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Radicals whose essential covers aresemisimple classesGary F. Birkenmeier aa Department of Mathematics, University of Southwestern Louisiana,Lafayette, Louisiana, 70504, U.S.A.Published online: 27 Jun 2007.

To cite this article: Gary F. Birkenmeier (1994): Radicals whose essential covers are semisimple classes,Communications in Algebra, 22:15, 6239-6258

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COMMUNICATIONS IN ALGEBRA, 22(15), 6239-6258 (1994)

RADICALS WHOSE ESSENTIAL COVERS ARE SEMISIMPLE CLASSES

Gary F. Birkenmeier

Department of Mathematics University of Southwestern Louisiana

Lafayette, Louisiana, 70504 U.S.A.

Dedicated to the memory of Leroy F. Birkenmeier

INTRODUCTION

Recall a s e m i s i m p l e class of rings (all rings in this paper are associative

but do not necessarily have a unity) is a class of rings having zero radical with

respect to some Kurosh-Amitsur radical. In [ll] or [XI a class of rings S has

been characterized as a semisimple class if and only if it satisfies the following

three conditions:

(i) S is hereditary;

(ii) S is closed under subdirect products;

(iii) S is closed under extensions.

As in [g], an ideal I of a ring R is an essential ideal if and only if B n I # 0

for each nonzero ideal B of R, and when this is the case R is said to be an

essen t ia l ex tens ion of I . Let X be a class of rings. The class EX, consisting

of all essential extensions of rings belonging to X, is called the essential cover

of the class X . The class X is called essential ly closed if X = EX. In [4,

Copyright 0 1994 by Marcel Dekker, Inc.

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6240 BIRKENMEIER

Theorem 1.11], we showed that if p, is the class of p-radical rings, where p is

a supernilpotent radical, then Ep, has the following properties:

(i) Ep, is hereditary;

(ii) Ep, is closed under extensions;

(iii) Ep, is closed under finite subdirect products;

(iv) Ep, is closed under arbitrary direct sums and direct products.

Thus Ep, is "almost," a semisimple class. Hence the following three ques-

tions were posed in [$I (the first two were due to a referee of [$I): (1) Is Ep, a semisimple class'!

(2) If Ep, is not a semisimple class, what are necessary and sufficient con-

ditions on p which would guarantee that Ep, is a semisimple class?

(3) What conditions can be put on a universal class W so that Ep, is a

semisimple class relative to W.

In this paper we investigate the essential cover of various classes of rings.

The interplay between the upper radical operator U, the semisimple operator

5, and the essential cover operator E is explored in section 1. The results

from section 1 are applied to the study of radicals whose essential covers are

semisimple classes in section 2. The results from section 2 enable us to show

that for the well known supernilpotent radicals (i.e., the prime radical, the

Jacobson radical, the nil radical, the Levitzki radical, the Brown-McCoy rad-

ical, and the generalized nil radical) the essential cover is not a semisimple

class. Thus question (1) is answered in the negative. However an example is

provided to show that there is a supernilpotent radical whose essential cover

is a semisimple class. Our last result provides a positive answer to question

(3) . Question (2) is still open.

Throughout this paper R denotes a ring and ( X ) denotes the ideal of R

generated by X , where X C R. The notation IaR and 1a.R symbolizes I is an

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SEMISIMPLE CLASSES OF RADICALS 624 1

ideal of R and I is an essential ideal of R , respectively. The ring of integers is

symbolized by Z. Greek letters will represent radical properties with the letter

p symbolizing a supernilpotent radical on the class of all rings unless indicated

otherwise. The class of all 0-radical rings and the sum of all 0-radical ideals in

R are denoted by 0, and 0(R), respectively. Recall a class X of rings is said to

be a regular class if X contains the ring consisting of zero and if 0 # I o R E X

then there exists X a I such that 0 # I /X E X. A class W of rings is said to

be universal class of rings if it is closed under taking ideals (i.e., hereditary)

and homomorphic images.

