PRINCIPAL IDEAL
J URBAN ALGEBRAS

   

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MICHIGAN STATE UNIVERSITY *
ROBERT MELVIN ANDERSON ;

' 1971 ”

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thesis entitled

PRINCIPAL IDEAL JORDAN ALGEBRAS

presented by

Robert Melvin Anderson

 

has been accepted towards fulfillment
of the requirements for

$2; degree in Mathematic S

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ABSTRACT
PRINCIPAL IDEAL JORDAN ALGEBRAS
BY

Robert Melvin Anderson

The purpose of this paper is to study the structure

of certain classes of principal ideal Jordan algebras.

AtL‘n ,

A Jordan algebra J over a commutative ring with identity

is said to have property A ( property B ), if J satisfies
the polynomial identities 2(x,y,xz) + (z,y,x2) = 0 and
(w,x,yz) + (z,x,yw) + (y,x,wz) = 0, J has an identity,

and each ideal A of J contains an element x such that A
is equal to the intersection of all the quadratic ideals
which contain X ( such that A = JUX ). The main results

of this paper are:

Theorem. If J is a Jordan algebra with property A, then

:1
J=eZ

Ji where each Ji has property A and is either a
1—1

u—prime algebra or a u-primary algebra containing a nonzero

nilpotent ideal.

 

Robert Melvin Anderson
Theorem. A Jordan algebra J has property B if and only
n
if J = 0 Z Ji where each summand is a simple Jordan
i=1

algebra with identity.

Theorem. Let J be a Jordan algebra over a field. J

n
has property A if and only if J = eiZlJi where each
summand has prOperty A, all but at most one of the summands
are simple Jordan algebras, and each summand is either a
simple Jordan algebra or contains only one proper u-prime

ideal, which is nilpotent.

 

    

‘ PRINCIPAL IDEAL JORDAN ALGEBRAS
: - By
\

Robert Melvin Anderson

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

A THESIS .
DOCTOR OF PHILOSOPHY ‘

|

Department of Mathematics l
|

1971

 

PRINCIPAL IDEAL JORDAN ALGEBRAS
BY

Robert Melvin Anderson

A THESIS
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
DOCTOR OF PHILOSOPHY
Department of Mathematics

1971

 

 

PLEASE NOTE:

Some Pages have indistinct
print. Filmed as received.

UNIVERSITY MICROFILMS

 

 

‘W‘-‘:._ i
'

TABLE OF CONTENTS

INTRODUCTION . . . . . . . . . . . . . . . . . . . 1
CHAPTER I. JORDAN ALGEBRAS WITH PROPERTY A . . . . 4

CHAPTER II. JORDAN ALGEBRAS WITH PROPERTY A OVER
A FIELD . . . . . . . . . . . . . . . . . . . 30

APPENDIX. POLYNOMIAL IDENTITIES IN JORDAN ALGEBRAS
OVER A COMMUTATIVE RING WITH IDENTITY . . . . . 36

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . 42

ii

 

 

 

 

INTRODUCTION

Let R be a commutative ring with identity. J is
a Jordan algebra over R if J is an R—module with a product

defined which satisfies the distributive laws, the polynomial

identities xy - yx = 0 and (x2,y,x) = (xzy)x - x2(yx) = 0,
and r(xy) = (rx)y = x(ry) for all x,y€ J and r6 R.
For any Jordan algebra J, the operator Ux y is
I

defined to be zUX y = {x,z,y} = x(zy) + (xz)y - z(xy) for
x,y,z€ J. Ux,x lS denoted by Ux' UX and UXIY are related

by the equation UX = U + U + 2 If A and B are

+y x y Ux,y'
subsets of J, then AUB is defined to be the set of all
finite sums of elements of the form an where as A and
b€ B.

In a Jordan algebra J, a subset A is called a
quadratic ideal if A is an R-module and JUA§.A. For any
x6 J, JUx is a quadratic ideal denoted by [x]. Rx + JUX
is a quadratic ideal equal to the intersection of all the
quadratic ideals which contain x and is denoted by <x>.
For example, if one considers the polynomial ring F[X] as
a Jordan algebra over a field F, then [X] is equal to the
principal ideal generated by X2, and <X> is equal to the

principal ideal generated by X.

 

2

A special Jordan algebra is a subspace of an
associative algebra over a field of characteristic not
equal to two which is closed under the Jordan product
x-y = %(xy + yx).

In order to use the polynomial identities which
arise from the study of Jordan algebras over a field of
characteristic not equal to two, one could study Jordan
algebras which have no elements of order two or, less

restrictively, those which satisfy the polynomial identities

h(x,y,Z) = 2(x,y,xz) + (z,y,x2) = 0 and d(w,x,y,z) =
(w,x,zy) + (z,x,yw) + (y,x,wz) = 0 where (x,y,z) = (xy)z —
x(yz) ( see appendix ). In this paper it shall be

assumed that all Jordan algebras satisfy the polynomial
identities h(x,y,z) = 0 and d(w,x,y,z) = 0. Thus identities
such as UbUa = UanUa ([2], p.52), and identities proved

by Macdonald's theorem for Jordan algebras over fields

hold in any Jordan algebra ([2], pp.40—47).

Macdonald's theorem. Any identity in three variables
which is of degree of at most one in one of these variables,
which holds for all special Jordan algebras, holds for

all Jordan algebras.

This paper deals with finding results for Jordan
algebras similar to the well known result for commutative,
associative rings with identity that a ring is a principal

ideal ring if and only if it is a finite direct sum of

 

3
principal ideal domains and special principal ideal rings
([5], pp.242-247). This result was generalized to
noncommutative principal ideal rings, where it is assumed
that each right ideal is principal [1]. In Jordan algebras,
quadratic ideals seem to correspond with the one-sided
ideals in the associative case; thus it would seem that a
similar result for Jordan algebras could be obtained if
it were assumed that each quadratic ideal were principal,
but actually one only need assume that each ideal is a

principal quadratic ideal of the form <x>.

 

CHAPTER I
JORDAN ALGEBRAS WITH PROPERTY A

In this chapter it will be understood that all
Jordan algebras are over an arbitrary commutative ring R

with identity.

Lemma 1—1. In a Jordan algebra XUy,zU = ZUy,xU for any
Y Y
x,y,z€ J.

Proof. If J is a special Jordan algebra then:

xU = {{y,z,y}lxry}

%(yzyxy + yxyzy)

1 z xU ) + (xU )z
2[y ( y y y)
= ZUy,xU
By Macdonald's theorem the identity holds in any Jordan

algebra.

Lemma 1—2. If C is an ideal in a Jordan algebra J and
an€ C for any a,b€ J, then aU<b>§;C. If A is a quadratic

ideal and AUbQ C, then AU<b>§;C.

Proof. For J an algebra over R, let rb + dUb be
an arbitrary element of <b>, where d6 J and r6 R.

aUrb+dUb = aUrb + aUdUb + 2aUrb,dU

b
_ 2
— r an + anUdUb + 2aUrb,dU

4

b

 

5

= r2(an) + (an)U + 2rdU by lemma l-l.

U
d b b,an

Thus aU C, and aU<b>Q C. The second statement

6
rb+dUb

follows from the first.

Lemma 1-3. In a Jordan algebra, bn(an) = (abn)Ub for n eDL
Proof. Let Rx be the operator which multiplies on
the right by x. bn(an) = a(2Rb2 - Rb2)Rbn
= aRbn(2Rb2 - sz) since
ijin = RXiRXj for any i,j€ N by ([3], p.92). Thus

bn(an) = (abn)Ub.

