) and

w(l-e)w=w. Hence, (1-e)ww=w. But (1-e)ww=(1-e). Hence,

l-e=w=(l-e)-ex(l-w). Thus, ex(1-e)=0 and therefore exe=ex.

Since exe=xe and exe=ex, we have ex=xe. •

Lemma 3.4. Let xeR. If x is unit regular, then x=eu for

some eeld(R) and ueU(R).

Proof. Suppose x is unit regular. Then for some veU(R)

we have xvx=x. Let e=xveId(R) and u=v"1. Then x=eu. B

Theorem 3.1. The following conditions on a ring R are

equivalent:

1. Id(R)cC(R).

2. Every regular element of R is semicommutative.

3. Every regular element in R is of the form eu for

some eeld(R) and ueU(R) such that eeC(R).

4. Every regular element of R is completely regular.

5. Every regular element of R is strongly regular.

Proof. (l)-+ (2). Let z be a regular element in R and xeR.

By Lemma 3.1, for some e,ceId(R) and heR, we have ez=z,

zc=z, hz=c, and zh=e. Hence, zxhz=zx(hz)=zx(c)=(zc)x=zx,

since ceC(R) and zc=z. Thus, let d=zxh. Then zx=dz. Further,

zhxz=(zh)xz=(e)xz=x(ez)=xz, since eeC(R) and ez=z. Let

f=hxz. Then xz=zf. Thus, z is semicommutative.

(2)->(3). Let a be a regular element of R. Since every

element in Id(R) is semicommutative, Id(R)cC(R) by Lemma

3.2. Hence, by Theorem 2.2, a is unit regular and by

14

Lemma 3.4 the claim is now clear.

(3)->(4). Let a be a regular element of R. By the hypothesis,

a=eu for some eeld(R) and ueU(R) such that eeC(R). Suppose

for some zeR we have aza=a. Then eu(z)eu=eu implies eu(z)e=

e. Now, (eu"1)eu(ze)= (eu"1eu) ze=ze, since eeC(R). Further,

(eu"1)eu(ze)=eu"1(euze)=eu"1(e)=eu"1. Thus ze=eu"1. Hence,

za=zeu=(ze) eu=(eu"1) eu=e. Further az=e , since ze=eu"1 and

a=eu and eeC(R). Thus, az=za and a is completely regular.

(4)->(5) . Clear.

(5) -+• (1). By Lemma 3.3. "

Corollary 3.1. Suppose R is regular. Then the following

are equivalent :

1. Id(R)cC(R).

2. Every element in R is semicommutative.

3. Every element of R is of the form eu for some

eeld(R), ueU(R) such that eeC(R).

4. Every element of R is completely regular.

5. Every element of R is strongly regular.

Theorem 3.2. Suppose Id(R)cC(R). Let xeR. Then x is

jr-regular if and only if there exist eeld(R) such that ex is

regular and (l-e)xeNil(R).

Proof. Since x is Tr-regular, for some n>l, xn is

regular. Hence, by Theorem 3.1, we have xn=eu for some

eeld(R) and ueU(R). Let y=xn"1u"1. Then ex(xn"1u"1)ex=

(exnu"1)ex= (euu"1) ex=ex. Hence, ex is regular. Now,

[ (l-e)x]n=(l-e)xn=(l-e)eu=0, Since (l-e)eC(R) and xn=eu.

15

For the converse, suppose for some eeld(R), ex is regular

and (1-e)xeNil(R). Then for some n>l, [ (l-e)x]n=(l-e)xn=0.

Hence, (*) exn=xn. Since ex is regular, by Theorem 3.1,

ex=cu for some celd(R) and ueU(R). Hence, (ex)n=(cu)n=cun,

since ceC(R). But (ex)n=exn=xn by (*) . Thus xn=cun. Let y=cu"

n. Then xnyxn=xn and hence x is yr-regular. m

In [5], it is shown that an element x in R is unit

regular if and only if there exist ueU(R) and group G in R

such that xeuG. Further, we have shown Corollary 2.1 if

Id (R) cC (R) , then all 7r-regular elements in R are unit

7r-regular. In the following Theorem we show that

a nonnilpotent element xeR is w-regular if and only if there

exist n>l and group G in R such that for all m>n, xra is in

G. The word "group" in this chapter means multiplicative

group.

Theorem 3.3. Suppose Id(R)<=c(R). Let x be a nonnilpotent

element of R. Then x is 7r-regular if and only if there exist

n>l and group G in R such that for all m>n, xmeG. Further, n

can be chosen to be the smallest positive integer such that

xn is regular.

Proof. Suppose x is 7r-regular. Let n be the smallest

positive integer such that xn is regular. Hence, xn=eu for

some eeld(R), and ueU(R) by Theorem 3.1. Observe that e*0,

since x£Nil(R). Let G be the maximal group of R with e in G

as the identity. It is known [2] and [7] that G =

{ eu : ueU(R) and eu=ue }. Hence, xneG. Now, suppose for

16

some m>n, xmeG. Then it suffices to show that x^eG. By the

proof of Theorem 3.2 , ex is regular. Hence, ex=cw for some

celd(R) and weU(R) by Theorem 3.1. First we show e=c. Hence

e(ex)=e(cw). But e(ex)=ex=cw. Thus, ecw=cw and therefore

(*) ec=c. Since e,ceC(R, we have (ex)m=exm= cw™. Since xmeG,

xm=ev for some veU(R) . Thus, ex^ev^w"1. Hence, e=cwmv"1.

Thus ec=cwmv"1c=cwmv"1, since ceC(R). Hence, ec=e. Since ec=c

by (*) and ec=e, e=c. Thus ex=ew and therefore exeG. Thus,

exxmeG. But exx"'=x(exB)=x(xm)=xlw1€G.