Finally we note that the upper radical operator U, the semisimple operator

S, and the essential cover operator E acting on a class X of rings are defined

U X = {R / R has no nonzero homomorphic image in X }, S X = {R / R has no nonzero ideal in X ), E X = {R / there exists I a R such that R is an essential extension

o f I a n d I ~ X } ,

respectively. Other terminology can be found in [fi] or [a].

In this section we develop some basic properties of essential covers of classes

of rings and consider the interplay between the operators 1, S, U over various

classes of rings. In particular we provide conditions, different than those of

Theorem 5 in [El, on a class X of rings which guarantee that U3C = UEX.

Kote from [l6] any nonempty class of rings X determines an upper radical

u where X is then contained in the a-semisimple class.

LEMMA 1.1 Let X and X be classes of rings such that X C X . Then:

(i) U3C c UX;

(ii) S X 5%;

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6242 BIRKENMEIER

(iii) E X EX;

(iv) If u is the upper radical determined by X , then a, = W if and only

if X is a regular class;

(v) X C S U X if and only if X is a regular class;

(vi) SILK is contained in every semisimple class containing X , hence S W

is contained in the semisimple class of the upper radical determined by X ;

(vii) X is a semisimple class if and only if X = SILK;

(viii) X is a radical class if and only if X = 'USX.

Proof. The proof is routine.

Since we are interested in essential covers of radical classes, let us observe

that the essent,ial cover of a radical class is cssentially closed [A, Lemma 1.41.

Also if X is a hereditary class of semiprime rings then EX = €EX [9, Theorem

41. However, as shown in [a, Theorem 51, EX = &EX does not necessarily

hold for an arbitrary class X of rings. The following definition will provide

a condition on a class X which insures that EX = EEX and extends the

ADS property of radical classes [i] to arbitrary classes of rings. Recall a

Plotkin radical [g] for rings can be defined as an assignment p which designates

to each ring R an ideal p(R) such that the class p, = {R I p(R) = R) is

homomorphically closed, p(R) contains every p,-ideal of R, and p(p(R)) =

p ( m

DEFINITION 1.2 We say a class 7-C of rings has the GADS property (i.e.,

generalized ADS property) rclative to a dass M of rings if wheriever X a I a R

with X g X and I E 3\12 t,here exists B a R such that B E X and X C B C I.

If the words "relative to a class M" do not follow the words "GADS property"

then we take n/C to be the class of all rings.

We note that Theorern 2 of [El is an immediatx corollary of the following

result.

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SEMISIMPLE CLASSES OF RADICALS

PROPOSITION 1.3 Let X be a class of rings.

(i) If X has the GADS property, then E X = EEX.

(ii) If X is a radical class, then X has the GADS property.

(iii) Let p be a Plotkin radical and X = {R I p(R) = R). Then X has the

GADS property if and only if X has the ADS property (i.e., p ( I ) a R for I a R ) .

(iv) If X is the class of nilpotent rings, then X is a nonradical class which

has the GADS property.

(v) If M is a class of semiprime rings, X a subclass of M, and X = E X ,

then X has the GADS property relative to M.

(vi) If X is the class of idempotent rings, then X has the GADS property.

(vii) If X contains all zero rings and is closed under extensions, then X has

the GADS property.

Proof. (i) Clearly E X C £EX. Let R E EEX. Then there exists X , I C_ R

such that X a .I 4 .R , where X E X . Since X has the GADS property, there

exists B a R such that B E X and X C B. Hence B a .R. So R E EX.

Therefore E X = £EX.

(ii) Let X = Q,, and X a I a R where X E X . Using the ADS property we

have X = Q(X) 5 Q ( I ) a R. Hence X has the GADS property.

(iii) Assume X has the GADS property. Let I a R. Then p ( I ) = C X,.

where X, a I and X, E X . Hence p ( I ) = C B,. where B, a R, B, E X, and

X C B, C I. Thus p ( I ) a R. So p has the ADS property. The converse is

immediate.

