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This is to certify that the

thesis entitled

Radicals in Standard Rings

presented by

Larry J. Zettel

has been accepted towards fulfillment
of the requirements for i

_P_h_._D_._degree inMflIhfimaliCS

 

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Major professor

 

Date 77?“‘1 [4-,IQ70 i
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0-169

 

ABSTRACT
RADICALS IN STANDARD RINGS
By
Larry J. Zettel

A standard ring is a non-associative ring,0l. , which
satisfies:

(1, y, z) + (z, x, y) - (x, z, y) = 0;

(x, y, wz) + (w, y, xz) + (z, y, wx) = 0; and

(x, y, x2) = O
for all w, x, y, z in a, where (a, b, c) = (ab)c - a(bc).

We consider standard rings 0!. for which if 16 Q, a
su‘bring of 02,, then there exists a unique y 6 3 fdr which
2y = x0

In this paper we consider various radicals for standard
rings.

A radical property is a prOpertwahich a ring may
possess which satisfies the following:

1, A homomorphic image of a ring which has propertyp

also has property 51
2s Every ring contains a P-ideal which contains all
other P—ideals. This ideal is called tbsp-radical.
3c Any ring modulo its P-radical has zero P—radical.

We begin by deriving the basic identities which we will
use throughout the paper. We then prove theorems which are
applicable for any choice of a radical property. Finally, we
list known results for the nil, Behrens, and Smiley radicals

for general non-associative rings.

Larry J. Zettel

In chapter 2, we define a prime ideal and show the
existence of a prime radical in a standard ring. We charac-
terize the prime radical of a standard ring CZLas the mini-
mal ideal 0 such that 01/0. has no non-zero nilpotent ideals.
We show that if the standard ring has a maximum nilpotent
ideal, then it is equal to the prime radical.

In chapter 3, we prove the equivalence of local solva-
bility and local nilpotence in a standard ring. This leads
t6 the existence of a maximal locally nilpotent ideal,
called the Levitski radical. We show that the Levitski
radical contains the prime radical and is contained in the
nil radical. Also, we show that the Levitski radical con-
tains all locally nilpotent one-sided ideals.

In chapter u, we give two generalizations of the
Jacobson radical for associative rings. The first stems
from the definition of an element, x, as being quasi-
regular if there exists an element y such that x + y - xy =
O, and follows that for Jordan rings. The radical obtained
by this process is called the Jacobson-MacCrimmon radical.
The second generalization is that of Brown's, and is true
for any non-associative ring. This is called the Jacobson-
Brown radical. We show that in a standard ring, it makes no
difference if the Jacobson-Brown radical is defined in terms
of left or right ideals. we show that the Jacobson-
MacCrimmon radical contains the nil radical, and is con-
tained in both the Behrens and Jacobson-Brown radicals.

Finally, we show that the Smiley radical contains all other

Larry J. Zettel
radicals considered.

In chapter 5, we consider the relationship of the
different radicals under various chain conditions. If the
ring satisfies the descending chain condition on right
ideals, then the prime and Levitski radicals are equal, as
well as the Jacobson-Brown and Smiley radicals. If every
subring of the ring satisfies the descending chain condi—
tion on right ideals, then in addition to the above, the
nil, Jacobson-HacCrimmon, and Behrens radicals are equal.
If the ring satisfies an ascending chain condition on sub-

rings, then the prime, Levitski, and nil radicals are equal.

RADICALS IN STANDARD RINGS

By
Larry Joseph Zettel

A THESIS
Submitted to
Michigan State University

In partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics
1970

ACKNOWLEDGEMENTS
I am extremely grateful for the suggestions, assis-

tance, and patience given by Professor C. Tlai during my

work on this problem.

ii

TABLE OF CONTENTS

Introduction

Chapter 1. Foundations

Chapter 2. The Prime Radical
Chapter 3. The Levitski Radical

Chapter h. The Jacobson-HacCrimmon and the Jacobson-
Brown Radicals.

Chapter 5. Chain Conditions

Bibliography

iii

Page

10
25
no

51
SS

INTRODUCTION

A standard ring 01 is a non-associative ring satis-
fying:

1) (x, y, z) + (z, x, y) - (x. z. y) = 0;

2) (x, y, wz) + (w, y, xz) + (z, y, wx) = O; and

3) (x, y, 12) = 0
for all x, y, z, w in 01, where (x, y, z) = (xy)z - x(yz).

It is easy to compute that in any non-associative
ring, the following identity holds:

(x,y,z) + (z,x,y) - (x,z,y) = [xy,z] - [x,a]y - x[y,a]
where lx,y] =axy - yx. Thus identity(l’is equivalent to:

h) [xy,z] = [x,i]y + xfy,z].

That is, for each 2 in 0L, the mapping xq[x,z] is a deriva-
tion of fit.

Identity (3) is known as the Jordan identity, for a
commutative ring which satisfies identity (3), is called a
Jordan ring. Also, identity (3) is equivalent to identity
(2) unless 0!. possesses an element a a‘ 0 such that
a + a + a = 0. Thus it is clear from the definition that
any associative ring and any Jordan ring are standard rings» A

In addition throughout the paper we shall assume that
if 3 is any subring of a and x 68, then there exists a

unique y 58 such that 2y = x.

2

The concept of a standard ring was first introduced
by Albert [1] where he considered a finite dimensional
standard algebra over a field. He showed that a finite
dimensional standard nil algebra is nilpotent and in a
finite dimensional standard algebra over a field of
characteristic # 2, an ideal is solvable if and only if it
is nilpotent. Then he called the maximal solvable ideal
of 01, the radical of 0! and showed that if’dl is any
finite dimensional standard algebra over a field of charac-
teristic if 2, then the quotient algebra al/R where R is the
radical of 01, is semisimple. Furthermore, any semisimple
standard algebra is a finite direct sum of simple ideals.

Kleinfeld [7] showed that a simple standard ring is
either associative or commutative Jordan. Thus, combining
these, we see that a finite dimensional standard algebra
modulo its radical is the finite direct sum of simple
associative or commutative Jordan algebras.

The problem we consider is derived by removing the
condition of an algebra over a field. Hith the relaxation
of this condition we have various choices for the radical
of the ring. All of our choices will be made so as to
furnish information about the structure of the ring in the
following manner. We introduce the radical as an ideal such
that the ring modulo its radical has zero radical. Then we
characterize rings which have zero radical. Since we have
dropped Albert's condition of a finite dimensional algebra,
we can no longer hope for a result such as a finite direct

sum of simple rings. However, we are able to represent

3

rings with zero radicals as subdirect sums of certain types
of rings. Thus we study the structure of a standard ring a,
by characterizing an ideal I, then investigating 62/1.

Several radicals have been studied for associative rings
and we can show the existence of similar ones in.a standard
ring. The prime radical, Q(0[), is the intersection of all
prime ideals (Chapter 2). The Levitski radical, Md), is
the maximal locally nilpotent ideal (Chapter 3). Corres-
ponding to the Jacobson radical of associative rings, we
have two radicals for standard rings. One, JH(0[), follows.
that considered by HacCrimmon [6] and Tsai [11] for Jordan
rings, and the other, JB(01), has been introduced by Brown[3]
for a general non-associative ring. In addition, there are
other radicals which have been studied for arbitrary rings:
3(a). the Smiley radical [10]; 3(a), the Behrens radical
[2]; and l(0l), the maximal nil ideal or nil radical [2].
For an arbitrary standard ring, the relationship of these is
shown by the following diagram where each radical is contained

in those above it in the diagram.

3(a)
/ \JB(a)
\Jn(a)/

B(0L)

h

When various chain conditions on the ideals of 61 are

present, we have the following.
If 01 satisfies a descending chain condition on right

ideals, then:
stat.) = JB(OZ)
B(dZ)
mm)

N(0£)

Ltoz.) = QM.)

If 0! satisfies a strong descending chain condition [9],
that is each subring of OZ satisfies a descending chain con-
dition on right ideals, then:

3(a) = JBML)

B(dll = JH(€R) = N(61)

Lm.) = am.)

1:02 satisfies an ascending chain condition on sub-

rings, than:

Mm) = Mon) = QM!)

 

Chapter 1. Foundations

A. Basic Identities

We begin by deriving some identities which will be used
in the remainder of the paper. Recalling the defining
identities, (1-3), we see that by setting x = z in (1) we
obtain that a standard ring is flexible and this combined
with (2) gives that a standard ring is a non-commutative
Jordan ring, in particular it is power-associative.

Using flexibility, we may rewrite (2) as

5) (wx,y,z) - (w,y,xz) + (wz,y,x) = O or as

6) (wx,y,z) + (xz,y,w) - (x,y,wz) = O.
Interchanging x and a in (2) and subtracting, we see that

7) (w,y,[x,z]) = 0.
Using (7) and flexibility on (6), we see that

8) (x,y,zw) - (xz,y,w) + (z,y,xw) = 0.
Now by considering (6) and (8) as operations of left and
right multiplications, we have:

9) Ry(xz) = Rnyz + Rx(Ryz - Rsz) + 32(Ryz'Rny) and

10) L(xz)y = Lnyz * Lx(Lzy - LzLy) + Lz(ny~Lny).

Identity (1) considered as operating on 3 gives:
- + - - =
11) ny Lny Rny ny Lny + Rny O.
B. General Radical Theory
Before studying standard rings, we discuss some general
properties of radicals. Let P be a preperty which a ring

may possess. A ring is a P-ring if it possesses property P.
An ideal I of at is aP-ideel if I is aP-ring. A ring

5

 

6
which does not contain any nonpzero FLideals is itsemisimple.
"is called a radical property if the following hold:

1) A homomorphic image of aP-ring is a P-ring.