For the converse. Suppose for some n>l, xneG, for some

group G in R. Let e be the identity in G. Then Gc

{eu : ueU(R) and eu=ue >. Hence xn=ew for some weU(R). Hence

xn is regular by Theorem 3.1. m

Corollary 3.2. Suppose Id(R)cC(R). Let xeR. Then x is

7r-regular if and only if there exist n>l such that for all

m>n, xm is regular.

Proof. If x is nilpotent, then the claim is clear.

Hence, suppose x is nonnilpotent. By the above Theorem there

exist n>l and group G in R such that for all m>n, xmeG. Now,

let m>n. and let y be the inverse of x" in G. Then x^x^x"*.

Hence, xm is regular.

The converse is clear. •

Suppose R is semicommutative 7r-regular. It is shown

Lemma 2.1 that Nil(R) is a two sided ideal of R and

therefore Nil(R)cj(R). Now, let xeJ(R). Then for some n>l

and eeld(R) and ueU(R), xn=eu. Hence, eeJ(R) and since J(R)

17

contains no nonzero idempotent we have e=0. Thus, xeNil(R).

Hence J(R)=Nil(R).

It is well-known that if Id(R)cC(R) and R is regular and

I is a prime ideal of R, then R/I is a division ring. Now,

suppose R is semicommutative and let P be a prime ideal of

R. Then it is easy to see that R/P is a semicommutative

domain and for every a,beR such that abeP implies aeP or beP

( some authors call such P completely prime ). However, if R

is semicommutative 7r-regular, then we have the following.

Theorem 3.4. Suppose R is semicommutative 7r-regular. Let

P be a prime ideal of R. Then R/P is a division ring.

Further, for every m>l, R/Pm is a local ring.

Proof. Let xeR such that x£P. Then for some eeld(R) and

ueU(R) and n>l, we have xn=eu by Theorem 3.1. observe that

e#0, since x£P. Thus, euCP and therefore e£P. Since (l-e)eeP

and eCP, (l-e)eP. Thus [e]=[l] in R/P. Thus [xn] = [eu] = [u] in

R/P. But [xn] = [u] in R/P implies [xn] is a unit in R/P and

therefore [x] is a unit in R/P. Hence, R/P is a division

ring. Now, let m>l and K=R/Pm. Since K is semicommutative

7r-regular, J(K)=Nil(K). But J(K)=Nil(K)=P/Pm. Hence,

K/J(K)«R/P a division ring. Thus, K is a local ring. •

REFERENCES

1. Ehrlich, Gertrude, " Unit Regular Rings," Portugal Math..

27 (1968), 209-212.

2. Farahat, H.K and Mirsky, " Group Membership in Rings of

Various Types," Math. Zeitschr.. 70 (1958), 231-244.

3. Fisher, Joe and Snider, Robert L., " On The Von Neumann

Regularity of Rings with Regular Prime Factor Rings, "

Pac. J. Math.. 54 (1974), 135-144.

4. Goodearl, K. R. , Von Neumann Regular Rings. London,

Pitman Publishing, Inc., 1979.

5. Hartwig, Robert E. and Luh Jiang, " A Note on The

Structure of Unit Regular Ring Elements, " Pac. J.

Math.. 71 (1977), 449-461.

6. Henriksen, Melvin, " Two Classes of Rings that are

Generated By Their Units, " J. Algebra. 31 (1974),

182-193.

7. Losey, Gerald and Schneider, Hans, " Group Membership in

Rings and Semigroups," Pac. J. Math.. 11 (1961),

1089-1098.

8. Lam T. Y., A First Course in Noncommutative Rings.

New York, Springer-Verlag, Inc., 1991.

18

CHAPTER IV

WHEN A fl"-REGULAR RING IS SEMI COMMUTATIVE

Let K be an associative ring with 1. Let S,D be nonempty

subsets of K and let seS. If for every deD, there exist f,y

in K such that sd=ys and ds=sf, then we say s restricted to

D is semicommutative. If every element in S restricted to D

is semicommutative, then we say S restricted to D is

semicommutative.

Throughout this chapter, SC(R) denots the set of all

semicommutative elements of R, Reg(R) denotes the set of all

regular element of R

In the proof of Theorem 3.1, it is shown if Id(R)cC(R)

and xeR such that for some m>l, eeld(R), and ueU(R) we have

xm=eu, then ex=ev for some veU(R) and (1-e)xeNil(R). We

state this result as a Lemma without a proof.

Lemma 4.1. Suppose Id(R)cC(R). Let xeR such that x is

7T-regular, that is for some m>l, eeld(R), and ueU(R) we have

xm=eu. Then ex=ev for some veU(R) and (l-e)xeNil(R).

Theorem 4.1. Suppose R is 7r-regular such that Id(R)cC(R)

and Nil(R) restricted to U(R) is semicommutative. Then

Nil(R) restricted to Reg(R) is semicommutative and Nil(R) is

a two-sided ideal of R.

Proof. Let xeReg(R). Since Id(R)cC(R), x=eu for some

eeld(R) and ueU(R) by Theorem 3.1. Let weNil(R). Then for

19

20

some f,hcR, we have wu=fw and uw=wh. Since eeC(R), we have

w(eu)=efw and (eu)w=weh. Hence, wx=efw and xw=weh. Thus,

Nil(R) restricted to Reg(R) is semicommutative. To show that

Nil(R) is an additive subgroup of R, it suffices to show

that if n,weNil(R), then n+weNil(R). Let y=n+w. Since y is

7 r - r e g u l a r , for some celd(R) and ueU(R), cy=cu and

(l-c)yeNil(R) by Lemma 4.1. Hence, cu=cy=cn+cw and therefore

cu-cn=cw. But cu-cn=cu(l-u"1n) and u"1ncNil(R) since n

restricted to U(R) is semicommutative. Thus, l-u"1n=veU(R) .

