Internat. J. Math. & Math. Sci.
VOL. 13 NO.
(1990) 87-92
87
ON COMMUTATIVITY THEOREMS FOR RINGS
H.A.S. ABUJABAL and M.S. KHAN
Department of Mathematics
Faculty of Science
King Abdul Az[z Unlvers[ty
P.O. BOX 9028, Jeddah- 21413
Saudi Arabia
(Received February 2, 1989)
Let R be an associative rlng with unity. It is proved that if R satisfies
0 (m
x]
I, n > I), then R is commutative.
Ohe polynomial identity Ix n y
ABSTRACT.
ymxn,
Two or more related results are also obtained.
KEY WORDS AND PIIRASES.
Commutative
rings,
torsion
free
rings,
center
of a ring,
commutator ideal.
16A20, I16A70.
1980 AMS SUBJECT CLASSIFICATION CODES.
I.
INTRODUCTION.
Throughout this paper, R will be an associative ring, Z(R) the center of R, N the
set of all nilpotent elements of R, N’ the set of all zero divisors of R, and C(R) the
commutator ideal of R.
Ix,y]
xy
For any pair of elements
,
y in R, we set as usual
yx.
Recently, generalizing some results from Bell [1] Quadrl and Khan [2,3]) proved
that if R is a ring satisfying the polynomial identity [xy
(m >I, n > I), then R is commutative.
ym x m,
x]
0
In [4], Psomopoulos has shown that an s-unital
[xny y x,x] 0 (m > I, n > 0) holds, must
ring R in which the polynomial identity
be commutative.
In this paper, motivated by the above polynomial identities, we intend to prove
results on commutativlty of a ring R with unity satisfying the fo]lowin property:
m n
n
0 for
x]
such that Ix y- y x
and n >
(i) "there exist positive integers m >
all x,y in
R".
Our property (i) can be regarded as an amalgam of those constder,.t by the above
authors.
H.A.S. ABUJABAL AND M.S. KHAN
BB
2.
PRELIMINARIES.
In preparation for the proof of our results, we first state the following wellknown results.
LEMMA
(Psomopoulos [4]).
2.|.
ix k, y]
positive Integer k,
,
Let
y e R.
(Nicholson and Yaqub [5]).
k
0
that for some positive integer k, x y
LEMMA 2.2.
(Bell [6]).
LEMMA 2.3.
Indetermlnates
commuting
ix,y], x
If
0, the for any
xk-l[x,y].
k
Let
Let R be a ring with unity I.
k(x+|)
y for all x, y In R.
Suppose
The y
0.
be a polynomial [dentlty in a finite number of noni,tegral
with
Then
coefficients.
the
are
following
equivalent
For any ring R satisfying
0, C(R) is a nll ideal.
(ll) For every prime p, (GF(p)) 2 fails to satisfy f
0.
(i)
(Tong [7]).
LEMMA 2.3.
let I (x)
It(x).
3.
Ik_l(X+l)
0 for j
>
Let R be a ring with unity I.
Ik_l(X).
Then I
r
r_l(X) =I/2(r-l)
Let
It(x)
o
r
r! + r!x" I (x)
r
x
r
If k
r!
>
and
r.
RESULTS.
Throughout the rest of the paper, R stands for a ring with unity I, and satisfies
the property (I).
Let us first note that for any x, y in R, the property (i) can also
be expressed as:
n
x ix,y]
ix,
ym
x
n
(3.1)
Then for any positive integer t, we obtain
x
tn
ix,y]
[x,ym
x
2
x
(t-|)n
2n
x
ix,
ym
(t- 3)n
[x,
x
n
ym
x
(t-2)n
3
x
3n
By repeating the above process and using (3.1), we get
x
tn
[x,y]
[x,
ym
t
x
tn
(3.2)
We also need the following two results for the proof of our main theorem.
LEMMA 3.1. Let R be a ring with unity which satisfies the property (P).
Z(R).
PROOF.
Let u e N.