Lemma 1—4. In a Jordan algebra with identity,

(ann)(cUbn) = ban n where a,b,c€ J and n6 N.

a,cUb

Proof. If J is a special Jordan algebra then:

U = §(U — U — U
an,cUb (a+b)Ub an

l _ _
(UbUa+cUb UbUan UbUcUb)

2
l _ _
2(Ubwa + Uc + 2Ualc)Ub UbUan UbUCUb]

cUbJ

= UbUa,CUb'
Thus in any Jordan algebra Uan,cUb = UbUa,cUb and
- = =2

(an)(cUb) — anUb'CUb lUbUa,cUb b Ua,cUb' Assume the
statement is true for all n i k.
(ank+1)(CUbk+l) = ((an)Ubk]((CUb)Ubk)

=2k

b UaU ,aU Ubk
b b
= b2k<U

bUa,cUb)Ubk
= b2<k+‘)U U k+ . Thus the lemma
a,c b 1

holds by induction.

 

6
i
Lemma 1-5. If J is a Jordan algebra then (an)c = i
|
i

2an,cb - (ac)Ub for any a,b,c EJ.

Proof. (an)c - 2aU + (ac)Ub

b,cb

2((ab)b)c - (ab2)c — 2(ba)(cb) - 2((cb)a]b + 2((cb)b)a

+ 2((ac)b)b — (ac)b2

Ii

2(((ab)b)c + ((cb)b)a + ((ac)b)b) - (ab2)c — 2(ba)(cb)

— 2((cb)a)b — (ac)b2

2(2(ab)(bc) + b2(ac)] — 2(ba)(cb) - (ab2)c — 2((cb)a)b
- (ac)b2 by the polynomial identity d(a,b,b,c) = O,
= 2(ab)(bc) + b2(ac) — (ab2)c - 2((cb)a]b = 0 by the

identity d(b,a,b,c) = 0.

Corollary 1-6. If A and B are ideals of a Jordan algebra
J, then AUB is an ideal of J.
Proof. For any aE A, b EB, and c6 J, then

(aU )c = —(ac)Ub +2aU by lemma 1-5. 2

an,cb =
)c EAUB and

b b,cb

an+cb - an —aUcb€ AUB‘ Therefore (an

AUB is an ideal of J.
Lemma l—7. Let J be a Jordan algebra and A be an ideal
of the form A = <a>. If b €J, then ban€ An when n is an
odd positive integer, and bane An"1 when n is an even
positive integer.

Proof. If n is an odd integer then n = 2m + l
for some integer m. an = aUam€ A(UA)m which is an ideal
by corollary 1-6. Therefore ban6 A(UA)m§;An. If n is

an even integer then n = 2r for some integer r. azr'lé

 

7
A(UA)r-1 where A(UA)r‘l is an ideal by corollary 1—6.

ban = ((azr‘1)a)b 6 ((A(UA)r")a)bCA(UA)r‘1CAn".

Lemma 1-8. Let J be a Jordan algebra and C be an ideal
of J of the form <c> for some c €C. If m and n are
positive integers such that n i 2, then (aUCm)cne

n—z

CnUCm — (aUcm+1)c

Proof. Suppose J is a special Jordan algebra.

(aUcm)°n = %(cmacm+n + cm+nacm)
= cm(C°a)cm+n—1 + cm+n—1(c'a)cm
_ %(cm+1acm+n—1 + cm+n—1acm+1)
= ((a°C)Ucm)cn-1 + cn"((a-C)Ucm)
- §((aUCm+1)cn'2 + cn'2(aUcm+x))
= 2((a-c)UCm)-cn'1 — (aUCm+1)-cn'2.
Thus (aUCm)cn = 2((ac)UCm)cn_1 - (aUcm+1)cn"2 in any

Jordan algebra by Macdonald's theorem. Therefore

n _ n-1 n-2
(aUCm)c — 2((ac)c )Ucm - (aUcm+1)c by lemma 1—3,
n n—z
€ C Ucm - (aUcm+1)c .
Let 6:N + N be a function defined as d(n) equal to the
greatest integer function of §(n + 2). For example

5(1) = 1; 6(2) = 2: 6(3) = 2; 5(4) = 3, and 6(5) = 3.

Lemma 1—9. If J is a Jordan algebra and C is an ideal
of J of the form <c> for some c 6C, then (aUc5(n))cm €

CmUc5(n) + [c6(m+n)] for any positive integers n and m.

 

8
Proof. If m is an odd integer then:
(aUc5(n))cm = (acm)Uc5(n) by lemma 1-3,
€ CmUc5(n) by lemma 1—7,
s CmUc5(n) + [C6(m+n)].
If m is an even integer where m = 2t then:

(aUc6(n))cm€ CmUc5(n) + Cm_2Uc5(n)+1 + ... + CZUc6(n)+t-1

6(n)+t

+ [c ] by lemma 1-8,

-_ 6 (n)+t
E(Cm + Cm 2Uc + ... + CzUct—1)UC6(n) + [C

6(m+n)]

]

€CmUC5(n) + [c

Lemma 1-10. Let J be a Jordan algebra with identity
and C be an ideal of J of the form <c>. For any positive

odd integer n, if Cng [c6(n)] + Z chl'i’1 then Cn is

i=1
an ideal.
Proof. If X€ Cn then x = aUCS (n) + Z riCn+i-1-
i=1
For any b6 J, xb = (<‘=1UC<S(n))b + X ribcn+i—1 e

(aUc5(n))b + Cn by lemma l—7, since n is an odd integer.

d(n)

aUC5(n)€ J(UC) which is an ideal by corollary 1—6,

and thus (aUC5(n))b€ J(UC)6(n)
6(n)-1
€ JUC (UC)
e C2(6(n)-1)+1
6 025(n)-1
6 c“.

Therefore XbE'Cn, and Cn is an ideal.

9
Lemma l—ll. Let J be a Jordan algebra with identity
and C be an ideal of the form <c> for some c6 C. Let
i and m be positive integers such that l i i i m. If

(f1§[c6(n)] + X ch+i'1 for all n i m, then
i=1

m

c25<i> c .

Ua.bch(m)-5(i)e

P f. 25(i)
roo c UarbUcd(m)-5(i)

<c=5(i)a)<bucs(m)—a(i)> + (cza‘i)<bUca(m)-5<i)>>a
- c25(i)(a(bucs(m)-6(i)))
= - (((bUC6(m)-6(i))c)aJ°26(i)-1

(((bUca(m)-5(i))czé(i)_1)aic

+ c25(i)(a(bUC5(m)-6(i))l + (a0)((bUC6(m)-6(i))026(i)_l)
+ <ac25<i)“)((bUc5(m>—5(i))c) + (026(i)(b0c5(m)‘5(i’)]a
- c25(i)(a(bucé(m)_5(i))) by the use of the identity
d(buc5(m)_5(i,,a,c,e25(i)‘1) = o,
= - [((bC)Uc6(m)—5(i))a)°26(i)_1

— (((bc25(i)'1)Uc5(m)-5(1))a]C

+ (ac)((bc25(i)“)uca(m)-6(i>)

+ (ac25(i)“)((bc)Uc5(m)~5(i)) + (<b625‘i))Uca(m>—s(i))a

by lemma 1-3,
€c25(m)'1 by lemma 1—7 and lemma 1-10,

ecm.

Theorem 1-12. If J is a Jordan algebra with identity
and C is an ideal of J of the form <c> for some c6 C, then
26(n)—n _
] + E ch+l'1.
i=1

 

10
Proof. The statement is true for n = 1 since
C = <c> = [c] + Rc. Assume that the statement is true
for all n i k. Let i and m be positive integers such

that l i i i m and i + m = k + 1. Let we Cl and 26 d“.