Hence, cu-cn=cuv=cw. But cweNil(R). Thus, cuveNil(R). Hence,

c=0, since uveU(R). But (l-c)y Nil(R). Hence, (l-c)y=

yeNil(R). Thus, n+weNil(R). Now, let w,yeNil(R). Then y+l=

veU(R). Hence, w(y+l)=wy+w=wv. Thus, wy=wu-w. Since w

restricted to U(R) is semicommutative, wueNil(R). Hence,

wy=wu-weNil(R). Next, let xeR and weNil(R). Then fx is

regular for some feld(R) and (1-f)xeNil(R) by Lemma 4.1.

Hence, (l-f)xw, w(l-f)xeNil(R), since Nil(R) is a subring of

R. Since Nil(R) restricted to Reg(R) is semicommutative,

fxweNil(R) and wfx=wxfeNil(R). Since w(l-f)x, (l-f)xw, xfw,

fxweNil(R), we have xweNil(R) and wxeNil(R). Hence, Nil(R)

is a two-sided ideal of R. •

Question. Does there exist a ring R satisfying the

hypothesis of the above Theorem which it is not

semicommutative ?.

In the following Theorem we state a necessary and

sufficient condition on a 7r-regular ring R such that R is

21

semicommutative.

Theorem 4.2. Suppose R is jr-regular. Then R is

semicommutative if and only if Nil(R)csc(R) and Id(R)csc(R).

Proof. We only need to prove the converse. Hence,

suppose Nil(R)cSC(R) and Id(R)cSC(R). Then Id(R)cC(R) by

Lemma 3.2. Let xeR. Since x is 7r-regular, for some eeld(R),

ex is regular and (1-e)x=x-ex=weNil(R) by Theorem 3.2. Let

feR. Then xf-exf=(x-ex)f=dx-dex for some deR, since (x-ex)=

weNil(R) and weSC(R). Thus, xf=exf+dx-dex. But exf=kex for

some keR, since ex is regular by Theorem 3.1. Thus, xf=

[ke+d-de]x. It is easy to verify that fx=xh for some heR.

Hence, x is semicommutative. 11

In Lemma 1.1 it is shown that if R is semicommutative,

then Nil(R) is a two-sided ideal of R. In fact, if

Nil(R)cSC(R), then Nil(R) is a two-sided ideal of R. Since

the proof of this claim is similar to that in 1.1, we state

this fact as a Lemma without a proof.

Lemma 4.2. Suppose Nil(R)cSC(R). Then Nil(R) is

a two-sided ideal of R.

Let I be a two-sided ideal of R and xeR. Then [x]

denotes the elemwnt x+i in R/I.

Theorem 4.3. A ring R is semicommutative 7r-regular if

and only if Nil(R)cSC(R), R/Nil(R) is regular, and

Id (R) cC (R) .

Proof. Suppose R is semicommutative 7r-regular. Then

Id(R)cC(R), by Lemma 2.2 and R/Nil(R) is regular by

22

Theorem 2.1.

For the converse, suppose Nil(R)cSC(R), K=R/Nil(R) is

regular and Id(R)cC(R). Let xeR. Since [x] is regular in K,

[x]=[e][u] for some [e]eId(K), and [u]eU(K) by Theorem 3.1.

But it is well-known that there exists feld(R) such that

[f]=[e] in K. Further, it is easy to see that ueU(R). Hence,

x=fu+w for some weNil(R). Thus, fx=fu+fw=f(u+w). But

Nil(R)cj(R) and therefore u+w=zeU(R). Hence, fx=fz. Thus, fx

is regular in R. Now, (1-f)x=x-fx=fu+w-fu-fw=w-fw=

(1-f)weNil(R) , since (l-f)eC(R). Hence, x is 7r-regular by

Theorem 3.2. By Theorem 4.2, R is semicommutative. a

Theorem 4.4. Suppose R is a finite reduced ring. Then R

is regular if and only if R is commutative.

Proof. Let J(R) denotes the Jacobson radical of R.

Suppose R is regular. Since R is Artinian, J(R) is

nilpotent. But R is reduced and hence J(R)=0. Since R is

Artinian, it is well-known that R/J(R)=R is isomorphic to a

finite direct sum of simple rings. But every reduced simple

ring is a division ring. Hence, R is isomorphic to a finite

direct sum of division rings. But every finite division ring

is commutative. Thus, R is commutative.

For the converse, suppose R is commutative. Then

R/J(R)=R isomorphic to a finite direct sum of finite fields.

Since every field is regular, the claim is evident. •

REFERENCES

1. Goodearl,K. R., Von Neumann Regular Rings , London,

Pitman Publishing, Inc., 1979.

2. Lipschitz, L., " Commutative Regular Rings With Integral

Closure Trans. Amer. Math. Soc.. 211 (1975),

161-170.

3. McCoy, N., H., 11 Generalized Regular Rings," Bull. Amer.

Math. Soc.. 45 (1939), 175-178.

23

CHAPTER V

PARTITION RINGS

Suppose the idempotents of R are in the center of R and

suppose for some a,beR we have ba=0 and aR+Rb=R. In this

chapter, we show that G = { xeaR : xR+Rb=R } is a maximal

multiplicative group of R. Further, we show that R-{0> is a

union of disjoint multiplicative groups if and only if R is

semicommutative regular.

