I nternat. J. Math.
Math. Sci.
121
Vol. 2 (1979) 121-126
RINGS WITH A FINITE SET OF NONNILPOTENTS
MOHAN $. PUTCHA
Department of Mathematics
N.C. State University
Raleigh, N.C. 27607
and
ADIL YAQUB
Department of Mathematics
University of California
Santa Barbara, California 93106
(Received October 16, 1978)
ABSTRACT.
Let
n
Let
be a ring and let
be a nonnegative integer.
of elements in
proved:
R
If
R
R
N
denote the set of nilpotent elements of
The ring
R
is called a
which are not in
N
is at most
@ -ring, then
n
R
is nil or
is a
is a nil ring or a finite ring, then
R
is a
n.
R
@ -ring if the number
n
The following theorem is
is finite.
8 -ring for some
n
Conversely, if
n.
The proof of
this theorem uses the structure theory of rings, beginning with the division ring
case, followed by the primitive ring case, and then the semisimple ring case.
Finally, the general case is considered.
KEY WORDS AND PHRASES.
Semigroup.
R.
Nil ring, Division ring, Primitive ring, Semisimple ing
R
M. S. PUTCHA & A. YAQUB
122
Prary 16A48, Secondary 16A22.
AMS (MOS) SUBJECT CLASSIFICATION (1970) CODES.
i.
INTRODUCTION.
Let R be a ring and let N denote the set of nilpotent elements of R.
a nonnegative integer.
suggests itself is the following:
a given ring R to be a
2.
0 -ring if the number of elements
n
The ring R is called a
in R which are not in N is at most n;
Let n be
IR/NI
that is,
< n.
One question which
what are necessary and sufficient conditions for
The answer is given in the following.
0 -ring?
n
MAIN RESULTS.
THEOREM I.
If R is a
n
-ring, then R is nil or R is finite.
R is a nil ring or a finite ring, then R is a
n
Conversely, if
-ring for some integer n.
In the proof of Theorem I, we use the structure theory of rings, beginning with
the division ring case, followed by the primitive ring case, and then the semisimple
ring case.
Finally, we consider the general case.
In preparation for the proof of Theorem I, we first prove the following lemmas.
LEMMA I.
Any subring and any homomorphic image of a
This follows at once from the definition of a
LEMMA 2.
direct sum R
R
1
PROOF.
RI,R2,...,Rn+ 1
Rn+ 1
2
Let u
I
is not a
n
-
n
i
has an identity.
(0,I,0,...,0),
1
R
2
Un+ I
+
Rn+ 1.
(0,0,
Then u
1,2,...,n + I, where N denotes the set of nilpotent elements of R.
and thus R
I
$ R
2
LEMMA 3.
Rn+ 1
R
Rn+ I}\N
2
is not a
Then the
-ring.
the direct sum R
I{R 1
-ring.
0 -ring.
n
be rings where each R
(I,0,0,...,0), u 2
u.. is an element of
where each
each i
Let
8 -ring is a
n
0,I),
i
’N
for
Hence
> n
8-ring.
n
If D is a division ring which is also a
This follows at once from the difinition of a
8 -ring, then D is finite.
n
8-ring.
n
123
RINGS WITH A FINITE SET OF NONNILPOTENTS
LEMMA 4.
If R is a primitive ring which is also a 8 -ring, then R is finite.
n
In fact, R is isomorphic to a complete matrix ring D
m
PROOF.
Suppose R
over a finite division ring D.
for any division ring D and any positive integer m.
D
m
(We shall show that this leads to a contradiction.)
Then, by Jacobson’s Density
Theorem [I; p. 33], for each positive integer q, there exists a subrlng R of R
0
which maps homomorphlcally onto D
exists a subring R
for some division ring D.
I of R which maps homomorphically onto
Dn+l
Now, define
q
AI,A2,...,An+ I
a
is
Dn+ I.
In particular, there
Hence, by Lemma I,
(2.1)
0 -ring.
n
to be the following (n
+ I)
(n + I) matrices in
Dn+l:
A. has zero entries in all rows except row i,
but A. has entries of I throughout row i
(i
1,2,...,n + I).
It is readily verified that each A. is idempotent and thus each A is not nilpotent.
i
l
Hence,
fore,
R
D
IDn+I\NI > n, where N denotes the set of nilpotent elements of Dn+ I. ThereDn+ 1 is not a 8n-ring, which contradicts (2.1). This contradiction shows that
m
for some division ring D and some positive integer m.
ring D is a subring of the
8n-ring Dm,
Now, since the division
D is finite, by Lemmas 1 and 3.
Thus, R is
finite, and the proof is complete.
LEMMA 5.
PROOF.
