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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
Books
Goodearl, K., R., Von Neumann Regular Rings. London, Pitman
Publishing, Inc., 1979.
Hungerford Thomas W., Algebra. New York, Springer Velag,
Inc. 1974.
Hutchins Harry C.. Examples of Commutative Rings. Harry C.
Hutchins 1981.
Jacobson Nathan, Basic Algebra I. New York, Freeman, Inc.
1985.
Lam, T., Y., A First Course in Noncommutative Rings. New
York, Springer Verlag, Inc., 1991.
McCoy Neal H., The Theory of Rings. New York, Bronx, Inc.
1973.
Rowen Louis, Ring Theory I. York, Acaddemic Press, Inc.,
1988.
Articles
Beard Jocob T., Jr. and McConnel Robert," Matrix Fields
Over The Integers Modulo m," Linear Algebra
Applications. 14 (1976), 95-105.
Chacron, M., "On Algebraic Rings," Bull. Austral. Math.
Soc.. 1 (1969), 385-389.
Ehrlich, Gertrude, " Unit Regular Rings," Portugal. Math..
27 (1968) 209-212.
45
46
Ehrlich, Gertrude," Units and One-sided Units in Regular
Rings," Trans. Amer. Math. Soc.. 216 (1976), 81-90.
Farahat, H.K and Mirsky, " Group Membership in Rings of
Various Types," Math. Zeitschr.. 70 (1958), 231-244.
Fisher, Joe W. and Snider, Robert L., " Rings Generated by
Their Units," J. Algebra. 42 (1976), 363-368.
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.
Foster, A.L., " Maximal Idempotent Sets In A ring With
Unit," Duke Math. J.. 13 (1946) 247-258.
Gilman L. and Henriksen M., "Some Remarks About Elementary
Divisor Rings," Tran. Amer. Math. Soc.. 82 (1956)
363-365.
Hartwig, Robert E. and Luh Jiang, " A Note on The Structure
of Unit Regular Ring Elements, " Pac. J. Math.. 71
(1977), 449-461.
Henriksen, Melvin, " two Classes of Rings That are
Elementary Divisor Rings," Arch. Math. (Basely. 24
(1973), 133-141.
Henriksen, Melvin, " Two Classes of Rings that Are Generated
By Their Units," J. Algebra. 31 (1974), 182-193.
Kaplansky I., "Elementary Divisors And Modules," Trans.
Amer. Math. Soc.. 66 (1949), 464-491.
Krupnik Naum, " Minimal Number of Idempotent Generators of
Matrix Algebra Over Arbitrary Field," Communication in
47
Algebra. 20 (1992), 3251-3257.
Losey, Gerald and Schneider, Hans, " Group Membership in
Rings and Semigroups," Pac. J. Math.. 11 (1961),
1089-1098.
Luh, J., " A characterization of Regular Rings," Proc. Japan
Acad.. 39 (1963), 741-742.
Luh, J., " A note on Strongly Regular Rings," Proc. Japan
Acad.. 40 (1965), 74-75.
McCoy, N., H., " Generalized Regular Rings," Bull. Amer.
Math.SOC.. 45 (1939), 175-178.
Park, J., K. and Hirano, Y., " On Self-Injective Strongly
7r-regular Rings," Communication in Algebra. 21 (1993),
85-91
Seshadri C. s., " Triviality of Vector Bundles Over Affine
Space K2," Proc. Nat. Acad, of Sci. U.S.A. 44 (1958),
456-458.
Steger A., " Diagonability of Idempotent Matrices," Pac. J.
Math.. 19 (1966), 535-542.
Xue, W., 11 Three Questions on Strongly 7r-Regular Rings and
Regular Rings," Communication in Algebraf 21 (1993),
699-704.