25(i)-i .
By the induction hypothesis, w = aUc5(i) + Z rncnl'l'1
n=l
25(m)—m + 1
and z = bUc6(m) + nil sncn m .
26(i)-i _ 26(m)-m
wz = (aUc5(i) + E rncn+l'l)(bUc5(m) + 2 sncn+m")
n=l n=l
25(i)-i +,_1
€(3Uc6(i))(bUcd(m)) + ( nil rncn l )(bUc6(m))
26(m)-m w
+ (aU 6(i))( X s cn+m-1) + E Rck+n
c n
n=l n=l
25(1) 25(i)’i n+i 1
6c U U ' + r ((bU )c _ )
aIbUC6(m)-6(i) c5(l) nil n C6(m)
26(m)—m w
+ E s ((aU 5(1))cn+m'1) + E Rck+n
n=l n c n=l

by lemma 1-4, where Uc5(m)'5(i) is the

identity operator when 6(m) = 6(i),

€CmUeMi) + ClUc5(m) + [06mm] + £1 Rck+n
n:

by lemma 1—9 and lemma l—ll,

26(m)-m
6([C6(m)] : RCm+n-1iUc6(i)
n=l
. 26(i)-i _
+ ([C6(l)] + Z RCl+n-1]Uc5(m)
n=l
+ [C6(m+i)] + RCk+n

 

11

by the induction hypothesis,

6 [05(m)+5(i)] + of Rcm+26(i)+n-1+ of Rci+26(m)+n-1
n=l n=l

+ [c6(m+i)] + Z RCk+n
n=l

e[c6(m+i)] + 2 ch+m+i—1 + Z Rck+n
n=l n=l
6[C<S(m+i)] + l RCk+n
n=l
26(k+l)—(k+l)
e[c5(k+1)] + f Rc<k+1)+n'1.
n=l

26(k+l%—(k+l)

Thus Cicm ; [06 (k+1)] RC (k+1)+n—l.

n=l

26(k+l)—(k+l)
+ i

Ck+1 = Cicmg [C6 (k+l)]

i+m=k+l n=l

RC(k+l)+n-l.

Therefore the theorem holds by induction.

Corollary l-l3. If J is a Jordan algebra with identity;
A and C are ideals of J; C is of the form <c> for some

c 6C, and cn€ A, then CanA.

Lemma 1-14. If J is a Jordan algebra with identity and
A is an ideal of J of the form <a>, then An is an ideal
of J for any odd positive integer n.

Proof. The lemma follows immediately from

lemma 1—10 and theorem 1—12.

Lemma 1—15. If B is an ideal of a Jordan algebra J,

and B is of the form <b> for some b6 B, then 2an cb€

 

12
aUB for any a,c€wJ.
Proof. b + ch B since b,ch B. 2an =

an+cb ~ an — aUcb€ aUB.

Lemma 1-16. Let C and Bi be ideals of the Jordan algebra
J for i = l,2,...,n, where Bi is of the form <bi> for

.€ .. ...
some bl Bl If an1Ub2 Ubne C, for some a€.J then

AUB UB ...UB SlC, where A is the principal ideal generated
1 2 n
by a. In particular if B is an ideal of the form <b>

and aU n6 C then AUanC.

b

Proof. For n = l, aU G C implies (aU )d =

b1 b1

2an ,dbl - (ad)U

6 C for any d€.J by lemma 1-5.
1 1

b

2 aU Q C by lemma 1—15 and lemma l-2. Therefore
1

aU G
b1,db1 B

(ad)Ub

6 C and AUb Q C. Thus by lemma 1—2, AUB Q C.
1 1

1

Assume the lemma holds for all n < k and aU U ...U E
" b1 b2 bk+1

C. (aU U U )U E C and b the case n = l
A U S C where A is the principal ideal generated by

6 . .
anlsz...Ubk. anlsz...Ubk Ak and by the induction

' c
hypotheSis AUBIU B _ Ak. Therefore AUB

U
k 1

...U Q
32 B2 Bk+1

(AU )U Q A U Q
Bk Bk+1 k Bk+1

B UB ...U C. Thus the lemma holds
1 2

by induction.

Lemma l-l7. If J is a Jordan algebra and A is an ideal

of the form A = <a>, then [a] is an ideal of J.

13
Proof. [a] is closed under addition by definition
and under multiplication by lemma 1-4. If x €[a], then
= e . .
x dUaé JUA‘ For any b J, xb€ JUA Since JUA is an

ideal by corollary 1—6. Thus:

.n
Xb = .2 (diU(r.a + c.U )]
1-1 1 l a
n
= 2 (diUr a + diU U + 2diUr a c U )
i=1 i Ci a i ' i a
n
= Z(r.2d.u +d.UU U +2r.d.U )
._ l i a l a c. a l l a,c U
1-1 1 1 a

E [a] since AZQ [a] by theorem 1-12. Therefore xb €[a],

and [a] is an ideal of J.

An ideal of a Jordan algebra is said to be
irreducible if it is not equal to a finite intersection

of ideals strictly containing it.

Theorem 1-18. In a Jordan algebra with the ascending chain
condition on ideals, every ideal is a finite intersection
of irreducible ideals.

Proof. Let C be the family of ideals which are
not the finite intersection of irreducilbe ideals. Suppose
C # O, then there is a maximal element A in C. Since A
cannot be irreducible, A = fEHAi where the A.l are ideals

l:

of J strictly containing A. By the maximality of A in C,

n.
A. = (i A.

. where the A.. are irreducible ideals. A =
i 3:1 l] I]

 

14
n.

Aij contradicting A belonging to C. Therefore

3::

i=1 j=1
C = ¢ and every ideal of J is a finite intersection of

irreducible ideals.

Lemma 1—19. If J is a principal ideal Jordan algebra with

identity, then J has the ascending chain condition on ideals.

Proof. If A1§1A2C ... is any ascending chain of
ideals, then A = L) Ai is an ideal. Therefore A is gen-
i=1

erated by some element a. a6 iii Ai which implies that
i=1

aé A for some n. Thus A = A for k > n.
n n k —

An ideal C in a Jordan algebra J is said to be
u—primary if for any ideals A and B such that AUBC C
then AQC or BnQC for some positive integer n. A Jordan
algebra is said to be u-primary if the ideal (O) is
u—primary. A Jordan algebra is said to have property A
if the algebra has identity and every ideal is of the form

<x> for some element x in the ideal.

Theorem 1-20. If a Jordan algebra J has property A,
then any irreducible ideal of J is u—primary.

Proof. Let C be any irreducible ideal and A and
B be any ideals of J such that AUBQQC. A = <a> and B =
<b>. Let Mi = { ye J : VUbiQ C where V is the principal

. ' ' C
ideal generated by y }. Mi 18 an ideal of J and Mi"Mi+l'

 

15
Therefore there exists a positive integer k such that
Mk = Mn for n i k since J has the ascending chain condition
on ideals by lemma 1—19.
B2k+‘£;[bk+‘] + Rb2k+1 by theorem 1—12,

_§[bk+11 + RbU

bk
gtbk].
Let 2 €(sz+1 + C)rA(A + C). Therefore 2 = dUbk + c =
a' + c' where c,c'€ C, a'é A, and d6 J. zUb = dUbk+1 +
= l l l l e
cUb a Ub + c Ub’ but (a Ub + c Ub), cUb C. Therefore

dUbk+16 C. By lemma 1-16, DUbk+1§;C where D is the

principal ideal generated by d. Thus de Mk+1 = Mk’ and
dUbk ec. Therefore 0 = (52k+1 + C)rW(A + C), but 32k+1
is an ideal by lemma 1-14. Since C is irreducible,

BZk'HQC or AQC, and C is u-primary.

Lemma 1-21. If J is a Jordan algebra, then Ub2(Uan2)nUa =

2n+1
(szUa) -
Proof. For n = 0, szUa = UbZUa. Assume the

statement holds for n = k, that is Ub2(Uan2)kUa =

2k+1
(UbZUa) I
= U

U 2 k+1U 2 k U
b (Uanz) a b (Uan2) Uan2 a

= U 2U 2 k U 2U
b b (Uanz) Ua b a

= UbZUan2(Uan2)kUan2Ua
_ 2k+1
‘ szUa(Ub2Ua) szUa
= 2(k+1)+1
(szUa) .