2) Every ring R contains a P-ideal S which contains

every other P-ideal of the ring.

3) The quotient ring, R/S, is P-semisimple.

We will use two methods for specifying a radical
preperty. The first is to explicitly define the property.
This will be done in Chapters 3 and h. The second is to
begin with a class of rings N and construct a new class of
rings N'as follows. A ring is said to be of degree one over
N if it is zero or a homomorphic image of some ring in N. A
ring, R, is said to be of degree two over N if every homo-
morphic image of R contains a non-zero ideal which is a ring
of degree one over N. For any ordinal number [3, if fi-l
exists, a ring is said to be of degree 3 over N if every
non-zero homomorphic image of R contains a non-zero ideal
which is a ring of degree 5-1 over N. Iffi is a limit
ordinal, then R is of degree 3 over N if it is of degree
over N for some “(fl. If N is the class of all rings which
are of any degree over N, then define a ring to be 3 Pi]- ring
if it belongs to N. It can be shown [11,, p. 12] that pfi is a
radical preperty. The radical determined by this property is
called the lower radical determined by the class N. We shall
use this approach at the end of Chapter 2 with N as the class
of nilpotent rings. We shall show that this process yields
the same result as the prime radical and thus in this special

case is a radical preperty.

7

Using the concept of a radical property, we can avoid

reproving the same theorems several times.

Theorem 1.1: If Rad(m) is the P-radical of OZ and if

 

I is any ideal of OZ such that “/1 is P-semisimple, then
IQ Rad(oz).
2.13292:
Let R = Rad(01). Since (I + R)/I i; isomorphic
to R/(IDR), it is aP-ring by preperty 1) and thus is
aP-ideal of 01/1. However, since “/1 is P—semisimple,
(I + R)/I = 0, that is, 12 a.

We will characterize semi-simple rings as subdirect
sums. A subdirect sum is a subring of the complete direct
sum such that the natural projections are onto each co-
ordinate summand. The following theorem which we use to

obtain our representations is well-known [13,, p. 6h].

Theorem 1.2: A ring 0L is isomorphic to a subdirect

 

sum of rings 01‘.‘ if and only iffll contains a class of
ideals {8‘} such that max = o and m/gfi sag.

Theorem IQ: A subdirect sum ofP-semisimple rings is

 

‘lsemisimple.
m:

Let ”L be a subdirect sum of ”(it where each 01“
contains no non-zero p-ideals. Let 77“ be the pro-
jection map onto 0L4. Let R be the p-radical of 0L.
Then 77“(R) is the homomorphic image of aP-ideal and
thus is ap-idesl of a“ . Thus 77“(R) = O for cachet and

8
so R = 0 and 01. is P-semisimple.

C. Known Results

Behrens [2] showed that being nil is a radical prop-
erty, that is, if NML) is the sum of all nil ideals of a
ring 01, then N(OZ) is a nil ideal and N(0Z/N(az)) = 0.

One method of obtaining radical properties is to
assign to each element a of a ring 61 an ideal F(a) for
which F(a+b)§. F(a) + F(b) for each a and b in a. An
element, a, is called F-regular if a E F(a) and an ideal is
F-regular if all its elements are F-regular. For each
choice of F(a), there is an F-radical R(0l), which con—
sists of all elements, a, of 6n.for which the ideal (a)
is F-regular, where (A) is the smallest ideal of containing
the set A.

Behrens showed that the choice of F(a) = (a2 - 8)
satisfies the desired property and leads to the Behrens
radical 3(01), which is the set of all elements, a, such
that b G (b2 - b) for each b e’ (a). He also showed that a
ring 61.18 B-semisimple if and only if it is a subdirect sum
of rings 6%“ where U@‘ has a non-zero idempotent generating
its minimal ideal. In his paper, it follows easily that
8(01) contains no non-zero idempotents.

Smiley [10] chose for F(a), the ideal (f ax — x +
ya - y I x, y€ 0?. }). Thus 8(a), the Smiley radical of
a is the set of all elements a of m such that if ‘66 (a),
thenb£( {bx-x+yb-yl x, yéQI). He showed
the following:

 

Theorem 1.4:
a) 3(02/S(OZ)) = 0.
b) azis S-semisimple if and only if it is isomorphic

 

to a subdirect sum of simple rings with identity.

c) If the descending chain condition holds for ideals
of an S-semisimple ring 61, then 62 is isomorphic
to the full direct sum of a finite number of

simple rings with identity.

Behrens also showed [2] that, for any non-associative

ring mans, swag 3(a).

 

Chapter 2. The Prime Radical

In this chapter, we shall first introduce the concept
of prime ideals which generalizes that of associative rings
and that of Jordan rings given in {13]. The prime radical
of a standard ring 01, which is defined to be the inter-
section of all prime ideals of 6!, will be investigated.
Finally, a characterization of the prime radical will be

given.

Lemma 2.1: If A and s are ideals of oz, A*B = A82 +

 

B2A + B(BA) + B(AB), and A#B = A32 + 13% + (AB)B + (BA)B,
then A*B = A#B.
319.93.:
It suffices to show that B(BA) and B(AB) are con-
tained in A#B, and that (AB)B and (BA)B are contained
in A*B.
Let a 6 A, b, b'é B. Then by the flexible law,
(b, b', a) + (a, b', b) = 0, we have b(b'a)€ 132A +
(AB)B + AB2_C_ A#B, i.e. B(BA)§ A#B. Furthermore, by
identity (1), (b, a, b') + (b', b, a) - (b, b', a) = 0,
thus we have b(ab')é (BA)B + 82A + B(BA) + (AB)BQ
A#B. Thus B(AB) is also contained in A#B.
The same identities also show that (AB)B and (BA)B

are contained in AfiB.

Theorem 2.1: If A and B are any ideals of 62, then

 

AeB is an ideal of’OZ.

10

 

11
3.13.292:

We first show that A*B is a left ideal of 02.
Let x601, aé A, and b, b' 6 B. Then x(a(bb')) =
1R8(bb') = ”‘(Ranbb' ” RbRab' ' RbRaRb' * Rb'Rab "
Rb'RaRb) by identity (9). Hence, x(a(bb')) = F(bb') +
'b'(ab') — (ba)b' + b"(ab) - (b'a)b A-aB where 3 =
xa, b = xb, and "5' = xb'. Thus x(ABZ)_C_ A-N-B. Using
the same identity with the apprOpriate substitutions,
we obtain x(B(AB)) S A-R-B and x(B(BA)) .5 Ail-B.

In order to show that AsB is a left ideal, it
remains to show that x(BZA) _C_ A*B. The following two
steps are used for that purpose:
(a) x(Rbb.Ra - Lbb'Ra)€ A-n-B; and
(b) xL(bb,)a€A*B for all b, b'€ B and a 6 A.
We can verify (a) by direct computation using identity
(1), for «(va38 - Lbb,Ra) = -[x(bb') - (bb')x]a
=[+ (x, b, b') 4» (xb)b' + (b, b', x) + b(b'x)] a =
+[(h, x, b') d» (xb)b' + b(b'x)] a = {[(bxlb' - b(xb')
d- (xb)b' + b(b'x)]a 6 82A S A*B. Identity (b) can be
verified using identity (10), for xL(bb')a = x(LaLbb,
+ Lear. - the. + vaLba - Leigh) = (W)? +
(b'afi; - a(b'b) + (bal'b' - b(a"5')€ A-X-B where 'a- = ex,
"5 = bx, and '5' = b'x.

We now return to showing that x(BzA )S A-K'B. Let

b, b' e B and at A. Then, by identity (11), we have

x((bb')a)l= JLR(bb')a = x l:‘(bb')a ' Lat‘bbt I Rbb'Ra

' Lbb'Ra " RaI‘bb' = XI«time " x(Rbb' " Lbb'Ra) "
(bb')? + (bb')a* where '5 = ex and a* = xa. Thus by

 

12

(a) and (b) we see that x((bb')a)€ A*B.

A*B can be shown to be a right ideal by similar

techniques.

Corollary 24; If A is an ideal of at, then A3 = Asa

 

and A3 is an ideal of 01.

We now proceed to define a prime ideal and the prime

radical for a standard ring.

Lemma 2.2: Let P be an ideal of 01. Then the follow-

ing are equivalent:

a) Whenever A and B are ideals of dtlwith A*B S P, then

A S P or BEE P.

b) Whenever A and B are ideals of a with A n c(P) #¢
and B n c(P) ,9 ¢, then (Ail-B) n c(P) #¢where c(P)

is the complement of P.

c) If a, b e c(P), then ((a)*(b))n c(P) at ¢, where

(x) is the smallest ideal of 01 containing x.

PROOF:

aléb):

 

Mfr-)0):

c)§ b):

One is merely the contrapositive of the
other.

Let A = (a), B = (h). Then a 6 A I) c(P)
M and b€ s n c(P) 26¢. Thus ((a)*(b))
n em = (are) n c(P) as ¢.

Let A and B be ideals with an c(P) # ¢ and
En c(P) # ¢. Thus we may choose a G An
c(P), he s n c(P). Then by c)¢# ((a)s
(tn/t c(P) g Ass 0 c(P). Thus b) holds.