(iv) Let X a I a R with X E 3C By the Andrunakievic lemma [i, p.1071,

(X) E X . Hence X has the GADS property.

(v) Let X a I a R with X E 3C and I E M. Since I is a serniprime ring a

straightforward argument yields tha t (X) is a semiprime ring and X a . ( X ) .

Hence (X) E E X = X. Therefore X has the GADS property relative to M.

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6244 BIRKENMEIER

(vi) Let X a 1 a R with X E X . Since X 2 = X , then ( X ) 2 = ( X ) . Hence

( X ) E X , so X has the GADS property.

(vii) Let X a1 aR with X E X . By thc Andrunakievic lemma [fi, p.1071, Y 2

is a zero ring, where Y = ( X ) / X . Since Y / Y 2 is a zero ring and X is closed

under extensions, ( X ) / X E X . Hence ( X ) E X . Therefore X has the GADS

property.

PROPOSITION 1.4 Let X be an abstract class (i.c., closed under isomorphic

copies and containing the one-clement ring). Then:

(i) if 11 4 S X , then R has a nonzero homomorphic image in E X ;

(ii) U E X c S X ;

(iii) if X is a regular class which is closed under homomorphic images, then

X c U E U X c U E U E X ;

(iv) if X is a hereditary class then EX is a regular class.

Proof. (i) Assume R @ S X . Then there exists 0 # l a R such that 1 E X . Let

X be an ideal of R which is maximal with respect to having zero intersection

with 1. Hence - + is essential in R / X . But - X

+

E x. Therefore R / X t X

E X .

(ii) This part is a consequence of part (i).

(iii) By Lemma 1.1, U E U X E U E U E X . So let R E X and assume R @

U E U X . Then there exists j ( R ) a nonzero homomorphic image of R such that

J(R) E E U X . Then there exists an ideal 1 of f ( R ) which is essential in f (R)

and 1 E U X . But this contradicts the regularity of X since f ( R ) E X . Thus

n E U E U X .

(iv) Let R E EX and 0 # I a R. There exists X a R such that X E X and

X is essential in R. Then 0 # X n 1 E X . Therefore 0 # 14: S X . By part ( i ) ,

1 has a nonzero homomorphic image in E X . Consequently, EX is regular.

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SEMISIMPLE CLASSES OF RADICALS 6245

Observe tha t Proposition 1.4 (ii) cannot be improved to UX C SX. For

example, let X be the class of prime radical rings. Then 2 / 4 2 E u3C but

2 / 4 2 @ SX.

Let X be a class of rings. The fact that UEX C UX and X is a "large"

subclass of EX naturally motivates the following question: What conditions

guarantee that UEX = UX? An answer was provided by Theorem 5 of [El:

Let M be a regular class of rings containing no or all zero rings. Then UEM

= UM if and only if UM is hereditary. The next result yields an alternative

answer t o the aforementioned question.

COROLLARY 1.5 Let X be an abstract class which is regular and closed

under homomorphic images. If UX is regular then UX = UEX.

Proof. Since X C EX, then UEX C U X . From Proposition 1.4 (iii), X C

UEUX. Hence EX E EUEUX, so UEUEUX C UEX. Again using Proposi-

tion 1.4 (iii) yields UX 2 U E U E W . So UX C UEX. Therefore UX = UEX.

Recall from [a, p.771, a radical P is called the radzcal supplementzng the

radical a , if a ( R ) n P(R) = 0 in every ring R and 1C, 5 P for all radicals 1C,

for which a(R) n $(R) = 0 in every ring R. The following definition will be

used t o extend the supplementing concept t o classes of rings which include the

radical rings of hereditary radicals.

DEFINITIOIV 1.6 L P ~ A and 'B be classes of rings. We say 'B satisfies

condttzon (s) r ~ l l ~ t z v e to A if for any ring R with ideals X and Y then X E

A and Y E B implies X n Y = 0. We say 'B h-supplements A if 23 satisfies

condition (s) relative t o A , and if X is any homomorphically closed class which

satisfies condition (s) relative t o A then X C_ 'B.