Therefore the lemma holds by induction.

 

16

Lemma 1—22. If J is a Jordan algebra then (an)2n =

b2(U U )rnU U where r = 2n-1 - l
a b2 a b n '
_ 2 _ _ _
Proof. For n — l, (an) — anUb — lUbUaUb —
2 _ 2 o
b Uan - b (Uanz) Uan. Assume the statement holds for
k r
= . ' 2 = 2 k
n k, that is (an) b (Uanz) Uan.
k+1 k
2 _
(an) - ((an)2 )2

_ 2 rk 2
- [b (Uanz) Uaub)

= 2 r
b U ) kU Ub

2
b (Uan2 a

rkU U

_ 2
_ b Uan2(Uan2) a b

= sza(Ub2Ua)2rk+lUb by lemma 1—21,
= 2 2r +1

b (Uanz) k Uan
= 2 rk+1 - =

b (Uanz) Uan Since 2rK + l rk+l‘

Therefore the lemma holds by induction.

An ideal C of a Jordan algebra J is said to be a
u—prime ideal if for any ideals A and B such that AUBQ(Z
then either A€;C or BQIC. A Jordan algebra is said to
be u-prime if (0) is a u-prime ideal. For any ideal A
of J, the u-prime radical P(A) of A is defined to be the

intersection of all the u-prime ideals in J which contain A.

Lemma 1—23. Let A and C be ideals of a Jordan algebra
with identity, and let A be of the form <a>. If an€ c

for some positive integer n then A§;P(C).

 

 

 

17
Proof. By corollary l-l3, AanZC. A2“ = (5)
in J = J/C. Since it is proved in [4] that P((U))
contains all nilpotent ideals, A£;P((5)). Therefore

A§P(C).

Theorem 1—24. If J is a Jordan algebra with property A
and C is a u-primary ideal, then P(C) is a u-prime ideal.
In particular, the u—prime radical of an irreducible
ideal is u-prime.

Proof. Suppose P(C) is not a u-prime ideal,
therefore there exists ideals A and B such that AUB;.P(C)
while neither A nor B is contained in P(C). Let A = <a>
and B = <b>. anG P(C). (an)2n€ C for some integer 2n
by theorem 7 in [4].

r
2 n _
b (Uan2) Uan6 C by lemma 1 22,
l r“ ec
Ub(Uan2) Uan
In 6C
lUb(UanUb)r Uan
n
C _
JUB(UAUBUB) UAUB__C by lemma 1 16.
n . .
Now JUB(UAUBUB) UA is an ideal of J by corollary 1-6.
r
Therefore JUB(UAUBUB) nUA£;C or BmS;C for some positive
integer m since C is a u—primary ideal. BHHiC for if
BmSQC then B£;P(C) by lemma 1—23, contradicting the choice
r
of B. Therefore JUB(UAUBUB) nUAQ C. By repeating this
argument and by using the fact that no power of A or B
can be contained in C, one obtains J§;C. This contradicts
the fact that P(C) is not a u-prime ideal. Therefore

P(C) is a u—prime ideal.

 

18
Lemma 1—25. Let J be a Jordan algebra; P be a u—prime
ideal, and B be an ideal of the form <b>. If aé.J such
that an€ P then a6 P or b6 P.

Proof. If ané P then AUbC P where A is the
principal ideal generated by the element a, by lemma l-l6.
AUBQ P by lemma 1-2. Therefore ASP or BQP and a5 P or
b 6?.

Lemma 1—26. If { Ai }?=1 is a set of ideals of a Jordan

algebra J such that {% Ai = (0) and Ai + Ji = J where
i=1
n i
J.= A,thenJ=9 J..
l k¢i k i=1 1

Proof. For any ><GJ, x = xi + yi where xi€ Ji’

n
yi6 Ai. x — kélxk = (x - Xi) - kgixk = yi — H; xk 6 Ai,
n
since xké JkSQAi for i # k. Thus x —k21xkéfjfi= (0),
n n
and x = Z xk. Therefore J = '2 Ji. The sum is direct

since Jifl kgiJngifl Ai = (0).

Lemma 1-27. Let J be a Jordan algebra with identity, B

and { Ai }I=l be ideals of J. If B is comaximal with each

Ai then B is comaximal with [A Ai.
i=1

n n n

__ n _ C

Proof. J — J - U (B + A1); B +_HA.1§B + L) Ai-
1—1 1—1 1—1

n
J. Therefore B is comaximal with [1 Ai.
i=1

 

 

19
Lemma 1—28. Let J be a Jordan algebra with identity and

n
{ Ai }.

1:1 be a set of pairwise comaximal ideals. If

n n
(j A. = (0), then J = e Z J. where J. = J/A..
. . l 1 1
i=1
Proof. By lemma 1-26 and lemma 1—27, Ai + J. =
i

n n
J and J = e 2 Ji where Ji = (W Ak. Let 1 = 2 e.
=1 k¢i i=1 1

where eie Ji' Let Yi(a) = aei, which is a homomorphism
of J onto Ji' The kernel of Yi’ Ki = { x6 J : xei = 0 } =

KgiJkQ Ai, but AirWJi = (0) and Yi(Ai) = (0). Therefore

K. = A., and J. 2 J/A..
i l i 1

Lemma 1-29. Let J be a Jordan algebra with identity and

{ Ai }I=l be a set of ideals of J. The Ai are pairwise
comaximal if and only if the P(Ai) are pairwise comaximal.

Proof. If Ai + Ak = J, then P(Ai) + P(Ak) = J

since an ideal is contained in its u—prime radical. If

P(A ) + P(A = J, then 1 = a. + a where a.6 P(A.) and
i i i i

k) k
aké P(Ak). a.n€ Ai for some positive integer n. 1n =
n
(ai + ak)

a positive integer m such that am €Ak. l =

i
= ain + a for some aGIP(Ak). There exists

lnm =

(a.n + a)m€ (A. + a)m§ A. + amQA. + A . Therefore
1 i i k

Lemma 1-30. Let J be a Jordan algebra with property A.

If P is a u—prime ideal of J, and ngA, where A is an

ideal of the form A = <a>, then F’Q (1 [a1].
i=1

 

20
Proof. Let P = <p> and p = ra + dUa.
ap = ra2 + (dUa)a = rUa + (da)Ua by lemma 1-3,
= (r + da)Ua€ P.
a f P, for if a6 P then A QP contradicting PSA. Therefore
by lemma 1-25, r + da€ PC3A, and r6 A. Let r = sa + bUa'
Thus p = (sa + bUa)a + dUa
= sa2 + (bU )a + dU
a a
= sU + (ba)U + dU by lemma 1-3,
a a a
= (s + ba + d)U
a
E [a].
Therefore P Q [a]. Assume PS;[ak], then p = cUak for some
c EJ. cUak = C(Ua)k€ P, therefore by lemma 1-16, C(UA)k§
P where C is the principal ideal generated by c. Since
P is a u—prime ideal and A¢P then CCP. c 6P9 [a].

_ _ _ _ k+1
_ . —— 6 0
Thus c eU p cU k eU U k eU k+1 [a ]

Thus P £[ak+1] and by induction PC _fil[ai] .
1:

Lemma 1-31. Let J be a Jordan algebra with property A,

and let P1,P2 be u-prime ideals of J. If Pl¢.P2 and

Pail- P1 then Pl + P2 = J.

Proof. Let Pl + P2 = P = <p>. P19 P and P2g_P
since P1¢ P2 and 15¢ P1. By lemma 1-30, P1; [p] and P29
[p]. Thus Pl + Pzg [p]§<p> = P1 + P2, and [p] = <p>.

p = aUp for some a6 J. lUp = anU = lUpUaUP = szaU

p.
p
Thus (1 - sza)Up = oePl. p¢ P1, for if pep1 then

PCP1 contradicting P2 ¢P1. Therefore by lemma 1-25,

1 - sza6 P1: P, and lE P = P1 + P2. Therefore Pl +

 

 

21

Lemma 1-32. Let J be a Jordan algebra with property A.

