BULL. AUSTRAL. MATH. SOC.
MOS 1630, 1649
VOL. 2 (1970), 95-99.
A commutativity theorem for rings
D. L Outcait and Adi I Yaqub
Let
ft be an associative ring with identity in which every
element is either nilpotent or a unit.
established.
ideal.
If
The set
R/N
N
is finite and if
x 2 = J/2
or both
then
is commutative.
R
x
and
y
x = y
(mod
N)
R
is an
implies
commute with all elements of
Examples are given to show that
need not be commutative if
for any integer
The following results are
of nilpotent elements in
k > 2 .
"x2 = y2"
The case
is replaced by
N = (0)
"x
N ,
R
= y "
yields Wedderburn's
Theorem.
1.
Introduction
Wedderburn's Theorem, asserting that a finite associative division
ring is necessarily commutative, has recently been generalized in [I].
Our purpose is to extend Wedderburn's Theorem to the case where fl is an
associative (but not necessarily finite) ring with identity in which every
element is either nilpotent or a unit.
alone need not force
are needed.
ideal in
R
The quaternions show that this
to be commutative, hence additional conditions
We prove that the set
N
of nilpotent elements of
R , and then impose conditions on
commutativity of
THEOREM.
R .
Let
R
the set
N
and
R/N
is an
which yield the
Indeed, we establish the following
be an associative ring with identity in which every
element is either nilpotent or a unit in
(a)
N
R
R .
of nilpotent elements in
Then
R
is an ideal;
Received 18 September 1969. The first author was supported by Air
Force Office of Scientific Research Grants AFOSR 698-65 and 698-67, and
the second author by National Science Foundation Grant GP-5929.
95
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96
D.L. Outcalt and Adi I Yaqub
(b) if (i) R/N
2
x
= y
2
is finite, and (ii) x = y
or both
then R
x
and y
(mod
N) implies
commute with all elements of
N ,
is aammutative.
Note that the case
N = (0) recovers Wedderburn's Theorem.
We also show that this theorem need not hold if either hypothesis (i)
or (ii) is deleted.
Moreover, it turns out, perhaps somewhat surprisingly,
"x2 = y2" in (ii) is replaced
that this theorem is not necessarily true if
by
k
k
"x = y "
for any
k > 2
(see examples below).
2.
Proof of part (a). Let
c $ N . Then
a
o)o~1 = a
Then
u
is a unit and
au
= 0 , k
Then
, contradicting the minimality of
and
b d R = ab € N
by (1). Let
and
k .
ba i N .
a - b = u . Suppose
a = u + b . Hence
au~
yn = 0 , and let
u \N .
= J. - y , where
W = l + Y + Y
= 1 - yn = 1 . Hence
w = w au
minimal.
ab = a . Suppose
similarly, we have
a € N , b f N , and set
y = -bu'1 t N
Then
ba
a (. N
Next, let
Let a
1
Thus, upon considering
(1)
a (. N , b (. R , and let
is a unit.
0 = [aCb)c~^~ = (a
Main section
au~
2 +
••• + Y^"1-
is a unit, a
contradiction since, by (l), au~
i N • We have thus shown that
(2)
b (. N => a-b (. N .
a € N
and
Part (a) readily follows from (l) and (2).
COROLLARY.
R/N
Let
R
and
N
be as in part (a) of the theorem. Then
is a division ring.
Proof.
First, by part (a), N
makes sense.
For any
is an ideal in R
a i R , let a = a+N 6 R/N . Suppose that
Then
a \ N , and hence
a
some
a 6 R , and hence
ac = aa = I . Thus
hence
R/N
and hence
is a unit in R . Therefore
a
R/N
a | 0 .
ao = aa = 1
is a unit in
for
R/N , and
is a division ring.
Next we prove two lemmas leading up to part (b) of the theorem.
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A commutativi+y theorem for rings
LEMMA
1.
Let
R
and
•is a commutative subring of
Proof.
(mod
Let
N
satisfy part (b) of the theorem. Then N
R .
a (. N , b (. N . Suppose
N) , b = 0
(mod
a2 = 0 , b2 = 0 ,
N) , a + b = 0
(mod
N) ,
LEMMA
Proof.
(a+1) = 1 , and hence
a = 0
ab + ba = 0 . This follows,
2a = 0 . Therefore
a + 1 = 1
i>a = -ab = ab , a
This proves the lemma.