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PROPOSITION 1.7 Let X be a hereditary class containing no or all nonzero

zero rings, and let 0 be the upper radical determined by EX. Then:

(i) 0, = UEX;

(ii) 0, is a hereditary class;

(iii) 0, h-supplements X .

Proof. (i) By Proposition 1.4 (iv) and Lemma 1.1 (iv), 8, = UEX.

(ii) Theorem 4 of [El yields 0, is a hereditary class.

(iii) Let X and Y be ideals of R with X E X and Y E 8,. Since X and

8, are hereditary dasses, X n Y E X no,. Hence X n Y = 0. So 0, satisfies

condition (s) relative to X . Let FC be any homomorphically closed class which

satisfies condition (s) relative to X . Assume R E FC, but R $ 0,. Then there

exists an ideal I of R such that 0 # R / I E EX. Then there exists V a R

such that V/I E X and V/I a .R / I . Thus ( R I I ) n (V/I) = V/I # 0. This

contradicts the fact that FC satisfies condition (s) relative to X. So R E 8,.

Therefore 0, h-supplements X.

Let LK denote the lower radical determined by a class of rings X .

T,F,MMA 1.8 Let a be a hereditary radical. Then P is the radical supple-

menting a if and only if @, h-supplements a,.

Proof. Assume /3 is the radical supplementing a . Clearly, PC satisfies condi-

tion (s) relative to a,. Let X be a.ny class which satisfies condition (s) relative

to a,. Let 0 # I = a ( R ) n L,(R). From the A.D.S. construction of LIi [l9]

there exists a descending chain

where 0 # I, E X . Since a is hereditary, 0 # I , E a, n X , a contradiction!

Thus a ( R ) n LK(R) = 0. Consequently LIi 5 P . Hence X C PC. Therefore PC

h-supplements a,. The converse is immediate.

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SEMISIMPLE CLASSES OF RADICALS 6247

Recall [a, p.791, a and P form a d u a l pa i r of s u p p l e m e n t i n g radicals, if P

is the radical supplementing a, and cu is the radical supplementing P .

THEOREM 1.9 Let X be a universal class of rings containing no or all

nonzero zero rings, and let 0 be the upper radical determined by EX. The

following statements are equivalent:

(i) X h-supplements 0,;

(ii) 3C = UEUEX;

(iii) ( 0 , L K ) is a dual pair of supplementing radicals, where X is the class

of LK-radical rings.

Proof. (i) (ii) By Proposition 1.7 (ii), 0, is hereditary. A routine argument

shows that if X contains no (or all) nonzero zero rings then 0, contains all (or

no) nonzero zero rings. Using Proposition 1.7, we have UE0, = UEUEX and

UEO, h-supplements 8,. Since UEO, is a radical class, it is homomorphically

closed. Definition 1.6 yields UEO, C X. From Proposition 1.4 (iii), X =

UEUEX.

(ii) + (iii) Since X = UEUEX = UEQ,, X is a radical class. So X is

the class of LK-radical rings. Using Proposition 1.7 and arguments similar

to those in the above paragraph, we have that 0, h-supplements X and X

h-supplements 0,. From Lemma 1.8. ( 0 , LK) is a dual pair of supplementing

radicals.

(iii) + (i) Since 0 and LK are hereditary radicals, Lemma 1.8 yields this

implication.

The following corollary is a direct application of Theorem 1.9.

COROLLARY 1.10 Let p be a supernilpotent radical. Then ( 0 , p ) is a dual

pair of supplementing radicals, where 0, = UEp,.

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6248 BIRKENMEIER

Alternatively, the above corollary could have been derived using Theorem

2.1 and 2.2 of [u] instead of Theorem 1.9.