1’1
If{Ci}

i=1 is a set of u-primary ideals of J such that

_ n
P = P(Ci) for i = l,2,...,n, then C = (7 C. is a

u-primary ideal and = P(C).

P
Proof. Let P = <p>, thus pni€ Ci for some
positive integer ni where i = l,2,...,n. Let m =

max{ ni }I=l' Thus pmé C and P§;P(C) by lemma 1—23.
Therefore P = P(C). Let A and B be any ideals of J such
that AUBQ C. Let A = <a> and B = <b>. If AQCi for

i = l,2,...,n, then A QC. If ASACj for some je

{ l,2,...,n }, then Bkgcjgp, implying that BQP, since
Cj is a u—primary ideal and P is a u-prime ideal. Since
pme c, szgc by corollary l—l3. Thus bk(2m) 6 Pm;

C, and B“km£;C by corollary 1-13. Therefore C is a

u—primary ideal.

Lemma 1-33. If J is a Jordan algebra with property A,
and C1,C are u—primary ideals such that P(C1)§;P(C),
then clgc.

Proof. Let P(Cl) = P1 and P(C) = P = <p>.
P1 and P are u-prime ideals by theorem 1-24. There

exists and integer m i 2 such that pmézC. By theorem 1-12,
szgmmfll + szm + Rp2m+1;c. But C19 P g 0 [pl] ;

[pzm]§rP2mS;C by lemma 1-30. Suppose C1 = C, then

 

22
pm6 Pl- By lemma 1—23, Pg?l contradicting P1; P.

Therefore Cl; C.

Lemma 1-34. Let J be a Jordan algebra with property A.
If A is an ideal of J, then there exists a finite set of

pairwise comaximal, u—primary ideals { Ci }?

1:1 such that

Proof. By theorem 1—18 and theorem 1-20, A =
m
{1 Ai where the Ai are u-primary ideals. Let M =
i=1

{ l,2,...,m }, and I = { i€lw : P(Ai) = P(A1) }. Let

D1 = -C21 Ai. 01 is a u-primary ideal with p(A1) = P(Dl)
l

by lemma 1—32. Let r€M — I, and H = { iEM 2 P(Ai) =

P(Ar) }. Let 02 = (A Ai. 02 is a u—primary ideal with
i6 H

P(Dz) = P(Ar) # P(Dl) by lemma 1—32. In this manner, one

can construct from { Ai }?=l the set { Di )k

. such that
i=1

k
A = {1 Di and the vi are u-primary ideals with the P(Di)
i=1

distinct. Let { Ci }? be the set of all minimal ideals

:1
}k

n
of the set { D. ._ . Therefore A = {A C.. Suppose
i i—l i=1 i

P(Ci) + P(Cj) # J for some i # j, then without lose of
generality one may assume by lemma 1—31 that P(Ci)§:P(Cj).
By lemma 1—33, CiQCH contradicting the choice of the Cj

being minimal in the set { Di } Therefore P(Ci) +

i=l'

p(c.) = J for i a j. By lemma 1-29, { c. }n

J 1 i=1 is a set

 

23

of pairwise comaximal ideals.

Theorem 1—35. If J is a Jordan algebra with prooertv A,
n
then J = 0 Z Ji where each Ji has property A and is
‘ i=1
either a u-prime algebra or a u-primary algebra containing
a nonzero nilpotent ideal.

n
Proof. By lemma 1-34, (0) = {W Ci where the Ci
i=1

are pairwise comaximal, u-primary ideals. Thus J =
n
0.2 Ji where Ji = J/Ci by lemma 1-28. Each Ji has
131
property A since it is a homomorphic image of J. If Ci =
P(Ci), then Ji is a u-prime algebra, if not then Ji is a

u-primary algebra with a nilpotent ideal P(Ci)/Ci'

Lemma 1-36. Let J be a u—prime Jordan algebra with
property A. If the u-prime ideals are pairwise comaximal,

then J is simple.

Proof. Suppose J is not simple, then J contains

a nontrival ideal A. A =

:35

Ci where the Ci are u-primary

i l
ideals by lemma 1-34. Thus J contains a nontrival
u—primary ideal C. P(C) is a u-prime ideal by theorem
1-24, but J has (0) and J as its only u-prime ideals.
If P(C) = (0), then C = (0) contradicting C being non—

trivial. If P(C) = J, then 1 = lné c for some positive

integer n: thus C = J contradicting C being nontrivial.

 

 

24

Therefore J is simple.

A Jordan algebra J is said to have property B,
if J has an identity and every ideal is of the form
[x] for some x in the ideal. Property B is a special

case of property A.

Lemma 1—37. If J is a Jordan algebra with property B,
then the u-prime ideals are pairwise comaximal.
Proof. Suppose that P1, P2 are distinct u-prime

ideals. Suppose Pl + P2 f J. By lemma l—3l, one may

assume without loss of generality that Plgig. P1 = [p]
E = E =
for some p P1. p aUp for some a J. 1U_ anU
= 2 — 2 = 6
lUpUaUp p UaUp' Thus (1 p Ua)Up 0 P2. By lemma

l-25, p6 P2 or (1 — p‘Ua)E P2. pg P2, for if p‘EP2 then
PIE P2 which is a contradiction. Thus (1 — sza)€ P2
and P1 + P2 = J. Therefore the u-prime ideals of J

are pairwise comaximal.

Lemma 1—38. If J is a Jordan algebra with property B,
then J contains no nonzero nilpotent ideals.

Proof. Suppose A is a nilpotent ideal of J.

A = [a] for some a6 A. Thus a = bUa and an = 0 for some
pOSltlve integer n. a = bUa = bUbUa = bUanUa =
bU U U b(Uan) Ua ... b(Uan) Ua A (0)

a b bU
a

by theorem 1-12. Therefore A = (0).

 

 

25
Theorem 1-39. A Jordan algebra J has property B if and
n
only if J = e E Ji where each summand is a simple
i=1
Jordan algebra with identity.

n
Proof. If J has property B, then J = e Z Ji
i=1

where each summand is either a u—prime algebra or contains
a nonzero nilpotent ideal by theorem 1—35. Since each
summand has property B, it must be a u-prime algebra by
lemma 1-38. Thus by lemma 1-37 and lemma 1-36, each
summand is a simple Jordan algebra with identity. If

n
J = eigl Ji where each summand is a simple Jordan algebra

with identity, then J has identity. Let A be any ideal,

then A = 9 Ai. Let e = e

l i

II-Mb

lib/1'3

e. where e. = 1. if A. =
l l i i

i l

Ji and ei = 0 if A. = (0). Thus e€ A. [e] = A since
n

if as A then a = e Z b.U where b.€ J., and a =
= e. i i

n
(j bi)Ue€ [e].
1:
An element 2 of a Jordan algebra J is said to be

an absolute zero divisor if JUz = (0).

Lemma 1-40. If J is a Jordan algebra with property A
and J contains a nonzero nil ideal, then J contains a

nonzero absolute zero divisor.

 

26

Proof. If A = <a> is a nonzero nil ideal then
an = 0 for some positive integer n, and A2n = (O) by
corollary 1—13. By lemma 1-14, AI“ is an ideal for
any positive odd integer m. Thus J contains a nilpotent
ideal of index 2 or 3, since either Ak-l or Ak—2 is an
ideal of J where k is the nilpotent index of A. Let
B = <b> be any nilpotent ideal of J of nilpotent index

2 or 3. If b2 = 0 and c€J then:

cU = 2(cb)b — cb2

b

= 2(cb)b since b2 = O,
= 2(rb + dUb)b since cb€ B = <b>,
= 2(db)Ub + 2rb2 by lemma 1—3,
= 2(db)Ub since b2 = 0,
= 2(sb + eUb)Ub since db€ B = <b>,
= 2sb3 + 2eUbUb
= 2sb3 + 2eUb2
= 0 since b2 = 0.

