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 Title:
 Properties of twisted semigroup rings
 Added title page title:
 Twisted semigroup rings, Properties of
 Added title page title:
 Semigroup rings, Properties of twisted
 Creator:
 Quesada Rettschlag, Antonio, 1948
 Publication Date:
 1978
 Language:
 English
 Physical Description:
 vii, 91 leaves : ; 28 cm.
Subjects
 Subjects / Keywords:
 Abstract algebra ( jstor )
Algebra ( jstor ) Homomorphisms ( jstor ) Integers ( jstor ) Kegs ( jstor ) Mathematics ( jstor ) Polynomials ( jstor ) Semigroups ( jstor ) Subrings ( jstor ) Torsion ( jstor ) Associative rings ( lcsh ) Dissertations, Academic  Mathematics  UF Group rings ( lcsh ) Mathematics thesis Ph. D Rings (Algebra) ( lcsh ) Semigroups ( lcsh )
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 bibliography ( marcgt )
nonfiction ( marcgt )
Notes
 Thesis:
 ThesisUniversity of Florida.
 Bibliography:
 Bibliography: leaves 8990.
 General Note:
 Typescript.
 General Note:
 Vita.
 Statement of Responsibility:
 by Antonio Quesada Rettschlag.
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 University of Florida
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 University of Florida
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 Copyright [name of dissertation author]. Permission granted to the University of Florida to digitize, archive and distribute this item for nonprofit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.
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Full Text 
PROPERTIES OF TWISTED SEMIGROUP RINGS
By
ANTONIO QUESADA RETTSCHLAG
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1978
ACKNOWLEDGEMENTS
I would like to express my gratitude to all the people
that in some way have contributed to make this work possible;
they are:
Dr. Jorge Martinez, Dr. Gerhard Ritter, Dr. Sergio
Zarantonello and Dr. Elroy Bolduc, Jr., who served as members
of my committee;
Dr. Bruce Edwards whose help in the computer work was
invaluable;
Dr. Bolling Farmer and Dr. Robert Long who assisted in
improving my background in related areas;
Dr. Mark Teply, who taught me most of the algebra that
I know, and who as an advisor was always ready to listen
with his limitless patience and unswerving interest. I
consider it a privilege to have had the opportunity to work
under his guidance.
In addition I would like to recognize the efforts and
sacrifices of my family throughout my years of study. I am
indebted to my parents for providing me the opportunity to
complete my graduate studies in Spain and to my wife and
daughter for having the patience and understanding which
made the burden of my doctoral studies considerably lighter.
Without the faith and love of these four people I could never
have finished. To them I dedicate this work.
ii
Finally, my thanks to Sharon for her fine and fast
job of typing.
This work was performed during a sabbatical granted to
me by the Catholic University of Puerto Rico.
iii
TABLE OF CONTENTS
Page
ACKNOWLEDGEMENTS .......................................ii
ABSTRACT ............................................... v
CHAPTER
I PRELIMINARIES AND GENERAL PROPERTIES OF SEMIGROUP
RINGS ............................................... 1
1.1 Semigroups .................................... 1
1.2 Rings ......................................... 4
1.3 Semigroup Rings ............................... 8
1.4 Twisted Group Rings ........................... 9
1.5 Twisted Semigroup Rings ....................... 21
II ON NONTRIVIAL UNITS OF TWISTED GROUP RINGS AND
TWISTED SEMIGROUP RINGS ............................ 24
2.1 Nontrivial Units in Twisted Group Rings ....... 26
2.2 Nontrivial Units in Twisted Semigroup Rings ...32
2.3 Nontrivial Units in Twisted Semigroup Rings
Over Nontorsion Free Semigroups ............. 35
2.4 Zero Divisors in Twisted Group Rings .......... 38
III NILPOTENT RADICALS OF NONCOMMUTATIVE SEMIGROUP
RINGS .............................................. 41
3.1 An Example .................................... 42
3.2 Semigroup Rings Over Right (Left) Zero
Semigroups and Their Nilradicals ............ 46
3.3 The Nilpotent Radical of the Semigroup Ring
of a Right (Left) Group ..................... 50
3.4 Semigroup Rings Over Semigroups with Univer
sally Minimal Ideals, and Their Nilradicals.60
IV SEMIPRIME SEMIGROUP RINGS .......................... 67
4.1 Semiprime Twisted Semigroup Rings ............. 69
4.2 On Semigroup Rings Over Semigroups Admitting
Relative Inverses ........................... 75
BIBLIOGRAPHY ........................................... 89
BIOGRAPHICAL SKETCH .................................... 91
iv
Abstract of Dissertation Presented to the Graduate
Council of the University of Florida in Partial Fulfillment
of the Requirements for the Degree of Doctor of Philosophy
PROPERTIES OF TWISTED SEMIGROUP RINGS
By
Antonio Quesada Rettschlag
August 1978
Chairman: Mark L. Teply
Major Department: Mathematics
Let R be a ring with identity, and let S be a multipli
cative semigroup with identity i. Let y be a function from
S x S into the group of central units of R such that
y(sls2)y(sls2,s3) = Y(s2,s3)Y(sls2s3) for all Sl,S2,S3 E S.
We define the twisted semigroup ring, denoted by Rt[S], as
an Ralgebra with basis {sls E S} and with multiplication
defined distributively such that for all SlS2 E S,
SlS2 = y(Sl,S2)SlS2.
Assuming that y is nonconstant, we prove the following
statements.
(A) If K is a field and G is a nontorsion free group,
then Kt[G] has nontrivial units.
(B) If S is a nontorsion free semigroup and char R is
finite, then Rt[Sj has nontrivial units.
(C) If S is a nontorsion free semigroup and K is a field
then Kt[S] has nontrivial units.
v
(D) If RtLSJ has no nontrivial units, then R has no
nilpotent elements and either R is indecomposable or the
largest subgroup of S containing i is trivial.
Next we let NLR[S]i denote the nilpotent radical of RES].
We prove the following statements.
(E) If R is a commutative ring with 1 and S is a right
(left) zero semigroup, then NERES]] = {Eaixi E R[S1Iai = 01.
(F) If R is a ring with 1 and S is a right (left)
group, then N[RES]] = N[R[H1]] + C, where H1 is any maximal
n n
subgroup of S and C = R[HI].W with W = { i aiei=i = 0,
ei is an idempotent in S and n E Z+}.
(G) Let R be a ring with 1, and let S be a semigroup
with a universally minimal ideal U. Let 8 be the semigroup
homomorphism from S onto U defined by 0(s) = s.l. Then
N[R[S]] = NER[U]] 0 NEK], where K = Ker 0 and e:R[S] R[U]
is the natural extension of 0.
(H) If R is a ring and S is a semigroup satisfying
Bauaschewski's condition on extremities, then R t[S] is semi
prime if and only if R is semiprime.
We say that a semilattice is of type P0 if it has a zero
element w and all the nonzero idempotents are primitive. A
semilattice is said to be of type P1 if it has a zero element
and satisfies the following condition: if F is a subset of
PI' then either g.l.b. {xjx E F} belongs to F or else there
exist x and y in F such that g.l.b. {x,y} is in P1 F.
Let S be a semigroup such that (i) S admits relative
inverses and (ii) the idempotents of S commute. If the
semilattice of subgroups Sa of S is of type P0' then we
decompose RLSJ as a direct sum of ideals which are completely
determined once RLS.1 is known for each a E P0. This allows
us to characterize the nilpotent and Jacobson radicals of
R[S]. For R field, we give necessary conditions for R[S]
to be semisimple and regular.
If the semilattice of subgroups SCC of S is of type P'.
then we prove that R[S] is semiprime if R[Se] is semiprime
for each a E P1"
vii
CHAPTER I
PRELIMINARIES AND GENERAL PROPERTIES OF SEMIGROUP RINGS
In this chapter we introduce some fundamental defnitions
and general results that will be used throughout this dis
sertation.
1.1 Semigroups
A semigroup S is a nonempty set closed under a binary,
associative operation that we will denote multiplicatively.
Let S be a semigroup. The cardinal number ISI of the set S
is called the order of S. If ISI is finite, we can exhibit
the binary operation in S by means of its Cayley multipli
cation table as for finite groups. An element t of S is
said to be left (right) cancellable if, for any x and y in S,
tx = ty (xt = yt) implies x = y. S is called left (right)
cancellative if every element of S is left (right) cancel
lable. We say that S is cancellative if it is both left and
right cancellative. An element u of S is called a left (right)
identity of S if us = s (su = s) for all s E S. An element
u of S is a twosided identity, or simply identity, if it is
both a left and a right identity of S. Exactly one of the
following statements must hold for a semigroup S:
(1) S has no left and no right identities;
(2) S has one or more left (right) identity, but no
right (left)identity;
(3) S has a unique twosided identity, that we denote
by i, and no other right or left identity.
An element z of S is called a left (right) zero if
zs = z (sz = z) for every s in S. An element z is called a
zero element of S if it is both a left and a right zero of
S. Clearly the foregoing trichotomy holds if we replace the
word identity by zero. S is said to be a right zero semi
group if xy = y for every x,y in S; that is, every element of
S is a right zero and a left identity. Left zero semigroups
are defined dually. A semigroup S with a zero element z is
called a zero or null semigroup if xy = z for all x,y in S.
An element e of S is called idempotent if e = e. One
sided identity and zero elements are idempotent. S is called
a band if all its elements are idempotent. By a subgroup of'
a semigroup S, we mean a subsemigroup T of S that is also
a group with respect to its binary operation; that is,
xT = Tx = T for every x in T. By a left (right) ideal of S
we mean a nonempty subset X of S such that SX c X (XS c X).
By twosided ideal, or simply ideal, we mean a subset of S
which is both a left and a right ideal of S. A semigroup S
is called left (right) simple if S itself is the only left
(right) ideal of S. Likewise S is called simple if it contains
no proper ideal. It follows from the definition that S is
right (left) simple if and only if aS = S (Sa = S) for every
a in S; moreover, S is a group if and only if it is both a
left and a right simple semigroup.
If s is an element of S, then the cyclic subsemigroup
generated by s, is the set (s) of all the positive powers of S.
Theorem 1.1. Let s c S. If J(s)I is infinite all the
powers of s are distinct. If I(s)l is finite, there exist
two positive integers, the index r and the period m of s,
such that sm+r = sr and (s = {s,s2 m+ri The order
of (s) is m+rl. The set Ks = {sr,sr+l,...,sm+rl} is a
cyclic subgroup of S of order m.
An element s E S has finite order if I(s)l is finite.
S is called torsion if every element of S has finite order.
If s is an element of finite order in S, then (s) contains
exactly one idempotent, namely the identity of Ks.
Let the semigroup S have an identity i. If x and y are
elements of S such that xy = i, then x is called a left inverse
of y, and y is called a right inverse of x. A right (left)
unit in S is defined to be an element of S having a right
(left) inverse in S. By a unit in S we mean an element of
S having both right and left inverse in S.
Theorem 1.2. Let S be a semigroup with identity i.
(1) The set P(Q) of all right (left) units of S is a
right (left) cancellative subsemigroup of S containing i.
(2) The set G of all units of S is a subgroup of S, and
G = P n Q. Each unit has a unique twosided inverse in G, and
has no other left or right inverse in S.
(3) Every subgroup of S containing i is contained in G.
If e is an idempotent element of S, then eSe is the set
of all elements of S for which e is a twosided identity.
Theorem 1.3. Let e be an idempotent of S, and let He
e
be the group of units of eSe. Then He contains every sub
group of S that meets H
e
It follows that the groups He are just maximal subgroups
of S; that is, no He is properly contained in any other sub
group of S. Moreover, if e and f are idempotent elements of
S, e ; f, then He and Hf are disjoint.
The preceeding terminology and results can be found in
[ 5 J, where the reader can find considerable additional
material on semigroups.
1.2 Rings
Let R denote an associative ring. If there is a least
positive integer n such that na = 0 for all a E R, then R is
said to have characteristic n. If no such n exists, R is
said to have characteristic 0. We will denote characteristic
of R by char R. A ring of characteristic pn, for a prime p,
is called a pring [24]. (Note that there are other conflict
ing uses of the term pring; for example, see [16]). For a
ring R and a prime integer p, define Rp = {r E Rlpnr = 0 for
some positive integer n}. The center of R is the set
C = {c E Rjcr = rc for all r E R}.
If R has identity 1, the subring of R generated by 1 will
be called.the prime subring. It is an easy consequence of
the definition that if ff is the prime subring of a ring R,
then n Zn whenever char R = n > 0, and nr Z otherwise.
Theorem 1.4 [13]. Let AlA2,...,An be ideals in a ring
R such that (i) A1 + A2 + ... + An = R and (ii) for each
k (1 < k < n), Ak n (A1 +... + Akl + Ak+l + ... + An) 0.
Then there is an isomorphism R i A1 $ A2 n ... ( An'
Lemma 1.5. Let R be a ring with nonzero characteristic
n, and let {plP2,...,pkI be the set of distinct prime divisors
of n. Then R = R 1 Rp2 ( ... E Rpn.
A nonzero element a in a ring R is said to be a left
(right) zero divisor if there exists a nonzero element
b E R such that ab = 0 (ba = 0). A proper zero divisor, or
simply zero divisor, is an element of R which is both a left
and a right zero divisor. A commutative ring with identity
1 x 0 and no zero divisors is called an integral domain.
Let a be an element of R. The principal ideal generated
n
by a is the set (a) = {ra + as + na + riasilr,s,ri, siE R
i=l
and n E Z}. An ideal P in R is said to be prime if P 9 R and,
for any ideals A,B in R, AB c P implies that A c P or B c P.
An ideal M in R is said to be maximal if M e R and, for every
ideal A such that M c A c R, either M = A or A = R.
Definition 1.6 [9 ]. Let L be a certain property that
a ring may possess. We shall say that the ring R is an Lring
if it possesses the property L. An ideal I of R will be
called an Lideal if I is an Lring. A ring which does not
contain any nonzero Lideals will be called Lsemisimple. We
shall call L a radical property if the following three conditions
hold.
(A) A homomorphic image of an Lring is an Lring.
(B) Every ring contains an Lideal A which contains
every other Lideal of the ring.
(C) The factor ring R/A is Lsemisimple.
n
An element x of R is said to be nilpotent if x = 0 for
some positive integer n. A ring R (an ideal A) is said to
be nil if every element x of R (of A) is nilpotent. An
ideal is called nilpotent if some power of it is zero. We
say that n is the index of nilpotency of a nilpotent ideal
n
A, if n is the smallest positive integer such that A = 0.
Lemma 1.7 [ 9 ]. The sum of any finite number of
nilpotent ideals of a ring R is again a nilpotent ideal.
The union of all the nilpotent ideals of a ring R isa nil
ideal.
In [ 9 1 we find the following construction. Denote by
N0 the union of all the nilpotent ideals of R. Then R/N0
may have nilpotent ideals. Let N1 be the ideal of R such
that NI/N0 is the union of all the nilpotent ideals of R/N0.
In general, for every ordinal a that is not a limit ordinal,
we define N to be the ideal of R such that N /N _1 is the
union of all the nilpotent ideals of R/N _I. If a is a limit
ordinal, we define N = u6< N In this way we obtain an
ascending chain of ideals Noc NC ... N ... Since
R is a set, we may then consider the smallest ordinal T such
that NT = NT+l = ... We denote NT by N[R], and NER] is
called the nilpotent radical of R. (N[RI is also called the
Baer lower radical of R in the literature [ 9 1). It is
characterized by the fact that R/NLR] is semiprime and N[RI
is the smallest ideal of R that gives such a factor ring.
w
Lemma 1.8. Let R = E XAR, be the weak direct sum of
the family {ROIXA of rings. Then the nilpotent radical of
R is Ew
We define the prime radical of R as the intersection'of
all prime ideals of R, and we denote it by P[R].
Propositon 1.9 [151. The following conditions con
cerning the ring R are equivalent.
(i) 0 is the only nilpotent ideal of R.
(ii) 0 is an intersection of prime ideals; that is
PLR] = 0.
(iii) For any ideals A and B of R, AB = 0 implies that
A n B = 0. Under these conditions R is called semiprime.
The following definitions and results can be found in
[17J.
Let a E R. If there exists an element b of R such that
a + b ab = 0, a is said to be right quasiregular (r.q.r.).
The Jacobson radical JER] of a ring R is defined as follows:
J[RJ = {a c Riar is r.q.r. for every r E R}.
Theorem 1.10. Let R be a ring with identity, and let
{Aiji E I} be the set of all the maximal right ideals in R.
Then J[R] = n A..
iEI 1
A ring R is called Artinian if the descending chain
condition for right ideals holds in R; that is, every strictly
decreasing chain of right ideals in R contains only a finite
number of right ideals.
Lemma 1.11. If R is artinian, then the Jacobson
radical of R coincides with the prime radical of R.
1.3 Semigroup Rings
Let R be a ring, and let S be a multiplicative semigroup.
N. Jacobson F14] defines the semigroup ring of S over R to
be the set of functions from S into R that are finitely
nonzero, with addition and mulitplication defined as follows:
(f + g)(s) = f(s) + g(s)
(fg)(s) = Z f(x)g(y),
xy=s
where the symbol E indicates that the sum is taken over
xy=s
all ordered pairs (x,y) of elements of S such that xy = s.
We use R[S] to denote the semigroup ring of S over R. Semi
group rings are generalizations of polynomial rings: if Z0
denotes the additive semigroup of nonnegative integers, then
R[Z0] is isomorphic to the polynomial ring in one indeterminate
over R.
n
We think of the elements of RES] as "polynomials" E a.s.,
i=l 1 1
where ai E R and si E S. A typical nonzero element a of RES]
n
has a unique representation a = E a s where the a. are
1=1 1 1 1
nonzero elements of R and the s. are distinct elements of S.
1
n
The unique representation E a s of a is called the canoni
i=l1
cal form of a, and the set {s n of "variables" of the
ii=l
canonical form of a is called the support of a (denoted by
supp a); moreover, the ideal of R generated by {a }i= is
i il
called the content of a.
If the semigroup S has an identity i, then the ring R
is naturally embedded in the semigroup ring R[S] by mapping
an element r E R to ri.
1.4 Twisted Group Rings
Let K be a field, and let G be a multiplicative group
whose identity we denote by i. Passman [221 defines a
twisted group ring K t[G] of G over K as an associative
Kalgebra with basis {xlx c G} and with multiplication defined
distributively such that, for all x,y E G,
(1) ix y = y(x,y)xy, y(x,y) E K {O}.
The associativity condition is clearly equivalent to
(xy)z = x(yz) for x,y,z E G, and this is in turn equivalent to
(2) y(x,y)y(xy,z) = y(y,z)y(x,yz)
Since
(xy)z = y(x,y)xy z= y(x,y)y(xy,z)xyz
and
x(yz) = y(y,z)x yz = y(y,z)y(x,yz)xyz
We call y:Gx G K M0l the twist function. This
function has also been called factor set in representation
theory; see for instance Curtis and Reiner [ 8 1.
There are a number of obvious similarities between
twisted group rings and ordinary ones, but there are also a
number of important differences. Passman points out the
following. First, for fixed K and G there are many different
possibilities for K t[GI, depending on the choice of the twist
function. Indeed, one such choice is always K[G], which is
obtained by taking y(x,y) = 1 for all x,y. On the other
hand, of course, KEG] is uniquely determined by K and G.
Second, G is not embedded naturally in K t[GI; that is, the
t
map G K [GI given by x x is not a multiplicative homomor
phism in view of equation (1). Third, if H is a subgroup of
G, then the Klinear span of {hlh E H} is clearly a twisted
group ring of H over K, but we must be careful and note
that this Kt[H] has as a twisting the restriction of the
function y to H x H. Finally, if H is a normal subgroup of
G, there is no obvious homomorphism from K t[Gi to some
twisted group ring of G/H.
Lemma 1.12. The following relations hold in K t[GI.
(i) Y(x,i) = y(i,z) = y(i,i) for all x,z E G.
(ii) y(i,i) I is the identity element of K [Gi.
(iii) For all x E G,
x =)l y(i,i) x y(x1 x) Iy(i i)_i x_
Thus we see that the elements x E K t[G] are all units.
Proof. (i) Putting y = i in equation (2) yields
y(x,i) = y(i,z). Thus for all x,z c G we have
y(x,i) = y(i,z) = y(i,i).
(ii) By (i) and equation (1) we see that y(i,i) l is
the identity element of K t[G].
(iii) By (i), (ii) and (1) y(x ,x) y(i,i) x is a
1 1 l
left inverse for x, and y(x,x )y(i,i) x is a right
1.
inverse for x. Thus x is invertible and x is equal to
both of these expressions.
It is quite possible that some given K t[G] is just
K[G] disguised in a rather natural way. We start with Kt[G]
and make a diagonal change of basis by replacing each x by
x' = 6(x)x for some nonzero 6(x) E K. Observe that the word
"diagonal" arises here because, for G finite, the correspond
ing change of basis matrix would be diagonal with the 6's as
diagonal entries. From (1) and the foregoing we have
x'y' = 6(x)6(y)y = 6(x)6(y)y(x,y)xy
= 6(x)6(y)6(xy) y(x,y) (xy)'
In other words, with this change of basis, K t[G] is realized
in a second way as a twisted group ring of G over K, this
time with twist function:
(3) y'(x,y) = 6(x)6(y)6(xy) y(x,y).
tlG t2
Thus we see that if K EC] and K 2[G] are two twisted group
rings of G over K and if their corresponding twist functions
y and y' are related as in (3) for some 6:G K {0}, then
these rings are actually identical, and we say that they are
diagonally equivalent. Inparticular, Kt[G] is diagonally
equivalent to K[G] if and only if
Y(x,y) = 6(xy)6(x) 16(y)l
for some function 6:G K {0}.
Let K t[GI be given with twist function y. Consider
K t[GI with twist function y' defined by y'(x,y) =
y(ii) y(x,y). Clearly y' and y satisfy equation (3); hence
K t[G] and K t[G] are diagonally equivalent. Moreover by
Lemma 1.12 (i) and (ii)
(4) i is the identity of K t[Gl and y'(x,i) = y' (i,z) = y(i,i)=l
for all x,z E G. In view of this, we give the following
definition: a twist function is said to be normalized if
it satisfies condition (4).
From now on all the twist functions considered are
supposed to be normalized.
Let us consider now some classical examples. Let G be
a cyclic group of order n generated by g, and let
:K[xl Kt[G] be defined by f(x) = g. Then is a ring
homomorphism such that fixn) = ai, where a = y(g,g)y(g,g2)...
y(g,g n). Thus (xn) = p(a), and hence the kernel of is
the principal ideal generated by the polynomial x n a. It
follows that K t[G] K[x]/(x na). Conversely, for any
a e K {0}, we see that K[xl/(xn a ) has as a Kbasis the
images of l,x,x 2...,x ; so it is a twisted group ring of
Cn, the cyclic group of order n. Thus, for instance, if
K = Q is the field of rationals, then by the Eisenstein's
criteria the polynomial xn 2 is always irreducible. Hence
for a = 2 we see thatQ Cn I Q[x1xAn2) is, in fact, a
field extension of Q, namely Q[n/2]. Moreover, if K is a
field of characteristic p > 0 that is not perfect, choose
n
a E K with t',/aJ 4 K. Then for all n, xp a is irreducible
n
over K and the field K[x1/(xp a) = K[nval is a twisted
group ring of the cyclic group C n' A last application of
p
the foregoing example is obtained when we let K be an
algebraically closed field and G = Cn. Then x = ai and
nVa exists in K. Since (a /nx)n = I it follows now that
K t[G is diagonally equivalent to K[G].
In the case where G = (g) is infinite cyclic, then we
m in
can make a diagonal change of basis by replacing g by g
for all integers m. This clearly implies that K t[GI is
diagonally equivalent to KEG].
Finally, if we let G = {i,g1,g2,g3} be the Klein group,
and let K be a subfield of the real numbers, then by defining
y(gl,g2)= Y(g2,g3) = Y(g3,g1) = 1
and
y(g2,g1) = Y(g31g2) = y(g1,g3) = y(g1,g1) = y(g2,g2)
= y(g3,g3) = 1,
we obtain a twisted group ring that is easily recognized as
the quaternion division algebra.
Just to mention how the twisted group rings might arise
we give the next two lemmas, whose proofs can be found in [221.
Lemma 1.13. Let Z be a central subgroup of G, and let
L be a maximal ideal in K[Z]. If I = LK[G], then
K[G]/I = Ft[G/Z] is a twisted group ring of G/Z over the
field F = K[ZI/I.
It turns out that every twisted group ring arises in
this manner as we see next.
Lemma 1.14. Let K t[H] be a twisted group ring of H
over K. Then there exists a group G with a central subgroup
Z such that G/Z 'H and K[G]/I K t[H], where I = L.K[G]
and L is an ideal of K[Z] with K[Z1/L K.
So far, we have seen one example where, if K is an
algebraically closed field, then K t[GI is diagonally equiva
lent to KEG], namely, when G is a cyclic group. We state
two more results that give conditions in order for this to
happen.
Lemma 1.15. Let K t[G] be given with K algebraically
closed. Set U = {axla E K {0},x E G}, and let
W = {aija E K {0}}. Then Kt[G] is diagonally equivalent
to K[G] if and only if U', the commutator subgroup of U,
is disjoint from W.
As an immediate consequence, we obtain the following
result.
Corollary 1.16. Let K be an algebraically closed field,
and let G be abelian. Then K t[G is diagonally equivalent
to K[G] if and only if K t[G is commutative.
A final example, in which Kt [G] is always K[G], is pro
vided by the case where K is a perfect field of characteristic
p and G is a finite pgroup.
We remark that the material discussed so far in this
section has been taken from Passman [221. Next we give
some new results that we have obtained in order to be able
to multiply elements of a twisted group ring over a finite
cyclic group.
In the remainder of this section we let G = (g) be a
cyclic group of prime order q, unless otherwise specified.
Definition. 7a = q1 y(ga,gh ), where 1 < a q.
a h=l
Notice that since y is normalized, R = 1 and Ta = 7 ay(gagq)
q a"
Proposition 1.17. 7l1 . = 7i+jy(gig J)q, where
1 i,j : ql.
Proof. Putting x 9 g y = gJ and z = g in equation
(2), we have
(5) y(g ,g3+k)y(g jgk) = y(g i+jgk) y(gl gj).
Fixing i and j and taking products, we get
q1i q1 i+j ,gj
11 Y(gg3+k)y(gJ'gk) q y( ,gk)y(gi );
k=2 k=2
hence
q1 qi qi
[ R y(g i + I y(g g H I Y(ijgk)]y(gigi)q2
k=2 k=2 k=2
It follows that
It. Jr. .
1 T. T_ i+j Y g~ j) q2 .
y1(gig *y(gl,g +l y(gj,g) y (g1+ y~j g g)
Then
Y.(. = (g gj)qi y(gi gj+l)y(gjg)
1T J TT Y(gi+jg)
By letting k = 1 in (5), we get y(gi,gj+l)y(gj,g) =
y(gi+jg)y(ggi,j). Therefore, *i jTT. = 'i+j y(gigj)q,
where 1 < i,j < ql.
Lemma 1.18. Let y(gi g k) be given for k = 1,2,...,q1.
If 1 !< x < ql, x x i, then
7T = y(gi ,gqi)qy(gi gq2i) 2 ... (gi gqni)q/7n
x 1~ g Yg'gYgi
where n is the smallest positive integer such that x ni
(mod q).
Proof. First, we claim that if 1 < x !< ql, then
x E qni (mod q), where 1 < n !< ql. To prove this is
enough to show that any two elements of the set {qni 0 :< n <
ql} represent different residual classes of Z For con
q
tradiction, suppose that qnli_= qn2i (mod q), where
0 < nI < n2 5 q1. Then nI n2 (mod q), which contradicts
0 < n2n1 < q.
Let x qni (mod q). By setting j = qi in Proposition
1.17, we get n qi = Y(g, ,gqi)q/i since iT = i. Hence, if
q
we let j = q2i Proposition 1.17 yields Tq2i =
q iy(g ,gq2i)q/Ti = y ,gi qi)qy(g ,gq2i)q/ 2. Thus the
result follows by recursion.
Corollary 1.19. If K = GF(2 n), where n is a positive
integer such that 2n _1 = q, then i= Moreover,
if 7. = 1 for some i, 1 < i ql, then = 1 for all j.
1 J
Proof. By hypothesis K is a finite field of order q+l;
hence its group of units is cyclic of order q. It follows
that y(gi1 j)q = 1 for all i,j. Then by Proposition 1.17
ri 3. = Hi+j. If, moreover, 7i = 1, then the conclusion
follows from Lemma 1.18.
Next we determine some conditions that force the twist
function to be trivial.
ih
Proposition 1.20. If y(g ,g = 1 for h = 1,2,...,q1
and i fixed, then y is trivial.
Proof. By the hypothesis and equation (5) we get
y(gj,gk) =y(g i+jgk) with 1 :< j,k < ql. As in Lemma 1.18,
j qni (mod q) for some n. We use induction on n.
If n = 1, then Y(gqi gj) = y(i,gk) = 1 for all k. Assume
qti k q(t+l)i k
that y(gq ,gk) = 1 for all k. Then y(gq,gk) =
qti 1 all
y(g ,gk) 1 for all k. Therefore y(gj,gk) 1 for all
j and k.
Lemma 1.21. If G is a cyclic group generated by g,
n mn
then y(g ngi) = y(g ,g n) for all n,m.
n1 gm 1)
Proof. Since 9n = n y(g,gi )gn and gm = rny(ggigm
i=l j=l
then
ni mi
n mrng
gg = ( Y Y(g,g')g )( ri y(g, )gm) =
i=l j=l
ni i m1 m) n+m
= ( f y(g,g1))( I y(gqg))y(gn gg
i=l j=i
and
m n
g *g
mi ni
T I y(ggj)gm) ( T y(g, )gn) =
j=l i=l
mi nI yg g+n
= ( I y(g,gJ)( IT y(g,gi))y(gm,gn) .
j=l i=l
nm n+m mn nm n
Theng g = g = g g implies that y(gn,gm) = y(g ,gn
for all m,n.
Lemma 1.22. If 1 < x < ql, then
y(gX,gX) = y(g,gX)y(g,g
x1
y(g,g ).
x+l)...y(g,g2Xl)/y(g,g)y(g,g2 )...
Proof. Fix i = 1 and k = x in equation (5), and first
let j = x. Then y(gg2X)y(gX,gX) = y(gX+l,gX); hence
X 1X ~ x 2
y(gX,gX) = y(g,gX)y(gX+ ,g )/y(g,g ). If j = x + 1, then
equation (5) yields y(gX+l gX) = Y(g,gx+i)y(gx+2 gX)/y(gg2X+l)
Thus y(gX,gX) = Y(g,gx)y(g,gx+l)y(gx+2,gx)/y(g,g2x)y(g,g2x+l).
For induction, assume that
Y (gX ,gX)
 ~gX)y gx+l)..ygx+ml)
y(g,g )y(g,g ) ... (g'g
(g'g 2x)y (g,g 2X+l ) ...y (g'g 2X+ml)
y(gX+m,gX .
Letting j = x+m in equation (5), we get
y(gX+m gX) = y(g,gx+m)
y(g,g2X+m
x(gX+m+l x
y~g ,g ).
y(ggX)y(ggX+) y(gg x+m) x+m1 x
Y(g,g 2X)y(g,g2x+l) ... y(g,g2x+m)
y~g~g)y~gQ
Finally, since for j = ql, we have
y(gq1 gx)
Y (gX,gX) =
= y(g,gql) x(gq gX) = y(g,gql)
Y(g,gX+ql) y(g,gXl)
y(g,gX) Y(g,gx+l).. (g,gql)
y(g,g2X)y(g,g2X+l) ...y(g,gX1
71/y(g,g)y(g,g 2)..y(g,gX )
7T 1/y(g,gX)y(g,gX)y...y(g,g2x1
Y (2a gX) Y(ggX+l) .. Y (g, g2xl
y(g,g) (g,g2) ...y(g,gX1)
Proposition 1.23.
Let K = GF(2 n), where 2n1
a prime. If 1 < p ql, then
y(gp gP)2n Y(g2p g2p)2n2 (g2kpYg2kp 2 2nkl
Sgni 2nI
Y(g2 p,g ) =
1*
Hence
y (gX, gX)
Thus
= q is
Proof. For b,c E Z, b < c, let [b,c] denote the product
y(ggb )Y(g,gb+l)...y(ggC). by Lemma 1.22 we see that
y(g 2kp,g2k P) = [2 kp,2 k+pl]/[,2 kpil for k = 0,1,...,n1.