 

13
Definition 2.1: An ideal P of a standard ring is a

 

prime ideal if it satisfies any one of the statements in
Lemma 2.2. An ideal P of an associative ring 6! is an A-
prime ideal of 6! if whenever A and B are ideals of 0! such
that ABS P, then A S P or Big P. An ideal P of a Jordan
ringfll is J—prime if whenever A and B are ideals of a such
that AUB S P, then A S P or B S P where AUB is the set of
all finite sums of elements of the form aiUbi for s1 E‘ A
and b1€ B and Ux is the quadratic Operator defined by

Ux = 23x3 - 3x2.

Theorem 2.2: If 01 is an associative ring, then an
ideal P is a prime ideal if and only if it is an A-prime
ideal. If 0! is a Jordan ring, then an ideal P is a prime
ideal if and only if it is a J-prime ideal.

213.922:

a) Let 0! be an associative ring.

Assume that P is a prime ideal of’OZ and that
A and B are ideals of at such that AB S P. Now
(BAH-(BA) = (BA)3 = B(AB)ABA S“ BPABAS P. Thus
BAS P. So A*BS P and hence A g P or B S P. That
is, P is an A-prime ideal of 01.

Conversely, assume that P is an A-prime ideal
of OZ and that A and B are ideals of 0! such that
AaBS P. Then, since A1325 Are and both A and 132
are ideals, we have AS P or 329. P. But if BZS P
thenBS P, soAS P orBS P. That is, P is a

prime ideal.

 

1’4
b) Let 0! be a Jordan ring.
Assume that P is a prime ideal of 0L and A and
B are ideals of 01 such that AUBS P. Then C =
A (1 B is an ideal of 02. contained in P or else C3 =
Cit-C = Cch AUB S P contradicts the fact that P is
a prime ideal. However, if C = An BS P, then
Ail-BS CSP so either A S P or BS P. Thus P is a
J—prime ideal.
Now, assume that P is a J-prime ideal of 62 and
that A and B are ideals such that A-n-BS P. Then
AU EA‘R’B S. P and so either A S P or B g P. Hence P

is a prime ideal of 02.

Definition 2.2: A non-empty subset, M, of OZ is a 0,-
system if whenever A and B are ideals of 02. such that A I) M

and B n M are non-empty, then (As-B) n M is non-empty.

Theorem 2.3: If P is an ideal of 6?, then c(P) is a
Q-system if and only if P is a prime ideal.
PROOF:

This is part b) of Lemma 2.2.

Definition 2.3: Let A be an ideal. The radical of A

is AQ = (r ' any Q-system containing r meets A ).

Theorem 2.h: Let A be an ideal of’dl. The radical of
A is the intersection of all prime ideals containing A.
PROOF:

 

Let b E AQ and let P be a prime ideal containing

A. Then c(P) is a Q-system missing A and so bf c(P).

 

 

15
Thus b €‘P and hence b is in the intersection of all
prime ideals containing A.

Now assume b¢ AQ. Thus there exists a Q-system M
with b e M and M n A = :5. Let& be the family of
ideals of 01. containing A but disjoint from ‘the set
M. New A is an ideal withAS A and A!) M =¢ so
A SJ and‘g is non-empty. Next, let 11 S 12 S °°° - ’
be a chain of elements urge. Let I =(IIJ. Since A
S 11, A S I. Also if x G I n M, then there exists a
,1 such that x E 11 and the 11 n M ,5 ¢, so 1;} ¢J which
is a contradiction. Thus I 58, and; is an inductive
set, and Zorn's Lemma may be applied to obtain a maxi-
mal element P. We claim that P is a prime ideal. Let
B and c be ideals with B $ P and c g P, thus P; a +
P and P; C + P. So, by the maximality of P, (B + P)
n are ¢, and (c + P)/) M#¢. Then, since u is a
Q-system, (B + New + P)/) M r ¢. But ((B + P)*

(c + Png (Be-c) + P, thus (asc)$ P or else Phi.
But then P is a prime ideal and b 6 M and M n P = ¢
means that bf P. Thus b is not in the intersection of

all prime ideals containing A.

Definition 2.h: An ideal P is semi-prime if for any
ideal A with MIA 5 P, we must have A SIP. A non-empty sub-
set S odeis an SQ-system if for any ideal A with A f) S
5‘ ¢, then (seam s as ¢.

Note that any prime ideal is also semi-prime.

16
Lemma 2.3: If P is an ideal, the following are equiva—

 

lent:
a) P is semi-prime.
b) c(P) is an SQ-system.
c) If a E c(P), then (a)*(a) n c(P) 5‘ ¢.
M:

Similar to Lemma 2.2.

Definition 2.§: Let A be an ideal, AQ = [r G 02, any

SQ-system containing r meets A} .

Theorem 2.6: Let A be an ideal of 4!. Then

 

a) AQ is the intersection of all semi-prime ideals
containing A.

b) AQ is a semi-prime ideal.

c) A is semi-prime if and only if A = Aq.

2.13192:

a) If b E AQ and P is a semi-prime ideal with A S P,
then c(P) is an SQ-system missing A and thus b¢
c(P) or b i P. Thus b is in the intersection of
all semi-prime ideals containing A.

If b¢ Aq, let S be an SQ-system missing A with
b€ 3. Now let; be the family of all ideals con-
taining A which are disjoint from S. A $3 80.2 is
non-empty and we can also show as in Theroem 2.h
that Zorn's Lemma may be applied to get a maximal
element P. Since b€ S and PI] S = ¢ , bfP. To

show that P is a semi-prime ideal, suppose that B is

then

17
an ideal with B¥ P. Thus P; B + P, so by the
maximality of P, (B + P) ll S 5‘ ¢. Now since S is
an SQ-system, (B + P)*(B + P) I) S# ¢. If (Ba-B)
SP, we would have ¢¢ (B + P)*(B + P) n S§((B*B)
+ PM) 3 g Pn s, which would imply that P¢J.
Thus P is semi-prime.

b) By part a) , AQ is an ideal and if B is an ideal
with BfiB S AQ = 0 Ps- where Pfi, runs over all semi-
prime ideals containing A, then Bit-BS Pa, and thus
BS Pg, for each R}. Thus B S n P* = Aq, that is,
AQ is semi-prime.

c) Since AQ is semi-prime, AQ is the smallest semi-
prime ideal containing A. So A is semi-prime if

and only if A = Aq.

Lemma 2J4: If a C a and S is an SQ-system with a C S,
there is a Q-system M with a 6 M and MS S.
PROOF:

Construct M = lal, a2, m} as follows. Let a1 =
a. Choose a2€ (81)*(81)n 8. "’. 8k.“ € ((ak)*(8k)l
n S. This is always possible since S is an SQ-system.
Clearly, a = a1€ M and ME 3. We now show that M is
a Q-system, that is, ((ai)*(aj)) A M 75¢. Now °t+l
€ ((Btl‘flatll E (at). 80 that (at+1) S (at) and more
generally, (ak) S (at) whenever t S 1:. Now let r =
"181(1. 3). “n+1 € ((ar)*(ar))n S C ((81)*(aJ))n S
since (ar) S (a1) and (er) S (a3).

Theorem 2.7: For any ideal A of ”Z. AQ = AQ.

18

PROOF:

 

Each prime ideal is also semi-prime so, AQ = I) Pg,
S ()P* = AQ where P.” runs over all semi-prime ideals
containing A and P* runs over all prime ideals which
contain A.

Also, if x€ AQ and if S is an SQ-system containing
x, then by Lemma 2.li, there is a Q-system, M, with
x: M and M53. But chQ implies that ¢¢ An M

QA n S, that is,x€ AQ. Therefore, AQ1= Aq.

Definition 2.6: For any ideal A, AQ ( =AQ) is the prime
radical of A. The prime radical, CH“), of a standard ring 6
01 is the prime radical of the zero ideal in a. A standard

ring is Q-semisimple if and only if Q,(dl == 0.

Theorem 2.8: Ifdl is a standard ring, and I is an

 

ideal of a, the quotient ring, all is Q-semisimple only
if 12 Q(a,). I
PROOF:

 

Let‘7: gig-)i = all be the natural homomorphism.
Consider the correspondence between the ideals of
all and the ideals of a which contain 1. Suppose P
is a prime ideal of a: containing I. Let P = P/I.
Suppose A and B are ideals of ”a, with KfiB S P. Then

=7L'lfA'), B =7z'1(B) are ideals of d with A-aB =
’L’1(K)*1[1(§) =2'1(A*B)S 72'1”) =7Z'1(?z (PDQ P +
I S P. Now since P is prime, either A S P or B S P,

but then either As P or BS P, that is, P = P/I is a

prime ideal of 01/1. Now suppose P/I is a prime ideal

19
of 61/1 and A and B are ideals of 61 with A*B QIP.
Then (A + I)/I 9:- (B + I)/I S ((Ax-B) + I)/I S P/I and
so by the primeness of P/I either (A + I)/I SfP/I or
(B + I) S P/I, so either A S P that is,.(A + I)§P or
(B + I)SP, or B S P. Thus we have shown that if I S P
Q“, then P/I is a prime ideal in 4/1 if and only if
P is a prime ideal of“.

Now, MI is Q-semisimple if and only if /) P/I = 0
where the intersection runs over all P/I, prime ideals
of 61/1, if and only if’ltP = I where the intersection
is over all prime ideals P which contain I. Thus, if
Ol/I is Q-semisimple, (1(a), the intersection of all
prime ideals of a is contained in 1.