2 . RADICALS WHOSE ESSENTIAL COVERS ARE SEMISIMPLE CLASSES

We begin this section with three results which indicate that there is a "duality

principle" at work among the radicals whose essential covers are semisimple

classes. Next we provide an example which establishes the existence of a su-

pernilpotent radical whose essential cover is a semisimple class. The main

result (Theorem 2.7) of this section provides a useful criterion that indicates

when a supernilpotent radical does not have an essential cover which is a

semisimple class. Using this result we show that many well known supernilpo-

tent radicals have essential covers which are not semisimple classes. Our final

result yields a universal class in which every supcrnilpotent radical has an

essential cover which is a semisimple class.

LEMMA 2.1 Let .J, and 8 be radicals. Then E$, = SO, if and only if E8, =

S$,. If either one of these conditions are satisfied, then both the radicals $

and 8 are hereditary.

Proof. Assume Ed), = 58,. Let R E £0, and assume $(R) # 0. Then

I = $(R) n 8 ( R ) # 0. From Proposition 1.3, E$, is essentially closed. In

Theorem 3.1 [z] Armendariz shows that a radical class is hereditary if and

only if its semisimple class is essentially closed. Hence 8, is a hereditary class.

Thus @ ( I ) = I & $(R) . Hence 8 ( $ ( R ) ) # 0. But this is contradictory to

$(R) E &$, = SO,. So $ ( R ) = 0. Therefore £8, 54,. Now let B be a ring

such that B E 5 4 , . Observe that O(B) # 0, otherwise B E Ed),, contradicting

$ ( B ) = 0. Let I be a nonzero ideal of B such that I n O(B) = 0. Since 8,

is hereditary. @ ( I ) = 0. So I E E$,. Hence $(I) # 0. Thus $ ( B ) # 0, a

contradiction. Consequently, E8, = S$,. Again using Theorem 3.1 of [z], we

have that $, is a hereditary class. The converse is proved sirrdarly.

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SEMISIMPLE CLASSES OF RADICALS 6249

PROPOSITION 2.2 Let T) be a radical. The following are equivalent:

(i) £$, is a semisimple class;

(ii) there exists a radical 0 such that E0, = S$, and both 0, and $, are

hereditary;

(iii) E$, = SUE$,.

Proof. (i) j (ii) There exists a radical 0 such that E4, = SO,. The impli-

cation now follows from Lemma 2.1.

(ii) + (iii) From Lemma 2.1, E$, is a semisimple class. By Lemma 1.1

(vi), SUE$, C Eg,. From Lemma 1.1 (v) and Proposition 1.4 (iv), 84, =

SUEd,.

(iii) =+ (i) This implication follows from Lemma 1.1 (vii).

PROPOSITION 2.3 Let 4 and 0 be radicals such that £4, = SO,. Then:

(i) ($, 0) is a dual pair of supplementing radicals;

(ii) 4, = UE0, = UEUE$,, and 0, = UEdc = UEUEO,.

Proof. (i) By Proposition 2.2, 4, and 0, are hereditary classes. Hence $(R)

n O(R) = 0. in every ring R. Let T be a radical such that $(R) n T(R) = 0,

in every ring R. Let S be a ring such that S # (0) and O(S) = 0. Hence

S E E$,. If T ( S ) # 0, then $(S) n r ( S ) # 0, a contradiction. So SO, c ST,,

consequently T 5 0. 'I'hcrefore 0 is the radical supplementing $. Similarly, $

is the radical supplementing 0.

(ii) Since £$, = SO,, Lemma 2.1 yields £0, = S$,. As 0, and $, are radical

classes. Lemma 1.1 (viii) yields 0, = USO, = UE$, and I />, = US$, = UE0,.

Hence 4, = UEQ,= UEUEd!,, and 0, = UE4, = UEUE0,.

In contrast to questions (1) and (2) of the introduction one could ask:

When is E0, a radical class? This question is answered in the following result.