Hence
(6) 2ni 2 2n2 Y(g2nip g2nlp)
(p, 2pl] 2
[i.Elpi]
[p,2p12
nT 12i
i=0
[l,pl]
ii 2 2n2
(1[2p,2 2pI])2 
[1,2pl
I 2 22p_1 2n2
2 2'
[2p,2 p ] [2n
n 22i
i=0 2
[p,2pl] [2p,2 2pi]
.. nlp,2 npil]
[1,2 nl P]
2p,2 nlp] 2[2 nlp,2 np]
...[2n2p,2 nlpl]
na i na+l
Since E 2 2
i=O
in equation (6) equal to
"1
(7)
for a < n, we get the expressions
"9
[p,2p1] [2p,2 2pi] '[2 n2p,2 nlpi] [2 np,2 pl]
2nli1 2 2n21 ... n2 nl
[l,plJ [p,2pl] [2p,2 pi] [2 p,2 pi]
But from the hypothesis the multiplicative subgroup of K is
cyclic of order 2 n1; hence [l,pl]2nI = 1. Thus by can
cellation (7) becomes [p,2pl][2p,22pl].[2nlp,2npl] =
[p,2 npi]. Now for all b E Z, [b,b+qi] =y(g,g b)y(g,g b+l)
Y(g'gb+ql = Y(g,gb b+l )'. Y(g'i)Y(g'g)'''Y(g'gbi)=7I.
Since [p,2np11 is a product of (2 np) (p1) = (2 nl)p= qp
factors, then [p,2 npl]= [p,p+ql][p+q,p+2ql].*[p+(pl)q,
p+pql = Therefore the result follows.
Corollary 1.24. Under the hypothesis of Proposition
1.23, if uj = 1 for some j, 1 !5 j < ql, then
2ni 2 2n2 2ni1 2np
y(gpgp)2 y (g 2p,g 2p) ... y(g P ,g .P)
Proof. This follows directly from Corollary 1.19 and
Proposition 1.23.
1.5 Twisted Semigroup Rings
Let R be a ring with identity and let S be a multipli
cative semigroup. Let y be a twist function from S x S into
the group U of central units of R; that is, y(x,y)y(xy,z) =
y(y,z)y(x,yz) for all x,y,z c S. We define a twisted semi
group ring, denoted by Rt[S], as the Ralgebra with basis
{sIs c S} and with multiplication defined distributively
such that, for all Sl,S2 in S,
SlS2 = y(sls2)sls2
Since
(r s.r 2 s2)r3s3 = rlr2y(sls2)sls2*r3s3 =
r1r2Y(sl,s2)r3Y(sI s2 s3)sls2s3
and
r1s1(r2s2.r3s3) = r1s1.r2r3y(s2,s3)s2s3 =
r1r2r3y (s21's3)y(sl's2s3 )s 1s2s3
then r1r2Y(sls2)r3Y(s1s2,s3) = r1r2r3Y(s2,s3)y(sl1s2s3)
for all rl,r2,r3 c R and sts,2,s3 c S. This shows that the
condition that y be defined into the central units of R
is needed in order to preserve the associativity.
From now on we will abuse in our notation by identifying
the elements of S (of a group G) with their images in Rt[S]
(R t[GI) under the natural embedding; that is, we will write
s(g) for s(g).
n
Let A be an ideal of R. The set B = { aisilai A,
+ t i=lI
si E S, n E Z } is an ideal of R [S] that we will denote by
the (abusive) descriptive symbol At[S]. If the semigroup
S has an identity, then the ideal A t[SI is the extension
of the ideal A of R to Rt [S]; that is, ARt[S] = At [S].
Moreover, AR t[S] n R = A.
Let X be an ideal of S. Then Rt[X] is an ideal of
R t[SI, where the twist function of Rt [XI is the restriction
to X x X of the twist function of Rt[S].
The following results are generalizations of similar
results in [201 for semigroup rings.
Lemma 1.25. If {AIAEA is a family of ideals of the
ring R, then (n A)tES] = n[S(Ai =S).
Theorem 1.26. Let R1 and R2 be rings with identity,
and let 0:R1 R2 be a homomorphism of R1 onto R2. Let S
be a semigroup, and let yi:S x S Ri, i=l,2, be twist
functions such that ey1 = Y2. Then there is a unique
homomorphism 5:R1 [S] R2 [SI from the ring R1 [S] with
twist function yi onto the ring R1t [S] with twist function
Y2 such that O(rs) = 0(r)s for each r ( R1 and s c S. More
over if A is the kernel of 0, then the kernel of 0 is A [S].
It follows that if A is an ideal of the ring R, then
R t[SI/A t[SI L (R/A) t[S]. Moreover, 0 is onetoone if and
only if 5 is onetoone.
Finally, in the remainder of this dissertation, we
denote the nilpotent, prime and Jacobson radicals of R by N,
P and J respectively; we maintain our previous notation for
these radicals whenever we deal with twisted or ordinary
semigroup rings.
CHAPTER II
ON NONTRIVIAL UNITS OF TWISTED GROUP RINGS
AND TWISTED SEMIGROUP RINGS
Higman [12] defines a trivial unit of a group ring REG]
as any element of the form rg, where r is a unit of R and
g E G. He also characterizes the group rings that have
always nontrivial units in the case where R is the ring of
the rational integers and G is a torsion group. Since then,
the problem of the existence of nontrivial units in group
rings, together with the characterization of the group that
they determine, has been largely studied. We only mention
a few such results. Ayoub and Ayoub [1 1 determined the group
of units of ZG, where G is a finite abelian group. Raggi
Cardenas [231 calculates the group of units of K[G] for any
Galois field K and any finite abelian group G. Passman [21,
Lemma 26.11 shows that if K is any field and G is a nontorsion
free group, then K[G] has nontrivial units provided IKI > 3.
Finally, Gilmer and Teply [111 have shown that if S is a
torsion semigroup and R[S] has no nontrivial units, then the
characteristic of R is 2,3 or 0; moreover, if R is an algebra
over a field, if S is torsion and if R[S] has no nontrivial
units, then the characteristic of R is either 2 or 3. Finally,
they characterize the semigroup rings of characteristic 2
or 3 which have no nontrivial units.
Let K be a field, and let G be a nontorsion free group.
Moreover, let R be a ring with identity, and let S be a
semigroup with identity i. In this chapter we answer the
following questions:
(A) When does the twisted group ring have nontrivial
units?
(B) What are some necessary conditions for R t[S] to
have no nontrivial units?
(C) If S is a nontorsion free semigroup and R has
finite characteristic, when does R t[SI have nontrivial units?
(D) If S is a nontorsion free semigroup, when does
K t[SI have nontrivial units?
In section 2.1 we answer (A) by showing that K t[G]
has always nontrivial units, provided that the twist function
is nonconstant.
Question (B) is answered in Theorem 2.9, which states
that if Rt [SI has no nontrivial units, then R has no nilpotent
elements and either R is indecomposable or the largest
subgroup of S containing i is trivial.
In Theorem 2.15 we prove that if S is a nontorsion free
semigroup and R has finite characteristic, then R t[SI has
nontrivial units, provided that the twist is nonconstant.
This answers question (C).
Theorem 2.16 answers question (D) by showing that if S
is nontorsion free and the twist function is nonconstant,
then Kt[S] has nontrivial units.
Passman [22] proves the existence of nontrivial units
in KEG] by showing first that these group rings always have
proper zero divisors. This is why, though independent of
of the chapter's title, in section 2.4 we study proper zero
divisors in some twisted group rings and give some sufficient
conditions for their existence.
2.1 Nontrivial Units in Twisted Group Rings
Let K be a field, and let G be a nontorsion free group.
We denote by i the identity of G, and by char K the character
istic of K. Passman [22, Lemma 13.1.11 has proven that with
the exceptions of K = GF(2), IGI = 2 or 3, and K = GF(3),
IGI = 2, K[G] has nontrivial units. Our goal in this section
is to extend this result to twisted group rings by showing
that KtEG] has nontrivial units whenever the twist function
is nonconstant.
Lemma 2.1. Let K be a field of characteristic distinct
from 2. Let G be a nontorsion free group. Then the twisted
group ring K t[G always has nontrivial units, provided that
the twist function is nonconstant.
Proof. Since G is nontorsion free and IGI > 1, it
contains a subgroup H of prime order. Moreover, since any
unit of K t[HI is clearly a unit of K t[GI, then it is sufficient
to prove the result for G = Cq, where Cq is the cyclic group
of order q generated by g.
Let the characteristic of K be p, where either p = 0
or else p is a prime integer different from 2.
ql i
Let n= y(g,g).
i=l
ki
Set b ki H y(g,g ) for k > 2.
i=l
Notice that bq1 = bq = 7i' and g(bk = bk for
k > 2. We consider two cases.
First, if Ti f 1 (mod p), then 1 0 (mod p);
hence 1 1 has an inverse. Thus for q > 2, we have
i ql k
[ (i 7I) (ig)1(i + g + E bk1g ) =
k=2
i qi k qi k
(171)(i + g + E bkIg g bk1 ili)
k=2 k=2
(lrl)l(l_7rl)i = i.
If q = 2, [ (lffl)1(ig)](i+g) (lfl) (ig+gig 2
(l T l)l 7T 1l)i = i.
Secondly, if 1l 1 (mod p), then p t 2 implies that
there exists (1 + 7r1) Thus
q1 q 3
1 27 k 2 k
(1+ T) (i+g) (i+rig + L k19 E bk1 ) =
k=2n k=2n+l
q1 n=l n=l q1
2n b klg bk1 +
(1 + rri) (irrig + ~ t b1kg ~ b~g
k=2n k=2n+l k=2n
n=l n=l n=l
q3
k + 1+ i
2 bk1g + 'li) 1 (1 + T1)i + (1 I1)(g +
k=2n+l
n=l
q1
k
E bk1gk)] = i.
k=2n
n=l
Remark 2.2. Notice that by using 'i (i q 1)
instead of l in the proof of Lemma 2.1, we get the same
kind of units.
The following result provides an example where the
twisted group ring has nontrivial units and the usual group
ring does not.
Corollary 2.3. If K = GF(3) and G = C2, then K t[G]
has nontrivial units whenever the twist function is non
constant.
Next, we consider the case of fields of characteristic
2.
Lemma 2.4. Let K be a field of characteristic 2, and
let G be a nontorsion free group. If the twist function
is nonconstant, then K t[GI has always nontrivial units.
Proof. As in Lemma 2.1, it is sufficient to consider
G to be a cyclic group of prime order; say G = C = (g).
q
Assume first that q > 2. Let i1 and bk be as defined in
the proof of Lemma 2.1. We consider two cases.
If 7i t 1 (mod 2), then by proceeding as in Lemma 2.1,
we get
q1 k
[(1 rl) (ig)](i+g+ E bkgk) =
k=2
Now let i 1 (mod 2).
Suppose, first, that there exists r E K {O} such
that rq x 1; that is, 1 + rq has inverse. Then
(1+rq)l(i+rg) (i+rg + E r bkg) =
k=2
(1+rq)l(i+rg + k k+rg + qlEq r kig Tr ii)
k=2 k=2
(1+r q)i (1+rq)i = i.
Suppose next that rq = 1 for all r E K {0}. Then K
is a finite field. Since char K = 2, then K = GF(2n), where
2n 1 = q. By Corollary 1.19 = 1 for all i, 1 !5 i !5 ql,
1
so we cannot use the construction of the case where TI1 X 1.
To prove that in this case K t[G has nontrivial units, it is
sufficient to show that, for all v E K {W}, there exists
a E K t[GI such that a2 = va. For then we can choose
x,y E K {0} with x + y t 0, and set v = (x+y)/xy. Then
2
xya = (x+y)a; hence
2
(i xa)(i ya) = i (x+y)a + xya = i.
q1 I t u w
Let a = E aig E K [G]. By Lemma 1.21, y(g ,gW) =
i=l
y(gw,gU) for all u,w c Z. Thus since by hypothesis
char K = 2, then
2 q1 i 2 m 2 y i g2i m 2 igm+j)g2ji
a. i~ aig ) ~ aiy(g 'g)g + j=1 am+jy(gm+j )g2
i=l 1 il1 j=1
2
where m= q1/2. For a = va we must have
va2 if 1 i < q
(1) a2 y(gi,gi (
a va2j1l if 2q +1 i ql, where
2
Now let Q = {l,2,...,ql} and define x y if
y 2 hx (mod 2 n1), where x,y c Q, h E Z. Clearly is an
equivalence relation. Since 2np = p (mod 2nl) for all
p E Q, then each of the equivalence classes that determine
on Q has the form f2UpI 1 u 5 n }. Thus we have Q as the
union of (ql)/n mutally disjoint subsets of n elements.
Notice first, that the system (1) has q1 unknowns and
as many equations. Moreover, any unknown a is related to
P
a2p; in general, a k is related to a2kl in one equation
and to a 2k+l in another. Hence, by the previous argument
the set {a xx E Q} of unknowns is partioned into (ql)/n
subsets of the form {a Ji 1 u 5 n} for various p. Since
2Up
the unknowns of each of the q1 equations are contained in
one and only one of these subsets, then we have obtained
(ql)/n systems of n equations each of which has the form:
2 2k (g2kp g2kp)
a2 y(gP,gP) a2k2 p
(2) v = 2p
a2 a
2 2nI g2nlp)
a (g P
2 p
a
P
where 1 5 k 5 nl. Now let, for instance, a = cv, where
P
c E K {O}. Then we consider the set:
a =c
p v
2
a2p = c vy(gP,gp)
2k2ki22k2 2k2 2k2
a k = c vy(gP,gp) y(g2p,g 2p) ..y(g 2 g P)
2 p
a n = c2 vy(gPgp)2 Y(g2pg2p)23...Y(g2n2 pg 22P).
2 p
We claim that, in fact, this is a set of solutions for
(2). By direct substitution we see that they satisfy the
first ni equations of (2). Thus all we need to show is
2 nI 2nip
that a 2n (g 2 pg2 P)
2 p
a
P
that is,
2n 2 nI 2~n2 2nlp 2nI
v = c v y(gP',gP)2 y(g 2pg2p) ... y(g p g P)/cv.
2ni
But c = 1, and thus the equality follows from Corollary
1.24.
This shows that each of these systems (2) of equations
and hence (1) has a solution. In fact we obtain one set
of solutions for each of the q possible values of c. Thus
the existence of a is proven.
Finally, let q = 2; that is, let G = {i,g}. Since y is
normalized and nonconstant, y(g,g) x 1. Hence there exists
(iy(g,g)) and then [(iy(g,g))l(i+g)] is a nontrivial
unit of K t[G with inverse ig.
We can now put the pieces together to get the following
result.
Theorem 2.5. If G is a nontorsion free group and K is
a field, then Kt[G] always has nontrivial units, provided
that the twist function is nonconstant.
Proof. This follows directly from Lemmas 2*l and 2.4.
2.2 Nontrivial Units In Twisted Semigroup Rings
Let S be a semigroup (possibly infinite) with identity
i. Denote by G the largest subgroup of S containing i.
Let R be a ring with identity. In this section we find
conditions necessary to insure that the twisted semigroup
ring Rt[s] does not have nontrivial units.
Notice that in this section we have the twist function
normalized but not necessarily nonconstant.
Lemma 2.6. If R contains a nonzero nilpotent element,
then Rt [S has nontrivial units.
Proof. Let a be a nonzero nilpotent element of R, and
n
let n be the smallest positive integer such that a = 0. Set
an/2 if n is even,
b= fa(n+l)/2 if n is odd.
Then b is a nonzero element of R such that b2 = 0. Let
s c S with s t i. Then (i + bs) (i bs) = i.
Corollary 2.7. Let R = L1 e .... 0 Ln be a finite
direct sum of left ideals. If RtLS] has no nontrivial units,
then Li is a twosided ideal for i = 1,2,...,n.
Proof. Let 1 = el + ... + en, where ei E L. Then the
ei are orthogonal idempotent elements of R such that Li = Rei.
12
Thus if r E R, (eirej)2 = 0 for all i e j. By hypothesis
Rt[s] has no nontrivial units; thus Lemma 2.6 implies that
eire. = 0 for i x j. Hence LiLj = R(eiRej) = 0 for i g j.
Therefore Li is a twosided ideal of R for all i.
Lemma 2.8. Let Rt[S] be a twisted semigroup ring. Let
n
R = $ R. be a direct sum of nonzero ideals Ri, i = l,...,n.
i=l 1
Then there exist yi:S x S Ri, i = l,...,n, such that
n
Y(SlS2) = Z Yi(sls2) for all (sis2) E S x S, and
n t.
Rt[S]= 9 R.IS].
i=l 1
Proof. Recall, first, that y is defined on the central
units of R. Clearly u = uI + u2 + ... + un is a central unit
of R if and only if ui is a central unit of Ri, i = 1,2,...,n.
Let Oi stand for the canonical epimorphism from R onto Ri.
Define yi: S x S Ri by y(sls2) = eiy(sl,s2). Since y
and ei are well defined functions that preserve the associa
tivity, so does yi. Moreover, our first observation shows
that yi is defined onto the central units of R.. Finally,
1
n n
Si (s'2) = y(sls2) = (sls2) for all (slEs2)
n t.
S x S. It follows that Rt[S] = 0 Ri IS].
i=l
We remark that the functions yi defined in the former
lemma are normalized with respect to the identities of the
respective rings.
Proposition 2.8. Let R be a finite (ring) direct sum
of nonzero ideals Rif... ,Rn. Then the following conditions
are equivalent.
(i) Rt[SJ has no nontrivial units.
t.
(ii) Each Ri.1s] has no nontrivial units for i = 1,2,...,n,
and G = {il.
n t.
Proof. By Lemma 2.7 Rt[S] = D R. [S]; thus every
t i=l 1
element a E R [SJ can be written uniquely as a = a, + ... +
ti n
with a i E Ri [S]. Hence a is a unit of Rt[SJ if and only if
ti
each ai is a unit of Ri [S]. Write 1 = eI + ... + en, where
ei is the identity of Ri.
Assume that Rt [S] has no nontrivial units. If R.J[S]
J
has a nontrivial unit v. for some j, then eI + .. + vj +
+ en is a nontrivial unit of Rt[S]. Moreover, if there
exists g E G such that g e i, then elg + e2 + ... + en is a
nontrivial unit of Rt [S]. Therefore if Rt [S] has no non
t.
trivial units, then each Ri 1[S has no nontrivial units,
and G is trivial.
The converse follows easily from the fact that if
v = vi + ... + vn is a nontrivial unit of Rt[S], then for
some j, vj is a nontrivial unit of Rt [S], provided that
G = {i}.
Theorem 2.9. If R t[S has no nontrivial units, then
R has no nilpotent elements, and either R is indecomposable
or G is trivial.
Proof. This follows directly from Lemma 2.6 and
Proposition 2.8.
2.3 Nontrivial Units In Twisted Semigroup Rings Over
Nontorsion Free Semigroups
Gilmer and Teply ll] have proven that if S is a torsion
semigroup and RLSJ has no nontrivial units, then the charac
teristic of R is 2,3 or 0; moreover, if R is an algebra over
a field, if S is torsion and if R[S] has no nontrivial units
then the characteristic of R is either 2 or 3. Finally,
they characterize the semigroup rings of characteristic 2
or 3 that have no nontrivial units.
Let S be a nontorsion free semigroup with identity i,
and let G be the largest subgroup of S containing i. More
over, let R be a ring with identity, and let K be a field.
Denote by P the prime subring of R, and denote the charac
teristic of R by char R. If s E S, then (s) stands for the
subsemigroup of S generated by s. In this section we extend
the previous result to the twisted semigroup ring Rt[S]
of finite characteristic and to Kt[S], showing that they
have always nontrivial units.
We begin by proving some preliminary results, the first
of which is known.
Lemma 2.10. S G is an ideal of S.
Proof. Let sls2 ( S G, and suppose that s 1s2 = g E G.
Then sl(s2g1) = i; hence sI is invertible with respect to i
and therefore sI E G. This contradiction shows that S G
is closed. To show that S G is an ideal, suppose
sg = gl E G for some s E S G and g E G. Then s = g1g1 E G
which contradicts the choice of s. Hence (S G)S c S G,
and symmetrically S(S G) S S G.
Lemma 2.11. Assume S G contains a torsion element.
Then either char R = 2 or else R t[S] has a nontrivial unit
with support in S G.
Proof. Let s be a torsion element of S G. By Theorem
1.1 the subsemigroup (s" contains an idempotent e. By Lemma
2.10 e E S G. Set b = y(e,e). Then (i 2be)2 =
i + 4b[by(e,e)l]e = i. Hence i 2be is a nontrivial unit
of RtLS] with support in S G, unless char R = 2.
Lemma 2.12. Let char R = 2, and let e be an idempotent
of S different from i. If the twist function is nonconstant,
then e is contained in the support of a nontrivial unit of
Rt[Sj1.
Proof. Since y is nonconstant, then the group of units
of R is not trivial. Let y(e,e) = c, and let u and v be
different units of R. By recalling that c is a central unit
of R, then [ui + cl(u+v)e][uli+cl(ul+vl)e] =
i+cl[4+2ulv+2uvl]e = i.
Corollary 2.13. If all the units of RtiS! are contained
in R t[G], then S G is torsion free, provided that the
twist function is nonconstant.
Proof. Suppose S G contains a torsion element. If
char R 2, then by Lemma 2.11 Rt[S] contains a nontrivial
unit with support in S G, which contradicts our hypothesis.
On the other hand, if char R = 2, then by Theorem 1.1 S G
contains an idempotent e. Hence by Lemma 2.12 Rt[S] contains
a nontrivial unit with e in its support. Therefore S G is
torsion free.
Proposition 2.14. Let G be a nontorsion free group, and
let R have finite characteristic. If the twist function is
nonconstant, then R t[G] has nontrivial units.
Proof. We show that P t[G always has nontrivial units.
Let char P = n, and let D = {ql,q2,.. .qk} be the set of
distinct prime divisors of n. If k > 1, then P is decompos
able as a finite direct sum of qrings. Since G is nontorsion
free by hypothesis, then Theorem 2.9 implies that Pt[G] has
nontrivial units. On the other hand, if k = 1, then P is a
qring. Let n = q where m > 1. Then P has a nilpotent
element; so by Lemma 2.6 P t[G] has nontrivial units. Finally,
if n = q, then P = GF(q); hence by Theorem 2.5 Pt[G] contains
nontrivial units.
Theorem 2.15. Let S be a nontorsion free semigroup and
let R be a ring of finite characteristic with identity. If
the twist function is nonconstant, then Rt [SJ has nontrivial
units.
Proof. Let s be a torsion element of S different from i.
If s E G, then G is a nontrivial, nontorsion free group;
thus Proposition 2.14 shows that Rt[G] and hence Rt [S] have
nontrivial units. Let s E S G. Then, either char R z 2
and the result follows from Lemma 2.11, or else char R = 2.
By Theorem 1.1 and Lemma 2.10, (s) contains an idempotent
e z i. Hence, Lemma 2.12 shows that Rt ES contains a non
trivial unit when char R = 2.
If we follow verbatim the proof of Theorem 2.15,
except for changing the use of Proposition 2.14 to Theorem
2.5; then we have proven the following result.
Theorem 2.16. Let S be a nontorsion free semigroup
and let K be a field. If the twist function is nonconstant,
then K[tS] has nontrivial units.
2.4 Zero Divisors In Twisted Group Rings
It has been proven [22, Lemma 13.1.1] that if G is a
nontorsion free group and K is a field, then KrG] has
proper zero divisors. In passing to the twisted group rings
case, this is no longer true, as we see in the following
example. Consider Q t[C4],where Q is the field of rational
numbers and C4 is the cyclic group of order 4 generated
3
by g. Let 7i = y y(g,g1), and assume that
i=l
4
SK. Then x 7r is irreducible; hence
K tC4j is isomorphic to the field KLxJ/(x 4). On the
other hand, (ig2) (i+g2) = 0 gives an easy example of proper
zero divisors in KEC4j.
In this section we give some sufficient condtions for
the existence of proper zero divisors in Kt[G].
If g is a torsion element of prime order q in G, let
q1
7l(g) = H y(g,g ).
i=l
Theorem 2.17. Let K be a field, and let G be a non
torsion free group. If q 7l(g)EK for some element g E G of
prime order q, then Kt LG] has proper zero divisors.
Proof. Since any proper zero divisor of Kt[H], where
H is a subgroup of G, is a zero divisor of Kt[G], it is
sufficient to show that Kt[(g)] has proper zero divisors
for some g E G.
Let g E G be a torsion element of prime order q, and
kl
assume that q/i(g)EK. Set c = 1/qV 11(g) and bk = H y(g,gi).
i=l
Then
q1l
(icg) (i+cg + qE ckbk gk) =
k=2
i+cg + E cg E c (g)i = ( (g))i=0.
k=2 k=2
Corollary 2.18. If the characteristic of K divides the
order of some element of G and the twist function takes
values only in the prime field of K, then K t[G] has proper
zero divisors.
Proof. Let char K = q > 0. Then the prime field of K
is GF(q). Hence q/x E K for all x E GF(q). By Theorem 2.17
the result follows.
For reasons of completeness we give the following result,
the proof of which resembles that of the ordinary group rings.
Lemma 2.19. If G is a torsion free abelian group, then
K t[G] is an integral domain.
Proof. Since G is a torsion free abelian group, then it
admits a total order compatible with the product. Let a and
t n
8 be nonzero elements of K [G]. Write a = E aixi and
m i=l
= b iy. Since G is ordered, supp a and supp are totally
i=l
ordered subsets of G; hence the leading terms of a and 6 are
anXn and bmYm respectively. Thus c has a nonzero leading
term akbmY(xnlym)x nym, which proves the result.
CHAPTER III
NILPOTENT RADICALS OF NONCOMMUTATIVE SEMIGROUP RINGS
Passman [22, theorem 1.9] has determined the nilpotent
radical of a group ring over a field of finite characteristic
in the following result.
Let K be a field of characteristic p > 0 and let G be a
group. Then (i) N[K[G]J = J[KLAP(G)]J.K[G], where AP(G) =
(x E GI the index of the centralizer of x in G is finite and
the order of x is a power of p); (ii) .J[K[AP(G)]] = N[K[AP(G)IJ;
and (iii) N[KLGI] = 0 if and only if AP(G) = (i).
Gilmer and Parker [10] have completely characterized
the nilpotent radical of any commutative semigroup ring in
the result that follows.
Let R be a commutative ring and let {P X}1A be the
family of prime ideals of R. Let S be an abelian semigroup.
Then the nilpotent radical of R[S] is n EA {P XS] + I p},
where PX is the characteristic of R/Px and where I is the
PA
ideal of R[S] generated by the set {rxa = rxbf ab }; if
p
PX = 0, then I is the ideal of R[S] generated by the set
{rxa rxbl r E R and a is almost equal to b}.
The natural question that arises immediately is whether
this result is valid or not in the noncommutative semigroup
rings. In the first section of this chapter we analyze an
example of a noncommutative semigroup ring over a field,
showing that the former result is no longer true.
Then, what is the nilpotent radical of a noncommutative
semigroup ring? The rest of this chapter deals with this
question, whose answer cannot be unique, since it will depend
upon the particular semigroups and rings in consideration.
We consider some known families of noncommutative semigroups,
and we determine the nilpotent radical of semigroup rings
constructed with them.
Thus, in section 3.2 we study the nilpotent radical of a
semigroup ring R[SI, where R is a commutative ring with iden
tity and S is a right (left) zero semigroup. Theorem 3.6
shows that in this case N[R[S]1 = {aixi XE R[S]J Eai E N}
where N is the nilpotent radical of R.
In section 3.3 we let S be a right (left) group and R a
ring with identity. Theorem 3.19 handles this case by
showing that N[R[S]] = N[R[H1J]R[S] + C, where H1 is amaximal
n n
subgroup of S and C = R[HI]W with W = { Z aieij Z a. = 0,
i i=l i=l 1
ei idempotent of S and n E Z In section3.4we consider the
family S of semigroups having a universally minimal ideal,
and we let R be a ring with identity. Proposition 3.26 states
that N[RLS]J = N[R[Uj @ N[K], where U is the group of zeroids
of S and K the kernel of the canonical epimorphism O:R[S] R[U].
Then we apply this result to one subfamily of S, where the
product outside of U is known.
3.1 An Example
In this section we consider an example of a semigroup
ring F[S], where F is a field and S is a torsion free finite
43
semigroup. We calculate the radicals of F[S] to show that
by just removing the commutativity of S we may have a nonzero
nilpotent radical. This contrasts the following result of
Gilmer and Parker [10] on commutative semigroup rings. If
either R is a ring of finite characteristic n and S is a
ptorsion free semigroup for each prime divisor p of n, or R
is a ring of zero characteristic and S is torsion free, then
the nilradical of RES] is N[S], where N is the nilpotent
radical of R.
Let S be the semigroup defined by the following table.
i x
i i x y
x x x
y y x y
Notice that S has identity element i and is torsion free,
but S is not commutative.
Let F be a field. Denote by (x,y) the ideal generated
by x and y. Let L = {ai + bx + cyla,b,c E F, a + b + c = 01.
Clearly L2 c L, F[S]L c L and LF[S] c L; hence L is an ideal
of FES].
In order to determine the radicals of F[S] we prove the
following results.
Claim 1. L and
Proof. Let (L,c) = M denote the enlargement of L by an
element a = ui + vx + wy of F[SJ L; that is, u + v + w e 0.
Since ax = (u + v + w)x E M and since (u + v + w)(i x) c M,
then by adding both elements we get (u + v + w)i E M. But
u + v + w has an inverse; hence i E M. Therefore M = F[S].
Similarly, let ((x,y), ) = T be any enlargement of
(x,y) by an element 8 = ui + vx + wy not in (x,y), that
is, u ; 0. Since vx + wy E (x,y), then ui and hence i belong
to T. Therefore T = F[S].
Claim 2. The only ideals of FLSI generated by elements
which do not contain i in their support, are (x,y) and
(x y).
Proof. Let I be an ideal of FES] such that i i supp
for all a E I. If either x or y belongs to I, then xy = y and
yx = x imply that (x,y) = {ux + vyl u,v E F} c I.
Let bx + cy E I, b 0 Q, c e 0. We consider two cases.
First if b + c x 0 for some element of I, then
(bx + cy)x = (b + c)x E I. But b + c has an inverse; thus
x E I, and by the previous paragraph (x,y) c I. By Claim 1
(x,y) is maximal; hence I = (x,y).
On the other hand, if b + c = 0 for each element of I,
then bx + cy = b(x y) E I. Hence (x y) c I. Moreover,
(x y) x I implies that there exist ax + by E I such that
a + b x 0. This contradiction shows that x y = I.
Claim 3. An element of F[S], which contains i in its
support, generates the ideal FES] if the sum of its coefficients
is not zero.
Proof. Let I be an ideal of FISI. If ai + bx c I,
where a x 0 and a + b e 0, then (ai + bx)x = (a + b)x E I
implies that x r I. Since ai = (ai + bx) bx E I, it follows
that i E I. Therefore I = F[SJ. Similarly, if ai + cy E I,
where a x 0 and a + c e 0, or if ai + bx + cy E I, where
a 0 and a + b + c e 0, then i E I. Hence I = F[S].
So far we have found the ideals (x y), (x,y) and L.
We prove next that, in fact, these are the only proper ideals
of FLSJ.
Claim 4. Any element of F[S], which contains i in its
support and which has the sum of its coefficients equal to
zero, is a generator of L.
Proof. Let ai + bx E L, where a x 0 and a + b = 0. Then
ai + bx= a(i x) E (i x). Similarly, if ai + cy E L
with a 0 and a + c = 0, then ai + cy = a(i E i y).
Now let a = ai + bx + cy E L, where a 0, b 0, c x 0 and
a + b + c = 0. Since (i x)(i y) = i x and
(i y)(i x) = i y, then (i x = i y). Moreover,
from(i x) =a and (i x) a=a(i x), itfollows
(i x) = (a). Hence, eachoftlhe elementsof L containing i in
its support generatesthe ideal Ki x). By definition, any
other nonzero element of L is of the form bx + cy with b 0,
b + c = 0. Then bx + cy = b(x y). But x y = (i y) 
 (i x); hence (x y) c (i x). Therefore L = (i x .
Since no element of (x y) has i in its support, then
(x y) is strictly contained in L. By claims 1 through 4
the lattice of ideals of F[S] is
F[S]
L (x,y)
(x y)
2
Since (x y) = 0, then (x y) is a nilpotent ideal.
Also i x and x are idempotent elements of L and (x,y),
respectively. Hence these ideals are not nilpotent; thus the
nilpotent radical of F[S] is N[F[SJ] = (x y). Moreover,
since S is finite, then F[S] is Artinian. Hence J[F[S]] =
N[F[S11 = P[F[S]i = (x y).
3.2 Semigroup Rings Over Right (Left) Zero Semigroups and
Their Nilradicals
In E 5 ] the following definition is given.
A semigroup S is called a right zero semigroup if
xy = y for every x,y in S. The left zero semigroups are
defined by duality. In spite of their triviality, these
semigroups arise naturally in the characterization of some
larger families of noncomuutative semigroups that we consider
in further sections. Let R be a commutative ring with iden
tity, and let S be a right zero semigroup. Our goal in this
section is to characterize the nilpotent radical of RIS.I.
But first we need the following four lemmas.
Lemma 3.1. Any right ideal of RLSJ is a twosided
ideal.