Corollary 2.2: The quotient ring, 62/Q(6l), is Q-
semisimple, that is, Q(a/Q(a.)) = O.
25.99::

In the previous proof, it was shown that 62/1 is
Q-semisimple if and only if I contains the intersection
of all prime ideals which contain I. Now let I = Q(6Z).
Q(61) is the intersection of all prime ideals, so each

prime ideal of 0(- contains CH“) and thus a/Q(a) is
Q-semisimple.

Definition 2.7: A ring is a prime ring if and only
if (0) is a prime ideal.

Theorem 2.9: A prime ring is Q-semisimple.
PROOF:

“JP

20
If (0) is a prime ideal, then Q(Cl), the inter-
section of all prime ideals is zero, that is,6lis Q-

semisimple.

Theorem 2.10: An ideal P of“ is prime if and only if

 

is a prime ring.

PROOF:

 

If fil/P is a prime ring, suppose A and B are ideals
with A-n-BQ P. Then Adi-BS P = 0, so A = O or B = O,
that is, A.S P or B Q P.

If P is a prime ideal and K and B are ideals of
“JP with 1.41% = 0, then since EMS = U = P, we have A-aB

EPandthusASPorBSP, thatis,-A-=Oorl_3'=0.

Theorem 2.11: A ring is Q—semisimple if and only if

 

it is a subdirect sum of prime rings.

fc‘,

PROOF:

Apply Theorem 1.2.

Lemma 2.5: If A is an ideal 01302 and r 6 A then

 

Q)
A for some positive integer k.

PROOF:

 

We will show that M = {‘r, r3, r32,oo-,r3k-o-}
is an SQ-system. If C is an ideal and r336 C I) M,
then r3j+l € (C*C)I) M. So M is an SQ-system contain-

ing r and thus Mr) A # ¢, so rke A for some k.

Theorem 2.12: Q(OZ) is a nil ideal of 6C.

 

PROOF:

 

21
If r 6 (2(a) = (0)Q, then for some k, rké (O),
that is rk = 0.

Definition 2.8: An ideal I of 02 is nilpotent if

 

there exists a positive integer k such that the product of

any k elements of I in any association is equal to zero,

that is,lk = 0.

Theorem 2.13: 6l.is Q-semisimple if and only if

 

contains no non-zero nilpotent ideals.

that

mass:
61 is Q-semisimple if and only if (0) = (0)Q if and
only if (0) is semi-prime. Suppose K is a non-zero
nilpotent ideal. Then there is a t such that K3t = 0,
but x3t'1 r 0. Hence x3t-1sx3t-1 = (x3t-1)3 = x3t = 0,
which shows that (O) is not semi-prime. Now if (0) is
not semi-prime, then there exists a non-zero ideal K

with K3 = K-fiK = 0.

Corollary 2.3: OJ“) is the smallest ideal I of“ such
61/1 has no non-zero nilpotent ideals.
£39.92:

Since Q(6Z/Q(6[)) = O, 6z/Q(6Z) has no non-zero
nilpotent ideals. If 61/1 has no non-zero nilpotent
ideals, then 61/1 is Q—semisimple, and by Theorem 2.8,
I 2 (Ma).

Theorem 2.1g: A prime standard ring 62,13 either

 

associative or Jordan.

PROOF:

 

22

If in identity (2), we interchange x and z and then
subtract, we obtain: (w, y,[x, z] ) = 0. Thus each
standard ring is an accessible ring as investigated
by Kleinfeld [7], and thus we have:

(a) [(w, x, y), z] = O for all w, x, y, z ind.

Now,following Kleinfeld, let A be the set of all
finite sums of elements of the form (x, y, z) or of
the form w(x, y, 2). Let B consist of all finite sums
of elements of the form [x, y] or of the form [x, yjz.
Then in a standard ring A and B are ideals such that
a/A is associative and MB is Jordan. Kleinfeld
showed that if 0! possesses no non-zero nilpotent
ideals, then AB = 0.

Now suppose that AB = 0. We claim that BA = 0.

Let 9 represent any associator, that is, a = (x, y,
2). Then A consists of sums of elements of the form
either a or we. By (a) be = ab for any h EB, and
since AB = 0, we have Ba = 0. Also, b(wa) = [b, we] +
(wa)b = [b, we] = -[wa, b] = -w[a, b] - [w, b]a 6‘ Ba
= 0. Thus if AB = 0, BA = 0, and hence A*B = 0.

Now, since a prime ring is Q-semisimple, it has no
non-zero nilpotent ideals and thus AB = O = A*B. But
then the primeness of a means that A = 0 or B = 0.

If A = 0, then a is associative, and if B = O, filis

Jordan.

we now proceed to give another characterization of the

prime radical. Given a ring 6!, let N0 = 2:18 such that IS

23
is a nilpotent ideal of a» Next choose Nl such that NlNO
= {I‘d/NO where I" is a nilpotent ideal of a/NO. In
general, suppose that a is any ordinal number. If a is not
a limit ordinal, select Na such that Na/Na-l is the sum of
all nilpotent ideals of CZ/Na_1. If a is a limit ordinal,
let Na =z'Nfl for all p<a. If a has ordinal number Y, then
this process stops in at most Yateps. Let’z'be the smallest
ordinal such that N7= 11.8,]. ( = “1+2 = ...), Let NT: N.
UZ/N has no non-zero nilpotent ideals and is the smallest
ideal in this chain with this property.

Lemma 2.6: N = n Q1 where Qi is such that a/Qi has
no non-zero nilpotent ideals.

25903.:

am has no non-zero nilpotent ideals, so [1 Q1 Q N.
Let Q1 be an ideal such that a/Qi has no non-zero
nilpotent ideals. Thus No S Qi' We will show by
transfinite induction that N‘SQi. If a is a limit
ordinal, N.3 = 2P:8Nfl S Q1. 11' a is not a limit
ordinal, by induction Na-l S Q1 and if Na f Q1, then
there is a nilpotent ideal I/Na_1 S a/Na-l with
I/Na_1¢ Q1/N8_1, so I g Q1 and (I + Q1VQ1 ;£ 0. Now
I is nilpotent and thus so is I 0 Q1, but then (I +
Q1)/Q1 (92’ I/(I fl Qi) is nilpotent which contradicts
the choice of Q1. Thus Na S Q1 for each ordinal

number a. Hence NSf Q1 for each i, and thus Nani'

Theorem 2.15: The prime radical offl is equal to N.

2h
PROOF:

 

Q(0l) is the intersection of all prime ideals of 62.
If P is a prime ideal, then CZ/P is a prime ring and
hence Q-semisimple. Thus by Theorem 2.13, “JP has no
non-zero nilpotent ideals. So by the preceeding lemma,
N S P. Thus NS QM.) = 0?.

Also, if Qi is such that CR/Qi has no non-zero nil-
potent ideals, then by Theorem 2.13, OZ/Qi is Q-semi-
simple and then by Theorem 2.8, Q1 2 Q(m). Thus

ma) S ”Q1 = N.

Corollary 2.1+: The prime radical of Ulcontains all

 

nilpotent ideals of 01..
PROOF:

 

By definition, each nilpotent ideal ofidzis con-
tained in No S ma).

Corollary 2.5: If Olpossesses a maximal nilpotent

ideal, w(oz). then mar.) = mm).
PROOF:

 

 

U(6KJ = NO and CZ/NO has no non-zero nilpotent
ideals, that is, N1 = N0 = ch). So w(oz) = sz).

Chapter 3. The Levitski Radical

In this chapter, we show first that a finitely gener-
ated standard ring is nilpotent if and only if it is solvable.
This implies the existence of a maximal locally nilpotent
ideal called the Levitski radical. Thus local nilpotence is
a radical property in the sense of chapter 1. ‘We charac-
terize rings with zero Levitski radical and show the rela-
tionship between the Levitski and prime radicals. Finally,
it is shown that the maximal locally nilpotent ideal contains

all locally nilpotent one-sided ideals.

Let X be an ordered subset of a standard ring OZ, and
let 'X- be the set of all words generated by X. If a 6 i,
then let deg a be the X-length of a. Order X'as follows:
If a, a'e X, then a“ a' if

1) deg a < deg a'; or

2) deg a = deg a', a = a1a2' a' = al'az', and al<f al'

or a1 = a1' and az< az'.

e.g. ((x1x2)x3)xu) (y1y2)(y3yh) for all xi, yi in X since
deg (1112)X3> deg (ylyz). Let Y be the set of right and
left multiplications by elements of X, that is, Y = f.

R 1'1: ' x6 1' and let T be the set of all (associative)

x,
words generated by Y. For a e I, let Ta denote either R8 or
L e 3: 000

a If W’ Tleszy3 Tyn is an element of Y} then deg
H =Zrdeg yi and t(W) = n = the T-length of w. Order Y and
Y as follows:

1) If y1< y2 are elements of X, then Ry1< Ly1< Ryz;

25

26

= .g. ' = 00.
2) If w Tleyz Tyn and w Tlezz sz, then
W<W' if
a) n< m, or
b) m = n, Tyl = T21 = T21, Tyz = T22, .00, Tyk =
Tyk+1 and Tyk+1<Tzk+1 for some k = 0, 1, 2: '°
131-1.
A word W in’Y is normal if V = Tleleszzz'HTynTZn where

OOO<T

T yn

3'1 < Tye < and Tzl< T

does not appear.)

22

can be written as a sum of normal

< . . .< Tzn ( possibly Tyl

A word'w in Y'can be normalized if gzkw

words each of degree

equal to dog W, for some non-negative integer k.