If R is a semisimple ring which is also a 8 -ring, then R is finite.
n
Thus, suppose that R is a semisimple ring which is
By contradiction.
also a 8 -ring, and suppose R is not finite.
n
ideal I (eel) of R such that
[I; p. 14]
e I--
(0)’,
each
R/Ie
is primitive
(2.2)
Hence, by (2.2) and Lemma 4,
By Lemma I, R/I ( is a 8 n -ring.
R/I
Since R is semisimple, there exists
a
Dm
R/I
is finite, for any e
(2.3)
M. S. PUTCHA & A. YAQUB
124
,
Now, choose s I
and having chosen
k
n I
s.i
i=l
k
-= RI
’RII
s.
i=l
i
choose
k
n
such that
e
Sk+ 1
I
i= I
so that
s1,...,sk
I
s.i
is proved as follows: suppose no such
(2.)
That such
Sk+ I
Sk+ I
Sk+ 1
can always be so chosen
k
Then (0)
exists.
n
s
Is
n
i=l
I
si
and hence (see (2.4))
k
R/
R
k
---_’R/I
I
0
si
i 1
i= 1
si
Thus, using (2.3), we see that R is finite, a contradiction. This contradiction
k
shows that there exists
such that
I
I
Now, as we can see from
Sk+ I
(2.3), R/I
Since, moreover,
is simple.
Sk+ I
s.i
i=l
k
i= I
IS
Sk+ I
I
i
Sk+ 1
we have
k
+ IS
IS
i=l
R.
Hence, by applying the second isomorphism theorem, we readily
k+l
i
verify that
RI
k+l
n I
s.i
i=l
by (2.4).
-
k
R/
k+l
R/l
I
0
i=l
s1
Sk+l
-= Z’R/Is
i=l
i
In particular, we have
n+l
n+l
’R/l
R/ n
si
I
i=l
si
n+l
Hence, using Lemma I,
(see (2.3)).
E" R/Is is a 8 n -ring. This, however, contradicts Lemma 2
i=l
1
This contradiction shows that R is finite, and the lemma is proved.
We are now in a position to prove our main result, stated at the beginning.
PROOF of Theorem I.
of R.
Let R be a
We claim that J is nil.
not nil.
8n-ring,
and let J denote the Jacobson radical
We prove this by contradiction.
Then, for some a, a e J, a
Thus, suppose J is
N, where N denotes the set of nilpotent
RINGS WITH A FINITE SET OF NONNILPOTENTS
elements of R.
Let
2
{a,a ,a 3 ,...,an+l }, (aJ, aN)
S
i
N, a
Since a
125
N for each i
1,2
ISI
since R is a 8 -ring.
n
<
,n
IR\NI
+ I,
and hence S
=
R\N.
Therefore,
< n
(2.6)
In view of (2.5) and (2.6), we see that a
i
aj
for some
distinct positive integers i, j, and hence some power of a is idempotent, say a
Thus, a
k
is an idempotent
element of J, and hence, as is well known, a
0.
This contradiction proves that J is nil.
Therefore, a e N, a contradiction.
Now, if J
k
k
# R.
So suppose J
R, then R is nil.
Then R/J is a semisimple
Moreover, by Lemma i, R/J is a 8 n -ring, and hence
We have thus shown that R/J is a finite semisimple ring
ring with more than one element.
by Lemma 5, R/J is finite.
with more than one element, and hence, as is well known,
u be the identity element of
+
N
u
{n + u
R/J has an idenitity.
Let
R/J, and let
neN}; N
(2.7)
is the set of nilpotents of R.
We claim that
(N + u) n N
For, if a e (N + u) n N, then a
a
n
+
u
N and a
an invertible element of
R/J.
(2.8)
n
+
u for some n
N.
Hence
Thus, a is an invertible element of R/J,
which contradicts the hypothesis that a e N (and hence a is nilpotent).
contradiction proves (2.8).
since R is a 8
n-ring.
This
In view of (2.8), we have N + u c_q R\N, and hence
IN + u < IR\NI <_ n
But INI
IN + ul
(2.9)
and hence by (2.9)
N is finite.
net result, then, is that both N and R\N are finite, and hence R is finite.
have thus shown that, in any case, R is nil (if J
The converse part of Theorem 1 is trivial.
The
We
R) or R is finite (if J # R).
M. S. PUTCHA & A. YAQUB
126
We conclude with the following
REMARK.
The analogue of Theorem 1 is not true for semigroups.
To see this,
let N be an infinite nil semigroup, and let 1 { N be an identity element; that is,
1
Let R
x
N u {i}.
x
1
x
for all
x
N,
in
Then R is a semigroup.
Also,
and
1
IR\NI
1
1
i.
But R is neither nil
nor finite.
REFERENCE
I. Jacobson, N. Structure of Rings, rev. ed., Amer. Math. Soc. Cpll.oq..Publ.,
Vol. 37, Providence, R.I., 1964.