Proof. Let A be a right ideal of RES]. Hence AR[S1 S A.
Since R is commutative, then RA = AR c A. Moreover,
S supp a = supp a for all a E A; hence SA = A. Then R[SJA =A
shows that A is also a left ideal.
Lemma 3.2. Let A be an ideal of R, and let T be a non
empty subset of S. Then A[T] is a left ideal of R[S].
Moreover, ALSI is a twosided ideal of R[S].
Proof. Any subset T of S is a set of right zeros; hence
ST = T. Thus, RES]A[T] SRALT] = A[T]. Also A[SJS c A[S]
and AES] R c A[Sj imply that A[S]R[S] cA[S]. Hence AES]
is a right ideal, and the result follows from Lemma 3.1.
Lemma 3.3. Let a = Ea.x. be an element of RES] such
i i
that Eai = 0. Then the principal ideal generated by a,
is (a) = Ra. Moreover, (a) is a nilpotent ideal of index
two.
Proof. Since ax = (Eai)x = 0 for all x E S, then
RaRES] = 0; hence Ra is a right ideal of RES]. It follows
from Lemma3.1 that Ra is a twosided ideal; that is, (a) = Ra.
Moreover, since a2 = (a ixi )(Eaixi) = (Ea i )(Eaixi) 0 and
since R is commutative, then (Ra) (Ra) = Ra2 = 0.
Lemma 3.4. Let R be an integral domain. Let I be a
nilpotent ideal of R[S], and let a = Eaixi E I. Then
Ea. = 0.
1
Proof. If x E S, then ax = (Eai)x E I. Let n be the
index of nilpotency of I. Then 0 = [(Eai)xln = (Eai)nx; hence
(Eai) n = 0. Since R is a integral domain, then Eai = 0.
Theorem 3.5. Let R be an integral domain. Then the
nilpotent radical of RESJ is N[R[SJJ = { = Eaixil~ai = 01.
Moreover, NLR[Sil is a nilpotent ideal of index two.
Proof. Let L = {EaixijEai = 0}. By Lemma 3.3, if
a E L, then (a) = R is a nilpotent ideal of RES] of index
two. Hence L is a nilpotent ideal of index two.
On the other hand, if I is a nilpotent ideal of RES],
then by Lemma 3.4 I c L; so L is the unique largest nilpotent
ideal of R[S]. Thus N[R[S]] = L.
Using this result, we characterize next the nilpotent
radical when R is a commutative ring.
Theorem 3.6. Let R be a commutative ring with identity,
and let S be a right zero semigroup. Then the nilpotent
radical of R[S] is NLR[S]] = {laixil~ai E N}, where N is the
nilpotent radical of R.
Proof. Let a = Ea.x. N[R[SI1. Let P(R) denote the
1 1
prime .radical of R. Since R is commutative, then N = P(R).
If P is a prime ideal of R, then R/P is an integral domain.
By Theorem 3.5 NL (R/P)LSJJ = {Zbiyijbi = 0}. Let
0: R[SJ (R/P)[SI be the canonical epimorphism. Since
(o) = aixi E NL(R/P)[SJ], then Eai = 0. Hence
Zai + P = P; that is, Eai E P. Since Eai belongs to each
prime ideal P of R, then Eai E nP, = N. Therefore,
N[R[S]] c {caixil~ai E NI.
Conversely, let 8 = ybiyi, where Eb. E N. Since R is
commutative, there exists a nilpotent ideal A c N such that
Ebi c A. If p = Ecjxj E R[S], then p = (Ebiyi)(Ecjxj) =
T Eb.c.(xiy.) = E(Ebi)cjyj. Hence the sum of the coefficients
jil l i j" J
of P is (Ebi) (sc.) E A. By a similar argument the sum of
i
the coefficients of any element of ($) = {e + $p + nS + ZoiSp
Z+
aptiPi E R[S], n E Z } belongsto A. Moreover, if 6. E ($)
1
i = 1,2,...,n, then the sum of the coefficients of 6162...6n
belongsto A n. Let k be the index of nilpotency of A. Then
the sum of the coefficients of any element of ( )k belongsto
Ak = 0. It follows now from Lemma 3.3 that (W)2k = 0. Hence
(S) c N[R[S]1; and so, { aixilai E NI S N[R[S]].
Therefore, N[R[S]1 = { aixilJai E NI.
We remark that if S is a left zero semigroup and R is a
commutative ring with identity (integral domain), then by
duality Theorem 3.6 (Theorem 3.5) also holds in RES].
Corollary 3.7. Let S be a right (left) zero semigroup,
and let R be a commutative ring with indentity. Then R[SJ
is semiprime if and only if R is a semigroup ring and
ISl = 1.
Proof. Let R be a semigroup ring, and let ISj 1.
Then RIS R; hence N[R[S]] = 0.
Conversely, let R be a commutative ring, and let S be a
right zero semigroup. Assume that R[S] is semiprime. Then
N[R[SJ] = 0. Hence, if either the nilpotent radical N of R
is nonzero, or if ISI > 1, then by Theorem 3.6 N[R[S]] x 0.
This contradiction shows that N = 0 and ISI = 1.
Corollary 3.8. Let R be a commutative ring. Let the
semigroup S = E1 x E2 x ... x E where E. are right zero
2n 1
semigroups. Then the nilpotent radical of R[S] is
n
N[R[S]] = 0 {a. .x. R[E ]I a. E N}, where N is the
i=l 3 '3 11 1 j
nilpotent radical of R.
n
Proof. Since R[S] = e R[Ei1, then the conclusion
i=l
follows from Theorem 3.6.
3.3 The Nilpotent Radical of the Semigroup Ring of A Right
(Left) Group.
A semigroup S is called a right group if it is right
simple and left cancellative [5 ]. Left groups are defined
dually. Being right (left) simple means that the semigroup
contains no right (left) proper nonzero ideals; thus these
semigroups form a subfamily of the simple semigroups.
From now on we consider right groups, but the analogous
conclusions for left groups will follow by duality.
Let R be a ring with identity, and let S be a right
group. In this section we find the nilpotent ideals of
the semigroup ring R[SJ. This will allow us to completely
characterize the nilpotent radical of R[S].
We begin by stating two important results on right
groups, the proofs of which can be found in [5, 1.26, 1.27].
Lemma 3.9. Every idempotent element of a right simple
semigroup S is a left identity of S.
Lemma 3.10. The following assertions concerning a semi
group S are equivalent.
(i) S is a right group.
(ii) S is right simple and contains an idempotent.
(iii) S is the direct product G x E of a group G and a
right zero semigroup E; that is, there exists
:G x E S:(g,e) ge, such that is an
isomorphism.
Let E = {ei}iI stand for the set of idempotents of S.
By Lemma 3.10 (ii) E is nonempty. By Lemma 3.9 every element
of E is a left identity of S; thus E is a right zero semigroup
of S. Let e E E, then Se = eSe, and we see that Se is a sub
group of S; in fact, by Lemma 3.10 (iii) Se G x {e}. Let
Hi denote the subgroup Sei, where ei E E. It follows that
for any ei,ej E E, i x j, the subgroups Hi and Hj are disjoint
and isomorphic under ij:Hi H., where ij (xi) = x iej.
Moreover, by Lemma 3.10 (iii) we also see that S is the
disjoint union of its subgroups Hi, i E I. We have shown
the following restatement of the previous results.
Lemma 3.11. If S is a right group, and E = {ei}i I is
the set of its idempotent elements, then the following
statements hold.
(i) E is a right zero semigroup.
(ii) Hi = Sei is a subgroup of S for all ei E E, with
identity 'Hi = e.
(iii) If ei,ej E E with i z j, then Hi n H. = i and
Oij:Hi Hj defined by eij(x i) =x Xe = x j is
a group isomorphism.
(iv) S = UiEI Hi.
Let R be a commutative ring with identity, and let S be
a right group. As a direct consequence of Lemma 3.11 we get
the following result.
Lemma 3.12. The following statements hold in RES].
(i) ( E R[Hi1) n R[HjI = 0 for all j E I.
ixj
(ii) There exist a group ring isomorphism ij:R[Hi] R[H]
such that ij (R[li ]) = R[Hi]e. = R[Hj] for all
i,j E I.
w
(iii) RES] = YiI R[Hi1, a internal weak direct sum of
Rmodules.
(iv) ei is a left identity of R[S] for all i E I.
(v) RES]e. = R[Hji]; that is, each R[Hj] is a left
ideal of R[S].
Proof. (i) Follows clearly from H. n H. for all
1 j = i fon all
i,j E I, i j. (ii) Let 0.ij be the natural extension of 0.i;
n
that is, define Oij: R[IHil R[IjJ by T ij ( ixli
A =1
n n
ai 0 ij(xi) = aXi(xAiej. Since 0.. is a semigroup
A=1 A=1 1J
isomorphism between the basis, then 8.. is a group ring
isomorphism. Hence R[H.1 = 0..(R[Hi]) = R[Hile.. (iii) Fbllows
directly from (i) and Lemma 3.11 (iv). (iv) Is clear since
ei is a left identity of S for all i E I. (v) By (iii) and
(ii) we get R[Slej = EiEI R[Hile. = iI R[H.I = R[H..
From now on we fix an element of the index set I, say
i E I.
Lemma 3.13. Let B1 be a twosided ideal of R[HI]. Then
B = B lei is a left ideal of RES] for every i E I. Moreover,
B = B. is a twosided ideal of RES].
1EI 1
Proof. By Lemma 3.12 (ii) 0ij is a group ring isomorphism
for all i,j E I. Hence li(B = Blei = Bi is a twosided
ideal of R[Hi] for all i E I. Since R[H.]Bi = R[Hj](eiBi)
= R[HiBi c Bi; then R[SIJBi = [jEH.1B. S B.. Thus,
Bi is a left ideal of R[S] for all i E I. Therefore,
B = EiEI Bi is a left ideal of R[S]. Moreover, B.R[Hi]
= B.(eiR[Hi]) = BiR[Hi] c Bi for all i,j E I. Hence
B.R[S] = B.Z. R[Hi = E B.REH] c Ei B. = B. It follows
J J 1EI 1 lEI J 1 lEI 1
that BR[S] = EicIBiR[S] c EiI B = B; thus B is a right ideal
of R[S]. Therefore B is a twosided ideal of R[SJ.
Proposition 3.14. EiEI N[R[H.]] i c N[R[S]].
Proof. Let N0[R[Hi]] c NI[RHi]] c .. Nj[RHi]] c ..
i I, and N0[R[S]] c NI[R[S]] c ... c Nj[R[S]] c ... be the
chains of ideals used in the definition of the nilpotent
radical of R[IIi1 and RIS] respectively. We use transfinite
induction. Let B1 be a nilpotent ideal of R[H1]. By
Lemma 3.12 (ii) 0ij is a ring isomorphism for all i,j E I;
hence eli(Bl) = B1ei = B. is a nilpotent ideal of R[Hi],
1 1
i E I, with the same index of nilpotency as BI, say n. By
Lemma 3.13 B = E.iEI Bi is a twosided ideal of R[S]. We
claim that B is nilpotent. Since B.B. = Bi(eB.) = B, it
follows by induction that B. B. .. .B. = Bm for all
11 12 m im
positive integer m. Thus B' = 0 for all i E I implies
that Bn = ( Bi)n =i Bn = 0. Therefore B c N0[R[S]]
i EI 1 lEI 1 0
and consequently E iI N0[R[Hi]] c N0[R[S]]. This shows the
first step of the induction argument.
Assume that EiEI Nk[R[Hi]] E N k[RES]] for all the
ordinals k < j. Suppose j is not a limit ordinal and let
AI/N jIR[H1]] be nilpotent in R[H1]/Nj_1[R[H11. As in the.
previous paragraph Ai/NjI [R[Hi]] = ei [A1/NjI[R[H1]] is a
nilpotent ideal of R[Hi]/NjI[RHi]], i c I, with the same
index of nilpotency, say n. As a clear consequence of
Lemma 3.13 A/ iEI Nj_[R[Hi] = EiEI Ai/NjI[R[Hi] is an
ideal of R[S]/ iEI Nj_I[R[Hi]i which we claim is nilpotent.
Since
Ai/Nj_1[R[Hi]] At/Nj_[R[Ht]] = Ai/N j_1R[Hi]] etAt/NJEl[Ht]I
A /NjI[R[Ht]],
by induction we obtain
Ail/NJI[R[H.i]i A.2/r.j: IR[Hi] . A. m/Nj[R[im i]
A11 r J1 LH 1 11 A2 /11j 1ER1 2 11..A1m IN31 RH mII
=Am /N._[R[Hi ]].
m m
Thus (Ai/Nj_1[R[Hi1])n = 0 for all i E I implies that
(A/iEI N jIR[H i) n = iEI (Ai/Nj_1 R[H i])n = 0.
Hence A/ENj _[R[Hi]l is nilpotent in R[S]/Nj_[R[Hi]].
By induction hypothesis (A + Nj _1R[S]])/Nj_ [R[S]] is nil
potent in R[SI/NjI [R[S]1; hence A c Nj[R[S]. Therefore
ZiEI Nj[R[Hi]1 c Nj[R[S]1. If j is a limit ordinal, then
N [R[HI]]= u NhR[HI]]. Thus
h
iEl NjER[I1i]1 = El( u Nh[R[Hi]]) = u (EiiNh[R[Hi]])
h
c u Nh[R[S]] = N.[R[S]].
h
Since N[R[Hi]] = N T[R[Hi]] for some ordinal T, eventually
we get Ei IN[R[Hi]1 c NERESI].
So far we have obtained nilpotent ideals of RES] as
sums of nilpotent ideals of R[Hi], i E I. We now see that
this is not always the case.
Lemma 3.15. Let A be a nonzero proper ideal of RES].
Let Ai = A n R[Hi], i E I. Then the following statements
hold:
(i) Ai is an ideal of R[Hi] for every i c I, and thus
is a left ideal of RES].
(ii) A i A. for all i,j E I.
(iii) If Ai = 0 for some i c I, then Ai = 0 for all
i c I. In this case ARLS = 0; hence A is nilpotent
of index two.
Proof. (i) Since AiR[Hi] c AR[Hi] = A., and symmetrically
R[Hi]Ai c Ai, it follows that Ai is an ideal of R[Hi] for
every i E I. By Lemma 3.13 each Ai is a left ideal of R[S].
(ii) AiR[Hj] c A n R[Hj] = A.; hence A. 2 Aiej.
Symmetrically Ai 2 Ajei = 5ij (A.). Thus
Ai 2 eji(Aj) ji (Aiej) = Aiejei i Ae= Ai
Since ij are monomorphisms, we conclude that Ai Aj.
(iii) Let A. = 0 for some j E I. By (ii) it is clear
that A. = 0 for all i E I. By Lemma 3.12 (v) R[Hil is a left
ideal of RESJ; hence AR[Hi] c A n R[HiI = 0 for all i E I.
Then ARLS= AE iIR[Hi] = E IARLHi] = 0 by Lemma 3.12 (iii).
In particular, A2 _c AR[S] = 0.
n n
Definition. W = { Z ai eilei E E, Z ai = 0, ai E R,
and n E Z + .
Lemma 3.16. Let L1 be a left ideal of R[H1]. Then
Llw is a nilpotent ideal of R[S] for all w E W, w 0.
n
Proof. Let x E H., and let w = E aiei E W. Since
n n i=l
wx = ai(eix) = ( Z ai)x = 0, then wR[Hj] = 0 for all i E I.
i=l i=1
Hence wR[S] = EjEI wR[Hj] = 0. Thus LIwR[Sj S Llw. By
Lemma 3.12 (v) R[S]L1 = R[S](eL) = R[H1I]L1 = L1; so
R[SJLIw c L1W. Therefore, Llw is a twosided ideal of RLSJ.
Finally (Llw)2 = L1(wL1)w = L1*Q0w = 0 implies that Llw is
nilpotent.
tI t2
Proposition 3.17. Let C = { e a e + E i xi +
t i=la111 i2=l 2 2i2
tn ti t2 tn
+ Z ax inx i ai = a xi E ax = 0;
in=1 n n nin 1 =1 1 i2= 22 in=l nn
xx3 i H1 XX mek X 3k, m,k,i. E I, tj and n E ,
JJJ J J J J
j = 1,2,...,n}. Then Cisa nilpotent ideal of R[S] that con
tains every ideal A of R[SJ such that AR[S] = 0. Moreover,
C = R[H1 ]W.
Proof. Define 4: R[S] R[H1J by f(a) = ce. By Lemma
3.12 (v) p is clearly an Rmodule epimorphism. Moreover,
= (a6)eI = a e) = =(e e) (ael)(Bel) = f(a)f(B),
and so p is also a ring epimorphism.
Let a E Ker p. Write
tit 2tin
= ajlejl + E2 a2j j + ...+ m
Jl j2.= 2x12 Jn=la mmximm
where x c E H., xX nek = xX k; that is, the elements in the
iJ Jii
support of one summation belong to different subgroups, and
all of them have the same image under any of the group
isomorphisms 0nk for all n,k E I. Then
t t2
0 = i(a) = Ee = 1 a. (e. 31 el + a x (x 2Jiel) +
j 1211 22 2
tm ti t2
+ E a j. (x xm e1) = ( Z a. )e1 + ( E aA )x +
im m Jl 1 j2=1 2j2 2
t
m
JE=1aAmm) x m i
m m
tI t2 tm
Hence E a. = E a = ... =E a. =0. It follows
9 1 92=1 2j2 9mjl m m
that a E C, and therefore Ker c c C.
On the other hand, it is clear that O(C) = 0, thus
C c Ker 0, and we conclude that C = Ker 0 is an ideal of R[S].
From (R[HI]W) = R[H1](WeI) = R[H1] 0 = 0, we get
R[H1IW c C. Let a E C. By the definition of C we can write
t 12 tm
a= E a. e. + E a + + a A
91=i 31 31 j2=l 12J2/X2J2 "' I =Amn m Amjm
But x e. = xIj implies that we can rewrite
il3i i3i
tI t2
a = eI a. e. + x E=1 a e + ... +
Jl=l 31ej1 2 J2=i 2 232j
t
+m a *e.
xn j =1 Amim m
t. t.
and since EI a'j 0, i l,...,m, then EI a e E W.
Ji=l i i=I i i i
Hence a E R[HI]W, it follows that C c R[H1]W. Therefore
C = R[HI]W. Whence C2 = R[HI](WR[H1])W = 0 by Lemma 3.16.
Thus C is nilpotent.
Finally, if A is an ideal of R[SJ such that AR[SI = 0,
then Ae1 = 0. Hence A c Ker = C.
Proposition 3.18. Every element a of R[S] can be uniquely
expressed as a = a, + ac, where a, E R[H1] and Ec E C.
Proof. Let a E R[S]. As in the proof of Proposition
3.17 we can write
t 1t 2 m
a = E a. e. + Z= a + ...+ ma x A
jl=l 31 31 j2l 12J2X2J2 jm=l m M m m
where x Xe. =x . Set
ti t2 t
dl _E a d2 _2 al~ d E a
1=2 al J2=2 2j2 m m=m.
2 m=
Then we can rewrite a as al + ac, where
aI = [(aldl)e1 + (a 21 d2)xx 21 + + (a, 1dm)XX 11
22m m
and
ti t2
c E a. e. + dlel + E aX + d2x, 1 + +
j=2 31 1 J 2=2 2J2X,2J2 2
t
imA
j=2 m m m m m
Clearly a cxE R[H 11, and since by construction
t.
di + E aA. = 0 for i = 1,2,...,m, then c E C.
Ji=l Xi3i c
Suppose now that c = cx + ac and also xc + c,
where al,61 E R[HI1 and cc C E C. Since
l R11 1n C = 0, then a and
ec =8c' ,which proves that the representation is unique.
We remark that Proposition 3.18 can be restated by
saying that R[S] = R[H1] C is a direct sum of left ideals.
Theorem 3.19. Let R be a ring with identity, and let
S be a right group. Then the nilpotent radical of RES] is
N[R[S]] = Z iI N[R[H.11 + C = N[R[H1]]R[S] + C.
Proof. By Propositions 3.14 and 3.17 it follows that
Ii I N[R[Hi]] + C c N[R[S]1.
Hence N[R[S]/CI = N[R[S]]/C N[R[H1 ]] by Lemma 3.18. Thus
N[R[S]1= N[R[H1]] + C = N[R[H1]]R[S] + C.
Finally, by Proposition 3.12 (iii) N[R[H1]]R[S]
N[R[HI](EiEIR[Hil) = EiEIN[R[H1]] (eiR[Hi]) = E iEI N 1RHill
3.4 Semigroup Rings Over Semigroups With Universally Minimal
Ideals, and Their Nilradicals
An element u of a semigroup S is called a zeroid element
of S if, for each element s E S, there exist x,y E S such
that sx = ys = u. It follows that if S has a zero element,
this will be the only zeroid of S. Let A be an ideal of S.
Then A is called locally minimal if it contains no proper
subideal. A is called universally minimal in S if it is
contained in every ideal of S. It is an easy consequence of
these definitions that a semigroup S contains a universally
minimal ideal A if and only if it contains a zeroid element,
and then A consists of all the zeroids of S.
Clifford and Miller [4 1 have shown that the set U of
zeroid elements of S is either vacuous or a subgroup and two
sided ideal of S. Moreover, the identity element z of U
commutes with every element of S, and the mapping
8: S U: s zs(=sz) is an epimorphism. They define the
core of S to be the subsemigroup J = {x E Slxz = z}; thus J
contains z as zero element. Finally the subsemigroup U u J
of S will be called the frame of S.
An easy example of a semigroup with universally minimal
ideal is provided by the direct product of any group and any
semigroup with zero.
Let R be a ring with identity, and let S be a semigroup
without zero such that S contains a universally minimal ideal
U. In this section we first show that R[S] is a ring direct
sum of two known ideals. Using this decomposition, we find
the general form of the nilpotent radical of R[S]. Finally
we apply these results to semigroup rings over a subfamily
of semigroups with universally minimal ideals, where the
multiplication in the semigroups is known.
We begin by stating as a remark some factstaken from
the main theorem in [4 1.
Remark 3.20. The following conditions hold in S:
(i) e:S U: s sz, is a semigroup epimorphism, where
z = lu. Moreover, zs = sz for all s E S.
(ii) S = u J(u), where J(u) = 1 (u) = {scSlsz=u=zs}.
UcU
(iii) J(u) n U = u and J(u) n J(v) = $ for all
U,V E U, U X V.
(iv) J(z) = J = {S E SI sz = zs = z1.
(v) J(u)J(v) c J(uv) for all u,v E U.
Lemma 3.21. Let U: R[SJ* R[U] be the natural extension
of 8; that is, O(.aisi) = Zaie(s). Then 8 is a semigroup
ring epimorphism.
Proof. By definition U is clearly an Rmodule homomor
phism. Moreover, by Remark 3.20 (i) 0 is an epimorphism.
Since zs = sz for all s E S and z2 z, it follows that 8 is
also a ring epimorphism.
n n
Definition 3.22. Let K(u) = { Z aixil Z ai = 0,
i=l i=l
xi E J(u) and n E Z+J and set K = EuEU K(u).
Lemma 3.23. K is an ideal of RLSJ.
Proof. We show that K = Ker 0. Let Eaixi E K(u). Since
(Eaixi E= lai (xi) = Eai(xiz) = (Zai)u = 0, then U(K(u)) = 0.
Thus, 0(K) = 0, and therefore K c Ker U.
On the other hand, let a E Ker U, and write
a = Zaiui + Z ZE jkx k'
i j k j
where Xu.k E J(uj) {uj} for all k. Then
3J J
0 = U(a) = Eau. + E E b'k(XukZ) = Ea.u. + E(Ebk)Uj'
1j k 3 i jk .
After renaming conveniently the subindexes, wcan write
0 = 0(a) = E(at + btk )ut; hence at + kb = 0 for each t.
t k k =k
Thus
=Z(atut + E b tkXuk) E tK(ut) c K.
t k t t
Therefore Ker 0 = K.
We are now in the position to show that the semigroup
ring of a semigroup having a universally minimal ideal is
completely determined by the group ring of the group of
zeroids and the ideal K.
Theorem 3.24. R[S] = R[U] 0 K (ring direct sum).
Proof. It follows from the definition of K(u) and (ii)
of Remark 3.20 that R[UJ n K(u) = 0 for all u E U. Hence
R[U] n K = 0.
By Lemma 3.23 K is an ideal of R[Sj. Since U is an ideal
of S, we get readily that RLUJ is also an ideal of RES]. As
K is the kernel of the epimorphism 0, it follows that
R[Sj/K R[U]. Thus RLS] = R[U] + K.
Corollary 3.25. If the semigroup S coincides with its
frame, then R[S] = R[U] ( K(z).
Proof. Since S = U u J, then Ker 8 = K(z) and the
result follows from Theorem 3.24.
Proposition 3.26. The nilpotent radical of RES] is
N[R[S]] = N[R[U]] 0 N[K]. In particular, if S = U u J, then
NLRLS]] = N[RLU]] $ N[K(z)].
Proof. This follows directly from Theorem 3.24 and
Corollary 3.25.
Corollary 3.27. Let S = U u J, where J is a right zero
semigroup with zero, and let R be an integral domain. Then
N[R[S]] = N[REUJI E K(z)
Proof. Since K(z)z = z.K(z) = 0, then K(z) is an ideal
of R[J]. By Theorem 3.5 K(z) is nilpotent; hence the result
follows by Proposition 3.26.
So far we have seen that the nilpotent radical of RES]
is determined by the nilpotent radicals of R[U] and K. If
we want to find out what the nilpotent radical of K looks
like, we need to know how the multiplication among the elements
of the sets J(u) is defined. The classification for the
possible infinite semigroups with a given frame is an open
problem; next we sketch a simple construction due to Clifford
and Miller [ 4 1 that shows there exist an infinite number of
such semigroups, and we find their nilpotent radical.
Let the group U and the semigroup J with zero be given.
By identifying the identity of U with the zero of J and
defining xu = ux = u for all u E U and x e J, we get a
semigroup S0 whose frame coincide with itself. This S0 is a
special case of the following type of semigroup.
Associate with each element u c U a set J(u) in any way
subject to the conditions (iii) and (v) of the Remark 3.20.
For each u E U, determine a right and a left representation
of J by single valued mappings of the set J(u) into itself
such that (a) left and right mappings commute with each other,
(b) the element z E J maps every element of J(u) into u, and
(c) all the mappings leave fixed the element u E J(u). The
multiplication in the class sum S of all J(u) is defined as
follows. Within J(z) it is defined as originally given in J.
For k E J and a E J(u) we define ak(ka) to be the image of a
under the mapping corresponding to k in the right (left)
representation already determined. Finally for xu E J(u) and
xv E J(v) where u e z and v z, we define xuxv = uv, where
uv is the product of u and v as originally given in U. Then
each S = u J(u) is a semigroup with frame SO = U U J.
UEU
Lemma 3.28. The sets K(u) are nilpotent ideals of index
two in R[S] for all u < z.
Proof. Let a Ea.x. E K(u), let p = Eckxk E REJ[(v)]
i I k
where u x z and v z, and let 6 = dtYt E R[J(z)]. Since
t
Za. 0 ten = (aix k kZcak) = Z a~ck(xiak) = Zl(Zai)ck(uv)
Ea. = 0, then cip =(Ea.x.) (Zc ai E Eak(iaEZ
i1 1 1 k k k k iI k i )cuv
= 0. Hence K(u)R[J(v)] = 0 for all u,v E U with u z and
v Z. In particular, K2c K REJ(u)] = 0. On the other hand,
U u
a6 = (Zaixi) (Ed tyt) = E iadt (xY) i Since xiYt E J(u)J(z)cJ(u)
1 t t i
and since Z Eaidt = (ai )(Edt) = 0, then a6 E K(u). Therefore
t i ii tt
K(u) R[J(z)] c K(u). Moreover, from S = uJ(u) we get
R[S] = EuUu R[J(u)]. Hence
K(u)R[S] = K(u)[ E R[J(v)] + R[J(z)]1 = K(u)R[J(z)]c K(u).
VEU{z}
Symmetrically R[SIK(u) c K(u), and the result follows.
66
Theorem 3.29. The nilpotent radical of R[S] is
N[R[SI] = N[R[U1] E (EuEU_{z}K(u) + N[K(z)]).
Proof. By Proposition 3.26 N[R[SJ] = N[R[UI] ( N[K].
But N[K] = N[uEuK(u)], and since K(u) is a nilpotent ideal
of index two for u z by Lemma 3.28, then
N[K] = N[u K(u)] = uEU_{z}K(u) + (N[R[J]] n K(z) =
EuU _fz} K(u) + N[K(z)].
CHAPTER IV
SEMIPRIME SEMIGROUP RINGS
In commutative semigroup rings Gilmer and Parker [101
have proven the following result. Let R be a commutative
ring with identity, and let S be an abelian semigroup. If
either the characteristic of R is zero and S is torsion
free or else the characteristic of R is n and S is ptorsion
free for each prime divisor p of n, then N[S] is the nil
potent radical of R[S], where N is the nilpotent radical
of R. It follows clearly that under the conditions of this
result R[S] is semiprime if and only if R is semiprime.
In this chapter we are mainly concerned with the problem
of determining conditions on R or S that will give rise to
semiprime (twisted) semigroup rings.
We have already seen in Corollary 3.7 that if E is a
right (left) zero semigroup and R is a commutative ring
with identity, then R[E] is semiprime if and only if R is an
integral domain and IEI = 1. As a direct consequence of
Theorem 3.19 we have that if S is a right (left) group with
more than one idempotent (that is, S is not a group) and R
is a ring with identity, then R[S] is never semiprime.
Finally, if S is a semigroup with universally minimal ideal
U and R is a ring with identity, then it follows from
Proposition 3.26 that R[S] is semiprime if R[UI and K are
semiprime, where K is the kernel ideal of the canonical
epimorphism from RLSJ into R[U].
Now we let R be a ring. In Theorem 4.5 we prove that
if S is a semigroup satisfying a suitable condition, then
the twisted semigroup ring Rt[S] is semiprime if and only if
R is semiprime.
In section two we consider two types of semilattices.
The first is a semilattice P0 with a zero element and in
which all the nonzero idempotents are primitive. The second
type of semilattice P1 has also zero element and satisfies
the following condition: if F is a subset of PI' then either
g.l.b{xlx E F} belongs to F or else there exist x,y in F such
that glb{x,y} is in P1 F.
Next, we consider two families of semigroups which admit
relative inverses. We say that a semigroup S belongs to the
family S if (i) S admits relative inveres, (ii) the idempotents
of S commute, and (iii) the semilattice of subgroups Sa of S
is isomorphic to P0* Then in Theorem 4.11 we decompose RES]
as an internal weak direct sum of ideals which are completely
determined once RES.] is known for each a E P0. This allows
us to characterize the nilpotent and Jacobson radicals of
RESJ. Further, if we let F be a field and use results of
group ring theory, we can give necessary conditions for F[S]
to be semisimple and regular in Corollaries 4.15, 4.16 and
4.17.
Passman has related semiprime group rings with their
center in the following two results L22, Theorems 2.12 and
2.13J.
(A) Let KLGJ be a group ring over a field of character
istic 0. Then (i) KEGJ is semiprime, (ii) ZIK[GJ] is semi
prime, and (iii) ZLKLGJi is semisimple.
(B) Let KEG] be a group ring over a field of character
istic p > 0. Then the following are equivalent.
(i) K[G] is semiprime.
(ii) Z[KLGI] is semiprime.
(iii) ZLKLGJ] is semisimple.
(iv) G has no finite normal subgroups H with piHi.
(v) A(G) is a p'group, where A(G) = {x E Gj the index of
the centralizer of x in G is finite).
In Corollary 4.18 we apply (A) and (B) to the group rings
FLS J, a E P0' to obtain a similar result between F[S] and
Z[F[Sjj.
Finally, we say that a semigroup S belongs to the family
T if (i) S admits relative inverses, (ii) the idempotents of
S commute, and (iii) the semilattice of subgroups S. of S is
of type P1. Then in Theorem 4.21 we prove that RES] is
semiprime if RLS] is semiprime for all a E P1
4.1 Semiprime Twisted Semigroup Rings
In this section we first show that a twisted semigroup
ring RtISJ is semiprime if and only if the ring R is semi
prime, provided the semigroup S satisfies a suitable condition.
From this we are able to determine the nilpotent radical of
this semigroup ring.
The following definitions are due to Banaschewski 1 2 J.
Definition 4.1. Let X be a subset of a semigroup S. An
element x of X is called an extremity of X is for all positive
integer n, xn = SlS2 ...n si X, implies si = x for all i.
Definition 4.2. A semigroup S is said to satisfy con
dition (E) if every nonempty subset of S has an extremity.
It is clear that any totally ordered semigroup S sat
isfies condition (E), since the greatest and the least
element of any finite subset X of S, are extremities of X.
Moreover, if a semigroup S satisfies condition (E) and
xn yn for x,y E S, then x = y for n = 1,2,...; that is,
S is power cancellative. Banaschewski proved also in [ 2 1,
that any cancellative, power cancellative abelian semigroup
satisfies condition (E).