Therefore b is an absolute zero divisor. If b3 = 0,

and b2 # 0 then for c €J:
csz = 2(cb2)b2 - cb“
€ 83 = (0).

Therefore b2 is an absolute zero divisor. Thus J contains

a nonzero absolute zero divisor.

Theorem 1—41. Let J be a Jordan algebra with property A.
If J contains no nonzero absolute zero divisors, then

J is a direct sum of u-prime algebras.

 

27

n
Proof. By theorem 1—35, J = e Z Ji where
i=1

each Ji is either a u-prime algebra or a u-primary

algebra containing a nonzero nilpotent ideal. By lemma
1-40, J contains a nonzero absolute zero divisor if

some Ji is not a u-prime algebra. Therefore if J contains
no nonzero absolute zero divisors, then J is a direct

sum of u—prime algebras.

Any associative, commutative principal ideal ring
with identity is a Jordan algebra with property A when
it is considered as an algebra over itself, and thus it
could appear as a summand in theorem 1—35.

Another example of a possible summand can be
constructed from the vector space J over a field F, of
characteristic not two, with a basis { e11,e12,e21,e22,z }.
Under the multiplication given below, which is obtained

by putting the Jordan product on the noncommutative

 

 

e11 e12 e21 e22 Z

..l. l 1-.

e11 911 2912 2921 O 22

er le. 0 A(e +e ) 1e O

12 2 12 2 ll 22 2 1.2

. i

621 2e21 z(ell+e22> 0 2e21 0

1 i 1

e22 0 2912 2921 e22 2z
1. A

z 22 0 O 22 0

 

 

28
Jordan algebra described in ([3], p.147), J becomes
a Jordan algebra over F. The ideals of J are: (0);
<z> Q <e11+e22> = J, and thus J is a u—primary Jordan
algebra with property A. Also the subalgebra generated
by the set { e11, e22, e12 + e21, z } has property A,
and its ideals are: (0)S;<z> §<e11 + e22>.

The subalgebra T of the preceeding example,
generated by the set { l, e, z }, where l = e11 + e22
and e = e12 + e21, is a principal ideal Jordan algebra
without property A. Its ideals are (0), N, C, D, and
T generated by the elements 0, z, l+e, lwe, and 1
respectively ( the ideals C and D are not of the right
form for T to have property A ). (0) is an irreducible
ideal of T, and P((0)) = N, which is not u-prime since
CUDQ N and no power of either C or D is contained in N.
In view of this example, theorem 1—24 cannot be generalized
to include all principal ideal Jordan algebras.

Consider the Jordan algebra obtained by putting
the Jordan product on the upper triangular 2x2 matrices
over a field F of characteristic not two. This algebra is
a principal ideal algebra, but does not have property A
since the ideals A = Fe11 + Fe12 and B = Fe22 + Fe12
are not of the right form. (0) is not a finite
intersection of ideals properly containing it, and thus
it is irreducible. AUB = BUA = (0), but neither A nor
B is nilpotent. Thus (0) is irreducible but not

u-primary. In view of these last two examples, the

 

    

all principal ideal Jordan ‘-

 

CHAPTER II
JORDAN ALGEBRAS WITH PROPERTY A OVER A FIELD

In this chapter it will be understood that all

Jordan algebras are over an arbitrary field F of
characteristic not two. Let J be a Jordan algebra and A

an ideal of the form <a>. It was proved in Chapter I
that AZS;[a], and a question which arises is whether or
not [aJQ A2. If [a]£;A2 then all the powers of A are
ideals which would simplify many of the proofs. When J
is an algebra over a field, this question is partially

answered in the following lemma.

Lemma 2—1. If J is a Jordan algebra with property A,
and A is an ideal of J of the form A = <a>, then either
a26 [a2] or A2 = [a] which is an ideal of J.

Proof. By lemma 1—17, [a] is an ideal of J;
therefore [a] = <cUa>. Let 3 = J/A3. Thus TUE = 52 =
fEUg + EU-EUE = fEUa + EUEUEUa = £505 since bUaUCUa e
A3. If f 76 0, then EUa = F152, and cUaE f‘ia2 + A‘QAZ.
Thus [a]; A2. A29 [a] by theorem 1—12. Therefore A2 =

[a] which is an ideal of J. If f = 0, then a26 A‘.

30

 

 

31

a2 = dUaz + ra3 by theorem 1—12,

dUaz + ra(a2)

dU 2 + ra(dUa2 + ra’)

_ 2 _
— dUaz + r(ad)Ua2 + r Uaz by lemma 1 3,
6 [a2].

Lemma 2-2. Let J be a Jordan algebra with an ideal A
of the form A = <a>. If aze [a2], then a2€ An for any

n 1 1. In particular, if A is a nilpotent ideal, then

2 — 2 = =
Proof. Let a — bUaz. Thus a bUbUa2

1 _ ‘1 __ 2 _ _
b(Ua2Ub) Uaz — b(Ua2Ub) UbUaz — h(UazUb) Uaz — ... —

n E n
b(UazUb) Uaz A .

Lemma 2-3 Let J be a Jordan algebra with property A.
If A is a nilpotent ideal of J then A“ = (0).

Proof. Let A = <a>. If a2€ [a2], then a2 = 0
by lemma 2-2, and A3 = (0) by theorem 1—12. In View

of lemma 2-1, it suffices to show that the lemma holds in

the case when [a] = A2 is an ideal. Let n be the smallest
positive integer such that an = 0. Suppose n > 4.
Without loss of generality it may be assumed that n = 5,

since one could pass to the quotient algebra J = J/AS,
since A5 is an ideal by lemma 1—14.
A3 Q Fa‘ + [a2] by theorem 1-12,

g Fa3 + A“ since [a2]§;A“ when [a] = A2 is an ideal,

gFa3 + Fa” by theorem 1—12 and by the fact that

 

 

32
a = 0. A3 = <f1a3 + fza“> since J has property A.

be 3 H: 3 ‘6
a A thus a f3(fla + fza ) + bUf1a3+f2a“

faflaa + fafza“ since bU .6 A5 = (0). o =

fla3+f2a
a5 = fafla“ + f3f2a5 = fafla”, thus fgfl = o and f1 =

0 since a“ # 0. Therefore A3 = <f2a“>. a36 A3 and

3 _ ‘i ._ LO ' 5 _
a — fufza + chza. - fkfza Since CUfza“€ A — (0).
a“ = (a3)a = (fgf2a“)a = fkfza5 = 0, contradicting the

choice of n. Therefore n i 4. By theorem 1—12,
A“ = [a3] + Fa” + Fas, but [a3]S.A5 = (0), when [a] = A2.

A“ = Fa“ + Fa5 = (0) since a“ = 0.

Corollary 2—4. Let J be a Jordan algebra with property A.
If A is an ideal of J, then A“ = An for n i 4.

Proof. Let J = J/Am where m is an odd integer
greater then n. A is a nil ideal of J, and by lemma 2u3

A“ = ('6). Therefore A“CAm;AnQA“, and A“ = An.

Lemma 2-5. Let J be a Jordan algebra with property A,

and let A be an ideal of J. If an = 0, then An = (0)
where A = <a>.
Proof. If n = l, A = <O> = (0). If n = 2,
A2§;Fa2 + A3 by theorem 1—12,
crez + Fa3 + [a2] = (0) by theorem 1—12.