Throughout this chapter, the word " group" means

multiplicative group.

Lemma 5.1. Suppose Id(R)cC(R). Let e€ld(R), e*0. Suppose

for some x,yeeR, xy=e. Then yx=e, that is xeU(eR).

Proof. Consider, (yx)(yx) =y (xy) x=y (e) x= (e) yx=yx, since

x,yeeR and eeC(R). Hence, yxeld(R) and therefore yxeC(R).

But e=(xy)(xy)=x(yx)y=(yx)xy=(yx)e=yx. The proof is

completed. B

Theorem 5.1. Suppose Id(R)cC(R) and for some a,beR, a#0,

we have ba=0 and aR+Rb=R. Then G = { xeaR : xR+Rb=R } is a

maximal group of R.

Proof. Since aR+Rb=R, for some f,heR we have (*) 1=

af+hb. Hence, afeG and af=af(af)+hb(af)=afaf, since ba=0.

Thus, afeld(R) and by the hypothesis afeC(R). Observe that

af*0 ( for if af=0, then a=hba=0 ). Let xeG. Then by (*) we

have x=afx+hbx=afx, since xeaR and ba=0. Hence xeafR.

24

25

Further, for some d,zcR we have l=xd+zb. Thus, af=

xd(af)+zb(af)= xd(af)= x(af)d. Since x, afdeafR and x(afd)=

af, we have xeU(afR) by Lemma 5.1. Thus GcU(afR). Next, let

yeU(afR). Then for some keU(afR) we have yk=af. But af+hb=l

by (*). Hence, yeG. Thus, G=U(afR) a maximal group of R with

af in U(afR) as the identity. a

Corollary 5.1. Suppose Id(R)cC(R). Let eeld(R), e#0.

Then G={ xeeR : xR+R(i-e)=R } = U(eR).

Proof. Let a=e and b=l-e and observe that a and b

satisfy the hypothesis of the above Theorem. m

In [4], it is shown that R is strongly regular if and

only if every element in R-{0} is contained in a group of R.

We will give an alternative proof of this fact.

Lemma 5.2. Suppose R is reduced ( i.e. R contains no

nonzero nilpotent elements ) and for some x,y, we have xy=0.

Then yx=0.

Proof. Consider, (yx)(yx)=y(xy)y=0. Thus, yx=0, since R

is reduced. •

A ring R is called a partition ring if R-{0} is a union

of disjoint groups.

Lemma 5.3. Suppose e,feId(R), e#f. Then U(eR), U(fR) are

disjoint groups in R.

Proof. Suppose xeU(eR) and xeU(fR). Then for some

yeU(fR), xy=f. Consider, exy=e(xy)= ef. Also, exy=(ex)y=

(x)y=f. Thus (*) ef=f. Further, there exists zeU(eR) such

that zx=e. Hence, zxf=(zx)f=ef=f by (*). But zxf=z(xf)=

26

z(x)=e. Hence e=f, a contradiction. u

It is well-known that a strongly regular ring R is

reduced and therefore Id(R)cC(R).

Theorem 5.2. A ring R is a partition ring if and only if

it is strongly regular.

Proof. Suppose R is a partition ring. Let xeR, x*0.

Then x2eU(eR) for some eeld(R). Let y be the inverse of x in

U(eR). Then x2y=x.

For the converse, suppose R is strongly regular. Let

xeR, x*0. Then for some yeR we have x2y=x. Hence, x(l-xy)=0.

Since R is reduced, (l-xy)x=0 by Lemma 5.2. Since Id(R)cC(R)

and x, (1-xy) satisfy the hypothesis of Theorem 5.1, we have

xeG = { dexR : dR+R(l-xy)= R } a maximal group of R. •

Lemma 5.4. Let xeR such that x is regular. Then there

exist e,ceId(R) and heR such that ex=x, xc=x, xh=e, and

hx=c.

Proof. Since x is regular, for some yeR we have xyx=x.

Let e=xy, c=yx, h=yxy. Then e,c,h satisfy the claim of the

Lemma. m

Theorem 5.3. A ring R is a partition ring if and only if

R is semicommutative regular.

Proof. Suppose R is a partition ring. Then R is reduced

( i.e. R contains no nonzero nilpotent elements) and

therefore Id(R)cC(R). Let xeR, x*0. Then xeU(eR) for some

eeld(R). Let y be the inverse of x in U(eR). Then xyx=x and

hence x is regular. Let beR. Then xb(yx)=xb(e)=(e)xb=

27

(ex)b=xb. Hence, let g=xby. Then xb=gx. Further, (xy)bx=

(e)bx=b(ex)=bx. Thus, let w=ybx. Then bx=xw. Hence, x is

semicommutative.

For the converse, suppose R is semicommutative regular.

Let xeR, x#0. Then for some yeR we have xyx=x. Since R is

semicommutative, for some deR we have x=xyx=x2d. Thus R is

strongly regular and R is a partition ring by Theorem 5.2. ™

In light of Theorems 5.2 and 5.3, we have the following

Corollary.

Corollary 5.2. The following are eguivalent :

1. R is a partition ring.

2. R is semicommutative regular.

3. R is strongly regular.

An element meR is called square-free if for every prime

divisor p of m we have p2 does not divide m.

Theorem 5.4. Suppose R is a principal ideal domain. Let

m be a nonzero nonunit element of R. Then P=R/(m) is a

partition ring if and only if m is square-free.

Proof. Suppose m is square-free. Then P is isomorphic

to a finite direct sum of fields by the Chinese remainder

Theorem. Hence, P is commutative regular and therefore it is

a partition ring by Theorem 5.3.