We might think of the possibility of defining a weaker
condition than (E) on S, both conditions related in the same
way as the ptorsion free condition is related to the torsion
free one. Namely, we would define a pextremity of a subset
n
X of S as an element a such that aP =Sl .s 2***n' si E X,
p
implies that a = sl = s2 s n Then we could set that
p
a semigroup S satisfies condition (E ) if any nonempty finite
subset X of S has a pextremity. This would allow us to
differentiate between twisted semigroup rings over rings of.
finite characteristic from those whose rings have characteris
tic 0, as Gilmer and Parker [10 did in commutative semigroup
rings.
Now assume that S satisfies condition (E ), and let a
n
be a pextremity of a finite subset X of S. Let a = S 2**Sn
Z+m
sI E X, and choose m c Z such that p > n. Then
m nmn m 
ap np p
ap ana Sl2**'sn a Hence condition (E p) implies
that a = sI = S2 = ... = Sn; that is, a is an extremity. This
shows that (E) and (E p) are equivalent conditions in S.
We begin by emphasizing a direct consequence of
Definition 4.1.
Lemma 4.3. Let S be a semigroup satisfying condition
(E), and let X be a nonvoid finite subset of S. If a is an
nn
extremity of X and n is a positive integer, then an is an
extremity of Xn = {x1x2. .XnlXi E X}.
Proof. Let (an)m = yly2. ym' where m E and each
y. = xilxi..x.n is an element of Xn for 1 i m. Then
Yi =ili2"'in
(an)m anm =(xX2 Xln)... (xmlXm2 x ...xmn ). But a is an
extremity of X; hence a = xij for all i and j. Therefore
n
a Yl Ym"
Proposition 4.4. Let R be a ring, and let S be a
semigroup. If the twisted semigroup ring Rt[S] is semiprime,
then R is semiprime.
Proof. Assume that R t[S] is semiprime. By way of con
tradiction, suppose that there exists a nonzero nilpotent
ideal A of R. By section 1.5 A t[s] is a nonzero ideal of Rt [s].
k
Let m be the index of nilpotency of A, and let a = E a.s.
i=l11
k
be a nonzero element of AtI S I. Then a = ( as E (a s
i=l1 111
(a. s s. ) with L = 1,2,...,k and (il,i2, ...i)
1a21l2 'm 'm
Lx... x L = Lm But (a.l s )(a. s. )(a. s. ) =a. a. y(s.
i2 )a.3 .. y(s. si. s. ism) slS2***Sm = 0, since
12 )a3 "'am 1 2 iri
ilai2 y(sl s .a. y(s. s. ...s. ,s ) EA = 0.
111 '1 2 3 im 1 12 rim
Hence m = 0. Therefore A [S] is a nilpotent ideal of R [S],
which contradicts our assumption; thus R is semiprime.
Theorem 4.5. Let R be a ring, and let S be a semigroup
satisfying condition (E). Then the following conditions are
equivalent:
(i) The twisted semigroup ring R t[S] is semiprime.
(ii) R is semiprime.
Proof. If Rt [S] is semiprime, then by Proposition 4.4
R is semiprime.
Conversely, assume that R is semiprime. By way of con
tradiction, suppose that Rt[Sl contains a nonzero nilpotent
ideal B. Let h be the index of nilpotency of B, and let
r
= b s. be a nonzero element of B. Then
i=l11
h r h rr
E =( bisi) = l(bisi) + (bilS).(b i s )
i=l i=l 1n n
.., Mh
with M = {l,2,...,ri and ~ ) E {x E M I not all
coordinates of x are equal}. Now supp 8 is a finite subset
of S, and since by hypothesis, condition (E) holds in S,
then there exists s. extremity in supp 6. Thus s. h S. S. s
J 2 l n
whenever ik x j for some k = 1,2,...,h. It follows that
h 2 hi h h h
o = (bjsj) bjy(sjsj)bj ... y(sj ,s.)sh = bcsh, where
2 hi h
c = y(sj,sj)y(s.,s.) ... y(si ,s) Hence b c = 0. Since
the images of y belong to the units of R, then c is a unit.
Therefore bh = 0; that is, b. is a nilpotent element of R.
J J
But B is an ideal; so the set D = {(xy(s2.s )s +
21s2 2 12
,(y~ j)si + (y(S s, j s )n +
+il (xYS'jlsj YYSj jIs
+ j i s i i sj)IXIYiXiYi E R, n E Z,
m E Z+} is contained in B. Since s. is an extremity of
3
supp f, it follows from Lemma 4.3 that s3 is an extremity
J
of the support of any element of D. Hence all the elements
of the principal ideal
m
(b.) = {xb.+bjy+nbj + i xibjy ilx,y,x iYi R,n E Z, m E Z}
3 3
of R are coefficients of s. in elements of B having s. as
J J
an extremity of their support. By the argument in the
previous paragraph, the hth power of each of these elements
is zero. Therefore (b.) is a nonzero nilpotent ideal of R,
which contradicts our hypothesis.
The preceding result allows us to determine the nilpo
tent radical of R [S].
Theorem 4.6. Let R be a ring, and let S be a semigroup
satisfying condition (E). Then the nilpotent radical of
the twisted semigroup ring Rt[S] is N[Rt[S]] = N TS], where
N is the nilpotent radical of R.
Proof. Let N0 N1 C ... c N ...and
N0 [Rt[S] c N1IR t[SI]] c ..., be the chains of ideals
used in the definition of the nilpotent radical of R and
R t[S], respectively. We show that N tiS] c NER t[S]] by
transfinite induction. We have seen in the proof of Pro
position 4.4 that Bt[S] is a nilpotent ideal of R tS] for
each nilpotent ideal B of R. Hence NtiS] c N ERtES]]. This
0 0
shows the first step of the induction argument.
Assume that NTIS] c N.[R t[S]] for all the ordinals
1 
i < j. Suppose j is nota limit ordinal and let A/Nj_1 be
a nilpotent ideal of R/Nj_1 with index of nilpotency m.
k t k m
Let f = Z arsr E A [S]. Then f = ( a s ) =
r=l r=lr
E(arls r) (asr2s)''(ar s r) with X {l,2,...,k} and
mmm
(rlr2,...,rm) E X x ... X X But
(arlsrl (ar2S (ar s aar2Y(s S)a "
Y(s s s )s s E N a S
r r r r r r
1 m l m M
Since a a y(s rs )a ... a y(s rl .s ,s ) E = 0.
rinc alr2Y r )rar3 arm r rmi rm
Thus (A/NjI.)t is a nilpotent in (R/Nj_1)tFS]; hence
At[S]/Njt[S] is a nilpotent ideal of Rt[S1/NIt[S]. By
induction hypothesis At[S] + Nj_ [Rt[S]]/Nj_ [Rt[S]] is
nilpotent in Rt[S]/Nj_1 Rt[s]]; so A t[S] c Nj[R t[S]].
Therefore Nt[S] c Nj[Rt[S]]. If j is a limit ordinal, then
J J
N. = u N.. Since by induction hypothesis Nt[S] c NER t[S]]
3 i
for all i < j, then
N.$LS = u N.LS] c u N.R t[S]] = N.[Rt[S]].
J i
Finally, since N = NT for some ordinal T, we have that
Nt[SJ S N[Rt[S].
Conversely, since R/N is a semiprime ring, it follows
from Theorem 4.5 that (R/N) t[S is semiprime; thus Rt[S]/Nt[S]
is semiprime. Since N[R t[S]/Nt[S] N[(R/N) t[S]] = 0, then
N[R t[S]] S Nt[SJ.
Proposition 4.7. Let R be a ring with identity, and
let S be a semigroup with identity in which condition (E)
holds. Denote by Z[Rt[S]] the center of Rt[S]. If R is
semiprime, then Z[Rt[SJI is semiprime.
Proof. By Theorem 4.5 and the hypothesis, it follows
that Rt [S] is semiprime. Suppose that there exists an ideal
2
A of Z[Rt[S]J such that A2 = 0. Let a E A, then a= 0.
Moreover, since a is in the center, aRt[S] is a twosided
ideal of Rt[S]. Now (aRt[Si)2 = a2Rt[SJ = 0. Since Rt[s]
is semiprime, it follows that aRt[S] = 0; hence a = 0, and
consequently Z[R t[S]] is semiprime.
4.2 On Semigroup Rings Over Semigroups Admitting Relative
Inverses
In this section we apply the results of section 3.4 to
semigroups of a type first discussed by Clifford [3 1 and
largely studied since then [5 1, [6 1, [19].
A semigroup S is said to admit relative inverses if, for
any element s in S, there exist elements e and s' in S such
that es = se = s and ss' = s's = e. By a semilattice we
mean a commutative semigroup all of whose elements are idem
potents. Let a and 3 belong to the semilattice P; then if
a6 = 6 we write a 8. This defines a partial order on P
with the property that any two elements have a greatest
lower bound, namely its product.
We are interested in semigroups admitting relative
inverses and such that every pair of idempotents commute.
Clifford's main result [ 3 ] on these semigroups may then
be stated thusly:
Every semigroup which admits relative inverses and
such that every pair of its idempotents commute, is isomorphic
with a semigroup S constructed as follows.
Let P be any semilattice, and to each a in P assign a
group Sa such that no two of them have an element in common.
To each pair of elements a > 6 of P assign a homomorphism
qw: Sa S6 such that if a > 8 > y, then 86By= .
We call this a transitivity condition on the homomorphisms.
Let be the identity automorphism of Sa. Let S be the
class sum of the groups Sa and define the product of any
two elements aa,b of S(aa E Sc and b8 E S8) by
aab = (aaaY) (b6P y), where a6 = y is the product of a and
8 in P.
Conversely, any semigroup S constructed in this fashion
admits relative inverses, and every idempotent element of S
is in the center of S.
The semilattice P, the groups S and the homomorphisms
1s8 together constitute a complete set of invariants of S.
The only result that we know on semigroup algebras over
this type of semigroup is due to Munn [18]. He proved
that under suitable conditions on the characteristic of the
field and the order of the subgroups, the semigroup algebra
of a finite semigroup which admits relative inverses is semi
simple if and only if the idempotents commute.
Let P0 be a semilattice with zero elements w such that
every element of P0 {w} is primitive; that is, for all
cS E P0 {w}, c = w. Denote by S the family of all semi
groups S with the following properties: (i) S admits relative
inverses; (ii) the idempotents in S commute; and (iii) S has
P0 as the semilattice of subgroups.
Let S c S, and let {S,} and {80w: Sa SwIa E P0}
stand for the sets of semigroups and homomorphisms of S
respectively. Finally, let R be a ring not necessarily with
1. Our first goal in this section is to characterize R[S]
as a weak direct sum of ideals which are completely determined
once the R[S.] are known.
We begin by relating the semigroup S with the semigroups
having a universally minimal ideal.
Lemma 4.8. Let S e S. The following conditions hold
in S.
(i) S is a semigroup with universally minimal ideal Sw.
(ii) The frame of S is Sw u J, where J = UEp0 Ker 0 w.
(iii) The classes J(u) are defined by J(u) = u 0 (u)
for each u c S
w
(iv) Let Sa (u) = S n J(u); then S (u)S (v) = uv, and
S (u)S (v) C S (uv).
Proof. (i) Let a E Sa and u E Sw. Then a u=
w (a )ew (U) E S w; that is, SSc Sw for all a E P0. Hence
SSw C Sw, and symmetrically SwS c Sw. This shows that Sw is
an ideal of S. Moreover, Sw is a group and therefore is a
simple semigroup. Finally, since xy E Sw for x E Sa, y E SE ,
where a and are distincts and a P w B, it follows that
Sw is universally minimal.
(ii) By definition J = {s E Sisew = e Let x E J n S0
for some a in P0. Then ew = xew = 8 w(X)ew implies that
e (X) = ew; hence x E Ker 0 w. On the other hand, if
y E Ker 0 w, then yew = e w(y)ew = ew shows that y E J, and
the result follows.
(iii) By definition J(u) = {s E Sisew = ul. Let
s E Sc n J(u). Then u = sew = 804w (s)ew implies that
l i
0W (s) = u; that is S E 0 1(u). Moreover, if t E 0 (u),
w w .W
then teW = 0 w(t)ew = u shows that t E J(u). Therefore
J(u) = U 1 u ).
(iv) Let x,z E S (u) and y E S (v) with a,6 E P and
u,v E S w. Then 0 w(X) = u and 0 w(y) = v. So for a r 6
xy = e a W(X) w(y) = uv implies that S (u)S (v) = uv. More
over, 004W (xz) = 0 (X) w(Z) = uv; so XZ E S n J(uv) =
S (uv). Thus S a4(u)S (v) c S C(uv).
n n
Recalling definition 3.22, we let K(u) = I Z aixi E a.=0
i=l i=l 1
and xi c J(u)} and K = ZuS K(u).
w
Now we can apply the results of section 3.4 to obtain
the following result.
Proposition 4.9. R[S] = R[S w] K as a ring direct
sum.
Proof. This follows directly from Lemma 4.8 and
Theorem 3.24.
m
Define C (u) = { ri(xiu)jr. E R, x. E S (u) and
il 1 1
m E Z+} and C C=EU w c(u
mEC(u).
Lemma 4.10. C is an ideal of R[S] for each a E P0.
Proof. First notice that C = 0, since S (u) = u for
w w
each u E Sw. Assume a g w, a E P0. By definition it is
clear that Cac K. Hence C UR[Sw] KR[Sw I = 0 and
R[Sw IC ac R[S wK = 0. To show that C is closed under
multiplication, it is enough to prove that the product of
the generators of C is in C Let xu and yv belong to
C (u) and C (v), respectively. Then (xu)(yv) =
xyxvuy+uv = xyO aw (X)Vuw (y)+uv = xy2uv+uv = xyuv.
By Lemma 4.8 (iv) xy E S (uv); hence xyuv E C (uv). There
fore, C (u)C (v) C C (uv) for all u,v E Sw. It follows
that C C C C
a=b
Now let xv E C (v), and let au = a.x. + b belong
a y ill Then
to K U, where xi E S ,(u) and yj E J(u)S,(u). Then
a = Eai(xiu) + Xb y. + (ai)u. After remaining conveniently
we an rita ai(Xi U) + Ec zk where each zk E S (u)
we can write k
for some a a, E E P0. Then zk(xv) = ZkXZkV =
O w(Zk)eaw (X)8w (Zk)V = uvuv = 0, and (xiu)(xv) =
xixxivux+uv = xixOaw (Xi)vue w (x) + uv = xix2uv+uv
xixuv E C a(uv) since xix E S (u)S a(v) E S (uv) by lemma
4.8 (iv). Hence a u(xv) = Eai(xiu)(xv) + kzCkZk(XV) =
i k
Eai(xixuv) E C (uv). Therefore K(u)Ca (v) S Ca (uv). It
follows now that KC = (EK(u))(C a(v)) = ZK uC (v) c EC (uv)=
U v vu V
C a. Symmetrically C K c C Thus C is an ideal of K. Then
by Proposition 4.9 C is an ideal of R[S].
a
Theorem 4.11. The ideal K = Zw C is the internal
aEP a
0
weak direct sum of the ideals C Hence R[S] = R[Sw I a E @ P C
Proof. Let f E K. From K = uES K(u) and
w
E K(u))n K(v) = 0, it follows that f can be written
u~v
uniquely as f = ZuS fu with fu E K(u). Let
nI1 w nk
f E a. x + ' + E a. x where x E S (u).
u Jl=l 31 alj Jk=I 3 k a3k ,3i i
n 1
Z al(Xij u) .
Since fu e K(u), we may write f u E a. (x.u) + +
nk
E a. (Xa U). Suppose that fucan be expressed in a
jk Jk ak
different way; say fu =l b (y u) + +
m kt 1=l1 1 1 1
mk
bt(~~u) withy E S (u). Then
tk=Ibtk ktk Yit 1i
nI m1
Z a. (x u) E bt l(Yo tU) + "'" +
j 1 1l tl1l=t 1 1 1
1l~ 1111 t~
nk nk
E a.k (x a. u) tkE bt (Yktk u)) = 0.
jl=1kk k
But the support of any two sets of terms in parenthesis are
disjoint; hence each term inside of a parenthesis must be 0.
ni mi
11
Then E a. (x .u) E b t(y,.tu) in C (u) for all i.
It follows that, except by possibly for order, the elements
are identical. This shows that each f has a unique repre
U
sentation in E EPo C(u); hence f can be uniquely expressed
in EEC (u) = ECC.
ucx
Therefore, K is the internal weak direct sum of its
ideals C.. By Proposition 4.9 we conclude that R[S] =
R[S I .woC
W O GE P 0C.
Lemma 4.12. For each U E P0 {w}, the ideal C is
isomorphic as an algebra with R[S 1.
Proof. Let f E RES aI with a e w. Since S = u UES S (u),
w
we can write f = Eail xil + ... + Eaik xi where xit ES (ut),
1 t k.
Define 0: RES ] C( by 0(f) = Eai(xi u) + +
a ik(xi kuk). Notice that if x E S,(u), then O(x) = xu.
That is, 0 takes each generator of RES ( to one generator of
C U. By the definition of C a this correspondence is clearly
onetoone and onto. Hence 6 is a well defined bijection.
Now let f,g E R[S I and let r E R. By definition of 0
it follows that 0(f+g) = e(f)+e(g) and 8(rf) = r6(f). To
show that e preserves products it is sufficient to prove it
for the generators. Let x E S (u) and y E S (v). By Lemma
4.8 (iv) xy E S (uv); hence e(xy) = xy uv = xyuvuv+uv =
xyeO (x)vue aw (y)+uv = (xu)(yv) =(x)e(y). Therefore 8
is an algebra isomorphism.
Theorem 4.13. N[R[SI] = N[R[Sw 1] a 0 N[C a, and
0
J[R[SI] = J[R[Sw1 1 a p0J[C 1, where NEC ] N[R[S 1]
w cxcP 0 x cx a
and J[C I z J[RES I].
Proof. This follows directly from Theorem 4.11 and
Lemma 4.12.
Corollary 4.14. The following conditions are equivalent.
(i) R[S I is semiprime (semisimple) for each a c P0.
(ii) RES] is semiprime (semisimple).
Proof. By Theorem 4.13 NERESI] = 0 (JER[S]] = 0) if
and only if N[R[S a1 = 0 (J[RES a] = 0) for all a E P0.
Corollary 4.15. If F is a field of characteristic 0 or
a prime not dividing the order of any of the groups S 0
where Is i < for all a E P0' then FES] is semisimple.
Proof. By Maschke's theorem [131 each F[S I is semi
simple; so the result follows from Corollary 4.14.
Corollary 4.16. (1) If F is a field of characteristic
0 that is not algebraic over the rational field Q, then
FLSJ is semisimple. (2) If F is a field of characteristic
p > 0 that is not algebraic over the prime field GF(p), and
if Sa contains no elements of order p for all a P0 then
FES] is semisimple.
Proof. Amitsur's theorem [221 and the hypothesis in (1),
or Passman's theorem [22] and the hypothesis in (2) show that
in either case FES.] is semisimple for each a E P 0 Hence
the results follow from Corollary 4.14.
Corollary 4.17. Let F be a field. The following con
ditions are equivalent:
(i) F[SII is regular;
(ii) for each a E P0 the group Sa is locally finite and
has no elements of order p in case char F = p.
Proof. By Theorem 4.11 and Corollary 4.12 FES] is
regular if and only if F[S.] is regular for all a E P0. Now,
Villamayor [25J and Connell L 7 I have shown that FESa] is
regular if and only if S. is locally finite and has no
elements of order p in case char F = p; hence the result
follows.
Corollary 4.18. Let F be a field. Assume either
(1) char F = 0 or else (2) char F = p > 0 and, for each
U E P0 Sa has no finite normal subgroup of order divisible
by p. Then the following conditions hold in F[SJ.
(i) F[S] is semiprime.
(ii) The center of F[S'j, Z[FLSjl, is semiprime.
(iii) Z[F[SJJ is semisimple.
Moreover, if (2) happens, then these three conditions are
equivalent.
Proof. From Theorem 4.13 and Lemma 4.12, it follows
that Z[FISJI is semiprime (semisimple) if and only if
ZLF[S.J]i is semiprime (semisimple) for all a E P0. Hence
the conclusion follows by applying Passman's results [22,
Theorem 2.12 and Theorem 2.13J.
Next we consider a second family of semigroups which
admit relative inverses.
Definition. A semilattice P is said to be of type P1
if the following conditions hold in P:
(i) P has a zero element;
(ii) if F is a subset of P, then either g.l.b. {xlxEF}
belongs to F or else there exist x,y in F such that
g.l.b. {x,y} is in PF.
Denote by T the family of all semigroup S with the
following properties: (i) S admits relative inverses;
(ii) the idempotents in S commute; and (iii) S has a semi
lattice of subgroups of type P1.
Let S E T and let {S },, be the family of subgroups of
S. Denote the zero element of P by a O. Let R be a ring
not necessarily with 1. Our goal now is to prove that the
semigroup ring R[S] is semiprime if R[S ] is semiprime for
each a c P. But first we need two preliminary results.
Lemma 4.19. There exists {Ti}ii, a family of subsets
of P, where P =uiiTi and the index set I is an initial
segment of the ordinals such that To c T1 c c T. c ....
Moreover, if we let S t = UET Sa then S t0c St c'...CSt C''
is a series of ideals of S with S = ui ISt.
1
Proof. The sets Ti will be obtained by transfinite
induction. Define TO = {c0}. Assume T. is defined for all
3
the ordinals j < i. If uj
there exists yEPu. .T. such that g.l.b. {x,y} E u. .T.I.
J<1 J J<1 J
If T! #, set T. = T! u (u. iT.). If T! = let
1 1 1 j<1j i
TV = g.l.b.
1 j
Notice that since P is of type PI. then Ti is defined.
Clearly T0 c T1 c c T c ... and P = uiiT. It follows
that S c c c Sti ... with S = u i St. Now let
a E T. and 6 c Ti, where i,j E I. By construction ax =
g.l.b. {a,6} belongs to Tk, where k = min{i,j}. Hence
S(XS cStk CS St and therefore St S S Symmetrically,
k J J 1 J
St St 3 S Stj This shows that S S and SS are contained
in St. for all j E I; that is, each St. is an ideal of S.
3 J
Lemma 4.20. Let R be a ring not necessarily with 1.
Let I be an initial segment of the ordinals, and let {A}I
1 iEI
be a family of ideals of R such that A0 C A1 c ... c A. c ...
and R = EiEiAi. Then the following conditions are equivalent.
(i) The ideals Ai are semiprime for all i E I.
(ii) R is semiprime.
Proof. Assume A. is semiprime for each i E I. For
1
contradiction, suppose that R is not semiprime, and let B
be a nonzero nilpotent ideal of R. Since the ideals Ai form
a chain, then B n A. 0 for some j E I; so B n A. is a nil
potent ideal of Aj, which contradicts our hypothesis.
Conversely, assume that R is semiprime, and suppose that,
for some i E I, the ideal A. is not semiprime. Let C be
1
a nonzero nilpotent ideal of A. For each j,k in I, we have
CAjCAk S CA AiAk c CAi c C. Hence (CR) 2 c C. Thus CR is a
nilpotent right ideal of R. Since R is semiprime, CR = 0.
Thus C is a nonzero nilpotent right ideal of R, which
contradicts our hypothesis. Therefore Ai is semiprime for
all i c I.
Theorem 4.21. Let S E T. If RES I is semiprime for
all a E P, then R[S] is semiprime.
Proof. Assume R[S ] is semiprime for all a E P. Let
{St }iI be as defined in Lemma 4.19. Using transfinite
t.I
1
induction, we show first that the ideals R[St I are semiprime
for all i E I. Then by Lemma 4.20 the result follows.
The first step of the induction argument is true by
hypothesis. Assume that RESt I is semiprime for all the
ordinals j < i, i,j E I. Let St! = TIS and let
1 1
S t? = uTV S a. By the proof of Lemma 4.17 we see that
1 1
either St =S u (ujS or else S =(u St )u S
t J 1j
Suppose St. =(u S.) u S By definition TVl [a}
tJ(1 t. t. 1
for some a c P uj
1 1
semiprime by hypothesis. By Lemma 4.19 each S and hence
t.
J
uj
J
hypothesis and Lemma 4.20 that R[u j
A be a nilpotent ideal of R[S t]. Then A + R[u j
ti J< <
being a nilpotent ideal of R[St ]/R[u j
i J
to a nilpotent ideal of RESt,]. Since R[S t,] is semiprime,
1 1
it follows that Ac R[u j
J 3
semiprime; hence A = 0. Therefore R[St I is semiprime.
Now letS = (u. iS ) u St!. By definition of T! if
t J< t.
Ct, E T!, then c = g.l.b. {a,6} belongs to uj iT. Hence
1 j~j j"
S Sc u S for all E Moreover, ( E R[S ]) n
j
R[S1 = 0 for all 6; thus we can write RESt] =RES It
1 1
where all R[S 1 are semiprime by hypothesis. As in the first
case R[uj
R[uj
J 1 1
R[uj
3 1 J 1
direct sum of Rmodules. Let f E RESt ]. We may write
i
f f +f + + f where f. E R[uj
f~i1R[Sn 1n
]E RESI for some l,...,f E T1.
1i 1 n 1
Fix p T and define IT : RI S I R[S I by
1 P P
1p
SP(f) = f Clearly ITP is an Rmodule epimorphism. To
show that iTp is a ring epimorphism, let f,g c R[S t1 such
that f = fi + f6 + f and g = gi + g6 + g ,where f6 and g6
belong to E cTi{p}R[S ]. Then fg = figi + fi(g6 + gp) +
(f6 + f )g+ + f6g + fg6 + f6g5 + f gp. Since RFu .
j
is an ideal, and since R[S6]R[Sp] c R[uj
j
term in the product fg that belongs to RS oI is f Pg" Hence
ITP (fg) = fpgP = IP(f)r (g); so Ip preserves products.
Suppose R[St ] contains a nonzero nilpotent ideal B.
1
Since R[u. .S I is semiprime, then for some p E T!, it (B);0.
j~1 t 1 p
Since it is a ring epimorphism, then it (B) is a nonzero
P P
nilpotent ideal of R[S ], which contradicts the hypothesis.
p
Corollary 4.22. Let P be an initial segment of the
ordinals. Let S be a semigroup which admits relative inverses
and such that every pair of its idempotents commute. Let
{S a } be the family of subgroups of S. Let R be a ring
not necessarily with identity. If R[S I is semiprime for
a E P, then R[S] is semiprime.
Proof. Since P is of type P1, the result follows from
Theorem 4.21.
BIBLIOGRAPHY
1. R. G. Ayoub and C. Ayoub, On the Group Ring of a Finite
Abelian Group, Bull. Austral. Math. Soc. 1 (1969),
245261.
2. B. Banaschewski, On Proving the Absence of Zero Divisors
for Semigroup Rings, Canad. Math. Bull. 4 (1961), 225232.
3. A. H. Clifford, Semigroups Admitting Relative Inverses,
Ann. of Math. 42 (1941), 10491307.
4. A. H. Clifford and D. D. Miller, Semigroups Having
Zeroid Elements, Amer. J. of Math. 70 (1948), 117125.
5. A. H. Clifford and G. B. Preston, Algebraic Theory of
Semigroups, Volume I, Math. Surveys of the Amer. Math.
Soc., Rhode Island, 1961.
6. A. H. Clifford and G. B. Preston, Algebraic Theory of
Semigroups, Volume II, Math. Surveys of the Amer. Math.
Soc., Rhode Island, 1967.
7. I. G. Connell, Zero Divisors in Group Rings, Comm. Algebra
2 (1974), 114.
8. C. W. Curtis and I. Reiner, Representation Theory of
Finite Groups and Associative Algebras. John Wiley and
Sons, New York, 1962.
9. N. J. Divinsky, Rings and Radicals, Univ. of Toronto
Press, 1965.
10. R. Gilmer and T. Parker, Nilpotent elements of Commutative
Semigroup Rings, Michigan Math. J. 22 (1975), 97108.
11. R. Gilmer and M. L. Teply, Units of Semigroup Rings,
Comm. Algebra 5 (1977), 12751303.
12. G. Higman, The Units of Group Rings, Proc. London Math.
Soc. 46 (1940), 231248.
13. T. W. Hungerford, Algebra, Holt Rinehart and Winston,
New York, 1974.
90
14. N. Jacobson, Lectures in Abstract Algebra, Volume I:
Basic Concepts, Van Nostrand, New Jersey, 1951.
15. J. Lambeck, Lectures in Rings and Modules, Blaisdell,
Toronto, 1966.
16. N. H. McCoy and D. Montgomery, A Representation of
Generalized Boolean Rings, Duke Math. J. 3 (1937),
445459.
17. N. H. McCoy, The Theory of Rings, Chelsea, New York,
1973.
18. W. D. Munn, On Semigroup Algebras, Proc. Cambridge
Phi. Soc. 51 (1955), 115.
19. W. D. Munn, Matrix Representations of Inverse Semigroups,
Proc. London Math. Soc. 14 (1964), 165181.
20. T. G. Parker, Infinite Group Rings, Dissertation, Florida
State University, 1973.
21. D. S. Passman, Infinite Group Rings, Marcel and Dekker,
New York, 1971.
22. D. S. Passman, The Algebraic Structure of Group Rings,
John Wiley and Sons, 1977.
23. F. Raggi Cardenas, Units in Group Rings, An. Inst. Mat.
Univ. Nac. Autonoma Mexico 7 (1967), 2735.
24. K. Shoda, Uber die Einheitengruppe Eines Endlicher Ringes,
Math. Ann. 102 (1930), 273282.
25. 0. E. Villamayor, On Weak Dimension of Algebras, Pac. J.
Math. 9 (1959), 941951.
BIOGRAPHICAL SKETCH
Antonio Quesada was born in Melilla, Spain, on January
29, 1948. He attended the University of Granada where he
received the "Licenciatura" in Mathematics in June 1971. In
September 1975, he enrolled at the University of Florida to
continue his mathematical education.
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Mark L. Teply
Associate Professor of Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Jo/ge Martinez
Associate Professor of Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
G < 2ard x. Ritter
sociate Professor of Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Sergio E. Zarantonello
Assistant Professor of Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Specialization Teacher Education
This dissertation was submitted to the Department of
Mathematics in the College of Arts and Sciences, and to
the Graduate Council, and was accepted as partial fulfill
ment of the requirements for the degree of Doctor of
Philosophy.
August 1978
Dean, Graduate School

Full Text 
38
the twist function is nonconstant, then RtLSJ has nontrivial
units.
Proof. Let s be a torsion element of S different from i.
If s e G, then G is a nontrivial, nontorsion free group;
thus Proposition 2.14 shows that Rt[G] and hence Rt[S] have
nontrivial units. Let s e S G. Then, either char R 2
and the result follows from Lemma 2.11, or else char R = 2.
By Theorem 1.1 and Lemma 2.10, (s) contains an idempotent
e i. Hence, Lemma 2.12 shows that Rt[S] contains a non
trivial unit when char R = 2.
If we follow verbatim the proof of Theorem 2.15,
except for changing the use of Proposition 2.14 to Theorem
2.5, then we have proven the following result.
Theorem 2.16. Let S be a nontorsion free semigroup
and let K be a field. If the twist function is nonconstant,
then Kt[S] has nontrivial units.
2.4 Zero Divisors In Twisted Group Rings
It has been proven [22, Lemma 13.1.1] that if G is a
nontorsion free group and K is a field, then K[G] has
proper zero divisors. In passing to the twisted group rings
case, this is no longer true, as we see in the following
example. Consider Q^LC^],where Q is the field of rational
numbers and C. is the cyclic group of order 4 generated
3 i
by g. Let tt. = n y(g,g ), and assume that
i=l
/n"^ K. Then x4 tt^ is irreducible; hence
50
Proof. Let R be a semigroup ring, and let s =1.
Then R[S] ~ R; hence N[R[S]] = 0.
Conversely, let R be a commutative ring, and let S be a
right zero semigroup. Assume that R[S] is semiprime. Then
N[R[SJ] = 0. Hence, if either the nilpotent radical N of R
is nonzero, or if s > 1, then by Theorem 3.6 N[R[S]] 0.
This contradiction shows that N = 0 and s = 1.
Corollary 3.8. Let R be a commutative ring. Let the
5
semigroup S = E^ x x ... x E^, where E^ are right zero
semigroups. Then the nilpotent radical of R[S] is
n
N[RLS]] = {fa..x.. e R[E.]Za e N}, where N is the
i=l j !3 13 i 'j ij
nilpotent radical of R.
n
Proof. Since R[S] = R[E.], then the conclusion
i=l 1
follows from Theorem 3.6.
3.3 The Nilpotent Radical of the Semigroup Ring of A Right
(Left) Group.
A semigroup S is called a right group if it is right
simple and left cancellative [5 .1. Left groups are defined
dually. Being right (left) simple means that the semigroup
contains no right (left) proper nonzero ideals; thus these
semigroups form a subfamily of the simple semigroups.
