Internat. J. Math. & Math. Sci.
VOL. 12 NO. 3 (1989) 467-472
467
COMMUTATIVITY OF RINGS WITH CONSTRAINTS
ON NILPOTENTS AND NONNILPOTENTS
MOHAMAD HASANALI
Department of Mathematics
College of Science and Technology
Abudies, Jerusalem, Israel
ADIL YAQUB
Department of Mathematics
University of California
Santa Barbara, California
93106
(Received August 27, 1987)
ABSTRACT.
and n
>
N,
y
Let R be a ring (not necessarily with identity), N the set of nilpotents,
a fixed integer.
Suppose that (i) N is commutative; (ii) If x N and
n
n
N and b
(iii) For a
then x y
0, then [a,b]
R, if n![a,b]
xy
[a,b]
where
ab
generalizes the
ba denotes the commutator.
n
"x --x" theorem of Jacobson.
Then R is commutative.
0,
This theorem
It is also shown that above theorem
need not be true if any of the hypotheses is deleted, or if
"n!"
in (iii) is replaced
by "n".
KEY WORDS AND PHRASES.
Commutator, nilpotent, Vandermonde determinant.
1980 AMS SUBJECT CLASSIFICATION CODE.
I.
16A70.
INTRODUCTION.
A well known theorem of Jacobson [2] states that a ring R satisfying the identity
x
n
x, n
nilpotents.
satisfying
>
is
fixed,
With
the
this
condition
is
as
commutative.
Such rings,
of
course, have no nonzero
motivation, we consider the commutativity of a ring
n
n
is fixed, and
x y
xy for all x E R\N, y
R\N, n >
where N is assumed to be commutative.
That such a ring R need not be commutative can
be seen by taking
R
n= 2.
468
M. HASANALI AND A. YAQUB
This naturally raises the following question:
What extra conditions are needed to
ring R?
Here we give one such extra
As a corollary of our main theorem, we obtain
Jacobson’s Theorem (quoted above).
We also give examples which show that all the
guarantee
the
condition
involving
commutativity of
the ground
commutators.
hypotheses of our main theorem are essential.
2.
MAIN RESULTS.
Our main result may be stated as follows:
MAIN THEOREM.
LetR be a ring (not necessarily with identity), N the set of
nilpotents, and n >
a fixed integer,.
Suppose that (i) N is commutative; (il) If
n
n
N and
x
N and
y
(iii) For a
N, then x y-- xy
b E R, if n![a,b]z0, then
[a,b]=0.
Then R is commutative.
PROOF.
n
x (x
2)
x e R, x
Let
x2)n
x(
x
x
Then
which implies x
(x-xn) n+2
and hence
N.
(x-xn)
n
x
2n+l
2
N, and hence by (ii),
x
n+2
(x-xn)xn+lg(x)
(x-xn) n+l
N for all x
Thus,
O,
N. Since, trivially, this is also true if
N,
x
therefore
n
x- x e N for all x
(2.1)
R.
Next, we prove that
[a,b]
(n!)
0 for some positive integer
Since N is commutative, by
u
a
o, (a
e
(i), to prove (2.2) we may assume that
+ b, (a e N, b
(2.2)
N, b e R).
b
N.
Let
(2.3)
N).
We now distinguish three cases.
CASE I.
Since
{l,...,n}.
ku e N for some k
N and N is commutative by (i), therefore (ku)a
a
(2.3), k(a+b)a
ka(a+b).
0 and hence n![a,b]
Thus, k[a,b]
a(ku) and hence by
0, which proves (2.2)
in this case.
CASE 2.
b + ku
N for some k
E
{l,...,n-l}.
Arguing as in Case I, we see that [b + ku, a] --0.
which implies
(k+l)[a,b]
0,
k +
Hence, [b + k(a+b), a]
n, and thus n![a,b] --0.
in this case.
CASE 3.
ku
N for k
n
and
b
+ ku
N for k
0,
Again (2.2) is proved
l,...,n-l.
COMMUTATIVITY OF RINGS WITH CONSTRAINTS ON NILPOTENTS AND NONNILPOTENTS
N, [see (2.3)],
b
Recall that
(ku)
n
b-- (ku)b
(b+ku)nb
for k
b + ku
N and
b
Similarly, since
n
and
(b+ku)b
n
ku
N for k
.....
n.
b
b
n+l
+
Alb + A2b
n+l
+
2Alb + 22A2b
b
n+l
+
+
+
(n-l)Alb
+
+ 2
+ (n-l) 2
A2b
+ ((n-l)u)
A
where each
b + u
An_
n
n-I
A
Alb + A2b
2Alb +
a
Moreover,
A(b A I)
+
+
n-I b
b
+
n+l
n+l
+ ub n
(2u)n
n-1
bn+l + (2u)bn
b
A
+ ((n-l)u)b
n-I b
n,
(2.6)
+ 2
+ (n-l) 2
A2b
A of
determinant
n-I
An_l
0
b
+ (n-l) n-! A n-I b
+
the
Hence, by (2.4) and (2.6), we get
0
b
An_l
22A2b +
(n-1)Alb
equations in
b
is a sum of terms each of which is a product in which u appears exactly
i
i times and b appears exactly (n-i) times.