For the following, let x be a generating set for“, .

Lemma 3.1:

 

If a' is a product of at least 2n + 1

elements of X, then a' = azwi for some a ed, W1€ Y where

t‘"i)2f no
PROOF:

It is possible to choose a such that a' = aw where

deg H'= 2n. Thus it suffices to show that if b is a

product of three elements of x, then Tb can be written

as a combination of words of T-length 2 or 3, each of

degree equal to deg b.

Rewriting identities (9) and (10), we have:

Rx(yz) = Rnyz + Ry(sz- Rsz) + Rzmxy - Rny) and
L(xy)z = Lzny + Lk(LYZ - LzLy) + Ly(sz - LzLx)°
From (11) we obtain: ny - ny = (Lx - Rx)Ry + (Ly -
Ry)Lx.

Replacing y by yz we obtain:

27

+ (Lx - Rx)Ryz + (L - Ryz)Lx and

Lx(yz) = Rx(yz)
replacing x by xz:

R(xz)y = L(xz)y + (sz ' sz)Ry + (Ly ' Ry)sz'
Thus the first two cases imply the desired result for

yz

the remaining two cases.

Lemma 3.2: For every 9, b, c 561, we have the follow—

 

ing:

a) RaLch = ’LcLbLa + RbRca ' Rb(ca) + Rach + LcLab3
b) RaRb - Rab = LaLb " Lbs;

c) LaR‘b ’ RbLa = Lab ” Rab + RaRb ‘ LbLa°

PROOF:

 

a) This is (2) considered as operating on x with the
following substitutions: z = a, y = b, w = c.
b) This is the linearized flexible law.

c) This is (1) operating on 2 with x = a and y = b.

Lemma 3.3: If we Y, then w =2 wiL +z 881 +231: where

WiL is a word involving only Ly's and Sk is either a RyLz
or a Ly 2’
we shall prove this by induction on the T-length of
W.
If‘H = Rny, we use Lemma 3.2b. For all other cases
where t(W) equals 1 or 2, the result is immediate.
Now assume the result is true for all words of T-
length less than n. Let S denote any word (or sum of

words) of T-length less than n, that is, words to

which the induction hypothesis may be applied.

28

Let w = Tszc-oTn, where nqz 3 and each Ti is either

8i

L or R .
81

CASE

CASE

CASE

CASE

1: Tn-lTn =LyLz
If each Ti is equal to an in, then we are done
immediately. If not, let Tk = R8 be the last
right multiplication occuring in w, that is, Tt =
Rd implies t: k, k< n-l. Using Lemma 3.2a, we
have
V = T1T2"'Tka+lTk€2°°'Tn =

= T1''°Tk-1(RaLch)Tk+3”'Tn

= Tl'°°Tk-l(‘LcLbLa + RbRca ’ Rb(ca) + Rach

+ LcLab)Tk+3"'Tn

= Tl'°'Tk-1LcLbLaTk+3°”Tn + S
New repeat the above process on the first term of
the above mud continue until k = 0, that is

= L11L12"'an + S.
2: TNT. = 15R.
Apply Lemma 3.2b.
w = T1---Tn-1(LyLz - Lzy + Byz) = Tl-ooTn_2LyLz+ s.
New apply Case 1.
3: Tn-2T ~1Tn = RnyLz
Apply Lemma 3.2b.
H'zT 1n”H.11- 3(LXLY- 911+ ny’Lz

= T:~-v n-BLxLyLz + 8.
Apply Case 1.
h: Tn-zT -1Tn = LnyLz
Apply Lemma 3.2c.

w’= Tlooo n-3‘Rny + ny - ny + Rny- Lny)Lz

29
= T1---Tn-3(Rny + Rny - Lny)Lz + S.
Apply Cases 1 and 3.
CASE 5: Tn_2Tn_1Tn = LxLsz
Apply Lemma 3.2b.
W = Tlo-oTn_3(Rny - ny + Lyx)Rz
Apply Case 2.

CASE 6: Tn-2Tn_1Tn = RxLsz
Apply Lemma 3. 2c .

w :2 T1. 0 oTn-3

Apply Cases 2 and 5.

(LyRx - Lyx + Ryx - Rny + LxLy)Rz
Thus we have considered all possibilities for the
final factors of W and have either proven theses cases
or reduced them to ones already proven, and therefore

the proof is complete.

Lemma 3.Q: Any word W'in Y involving only left multi-

 

plications can be normalized.

PROOF:

 

This follows by induction on t(W) by using:
LxLyLz = ’LzLny * Lnyz + LxLzy + Lzny ' L(xz)y and
ZLxLny = Lnyz + 2Lxny - szy-
These are both merely restatements of identity (10).
Note that in the resulting words which are of the same
T-length as that we started with, none of the subscripts
increase in degree. Thus if H = LalLblLasz2'°°Laann,
suppose that by is the largest element of b1, b2,---bn
with respect to the ordering of 3. Then using the
above identities we obtain 12kw’= LclLdl"‘Landn + S

where c1, c2, ---,cn is a reordering of a1,--',an,

3O
d1,...,dn is a reordering of b1, ---, bn, (1n = bN and
S is a sum of words of T-length less than t(W) (perhaps
the first word does not appear.) Then the induction

hypothesis may be applied to LclLd1"'Ldn L Since

-1 cn'
(1D is larger than any other d1, we have the desired

result.

Theorem 3.1: Any word in Y’can be normalized.

 

PROOF:

 

Any word of length l or 2 is normal, so the

result follows from the two preceding lemmas.

Definition 3.1: 010 = a,a1 =.adԤ, ---a1,1 =az13.
Also, let “(0) =6! and dun-)4; a”) a”). For any
integer k.2 l, tlk is the set of all sums of elements which
are products of 1: elements of 4. Thus in agreement with
Definition 2.8, a is nilpotent if there exists a positive
integer k such that CIR = O. 61 is solvable if there
exists a positive integer t such that 61(t) = 0. Clearly,
iffiz.is nilpotent, it is solvable. Also for each n,

dznsgdz(n)'

Note: By Theorem 2.1, each ‘11 is an ideal of 62L

Theorem 3.2: Ifmis a standard ring generated by a

 

finite set X, then for each integer m, there exists an
integer f(m), depending onIXI, such that Ulf(m)£am.
28.922:
By induction on m.

If m = 1, let f(1) = 3.

31
Now assume the existence of f(m-l), so that

“f(m'lg Olm_1. Let Pk be the number of distinct
words of length 1:: generated by X, and let M =z{(m"1)Pk
and let N = az§(m‘1)kPk. Thus M is the number or dis-
tinct words of length less than or equal to f(in-l) and
N is twice the number of elements of X used in the
expression of these M words. Suppose W 6' Y is a word
with t(W)2_ [2N(M+2) + f(m-l)] . By inserting paren-
thesis, write W as WOW1W2°HWM+2 where t(wi) = 2N,

i = 1, 2, N, n+2 and t(w0)2 f(m-l). Each wi can be

normalized by words of the form u = TleleszzZ’
TYszk where deg u = deg W12 t(Wi) = 2N. ThusZ' deg y1
+[deg 21 2 2N, so eitherfdeg yi 2 N orz deg 212 N.

21 < T22 < "'(Tzk and

because of the ordering of Y each yi and each 21 can

Now T31 < Tyz ( "-(Tyk and T

occur at most twice in each sequence. If we assume
that each sequence contains only elements offll which
are of degree less that f(m-l) and that each of these
elements actually occurs twice in each sequence, we
have deg yi =f {(m-1)-1 2kPl,( N and similarly: deg
21< N, which is a contradiction. Thus either deg
ykz f(m-l) or deg 21:2 f(m-l)., that is, ykeam-l
or 2,, a 62,“. Hence u = Ttth2"’TtsTths+l with b
6 aim-l and where Ttsfl need not appear. So the sub-
word, wle-uwmz is a sum of words of the form J =
2n(J)Tt(O,l)Tt(O,2)"'Tt(0,kO)Tb1Tt(l,l)Tt(l,2)'°'TbZH'

TbM+2Tt(M+2’1) where t(i, NS“ , bkEdm-l and
possibly Tt(M+2,1) does not appear.

32
Lemma 3.5: TbTxTy' b an element of an ideal B of OZ,
x, y EWOZ can be written as a sum of words of the form

 

Tvaytht, TbsTxI, Txthv, Tb' where x'Ed , b'e B.
PROOF: I

 

This is proven by considering all combinations of
left and right multiplications possible in TbTxTy'
‘He do this by first considering the case involving
only left multiplications or only left multiplications.
This is done by using (9) and (10).
RxBsz z'RzRny + Rnyz + Rnyz + RzRyx ' Ry(xz)
LxLyLz = 'LzLny + Lnyz + LxLzy + Lzny ' L(xz)y'
We check the remaining cases by reducing them to

ones already considered by means of the identities

used in the proof of Cases 3-6 in Lemma 3.3.

Now apply this lemma repeatedly toe), starting with
T T T th t J
bM+l t(H+l,l) t(M+1,2)’ so a may be written as a sum
of words of the type:

I. :J' = STbTb.S'; or

I . I! = S T see [

I J lel 31Tb2T32 TbM+1T3H+lTbM+25
where S, S' G Y and b, b', bk 6 dm_1, 8k¢dm-1' Thus
W’= “0(‘W1' + NJ”) where "i. are of type I and W3“ are
of type II.

Let J be a word of type II. (J = STb1T31---Tbn+2S',
31 ¢ $314.. By the choice of H and since ”(NM-1E aim-1’
there are less than M possible distinct choices for the 3i»

so two of the Si must be equal.