From now on we consider right groups, but the analogous
conclusions for left groups will follow by duality.
Let R be a ring with identity, and let S be a right
group. In this section we find the nilpotent ideals of
85
semigroup ring RL5J is semiprime if RL J is semiprime for
each a e P. But first we need two preliminary results.
Lemma 4.19. There exists {T^}^ j, a family of subsets
of P, where P = u. and the index set I is an initial
i e I r
segment of the ordinals such that tq c Tl c c T^ c ... .
Moreover, if we let S. = u S then S. c S. c...cS, c...
01 Â£ 1 CL L. t..
11 oil
is a series of ideals of S with S = ueIst
i
Proof. The sets T^ will be obtained by transfinite
induction. Define T~ = {a}. Assume T. is defined for all
0 0 j
the ordinals j < i. If u^T P, let T = {xePu^.Tl
J J 3 ^ 3
there exists yePu. .T. such that g.l.b. {x,y} e u. ,T.}.
D
If T! , set T. = T! u (u . T. ) If T! = , let
l li j
T'.' = g.l.b. {x  xe Pu . T.}, and set T. = (u. T ) u T'.'.
1 l
Notice that since P is of type P^, then T^ is defined.
Clearly Tn c T, c ... c t. c ..., and P = u. _T.. It follows
01 i iel i
that S, c c ... c s c ... with S = u. Now let
t0 t ti lei t.
a e T. and 8 e T^, where i,j e I. By construction aB =
g.l.b. {a,0} belongs to T^, where k = min{i,j}. Hence
SaSg c St Â£ st / and therefore S c s Symmetrically,
k j lil
St S c s This shows that S S and SS are contained
i j j tj
in S. for all j e I; that is, each S. is an ideal of S.
j 3
Lemma 4.20. Let R be a ring not necessarily with 1.
Let I be an initial segment of the ordinals, and let {A^}^eI
be a family of ideals of R such that A c a. c ... c a. c ...
= ^ieiAi* Then the following conditions are equivalent.
and R
37
Corollary 2.13. If all the units of RtLSj are contained
in RtLGj, then S G is torsion free, provided that the
twist function is nonconstant.
Proof. Suppose S G contains a torsion element. If
char R 2, then by Lemma 2.11 R**[S] contains a nontrivial
unit with support in S G, which contradicts our hypothesis.
On the other hand, if char R = 2, then by Theorem 1.1 S G
contains an idempotent e. Hence by Lemma 2.12 R^[S] contains
a nontrivial unit with e in its support. Therefore S G is
torsion free.
Proposition 2.14. Let G be a nontorsion free group, and
let R have finite characteristic. If the twist function is
nonconstant, then Rt[G] has nontrivial units.
Proof. We show that Pt[G] always has nontrivial units.
Let char P = n, and let D = {q^,q2,...,q^} be the set of
distinct prime divisors of n. If k > 1, then P is decompos
able as a finite direct sum of qrings. Since G is nontorsion
free by hypothesis, then Theorem 2.9 implies that Pt[G] has
nontrivial units. On the other hand, if k = 1, then P is a
qring. Let n = qm, where m > 1. Then P has a nilpotent
element; so by Lemma 2.6 P^[G] has nontrivial units. Finally,
if n = q, then P = GF(q); hence by Theorem 2.5 Pt[G] contains
nontrivial units.
Theorem 2.15. Let S be a nontorsion free semigroup and
let R be a ring of finite characteristic with identity. If
62
(ii)S = u J(u), where J(u) = 0 (u) = {seS  sz=u=zs}.
UeU
(iii)J (u) n U = u and J (u) n J (v) = 4) for all
u,v e U, u v.
(iv) J(z) = J = {s e S sz = zs = z}.
(v) J(u)J(v) c J (uv) for all u,v e U.
Lemma 3.21. Let 0: R[S]'+ R[U] be the natural extension
of 0; that is, 0(Ea^s^) = Ea^0(s). Then 0 is a semigroup
ring epimorphism.
Proof. By definition 0 is clearly an Rmodule homomor
phism. Moreover, by Remark 3.20 (i) 0 is an epimorphism.
Since zs = sz for all s e S and z = z, it follows that 0 is
also a ring epimorphism.
n n
Definition 3.22. Let K(u) = { E ax E a. = 0,
i=l i=l
x^ e J(u) and n e Z+} and set K = fUeU K(u).
Lemma 3.23. K is an ideal of RLS].
Proof. We show that K = Ker 0. Let Ea^x^ e K(u). Since
0(Ea^x^) = Ea^0(x^) = Ea^(x^z) = (Ea^)u = 0, then 0(K(u)) = 0.
Thus, 0(K) = 0, and therefore K c Ker 0.
On the other hand, let a e Ker 0, and write
= ^iui + S l bjkxu,k'
1 J K 3
where xu k e J(Uj) {uj} for all k. Then
0 = 0(a) = Ea.u. + E E b',(x ,z) = Ea.u. + E(Eb.,)u^.
i j k Ujk i 1 1 j k J
22
for all r'r2'r2 e R and sl's2's3 e This shows that the
condition that y be defined into the central units of R
is needed in order to preserve the associativity.
From now on we will abuse in our notation by identifying
the elements of S (of a group G) with their images in R*"[S]
(Rt[G]) under the natural embedding; that is, we will write
s(g) for s(g).
n
Let A be an ideal of R. The set B = { E a.s.la. e A,
i_l i i1 i
Sj. e S, n e Z+} is an ideal of Rt[S] that we will denote by
the (abusive) descriptive symbol A^[S]. If the semigroup
S has an identity, then the ideal Afc[S] is the extension
of the ideal A of R to Rt[S]; that is, ARt[S] = Afc[S].
Moreover, AR^CS] n R = A.
Let X be an ideal of S. Then Rt[X] is an ideal of
R*"[S], where the twist function of R^[X] is the restriction
to X x X of the twist function of R^[S].
The following results are generalizations of similar
results in [20] for semigroup rings.
Lemma 1.25. If {A^}^ ^ is a family of ideals of the
ring R, then (n^Â£^A^)t[S] = n^g^(A^[S]).
Theorem 1.26. Let R^ and R2 be rings with identity,
and let 0: R^ * R2 be a homomorphism of R^ onto R2. Let S
be a semigroup, and let y^:S x S * R^, i=l,2, be twist
functions such that 0y^ = Y2. Then there is a unique
homomorphism 0 : R^ [S] R^ [S] from the ring R^ [S] with
twist function y^ onto the ring R^ [S] with twist function
CHAPTER I
PRELIMINARIES AND GENERAL PROPERTIES OF SEMIGROUP RINGS
In this chapter we introduce some fundamental defnitions
and general results that will be used throughout this dis
sertation.
1.1 Semigroups
A semigroup S is a nonempty set closed under a binary,
associative operation that we will denote multiplicatively.
Let S be a semigroup. The cardinal number s of the set S
is called the order of S. If s is finite, we can exhibit
the binary operation in S by means of its Cayley multipli
cation table as for finite groups. An element t of S is
said to be left (right) cancellable if, for any x and y in S,
tx = ty (xt = yt) implies x = y. S is called left (right)
cancellative if every element of S is left (right) cancel
lable. We say that S is cancellative if it is both left and
right cancellative. An element u of S is called a left (right)
identity of S if us = s (su = s) for all s e S. An element
u of S is a twosided identity, or simply identity, if it is
both a left and a right identity of S. Exactly one of the
following statements must hold for a semigroup S:
(1) S has no left and no right identities;
(2) S has one or more left (right) identity, but no
right (left) identity;
1
40
Proof. Let char K = q > 0. Then the prime field of K
is GF(q). Hence ^/x e K for all x e GF(q). By Theorem 2.17
the result follows.
For reasons of completeness we give the following result,
the proof of which resembles that of the ordinary group rings.
Lemma 2.19. If G is a torsion free abelian group, then
Kt[G] is an integral domain.
Proof. Since G is a torsion free abelian group, then it
admits a total order compatible with the product. Let a and
t n
8 be nonzero elements of K [G]. Write a = E ax and
m i=l
8 = E b.y.. Since G is ordered, supp a and supp 8 are totally
i1 1 1
ordered subsets of G; hence the leading terms of a and 8 are
anxn and t>mym respectively. Thus a8 has a nonzero leading
term akbmy(xn,ym)xRym, which proves the result.
72
be a nonzero element of A^ISJ. Then um = ( Y. a. s. )ni = Ka. s. )
i_l ii X1 11
(a. s. s. ) with L = 1,2,...,k and (i.,i,...i ) c
i~ i li 1 2 m
2 2 mm
L x... x l = Lm. But (a. s. )(a. s. )(a. s. ) = a. a. y(s. ,
1,1. 11 i 1 1.1 1.
11 22 mm 12 1
s. )a. a. y(s. s. s. ,s ) s.s~***s = 0, since
i i i 1 i. i i m 12 m
2 3 m 12 m1
a. a. y(s. s. )a. a. y(s. s. s. ,s. ) e Am = 0.
i. i' i. i i i 1 i. i i i
12 123 m 12 m1 m
Hence am = 0. Therefore At[S] is a nilpotent ideal of Rt[S],
which contradicts our assumption; thus R is semiprime,
Theorem 4.5. Let R be a ring, and let S be a semigroup
satisfying condition (E). Then the following conditions are
equivalent:
(i) The twisted semigroup ring Rt[S] is semiprime.
(ii) R is semiprime.
Proof. If Rt[S] is semiprime, then by Proposition 4.4
R is semiprime.
Conversely, assume that R is semiprime. By way of con
tradiction, suppose that R^'CS] contains a nonzero nilpotent
ideal B. Let h be the index of nilpotency of B, and let
r
6 = E b.s. be a nonzero element of B. Then
i=l 1 1
IT IT
0 = Bh = ( l b.s.)h = Z (b.s.)r + I(b. s. )(b. s. )
i=l 11 i=l 1 1 X1 11 1n xn
with M = {1,2,...,r} and (i^,i2,...,i^) e (x e Mn not all
coordinates of x are equal}. Now supp 8 is a finite subset
of S, and since by hypothesis, condition (E) holds in S,
then there exists s. extremity in supp 8. Thus s*1 s. s. s.
1 3 11 12 1n
15
Corollary 1.16. Let K be an algebraically closed field,
and let G be abelian. Then Kt[G] is diagonally equivalent
to K[G] if and only if Kt[Gj is commutative.
A final example, in which Kt[G] is always K[G], is pro
vided by the case where K is a perfect field of characteristic
p and G is a finite pgroup.
We remark that the material discussed so far in this
section has been taken from Passman [22]. Next we give
some new results that we have obtained in order to be able
to multiply elements of a twisted group ring over a finite
cyclic group.
In the remainder of this section we let G = (g) be a
cyclic group of prime order q, unless otherwise specified.
Definition. tt
a
M ah
II y (g ,g ) where 1 < a < q.
h=l
a
= "aY (ga,gq)
Proposition 1.17. = nj_+ j Y (g1 > g~* ) q r where
1 < i,j < q1.
i "i k
Proof. Putting x = g y = gJ and z = g in equation
(2), we have
(5) y (g1fg^+k) y (g*,gk) = Y(g1+^,gk) Y(g1/g^).
Fixing i and j and taking products, we get
BIOGRAPHICAL SKETCH
Antonio Quesada was born in Melilla, Spain, on January
29, 1948. He attended the University of Granada where he
received the "Licenciatura" in Mathematics in June 1971. In
September 1975, he enrolled at the University of Florida to
continue his mathematical education.
91
78
(iii) The classes J(u) are defined by J(u) = u 0 (u)
aePQ aw
for each u e S .
w
(iv) Let S (u) = S n J(u); then S (u)SD(v) = uv, and
a a a p
S (u) S (v) c s (uv) .
a a a
Proof.
(i)Let a e S and u e S Then a u =
a a w a
0 (a )0 (u) e S ; that is, S S c S for all a e Pn. Hence
aw a ww w a w w 0
SS c s and symmetrically S S c s This shows that S is
w w 1 J w w w
an ideal of S. Moreover, S is a group and therefore is a
w
simple semigroup. Finally, since xy e Sw for x Â£ Sfl, y e S^,
where a and 6 are distincts and a w 6, it follows that
S, is universally minimal,
w
(ii)By definition J = {s e Sse = e }. Let x e J n S
w
w
for some a in P. Then e = xe =0 (x)e implies that
(j 1*7 T7 /"VI T*7 T.7 ^
W
6aw(x) = ew; hence x e Ker 0aw
w aw w
On the other hand, if
a
y e Ker 0aw, then yew = aw(y)ew = ew shows that y e J, and
the result follows.
(iii)By definition J(u) = {s e Sse = u}. Let
w
s e Sa n J(u). Then u = se^ = 0aw(s)ew implies that
0 (s) = u; that is s e 0 ^(u). Moreover, if t e 0 ^(u),
U* V/ 0tvy CXW
then te =0 (t)e = u shows that t e J(u). Therefore
w aw w
J (u) = u D 0 ^ (u) .
aeP^ au
(iv)Let x,z e Sa(u) and y e (v) with a,6 e P and
u,v e S Then 0 (x) = u and 0D (y) = v. So for a 8
W CtW pW
xy = eaw(x)e6w(y) = uv implies that Sa(u)Sg(v) = uv. More
OVer' 0aw(xz) = 6aw(x)0aw(z) = uv; so xz e sa n J(uv) =
Sa(uv). Thus Sa(u)Sa(v) c Sa(uv).
20
Proof. For b,c e Z, b < c, let [b,cj denote the product
Y(g/gb)Y(ggb+1) *Y(g,gc). By Lemma 1.22 we see that
k k
Y(g2 P,g2 P) = C 2kp,2k+1pl]/[1,2kpl] for k = 0,l,...,nl.
Hence
nnl ,, ~n2 ~nl n1
(6) Y(gp,gp) Y(g pfg p) ***Y(g2 p,g2 p) =
f _nl ? n2
[p,2pl] y /[2p,2 pnr
[l,pl] / \[1,2pl] /
... (
\Cl,2n 1pl]
r o ,i2n _2 in2n ^ r~n2 ,nl in2rnl _n
[p,2pl] [2p,2 p1] [2 p,2 p13 [2 p,2 p1]
n^l2i n22i i"
i0 i=0 i = 0 _
[1,P1] [p,2pl] [2p,2pl] *[2n p,2n p1]
n_a21 +1
Since E = 2n a 1 for a < n, we get the expressions
i=0
in equation (6) equal to
nl
.n2
(7) [p,2pl]
[2p,2 p1]
r n n 2 ri 1 ^ r 'in 1
[2 p, 2 p1] [2 p, 2 p1]
1,p1J
2nl
2n 1 ?
[p,2pir L [ 2p, 2 p1 3
2n2_i
r _n2 nl ,
[2 p,2 p1]
But from the hypothesis the multiplicative subgroup of K is
n 2nl
cyclic of order 2 1; hence [l,pl] = 1. Thus by can
cellation (7) becomes [p, 2pl ][ 2p, 22pl] [ 2n *p,2npl] =
[p,2npl]. Now for all b e Z, [b,b+ql] = y(g,gb)Y(g/gb+1)* *
Y(g,gb+q1) = y (g,gb)y (g^gb+1) *y (g/i)y(gg) * *y(g/gb 1)=tt1.
Since [p,2npl] is a product of (2npl) (p1) = (2nl)p = qp
factors, then [p,2npl] =[p,p+q1][p+q,p+2ql][p+(p1)q,
p+pq1] = tt^. Therefore the result follows.
decompose R[S] as a direct sum of ideals which are completely
determined once RLSa] is known for each a PQ. This allows
us to characterize the nilpotent and Jacobson radicals of
R[S]. For R field, we give necessary conditions for R[S]
to be semisimple and regular.
If the semilattice of subgroups SQ of S is of type P^,
then we prove that R[S] is semiprime if R[SaJ is semiprime
for each a e P^.
Vll
79
n n
Recalling definition 3.22, we let K(u) = { 1 a.x. E a.=0
i 1 1 i 1
1=1 1=1
and x. e J(u)} and K = Â£ K(u).
i u a o
w
Now we can apply the results of section 3.4 to obtain
the following result.
Proposition 4.9. R[S] = R[Sw] K as a ring direct
sum.
Proof. This follows directly from Lemma 4.8 and
Theorem 3.24.
m
m e
Define C (u) = { E r.(x.u)
a .,ii
f 1
Z } and C = E C (u).
a ueS a
w
r. e R, x. e S (u) and
i i a
Lemma 4.10. C is an ideal of R[S] for each a e P.
a 0
Proof. First notice that C =0, since S (u) = u for
w w
each u e S Assume a w, a e P. By definition it is
w u
clear that C c k. Hence C R[S ] c KR[S ] = 0 and
Ot O W w
R[Sw]Ca c R[Sw]K =0. To show that Ca is closed under
multiplication, it is enough to prove that the product of
the generators of is in C Let xu and yv belong to
Ca(u) and C (v), respectively. Then (xu)(yv) =
xyxvuy+uv = xy0aw(x)vu0(y)+uv = xy2uv+uv = xyuv.
By Lemma 4.8 (iv) xy e Sa(uv); hence xyuv e Cq(uv). There
fore, C (u)C (v) c c (uv) for all u,v e S It follows
a a ~ a w
that C C c c .
a a a
Now let xv C (v), and let a = Ea.x.
a u i i
i
to K where x c S (u) and y. e J(u)S (u).
1 O J ct
+ Eb.y. belong
i 3 3
Then
17
Let x = qni (mod q). By setting j = qi in Proposition
1.17, we get tt = Y(gX,gq 1)"Vtt since tt =1. Hence, if
^ qi 1 ^ ^ i q '
we let j = q2i Proposition 1.17 yields 7Tg_2i =
, i q2i.q, i qi*q i q2i.q. 2 ..
TTq_iY(g ,9 ) /ffi = Y(g ,9 ) Y (g ,9 ) v^. Thus the
result follows by recursion.
Corollary 1.19.
integer such that 2n
if tt =1 for some i
i
If K = GF(2n), where n is a positive
 1 = q, then tt tt = tt . Moreover,
M i j i+J
1 < i < q1, then tt^ = 1 for all j.
Proof. By hypothesis K is a finite field of order q+1;
hence its group of units is cyclic of order q. It follows
that y(g1/gD)q = 1 for all i,j. Then by Proposition 1.17
tt i tt j = Iff moreover, tk = 1, then the conclusion
follows from Lemma 1.18.
Next we determine some conditions that force the twist
function to be trivial.
Proposition 1.20. If YgSg^) = 1 for h = 1,2,...,q1
and i fixed, then y is trivial.
Proof. By the hypothesis and equation (5) we get
~l k i f ~i k
Y(g ,g ) = y(g ,9 ) with 1 < j,k < q1. As in Lemma 1.18,
j 5 qni (mod q) for some n. We use induction on n.
If n = 1, then y(gg ^g^) = = 1 fr all k. Assume
that y(gq tgk) = 1 for all k. Then y (gq_ ,g^) =
Y(gq t:L,gk) = 1 for all k. Therefore y(gi,gk) = I fr all
j and k.
30
Now let Q = (l,2,...,ql) and define x ~ y if
y E 2hx (mod 2nl), where x,y e Q, h e Z. Clearly ~ is an
equivalence relation. Since 2np = p (mod 2nl) for all
p e Q, then each of the equivalence classes that ~ determine
on Q has the form {2up 1 < u < n }. Thus we have Q as the
union of (ql)/n mutally disjoint subsets of n elements.
Notice first, that the system (1) has ql unknowns and
as many equations. Moreover, any unknown a is related to
P
a9 ; in general, a is related to a , in one equation
2Kp 2Kp
and to a ^ in another. Hence, by the previous argument
2 P
the set (a x e Q} of unknowns is partioned into (ql)/n
subsets of the form {a l < u < n} for various p. Since
2^1
P
the unknowns of each of the ql equations are contained in
one and only one of these subsets, then we have obtained
(ql)/n systems of n equations each of which has the form:
apY(gp,gp)
a2p
q qk qk
2 2 p 2 px
a k y(g ,g H
2 p
~k+l
2 P
n1
P
r
/
where 1 < k < n1. Now let, for instance, a = cv, where
P
c e K {0}. Then we consider the set:
47
section is to characterize the nilpotent radical of R1 SI.
But first we need the following four lemmas.
Lemma 3.1. Any right ideal of RLSj is a twosided
ideal.
Proof. Let A be a right ideal of R[S]. Hence ARCS] Â£ A.
Since R is commutative, then RA = AR c a. Moreover,
SSupp a = supp a for all a e A; hence SA = A. ThenRLSjA =A
shows that A is also a left ideal.
Lemma 3.2. Let A be an ideal ofR, and let T be a non
empty subset of S. Then ALT] is a left ideal of R[S].
Moreover, A[S] is a twosided ideal of R[S].
Proof. Any subset T of S is a set of right zeros; hence
ST = T. Thus, R[S]A[T] cRACT] = A[T]. Also A[S]S Â£ A[S]
and A[S] R c A[SJ imply that A[S]R[S] cA[S]. Hence A[S]
is a right ideal, and the result follows from Lemma 3.1.
Lemma 3.3. Let a = Ea.x. be an element of R[S] such
i i
that Ea^ = 0. Then the principal ideal generated by a,
is (a) = Ra. Moreover, (a) is a nilpotent ideal of index
two.
Proof. Since ax = (Za^)x = 0 for all x e S, then
RaRLS] = 0; hence Ra is a right ideal of R[S]. It follows
from Lemma 3.1 that Ra is a twosided ideal; that is, (a) = Ra.
Moreover, since a2 = (la.^) (Ea.^) = (Ea.^) (Ea^. ) = 0 and
since R is commutative, then (Ra)(Ra) = Ra2 =
0.
32
We can now put the pieces together to get the following
result.
Theorem 2.5. If G is a nontorsion free group and K is
a field, then Kt[G] always has nontrivial units, provided
that the twist function is nonconstant.
Proof. This follows directly from Lemmas 2.1 and 2.4.
2.2 Nontrivial Units In Twisted Semigroup Rings
Let S be a semigroup (possibly infinite) with identity
i. Denote by G the largest subgroup of S containing i.
Let R be a ring with identity. In this section we find
conditions necessary to insure that the twisted semigroup
ring Rt[S] does not have nontrivial units.
Notice that in this section we have the twist function
normalized but not necessarily nonconstant.
Lemma 2.6. If R contains a nonzero nilpotent element,
then Rt[S] has nontrivial units.
Proof. Let a be a nonzero nilpotent element of R, and
let n be the smallest positive integer such that a11 = 0. Set
b
Then b is a nonzero element of R such that b^ = 0. Let
s e S with s i. Then (i + bs) (i bs) = i.
68
semiprime, where K is the kernel ideal of the canonical
epimorphism from RLS] into R[U].
Now we let R be a ring. In Theorem 4.5 we prove that
if S is a semigroup satisfying a suitable condition, then
the twisted semigroup ring Rt[S] is semiprime if and only if
R is semiprime.
In section two we consider two types of semilattices.
The first is a semilattice PQ with a zero element and in
which all the nonzero idempotents are primitive. The second
type of semilattice has also zero element and satisfies
the following condition: if F is a subset of P^, then either
g*l*b{xx e F} belongs to F or else there exist x,y in F such
that g*l*b{x,y} is in P^ F.
Next, we consider two families of semigroups which admit
relative inverses. We say that a semigroup S belongs to the
family S if (i) S admits relative inveres, (ii) the idempotents
of S commute, and (iii) the semilattice of subgroups Sa of S
is isomorphic to Pq. Then in Theorem 4.11 we decompose R[S]
as an internal weak direct sum of ideals which are completely
determined once R[Sa] is known for each a e Pq. This allows
us to characterize the nilpotent and Jacobson radicals of
R[SJ. Further, if we let F be a field and use results of
group ring theory, we can give necessary conditions for F[S]
to be semisimple and regular in Corollaries 4.15, 4.16 and
4.17.
Passman has related semiprime group rings with their
center in the following two results [22, Theorems 2.12 and
2.13J.
16
hence
*3 1 ; i+v 9 1 i k *3 ^ i+ k i i a2
[ n y (g ,g3 K)][ IT y(g^,g ) ] = t IT y(g ]>g )]y(g ,gD)q
k=2 k=2 k=2
It follows that
IT .
1
y (g1fg^)y(g1,g^+1)
y(gl,g) y(g1+;]rg)
i+j i j\q2
ir y(g g )
Then
it tt .
i 1
TT .
1+J
y(gx,g^)q Xy(g1/g^+1)y(g^>g)
y (g1+^g)
By letting k = 1 in (5), we get y (g1,g^+q) y (g1,g) =
r. tt = tt
i j x+D
y (g1+:i ,g) Y (g1 ,g^ ) Therefore, tt tt = tt . Y^Sg'1)^
where 1 < i,j < q1.
i v
Lerana 1.18. Let y(g ,g ) be given for k = 1,2,...,q1,
If 1 < x < q1, x i, then
, i qi.q i q2i,2 i qni.q, n
= y(g g ) y(g ,g ) y(g ,g ) vtt. ,
where n is the smallest positive integer such that x = ni
(mod q).
Proof. First, we claim that if 1 < x < q1, then
x E qni (mod q), where 1 < n < q1. To prove this is
enough to show that any two elements of the set {qni 0 < n <
q1) represent different residual classes of Z For con
tradiction, suppose that qn^i= qn2i (mod q), where
0 < n^ < n2 < q1. Then n^ E n2 (mod q), which contradicts
0 < n2~n^ < q.
65
such that (a) left and right mappings commute with each other,
(b) the element z e J maps every element of J(u) into u, and
(c) all the mappings leave fixed the element u e J(u). The
multiplication in the class sum S of all J(u) is defined as
follows. Within J(z) it is defined as originally given in J.
For k e J and a e J(u) we define ak(ka) to be the image of a
under the mapping corresponding to k in the right (left)
representation already determined. Finally for xu e J(u) and
x e J(v) where u z and v z, we define x x = uv, where
V U V
uv is the product of u and v as originally given in U. Then
each S = u J(u) is a semigroup with frame S = U u J.
ueU
Lemma 3.28. The sets K(u) are nilpotent ideals of index
two in R[S] for all u z.
Proof. Let a = Ea.x. e K(u), let p = Ec.x, e R[J[(v)]
i 1 1 k K k
where u z and v z, and let 6 = Ed y e R[J(z)]. Since
t t t
Ea. = 0, then ap = (Ea. x. ) (Ec, a, ) = E Eac, (xa.) = E(Ea.)c,(uv)
i 1 i 1 1 k k k k i 1 k 1 k kilk
= 0. Hence K(u)R[J(v)] = 0 for all u,v e U with u z and
2
v z. In particular, Ku c KuR[J(u)] = 0. On the other hand,
a = (Ea.x.)(Ed y ) = E Ea.d (x.y ). Since x.y e J(u)J(z)cj(u)
i 1 x t ^ ti 1 t
and since E Ea.d = (Ea.)(Ed.) = 0, then a6 e K(u). Therefore
t i i1 t
K(u) R[J(z)] c K(u). Moreover, from S = uJ(u) we get
R[S] = EUeU R[J(u)]. Hence
K (u) R [S ] = K (u) [ E R[ J (v) ] + R[J(z)]] = K(u) R[ J (z) ] c K(u) .
ve U{z}
Symmetrically R[S]K(u) c K(u), and the result follows.
45
Proof. Let I be an ideal of FlSj. If ai + bx e I,
where a 0 and a + b 0, then (ai + bx)x = (a + b)x e I
implies that x e I. Since ai = (ai + bx) bx e I, it follows
that i e I. Therefore I = F[Sj. Similarly, if ai + cy e I,
where a 0 and a + c 0, or if ai + bx + cy e I, where
a 0 and a + b + c 0, then i e I. Hence I = F[S],
So far we have found the ideals (x y ), (x,y> and L.
We prove next that, in fact,these are the only proper ideals
of FLSJ.
Claim 4. Any element of FLSJ, which contains i in its
support and which has the sum of its coefficients equal to
zero, is a generator of L.
Proof. Let ai + bx e L, where a 0 and a + b = 0. Then
ai + bx = a(i x) e (i x). Similarly, if ai + cy e L
with a 0 and a + c = 0, then ai + cy = a (i y) e (i y) .
Now let a = ai + bx + cy e L, where a 0, b 0, c 0 and
a + b + c = 0. Since (i x)(i y) = i x and
(i y) (i x) = i y, then (i x) = (i y). Moreover,
from a(i x) = a and (i x) a = a(i x), it follows
(i x) = (a). Hence, each of the elementsof L containing i in
its support generatesthe ideal (i x). By definition, any
other nonzero element of L is of the form bx + cy with b 0,
b + c = 0. Then bx + cy = b(x y). But x y = (i y) 
 (i x) ; hence (x y) c (i x). Therefore L = (i x^ .
18
Lemma 1.21. If G is a cyclic group generated by g,
then y(gn,gm) = y(gm,gn) for all n,m.
n1 m1
Proof. Since gn = IT y(g,g1)gn and gm = II y(g,g^)gm,
i=l j=l
then
n1 m1
gn.gm = ( n y(g,g1)gn) ( H y(g,g3)gm) =
i=l j=l
n1 m1
= ( H Yig/g1))( n Y{q,g3))y(gn,gm)gn m
i=l j=l
and
_ m1 ' n1
g *g = ( n y (g^1) gm) ( n Yigrg'^g11) =
j=i i=i
m1 n1 ^7
= ( n y(g,gJ)( II y (g g1) ) y (gm, gn) gm n .
j=l i=l
Then gngm = gn+m = gmgn implies that y(gn,gm) = y(gm,gn)
for all m,n.
Lemma 1.22. If 1 < x < q1, then
y(gx,gx) = y(g,gX)y(g,gX+1)...y(g,g2x_1)/y(g,g)y(g,g2)...
/ x1,
Y(g,g )
Proof. Fix i = 1 and k = x in equation (5), and first
let j = x. Then y(g,g2x)y(gx,gx) = y(gX+1,gX); hence
Y(gx,gx) = y(gfgx)y(gx+1,gx)/y(g,g2). if j = x + l, then
equation (5) yields y(gx+1,gx) = Y(g,gx+1)y(gx+2,gx)/y(g,g2x+1).
Thus y(gx,gx) = Y (g,gx)y (g,gx+1)y(gx+2,gx)/y(g,g2x)y(g,g2x+1).
For induction, assume that
33
Corollary 2.7. Let R = L^ .... be a finite
direct sum of left ideals. If Rt[Sj has no nontrivial units,
then is a twosided ideal for i = l,2,...,n.
Proof. Let 1 = e^ + ... + en, where e^ e L^. Then the
e^ are orthogonal idempotent elements of R such that L^ = Re^,
Thus
if r e R, (e^rej)^ = 0 for all i j. By hypothesis
Rt[S] has no nontrivial units; thus Lemma 2.6 implies that
e. re = 0 for i j. Hence LL' = R(ejRe) = 0 for i j.
ij i j i j J
Therefore L^ is a twosided ideal of R for all i.
Lemma 2.8. Let Rt[S] be a twisted semigroup ring. Let
n
R = R. be a direct sum of nonzero ideals R> i = l,...,n.
i=l 1
Then there exist y^:S x S + R., i = l,...,n, such that
n
y(s1,s2) = i^1Yi(s1,s2) for all (s1,s2) e S x s, and
n t.
Rt[SJ= R.11S].
i1 1
Proof. Recall, first, that y is defined on the central
units of R. Clearly u = u^ + u2 + ... + ur is a central unit
of R if and only if u^ is a central unit of R^, i = l,2,...,n.
Let 0^ stand for the canonical epimorphism from R onto R^.
Define y^ : S x S + R. by y(s1,s2) = 0^y (sj_, s ) Since y
and 0^ are well defined functions that preserve the associa
tivity, so does y^. Moreover, our first observation shows
that y^ is defined onto the central units of R.. Finally,
n n
Â£ Yi (si,s2) = .^ 6Y(s1/s2) = Y(s^,s2) for all (s^,s2) e
n fci
It follows that RC[S] = R. [S].
i=l 1
S x S.
80
a = Za. (x.u) + Zb.y. + (Za.)u. After remaining conveniently
u i11 i
we can write au = Za^ (x^u) + Zc^z^., where each e SJu)
i k
for some 6 a, 8 e P^. Then z^(xv) = z^xz^v =
e6w(zk)0aw(x)_06w(zk)v = uv_uv = ' and (xi~u) (X'V) =
xxx.vux+uv = x.x0 (x.)vU0 (x) + uv = x.x2uv+uv =
ii i aw i aw i
x.xuv e C (uv) since x.x e S (u)S (v) e S (uv) by lemma
i a i a a a
4.8 (iv) Hence a (xv) = Za. (x.u) (xv) + Zc z, (xv) =
u i 1 1 k K K
Za.(x.xuv) e C (uv). Therefore K(u)C (v) c c (uv). It
^ 1 1 ot ct ct
follows now that KC = (ZK(u))(ZC (v)) = ZZK C (v) c ZC (uv)=
a a u a a
u v vu v
C Symmetrically C K c c Thus C is an ideal of K. Then
a a a a
by Proposition 4.9 Ca is an ideal of R[S].
