Sep 16, 1986 - other hand, in a nil ring, every element is trivially a product of nilpotents. ... multiplicatively generated by its nilpotents is nil if it is Artinian or if it.
Internat. J. Math. & Math. Sci. (1988) 5-8 VOL. ii NO.
A NOTE ON RINGS WHICH ARE MULTIPLICATIVELY GENERATED BY IDEMPOTENTS AND NILPOTENTS HAZAR ABU-KHUZAM Dp_rtment of Mathematical Sciences University of Petroleum and Minerals Dhahran, Saudi Arabia
(Received June 2, 1986 and in revised form September 16, 1986)
ABSTRACT.
We give the structure of certain rings which are multiplicatively
generated by nilpotents or multiplicatively generated by idempotents and nilpotents.
KEY WORDS AND PHRASES. Boolean ring, Nil ring, polynomial identy. 1980 AMS SUBJECT CLASSIFICATION CODE. 16A70. 1.
INTRODUCTION. On the
In a Boolean ring, every element is trivially a product of idempotents.
other hand, in a nil ring, every element is trivially a product of nilpotents.
This
motivates the study of the structure of a ring, which as a seml-group, is generated
by its idempotents, or is generated by its nilpotents, or more generally, is Indeed, we prove that a ring which is
generated by its idempotents and nilpotents.
multiplicatively generated by its nilpotents is nil if it is Artinian or if it m+l m f(x) (f(x) is a polynomial with integer x satisfies the polynomial identity x
coefficients).
We also prove that if
R
is a ring which is multiplicatively
generated by its idempotents and nilpotents such that the set N
elements is commutative, then
R
forms an ideal of
N
R/N
and
of nilpotent
We
is Boolean.
also give examples to show that our conditions are essential for the validity of our theorems.
We start with the following definitions, the first of which was introduced in [I].
DEFINITIONS.
A ring
by its idempotents.
R
A ring
generated by its nilpotents.
is called an 1-ring if as a semigroup
R
R
is called an N-ring if as a semi-group
R
is generated
R
is said to be an Nl-ring if as a semlgroup
is
R
is
generated by its idempotents and nilpotents.
I].
The following two theorems were proved in
THEOREM A.
Let
R
be an 1-ring with identity.
THEOREM B.
Let
R
be a finite 1-ring.
Then
Then
R
R
i__s Boolean.
i__s Boolean.
REMARKS. i.
A homomorphic image of an 1-ring, N-ring, or an Nl-ring is an 1-ring, N-ring, or
an Nl-ring.
R
R
{0}.
2.
If
3.
Trivially, every 1-ring and every N-ring is an Nl-ring.
is an N-ring with identity, then
H. ABU-KHUZAM
6
An 1-ring need not be Boolean as shown in [I].
4.
An N-ring need not be nil (see
An Nl-ring need not be neither Boolean nor nil (see Example 2
below).
Example
below).
MAIN RESULTS.
2.
In preparation for the proofs of our theorems, we start with the following lemmas. Lemma
is known but we give its proof for completeness.
LEMMA 1.
xm(f(x)) TM PROOF.
(f(x))
m
R
Let
f(x)
polynomial
x
x
m+l
If a
J.
x
m
H
for all
xm+2
xf(x)
xm(f(x)) m
e
x
Then
R.
i__n
f(x)).
m
x
of
R
is nil.
x
xm+If (x)
m
xm(f(x)) TM is an idempotent element in m 0 for (f(x) )m (Lemma I), we obtain x
then the
By Lemma I, x
m
x
2m
2m
is an idempotent.
rin J
x
Continuing we get
satisifes the polynomial identity
0 and since
So
m
R
R
x e J.
Let
PROOF.
in
f(x)
which implies that
Jacobson radical
x (f(x))
idempotent of
is an
m
LEMMA 2.
m
be a ring such that for some positive integer m, and some m m+l x f(x) for all x i__n R.
with integer coefficients, x
J.
So
every
x
is nil.
In the following two
In [lJ, it is proved that a finite 1-ring is Boolean.
Indeed, we prove that an N-ring R m m+l x satisfies the polynomial identity x f(x).
theorems we study the analogous case for N-rings. is nil of
R
PROOF.
R
is Artinian or if
THEOREM I.
R
Let
Let
J
be an Artinian
R/J
{0}, by Remark 2.
J, and hence
R
THEOREM 2.
(m i_s
R.
R
R
Let
be an
f(x)
R
nilpotent.
is
N-rin
is a
R/J
J # R, then
Suppose
(being
is an N-ring with identity (Remark I).
J # R.
This contradicts our assumption that
So
J
is nilpotent, since
positive and
R/J
So
semisimple Artinian) has an identity.
Thus
Then
N-rin.
be the Jacobson radical of
is nilpotent in an Artinian ring. m+l m x f(x) satisfying th___e polynomial identity x
polynomial with integer coefficients).
Then
R
is
nil.
PROOF.