Then N
Then by (3.2) for any x e R and a positive integer t, we have
xtn[x,u]
ix, um
x
tn.
t
But
we
have
u
0, for sufficiently large t.
0 for all x in R.
Therefore, x ix,u]
0 for all x in
Then we have (x+l) tn ix,u]
R. By Lemma 2.2, this implies that [x,u]
0, which forces N Z(R).
LEMMA 3.2.
Let R be a ring with unity which satisfies the property (i). Then
C(R) Z(R).
tn
as
a
nilpotent
element,
then u
89
COMMUTATIVITY THEOREMS FOR RINGS
PROOF.
n
Replacing x by (x + 1) in (3.1) and multiplying both sides by x on the
right and again using (3.1), we get
n
(x+l) n [x,y] n
x [x,y] (x+l)
n,
(3.3)
for x,y e R.
Define
I),
/o
Let x
Eli
El2 + E12
+
for a prime p.
C(R)
Z(R).
and y
0
and E
2
( 0ol-
Then x and y fail to satisfy (3.3) in (GF(p)) 2,
Eli.
So by Lemma 2.3, C(R) is a rill ideal, and hence by Lemma 3.1,
This ends the proof.
In view of Lemma 3.2, it is guaranteed that the condition of Lemma 2.1 holds for
each pair of elements x, y in a ring R wlth unity which satisfies the property (i).
satisfying property
THEOREM 3.1. Let R be a rlng with unity
Then R Is commutative.
(i)
PROOF.
subdlrect
So R is isomorphic to a
We are given that R is a ring with unity.
sum of subdlrectly irreducible rings R (i e I), each of which as a
Thus we may assume that R is a
homomorphlc image of R, satlsfing the property (i).
subdirectly irreducible ring satisfying (I).
Let S be the Intersection of all non-zero ideals of R i.
Then clearly S
$
(0).
(x+l) Ix,y]
in the polynomial identity (3.1), we obtain [x,y-y m]
Now, if n
R. Thus R is commutative by
0 for all x,y
[x,y m] (x+l)
[x,ym] x
x[x,y]
Herstein [8, Theorem 18].
Henceforth we assume n
>
Consider the positive Integer k
I.
k x [x,y]
pmxn [x,y]
pxn[x,y]
n
n
m
p [x,y m] x
Ix,
(py)m]
n
[x,(py)]
x
p, where p is
Then by (3.1) we get
a positive integer greater than I.
n
p
p x Ix,y]
x
n
xn[x,
x
n
(py)]
[x,(py)].
xn[x,y)
O. Now combining
O, whclh on replacing x by (x+l) yields k Ix,y]
k-I
k
0.
[x,y]
Therefore, it follows
k x
we have Ix ,y]
Lemma 3.2 with Lemma 2
Thus k
that
Now, replaclng y by
k
x
E
ym
in (3.1) we get
x
n
Z(R) for all x In R.
Ix, ym
Ix (ym
n
m
x
(3.4)
(3.5)
Then by Lemma 3.2, we have
x
n
[x,
ym
[x,
ym
x
n
(3.6)
But
[x,ym]
m y
m-I [x,y].
(3.7)
H.A.S. ABUJABAL AND H.S. KHAN
90
So using (3.6) and (3.7) we obtain
n
[,
ym
m y
m-I
Ix, ym
x
n
and
xn
[x,(ym)m]
m(ym) m-l[x,y m]
m y
Thus (3.5) gives
my
m-I (m-l)
n
[x,
y
m-I (1-y (m-l)2) [x,y m]
x
2
xn=
ym
x
n
(3.8)
O.
ym-I
(m-1)
2
[x,y m] (x+l)n=0. So
(l-y
by (x+l) in (3 8). Then we get m
2
m-I (l-y (m-l)
0.
Therefore, by Lemma 2.3 of Quadrl and
[x,y m]
by Lemma 2.2, m y
Let us replace
Khan
[3],
we have
my
Now, let u e N’.