The
n
+ (n-l)
b
(2.5)
n-I
n-I in (2.5), we obtain
+
b
(2.4)
I,
I, then k-- 2,..., and finally k
Setting k
Hence by (li),
n.
l,...,n-l, therefore by (ll) again,
N for k
for k
.....
matrix
of
(2.7)
0.
coefficients
of
the
system
Alb A2b,...,An_ib
can be
seen
0, and hence
u
that
A[AI,b]
bn-1
+ bu
A(AIb)
0.
bn-2
and hence
0
A[AI,b]
of
linear
in (2.7) is a Vandermonde determinant, and hence
a product of positive integers each of which is less than n.
it
469
A[u,b n].
0.
Recalling the definition of
+ ...+b n- u,
(2.8)
A similar argument also shows that
Ai,
we see that
M. HASANALI AND A. YAQUB
470
u
Since
a
+ b, [see (2.3)], therefore the above equation yields
A[a,b n]
0, (a
N).
N, b
(2.9)
Combining (2.9) and (2.1), keeping (i) in mind, we se that
A[a,b-b n]
0
A[a,b n]
A[a,b]
A[a,b],
and hence
A[a,b]
0, (a
N).
N, b
(2.10)
Now, combining (2.10) and (2.8), we obtain (taking into account repeated factors of
A),
(n!)
[a,b]
0 for some positive integer
which proves (2.2) in this case also.
Thus completes the proof of (2.2).
Returning to the proof of the theorem,
o
note that if
>
then (2.2)
implies
that
n![a, (n!)
b]
O,
O-I
b]
[a, (nt.)
0, that is,
this process, we eventually obtain [a,b
0 for all
commutative, by (i), therefore,
and hence by
(ill),
[a,b]
0 for all a
E
N, b
(nt.)-l[a,b]
a
g
N, b # N.
0
Continuing
But, since N is
(2.11)
R.
Combining (2.1) and (2.11), we see that
x
x
n
is in the center of R, for all x in
R,
and hence R is commutative, by a well known theorem of Herstein [I].
This proves the
theorem.
COROLLARY I.
Le___t
R be a ring, N the set of nilpotents.,
n
Suppose that (1) N is commutative; (ii) If x
N, then x
b
R,
if
n![a,b]
0, then [a,b]
0.
...and
n
>
a fixed integer.
x; (iii) For
a E N and
Then R is commutative.
As a further corollary, we obtain Jacobson’s Theorem [2]:
COROLLARY 2.
n
x
Let R be a ring and suppose n
Then R is commutative.
>
is a fixed integer such that
x for all x in R.
We conclude with the following examples which show that our Main Theorem need not
be
true
if,
in
hypothesis
(iii), "n!"
is
replaced by "n", or if any one of the
COMMUTATIVITY OF RINGS WITH CONSTRAINTS ON NILPOTENTS AND NONNILPOTENTS
471
hypotheses (i), (ii), (ill) is deleted.
EXAMPLE I.
Let
R
R satisfies hypothesis
that
Observe
GF(4)
a,b,c
(1)
of
and
our Main Theorem,
also
satisfies
But hypothesis (ill) is not satisfied for this value of
n.
However, if n! is replaced by n in hypothesis (iii), then R would satisfy this new
7). This example shows that "n!" cannot be replaced by "n" in (iii).
hypothesis, (n
hypothesis (ii) with n
EXAMPLE 2.
7.
Let
a,b,c
E
GF(3)
It is easily checked that R satisfies all the hypotheses of our Main Theorem except
Hence, (1) cannot be deleted.
hypothesis (i), but R is not commutative.
EXAMPLE 3.
integer.
Note
R
R
that
satisfies
all
hypotheses
the
of
our
Main
be any positive
Theorem
except
Hence, (ii) cannot be deleted.
hypothesis (ii).
EXAMPLE 4.
>
Let R be the ring of quaternions, and let n
Let
i(0
0
i
0
i),
0,I
GF(2)
n=2.
It is readily verified that all the hypotheses of our Main Theorem are satisfied
hypothesis (iii).
Hence, (iii)
cannot be deleted, since R is not commutative.
REFERENCES
i. HERSTEIN,
I.,
A generalization of a theorem of Jacobson, I., Amer. J. Math, 73
(1951), 755-762.
2. JACOBSON, N., Structure of rings, A.M.S. Colloq. Publ. (1964).