 

Lemma 3.6: TblTsTstthfl = TbuTslTstTbfl plus words

33
of type I.
2.13292:
Similar to the proof of Lemma 3.5 by considering
possibilities for TsTst..

Lamma3.7: Tb:T8TstTbn is a sum of words of type I.

 

PROOF:

 

Similar to the above.

Applying the last two lemmas and the previous
comment about equality of two sk's, we see that W =
wag-vii), where each wi' is of type I.

Now, let f(m) = [;2(2N(M+2) + f(m—l))]+ l and let
a'Gaf(m). Then by Lemma 3.1, a' is a sum of words
of the form aw where t(W) 2 2N(M+2) + f(m-l). Now
applying the above, a' is a sum of words of the form
awowi' where the W1' are of type I and t(WO) 2 f(m-l).

Thus awo Ed Now let wi' = STbTb:S', b, b'C

m-l'
%.1. Since and is an ideal, (aw0)S E dm_1, so a'
fawowi' 6 am_1TbTb:S'S (madam-1’ fl.m-1+ flm-1
(am-l film-lm' S “ms. Q “m'

Thus a f(m) Q mm.

Theorem 3.3: If“ is a finitely generated standard
ring, then 02, is nilpotent if and only if 02 is solvable.

PROOF:
0: “Mg 02m €d(m)§012"’.

Definition 3.2: A standard ring 0?. is locally nil-

 

potent if every finitely generated subring is nilpotent.

3h

It is locally solvable if every finitely generated subring

is solvable. An ideal I oral is locally solvable (nil-

potent) if it is locally solvable (nilpotent) when con-

sidered as a ring.

Lemma 3.8: Let 61 be e finitely generated standard

 

ring. Then for any m, 61m is finitely generated.

“VI

23.9%.:

It suffices to show that 621 is finitely generated,
for then the result follows easily by induction.

Let X = {'31: x2, ---, xnl' be a finite generating
set for 61. By Theorem 3.2, there is an integer f(2)
such that armada, = (alalml + 621(021021).

Now let Y be the set of words in.621 of X-degree less
than f(2). Since X is finite, Y is a finite subset

of 07.1. Also, if aédll, a = u + v +...+ w is a finite
sum of X-words of all. Suppose deg u.2 f(2), then by
the choice of f(2), u Ede, so u =Z(risi)t1 +Zri'
(si'ti')' r1, si, ti, ri', ’1' ti'GECZl and deg r1 +
deg s1 + deg ti = deg ri' + deg 31' + deg ti' = deg u.
Continuing this process, we see that e is an element of
the subring of “1 generated by the elements of 031
which are of X-length less than f(2), that is, 621 is

finitely generated.

Theorem 3.1:: If I is locally solvable ideal oral and

 

is locally solvable, then CE is locally solvable.
PROOF:

Let B be a finitely generated subring of 61. B =

35
(B + I)/I is a finitely generated subring of 52/1 and

hence solvable. Thus, there is an N so that (B)N =‘5,
that is BNSI. By the preceding lemma, BN is a
finitely generated subring of the locally solvable
ideal I. Thus BN is solvable, so there exists an M

so that (BN)M = 0. Thus BN+M = 0.

Corollary 3.1: If I is a locally nilpotent ideal and

 

OL/I is locally nilpotent, then 52 is locally nilpotent.

Theorem 3.5: The sum of two locally solvable ideals

 

is a locally solvable ideal.
PROOF:

 

Let I and I' be locally solvable ideals. I/(I I')
is the homomorphic image of the locally solvable ring
I and hence is locally solvable. Thus (I + I')/I'

(3' III (I I') is locally solvable. Since I' is locally
solvable, by the preceding Theorem, we have that I + I'

is locally solvable.

Corollary 3.2: A finite sum of locally solvable ideals

is locally solvable.

Theorem 3.6: The sum of all locally solvable ideals is
the unique maximal locally solvable ideal of 62 .
23.99.11:
Let{I‘} be the collection of locally solvable
ideals of 01 . Let ma) =21“. Clearly, if L(J£)
is locally solvable, then it is the unique maximal

such ideal of 62. Let A be a finitely generated

36
subring of L(UZ). Since A is finitely generated,
n
A521 I!a1 for some choice of the all. But by the
above,2iJ I81 is locally solvable. Thus A is solvable.

Therefore, 1461) is a locally solvable ideal.

Definition 3,3: The maximal locally solvable (nil-

 

potent) ideal of a standard ring 02, L(0Z), is called the
Levitski radical of 01. OZ is L-semisimple if M6!) = 0.

Theorem 3.7: The quotient ring, 62/L462) is L-
semisimple, that is, L(62/L(0£)) = 0.
2119.92:

Suppose that I/L(dl) is a locally nilpotent ideal
of a/sz). Since M0!) is locally nilpotent, I is
locally nilpotent by Corollary 3.1. Hence by the
maximality of L(dl ), I E Md) or I/L(a) = 0.

Corollary 3.3: Local nilpotence is a radical property.

 

PROOF:

 

The only part of the definition given in Chapter 1
which we have not shown is the easily verified fact
that the homomorphic image of a locally nilpotent ideal

is locally nilpotent.

Corollary 34: If UZ/I is L-semisimple, then 12 HQ).

 

PROOF:
Since local nilpotence is a radical property, this

follows from Theorem 1.1.

Theorem 3.8: The Levitski radical of a standard ring

 

37
02. is equal to the intersection of all prime ideals P8 such
that 61/?8 is L—semisimple.

PROOF:

 

By the preceding corollary, L(DZ) is contained in
each Pa, so L(CY)S/) Pa“

Now assume that xf L(0l). We will find a prime
ideal P, such that a/P is L-semisimple and xfi‘ P.
By the maximality of ma ) and since x¢ L(az ), (x)
is not locally solvable, so there is a finitely
generated subring S S (x) with S not solvable. Now
let}. be the collection of all ideals Q, such that Sm;
Q for all m. Since (0) EXDJis non-empty and to see
that Zorn's Lemme may be applied, let A1 5 A25 be
a chain of elements ofJ. Let A =UA1. A is an
ideal and if SmS A, then since Sm is finitely gener-

ated, SmS Ak for some 1:, which is a contradiction.

Thus A £3. Now let Q be a maximal element of 4. SO
8 S (x), x; Q.

We claim that Q is a prime ideal, for if not, there

are ideals A and B with A$ Q and a; Q but sass Q.
TheanA +Q=A' andQSB +Q=B' so by the maxi-
mality of Q, A', B' i J. . Thus for some m and n,
3015 A' and SnS B'. Without loss of generality, m<
n so SnS SmS A' and thus Sn S A'n B'. Now Sn+1 =
(Sn)3 = Sn*SnS A'*B' S (A-x-B) + Q S Q which contra-
dicts the choice of Q.

We now claim that a/Q is L-semisimple. If not,
let (0) 75 N = N/Q be a locally solvable ideal oféz/Q.

38
The maximality of Q means that St S N for some t. St
is finitely generated by Lemma 3.8 and hence (St + Q)/Q
SN/Q = N is finitely generated and thus solvable. So
for some m, (St 4' Q/Q.)m = O, that is, (St + Q)m S Q

so (St)m S Q or St”, S Q which is a contradiction.

Corollary 3.5: For a standard ring 62, the prime

 

radical, Q(0[), is contained in the Levitski radical 1462).
PROOF:
Since Q(6[) is the intersection of all prime ideals,

the result follows from the above.

Corollary 3.6: Ulis L-semisimple if and only if it is

 

a subdirect sum of prime L-semisimple rings.
PROOF:

Theorem 1 e 2 e

LemmL3.9: If“ is L-semisimple, then OZ contains no

 

non-zero locally nilpotent one—sided ideals.
m:

Suppose A i O is a locally nilpotent one-sided
ideal of 02. Since Hal) = 0, “is a subdirect sum of
672/Pa where él/Pa is L-semisimple and prime wittha =
0. By Theorem 2.1a, each ”UPa is either associative
or Jordan. Now, (A + Pa)/Pa is a locally nilpotent
one-sided ideal of 63/P8. If 61/?8 is Jordan, then
A + PB/Pg is a locally nilpotent two-sided ideal of

the L-semisimple ring fil/Pa. Thus A.S P If6Z/Pa

is associative, then it contains no locally nilpotent

39
one-sided ideals, [14, p. 127], so ASPa. Thus for

eacha,ASPa orASflPa=O.

Theorem 3.9: The Levitski radical, L(OZ) contains all

 

locally nilpotent one-sided ideals of 01.
111923:

If A is a locally nilpotent one-sided ideal of 6?,
then (A + Han/M01) is a locally nilpotent one-
sided ideal of the L-semisimple ring fll/LMZ) so
(A + MOM/ma) = 0, or A S LMZ).

Chapter h. The Jacobson-MacCrimmon Radical and the Jacobson-

Brown Radical

In this chapter, we obtain two generalizations for the
Jacobson radical of associative rings. The first follows
that for Jordan rings given by MacCrimmon [6], and is true
for any non-commutative Jordan ring. The second is that
given by Brown [3] for any non-associative ring. We show
that the possibility illustrated by Brown for any arbitrary
non-associative ring, the inequality of the radical defined
in terms of right ideals with that defined by left ideals,
cannot hold in a standard ring. Finally, the position of
each of these radicals in relation to other radicals is

Shown 0

Definition u.l: If 01 is a non-commutative Jordan
ring with identity, 1, then an element a 601 is regular
with inverse b G'Cl if ab = be = l, and (a, a, b) = 0.