Theorem 4.11. The ideal K = Zw ^ C is the internal
aePQ a
weak direct sum of the ideals C Hence R[S] = R[S ] Z ^ C
a w aePQ a
Proof. Let f e K. From K = Z K(u) and
ue S
w
( Z K(u)) n K(v) =0, it follows that f can be written
u*v
uniquely as f =
Z e f
ueS u
with
f
u
nl
w
\
f = Z a x
+
+ Z
a .
U j1=l 31
1
=>k=
=1 >
c e S (u)
a i a .
iJi i
n
Since f e K(u), we may write f = Z a. (x u) +
U U j1=l =>1 al3l
nk
Z a. (x u). Suppose that f can be expressed in a
jk ^k ajk u
mi
m.
t =1 1 11
different way; say f = Z h (y u) + +
u t, J a11,
\
Z
V
E bt (ya t ~u) with ya t 6 Sa
t, =1 k akCk aiti ai
(u). Then
71
Now assume that S satisfies condition (E ), and let a
P
be a pextremity of a finite subset X of S. Let a11 = s^s2...sn,
j ^
s^ e X, and choose m e Z such that p > n. Then
m m_ m_n
ap = ana^ n = s^s2...sn aP n. Hence condition (E ) implies
that a = s^ = s2 = ... = sn; that is, a is an extremity.
shows that (E) and (E ) are equivalent conditions in S.
P
We begin by emphasizing a direct consequence of
Definition 4.1.
This
Lemma 4.3. Let S be a semigroup satisfying condition
(E), and let X be a nonvoid finite subset of S. If a is an
extremity of X and n is a positive integer, then a11 is an
extremity of Xn = {x^x2...x^x^ e X}.
Proof. Let (an)m = Y]_y2***ym w^ere m e Z+ and each
y. = x..x.0...x. is an element of Xn for 1 < i < m. Then
Ji ll i2 in
(a ; = a = (x.. x, 0.. .x, ) . (xm1 x 0...x, ). But a is an
11 iz in ml mz mn
extremity of X; hence a = x^^ for all i and j. Therefore
. = y
n
a = Yf =
m
Proposition 4.4. Let R be a ring, and let S be a
semigroup. If the twisted semigroup ring Rt[S] is semiprime,
then R is semiprime.
Proof. Assume that Rt[S] is semiprime. By way of con
tradiction, suppose that there exists a nonzero nilpotent
ideal A of R. By section 1.5 A^TS] is a nonzero ideal of Rt[S].
k
Let m be the index of nilpotency of A, and let a = Z a.s.
i=l
5
Theorem 1.4 L 13J Let A]_, A2 . ,An be ideals in a ring
R such that (i) A^ + A2 + . + An = R and (ii) for each
k (1 < k < n), Ak n (A^ + ... + Ak1 + Ak+1 + ... + An) = 0.
Then there is an isomorphism R ~ A^ A2 ... An.
Lemma 1.5. Let R be a ring with nonzero characteristic
n, and let (p^,P2//Pk} be the set of distinct prime divisors
of n. Then R = R_ R^ ... R
P1 P2 pn
A nonzero element a in a ring R is said to be a left
(right) zero divisor if there exists a nonzero element
b e R such that ab = 0 (ba = 0). A proper zero divisor, or
simply zero divisor, is an element of R which is both a left
and a right zero divisor. A commutative ring with identity
1*0 and no zero divisors is called an integral domain.
Let a be an element of R. The principal ideal generated
n
by a is the set (a) = {ra + as + na + Z r^as^r,s,r^, s^ e R
i=l
and n e Z}. An ideal P in R is said to be prime if P R and,
for any ideals A,B in R, AB c p implies that A c p or B c p.
An ideal M in R is said to be maximal if M R and, for every
ideal A such that M c a c r, either M = A or A = R.
Definition 1.6 [9 ]. Let 1 be a certain property that
a ring may possess. We shall say that the ring R is an Lring
if it possesses the property 1. An ideal I of R will be
called an Lideal if I is an 1ring. A ring which does not
contain any nonzero 1ideals will be called 1semisimple. We
shall call 1 a radical property if the following three conditions
hold.
35
Theorem 2.9. If RtLSJ has no nontrivial units, then
R has no nilpotent elements, and either R is indecomposable
or G is trivial.
Proof. This follows directly from Lemma 2.6 and
Proposition 2.8.
2.3 Nontrivial Units In Twisted Semigroup Rings Over
Nontorsion Free Semigroups
Gilmer and Teply [11] have proven that if S is a torsion
semigroup and RLS] has no nontrivial units, then the charac
teristic of R is 2,3 or 0; moreover, if R is an algebra over
a field, if S is torsion and if R[S] has no nontrivial units
then the characteristic of R is either 2 or 3. Finally,
they characterize the semigroup rings of characteristic 2
or 3 that have no nontrivial units.
Let S be a nontorsion free semigroup with identity i,
and let G be the largest subgroup of S containing i. More
over, let R be a ring with identity, and let K be a field.
Denote by P the prime subring of R, and denote the charac
teristic of R by char R. If s e S, then (s) stands for the
subsemigroup of S generated by s. In this section we extend
the previous result to the twisted semigroup ring Rt[S]
of finite characteristic and to K^'CS], showing that they
have always nontrivial units.
We begin by proving some preliminary results, the first
of which is known.
Lemma 2.10. S G is an ideal of S.
61
minimal ideal A if and only if it contains a zeroid element,
and then A consists of all the zeroids of S.
Clifford and Miller [4 ] have shown that the set U of
zeroid elements of S is either vacuous or a subgroup and two
sided ideal of S. Moreover, the identity element z of U
commutes with every element of S, and the mapping
0: S > U: s * zs( = sz) is an epimorphism. They define the
core of S to be the subsemigroup J = {x e Sxz = z}; thus J
contains z as zero element. Finally the subsemigroup U u J
of S will be called the frame of S.
An easy example of a semigroup with universally minimal
ideal is provided by the direct product of any group and any
semigroup with zero.
Let R be a ring with identity, and let S be a semigroup
without zero such that S contains a universally minimal ideal
U. In this section we first show that R[S] is a ring direct
sum of two known ideals. Using this decomposition, we find
the general form of the nilpotent radical of R[S]. Finally
we apply these results to semigroup rings over a subfamily
of semigroups with universally minimal ideals, where the
multiplication in the semigroups is known.
We begin by stating as a remark some facts taken from
the main theorem in [4 ].
Remark 3.20. The following conditions hold in S:
(i) 6:S * U: s * sz, is a semigroup epimorphism, where
z = ly. Moreover, zs = sz for all s e S.
52
Lemma 3.11. If S is a right group, and E = {e^}^ is
the set of its idempotent elements, then the following
statements hold.
(i)E is a right zero semigroup.
(ii) = Se^ is a subgroup of S for all e^ e E, with
identity 1H^ = e^.
(iii)If e.,e. e E with i j, then H. n H. = d> and
l j J 1J
0. : H.  H. defined by 0..(x,.) = x. e = x, is
i "J u 13 A1 Alj AJ
a group isomorphism.
(iv)S = uieI Ht.
Let R be a commutative ring with identity, and let S be
a right group. As a direct consequence of Lemma 3.11 we get
the following result.
Lemma 3.12. The following statements hold in R[S].
(i) ( Z R[H.]) n R[H.] = 0 for all j e I.
i*j 1 3
(ii)There exist a group ring isomorphism 0^:R[H^] R[Hj ]
such that 0l j (R[ Hl H ) = RCl^le^ = R[ H ^ ] for all
i,j e I.
w
(iii)R[S] = E^eIR[H^], a internal weak direct sum of
Rmodules.
(iv)e^ is a left identity of R[S] for all i e I.
(v)RCSje^ = R[Hj]; that is, each R[Hj] is a left
ideal of R[S].
Proof. (i)
i,j e I, i j .
Follows clearly from H. n Hj = (j) for all
(ii) Let 0\ j be the natural extension of
77
The semilattice P, the groups Sa, and the homomorphisms
4,ap together constitute a complete set of invariants of S.
The only result that we know on semigroup algebras over
this type of semigroup is due to Munn [18]. He proved
that under suitable conditions on the characteristic of the
field and the order of the subgroups, the semigroup algebra
of a finite semigroup which admits relative inverses is semi
simple if and only if the idempotents commute.
Let Pq be a semilattice with zero elements w such that
every element of Pq {w} is primitive; that is, for all
a,8 e Pq {w}, a8 = w. Denote by S the family of all semi
groups S with the following properties: (i) S admits relative
inverses; (ii) the idempotents in S commute; and (iii) S has
Pq as the semilattice of subgroups.
Lets e S, and let {Sa)a(.p and (0aw: Sa + Sw  a e PQ}
stand for the sets of semigroups and homomorphisms of S
respectively. Finally, let R be a ring not necessarily with
1. Our first goal in this section is to characterize R[S]
as a weak direct sum of ideals which are completely determined
once the R[Sa] are known.
We begin by relating the semigroup S with the semigroups
having a universally minimal ideal.
Lemma 4.8. Let S e S. The following conditions hold
in S.
(i) S is a semigroup with universally minimal ideal Sw.
(ii) The frame of S is STI u J, where J = u Ker 0
w aePn aw
9
nonzero elements of R and the s. are distinct elements of S.
n
The unique representation Â£ aisi f a *s called the canoni
cal form of a, and the set {s^}*?_^ of "variables" of the
canonical form of a is called the support of a (denoted by
supp a); moreover, the ideal of R generated by {a^}?_^ is
called the content of a.
If the semigroup S has an identity i, then the ring R
is naturally embedded in the semigroup ring R[S] by mapping
an element re R to ri.
1.4 Twisted Group Rings
Let K be a field, and let G be a multiplicative group
whose identity we denote by i. Passman [22] defines a
twisted group ring K^G] of G over K as an associative
Kalgebra with basis xx e G) and with multiplication defined
distributively such that, for all x,y e G,
(1) x y = y(x,y)xy, y(x,y) e K {0}.
The associativity condition is clearly equivalent to
(xy)z = x(yz) for x,y,z e G, and this is in turn equivalent to
(2) Y(x,y)y(xy,z) = y(y,z)y(x,yz)
Since
(xy) z = y(x,y)xyz= y (x,y) y (xy, z) xyz
x(yz) = y(y z)x yz = y(y,z)y(x,yz)xyz .
and
13
images of 1, x, x2, . x11 ^; so it is a twisted group ring of
C the cyclic group of order n. Thus,for instance, if
K = Q is the field of rationals, then by the Eisenstein's
criteria the polynomial xn 2 is always irreducible. Hence
for a = 2 we see thatQ^TC ] ~ Q[xT/(xn2) is, in fact, a
field extension of Q, namely Q[n/2], Moreover, if K is a
field of characteristic p > 0 that is not perfect, choose
n
a e K with j1 /a K. Then for all n, x^ a is irreducible
n
over K and the field K[x]/(x^ a) = K[n/a] is a twisted
group ring of the cyclic group C A last application of
P
the foregoing example is obtained when we let K be an
algebraically closed field and G = C Then xn = ai and
n/a exists in K. Since (a ^nx)n = i it follows now that
K*"[G] is diagonally equivalent to K[G].
In the case where G = (g) is infinite cyclic, then we
can make a diagonal change of basis by replacing gm by gra
for all integers m. This clearly implies that Kt[G] is
diagonally equivalent to K[G].
Finally, if we let G = {i,g^,g2,g^} be the Klein group,
and let K be a subfield of the real numbers, then by defining
y(g1.g2)= y(g2,g3) = y(g3rg1) = 1
and
y(g2igx) = y(g3,g2) = y(g1g3) = Ytg^gL) = y(g2/g2) =
= Y(g3iq3) = 1,
we obtain a twisted group ring that is easily recognized as
the quaternion division algebra.
CHAPTER III
NILPOTENT RADICALS OF NONCOMMUTATIVE SEMIGROUP RINGS
Passman [22, theorem 1.9] has determined the nilpotent
radical of a group ring over a field of finite characteristic
in the following result.
Let K be a field of characteristic p > 0 and let G be a
group. Then (i) N[K[G]J = J[K[A^(G)]]K[G], where AP(G) =
(x e G the index of the centralizer of x in G is finite and
the order of x is a power of p); (ii) J[K[AP(G)]] = N[K[AP(G)]];
and (iii) N[K[G]j = 0 if and only if Ap (G) = (1).
Gilmer and Parker [10] have completely characterized
the nilpotent radical of any commutative semigroup ring in
the result that follows.
Let R be a commutative ring and let ke t*ie
family of prime ideals of R. Let S be an abelian semigroup.
Then the nilpotent radical of R[S] is n{P^[S] + Ip },
A
where p^ is the characteristic of R/P^ and where Ip is the
' A
ideal of R[S] generated by the set (rxa = rx^ }; if
P
p, = 0, then I is the ideal of R[S] generated by the set
PA
(rxa rx*31 r e R and a is almost equal to b}.
The natural question that arises immediately is whether
this result is valid or not in the noncommutative semigroup
rings. In the first section of this chapter we analyze an
example of a noncommutative semigroup ring over a field,
showing that the former result is no longer true.
41
3
Theorem 1.1. Let s e S. If (s) is infinite all the
powers of s are distinct. If (s) is finite, there exist
two positive integers, the index r and the period m of s,
such that sm+r = sr and (s) = {s,s2,...,sm+r ^}. The order
of (s) is m+r1. The set Kg = {sr,sr+^,...,sm+rM is a
cyclic subgroup of S of order m.
An element s e S has finite order if (s) is finite.
S is called torsion if every element of S has finite order.
If s is an element of finite order in S, then (s) contains
exactly one idempotent, namely the identity of Kg.
Let the semigroup S have an identity i. If x and y are
elements of S such that xy = i, then x is called a left inverse
of y, and y is called a right inverse of x. A right (left)
unit in S is defined to be an element of S having a right
(left) inverse in S. By a unit in S we mean an element of
S having both right and left inverse in S.
Theorem 1.2. Let S be a semigroup with identity i.
(1) The set P(Q) of all right (left) units of S is a
right (left) cancellative subsemigroup of S containing i.
(2) The set G of all units of S is a subgroup of S, and
G = P n Q. Each unit has a unique twosided inverse in G, and
has no other left or right inverse in S.
(3) Every subgroup of S containing i is contained in G.
If e is an idempotent element of S, then eSe is the set
of all elements of S for which e is a twosided identity.
73
whenever i^ j for some k = 1,2,It follows that
0 = (bs.)*1 = b2y(s.s.)b.
3 3 3 3 3 3
, h1 h h h ,
y(s. ,s.)s. = b.cs., where
3 3 3 3 3
c = y(s .,s )y(s2,s .) ... y(s^ 1,s .) Hence bhc = 0. Since
3 3 3 3 3 3 3
the images of y belong to the units of R, then c is a unit.
Therefore b? = 0; that is, b^ is a nilpotent element of R.
2 12
But B is an ideal; so the set D = {(xY(s.,Sj) s^)B +
+ 6(yY(s^,s2)_1Sj) + (y(sj,sj) 1s2)nB +
m 1 O I
+ l (x Y (s ., s .) S .) B (y Y (sf s ) s .)  x, y, x ,y e R, n e Z,
J LJJ J L1
m e Z+} is contained in B. Since s^ is an extremity of
3
supp Br it follows from Lemma 4.3 that s^ is an extremity
of the support of any element of D. Hence all the elements
of the principal ideal
m +
(b.) = {xb.+b.y+nb. + X x b y  x, y, x y eR, neZ, meZ}
J J J J i=l 1 J 1 1
3 3
of R are coefficients of s^ in elements of B having s^ as
an extremity of their support. By the argument in the
previous paragraph, the hth power of each of these elements
is zero. Therefore (b^) is a nonzero nilpotent ideal of R,
which contradicts our hypothesis.
The preceding result allows us to determine the nilpo
tent radical of R^[S].
Theorem 4.6. Let R be a ring, and let S be a semigroup
satisfying condition (E). Then the nilpotent radical of
the twisted semigroup ring Rt[S] is N[Rt[S]] = Nt[S], where
N is the nilpotent radical of R.
Finally, my thanks to Sharon for her fine and fast
job of typing.
This work was performed during a sabbatical granted to
me by the Catholic University of Puerto Rico.
in
29
(l+rq) 1(i+rg) (i+rg + \ rkb,,gk) =
k=2 k l
 ^3 If ^ ^ f f
(l+rq) x(i+rg + E r b^_^g + r
k=2 k=2
(l+rq)1(l+rq)i = i.
Suppose next that rq = 1 for all re K {0}. Then K
is a finite field. Since char K = 2, then K = GF(2n), where
2n 1 = q. By Corollary 1.19 tt^ = 1 for all i, 1 < i < q1,
so we cannot use the construction of the case where tt^ 1.
To prove that in this case Kt[G] has nontrivial units, it is
sufficient to show that, for all v e K {0}, there exists
t 2
a e K [G] such that a = va. For then we can choose
x,y e K {0} with x + y 0, and set v = (x+y)/xy. Then
2
xya = (x+y)a; hence
(i xa)(i ya) = i (x+y)a + xya = i.
ql .
Let a = E a.g1 e K [G]. By Lemma 1.21, y(gU,gW) =
w u i=1 1
y(g ,g ) for all u,w e Z. Thus since by hypothesis
char K = 2, then
q 1 *o m 0 m
a = ( Z aig1) = Z a.y (g1, g1) g21 + Z a2 y (giUT J gm'r J ) g
i=l 1 i=l 1 j=l m+J
m+j m+j> 2 j1
where m = q1/2. For a = va we must have
/va2., if 1 < i < SÂ£i
(l) a2 y (g1,g1) =
CT 1
va2jl' 2~ + 1 < i < ql, where
ql
J = 1 2 *
54
chains of ideals used in the definition of the nilpotent
radical of RlILl and Rl S ] respectively. We use transfinite
induction. Let be a nilpotent ideal of REH^]. By
Lemma 3.12 (ii) 0^^ is a ring isomorphism for all i,j e I;
hence 0. (B. ) = B.e. = B. is a nilpotent ideal of R[H.].
li 1 1 i i ^ i
i e I, with the same index of nilpotency as B^, say n. By
Lemma 3.13 B = Z. T B. is a twosided ideal of R[S]. We
le I i
2
claim that B is nilpotent. Since B.B. = B.(eB.) = B., it
i j i j j j
follows by induction that B. B. ...B. = Bm for all
2 i. i i i
12 m m
positive integer m. Thus B? = 0 for all i e I implies
that Bn = B^)n = = 0* Therefore B c Ng[R[S]]
and consequently Z^Â£l NgEREH.^]] Â£ NQ[R[S]]. This shows the
first step of the induction argument.
Assume that N^EREIL]] c ITERES]] for all the
ordinals k < j. Suppose j is not a limit ordinal and let
A^/N j_^E RE ] ] be nilpotent in RE H1 ]/N_. _^E RE ] ] As in the
previous paragraph A^/N^_^E REH^] ] = O^E Aj/N^RE H1 ] ] is a
nilpotent ideal of RE 1L ]/N^_^ERE EL] ] i e I, with the same
index of nilpotency, say n. As a clear consequence of
Lemma 3.13 A/ZieI Nj_1EREHi]] = ZieI Ai/N RE Hi ] ] is an
ideal of RES]/E^eI Nj_^EREH^]] which we claim is nilpotent.
Since
fli/Nj_i[R[Hi]] At/N._1[R[Ht:] = .Â¡/N j_j_C R[ ] ] e^/N WH,. ] ]
by induction we obtain
53
n
that
n
A
is, define 0..: RLHJ y RT.I1. J by 0 ( E a, .x, .)
1J 1 J 1 J A1
n
E a, 0 .(x, ) = E a, (x, e .) Since 0 . isa semigroup
Ai 13 Al Al Ai j 13 ? r
isomorphism between the basis, then 0^^ is a group ring
isomorphism. Hence R[Hj] = 0^j(R[H^]) = RCH^De^. (iii) Fbllows
directly from (i) and Lemma 3.11 (iv). (iv) Is clear since
e^ is a left identity of S for all i e I. (v) By (iii) and
(ii) we get RCSle^ = EieJ RCH^e^ = EifiI R[Hj] = R[H^].
From now on we fix an element of the index set I, say
i e I.
Lemma 3.13. Let be a twosided ideal of R[H^]. Then
B = B^e^ is a left ideal of R[S] for every i e I. Moreover,
B = E^^B^ is a twosided ideal of R[S].
Proof. By Lemma 3.12 (ii) 0^ is a group ring isomorphism
for all i,j e I. Hence 0^(B^) = B^e^ = B. is a twosided
ideal of R[H^] for all i e I. Since R[H^]B^ = R[Hj](e^B^)
= R[H.]B. c b.; then R[S]B. = E. TR[H.]B. c B.. Thus,
ii_i l j e I j i l
B^ is a left ideal of R[S] for all i e I. Therefore,
B = B^ is a left ideal of R[S]. Moreover, B^RTH^]
= Bj(e^R[H^]) = B^R[H^] c for all i,j e I. Hence
B.R[S] = B.E. T RC H. ] = E. T B.R[H. ] c l. _B. = B. It follows
3 j lei l lei 3 i lei i
that BR[S] = E. TB.R[S] c E. TB = B; thus B is a right ideal
of R[S]. Therefore B is a twosided ideal of R[S].
Proposition 3.14. E j N[R[H^1 ] c N[R[S]].
Proof. Let NQ[R[Hi]] c N1[R[Hi]] c c Nj[R[H^]] c
i e I, and Ng[R[S]] c N^[R[S]] c c N^[R[S]] c be the
48
Lemma 3.4. Let R be an integral domain. Let I be a
nilpotent ideal of R[S], and let a = Ea^x^ e I. Then
Ea. =0.
i
Proof. If x e S, then ax = (Ea^)x e I. Let n be the
index of nilpotency of I. Then 0 = C(Ea^)x]n = (Ea^)nx; hence
(Ea^)n = 0. Since R is a integral domain, then Ea^ = 0.
Theorem 3.5. Let R be an integral domain. Then the
nilpotent radical of R[SJ is N[R[SJ] = {a = Ea^x^Ea^ = 0}.
Moreover, N[R[SJ] is a nilpotent ideal of index two.
Proof. Let L = {Ea.x.IEa. = 0}. By Lemma 3.3, if
i i 1 i
a e L, then (a) = Ra is a nilpotent ideal of R[S] of index
two. Hence L is a nilpotent ideal of index two.
On the other hand, if I is a nilpotent ideal of R[S],
then by Lemma 3.4 I c L; so L is the unique largest nilpotent
ideal of R[S]. Thus N[R[S]] = L.
Using this result, we characterize next the nilpotent
radical when R is a commutative ring.
Theorem 3.6. Let R be a commutative ring with identity,
and let S be a right zero semigroup. Then the nilpotent
radical of R[SJ is NLR[S]] = {Ea^x^jEa^ e N}, where N is the
nilpotent radical of R.
Proof. Let a = Ea.x. e N[R[S]]. Let P(R) denote the
i i
prime .radical of R. Since R is commutative, then N = P(R).
If P is a prime ideal of R, then R/P is an integral domain.
42
Then, what is the nilpotent radical of a noncommutative
semigroup ring? The rest of this chapter deals with this
question, whose answer cannot be unique, since it will depend
upon the particular semigroups and rings in consideration.
We consider some known families of noncommutative semigroups,
and we determine the nilpotent radical of semigroup rings
constructed with them.
Thus, in section 3.2 we study the nilpotent radical of a
semigroup ring R[S], where R is a commutative ring with iden
tity and S is a right (left) zero semigroup. Theorem 3.6
shows that in this case N[R[S]] = {Ea^x^eR[S] Â£a^ e N}
where N is the nilpotent radical of R.
In section 3.3 we let S be a right (left) group and R a
ring with identity. Theorem 3.19 handles this case by
showing that N[R[S]] = N[R[H^]]R[S] + C, where is amaximal
n n
subgroup of S and C = R[H,]W with W = { E a.e. E a. = 0,
i=l i=l
e^ idempotent of S and n e Z }. In section 3.4 we consider the
family 5 of semigroups having a universally minimal ideal,
and we let R be a ring with identity. Proposition 3.26 states
that N[R[S]j = N[R[Uj] N[K], where U is the group of zeroids
of S and K the kernel of the canonical epimorphism 0: R[ S ] * R[U].
Then we apply this result to one subfamily of 5, where the
product outside of U is known.
3.1 An Example
In this section we consider an example of a semigroup
ring F[S], where F is a field and S is a torsion free finite
60
a, Bi = B t Rl H, I n C = 0, then a, = Bn and
1 1 c c 1 11
ac = 8C, which proves that the representation is unique.
We remark that Proposition 3.18 can be restated by
saying that R[S] = REH^] C is a direct sum of left ideals.
Theorem 3.19. Let R be a ring with identity, and let
S be a right group. Then the nilpotent radical of RES] is
NERES]] = EieI NEREIS]] + C = NEREH^lRES] + C.
Proof. By Propositions 3.14 and 3.17 it follows that
ZieI NEREEL]] + C c NERES]].
Hence NERES]/C] = NERES]]/C ~ NE RE H^] ] by Lemma 3.18. Thus
NE RE s ]] = NE RE H1 ] ] + C = NEREHj^] ]RES] + C.
Finally, by Proposition 3.12 (iii) NEREH^]]RES] =
NEREH1]](ZieIREHi]) = fieINEREH1]](eiREHi]) = ^NE RE Hi]].
3.4 Semigroup Rings Over Semigroups With Universally Minimal
Ideals, and Their Nilradicals
An element u of a semigroup S is called a zeroid element
of S if, for each element s e S, there exist x,y e S such
that sx = ys = u. It follows that if S has a zero element,
this will be the only zeroid of S. Let A be an ideal of S.
Then A is called locally minimal if it contains no proper
subideal. A is called universally minimal in S if it is
contained in every ideal of S. It is an easy consequence of
these definitions that a semigroup S contains a universally
14
Just to mention how the twisted group rings might arise
we give the next two lemmas, whose proofs can be found in [22].
Lemma 1.13. Let Z be a central subgroup of G, and let
L be a maximal ideal in K[Z]. If I = L*K[G], then
K[G]/I = Ft[G/Z] is a twisted group ring of G/Z over the
field F = K[Z]/I.
It turns out that every twisted group ring arises in
this manner as we see next.
Lemma 1.14. Let Kt[H] be a twisted group ring of H
over K. Then there exists a group G with a central subgroup
Z such that G/Z ^ H and K[G]/I s Kt[H], where I = L*K[G]
and L is an ideal of K[Z] with K[Z]/L ~ K.
So far, we have seen one example where, if K is an
algebraically closed field, then Kt[G] is diagonally equiva
lent to K[G], namely, when G is a cyclic group. We state
two more results that give conditions in order for this to
happen.
Lemma 1.15. Let Kt[G] be given with K algebraically
closed. Set U = {axa e K {0}, x e G}, and let
W = {aia K {0}}. Then Kt[G] is diagonally equivalent
to K[G] if and only if U', the commutator subgroup of U,
is disjoint from W.
As an immediate consequence, we obtain the following
result.
23
Y2 such that O(rs) = 0(r)s for each r t and s e S. More
over if A is the kernel of 0, then the kernel of 0 is A^[S].
It follows that if A is an ideal of the ring R, then
Rt[S]/At[S] Â£ (R/A)t[S]. Moreover, 0 is onetoone if and
only if 0 is onetoone.
Finally, in the remainder of this dissertation, we
denote the nilpotent, prime and Jacobson radicals of R by N,
P and J respectively; we maintain our previous notation for
these radicals whenever we deal with twisted or ordinary
semigroup rings.
Abstract of Dissertation Presented to the Graduate
Council of the University of Florida in Partial Fulfillment
of the Requirements for the Degree of Doctor of Philosophy
PROPERTIES OF TWISTED SEMIGROUP RINGS
By
Antonio Quesada Rettschlag
August 1978
Chairman: Mark L. Teply
Major Department: Mathematics
Let R be a ring with identity, and let S be a multipli
cative semigroup with identity i. Let y be a function from
S x S into the group of central units of R such that
Y (s^,S2)y (sqs2 s 3 ^ Y(s2^s^)y(s^,S2S^) for all Â£ S.
We define the twisted semigroup ring, denoted by R^[S], as
an Ralgebra with basis {ss e S} and with multiplication
defined distributively such that for all s^,S2 e S,
s^S2 y (sq s2 ) s]_s2 *
Assuming that y is nonconstant, we prove the following
statements.
(A) If K is a field and G is a nontorsion free group,
then Kt[G] has nontrivial units.
(B) If S is a nontorsion free semigroup and char R is
finite, then Rt[Sj has nontrivial units.
(C) If S is a nontorsion free semigroup and K is a field
then Kt[S] has nontrivial units.
v
55
A. /N. tRL H. JJ A. /N. ,[R[H. ]] A. /N. ,C R[H. ]]
1 3 1 2 3 2 m J m
. m
= Ai /N. 1[R[Hi ]].
m 3 m
Thus (Ai/Nj_1[R[Hi]])n
0 for all iel implies that
(A/Z.eINj_1[R[Hi]J)n = ZieI(A./N._1[R[Hi]])n = 0.
Hence A/ZNj_^[ R[~ ] 1 is nilpotent in R[ S ]/ZN^ _^[ R[ ] ] .
By induction hypothesis (A + N^_^[R[S]])/Nj_^CR[S]] is nil
potent in R[ S]/N j_^CR[S] ] ; hence A c Nj[R[S]]. Therefore
Z^eI Nj[R[H^]] c Nj[R[S]]. If j is a limit ordinal, then
N.[R[H^]] = u N^ERCH^]]. Thus
J h< j
Z
iel
Nj[R[Hi]]
'iel
( u N,[ R[ Fh ] ] )
h< j
h< j
(EieINhrR[Hi]])
c u N,[ R[ S ] ] = N [ R[ S 1 ] .
h< j 3
Since N[R[H^]] = N^CRtH^]] for some ordinal x, eventually
we get Zie];N[R[Hi]] c N[R[S]].
So far we have obtained nilpotent ideals of R[S] as
sums of nilpotent ideals of R[H^], i e I. We now see that
this is not always the case.
Lemma 3.15. Let A be a nonzero proper ideal of R[S].
Let A. = A n R[H.]( i e I. Then the following statements
hold:
(i) A^ is an ideal of R[H^] for every iel, and thus
is a left ideal of R[Sj.
19
, x. x+1. x+m1.
viax ax) = y (g>g )y (g>g ) y (g>g )
Y y ,y 2x> 2x+l. 2x+ml,
Y(g,g )y(g/g )* *y(g,g )
, x+m xx
Y(g ,g )
Letting j = x+m in equation (5), we get
, x+m Xv
Y(g ,g )
x+m x Y (g,gx+m)
2x+m.
Y(g,g )
, x+m+l x.
Y (g ,g )
Hence
,x x,
Y(g ,g )
y(g,gx)y(g>gx+1)Y(g>gx+m)
Y(g,g2x)Y(g,g2x+1)Y(g,g2x+m)
, x+m+1 x>
Y(g ,g )
Finally, since for j = q1, we have
(gq'1,gx) ^4^17 ^q.gx> 
Y(g,g M ) Y(g,g )
Thus
/X X.
Y(g ,9 )
_ y(g>gx)y(g>gx+1)y(g>gq 1)
, 2x> 2x+l. xl>
y(gg )y(g,g )*y(g,g )
2 x 1
"j/y(g,g)y(g,g )***Y(gg )
, x, x+1, 2xl.
^i/Yigrg )y(g^g )***Y(g#g )
= y(g>gx)y(g>gx+1)Y(g>g2x 1)
Y(gg) (g,g2)*y(g,gx_1)
Proposition 1.23. Let K = GF(2n), where 2n ^
q is
a prime. If 1 < p < q1, then
_ nl _ n2 _k k nkl
Y(gp,gp)2 y(g2p,g2p)2 y(g2 p,g2 p)2
_nl ~nl
,2 P 2 p. p
Y (g ,9 ) =
66
Theorem 3.29. The nilpotent radical of R[S] is
N[ R[ S ] ] = NT R[ U ] ] (^U( u { z }K (u) + N[K(z)]).
Proof. By Proposition 3.26 N[R[S]] = N[R[U]] N[K].
But N[K] = N[EueUK(u)], and since K(u) is a nilpotent ideal
of index two for u z by Lemma 3.28, then
N[K] = N[ZueUK(u)] = zUeU_{z}K(u) + (N[R[J]] n K(z) =
EueU{z)K(u) + N[K(z)]
PROPERTIES OF TWISTED SEMIGROUP RINGS
By
ANTONIO QUESADA RETTSCHLAG
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1978
44
Proof. Let (L,a) = M denote the enlargement of L by an
element a = ui + vx + wy of F[SJ L; that is, u + v + w 0.
Since ax = (u + v + w)x e M and since (u + v + w)(i x) e M,
then by adding both elements we get (u + v + w)i e M. But
u + v + w has an inverse; hence i e M. Therefore M = F[S].