PROPOSITION 2.4 The following statements are equivalent:

(i) £0, is a radical class;

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BIRKENMEIER

(ii) E0, is a radical semisimple class;

(iii) EB, is a variety;

(iv) £ 0 , is homomorphically closed;

(v) E0, = 0, .

Proof. We will only consider (iv) * (v), since the remaining implications

follow from Proposition 1.3 and [a, Theorem 11. So assume E0, is homo-

morphically closed. Let R € EB, and suppose R $ @,. Then 0 f R / O ( R ) E

EO,nSB, = 0, a contradiction. Hence E0, = 0 , .

Proposition 2.4 (iii) motivates one to ask: What happens if £ 0 , is a quasi-

variety? From [4, Theorem 1.111, if p is a supernilpotent radical then E p , is

a hereditary class which is closed under extensions and essential extensions.

Thus if Ep, is a quasi-variety, then Ep, is a se~nisimple class. Hence E p , =

S$,, for some radical $.

Analogously, since we are interested in the condition E0, = S7b,, one can

ask: If a semisimple class is a quasi-variety, is it essentially closed? However,

in this case, a negative answer is provided by considering the quasi-variety X

defined by

x 2 = x * z = 0 .

From [ I , Proposition 3.31, X is a semisimple class. However Lemma 1 of [9]

shows that X is not essentially closed.

The following examplc shows that radical semisimple classes provide exam-

ples of supernilpotent radicals whose essential covers are semisimple classes.

EXAMPLE 2.5 Let X be a radical semisimplc class which is not the class of

all rings [a] and [B]. Since X is a hereditary class of semisimple rings, there

exist radicals p and 0 such that S p , = X = O , , where p is a supernilpotent

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SEMISIMPLE CLASSES OF RADICALS 625 1

radical. From [l2, Theorem 11, 0, is essentially closed so £0, = 8, = X. Hence

0 is a radical such that EQ, is a semisimple class. By Proposition 2.2, Ep, =

SO,. Thus p is a supernilpotent radical whose essential cover is a semisimple

class. Observe that p, = ILK, Z E Ep,, but p(Z) # Z .

LEMMA 2.6 Let li, and O be radicals such that 8, 2 E$,. Then B 2 4.

Proof. Assume R E S+, but O(R) # 0. Then $(O(R)) f 0. Hence $(R) # 0 ,

a contradiction. Thus R E SO,. Consequently 0 < I).

Henceforth P will denote the prime radical.

THEOREM 2.7 Let p and 0 be radicals such that P I p and Ep, = SO,.

Then:

(i) p is a special radical, where p is the upper radical determined by the spe-

cial class of all subdirectly irreducible rings with a p-semisimple (equivalently,

8-radical) heart;

(ii) 0 is a subidempotent radical (i.e.. every ideal of O(R) is idempotent);

(iii) Z E Ep,;

(iv) if p(Z) = n Z (n # 1). then n = . .pi, is a product of distinct

primes, Z / n Z E O,, and the variety generated by Z/p,Z, i = 1 , 2 , . . . k , is a

radical semisimple class which is contained iri 8,. Furthermore if q is a prime

integer such that q @ {p l ,pz , . . . , p i , ) , then Z/qZ E p,.

Proof. (i) By Proposition 2.2, p, is hereditary. Hence p is a supernilpotent

radical. From Proposition 2.3, (p, 0) is a dual pair of supplementing radicals.

Sow Theorem 13.6 of [a] yields that p is a special radical d e t ~ r ~ n i n e d by the

indicated special class.

(ii) Let R E Oc and I a nonzero ideal of R. By Proposition 2.2, R and

R/12 are in Sp,. Hence R and R/12 are semiprime rings. So I/IZ = 0. Hence

I = I Z .

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6252 BIRKENMEIER

(iii) The ring of even integers, 22 , is a subdirect product of nilpotent rings

' 2Z/2k Z , for k 2 3. Since a semisimple class is closed under subdirect products

[ll, Theorem 11, then 2 2 E Ep,. Hence Z E Ep,.