By Lemma 2, the Jacobson radical J of R is nil. R/J being semisimple Each nonR/J is a subdirect product of prime rings R
is semiprime, and hence
R
zero prime ring
of [2], R
has a nontrivial center. m
By Lemma i, e idempotent of
m
satisfies the identity
ca(f(ca))
Ra. ea
#
Let
x
xm+if (x)
# 0
c
m
is an idempotent of m 2m 0, otherwise
and hence by Theorem 1.4.2
R
be a central element of
R
and hence
ca ca (f(ca))m
0
is a central
e
which contradicts the
fact that c a is a nonzero central element of a prime ring and cannot be a zero So divisor by Lemma 2.1.3 of [3]. But e R e x R x 0 for all x a a a a a a e x x 0 for all x and hence R has an identity element. So in R is an N-ring (Remark I) with identity.
R/J
{0}, and R
J
So
R
0
(Remark 2).
R
This implies that
is nil.
We now give an example to show that Theorem
need not be true if
R
is not
Artinian and Theorem 2 need not be true if R does not satisfy the identity m m+l x x f(x). The ring used in the following example was used in [I] to show that an 1-ring need not be Boolean.
EXAMPLE I. x
D
Let
matrices over
be any element of
D
R
be any ring with identity, and let
be the ring of all
in which at most a finite number of entries are nonzero.
R.
Then, for some positive integer
n
and some
nxn
Let
matrix
7
RINGS GENERATED BY IDEMPOTENTS AND NILPOTENTS A
D
over
we have
A
Let
are zero matrices.
n
n
S
O’s
nxn,
is
O’s
A
T
0
are zero matrices.
0
S
It is easy to verify that
is not nilpotent.
is an N-ring which is not nil since
ST.
X
are nilpotent elements, and
T
and
Thus
R
This example shows
0
0
or the hypothesis
that we cannot drop the hypothesis that R is Artinian in Theorem m+l m x f(x) in Theorem 2. that R satisfies the identity x
The following example shows that
Next we study the structure of certain Nl-rings. an Nl-ring need not be neither Boolean nor nil.
Let R
Example 2.
GF(2).
over
is a finite Nl-ring which is neither Boolean nor nil.
R
Trivially,
In example 2 above, the Nl-ring R
elements forms an ideal of
R
potent elements of
R
If
N
R
is an ideal of
R
So
N
R
2
exe)
exe) (xe ex
2
of nilpotent elements
is multiplicatively generated
has nonzero idempotents.
e(xe
Let
Clearly, (ex
R.
exe) (ex
exe)
e
be any nonzero
exe)
N
and
0.
0, and hence
-exexe
2 ex e.
Using induction, (I) implies that
2
n
(2)
(exe)
Let
a e N.
(3)
eae e N
Since
ea
N eae
ex
2
n e
for all positive integers
(2)
Then using
for every
a
N
and
ae
n.
we obtain
N.
N is a subring of eae e N we get
is commutative,
o__f
is Boolean.
is commutative, and the theorem
N
is nil since
So we may assume that
This implies that
(exe)
R
has no nonzero idempotents, then
e(ex
N
R/N
and
idempotent of R and let x be any element of exe) e N. Now, since N is commutative (xe
(i)
of nilpotent
This motivates the study in the
is Boolean.
be an Nl-ring such that the set
Then
by nilpotents only. follows.
R/N
N
commute.
Le__t R
THEOREM 3.
is commutative.
PROOF.
has the property that the set
Indeed, we prove that an Nl-ring will have this property if the nil-
next theorem.
R
and
R
R.
So using (3) and the fact that
8
H. ABU-KHUZAM ea e N, ae
(4)
R
Now since
N
N
Let
x
+ N
that either e
en.
2,
x
N
R.
be any nonzero element of
x e N
e
e.
(4) implies that
is an ideal of
x
and every idempotent
is multiplicatively generated by idempotents and nilpotnets and since
is commutative,
(5)
a e N
for every
x
or
e
e
en
2
R/N.
R
Since
is an Nl-ring, (5) implies
for some idempotent elements
So e
e
en + N
2
(el
+ N)
+ N)...(e n + N),
(e2
and hence
R/N
(6) If
is an I-ring.
R/N, then
is any idempotent element of
R/N.
nilpotent elements of for all
xe
x
ex
exe
(7)
The idempotents of
is Boolean.
R/N
R/N
Now, by (6) and (7),
But in
R/N
R/N
(
)
and
(
e)
has no nonzero nilpotent elements.
are
Thus
and hence
are central.
R/N
is 1-ring with central idempotent elements, and hence
This completes the proof of Theorem 3.
We now give an example to show that Theorem 3 need not be true if the nilpotents of
R
do not commute.
EXAMPLE 3. Nl-ring.
Let
Clearly,
R
te
be the ring of Example i. set
N
Then
of nilpotent elements of
R, being an N-ring, is an R
is not an ideal of
R.
This example shows that we cannot drop the hypothesis that the nilpotent commute in
Theorem 3.
REFERENCES
I.
PUTCHA, M.S. and YAQUB, A.
Idempotents,"
"Rings which are Multiplicatively Generated by Communications in Alsebra 8(1980), 153-159.
2.
HERSTEIN, I.N.
3.
HERSTEIN, I.N. Noncommutative Rinss, The Carus Math. Monographs, John Wiley and Sons, Inc., 1968.
Rings With Involution, University of Chicago Press, Chicago, 1976.