For the converse, assume P is a partition ring and m is

not square-free. Then m=nqr, where r>2, n,qeR and

g.c.d(n,q)=l. Hence, y=nq+(m) is a nonzero element in P, but

yr=0 in P, a contradiction, since R is reduced. B

28

Corollary 5.3. Let F be a field, and Z be the set of all

integers. Further, let m,n be nonzero nonunit elements of

F[x], Z respectively. Then F[x]/(m), Z/(n) are partition

rings if and only if m,n are square-free.

REFERENCES

1. Goodearl, K. R., Von Neumann Regular Rings. London,

Pitman Publishing, Inc., 1979.

2. Lam, T. Y., A first Course in Noncommutative Rings.

New York, Springer-Verlag, Inc., 1991.

3. Luh, J., 11 A note on Strongly Regular Rings," Proc. Japan

Acad.. 40 (1964), 74-75.

29

CHAPTER VI

A RELATION BETWEEN THE IDEMPOTENTS AND

A CLASS OF UNITS IN A RING

Let R be an associative ring with 1 such that 2 is not a

zero divisor in R. Let SU(R) denotes the set { ueR:u2=l }

and SQ(R) denotes the set { ueSU(R): 2x=l-u has a solution

in R).

Lemma 6.1. Suppose for some ueSU(R), there exists eeR

such that 2e=l-u. Then eeld(R) and e is the unique solution

to the equation 2x=l-u.

Proof. Since 2e=l-u, we have (2e)2=(l-u)2=2 (l-u) = . Hence

4e2=2(l-u). Since 2 has no zero divisors, 4e2=2(l-u) implies

2e2=(l-u)=2e. Hence 2e2=2e implies e2=e and therefore

eeld(R). The uniqueness of e is evident, since 2 has no zero

divisors. m

Theorem 6.1. There exists a one to one correspondence

between Id(R) and SQ(R).

Proof. Let F be a map from Id(R) into SQ(R) given by

F(e)=l-2e. Then (l-2e)2=l. Since 2 has no zero divisors,

l-2e is distinct for every eeld(R). Thus F is a one to one

map. Let G be a map from SQ(R) into Id(R) given by G(u)= e

where e is a solution to 2x=l-u. By Lemma 6.1, G is well-

defined and is one to one. B

Corollary 6.1. Suppose 2eU(R). Then there exists a one

30

31

to one correspondence between Id(R) and SU(R).

Proof. If 2eU(R), then for every ueSU(R), the equation

2x=l-u has a solution in R. Hence SU(R)=SQ(R). m

Remark: If 2eU(R), then the map G from SQ(R) into Id(R)

in Theorem 6.1 can be defined as G(u)= w(l-u) where w is the

multiplicative inverse of 2 in R.

Theorem 6.2. Suppose SQ(R) is a subgroup of U(R). Then

for some u,veSQ(R) we have u+veU(R) if and only if u=v and

2eU(R).

Proof. Suppose for some u,veSQ(R), u+veU(R). Then

u+v=[l-(-v)u]u and hence [l-(-v)u]eU(R). Since SQ(R) is a

subgroup of U(R) and -v,ueSQ(R), we have (-v)ueSQ(R). Hence

by the proof of Theorem 6.1, for some eeld(R) we have 2e=

[l-(-v)u]. Since [l-(-v)u] is unit in R, we have e=l. Thus

l-(-v)u=2 and therefore uv=l. Since u,veSQ(R), we have u=v.

Thus 2ueU(R) and therefore 2eU(R). The other direction is

evident. •

Lemma 6.2. Let S be a ring with 1. Then Id(S)cC(S) if

and only if for every e,feId(S), e+f-2efeld(S).

Proof. If Id(S)cC(S), then the claim is trivial.

Assume that for every e,feId(S), we have e+f-2efeld(S). Let

eeld(S) and xeS. Then e+(l-e)xeeId(S). Hence, by the hypoth-

esis, we have e+(1-e)xe+e-2(e+(l-e)xe)e=-(l-e)xeeId(S).

Since [-(l-e)xe]2=0 and -(l-e)xeeld(S), we have -(1-e)xe=0

and therefore (l-e)xe=0. Also; e+ex(l-e)eld(S). In a similar

argument as above, we have ex(l-e)=0. Thus (1-e)xe=ex(l-e)=0

32

implies exe=xe and exe=ex. Hence ex=xe and e is in C(R). B

Theorem 6.3. Id(R)cC(R) if and only if SQ(R) is closed

under multiplication.

Proof. Suppose Id(R)cC(R). Let u,weSQ(R). By the proof

of Theorem 6.1, there exist e,feId(R) such that 2e=l-u, and

2f=l-w. Hence 4ef=(l-u)(l-w)=(l-u)+(l-w)+(uw-l). Thus,

(l-u)+(l-w)-4ef=l-uw. Since 2e=l-u and 2f=l-w, we have

2e+2f-4ef=2(e+f-2ef)=l-uw. Since e+f-2efeId(R),

[1-2 (e+f-2ef) ]2=(uw)2=l. Thus uweSU(R) and since

2(e+f-2ef)=l-uw, we have uweSQ(R). Now, Assume SQ(R) is

closed under multiplication. Let e,feId(R). Then for some

u,weSQ(R), we have 2e=l-u and 2f=l-w. Since uweSQ(R), by

Lemma 6.1 for some celd(R), 2c=l-uw. From the proof of the

first direction, observe that c=e+f-2ef. Hence by Lemma 6.2,

Id(R)cC(R). •

Corollary 6.2. Id(R)cC(R) if and only if SQ(R) is a

multiplicative subgroup of U(R).