(3.9) we obtain
m-1 (l
Then by
_yk(m-1)
(3.4), u
2
[x,ym]
k(m-l)
(3.9)
O.
2
e N’
Z(R), and S u
k(m-1)
2
(0).
Hence using
2
m-I Ix, u m (I _uk(m-l)
0.
k(m-l )2
k(m-1)’
E N’ and so S=S (l-u
0, then (l-u
mu
If m u
m-I
[x,u m]
(0).
a contradiction as S
m u
Now, from (3.1)
m-1
Therefore
Ix, u m
0.
we have
x
2n
Ix, um
Ix,u]
=mu
m u
=mu
This implies that x
u
2n
[x,u]
Therefore, N’
Z(R).
(0), which gives
2
x
m(m-
2n
Ix
u
2
m
x
2n
m-1 u (m-l) [x,um] x 2n
2
(m-I) 2n
m
m-I
Ix,
u
u
x
Hence by Lemma 2.2, we obtain [x,u]
0.
0, that
is
Z(R).
km
Clearly, for any x e R, xk and x
Z(R).
Then by (3.1) for any y e R, we have
the identity
(xk-xkm)x n
[x,y]
(xk
x
xk(x n [x,y]) xkm (xn[x,y])
n
(xkm [x,y]) xn
x (xk[x,y])
n
[x,(xky) m] x n
x [x,xky]
N
n
X [x,xky].
x [x,xky]
Therefore,
(-x t)
where t
km
x
n
[x,y]
xS[x,y]
km-k+l, and s=n+k-l.
0,
0, and
(3.to)
91
COMMUTATIVITY THEOREMS FOR RINGS
Now,
if
x-
t
xS[,y]
N’
O.
0, then Lemma 2.2 yields ix,y]
But
xS[x,y]
0 gives
Z(R).
Therefore, ix
Theorem 18 of [9]
t,
0, for all x, y in R, which implies that ix,y]
O.
Thus in evey case ix,y]
x
y]
0, by
This proves that R is commutative.
THEOREM 3.2.
Then R is
Let R be an s-unltal ring satisfying the property (i).
commutative.
PROOF
This
follows
of Hirano, Kobayashi and Tomlnaga
from Proposition
since R with unity satisfying (i) is commutative by Theorem 3.1.
[9]
Finally, we present
a short and easy proof of Theorem 3.1, but under an extra condition on the commutators
We use an Iteration technique as given in Tong [7].
Let R be a ring with unity satisfying the property (1).
commutator in R is m!
torsion free, then R must be commutative.
PROOF. The ring R satisfies the identity
in the ring R.
THEOREM 3.3.
x
n
[,y]
ix,
We shall apply the iteration on
ym
x
ym.
As in [7], let
n
l, m
n
>
If every
I).
lj(y)
IT(y)_
for
0,I,2,...
Then the above identity can be rewritten as
x
n
ix y]
Replacing y by (y+l) in
x
n
n
I (y)] x.
ix
(3.11),
(3.11)
we obtain
[x, I o (y+l)] x n
[x,y]
Now, using Lemma 24, we get
n
x [x
Ix, I o (y) +
y]
ll(Y)]
x
n
(3.12)
Equations (3.11) and (3.12) when combined, give
0
Ix,
ll(Y)]
Again, let y=y+l In (3.13).
0
[x,
x
n
(3.13)
Then using Lemma 2.4 we have
12(Y)]
x
n
Repeating the above process (m-l) times, we reach the identity
0
(y)] x n
Ix, I
With an application of Lemma
2.4,
m![x,y] x n
(3.14)
we end up with
0.
Now replacing x by (x+l) in the above identity, and making ue of Lemma 2.2, we have
m![x,y]
0, for all x,y e R.
By every commutator in R is m!-torslon free, so ix,y]
Therefore, R is commutative.
This completes the proof.
0 for a11 x and y in R.
92
H.A.S. ABUJABAL AND M.S. KHAN
REFERENCES
I.
2.
B.
4.
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