Similarly, let (
(x,y> by an element 3 = ui + vx + wy not in (x,y), that
is, u 0. Since vx + wy e (x,y), then ui and hence i belong
to T. Therefore T = F[S].
Claim 2. The only ideals of F[S] generated by elements
which do not contain i in their support, are (x,y) and
Proof. Let I be an ideal of F[S] such that i suppa
for all a e I. If either x or y belongs to I, then xy = y and
yx = x imply that (x,y) = {ux + vy u,v e F} Â£ I.
Let bx+cye I, b 0, c 0. We consider two cases.
First if b + c 0 for some element of I, then
(bx +cy)x= (b+c)xe I. But b + c has an inverse; thus
x e I, and by the previous paragraph (x,y) Â£ I. By Claim 1
(x,y) is maximal; hence I = (x,y>.
On the other hand, if b + c = 0 for each element of I,
then bx +cy=b(xy) e I. Hence (x y) Â£ I. Moreover,
a + b 0. This contradiction shows that x y =1.
Claim 3. An element of F[S], which contains i in its
support, generates the ideal F[S] if the sum of its coefficients
is not zero.
63
After renaming
0 = 0(a) = Â£(a.
t
Thus
convcnien
+ Â£bt )ut
k
tly the
; hence
sub .indexes, we can write
a, + Eb^, = 0 for each
t.
a = E (a. u. + Â£ b^.x v) e ZK(ut) Â£ K.
t k K UtK t
Therefore Ker 0 = K.
We are now in the position to show that the semigroup
ring of a semigroup having a universally minimal ideal is
completely determined by the group ring of the group of
zeroids and the ideal K.
Theorem 3.24. R[S] = RLU] K (ring direct sum).
Proof. It follows from the definition of K(u) and (ii)
of Remark 3.20 that R[U] n K(u) = 0 for all u e U. Hence
R[U] n K = 0.
By Lemma 3.23 K is an ideal of RlSj. Since U is an ideal
of S, we get readily that RLUJ is also an ideal of R[SJ. As
K is the kernel of the epimorphism 0, it follows that
RLSj/K ~ R[U]. Thus RL S] = R[U] + K.
Corollary 3.25. If the semigroup S coincides with its
frame, then R[S] = R[U] 0 K(z).
Proof. Since S = U u J, then Ker 0 = K(z) and the
result follows from Theorem 3.24.
Proposition 3.26. The nilpotent radical of R[S] is
N[R[S]] = N[R[U]] N[K]. In particular, if S = U u J, then
NLRLS]] = N[RL U]] N[K(z)].
CHAPTER II
ON NONTRIVIAL UNITS OF TWISTED GROUP RINGS
AND TWISTED SEMIGROUP RINGS
Higman [12] defines a trivial unit of a group ring R[G]
as any element of the form rg, where r is a unit of R and
g e G. He also characterizes the group rings that have
always nontrivial units in the case where R is the ring of
the rational integers and G is a torsion group. Since then,
the problem of the existence of nontrivial units in group
rings, together with the characterization of the group that
they determine, has been largely studied. We only mention
a few such results. Ayoub and Ayoub [ 1 ] determined the group
of units of ZG, where G is a finite abelian group. Raggi
Cardenas [23] calculates the group of units of K[G] for any
Galois field K and any finite abelian group G. Passman [21,
Lemma 26.1] shows that if K is any field and G is a nontorsion
free group, then K[G] has nontrivial units provided k >3.
Finally, Gilmer and Teply [11] have shown that if S is a
torsion semigroup and R[S] has no nontrivial units, then the
characteristic of R is 2,3 or 0; moreover, if R is an algebra
over a field, if S is torsion and if R[S] has no nontrivial
units, then the characteristic of R is either 2 or 3. Finally,
they characterize the semigroup rings of characteristic 2
or 3 which have no nontrivial units.
24
ACKNOWLEDGEMENTS
I would like to express my gratitude to all the people
that in some way have contributed to make this work possible;
they are:
Dr. Jorge Martinez, Dr. Gerhard Ritter, Dr. Sergio
Zarantonello and Dr. Elroy Bolduc, Jr., who served as members
of my committee;
Dr. Bruce Edwards whose help in the computer work was
invaluable;
Dr. Bolling Farmer and Dr. Robert Long who assisted in
improving my background in related areas;
Dr. Mark Teply, who taught me most of the algebra that
I know, and who as an advisor was always ready to listen
with his limitless patience and unswerving interest. I
consider it a privilege to have had the opportunity to work
under his guidance.
In addition I would like to recognize the efforts and
sacrifices of my family throughout my years of study. I am
indebted to my parents for providing me the opportunity to
complete my graduate studies in Spain and to my wife and
daughter for having the patience and understanding which
made the burden of my doctoral studies considerably lighter.
Without the faith and love of these four people I could never
have finished. To them I dedicate this work.
11
58
1 2
O = ij>(a) = ote, = E a. (e. e, ) + E a, (x, e, ) + ...
ji=l 31 31 o=l 2J2 A2D2
?2Z
m 12
+ E a (x e.) = ( E a. ) e. + ( E a ) x, 1 + ..
j =1 m'm m^m j,=l ^1 j~=l A252 A2X
J1 J2
m
+ (.E aA j )xA 1*
] =1 mJm ra
Jm
t. t t
1 m
Hence E a. = E a, =...= E a
jx=l 31 j2=1 232
i ^ j
3 =1 mJm
Jm
= 0. It follows
that a e C, and therefore Ker ip c c.
On the other hand, it is clear that iMC) = 0, thus
C c Ker ip, and we conclude that C = Ker ip is an ideal of R[S]
From iMRCH^W) = RCH^fWe^ = RCH^ 0 = 0, we get
R[H,]W c c. Let a e C. By the definition of C we can write
t. t0 t
a = l a. e. + E a, x, + ... + E a. x, .
j,=l 'l "l j9 = l 2^2 232 j =1 mV m^m
1 2 Jm
But x^ ^e. = x^ implies that we can rewrite
i *i Ai)i
fcl fc2
a = e. E a. e. + x. E a, e + ... +
1 jL=l H H V j2=l X2]2 ^2
m
+ x, E a, e. ,
A 1 A j j
m j =1 mm Jm
Jm
and since E a, = 0, i = 1 ,m, then E1 a, ^ e. e W.
jH=l Vi j,=l Vi 3i
Hence a e RtH^W, it follows that C c RCH^W. Therefore
C = R[H^]W. Whence = R[H^](WR[H^])W = 0 by Lemma 3.16.
Thus C is nilpotent.
12
Y and y' are related as in (3) for some 6 :G > K {0}, then
these rings are actually identical, and we say that they are
diagonally equivalent. In particular, Kt[G] is diagonally
equivalent to K[G] if and only if
Y(x,y) = 6(xy)6(x) 16(y) 1
for some function 6:G * K {0}.
Let K^[G] be given with twist function y. Consider
*
K [G] with twist function y' defined by y'(x,y) =
y(i,i) ^y(x,y). Clearly y' and y satisfy equation (3); hence
t t'
K [G] and K [G] are diagonally equivalent. Moreover by
Lemma 1.12 (i) and (ii)
(4) i is the identity of [G]and y'(x,i) = y'(i,z) = y(i,i)
for all x,z e G. In view of this, we give the following
definition: a twist function is said to be normalized if
it satisfies condition (4).
From now on all the twist functions considered are
supposed to be normalized.
Let us consider now some classical examples. Let G be
a cyclic group of order n generated by g, and let
4>: K [ x 3 + Kt[G] be defined by cp (x) = g. Then iji is a ring
n 2
homomorphism such that (xn) = ai, where a = y (g g) y (g, gz) . .
y(g,gn ^) Thus cj)(xn) = (a) and hence the kernel of is
the principal ideal generated by the polynomial x11 a. It
follows that K^[G] ^ K[x]/(xna). Conversely, for any
a e K {0}, we see that K[x]/(xna ) has as a Kbasis the
82
Ca. By the definition of this correspondence is clearly
onetoone and onto. Hence 0 is a well defined bijection.
Now let f,g e Rt 3 and let r e R. By definition of 0
it follows that 0(f+g) = 0(f)+0(g) and 0(rf) = r0(f). To
show that 0 preserves products it is sufficient to prove it
for the generators. Let x e Sa(u) and y e Sa(v). By Lemma
4.8 (iv) xy e (uv); hence 0(xy) = xy uv = xyuvuv+uv =
xy0 (x)vu0 (y)+uv = (xu)(yv) = 0(x)0(y). Therefore 0
is an algebra isomorphism.
Theorem 4.13. N[R[S]] = N[R[S ]] I N[C ], and
w aeFq a
J[R[S]] = J[R[S ]] E J[C ], where N[C ] ~ N[R[S ]]
V* CX Â£ t q cx cx cx
and J[Ca] s, J[R[Sa]].
Proof. This follows directly from Theorem 4.11 and
Lemma 4.12.
Corollary 4.14. The following conditions are equivalent.
(i) R^Sal is semiprime (semisimple) for each a e Pq.
(ii) RCS] is semiprime (semisimple).
Proof. By Theorem 4.13 N[R[S]] = 0 (J[R[S]] = 0) if
and only if N[R[Sa]] = 0 (J[R[Sa]] = 0) for all a e PQ.
Corollary 4.15. If F is a field of characteristic 0 or
a prime not dividing the order of any of the groups S^,
where  Sa  < 00 for all a e PQ, then F[S] is semisimple.
Proof. By Maschke's theorem [13] each F[S ] is semi
simple; so the result follows from Corollary 4.14.
8
A ring R is called Artinian if the descending chain
condition for right ideals holds in R; that is, every strictly
decreasing chain of right ideals in R contains only a finite
number of right ideals.
Lemma 1.11. If R is artinian, then the Jacobson
radical of R coincides with the prime radical of R.
1.3 Semigroup Rings
Let R be a ring, and let S be a multiplicative semigroup.
N. Jacobson [14] defines the semigroup ring of S over R to
be the set of functions from S into R that are finitely
nonzero, with addition and mulitplication defined as follows:
(f + g)(s) = f (s) + g (s)
(fg) (s) = E f(x)g (y) ,
xy=s
where the symbol E indicates that the sum is taken over
xy=s
all ordered pairs (x,y) of elements of S such that xy = s.
We use R[S] to denote the semigroup ring of S over R. Semi
group rings are generalizations of polynomial rings: if ZQ
denotes the additive semigroup of nonnegative integers, then
R[Zq] is isomorphic to the polynomial ring in one indeterminate
over R.
n
We think of the elements of R[S] as "polynomials" I a.s.,
i=l 1 1
where a^ e R and s^ e S. A typical nonzero element a of R[S]
n
has a unique representation a = E a.s., where the a. are
i=l 1 1 1
36
Proof. Let e S G, and suppose that s^S2 = g e G.
Then s^(s2g^) = i; hence s^ is invertible with respect to i
and therefore s^ e G. This contradiction shows that S G
is closed. To show that S G is an ideal, suppose
sg = g^ e G for some s e S G and g e G. Then s = g^g ^ e G
which contradicts the choice of s. Hence (S G)S c S G,
and symmetrically S(S G) c S G.
Lemma 2.11. Assume S G contains a torsion element.
Then either char R = 2 or else R^[S] has a nontrivial unit
with support in S G.
Proof. Let s be a torsion element of S G. By Theorem
1.1 the subsemigroup (s') contains an idempotent e. By Lemma
2.10 e e S G. Set b = y(e,e)i. Then (i 2be)^ =
i + 4b[by(e,e)1]e = i. Hence i 2be is a nontrivial unit
of Rt[S] with support in S G, unless char R = 2.
Lemma 2.12. Let char R = 2, and let e be an idempotent
of S different from i. If the twist function is nonconstant,
then e is contained in the support of a nontrivial unit of
Rt[S].
Proof. Since y is nonconstant, then the group of units
of R is not trivial. Let y(e,e) = c, and let u and v be
different units of R. By recalling that c is a central unit
of R, then [ui + c1(u+v)e][u1i+c1(u1+v"l)e] =
i+c_iL4+2uiv+2uvi]e = i.
69
(A) Let KLGJ be a group ring over a field of character
istic 0. Then (i) K[G] is semiprime, (ii) Z[K[G]] is semi
prime, and (iii) ZLKLGJJ is semisimple.
(B) Let KlG] be a group ring over a field of character
istic p > 0. Then the following are equivalent.
(i)K[G] is semiprime.
(ii) Z[KLG]] is semiprime.
(iii)ZLKLG]] is semisimple.
(iv)G has no finite normal subgroups H with pH.
(v)A(G) is a p'group, where A(G) = {x e GÂ¡ the index of
the centralizer of x in G is finite}.
In Corollary 4.18 we apply (A) and (B) to the group rings
FLSaJ, a e Pq, to obtain a similar result between F[SJ and
ZCFCSj].
Finally, we say that a semigroup S belongs to the family
i if (i) S admits relative inverses, (ii) the idempotents of
S commute, and (iii) the semilattice of subgroups SQ of S is
of type P^. Then in Theorem 4.21 we prove that R [S ] is
semiprime if R[Sa] is semiprime for all a e .
4.1 Semiprime Twisted Semigroup Rings
In this section we first show that a twisted semigroup
ring RtiSj is semiprime if and only if the ring R is semi
prime, provided the semigroup S satisfies a suitable condition.
From this we are able to determine the nilpotent radical of
this semigroup ring.
BIBLIOGRAPHY
1. R. G. Ayoub and C. Ayoub, On the Group Ring of a Finite
Abelian Group, Bull. Austral. Math. Soc. 1 (1969),
245261.
2. B. Banaschewski, On Proving the Absence of Zero Divisors
for Semigroup Rings, Caad. Math. Bull. 4 (1961), 225232.
3. A. H. Clifford, Semigroups Admitting Relative Inverses,
Ann. of Math. 42 (1941), 10491307.
4. A. H. Clifford and D. D. Miller, Semigroups Having
Zeroid Elements, Amer. J. of Math. 70 (1948), 117125.
5. A. H. Clifford and G. B. Preston, Algebraic Theory of
Semigroups, Volume I, Math. Surveys of the Amer. Math.
Soc., Rhode Island, 1961.
6. A. H. Clifford and G. B. Preston, Algebraic Theory of
Semigroups, Volume II, Math. Surveys of the Amer. Math.
Soc., Rhode Island, 1967.
7. I. G. Connell, Zero Divisors in Group Rings, Comm. Algebra
2 (1974), 114.
8. C. W. Curtis and I. Reiner, Representation Theory of
Finite Groups and Associative Algebras. John Wiley and
Sons, New York, 1962.
9. N. J. Divinsky, Rings and Radicals, Univ. of Toronto
Press, 1965.
10. R. Gilmer and T. Parker, Nilpotent elements of Commutative
Semigroup Rings, Michigan Math. J. 22 (1975), 97108.
11. R. Gilmer and M. L. Teply, Units of Semigroup Rings,
Comm. Algebra 5 (1977), 12751303.
12. G. Higman, The Units of Group Rings, Proc. London Math.
Soc. 46 (1940), 231248.
13. T. W. Hungerford, Algebra, Holt Rinehart and Winston,
New York, 1974.
89
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Associate Professor of Subject
Specialization Teacher Education
This dissertation was submitted to the Department of
Mathematics in the College of Arts and Sciences, and to
the Graduate Council, and was accepted as partial fulfill
ment of the requirements for the degree of Doctor of
Philosophy.
August 1978
Dean, Graduate School
46
Since no element of (x y) has i in its support, then
(x y) is strictly contained in L. By claims 1 through 4
the lattice of ideals of FIS] is
F [ S J
0
2
Since (x y) = 0, then (x y) is a nilpotent ideal.
Also i x and x are idempotent elements of L and (x,y),
respectively. Hence these ideals are not nilpotent; thus the
nilpotent radical of F[S] is N[F[SJ] = (x y). Moreover,
since S is finite, then F[S] is Artinian. Hence J[F[S]] =
N L F[S ] ] = P[F[S]] =
3.2 Semigroup Rings Over Right (Left) Zero Semigroups and
Their Nilradicals
In L 5 ] the following definition is given.
A semigroup S is called a right zero semigroup if
xy = y for every x,y in S. The left zero semigroups are
defined by duality. In spite of their triviality, these
semigroups arise naturally in the characterization of some
larger families of noncommutative semigroups that we consider
in further sections. Let R be a commutative ring with iden
tity, and let S be a right zero semigroup. Our goal in this
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Mark L. Teply / S'
Associate Professor of' Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
'Os/rh' r
Jo^ge Martinez ]
Associate Professor of Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.
Gerhard X. Ritter
rsociate Professor of Mathematics
I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree^of Doctor of Philosophy.
Sergio I'E. Zarantonello
Assistant Professor of Mathematics
RLSjL^w c L^w. Therefore, L^w is a twosided ideal of RLSJ.
Finally (L^w)2 = L^iwL^w = L^
0*w = 0 implies
that Lj^w is
nilpotent.
Proposition 3.17. Let C =
. fcl
v2
{ E a, e, +
1 aA i XA i.
i =1 1 1
i9=l 22 A2 2
fcn h
tn
'* + .Z aA i XA i 1 ,E nai, =
Â£ a > =
A ^ l
.E aA i
i=l n n nxn i, =1 1
n 1
i2=i 2
1n=1 n n
x, e H. x, ^e, = x, , m,k
A.i. i. A.mk A.k'
J 3 3 3 3
,ij e I, tj and
n e Z+,
j = l,2,...,n}. Then C is a nilpotent ideal of R[S] that con
tains every ideal A of R[S] such that ARCS] = 0. Moreover,
C = R[ H1 ]W.
Proof. Define ip: R[S] * R[H^] by i^(a) = ae^. By Lemma
3.12 (v) ip is clearly an Rmodule epimorphism. Moreover,
>MaB) = (aBJeL = aiBe^) = a(e1Be1) = (ae^) (Be^) = i^(a)ip(B)
and so ip is also a ring epimorphism.
Let a e Ker ip. Write
^1 t2
a = E a e + E a^ x> +
ji1 D1 31 j2=l A2^2 X2^2
I'm
+ A aA j XA j '
jn=l mJm mJm
where x* e H. x, e. = x, v; that is, the elements in the
LA. j .
iJ i
"A n k AA.k'
i i
support of one summation belong to different subgroups, and
all of them have the same image under any of the group
isomorphisms 6nk for all n,k e I. Then
25
Let K be a field, and let G be a nontorsion free group.
Moreover, let R be a ring with identity, and let S be a
semigroup with identity i. In this chapter we answer the
following questions:
(A) When does the twisted group ring have nontrivial
units?
(B) What are some necessary conditions for Rt[S] to
have no nontrivial units?
(C) If S is a nontorsion free semigroup and R has
finite characteristic, when does Rt[S] have nontrivial units?
(D) If S is a nontorsion free semigroup, when does
Kt[S] have nontrivial units?
In section 2.1 we answer (A) by showing that K^[G]
has always nontrivial units, provided that the twist function
is nonconstant.
Question (B) is answered in Theorem 2.9, which states
that if R^CS] has no nontrivial units, then R has no nilpotent
elements and either R is indecomposable or the largest
subgroup of S containing i is trivial.
In Theorem 2.15 we prove that if S is a nontorsion free
semigroup and R has finite characteristic, then Rt[S] has
nontrivial units, provided that the twist is nonconstant.
This answers question (C).
Theorem 2.16 answers question (D) by showing that if S
is nontorsion free and the twist function is nonconstant,
then Kfc[S] has nontrivial units.
Passman [22] proves the existence of nontrivial units
in K[G] by showing first that these group rings always have
28
Remark 2.2. Notice that by using (i < q 1)
instead of in the proof of Lemma 2.1, we get the same
kind of units.
The following result provides an example where the
twisted group ring has nontrivial units and the usual group
ring does not.
Corollary 2.3. If K = GF(3) and G = C2, then Kfc[G]
has nontrivial units whenever the twist function is non
constant .
Next, we consider the case of fields of characteristic
2.
Lemma 2.4. Let K be a field of characteristic 2, and
let G be a nontorsion free group. If the twist function
is nonconstant, then K^'CG] has always nontrivial units.
Proof. As in Lemma 2.1, it is sufficient to consider
G to be a cyclic group of prime order; say G = C = (g).
Assume first that q > 2. Let and b^ be as defined in
the proof of Lemma 2.1. We consider two cases.
If tt^  1 (mod 2) then by proceeding as in Lemma 2.1,
we get
1 q_1 k
[ (1 ) (ig)](i+g+ E b^^g ) = i
k2
Now let tt^ E 1 (mod 2) .
Suppose, first, that there exists r e K {0} such
that rq 1; that is, 1 + rq has inverse. Then
74
Proof. Let Nq c c ... c n. Â£ ... and
Nq[R^[S]1 c nl Rt[S]] c ..., be the chains of ideals
used in the definition of the nilpotent radical of R and
Rt[S'J, respectively. We show that Nt[S] c N[Rt[S]] by
transfinite induction. We have seen in the proof of Pro
position 4.4 that Bt[S] is a nilpotent ideal of Rt[S] for
each nilpotent ideal B of R. Hence N^[S] c NQ[Rt[S]]. This
shows the first step of the induction argument.
Assume that N^CS] Â£ N^[Rt[S]] for all the ordinals
i < j. Suppose j is not a limit ordinal and let A/N^_^ be
a nilpotent ideal of R/N._, with index of nilpotency m.
k 31 k
Let f = Zas e A lS]. Then fm = ( Z a s )m =
r=l r r r=l r r
Z (a s ) (a s ) (a s ) with X = {l,2,...,k} and
1 1 Ti 2T o I IT
11 2 2 mm
(r.,r0,...,r ) e X x ... x x = Xm. But
1 2' m
(a s ) (a s ) (a s ) = a a y(s ,s )a a
r. r, r r_ r r rn r r.' r0 r, r
11 22 mm 12 123 m
y(sr ...sr ,sr )sr ...sr e N^CS] .
X1 m1 m rl m
Since a a y(s s )a
rl r2 rl r2 r3
.a y(s ...s ,s )eAm=0,
T 2T, IT XT
m 1 m1 m
Thus (A/N^^) t[S] is a nilpotent in (R/N^ S ] ; hence
At[S]/Nj_^t[S] is a nilpotent ideal of R^[S]/NS]. By
induction hypothesis At[S] + N Rfc[ S ] ]/N Rt[ S ] ] is
nilpotent in Rt[S]/N^._1 Rt[S]]; so Afc[S] c N^[Rt[S]].
Therefore Nj[S] c Nj[R*"[S]]. If j is a limit ordinal, then
N. = u N.. Since by induction hypothesis n!"[S] c N[Rt[S]]
3 i< i 1
86
(i) The ideals A^ are semiprime for all i e I.
(ii) R is semiprime.
Proof. Assume A^ is semiprime for each i e I. For
contradiction, suppose that R is not semiprime, and let B
be a nonzero nilpotent ideal of R. Since the ideals A^ form
a chain, then B n Aj 0 for some j e I; so B n A. is a nil
potent ideal of A^, which contradicts our hypothesis.
Conversely, assume that R is semiprime, and suppose that,
for some i e I, the ideal A^ is not semiprime. Let C be
a nonzero nilpotent ideal of A^. For each j,k in I, we have
CACA, c CA.A.A, c CA. c c. Hence (CR)2 c c. Thus CR is a
] k j i k ~ i
nilpotent right ideal of R. Since R is semiprime, CR = 0.
Thus C is a nonzero nilpotent right ideal of R, which
contradicts our hypothesis. Therefore A^ is semiprime for
all i e I.
Theorem 4.21. Let S e t. If RCS^] is semiprime for
all a e P, then R[S] is semiprime.
Proof. Assume R[Sa] is semiprime for all a e P. Let
(St be as defined in Lemma 4.19. Using transfinite
i
induction, we show first that the ideals R[S^_ ] are semiprime
for all i e I. Then by Lemma 4.20 the result follows.
The first step of the induction argument is true by
hypothesis. Assume that RCS^ ] is semiprime for all the
ordinals j < i, i,j e I. Let St, = uaeT,Sa, and let
75
for all i < j, then
Nt[Sj = u N^LSJ c u N.[Rt[S]] = N.[Rt[S]].
3 i
Finally, since N = for some ordinal t, we have that
Nt[S] c N[Rt[S].
Conversely, since R/N is a semiprime ring, it follows
from Theorem 4.5 that (R/N)t[S] is semiprime; thus Rt[S]/Nt[S]
is semiprime. Since N[ R^[ S ] ]/Nt[ S ] c N[ (R/N) *"[ S ] ] = 0, then
N[Rt[Sj] c Nt[SJ.
Proposition 4.7. Let R be a ring with identity, and
let S be a semigroup with identity in which condition (E)
holds. Denote by Z[Rt[S]] the center of Rt[S]. If R is
semiprime, then Z[Rt[S]] is semiprime.
Proof. By Theorem 4.5 and the hypothesis, it follows
that RtLS] is semiprime. Suppose that there exists an ideal
A of ZCR^CS]] such that A =0. Let a e A, then a =0.
Moreover, since a is in the center, aRt[S] is a twosided
ideal of R^S]. Now (aRt[S])2 = a2Rt[S] = 0. Since Rt[S]
is semiprime, it follows that aR^CS] = 0; hence a = 0, and
consequently Z[Rt[S]j is semiprime.
4.2 On Semigroup Rings Over Semigroups Admitting Relative
Inverses
In this section we apply the results of section 3.4 to
semigroups of a type first discussed by Clifford [3 ] and
largely studied since then [5 ], [5 ], [19].
81
n.
m
E j.'vju> ,E ,\l*a,tu) +
1=1 J1 11 t =1 1 11
n,
n.
( Â£ a. (x u) E b (y u)) = 0.
j =1 Jk kJk t =1 k k k
k k
But the support of any two sets of terms in parenthesis are
disjoint; hence each term inside of a parenthesis must be 0.
ni mi
Then E a. (x u) = E b, (y u) in C (u) for all i.
1 a j 4. t t. J a t. a.
1 =1 Ji iJi t.=li li i
Ji i
It follows that, except by possibly for order, the elements
are identical. This shows that each f has a unique repre
sentation in Â£aep Ca(u); hence f can be uniquely expressed
in EEC (u) = EC .
a a
ua a
Therefore, K is the internal weak direct sum of its
ideals C By Proposition 4.9 we conclude that R[S] =
R[S ] EW C .
w cteP0 a
Lemma 4.12. For each a e Pn {w}, the ideal C is
0 a
isomorphic as an algebra with R[Sa].
Proof. Let f e R[S ] with a w. Since S = u S (u),
a a ue S a '
w
we can write f = Ea. x. + + Ea. x. where x. e S (u,),
11 11 xk 1k xt
1 < t < k.
Define 0: * Ca by 0(f) = Ea^ (x^ u^) + * +
Ea. (x. u. ). Notice that if x e S (u), then 0(x) = xu.
xk k K a
That is, 0 takes each generator of R[SQ] to one generator of
70
The following definitions are due to Ban'aschewski [2 J.
Definition 4.1. Let X be a subset of a semigroup S. An
element x of X is called an extremity of X is for all positive
integer n, x11 = sjS^..s s^ e X, implies s^ = x for all i.
Definition 4.2. A semigroup S is said to satisfy con
dition (E) if every nonempty subset of S has an extremity.
It is clear that any totally ordered semigroup S sat
isfies condition (E), since the greatest and the least
element of any finite subset X of S, are extremities of X.
Moreover, if a semigroup S satisfies condition (E) and
xn = y11 for x,y e S, then x = y for n = 1,2,...; that is,
S is power cancellative. Banaschewski proved also in [2 ],
that any cancellative, power cancellative abelian semigroup
satisfies condition (E).
We might think of the possibility of defining a weaker
condition than (E) on S, both conditions related in the same
way as the ptorsion free condition is related to the torsion
free one. Namely, we would define a pextremity of a subset
X of S as an element a such that aPn = sns~...s s e X,
12 n' i
P
implies that a = s^ = S2 = = s Then we could set that
P
a semigroup S satisfies condition (E ) if any nonempty finite
IT
subset X of S has a pextremity. This would allow us to
differentiate between twisted semigroup rings over rings of
finite characteristic from those whose rings have characteris
tic 0, as Gilmer and Parker [10J did in commutative semigroup
rings.
56
(ii) Aj for all i,j e I.
(iii) If A^ = 0 for some i e I, then A^ = 0 for all
i e I. In this case ARCS] = 0; hence A is nilpotent
of index two.
Proof. (i) Since A^RLH^j c ARC ] = A_^, and symmetrically
RCH^JA^ c a^, it follows that A^ is an ideal of RCHj_] for
every i e I. By Lemma 3.13 each A^ is a left ideal of RCS].
(ii) A^R[Hj] c a n R[ H j ] = A j ; hence Aj 2 Afej*
Symmetrically A^ 2 Ajei = 0j(Aj). Thus
Ai
0 . (A ) 3
31 3 
0 . (A. e .) =
31 1 3
Aiejei
= Aiei
= Ai
Since 0^j are monomorphisms, we conclude that A^ s A j .
(iii) Let Aj = 0 for some j e I. By (ii) it is clear
that A^ = 0 for all i e I. By Lemma 3.12 (v) R[H^] is a left
ideal of RCSJ; hence ARCH^] c A n RCIL] = 0 for all i e I.
Then ARLSJ = AZ RCH.] = Z. tARCH.] = 0 by Lemma 3.12 (iii).
lell lei 1
In particular, A2 c ARCS] = 0.
Definition. W
and n e Z+).
( E aie e e E, E a
1=1 1=1
0, ai e R,
Lemma 3.16. Let L^ be a left ideal of R[H^]. Then
L^w is a nilpotent ideal of R[S] for all w e W, w 0.
n
Proof. Let x e H, and let w = Z a^e^ e W. Since
n n J i=l
wx = Z a^(e^x) = ( Z a^)x = 0, then wRCH.] = 0 for all i e I.
i1 i=l J
Hence wR[S] = ZjeI wRCHj] = 0. Thus L^RCS] c Lxw. By
Lemma 3.12 (v) RCSJLj^ = R[S](e1L1) = RCH^Lj^ = Lx; so
11
Proof. (i) Putting y = i in equation (2) yields
y(x,i) = y(i,z). Thus for all x,z e G we have
y (x,i) = y (i,z) = y (i,i).
(ii) By (i) and equation (1) we see that y(i,i) ^i is
the identity element of K^[G].
(iii) By (i), (ii) and (1) y(x 1,x)_1y(i,i)1x1 is a
left inverse for x, and y(x,x *") ^y(i,i) ^x ^ is a right
inverse for x. Thus x is invertible and x ^ is equal to
both of these expressions.
It is quite possible that some given Kt[G] is just
K[G] disguised in a rather natural way. We start with K^CG]
and make a diagonal change of basis by replacing each x by
x' = 6(x)x for some nonzero 6(x) e K. Observe that the word
"diagonal" arises here because, for G finite, the correspond
ing change of basis matrix would be diagonal with the 6's as
diagonal entries. From (1) and the foregoing we have
x 'y' = 6 (x) x (y) y = (x) (y) y (x,y) xy
= 6(x)6(y)6(xy) 1y(x,y)(xy)'
In other words, with this change of basis, K^'LG] is realized
in a second way as a twisted group ring of G over K, this
time with twist function:
(3) y'(x,y) = 6(x)6(y)6(xy) 1y(x,y).
, tl t2
Thus we see that if K [G] and K [G] are two twisted group
rings of G over K and if their corresponding twist functions
4
Theorem 1.3. Let e be an idempotent of S, and let IIg
be the group of units of eSe. Then He contains every sub
group of S that meets He.
It follows that the groups He are just maximal subgroups
of S; that is, no He is properly contained in any other sub
group of S. Moreover, if e and f are idempotent elements of
S, e f, then and are disjoint.
The preceeding terminology and results can be found in
[ 5 ], where the reader can find considerable additional
material on semigroups.
1.2 Rings
Let R denote an associative ring. If there is a least
positive integer n such that na = 0 for all a e R, then R is
said to have characteristic n. If no such n exists, R is
said to have characteristic 0. We will denote characteristic
of R by char R. A ring of characteristic pn, for a prime p,
is called a pring [24]. (Note that there are other conflict
ing uses of the term pring; for example, see [16]). For a
ring R and a prime integer p, define R = {r e Rpnr = 0 for
some positive integer n). The center of R is the set
C = (c e Rcr = rc for all r e R}.
If R has identity 1, the subring of R generated by 1 will
be called the prime subring. It is an easy consequence of
the definition that if tt is the prime subring of a ring R,
then it ~ Zn whenever char R = n > 0, and tt ^ Z otherwise.
39
Kt CÂ¡jJ is isomorphic to the field Kl xJ/ (x^tt^) On the
other hand, (ig2)(i+g2) = 0 gives an easy example of proper
zero divisors in
In this section we give some sufficient condtions for
the existence of proper zero divisors in Kt[G].
If g is a torsion element of prime order q in G, let
q_1 i
TT1 (g) = n Y (g,g ) .
i=l
Theorem 2.17. Let K be a field, and let G be a non
torsion free group. If q/iT^ (g) eK for some element g e G of
prime order q, then Kt[G] has proper zero divisors.