Proof. Suppose Id(R)cC(R). Since every element in SQ(R)

is the inverse of itself and SQ(R) is closed under multipli-

cation by Theorem 6.3, the claim is evident. The other

direction follows from Theorem 6.3. •

Theorem 6.4. The following conditions on R are

equivalent.

1. Id(R)cC(R).

2. SQ(R)cC(R).

3. The elements of SQ(R) commute with each other.

33

4. the elements in Id(R) commute with the elements

in SQ(R).

5. The product of any two idempotents of R is

idempotent.

6. The product of any two idempotents of R is zero

or nonnilpotent.

7. the elements in Id(R) commute with the elements

in Nil(R).

8. For every e,feId(R), we have e+f-2efeld(R).

9. SQ(R) is closed under multiplication.

Proof. (l) + (2). Let ueSQ(R), xeR. Then by the proof of

Theorem 6.1, there exists eel(R) such that 2e=(l-u). Since

ex=xe and 2eC(R), we have xu=ux.

(2) -• (3). Trivial.

(3)-f(4). Let eeld(R) and ueSQ(R) . Then for some veSQ(R), we

have 2e=(l-v). Since 2eC(R) and u commutes with v, we have

ue=eu.

(4)->(5). Let e,feld(R). Then for some veSQ(R), 2f=(l-v).

Since 2eC(R) and ev=ve, we have (ef) 2=e2f2=ef.

(5)->-(6) . Trivial.

(6) -• (7). Let eeld(R) and xeNil(R). Then [e+ex(l-e) ] (l-e) =

ex(l-e). Since [e+ex(l-e)] and 1-e are in Id(R) and

[ex(l-e)]2=0, we have ex(l-e)=0. Also; [(l-e)+(l-e)xe]e=

(l-e)xe. Since [(l-e)+(l-e)xe] and e are in Id(R) and

[(l-e)xe]2=0, we have (l-e)xe=0. Hence ex(1-e)=(1-e)xe=0

implies ex=xe.

34

(7)-*(l). Let eeld(R), xeR. Since [ (l-e)xej2=0, we have

0=e(l-e)xe=[(l-e)xe]e=(l-e)xe. Also; since [ex(l-e)]2=0, we

have 0=(l-e)[ex(l-e)]=[ex(l-e)](1-e)=ex(l-e). Thus

ex(1-e)=(1-e)xe=0 implies ex=xe.

8-*l. Lemma 6.2.

9-wl. Theorem 6.3. 11

Remark : If 2 is a zero divisor in a ring s, observe

that the conditions (1), (5), (6), (7), (8) are still equiv-

alent.

Related Results.

Let S be a ring with 1. Suppose Id(S)cC(S). It is well-

known that (Id(R),.,#) is a Boolean ring where . is the

normal multiplication operation as in R and # is given by

e#f=e+f-2ef.

Theorem 6.5. Suppose Id(R)<=c(R). Then (Id(R),#) is

isomorphic to (SQ(R),.).

Proof. Let F be a map from (Id(R),#) into (SQ(R),.)

given by F(e)=u where l-2e=u. Clearly, F is well-defined,

and by the proof of Theorem 6.1 it is one to one and onto.

Now, let e,feId(R). Then for some u,weSQ(R) we have F(e)=

l-2e=u and F(f)=l-2f=w. Hence F(e#f)=F(e+f-2ef)=

1-2(e+f-2ef)=(l-2e)(l-2f)=uw. Thus F(e#f)=F(e)F(f). •

Corollary 6.3. Suppose Id(R), U(R) are finite sets of R

and Id(R)cC(R). Let m be the order of Id(R), that it is m=2k

for some k>l, and let n be the order of U(R). Then m divides

n.

35

Remark: the fact that 2 is not a zero divisor of R is

crucial in the hypothesis, for example in S=Z6 we have

Id(S)={0,1,3,4} and U(S)={1,5}. Hence m=4 and n=2 but 4 does

not divide 2.

Proof. The proof follows from Corollary 6.2 and the

above Theorem. B

Let S be a ring with 1. Then S is called Dedekind-finite

( or directly finite or weakly 1-finite ) if whenever for

some x,yeR, xy=l then yx=l . Now, suppose for some x,yeR we

have xy=l. It is easy to verify that yxeld(R). Assume

yxeC(R). Then l=(xy)(xy)=x(yx)y=(yx)xy=yx and therefore

xeU(R). In light of this discussion, we have the following

result.

Theorem 6.6. If R satisfies one of the equivalent

conditions in Theorem 4, then R is Dedekind-finite.

Theorem 6.7. A ring R is Dedekind-finite if and only if

for some x,yeR and ueSQ(R) with xy=l and xuy=l, then u=l.

Proof. Soppose R is Dedekind-finite, x,yeR such that

xy=l, and for some ueSQ(R) we have xuy=l. Since x,yeU(R) and

xy=xuy, we have l^x^x) (yy"1)=u. Now, assume whenever for

some ueSQ(R) and xuy=l, then u=l. Since xy=l, l=(xy)(xy)=

x(yx)y. Hence 2=x(2yx)y. Since yxeld(R), for some weSQ(R) we

have 2(yx)=l-w. Thus 2=x(l-w)y=xy+x(-w)y. Since xy=l, we

have l=x(-w)y. Since 2(yx)=l-w, we have 2(l-yx)=l-(-w). Thus

(-w)eSQ(R). Hence by the hypothesis, we have -w=l. Thus

2(yx)=l-w=2. Since 2 is not a zero divisor of R, we have

36

yx=l. Hence, R is Dedekind-finite "

Let K be a ring with 1 and e1f e2 be idempotents of K.