In View of lemma 2—1 and lemma 2-2, it may be assumed that

A2 [a] is an ideal of J. If n = 3,

ll

A3

10

Fa3 + [a2] by theorem 1—12,

Q Fa3 + A” since [a2]€;A“ when A2 = [a] is an ideal,

 

33
gFa3 + Fa“ + Fas + [a3] = (0) by theorem 1-12.

If n i 4, then An = (0) by lemma 2—3.

Theorem 2-6. If J is a Jordan algebra with property A,
then all the u-prime ideals are pairwise comaximal.
Proof. Let P1 and P2 be any distinct u-prime

ideals of J. Without loss of generality one may assume
that P1$P2- Let P1: <p1>. P1“ = P19g[p15] + Fplgg
[p1“]£;Pl“ by theorem 1-12 and corollary 2—4. Thus

[p] = P1“ and p = aUp where p = p1“. 1Up = anU
p

2 — 2 = G — ED
p UaUp' (l p Ua)U 0 P2. By lemma 1 25, p .2 or

- 2 e = ”e C =
l p Ua P2. If p p1 P2, then P1- P(Pz) P2 by
lemma 1-23, which contradicts PIQ P2. Thus 1 — pZUae

P and P1 + P2 = J.

2 I

Theorem 2—7. Let J be a Jordan algebra. J has property

n
A if and only if J = $121 Ji where each summand has property
A, all but at most one of the summands are simple Jordan
algebras, and each summand is either a simple algebra or
contains only one proper u-prime ideal, which is nilpotent.

Proof. If J has property A, then by theorem
n
lm35, J = e Z Ji where each summand is either unprime

or contains a proper u-prime ideal which is nilpotent.

Suppose that there are two summands Ji’ Jk which do not

have property B. There exist ideals A of Ji and B of Jk

 

 

34
which are not of the form [x] for some x in the ideal.
A s B is an ideal of J, thus A c B = <a e b> for some
aEA and bEB. a€A e B, therefore a = r(a eb) +
dU(aeb)‘ (l - r)a = dUa and —rb = dUb since the sum is
direct. l — r # 0 or r f 0, without loss of generality
assume r # 0. Thus b = r‘ldUb and B = [b] contradicting
the choice of B. Therefore all but at most one of the
summands have property B, and by theorem 1—39, they are
a direct sum of simple Jordan algebras. If the remaining
summand is u-prime, then it is Simple by lemma 1-36
and theorem 2—6, otherwise it has only one proper

n

u-prime ideal, which is nilpotent. If J = $121 Ji where
the summands satisfy the conditions given in the statement
of the theorem, then J has identity since each summand

does. Let Ji for i = 2,3,...,n have property B. If C

II M5

is any ideal then C = 9 Ci where Ci is an ideal of Ji'

i 1
Thus C1 = <c1>, Ck = [ck] for k = 2,3,...,n and ci€ Ci'
n
Let c = 2 Ci. There exist elements dkE Jk such that
i=1
deck — Ok for k = 2,3,...,n. Thus deC = cké <c> for
n
k = 2,3,...,n, and c1 = c - E ciE <c>. Therefore
i=2

<c> = C and J has property A.

 

35
Theorem 2-8. Let J be a Jordan algebra with property A.
If Z(J) = { x EJ : JUX = 0 }, then Z(J) is an ideal of J
of the form Fx for some x€.J.
Proof. Let A = <a> be the nil radical of J.
Kevin McCrimmon has shown that Z(J)€;A. Let n be the
n

smallest positive integer such that a = 0. If n = l,

A = <o> = (0) = Z(J). If n = 2, A2§;[a] = JUa = (0) as

in lemma 1—40, and Z(J) = A = Fa. For n > 2, one may
assume [a] = A2 in View of lemma 2—1 and lemma 2-2. If
n = 3, A3 = (0) by lemma 2—5. For xe Z(J), x = ra + bUa
and 0 = 1UX = lUra+bUa
_ 2
— r Ua + 1UbU + 2Ura,bU
a a
= rzU + a2U U + 2r(ba)U by lemma 1-3,
a b a a
= 2 2 - 2 E 3 =
r a Since a UbUa + 2r(ba)Ua A (0).

Since a2 # O, r = 0 and Z(J) §[a] = A2. Therefore Z(J)

A2 = <a'> = Fa' for some a‘6 [a].

If n = 4, A“ = (0) by lemma 2—5. For x6 Z(J),

0 = lU = rza2 + azU U + 2r(ba)U as in the case for
x b a a

n = 3. Thus r2a2€ A3. If r # 0 then aZE A3 and a3€ A“
(0), which contradicts the choice of n. Therefore r = O

and Z(J) Q[aJ = A2. For d €[a], d = cUa, then eUd =

= ' E“: = :2:
eUGUa eUaUCUa A (0). Therefore Z(J) [a] A

<a'> = Fa' for some a' €[a].

 

 

 

APPENDIX

 

APPENDIX

POLYNOMIAL IDENTITIES IN JORDAN ALGEBRAS OVER

A COMMUTATIVE RING WITH IDENTITY

For any set X, let N(X)' be the free monad
generated by X ([2], pp.23—25), then the free module FR(X)
over Z generated by N(X)' is the free nonassociative ring
generated by X. Let T(X) be the ideal generated by all
the elements of the form f(x,y) = (x,y,xz) = (xy)x2 —
x(yxz) and g(x,y) = xy — yx where x,y €FR(X), then FJR(X) =
FR(X)/T(X) is the free Jordan ring generated by X.

Define h(x,y,z) = (x,y,xz + zx) + (z,y,x2) and
d(w,x,y,z) = (w,x,zy) + (z,x,yw) + (y,x,wz). Let A(X) =
{ifixd,m :XQMZENM)'}U{dOMXJIm =WRGYJ€
N(X)' }, and w(x) = A(X)U{ g(x,y) :x,y€N(X)' }. Let

M(X) be the ideal of FR(X) generated by W(X).

Lemma A—l. FR(X) = M(X) 6 P as modules, where P is a
free module over Z.

Proof. Let L(X) = { Ry : y€N(X)' }U{ Ly
ye N(X)' }LJ{ identity operator }. where Ry anf Ly are
right and left multiplication by y respectively. Let

B(X) = {wT1T2...Tn : wC-W(x), Tie L(X), and nEN }.

36

 

37
B(X) generates M(X) as a Z-module. Let Q(X)' be the
free nonassociative algebra over the rational numbers Q
( the vector space over Q with N(X)' as a basis ).
FR(X); Q(X)'. Let M(X)‘ be the subspace of Q(X)'
generated by B(X). Let H(X) be a basis of M(X)‘ such
that H(X)§;B(X). Extend H(X) to a basis V(X) of Q(X)'
by elements of N(X)'. H(X) is a basis over Z for FR(X).
Thus M(X) is spanned by H(X), and if P is the free
Z-module generated by V(X) - H(X), then FR(X) = M(X) o

P as modules.

Lemma A-2. If w,x,y,z €FR(X), then d(w,x,y,z), g(x,y)6
M(X).
Proof. The lemma holds by multilinearity of

d(w,x,y,z) and g(x,y) and by the definition of M(X).