Proof. Since any proper zero divisor of Kk[H], where
H is a subgroup of G, is a zero divisor of Kt[G], it is
sufficient to show that Kk[(g)] has proper zero divisors
for some g e G.
Let g e G be a torsion element of prime order q, and
k1
assume that (g) eK. Set c = I/^/tt. (g) and b, = II y^g1)
1 i=l
Then
q_1 k k
(icg) (i+cg + Â£ c b, g ) =
k=2 K 1
2^ cj 1
i+cg + Â£ ckbk_.gk eg Â£ ckb, gkcqir,(g)i = (lcqTr (g))i
k=2 k=2 K 1 1 1
Corollary 2.18. If the characteristic of K divides the
order of some element of G and the twist function takes
values only in the prime field of K, then Kt[G] has proper
zero divisors.
21
Corollary 1.24. Under the hypothesis of
1.23, if uj = 1 for some j, 1 < j < q1, then
,nl ^ nn2 nl nl
y(gp.gp)2 Y2
y (g"
P 2
Proposition
= 1.
Proof. This follows directly from Corollary 1.19 and
Proposition 1.23.
1.5 Twisted Semigroup Rings
Let R be a ring with identity and let S be a multipli
cative semigroup. Let y be a twist function from S x S into
the group U of central units of R; that is, y(x,y)y(xy,z) =
y(y,z)y(x,yz) for all x,y,z e S. We define a twisted semi
group ring, denoted by Rt[S], as the Ralgebra with basis
{ss e S} and with multiplication defined distributively
such that, for all s^,S2 in S,
s^S2 y(s^,S2)s^S2
Since
(rlsl*r2s2,r3s3 =
r1r2y (s1, s2) r3y ^
rlr2Y(sl,s2)sls2'r3s3 =
s2's3 S1S2S3
and
rlSl(r2S2 r3s3) rIs1* r2r3^ ^ s2 f s3 >S2S3
2^ 2 ^ 3 Y ^s2 s3^ ^ / S2S3) s ^ s 2 s 3
then r1r2Y(s^s^r^y(SjS^s^
rlr2r3Y^ s2's3 ^ Y(s1/S2s3)
90
14. N. Jacobson, Lectures in Abstract Algebra, Volume I:
Basic Concepts, Van Nostrand, New Jersey, 1951.
15. J. Lambeck, Lectures in Rings and Modules, Blaisdell,
Toronto, 1966.
16. N. H. McCoy and D. Montgomery, A Representation of
Generalized Boolean Rings, Duke Math. J. 3 (1937),
445459.
17. N. H. McCoy, The Theory of Rings, Chelsea, New York,
1973.
18. W. D. Munn, On Semigroup Algebras, Proc. Cambridge
Phi. Soc. 51 (1955) 115.
19. W. D. Munn, Matrix Representations of Inverse Semigroups,
Proc. London Math. Soc. 14 (1964), 165181.
20. T. G. Parker, Infinite Group Rings, Dissertation, Florida
State University, 1973.
21. D. S. Passman, Infinite Group Rings, Marcel and Dekker,
New York, 1971.
22. D. S. Passman, The Algebraic Structure of Group Rings,
John Wiley and Sons, 1977.
23. F. Raggi Cardenas, Units in Group Rings, An. Inst. Mat.
Univ. Nac. Autonoma Mexico 7 (1967), 2735.
24. K. Shoda, Uber die Einheitengruppe Eines Endlicher Ringes,
Math. Ann. 102 (1930), 273282.
25. O. E. Villamayor, On Weak Dimension of Algebras, Pac. J.
Math. 9 (1959), 941951.
51
the semigroup ring RLSJ. This will allow us to completely
characterize the nilpotent radical of RSJ.
We begin by stating two important results on right
groups, the proofs of which can be found in [5, 1.26, 1.27].
Lemma 3.9. Every idempotent element of a right simple
semigroup S is a left identity of S.
Lemma 3.10. The following assertions concerning a semi
group S are equivalent.
(i)S is a right group.
(ii)S is right simple and contains an idempotent.
(iii)S is the direct product G x E of a group G and a
right zero semigroup E; that is, there exists
:G x E * S: (g,e) ge, such that is an
isomorphism.
Let E = {e}^ej stand for the set of idempotents of S.
By Lemma 3.10 (ii) E is nonempty. By Lemma 3.9 every element
of E is a left identity of S; thus E is a right zero semigroup
of S. Let e e E, then Se = eSe, and we see that Se is a sub
group of S; in fact, by Lemma 3.10 (iii) Se ~ G x {e}. Let
IL denote the subgroup Se^, where e^ e E. It follows that
for any e^,ej e E, i j, the subgroups and Hj are disjoint
and isomorphic under 0.:H. H where 0 . (x, .) = x, e .
1J 1 J X] A1 A 1 J
Moreover, by Lemma 3.10 (iii) we also see that S is the
disjoint union of its subgroups H^, i e I. We have shown
the following restatement of the previous results.
10
We call y:G*G  K {0} the twist function. This
function has also been called factor set in representation
theory; see for instance Curtis and Reiner [ 8 ].
There are a number of obvious similarities between
twisted group rings and ordinary ones, but there are also a
number of important differences. Passman points out the
following. First, for fixed K and G there are many different
possibilities for Kt[G], depending on the choice of the twist
function. Indeed, one such choice is always K[G], which is
obtained by taking y(x,y) = 1 for all x,y. On the other
hand, of course, K[G] is uniquely determined by K and G.
Second, G is not embedded naturally in K*"[G]; that is, the
map G + Kt[G] given by x + x is not a multiplicative homomor
phism in view of equation (1). Third, if H is a subgroup of
G, then the Klinear span of {hh e H} is clearly a twisted
group ring of H over K, but we must be careful and note
that this Kt[H] has as a twisting the restriction of the
function y to H x H. Finally, if H is a normal subgroup of
G, there is no obvious homomorphism from K^[G] to some
twisted group ring of G/H.
Lemma 1.12. The following relations hold in K^[G].
(i)Y (x, i) = Y(i/Z) = Y(i/i) for all x,z e G.
(ii)y(i/i) 1i is the identity element of Kt[G].
(iii)For all x e G,
x 1 = y(x,x1) Xy(i,i)_1x~1 = y(x1,x)1y(i,i)_1x_1.
Thus we see that the elements x e Kt[G] are all units.
59
Finally, if A is an ideal of R SJ such that AR[SJ = 0,
then Ae^ = 0. Hence A c Ker \p = C.
Proposition 3.18. Every element a of R[S] can be uniquely
expressed as a = + ac, where a^ e R[H^] and ac e C.
Proof. Let a e R[SJ. As in the proof of Proposition
3.17 we can write
t. t~ t
a = Z1 a. e + l a. x. + ... + Em a, x. ,
j =1 ^l ^1 j0=l A2^2 A2^2 j =1 m^m m^m
1 J 2
where x, e. = x, .
a 1 i A j .
i Ji iJi
Set
fcl fc2
d, = l a. d = E a .
x' Ji=2 Z j2=2 A2^2
,m
. d = E a. .
m A j
j =2 mJm
Jm
Then we can rewrite a as ct^ + ac, where
al [(al dl)el + (aA2l d2)xA2l + + (aA_l dm)xA_l]
m
m
and
fcl t2
a = E a. e. + d.e, + E a, x, + dnx, ,+...+
C j.^2 ^1 d1 11 j2=2 A2^2 A2^2 2 A2i
t
+ E a. x, + d x, .
n A i A j m A 1
i =2 m'm mm m
m
Clearly a. e R[H.], and since by construction
t.
di + E1 a^ =0 for i = l,2,...,m, then a e C.
jj=l C
Suppose now that a = a, + a and also a = 6, + 8 ,
C o
where a^Bj e RCH^ and ac,Bc e C.
Since
31
a = c
P v
2 / P P\
a2p = c VY(g g >
k ~kl k2 k2 k2
2 p p, 2 2p 2px 2 2 p 2 p,
a k = c vY(g /g^) y(g >9 ) ...y(g g )
2 p
n1 n2 _n3 n2 n2
2 p p, 2 2p 2p. 2 2 p 2 p.
a = c vy (g gM y(g ^,g ..y(g ,g
2n xp
We claim that, in fact, this is a set of solutions for
(2). By direct substitution we see that they satisfy the
first n1 equations of (2). Thus all we need to show is
that
v =
 nl n1
2 2 p 2 p.
a,nl Y(9 '9 >
2 P
that is,
0n nl n2 n1 n1
2 2 p p, 2 / 2p 2p> 2 2 p 2 p. ,
v c v y(g*\g^) y(g ^,g **) ...y(g ,g F)/cv.
,nl
But c
1.24.
1, and thus the equality follows from Corollary
This shows that each of these systems (2) of equations
and hence (1) has a solution. In fact we obtain one set
of solutions for each of the q possible values of c. Thus
the existence of a is proven.
Finally, let q = 2; that is, let G = {i,g}. Since y is
normalized and nonconstant, y(g,g) 1. Hence there exists
(ly(g,g)) and then r(ly(g,g)) ^(i+g)] is a nontrivial
unit of Kfc[GJ with inverse ig.
43
semigroup. We calculate the radicals of FLSJ to show that
by just removing the commutativity of S we may have a nonzero
nilpotent radical. This contrasts the following result of
Gilmer and Parker ElO] on commutative semigroup rings. If
either R is a ring of finite characteristic n and S is a
ptorsion free semigroup for each prime divisor p of n, or R
is a ring of zero characteristic and S is torsion free, then
the nilradical of RES] is NES], where N is the nilpotent
radical of R.
Let S be the semigroup defined by the following table.
i
X
y
i
i
X
y
X
X
X
y
Y
y
X
y
Notice that S has identity element i and is torsion free,
but S is not commutative.
Let F be a field. Denote by (x,y) the ideal generated
by x and y. Let L = {ai + bx + cya,b,c e F, a + b + c= 0}.
Clearly L^ c L, FES]L c l and LFES] c L; hence L is an ideal
of FES].
In order to determine the radicals of FES] we prove the
following results.
Claim 1. L and (x^'^ are maximal ideals of FESJ.
CHAPTER IV
SEMIPRIME SEMIGROUP RINGS
In commutative semigroup rings Gilmer and Parker [10]
have proven the following result. Let R be a commutative
ring with identity, and let S be an abelian semigroup. If
either the characteristic of R is zero and S is torsion
free or else the characteristic of R is n and S is ptorsion
free for each prime divisor p of n, then N[S] is the nil
potent radical of RLS], where N is the nilpotent radical
of R. It follows clearly that under the conditions of this
result R[S] is semiprime if and only if R is semiprime.
In this chapter we are mainly concerned with the problem
of determining conditions on R or S that will give rise to
semiprime (twisted) semigroup rings.
We have already seen in Corollary 3.7 that if E is a
right (left) zero semigroup and R is a commutative ring
with identity, then R[E] is semiprime if and only if R is an
integral domain and E =1. As a direct consequence of
Theorem 3.19 we have that if S is a right (left) group with
more than one idempotent (that is, S is not a group) and R
is a ring with identity, then R[SJ is never semiprime.
Finally, if S is a semigroup with universally minimal ideal
U and R is a ring with identity, then it follows from
Proposition 3.26 that R[S] is semiprime if R[U] and K are
67
76
A semigroup S is said to admit relative inverses if, for
any element s in S, there exist elements e and s' in S such
that es = se = s and ss' = s's = e. By a semilattice we
mean a commutative semigroup all of whose elements are idem
potents. Let a and 3 belong to the semilattice P; then if
a3 = 3 we write a > 3. This defines a partial order on P
with the property that any two elements have a greatest
lower bound, namely its product.
We are interested in semigroups admitting relative
inverses and such that every pair of idempotents commute.
Clifford's main result [ 3 ] on these semigroups may then
be stated thusly:
Every semigroup which admits relative inverses and
such that every pair of its idempotents commute, is isomorphic
with a semigroup S constructed as follows.
Let P be any semilattice, and to each a in P assign a
group Sa such that no two of them have an element in common.
To each pair of elements a > 3 of P assign a homomorphism
4*013: sct ** s3 such that if a > 3 > y, then 4>agg Y=^r
We call this a transitivity condition on the homomorphisms.
Let aa be the identity automorphism of Sa. Let S be the
class sum of the groups Sa, and define the product of any
two elements a,bR of S(a e S and bD e SQ) by
u p ot ct p p
aab3 = ^ct^ay^ (bg'fgy) where a3 = y is the product of a and
3 in P.
Conversely, any semigroup S constructed in this fashion
admits relative inverses, and every idempotent element of S
is in the center of S.
34
We remark that the functions defined in the former
lemma are normalized with respect to the identities of the
respective rings.
Proposition 2.8. Let R be a finite (ring) direct sum
of nonzero ideals R^,...,R Then the following conditions
are equivalent.
(i) Rt[S] has no nontrivial units,
t.
(ii) Each R^1iS] has no nontrivial units for i = l,2,...,n,
and G = {i}.
n t.
Proof. By Lemma 2.7Rt[S]= R. [S]; thus every
t i=1 1
element a e R [Sj can be written uniquely as a = a^ + ... + an
with e R^1[S]. Hence a is a unit of Rt[Sj if and only if
fci
each a is a unit of R. [S]. Write 1 = en + ... + e, where
i i in'
e. is the identity of R..
1 t
t J
Assume that R [S] has no nontrivial units. If Rj [S]
has a nontrivial unit Vj for some j, then e^ + ... + vj +
+ en is a nontrivial unit of Rt[S]. Moreover, if there
exists g e G such that g i, then e^g + + + e^ is a
nontrivial unit of Rt[S]. Therefore if R^[S] has no non
t.
trivial units, then each R^ CS] has no nontrivial units,
and G is trivial.
The converse follows easily from the fact that if
v = v^ + ... + vR is a nontrivial unit of Rt[S], then for
some j, Vj is a nontrivial unit of Rt[S], provided that
G = {i}.
87
Su. = u S By the proof of Lemma 4.17 we see that
t. aeT. a 2 1
i i
either S, = u (u.,.S, ) or else S, = (u...S, ) u S. .
t. t i
iiJ3 i J j i
Suppose = (u .<:^St ) u St By definition TV = {a}
for some a e P u . T.; hence S = S Then R[S ,,] is
^ 1 ] "t Ot u
J J 1 1
semiprime by hypothesis. By Lemma 4.19 each and hence
1
u .^S is an ideal of S. It follows from the induction
] 1 j
hypothesis and Lemma 4.20 that R[u .S ] is semiprime. Let
3 1 j
A be a nilpotent ideal of R[ S ]. Then A + RCu^.S, 1/RCu^.S ],
i 3 1 j 3 1 j
being a nilpotent ideal of R[Sfc ]/R[u.<^St ], is isomorphic
i ] 1 tj
to a nilpotent ideal of R[S.]. Since R[S ,,] is semiprime,
i i
it follows that A c R[u .^S^ ]. But R[u <5, ] is also
1 1 tj j i tj
semiprime; hence A = 0. Therefore R[S ] is semiprime.
fci
Now let S = (u.,.S ) u S,,. By definition of T! if
t. i
i 3 i
a,B e Tj, then aB = g.l.b. {a,B} belongs to Uj<^Tj. Hence
S SR c u. S for all a,B e T!. Moreover, ( Â£ R[S ]) n
j a*B
R[Sg] = 0 for all B; thus we can write R[S ] = Â£geT, R[S],
i i B
where all R[Sg] are semiprime by hypothesis. As in the first
case R[u . S ] is a semiprime ideal of R[S ]. Moreover,
3 j i
R[u. .S ] n R[S ,] = 0; thus we can write R[S, ] =
J D i 1
R[Uj
direct sum of Rmodules. Let f e R[S ]. We may write
fci
Â£ = f + fR + ... + fa where f. e R[u S. ] and
l n J j
fR e R[S ] for some Bn,...,B e T!.
P; B^ 1 n i
27
q1 i
Let 7T = n Y (g,g ) .
i=l
Jci .
Set b, = IT y(g,g ) for k > 2.
k 1 i=l
k k + 1
Notice that b^_^ = b^ = ir^, and g(b^_^g ) = b^g for
k s 2. We consider two cases.
First, if tt^  1 (mod p) then 1 tt^ ^ 0 (mod p) ;
hence 1 tt^ has an inverse. Thus for q > 2, we have
i qi k
[(1TT..) (ig) ] (i + g + Â£ b .g ) =
1 k=2 Ki
1 q_1 k q_1 k
(1TT ) (i + g + E b. ,g g E b, ngK Trni) =
k=2 K x k=2 K~1 l
(1TTj^) 1(l7T1)i = .
If q = 2, [ (17T1) 1(ig)](i+g) = (17^) 1 (ig+gT^g2) =
(1tt1 ) 1(1n^^Ji = i.
Secondly, if tt^ = 1 (mod p), then p 2 implies that
there exists (1 + tt^) ^. Thus
qi q^3
[ (1 + Ti ) 1 (i+g) ] (i+TT. g + L b, _,gk E b, ,gk) =
k=2n K k=2n+l K 1
, n=l n=l
ql q3
(1 + 7T )1(tt g + \ b,_k
k=2n K X k=2n+l
n=l n=l
ql
ti 1 gJ^ ~ 'Â£ b, _,gk + g7T ~
1 k 1 1 k=2n
n=l
5klq +
k_!gk + = (1 + TI1) 1[ (1 + 7T) i + (1 7T1) (g
Â¥ b
k=2n+l
n=l
ql
E b g*)] = i.
k=2n K 1
n=l
6
(A) A homomorphic image of an Lring is an Lring.
(B) Every ring contains an iideal A which contains
every other Lideal of the ring.
(C) The factor ring R/A is Lsemisimple.
An element x of R is said to be nilpotent if xn = 0 for
some positive integer n. A ring R (an ideal A) is said to
be nil if every element x of R (of A) is nilpotent. An
ideal is called nilpotent if some power of it is zero. We
say that n is the index of nilpotency of a nilpotent ideal
A, if n is the smallest positive integer such that An = 0.
Lemma 1.7 [ 9 ]. The sum of any finite number of
nilpotent ideals of a ring R is again a nilpotent ideal.
The union of all the nilpotent ideals of a ring R isa nil
ideal.
In [ 9 ] we find the following construction. Denote by
Nq the union of all the nilpotent ideals of R. Then R/Nq
may have nilpotent ideals. Let be the ideal of R such
that N^/Nq is the union of all the nilpotent ideals of R/Nq .
In general, for every ordinal a that is not a limit ordinal,
we define N to be the ideal of R such that N /N is the
a a a1
union of all the nilpotent ideals of R/N ^. If a is a limit
ordinal, we define N = un^ N. In this way we obtain an
a 6
ascending chain of ideals N c n, c ... c n c .... Since
R is a set, we may then consider the smallest ordinal x such
that Nx = Nt+i = ... We denote N by N[R], and N[R] is
88
Fix p c T! and define n : Rl S. I > Rl S I by
i p t. p' 1
ir (f) = fp. Clearly is an Rmodule epimorphism. To
show
that
that it is a ring epimorphism, let f,g c R[ S, ] such
P ^
f = fi + f6 + fp and g = gi + g6 + gp, where and g6
belong to EgeT. _{p }R[ Sg ] Then fg = figi + f^g^ + gp) +
(f, + f )g. + f.g + f g. + frgr + f g Since Rf u .S. ]
0 p 3i o^p p^ O^U P^P j
is an ideal, and since R[Sr]R[S ] c R[u .], then the only
0 P ~ < 1 t .
1
term in the product fg that belongs to R[Sp] is f gp. Hence
TTp (fg) = fp9p = TTp (f) Tip (g) ; so ttp preserves products.
Suppose R[S ] contains a nonzero nilpotent ideal B.
i
Since R[ui .S. ] is semiprime, then for some p e T!, it (B)*0.
1<;l tj i p
Since Tip is a ring epimorphism, then tt (B) is a nonzero
nilpotent ideal of R[S ], which contradicts the hypothesis.
Corollary 4.22. Let P be an initial segment of the
ordinals. Let S be a semigroup which admits relative inverses
and such that every pair of its idempotents commute. Let
(Sa)afp be the family of subgroups of S. Let R be a ring
not necessarily with identity. If R[S ] is semiprime for
a e P, then R[S] is semiprime.
Proof. Since P is of type P^, the result follows from
Theorem 4.21.
PAGE 1
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TABLE OF CONTENTS
Page
ACKNOWLEDGEMENTS ii
ABSTRACT v
CHAPTER
I PRELIMINARIES AND GENERAL PROPERTIES OF SEMIGROUP
RINGS 1
1.1 Semigroups 1
1.2 Rings 4
1.3 Semigroup Rings 8
1.4 Twisted Group Rings 9
1.5 Twisted Semigroup Rings 21
II ON NONTRIVIAL UNITS OF TWISTED GROUP RINGS AND
TWISTED SEMIGROUP RINGS 24
2.1 Nontrivial Units in Twisted Group Rings 26
2.2 Nontrivial Units in Twisted Semigroup Rings...32
2.3 Nontrivial Units in Twisted Semigroup Rings
Over Nontorsion Free Semigroups 35
2.4 Zero Divisors in Twisted Group Rings 38
III NILPOTENT RADICALS OF NONCOMMUTATIVE SEMIGROUP
RINGS 41
3.1 An Example 42
3.2 Semigroup Rings Over Right (Left) Zero
Semigroups and Their Nilradicals 46
3.3 The Nilpotent Radical of the Semigroup Ring
of a Right (Left) Group 50
3.4 Semigroup Rings Over Semigroups with Univer
sally Minimal Ideals, and Their Nilradicals.60
IV SEMIPRIME SEMIGROUP RINGS 67
4.1 Semiprime Twisted Semigroup Rings 69
4.2 On Semigroup Rings Over Semigroups Admitting
Relative Inverses 75
BIBLIOGRAPHY 89
BIOGRAPHICAL SKETCH 91
IV
83
Corollary 4.16. (1) If F is a field of characteristic
0 that is not algebraic over the rational field Q, then
FLSJ is semisimple. (2) If F is a field of characteristic
p > 0 that is not algebraic over the prime field GF(p), and
if Sa contains no elements of order p for all a e Pg, then
F[SJ is semisimple.
Proof. Amitsur's theorem [22J and the hypothesis in (1),
or Passman's theorem [22] and the hypothesis in (2) show that
in either case FCS^J is semisimple for each a e P Hence
a 0
the results follow from Corollary 4.14.
Corollary 4.17. Let F be a field. The following con
ditions are equivalent:
(i) F[S] is regular;
(ii) for each a e PQ the group Sa is locally finite and
has no elements of order p in case char F = p.
Proof. By Theorem 4.11 and Corollary 4.12 F[S] is
regular if and only if FlSaJ is regular for all a e PQ. Now,
Villamayor [25J and Connell [ 7 ] have shown that F[SaJ is
regular if and only if Sa is locally finite and has no
elements of order p in case char F = p; hence the result
follows.
Corollary 4.18. Let F be a field. Assume either
(1) char F = 0 or else (2) char F = p > 0 and, for each
a e Pq, Sa has no finite normal subgroup of order divisible
by p. Then the following conditions hold in F[SJ.
7
called the nilpotent radical of R. (N[RJ is also called the
Baer lower radical of R in the literature 19]). It is
characterized by the fact that R/NLRJ is semiprime and N[R]
is the smallest ideal of R that gives such a factor ring.
w
Lemma 1.8. Let R = Â£. .R, be the weak direct sum of
AeA A
the family {Rx^AeA rin9s Then the nilpotent radical of
R is eLanlrx]
We define the prime radical of R as the intersection of
all prime ideals of R, and we denote it by P[R].
Propositon 1.9 [15]. The following conditions con
cerning the ring R are equivalent.
(i)0 is the only nilpotent ideal of R.
(ii)0 is an intersection of prime ideals; that is
P[R] = 0.
(iii)For any ideals A and B of R, AB = 0 implies that
A n B = 0. Under these conditions R is called semiprime.
The following definitions and results can be found in
[ 17 j .
Let a e R. If there exists an element b of R such that
a + b ab = 0, a is said to be right quasiregular (r.q.r.).
The Jacobson radical JLR] of a ring R is defined as follows:
J[RJ = (a e Rar is r.q.r. for every r e R}.
Theorem 1.10. Let R be a ring with identity, and let
{A^i e I} be the set of all the maximal right ideals in R.
Then JLR]
49
By Theorem 3.5 NL(R/P)LSJJ = (Eb^y^jEb^ = 0}. Let
0: RLSJ (R/P)fSl be the canonical epimorphism. Since
6(a) = Ea^x^ e NL(R/P)tSj], then Ea^ = 0. Hence
Ea^ + P = P; that is, Ea^ e P. Since Ea^ belongs to each
prime ideal P, of R, then Ea. e nP = N. Therefore,
A 1 A
N[R[SJ] c {Ea^x^J Ea^ e N}.
Conversely, let $ = Eb^y^, where Eb^ e N. Since R is
commutative, there exists a nilpotent ideal A c N such that
Ebj e A. If p = Ec.x. e R[S], then Bp = (Eby)(Ecx.) =
1 3 3 i 1 1 j 3 J
E Eb.c.(x.y.) = E(Eb.)cy_;. Hence the sum of the coefficients
j i 1 ^ 1 3 j i 1 33
of Bp is (Eb. ) (Ec ) e A. By a similar argument the sum of
i j 3
the coefficients of any element of (B) = (aB + Bp + nB + Ea^Bp. 
a,p,a^,p^ e R[S], n e Z + } belongs to A. Moreover, if 6^ e (B)
i = l,2,...,n, then the sum of the coefficients of 0^62* (5n
belongs to An. Let k be the index of nilpotency of A. Then
the sum of the coefficients of any element of (B) belongs to
k 7k
A = 0. It follows now from Lemma 3.3 that (B) = 0. Hence
(B) Â£ N[R[S]]; and so, {Ea^x^jEa^ e N} c N[R[S]].
Therefore, N[R[S]] = {Ea^x^jEa^ e N}.
We remark that if S is a left zero semigroup and R is a
commutative ring with identity (integral domain), then by
duality Theorem 3.6 (Theorem 3.5) also holds in R[S].
Corollary 3.7. Let S be a right (left) zero semigroup,
and let R be a commutative ring with indentity. Then R[S]
is semiprime if and only if R is a semigroup ring and
I S = 1.
(D) If RtLSj has no nontrivial units, then R has no
nilpotent elements and either R is indecomposable or the
largest subgroup of S containing i is trivial.
Next we let N[R[S]] denote the nilpotent radical of R[S].
We prove the following statements.
(E) If R is a commutative ring with 1 and S is a right
(left) zero semigroup, then N[R[S]] = {Za^x^ e R[S]Za^ = 0}.
(F) If R is a ring with 1 and S is a right (left)
group, then N[R[S]] = NCRlH^J] + C, where H^ is any maximal
n n
subgroup of S and C = R[Hj_]*W with W = { Z a.e.Z a. = 0,
i=l 1 i=l
is an idempotent in S and n e Z+}.
(G) Let R be a ring with 1, and let S be a semigroup
with a universally minimal ideal U. Let 0 be the semigroup
homomorphism from S onto U defined by 0(s) = s*ly. Then
N[R[S]] = N[RlU]] NLKJ, where K = Ker 0 and 0:R[S] *> R[U]
is the natural extension of 0.
(H) If R is a ring and S is a semigroup satisfying
Bauaschewski's condition on extremities, then Rt[S] is semi
prime if and only if R is semiprime.
We say that a semilattice is of type Pq if it has a zero
element w and all the nonzero idempotents are primitive. A
semilattice is said to be of type P^ if it has a zero element
and satisfies the following condition: if F is a subset of
P^, then either g.l.b. (xx e F} belongs to F or else there
exist x and y in F such that g.l.b. {x,y} is in P^ F.
Let S be a semigroup such that (i) S admits relative
inverses and (ii) the idempotents of S commute. If the
semilattice of subgroups Sa of S is of type PQ, then we
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(i)FLSJ is semiprime.
(ii)The center of FlS], ZlFCSjj, is semiprime.
(iii)ZCFL S]J is semisimple.
Moreover, if (2) happens, then these three conditions are
equivalent.
Proof. From Theorem 4.13 and Lemma 4.12, it follows
that ZLFiSJ] is semiprime (semisimple) if and only if
ZLF[SaJ] is semiprime (semisimple) for all a e Pg. Hence
the conclusion follows by applying Passman's results [22,
Theorem 2.12 and Theorem 2.13].
Next we consider a second family of semigroups which
admit relative inverses.
Definition. A semilattice P is said to be of type P^
if the following conditions hold in P:
(i) P has a zero element;
(ii) if F is a subset of P, then either g.l.b. {xxeF}
belongs to F or else there exist x,y in F such that
g.l.b. {x,y} is in PF.
Denote by T the family of all semigroup S with the
following properties: (i) S admits relative inverses;
(ii) the idempotents in S commute; and (iii) S has a semi
lattice of subgroups of type P^.
Let Sex and let {Sa)aep be the family of subgroups of
S. Denote the zero element of P by a^. Let R be a ring
not necessarily with 1. Our goal now is to prove that the
2
(3) S has a unique twosided identity, that we denote
by i, and no other right or left identity.
An element z of S is called a left (right) zero if
zs = z (sz = z) for every s in S. An element z is called a
zero element of S if it is both a left and a right zero of
S. Clearly the foregoing trichotomy holds if we replace the
word identity by zero. S is said to be a right zero semi
group if xy = y for every x,y in S; that is, every element of
S is a right zero and a left identity. Left zero semigroups
are defined dually. A semigroup S with a zero element z is
called a zero or null semigroup if xy = z for all x,y in S.
2
An element e of S is called idempotent if e = e. One
sided identity and zero elements are idempotent. S is called
a band if all its elements are idempotent. By a subgroup of '
a semigroup S, we mean a subsemigroup T of S that is also
a group with respect to its binary operation; that is,
xT = Tx = T for every x in T. By a left (right) ideal of S
we mean a nonempty subset X of S such that SX Â£ x (XS Â£ X).
By twosided ideal, or simply ideal, we mean a subset of S
which is both a left and a right ideal of S. A semigroup S
is called left (right) simple if S itself is the only left
(right) ideal of S. Likewise S is called simple if it contains
no proper ideal. It follows from the definition that S is
right (left) simple if and only if aS = S (Sa = S) for every
a in S; moreover, S is a group if and only if it is both a
left and a right simple semigroup.
If s is an element of S, then the cyclic subsemigroup
generated by s, is the set ( s) of all the positive powers of S.
26
proper zero divisors. This is why, though independent of
of the chapter's title, in section 2.4 we study proper zero
divisors in some twisted group rings and give some sufficient
conditions for their existence.
2.1 Nontrivial Units in Twisted Group Rings
Let K be a field, and let G be a nontorsion free group.
We denote by i the identity of G, and by charK the character
istic of K. Passman [22, Lemma 13.1.1] has proven that with
the exceptions of K = GF(2), g = 2 or 3, and K = GF(3),
G = 2, K[G] has nontrivial units. Our goal in this section
is to extend this result to twisted group rings by showing
that Kt[G] has nontrivial units whenever the twist function
is nonconstant.
Lemma 2.1. Let K be a field of characteristic distinct
from 2. Let G be a nontorsion free group. Then the twisted
group ring K*"[G] always has nontrivial units, provided that
the twist function is nonconstant.
Proof. Since G is nontorsion free and G > 1, it
contains a subgroup H of prime order. Moreover, since any
unit of Kt[H] is clearly a unit of Kt[G], then it is sufficient
to prove the result for G = C^, where is the cyclic group
of order q generated by g.
Let the characteristic of K be p, where either p = 0
or else p is a prime integer different from 2.
64
Proof. This follows directly from Theorem 3.24 and
Corollary 3.25.
Corollary 3.27. Let S = U u J, where J is a right zero
semigroup with zero, and let R be an integral domain. Then
N[ R[ S ] ] = N[R[U]] K (z) .
Proof. Since K(z)*z = z*K(z) = 0, then K(z) is an ideal
of R[J]. By Theorem 3.5 K(z) is nilpotent; hence the result
follows by Proposition 3.26.
So far we have seen that the nilpotent radical of R[S]
is determined by the nilpotent radicals of R[U] and K. If
we want to find out what the nilpotent radical of K looks
like, we need to know how the multiplication among the elements
of the sets J(u) is defined. The classification for the
possible infinite semigroups with a given frame is an open
problem; next we sketch a simple construction due to Clifford
and Miller [ 4 ] that shows there exist an infinite number of
such semigroups, and we find their nilpotent radical.
Let the group U and the semigroup J with zero be given.
By identifying the identity of U with the zero of J and
defining xu = ux = u for all u e U and x e J, we get a
semigroup SQ whose frame coincide with itself. This Sq is a
special case of the following type of semigroup.
Associate with each element u e U a set J(u) in any way
subject to the conditions (iii) and (v) of the Remark 3.20.
For each u e U, determine a right and a left representation
of J by single valued mappings of the set J(u) into itself