Then e,, e2 are called orthogonal ( nonorthogonal ) if e,,e2=0

( e,,e2#0 ) . Further, an idempotent e of k is called noncen-

tral if e is not in the center of k. Suppose for some x,yeK

we have xy=l and yx#l. Then Kaplansky has shown that x has

an infinite number of right inverses. On the other hand,

Jacobson [2] has shown that K contains infinitely many

orthogonal idempotents.

Theorem 6.8. Let K be a ring with 1. Suppose for some

a,beK we have ab=l and ba*l. Then K contains infinitely many

nonorthogonal noncentral idempotents. Further, K contains

infinitely many nilpotent elements.

Proof. Assume the hypothesis. For every i>l, let

e).= [b+(l-ba)a'"1]a. Then ei

2=[b+(l-ba]a,"1]a[b+(l-ba)ai"1]a=

[b+(l-ba)a1"1] [ab+(a-aba)al"1]a=[b+(l-ba)a1"1] a = e, since ab=l

and a-aba=0. Let i,j>l such that i*j. We show e^ej.

Suppose e^ej. Then ba+(l-ba)a1 = ba+(l-ba)aJ . Thus

(l-ba)a1 = (l-ba)aJ. Without loss of generality assume i>j.

Hence (l-baja'b1 = (l-ba)aJb'. Since ab=l, we have (1-ba) =

(l-ba)b1_j. Hence 1-ba = b'^-b^^O. Thus ba=l a contradic-

tion. One easily verifies that for every i,j>l we have

e^^e,. Hence {e,-} is an infinite set of nonorthogonal

noncentral idempotents of R. Now, for every i>l, let x. =

e,,-e.. Since e^^e, and e{e.,=e{, we have xi2=0. Thus {x,} is an

infinite set of nilpotents. •

37

Corollary 6.4. K contains an infinite set of idempotents

{e,-} such that for every i,j>l we have e^^e,. and

(e.-ej)2 = 0.

A regular ring S is called abelian regular if

Id(R)cC(R). It is well-known [see 3] that abelian regular

rings are unit regular rings.

Lemma 6.3. Let S be a unit regular ring with 1. Then for

every xeS, there exists eeld(S) and ueU(S) such that x=eu.

Proof. Let xeS. Then xux=x for some ueU(S). Let v be the

multiplicative inverse of u. Then x=(xu)v and observe that

xueld(S). •

Theorem 6.8. Suppose R is regular. Then R is commutative

if and only if U(R) is a commutative group.

Proof. If R is commutative, then U(R) is a commutative

group. Suppose U(R) is a commutative group. Then the ele-

ments of SQ(R) commute with each other and therefore by

Theorem 6.4 (condition #3), Id(R)cC(R). Hence R is abelian

regular and therefore it is unit regular. Let x,ye,R. By

lemma 6.3, for some e,feId(R) and u,veU(R) we have x=eu and

y=fv. Hence xy=yx and R is commutative. •

Theorem 6.9. Suppose R is regular. Then R is a division

ring if and only if SQ(R)={1,-1}.

Proof. If SQ(R)={-1,1}, then by Theorem 6.1, Id(R)=

{0,1}. Let xeR, x*0. Then for some yeR, we have xyx=x. Since

x*0 and xy,yxeId(R) and Id(R)={0,l}), we have xy=yx=l. Thus

xeU(R) and R-{0}=U(R). •

38

Corollary 6.5. Suppose R is regular and for some

eeld(R), we have e€C(R). Then the following are equivalent:

1. eR is a division ring.

2. SQ(eR)={-e,e}.

3. Id(eR)={e,0}.

Proof. Since eeC(R), It is easy to see that eR is a

regular subring of R. By Theorem 6.7 and Theorem 6.1 we have

(1) iff (2)iff (3). •

Let S be a ring with 1, and let Sn denotes the ring of

all nxn matrices with entries from S. Then AeSn is called

diagonable over S iff there exists an invertible matrix PeSn

such that PAP"1 is diagonal. If S is commutative, then S is

called ID-ring (basal) as in [8] ([2]) iff for every n>l,

the idempotents of Sn are diagonable.

Theorem 6.10. Suppose R is commutative. Then R is

ID-ring if and only if every element in SQ (Rn) is diagonable

for every n>l.

Proof. Let n>l. Suppose R is ID-ring, let AeSQ(Rn) . Then

by the proof of Theorem 6.1, there exists eeId(Rn) such that

2Ie=I-A, where I is the nxn identity matrix in Rn. By the

hypothesis, there exists an invertible matrix P in Rn such

that PeP"1 is diagonal. Thus P(2Ie)P"1=I-PAP"1 and therefore A

is diagonal. Next, assume that every element in SQ(Rn) is

diagonable. Let feId(Rn) . Once again, by the proof of

Theorem 6.1, there exists BeSQ(Rn) such that 2If=I-B. Hence,

But for some invertible matrix K in Rn, we have KBK"1 is

39

diagonable. Thus, K2IfK"1=K( I- B)K"1=I-KBK"1 is diagonable.

Thus, f is diagonable.

REFERENCES

1. Baer R., " Inverses and Zero-devisors," Bui1. Amer« Math.

SOC.. 48 (1942), 630-638.

2. Foster A.L.," The Idempotents Elements of A Commutative

Ring form A Boolean Algebra," Duke Math. J.. 12 (1945),

143-152.

3. Goodearl K. R., Von Neuwmann Regular Rings. London,

Pitman Publishers, Inc., 1979.

4. Jacobson Nathan, " Some Remarks on One-sided Inverses,"

Proc. Amer. Math. Soc.. 1 (1950) 352-355.