Note that is a is regular according to the above,
then a2b = (a, a, b) + a(ab) = a'l = a, and be2 = (ba)a -
(b, a, a) = 1-a + (a, a, b) = a. Thus this definition
corresponds to the definition given by MacCrimmon [8] where

he showed:

Theorem 4.1: If a G 02, then a is regular with in-

 

verse b if and only if a is regular in 01+ with inverse b.

For each element a of a non-commutative Jordan ring,

consider the operator Us = RaLa + Ra2 - R82. By identity

hO

ul
(11) we also have, Ua = LaRe + La2 - La2° In the com-
mutative Jordan ring 01*, we have the quadratic operator
U8+ = 2Ra+2 - R82+, where Rx+ is right multiplication by
x in 61f. Since Rx+ = !5(Rx + Lx)o we have: Ua+ = 2R8+2
- 392* = 2pm, + La)2] - Jame + L92) = emf + Rate +
Lea, + L,2 - 3,2 - L92) = Jamaal.a + 3,2 - R32) + (LaRa +
L82 - La2)] = Us: Thus the identities for quadratic
Operators in Jordan rings are also true in the non-

commutative JOrdan case.

Theorem h.2: Let a, b be elements of a non-commutative

 

Jordan ring with 1. Then the following are equivalent:
a) a is regular with inverse b;
b) l is in the range of Us;
c) U '1 exists;
d) $(ab + be) = 1, bU = a;

PROOF:

 

This follows since it is true in.0[*’(See Tsaifll] )
and each of the stated conditions is valid in 6? if
and only if it is valid in UL".

Lemma u.l: Lat 62 be a non-commutative Jordan ring
with 1. Let a, b E 01. Then an is regular if and only if
both a and b are regular.

PROOF:

 

Again this is true in 61*, thus it is true in 6!,

Definition h.2: An element a of 01 is quasi-regular

with quasi-inverse b if a + b = ab = ba and (a, a, b).

M2
Lemma Uzi: If 1 E CZ, a is quasi—regular with quasi~
inverse b if and only if (1 — a) is regular with inverse
(l - b).

PROOF:

(l-a)(l-b) = l - (a + b - ab) and (l-b)(l-e)

l - (a + b - be). Thus (l~a)(l-b) = (l-b)(1-a) 1
if and only if a + b = ab = be. Also, (l-a, l-a, l-b)
= —(a, a, b) since the associator is linear and any

associator involving 1 is equal to zero.

Given a standard ring 62, we can embed.62 in a ring
with identity, 02.1, as follows: Let 021 =Z@ OZ. Define
additicn component wise and multiplication by (n,a)(n',a')
= (nn’, na’ + n'a + 28'). Elements of 611 will be denoted

by n + a where 1162 and a 607..

EemerLLti. If K is an ideal of (5:, WU“). = (021/10

and (62+)1 2 (Jfll)+. If 62 is a non—commutative Jordan
(standard) ring, then éql is a non~commutative Jordan

(standard) ring.

The first isomorphism is given by: n + (a +'K)°9
(n - a) + K. The second statement is true since both
are composed of the same vector space and it can
easily be verified that the multiplication is the

same. The last part is also easily verified.

: If a<ECZ and a is quasi regular inéfl,

Lemma h.h

u.-. *u—Wv -_—

then a is quasi-recular in Cfis

h3
PROOF:

 

Since a is quasi-regular in 511, l-a is regular
with inverse x E (11. Now it = n + x‘ where x' 5 OZ.
1 = (l-a)(n+x') = n + x' - an - ax' so n = l and thus
the quasi—inverse of a, l-x = x' is inél, and so a is

quasi-regular in CZ.

Theorem h.3: If a 661, then a is quasi-regular in 01

 

if and only if a is quasi—regular in 6[+.

PROOF:

 

Since (02,1)+ ='(6[+)1, the following statements are
equivalent:

a is quasi-regular in fl;

e is quasi-regular in 021:

(l-a) is regular in Oil;

(l-a) is regular in (621)+ = (01+)13

a is quasi~regular in 62+l;

a is quasi-regular in 62+.

Definition h.3: An ideal is quasi-regular if each of

its elements is quasiuregular.

Theorem ugh: No non-zero idempotent of 01 is quasi-

regular. Every nilpotent element is quasi—regular. If for

n, 2“ is quasi-regular, then z is quasi-regular. The

O

m

(D

3
sum of two quasi-regular ideals is a quasi-regular ideal.

PROOF:

 

This follows from the preceding theorem and the
fact that they are all known to be true in a Jordan

ring [11].

M4

Corollary h.l: The sum of all quasi-regular ideals

 

of a standard ringc2.is the unique maximal quasirregular

ideal.

Definition h.h: The maximal quasi-regular ideal of

 

will be denoted by JM(61) and called the Jacobson-MacCrimmon

radical. OZ is JM-semisimple if JMKdZ) = 0.

Lemma h.5: If 1 5:62 and K is a quasi-regular ideal of

 

01., suppose that a"; is a quasi-regular element offi = al/K.
Let w be any pre—image of w'in 62. Then w is quasi-regular
in 61,

PROOF:

 

Since a is quasi-regular, (I:w)is regular in
so there exists 2 €352 with (1:5)Usz = T’or equiva-
lently (l—w)U1_.Z = l - y for some y€ Ki. Since K is
a quasi-regular ideal, l-y is regular and thus by
Lemma n.1, both l-w and l-z are regular, thus w is

quasi—regular.

Theorem ll.5: UZ/JMM’L) is JM-semisimple.

 

PROOF:

 

Let K'be a quasi-regular ideal offii = CZ/JM(62).
Let AS563 be the pre—image of A} ‘We will show that A
SEJM(01) and thus A = 0. Clearly, A is an ideal so

it suffices to show that A is quasi-regular. But A =
(A + Jh(ol))/JM(0Z) is quasi-regular in al/JM(6Z) and
hence in (fl/JM(€Z))lgflll/JM(JL). So by the preceding

lemma, A is quasi-regular in 621 and thus by Lemma h.h,

quasi-regular in 52.

1:5

Corollary h.2: Quasi-regularity is a radical prOperty

 

of standard rings.

PROOF:

 

This follows from Corollary n.1, Theorem h-S. and

the easily verified fact that the homomorphic image

of a quasi-regular element is quasi-regular. Thus all

conditions for a radical prOperty given in Chapter 1

are satisfied.

Theorem tho: If an is J'M-semisimple, then 12 JM

 

(01).

PROOF:

 

Corollary h.2 and Theorem 1.1.

We now turn to the generalization of theXJacobson

radical

given by Brown [3]. we begin by listing some of

his results.

Definition Qg5:

 

a)

b)

e)

d)

e)

Let I be a right ideal of 01. I' = [a 601 I (a)£ 1)}.
I} 01 =r{.a(E61'02a5; I}.

If a601, Q(a) is the minimal right ideal containing
{ax - x | xGOZ}.

An element a of'OZ is Brown quasi-regular (B.q.r) if
a E Q(a).

An ideal (right, left, or two-sided) is Brown quasi-

regular if each element is B.q.r.

JB(UZ) = ( a €01, (a) is B.q.r.}.

f) A right ideal I is modular if there is an element

eefl with ex - x6 I for each x6“.

h6
g) m is primitive if“ contains a modular maximal
right ideal M such that M' = 0.

Theorem h.7:

a) If I is a right ideal ofCZ, then I' is the largest
two-sided ideal of 61.contained in I.

b) If I is a right ideal and no: S I, then I' =
1:61.

c) If I is a modular right ideal, then I:6H.S I and
hence I:61.= I'.

d) JB(0l) = n H' such that M is a modular maximal
right ideal of 0!.

e) JB(0[) is the maximal B.q.r. ideal of'6Q.

f) JB(0L/JB(01)) = 0.

g) Ja(al) = O if and only if 62.1: isomorphic to a

subdirect sum of primitive rings.

Theorem h.8: Brown quasi-regularity is a radical

property.

Corollary h.3: If w(a/I) = 0, then JBWZ) S I.
PROOF:

Theorems 1.1 and h.9.

Theorem h.9: A primitive standard ring is either
primitive associative or a simple Jordan ring with indentity.
21329::
Kleinfeld [7] showed that a primitive standard ring
is either associative or commutative. If 61 is

associative, then it is primitive associative. If

h?

is commutative, it is thus Jordan and primitivity
implies that Clicontains a modular maximal right

ideal M containing no non-zero two-sided ideals. But
under commutativity the concepts of one-sided and two-
sided ideals are equivalent and thus (0) is a modular
maximal ideal of 01. (0) being maximal means that

has no non-trivial ideals, that is, 0?. is simple.

(0) being modular implies that there exists cell such
that ex - xi= O for all x631, that is, ex = x = xe,

or e is an identity for .

As Brown states, a similar theory can be derived in
terms of left ideals. He presents an example to show that
for an arbitrary non—associative ring, the left radical and
the right radical may not be equal. We now show that this

is not possible fer standard rings.

Theorem h.10: A right (left) JB-semisimple standard

 

ring is a subdirect sum of right (left) primitive standard
rings. A right (left) primitive standard ring is either
associative or Jordan.

PROOF:

 

The statement with "right" is merely Kleinfeld's

result and the proof for "left” is similar.

Lemma h.6: Let 0! be a standard ring. 6? is right

 

JB-semisimple if and only if it is left JB-semisimple.