(iv) Since Z l n Z E Sp,, then Z l n Z is a semiprime ring. Hence Z / n Z is a

finite direct sum of fields, Z/p;Z, where p; E {pl ,pz, . . . , pk) is a set of distinct

primes. Hence n = plpz . . . p k . By Proposition 2.2, Z l n Z E 80,. Since Z l n Z

has no proper essential ideals, Z / n Z E 6,. Let X = { ZlpiZI i = 1 , 2 , . . . , k).

Then X is a strongly hereditary class (i.e., K is closed under subrings [a, Def-

inition 3.61). Let V be the class of all isomorphic copies of subdirect products

of fields in X. By [l& Theorem 4.31, V is a radical semisimple class. From [l3]

or [B], V is a variety. Since every variety is closed under subdirect products and

isomorphic copies, then V is the variety generated by X. Moreover X 0, and

Ed, is closed under subdirect products (since E6, is a semisimple class). Hence

V 2 E6,. Let u be a radical such that a, =V. By Lemma 2.6, V 6,. Let h

be the natural homomorphism from Z onto ZJqZ. Then 0 # h(n) E p(ZlqZ).

Therefore Z/qZ E p,.

Part (iii) of Theorem 2.7 allows us to split the class of supernilpotent rad-

icals whose essential covers arc semisimple classes into two distinct subclasses

as is indicated in the following corollary.

COROLLARY 2.8 Let p and 0 be radicals such that P < p and Ep, = SO,.

Then:

(i) p(Z) = n Z (n # 1) if and only if there exists a radical semisimple class

contained in 0,;

(ii) P(Z) = Z if and only if V(Z) Ep,, where V(Z) is the variety generated

by Z.

Proof. (i) Let X be a radical semisimple class contained in 0,. By Theorem

2.7 (iii), either p(Z) = Z or p(Z) = n 2 for some n # 1. Assume P(Z) = Z.

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SEMISIMPLE CLASSES OF RADICALS 6253

So Z/qZ E p, C SO, for every prime integer q. Hence 0, contains no fields of

prime order. But Theorem 4.3 of [l8] implies that every radical semisimple

class contains fields of prime order, a contradiction. Therefore p(Z) = n Z for

some n # 1. The converse follows from Theorem 2.7 (iv).

(ii) Assume p(Z) = Z. Since radical classes are closed under homomorphic

images and Ep, is hereditary and closed under subdirect products, it follows

that V(Z) C Ep,. Conversely, if V(Z) & Ep,, then Z/qZ E p, for every prime

integer q. Thus Z/qZ @ 0, for every prime integer q. From part (i) and

Theorem 4.3 of [El, p(Z) # nZ (n # 1). Therefore p(Z ) = Z .

Theorem 2.7 (iii) allows us to quickly see that several supernilpotent rad-

ical classes do not have an essential cover which is a semisimple class. Thus,

in general. question (1) of the introduction has a negative answer. For exam-

ple, if p is the prime radical, Jacobson radical, Levitzki radical, nil radical,

generalized nil radical, or Brown-McCoy radical, then p(Z) = 0. Hence Z @

Epc.

In the following example, we will exhibit a supernilpotent radical p which

satisfies conditions (i) and (ii) of Theorem 2.7 and for which there exists a

radical 6' such that (P, 0) is a dual pair of supplementing radicals with p, =

UEO,, but Ep, is not a semisimple class.

EXAMPLE 2.9 Let p denote the antisimple radical (i.e., t h r upper radical

determined by the class of all subdirectly irreducible rings with idempotent

heart [a, p.751. It follows from the definition that p is a supernilpotent ~adica l .

Using Theorem 1.9 with p, in place of K, we obtain that (0, p) is a dual pair of

supplementing radicals with 0, = UEp, and p, = UEO,. Now Theorem 13.6 of

[a] yields that p satisfies Theorem 2.7 (i). The fact that p satisfies Theorem

2.7 (ii) is indicated in [a, p.791. However Proposition 12.5 of [a] shows that

p(Z) = 0. From Theorem 2.7 (iii), Ep, is not a semlsimple class.