2. Let
Let
(mod
ab 4 ba . Since
N) , we have by tii),
ab = ba . Similarly, since
2
R , W , R/N
Then every element of
a + b = b
(mod
(a+b)2 = 0 , and hence
since otherwise (ii) would force
contradiction.
97
N
satisfy all the hypotheses of Lemma 1.
commutes with every element of
a € N , b € R , and suppose
N) , therefore by (ii)>
R .
ab ^ ba . Since
(a+b)2 = b2 . Hence
0 = (a+b)(a+b)2 - (a+b)2(a+b) = (a+b)b2 - b2(a+b) = ab2 - b2a .
Therefore,
ab2 = b2a .
(3)
Now,
since
argument to
ab j ba , a(b+l) =(= (b+l)a , and hence we may repeat the above
b + 1
(instead of
b)
to get
a(b+l)2 = (Jb+l)2a . This
equation, when combined with (3), yields
(It)
2(ab-ba) = 0 .
But, by the corollary, R/N
is a division ring which, by til and
Wedderburn's Theorem, must be a finite field of characteristic
Hence pb t N , and thus by Lemma 1,
(5)
a(pb) = (pb)a . Therefore,
p(ab-ba) = 0 .
Now, if p 4 2 , then (h), (5) readily imply
contradiction.
exactly
thus
(6)
Now,
(7)
p , say.
b2
2
Next, suppose
ab - ba - 0 , a
p = 2 . Then the finite field
elements for some integer
k . Hence
(3)2
R/N
has
= 5 , and
- b € N . Therefore, by Lemma 1, we get
a[b2k-b) = [b2k-b)a .
iterative multiplication of both sides of (3) by
b2
yields
ab2k = b*ka .
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D.L. Outcalt and Adi I Yaqub
98
Hence, by (6), (7), we get
ab = ba , a contradiction.
obtained a contradiction whether p 4
2
or
P
=
We have thus
2 • This contradiction
proves the lemma.
We are now in a position to complete the proof of the theorem.
Proof of part (b).
x \ N
and
In view of Lemmas 1 and 2 we may assume that
y \ N . Now, by the corollary, hypothesis (i), and
Wedderburn's Theorem,
R/N
is a finite field, and hence the
multiplicative group of non-zero elements of R/N
% = E, + N
be a generator for
i, j , and some
is cyclic.
Let
R/N , £ € R . Then for some integers
a, a' f N , we have, x = £
t
+a,
y = Zp + a' . Hence,
by Lemmas 1 and 2, xy = yx , and the theorem is proved.
3. Examples
The following examples serve to show that part Cb) of the theorem
need not hold if either of the hypotheses (i), (ii) is deleted.
EXAMPLE
1.
Let
R
be any division ring which is not commutative
(e.g., the quaternions). Here
EXAMPLE
R
satisfies (ii), but (i) fails to hold.
2. Let
a b o
R =
0
a d
a, b, a, d i GF{2)
0 0 a
Here
R
is not commutative, and N
triangular matrices in
consists of the strictly upper
R . Moreover, R
satisfies (i) but (ii.) fails to
hold.
We remark that the equation
•al be replaced by
general
ring
R
p
k,
= y
for any
in (ii) of part (b) cannot in
k > 2 . For, consider the
defined by
I
where
"x
k
"x 2 = y2"
a b a
0 a d
a, b, c, d £ GF(p) , p = prime!
0 0 a)
j
is chosen as follows:
prime divisor of
k , and if
k
if k
is odd take
is even take
p
p
to be any fixed
to be any fixed prime
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99
A commutativity theorem for rings
divisor of
k/2 .
Since
k > 2 , such a prime
readily verified that the ring
(b),
except that
"x2 = y2"
Note, however, that
R
R
p
always exists.
It is
satisfies all the hypotheses of part
is now replaced lay
"xk = yk"
in (ii).
is not commutative.
We observe that the theorem applies to infinite rings and to rings
with non-zero nilpotent elements.
to
GF(p)
An example is furnished by adjoining
an infinite number of commuting nilpotent elements
rii, ^ 2 . ••• • R = GF{p) [rii, H2> — ]
(p
prime).
Another example is
obtained by taking
a. i GF[pk) , i = X
n
0
Here the nilpotent elements consist of the strictly upper triangular
matrices in
R .
It is readily verified that the rings in both of these
two examples satisfy all the hypotheses of the theorem.
Reference
[I]
D.L. Outcalt and Adi! Yaqub, "A generalization of Wedderburn's
theorem", Proa. Amer. Math. Soo. 18 (1967), 175-177-
University of California,
Santa Barbara, California.
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