EROOP:

 

The proof follows from the facts that for commuta-

tive rings the two concepts coincide and that for

1:8
associative rings the result is known [5, p. 13].
Thus if at is right JB-semisimple, it is a subdirect
sum of commutative and associative right primitive
rings which-are both right and left JB-semisimple.
Thus 62 is a subdirect sum of left JB-semisimple
rings and hence left semisimple itself. Similarly,
if 01 is left semisimple, it is right JB-semisimple.

 

Theorem hill: Let JBP be the radical in terms of
right ideals and let J31 be the radical in terms of left
ideals, then JBP = J31.

€39.93:

6l/JBr is right JB-semisimple and thus by the above
lemma, it is left JB-semisimple. Hence JBr:2 JBl by

Corollary h.3. Similarly, J31 2 JBP. Thus equality

11016.3 0

We now show the position of these two generalizations
in relation to the radicals we have previously considered.
Recall that the nil radical of OZ is the largest ideal of
5% containing only nilpotent elements. Also the Behrens
radical is the ideal such that“ is a subdirect sum of rings
in which every ideal contains a non-zero idempotent.
Finally, the Smiley radical of a ring is zero if and only

if the ring is a subdirect sum of simple rings with identity.

Theorem 14.12: 3““) contains Md), the nil radical
of 62.

PROOF:

 

1:9
By Theorem h.h, each nilpotent element is quasi-
regular. Thust) is a quasi-regular ideal and con-
tained in the maximal quasi-regular ideal, JFK“).

Theorem Ll}: J'Md) is contained in the Behrens
radical, B(d).

mast:

We first assume that a is B-semisimple. Then
is a subdirect sum of a. where a, has a non-zero
idempotent generating its minimal ideal. Thus each
ideal of d‘contains a non-zero idempotent and thus
cannot bequasi-regular. Thus Mada) = 0, and hence
mm = o. '

Now let. a. be an arbitrary standard ring. d/B(0L)
is B-semisimple and thus by the above, Jud/Na” =
0. So by Theorem 14.6, JIM“) S B(a).

Theorem h.15: J'B(a.) contains JM(flZ,).
sass:

“(01.) is a quasi-regular ideal and we will show
that a quasi-regular element is B.q.r., and thus the
result will follow from the maximality of JB(UZ).

Let a be a quasi-regular element of ”1. Then there
exists an element b €01, with a + b - ab = O, that is
a = ab - 13. Thus a is an element of Q(a), the right
ideal generated by {ex - x ' x 602}. Thus a is

B.q.r.

Theorem L13: (8(a) contains JB(OZ.).

SO
PROOF:

 

First, letabe S-semisimple, that is, 3(a) = 0.
Then 01. is a subdirect sum of a.“ , simple rings’with
identity. Since a“ is simple, JB(d-“) = O or ”(a .

But Q(l) 8 ( { l-x - x‘} ) = 0 so 1 is not B.qtr. Thus
B(m‘) = O and each ”I“ is J'B-semisimple. Therefore
01 is JB-semisimple. :

Now if 0! is an arbitrary standard ring, JB(6Y/S(62))

= O, and so JB(a) S 3(a).

Thus we have completed all parts of the diagram given

in the introduction and reproduced below.

/s(a )\

3(a) ' l JBML)
Rim)
LMU

«2(a)

Chapter 5. Chain Conditions

In this chapter, we study the relationship of the
radicals we have previously considered when the ring satis-
fies a chain condition. a is said to satisfy the DCC if
every descending chain of right ideals of 0L has a minimal
element. a is said to satisfy the strong DCC [9] if every
subring of a satisfies the DCC.

Theorem 5.1: If 4 satisfies the DOC, than QM!) =
L(01).
£13922:

We have that Q(a )1 5 ma).

Now let a = JUQMZ). fl is Q-semisimple and thus
is a subdirect sum of my; = a/P,‘ where 0P“ = O and
each 6”“ is either prime associative or prime Jordan.
Now since ”L, is Q-semisimple and satisfies the DCC, if
01,, is either associative [1;] or Jordan {121then LWZK)
= Q(62‘) = 0. Thus Oi is a subdirect sum of L-semi-
simple rings and thus L-semisimple itself, i.e. L(6Q)
= Ida/CHO!“ = 0. So, MUDS GHQ). Thus equality
holds.

Theorem 5.2: If ”2 satisfies the DCC, then JB(€Z) =
3(01).
m!
We have Jam) S 8(a).
Consider oz/JB(a). Since this is Brown semisimple,

it is a subdirect sum of simple commutative Jordan

51

, 52
rings with l, and primitive associative rings. But
since the DCC is preserved under homomorphisms, each
of thesubdirect summands satisfies the DCC. Also,
a primitive associative ring is a simple ring with 1.
Thus CE/JB(CZ) is a subdirect sum of simple rings
with 1 and thus by Theorem 1.3b, S(OZ/JB(0I)) = O,
or 8(a) S JB(a.). Thus equality holds.

Thus if 01 satisfies the DCC we have the following

diagram.
3(a) = JBMZ.)

3(a)

mm.)
mm.)
Md) = M02.)

Corollary_§;l: If 6! satisfies the strong DCC, then
szl = QM!) and 3(a) = Jame).
2.59.92:
This is true since if asatisfies the strong

DCC every subring of 62, including 67.1tself satis-
fies the DCC. Thus we merely apply the above.

Theoremg5.3: If 6! satisfies the strong DCC, then
a(fll) = JIM“) == Mac).
3.3.993:
we have ma) 5 mm )9. EMU.

53
Now suppose that xf Na ). t; We will show that

x¢ B(a). x¢ NW) implies thatw(x) is not a nil
ideal so there exists y Q (x) which is not nilpotent.
Since“ is power-associative, {yk} generates an
associative subring of“ which according to our
hypothesis satisfies the DCC. But then this subring
contains a non-zero idempotent,e, [5, p.221. Since

e E (y)-S (x) and 3(a) is an ideal which contains no
non-zero idempotents, x; B(OL). Thus B(fl)£ Mal),
and the stated equality holds.

Thus if ”L satisfies the strong DCC, we have the

following.
3(a) = JBwL)

mm) = JMML) =Nwz)
ma) = cm.)

Theorem 5-H: If a satisfies an ascending chain
condition on subrings, that is, every ascending chain of
subrings has a maximal element, then M“) = Mal) =
1““).

mass:

We have cud) S LMZ) S MOI).
Let 0.1.: a/Q(01). It suffices to show that Mil)
= O, that is i has no non-zero nil ideals. Since

0.1. is Q-semisimple, it is a subdirect sum ofdg‘:

fl/Pd where 02“ are either associative or Jordan.

Each 0?.“ is Q-semisimple and thus possesses no

Sh
non-zero nilpotent ideals. Let K be a nil ideal of

07.. (K + P.,)/P‘ is a nil ideal oral“. I: 02.: is
Jordan then (K +R, HP" is nilpotent [114.1 and thus
K + P = O or K,S)B‘. If 62‘ is associative then
(K + P)/P is nilpotent [u] and so again K S sP,‘ .
Thus for sachet, K S P,‘ or KS!) Pa = 0. Therefore,

Ma) = mat/cum) = 0.

Hence if a satisfies an ACC on subrings, we have:

8(a)
awz) JB(0L)
\ /

BIBLIOGRAPHY

13.

BIBLIOGRAPHY

Albert, A. A., ”Power-associative Rings", Transactions
of the American Mathematical Society, vol. 6h (l9u8)

pp. 552-593.

Behrens, E. A., "Nichtassoziative Ringe," Mathimatische
Annallen, vol. 127 (195u) pp. hul-hSZ.

Brown, B., "An Extension of the Jacobson Radical,"
Proceedings of the American Mathematical Society,
vol. 2 (1951) pp. 11h-ll7.

Divinsky,N., Riggs and Radicals, University of Toronto

Rerstein, I. N., Noncummutative Rin s, Carus Mathe-
matical Monograph§;no.l5, Amer can Mathematical
Society (196 ) ’

Jacobson, N., Structure and Representations g£_Jordsn
Al ebras, American MatfiematicaIDSociety Colloquium
Puglications, vol. 39, (1968).

Kleinfeld, Erwin, ”Standard and Accessible Rings",
Canadian Journal of Mathematics, vol. 8 (1956)
pp ' 335-31400

MacCrimmon, K., ”Norms and Noncummutative Jordan
Algebras," Pacific Journal of Mathematics, vol. 15

(1965) pp. 925-956.

Mathiak, K., "Zur Theorie nicht endlich-dimensionaler
Jordanalgebran uber Korper einer # 2," Journal fur
Methematik, vol. 22b (19 ) pp. 185-201.

Smiley, M. F., ”Application of R Radical of Brown and
McCoy to Non-associative Rings, American Journal of
Mathematics, vol. 72 (1950) pp.-93-lOO.

Tsai, C., Unpublished Lecture Notes, Michigan State
University, (1969).

Tsai, C., "The Levitzki Radical in Jordan Rings,"
Proceedings of the American Mathematical Society,
vol. 2t (1970) pp. 119-123.

Tsai, C., "The Prime Radical in a Jordan Ring",

Proceedings of the American Mathematical Society,
vol. 19 (1968) pp. 1171-1175.

55

S6

1%. Z?V18kfiv’ K. A., "Solvability and Nilpotence of Jordan
Rings, Algebra 1 Lo ika Seminar., vol. 5 (1966)
pp. 37-58. (Russian