Lemma A~3. If y€ FR(X) and z EN(X)', then f(y,z)€
M(X).

n
Proof. Let y = E n.y. where n.€ Z and y.€ N(X)'.
i=1 i l l i

For n - l, f(n1y1,z) = n13f(y1,z)€ M(X). Assume that

the statement holds for n i k. Then for n = k + l:

f(y,z) = (x + nk+1yk+1,z,(x + nk+1Yk+1)2) where x =
k
iglniyi,

(x,z,X2) + nk+1(x,zlxyk+1 + yk+1X)

+ nk+12(X,Z,Yk+12) + nk+1(Yk+llzIX2)

3 2
+ nk+12<Yk+uzIXYk+1 + Yk+1Xi + nk+1 (Yk+1'zlyk+1 )

 

 

38
x f(x,z) + nk+1'f(yk+l,z) + nk+12h(yk+l,z,x)

+ nk+1h(szIYk+l)

f(x,z) + nk+13f(yk+1'2) + nk+£2h(yk+l,z,x)
+ nk+1nkd(yk'z'yk+1'w) + nk+lnk2h(yk'zryk+1)
+ nk+1nkd(YklzlwIYR+l) + nk+1h(WIzIYk+1)

k-l
where w s '2 niyi,
i=1

k
f(YIZ) + nk+1’f(yk+1,z) + .Zlnk+yznih(yk+l,z,yi)
l:

+ nk+1nkd(YkerYk+1lw) + nk+1nk2h(YkIzIYk+1)
+ nk+1nkd(yk’z’w’yk+l) + f(nk+lyk+l + VIZ)
k-l
- f(W,Z) - nk+13f(yk+1,2) ‘ _Elnk+12nih(yk+l,z,yi)
1:
E M(X) by lemma A—2, the induction hypothesis, and
by the definition of M(X). Therefore the lemma holds

by induction.

Lemma A-4. If x,y €FR(X) then f(x,y)E'M(X).
n

Proof. Let y = X n.y. where y.6 N(X)' and n.€ Z.
i=1 i l i i

n n
f(x,y) = f(x, 2 niyi) = X nif(x,yi)€ M(X) by lemma A-3.
i=1 i=1

Lemma A-S. Let J be any nonassociative ring with a
generating set X. If the polynomial identities g(x,y) =
0, f(x,y) = 0, h(x,y,z) = 0, and d(w,x,y,z) = 0 are
satisfied on X, then J is a Jordan ring. In particular

if J is an algebra over a field of characterstic not

 

39
two or three, with the polynomial identities g(x,y) = O
and d(w,x,y,z) = 0 being satisfied on X, then J is a
Jordan algebra.

Proof. FR(X)/M(X) is a Jordan ring by lemma A-2
and lemma A—4. Since g(x,y) = 0. f(x,y) = 0, h(x,y,z) =
O, and d(w,x,y,z) = 0 are satisfied on the generating
set X of J, then J is a homomorphic image of FR(X)/M(X)
and is a Jordan ring. If J is an algebra over a field
of characteristic not two or three then: f(x,y) =
%d(x,y,x,x) = 0 and h(x,y,z) = d(x,y,Z,X) = 0 for x,y,ze x.

Thus by the first part of the lemma, J is a Jordan algebra.

Lemma A—6. If w,x,y,z eFR(X), then 2h(x,y,z), 2d(w,x,y,z)e

T(X).

Proof. Let h'(x,y,z) = h(x,y,z) + h(z,y,x).
For a,b,c €FR(X), h'(a,b,c) = f(a + C,b) - f(a,b) — f(c,b)€
T(X). Therefore 2d(w,x,y,z) = h'(w + y,x,z) — h'(w,x,z) —
h'(y,x,z) + g(y,x)LZw - g(y,x)LwLz + g(x,z)Lyw -
g(x,z)LwLy + g(z,y)LXW - g(z,y)LwLX€ T(X). For n€ N,
nh(a,b,c) + n2h(b,c,a) = f(a + nb,c) - f(a,c) - f(nb,c)€

T(X). Therefore 2h(z,y,x) + 4h(x,y,2), h(z,y,x) +
h(x,y,2)€ T(X), and 2h(x,y,z) = 4h(X,y,Z) + 2h(z,y,X) -

2[h<z,y,x> + h(x,y,z)) e T(x).

Lemma A—7. Let S be any Jordan ring, if t(x1,x2,...,xn) =

0 is a polynomial identity for FR(S)/M(S), then

2t(x1,x2,...,xn) = O is a polynomial identity for S.

 

40
If in addition h(x,y,z) = O and d(w,x,y,z) = 0 are
polynomial identities for S, then t(x‘,x2,...,xn) = O
is also a polynomial identity for S.

Proof. By lemma A-2 and lemma A—4, FR(S)/M(S)
is a Jordan ring, therefore there is a homomorphism
yzFJR(S) + FR(S)/M(S) such that y(s) = s for SE S. By
lemma A—6 and by the definition of T(S), it follows that
f(x,y), g(x,y), 2d(w,x,y,z), and 2h(x,y,z)€ T(S) for
w,x,y,zE FR(S). Thus the kernel K of Y is the Z—module
generated by { aTsz...Tn : ae A(S) and Tie L(S) } for
L(S) defined in lemma A-l. In FJR(S), t(x1,x2,...xn)€ K
and 2t(xl,x2,...xn) = 0. Since S is a homomorphic image
of FJR(S), then 2t(x1,xz,...,xn) = O is a polynomial
identity for S. If h(x,y,z) = 0 and d(w,x,y,z) = 0 are
polynomial identities for S, then there is a homomorphism
d:FJR(S) + S such that d(s) = s, and K is contained in
the kernel of a. Therefore there is a homomorphism

B:FR(S)/M(S) + S such that BY = d. Since S is a

homomorphic image of FR(S)/M(S), t(xl,x2,...,xn) = O
is a polynomial identity for S.
Lemma A-8. If t(x1,x2,...,xn) = O is a polynomial

identity, with integral coefficients, for every Jordan

algebra over the rational numbers Q, then it is a

polynomial identity for FR(X)/M(X) for any set X.
Proof. Since h(x,y,z) = 0 and d(w,x,y,z) = 0

are polynomial identities for Jordan algebras over fields

 

 

41
of characteristic not two ([3], p.91), there is a
homomorphism B:FR(X)/M(X) + FJ(X)' where B(x) = x for
x:€X ( where FJ(X)' is the free Jordan algebra over Q
generated by X ). Let yeFR(X) such that B(y) = 0.
FR(X) is contained in Q(X)', and FJ(X)' = Q(X)'/M(X)‘

where M(X)‘ is the subspace generated by M(X). Since

n
B(y) = 0, y€ M(X)‘ and y = X rizi where rie Q and
i=1
ziE M(X). Let m be a common denominator of the ri's,
n
thus ri = ni/m for m,ni€ Z. Therefore my = i=1nizie

M(X), and my = 0. By lemma A—l, FR(X)/M(X) is isomorphic
to a free module over Z, but this implies y = 0. Thus
8 is a monomorphism, and any polynomial identity with

integral coefficients for FJ(X)' holds in FR(X)/M(X).

Theorem A—9. Let S be any Jordan ring or Jordan algebra
over a commutative ring with identity. If
t(x1,x2,...xn) = O is a polynomial identity, with

integral coefficients, for every Jordan algebra over Q,

then 2t(xl,x2,...,xn) = O is a polynomial identity for
S. If in addition h(x,y,z) = 0 and d(w,x,y,z) = O are
polynomial identities for S, then t(x1,x2,...,xn) = O

is a polynomial identity for S.
Proof. Since any Jordan algebra over a commutes
tive ring with identity is a Jordan ring, the theorem

follows immediately from lemma A—7 and lemma A-8.

 

 

 

BIBLIOGRAPHY

 

BIBLIOGRAPHY

Goldie, A. W. "Non-commutative principal ideal rings,"
Archiv der Mathematik, 13(1962), 213-221.

Jacobson, Nathan. Structures and Representations of
Jordan Algebras. American Mathematical Society
Colloquium Publications. Vol. XXXIX. Providence,
Rhode Island: American Mathematical Society, 1968.

Schafer, Richard D. An Introduction to Nonassociative
Algebras. New York: Academic Prcss, 1966.

 

Tsai, Chester. "The Prime Radical in a Jordan Ring,"
Proceedings of the American Mathematical Society,
19(1968), ll7l—ll75.

Zariski, Oscar and Samuel, Pierre. Commutative

Algebra, Vol. I. Princeton, New Jersey: Van
Nostrand, 1958.

42

 

 

 

 

 

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