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6254 BIRKENMEIER

The following example is due to E.P. Arrnendariz. He used it to negatively

answer the following question posed in [z]: Is the direct product of an arbitrary

family of 2-primal rings a 2-primal ring? (a ring is 2-primal if its prime radical

equals its set of nilpotdent elements). In the next example N ( X ) denotes the

set of nilpotent elements contained in X .

EXAMPLE 2.10 There exists a supernilpotent radical p such that p(Z) =

Z but Ep, is not a semisimple class. Let R2 denote the class of all rings such

that every prime ideal is completely prime. Let Y,(R) be the sum of all ideals

of R which are in R 2 . Then PC is a supernilpotent radical [&I. Thus p , ( Z ) =

z 2 z Z . N o w l e t R = [ 2 z ] a n d A = [i: ;:I. ThenRisap r i rne r ingand

M

RIAk E R2. But n Ak = 0. Hence R is a subdirect product of elements of k = l

R 2 but R $ R 2 . Observe P ( P , ( R ) ) = N(Y,(R)) = 0 since R is prime. Thus

P , (R) is a reduced ideal of R. Then (P , (R)) ( (N(R))) = 0. Since N(R) # 0,

then p C ( R ) = 0. SO R 6 &(PC) . Thus & ( P C ) is not closed under subdirect

products. Consequently E(y,) is not a semisimple class.

The last two examples indicate that the results of t,his section do not pro-

vide enough information to completely characterize a supernilpotent radical

whose essential cover is a semisimple class.

However our final result does show that in a suitably restricted universal

class the essential cover of every supernilpotent radical is a semisimple class.

Thus question (3) of the introduction has a positive answer. To do this we

use the following. criteria [G, p.51 or [a]. An abstract class X of rings is a

semisimple class for some radical class, relative to some universal class W, if

and only if it has the following properties:

(a) If R E X, then every nonzero ideal of R has a nonzero homomorphic

image in X.

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SEMISIMPLE CLASSES OF RADICALS 6255

(b) If R E W but R cf X. then R has a nonzero ideal I E U w X , where

U E , ~ = {T E W I T has no nonzero homomorphic image in X ) .

For completeness we include the following definition. A class X C W is

said to be a radical class in W if it satisfies the following three properties:

(i) X is homomorphically closed.

( i i ) If R E W, R contains an ideal I E X such that I contains every other

X-ideal of R.

(iii) The factor ring R / I contains no nonzero X-ideals.

PROPOSITION 2.11 Let W be a universal class of rings such that if 0 # R E

W then there exists a nonzero minimal ideal of R. If p is a supernilpotent in

W , then Ep, is a semisimple class relative to W.

Proof. We will proceed by verifying conditions (a) and (b) mentioned above.

From [4, Theorem 1.111, Ep, is hereditary. Hence condition (a) is satisfied. So

assume 0 # R E W but R cf Ep,. Then there exists a nonzero ideal S of R

such that S n P(R) = 0. Consequently there exists a nonzero minimal ideal I

of S. Hence 0 # ( I ) ; C I . By the minimality of I in S, ( I ) ; = I and so I

is a minimal ideal of R. Then p ( I ) = 0. Hence I is a simple nontrivial ring

[e, p.1351. Therefore I E UwEpc, so (b) is satisfied. Consequently, Ep, is a

semisimple class relative to W.

Note that the class of finite rings and the class hereditarily artinian rings

(i.e., artinian rings whose Jacobson radical is artinian) [a] are universal classes

which satisfy t,he hypothesis of Proposit,ion 2.11.

ACKNOWLEDGEMENT

The author is grateful for the comments and suggestions made by the

referees which lead to the present improved version of this paper.

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Received: November 1993

Revised: May 1994

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