5. Krupnik, N., " Minimal Number of Idempotent Generators of

Matrix Algebra over Arbitrary Field," Communication in

Algebra, 20 (1992), 3251-3257.

6. Lam, T.Y., A first Course in Noncommutative Rings.

New York, Springer-Verlag, Inc., 1991.

7. Rowen Louis, Ring Theory I. New York, Academic Press,

Inc., 1988.

8. Steger A.," Diagonability of Idempotent Matrices," Pac.

J. Math.. 19 (1966), 535-542.

40

CHAPTER VII

Rn CONTAINS A DIVISION RING IF AND ONLY IF R DOES

Let R be a ring with 1, and let Rn denote the complete

matrix ring of all nxn matrices over R under the usual

matrix addition and multiplication. Recall A,BeRn are

similar iff there exists PeR„ such that A=PBP"1. If AeR„ is n n

similar over R to a diagonal matrix, then A is called [1]

diagonable over R. For BeRn, b̂ . denotes the entry of B in

the ith row and jth column.

We give an alternative proof of [1, Theorem 1] which is

quite shorter than that in [1]. We would like to point out

that our proof begins exactly like the original.

Theorem 7.1 ( [1, Theorem 1]). Let R be a ring with 1

for which each idempotent matrix in Rn is diagonable over R.

Then R contains a division ring if and only if Rn contains a

division ring.

Proof. If R contains a division ring, then clearly Rn

contains a division ring. Assume Rn contains a division ring

K. The division ring K has an identity -call it J- and by

the hypothesis PJP"1=I a diagonal matrix for some invertible

matrix PeRn- Since the conjugation of Rn by P ( that is a

map F from Rn onto Rn given by F(x)=PxP"1) induces a ring

automorphism of Rn, M=PKP"1 is a division ring of Rn and has

I as the identity. Hence I is a nonzero idempotent of Rn.

41

42

Let S={ AeM : A is diagonal }. Since IeS, S is not empty. It

is easy to see that (S,+) is an additive subgroup of M and

(S-{0}, . ) is a multiplicative subgroup of (M-{0), . ).

Hence, S ia a division subring of M. Since 3>0, there exists

l<j<n such that ijj is a nonzero idempotent of R. Let

D={ ajj : AeS }. Then D is a division ring of R with ijj as

the identity. ™

We end this chapter with some examples that satisfy the

hypothesis of the Theorem and with one example where the

hypothesis fails. Let R be a commutative ring with 1. Then R

is called ID (basal) as in [7] ([2]) if for every n>l the

idempotents of Rn are diagonable. Foster [2] has shown that

if R is a principal ideal domain, then R is ID. Seshadri [6]

has shown that if R is a principal ideal domain, then R[x]

is ID. In particular if F is a field, then F[x,y] is ID.

Steger [7] has shown that if R is an elementary division

ring (i.e., for every n>l and AeRn there exist invertible

matrices P,Q in Rn such that PAQ is diagonal) then R is ID.

Also; Steger has shown that if R is rr-regular ring

(i.e., for every x in R there exists n>l and y in R

( n and y depending on x ) such that xnyxn=xn ) then R is ID.

In particular for every m>l, Zm (i.e.,Z/mZ) is ID

( Foster has shown independently that Zm is ID ) .

Finally, Theorem 3 in [7] states that if R is ID, then

every invertible ideal of R is principal. Thus if R is a

Dedekind domain which is not principal, then R is not ID. In

43

particular, Let R=Z[^5] ( Z is the set of all integers ).

Then R is a Dedekind domain, see [4, EX. 37, P.70]. But R is

not a unique factorization domain, for example 21 does not

have unique factorization in R. Thus R is not principal and

therefore it is not ID.

REFERENCES

1. Beard Jocob T., Jr. and McConnel Robert," Matrix Fields

Over The Integers Modulo m," Linear Algebra and Its

Applications. 14 (1976), 95-105.

2. Foster, A.L., " Maximal Idempotent Sets In A ring With

Unit," Duke Math. J.. 13 (1946) 247-258.

3. Gilman L. and Henriksen M., "Some Remarks About

Elementary Divisor Rings," Tran. Amer. Math. Soc.. 82,

(1956), 363-365.

4. Hutchins Harry C.. Examples of Commutative Rings.

Harry C. Hutchins (1981).

5. Kaplansky I., "Elementary Divisors And Modules," Trans.

Amer. Math. Soc.. 66 (1949), 464-491.

6. Seshadri C. S., " Triviality of Vector Bundles Over

Affine Space K2," Proc. Nat. Acad, of Sci. U.S.A. 44

(1958), 456-458.

7. Steger A., " Diagonability of Idempotent Matrices," Pac.

J.Math.. 19 (1966), 535-542.

44

BIBLIOGRAPHY

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Goodearl, K., R., Von Neumann Regular Rings. London, Pitman

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Hungerford Thomas W., Algebra. New York, Springer Velag,

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Hutchins Harry C.. Examples of Commutative Rings. Harry C.

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Jacobson Nathan, Basic Algebra I. New York, Freeman, Inc.

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Lam, T., Y., A First Course in Noncommutative Rings. New

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McCoy Neal H., The Theory of Rings. New York, Bronx, Inc.

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Rowen Louis, Ring Theory I. York, Acaddemic Press, Inc.,

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Chacron, M., "On Algebraic Rings," Bull. Austral. Math.

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Ehrlich, Gertrude, " Unit Regular Rings," Portugal. Math..

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45

46

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Foster, A.L., " Maximal Idempotent Sets In A ring With

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Gilman L. and Henriksen M., "Some Remarks About Elementary

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McCoy, N., H., " Generalized Regular Rings," Bull. Amer.

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