Journal of Algebra 223, 477᎐488Ž. 2000 doi:10.1006rjabr.1999.8017, available online at http:rrwww.idealibrary.com on

Armendariz Rings and Reduced Rings

Nam Kyun Kim*

Department of Mathematics, Kyung Hee Uni¨ersity, Seoul 130-701, Korea View metadata, citation and similar papers at core.ac.uk brought to you by CORE

provided by Elsevier - Publisher Connector and

Yang Lee†

Department of Mathematics Education, Pusan National Uni¨ersity, Pusan 609-735, Korea

Communicated by Susan Montgomery

Received February 18, 1999

Key Words: Armendariz ring; reduced ring; classical quotient ring.

1. INTRODUCTION

Throughout this paper, all rings are associative with identity. Given a ring R, the polynomial ring over R is denoted by Rxwx. This paper concerns the relationships between Armendariz rings and reduced rings, being motivated by the results inwx 1, 2, 7 . The study of Armendariz rings, which is related to polynomial rings, was initiated by Armendarizwx 2 and Rege and Chhawchhariawx 7 . A ring R is called Armendariz if whenever s q q иии q m s q q иии q n polynomials fxŽ. a01ax axm , gxŽ. b01bx bxn g wx s s Rx satisfy fxgxŽ.Ž.0, then abij 0 for each i, j. Ž The converse is obviously true.. A ring is called reduced if it has no nonzero nilpotent elements. Reduced rings are Armendariz bywx 2, Lemma 1 and subrings of Armendariz rings are also Armendariz obviously. We emphasize the con- nections among Armendariz rings, reduced rings, and classical quotient rings. Moreover several examples and counterexamples are included for answers to questions that occur naturally in the process of this paper.

*E-mail: [email protected]. † E-mail: [email protected].

477 0021-8693r00 $35.00 Copyright ᮊ 2000 by Academic Press All rights of reproduction in any form reserved. 478 KIM AND LEE

2. ARMENDARIZ RINGS

First we consider some examples and counterexamples for Armendariz rings. Rege and Chhawchhariawx 7 showed that every n-by-n full matrix ring over any ring is not Armendariz, where n G 2. We have a similar result in the following.

EXAMPLE 1. Let R be a ring. We claim that n-by-n upper triangular matrix rings over R are not Armendariz, where n G 2. It is enough to show that the 2-by-2 upper triangular matrix ring over R is not Armendariz because each subring of an Armendariz ring is also Armendariz. Let S be

Ž.sqŽ.10 the 2-by-2 upper triangular matrix ring over R, and let fx 00 y Ž.1 10001Ž.sqŽ.Ž. wx Ž.Ž. 0 0 x and gx 01 01 x be polynomials in Sx. Then fxgx s Ž.Ž.10 01 / 0, but 00 01 0. So S is not Armendariz and consequently every n-by-n upper triangular matrix ring over R is not Armendariz. But we may find subrings of the 3-by-3 upper triangular matrix rings which may be Armendariz as in the following.

PROPOSITION 2. Let R be a reduced ring. Then

abc S sN0 ad a, b, c, d g R ½5ž/00a is an Armendariz ring. Proof. We employ the method in the proof ofwx 7, Proposition 2.5 . First notice that for

abc111 abc 222

0 ad11, 0 ad 22g S 00a 00a 0012 we can denote their addition and multiplication by q s q q q q Ž.Ž.Ža1111, b , c , d a 222, b , c , d 2a 1a 21, b b 21, c c 21, d d 2. and

Ž.a1111, b , c , daŽ. 222, b , c , d 2 s q q q q Ž.aa12, ab 12ba 12, ac 12bd 12ca 12, ad 12da 12, respectively. So every polynomial in Sxwxcan be expressed in the form wx ŽŽ.Ž.Ž.Ž..px0123, px, px, px for some pxi Ž.’s in Rx. ARMENDARIZ RINGS AND REDUCED RINGS 479

s s Let fxŽ. Žfx0123 Ž., fx Ž., fx Ž., fx Ž..and gx Ž. Žgx0 Ž., gx 1 Ž., gx 2 Ž., wx s s gx3Ž..be elements of Sx. Assume that fxgxŽ.Ž.0. Then fxgx Ž.Ž. q q q ŽŽ.Ž.fxgx00, fxgx 01 Ž.Ž.fxgx 10 Ž.Ž., fxgx 02 Ž.Ž.fxgx 13 Ž.Ž. q s fxgx20Ž. Ž., fxgx 03 Ž. Ž.fxgx 30 Ž. Ž.. 0. So we have the following system of equations: s Ž.0 fxgx00Ž. Ž. 0; q s Ž.1 fxgx01Ž. Ž.fxgx 10 Ž. Ž. 0; q q s Ž.2 fxgx02 Ž. Ž.fxgx 13 Ž. Ž.fxgx 20 Ž. Ž. 0; q s Ž.3 fxgx03Ž. Ž.fxgx 30 Ž. Ž. 0. wx s Use the fact that Rx is reduced. From Eq.Ž. 0 , we see that gxfx00 Ž. Ž. 0. q If we multiply Eq.Ž. 1 on the right side by fx0010Ž., then fxgxfx Ž. Ž. Ž. s s s fxgxfx100Ž. Ž. Ž.0. So fxgx01 Ž. Ž. 0 and hence fxgx10Ž. Ž. 0. Also if q we multiply Eq.Ž. 3 on the right side by fx0030Ž., then fxgxfx Ž. Ž. Ž. s s s fxgxfx300Ž. Ž. Ž.0. So fxgx03 Ž. Ž. 0 and hence fxgx30Ž. Ž. 0. Now if q we multiply Eq.Ž. 2 on the right side by fx0020Ž., then fxgxfx Ž. Ž. Ž. q s s fxgxfx130Ž. Ž. Ž.fxgxfx 200 Ž. Ž. Ž.0. So fxgx02 Ž. Ž. 0 and hence Eq. Ž.2 becomes Ј q s Ž.3 fxgx13Ž. Ž.fxgx 20 Ž. Ž. 0. Ј If we multiply Eq.Ž. 3 on the right side by fx11Ž., then we have fxgxŽ.3 Ž. s s 0 and so fxgx20Ž. Ž. 0. Now let

X X X a b c nmabciii jjj XX s ijs fxŽ. ÝÝ0 adiix and gxŽ. 0 adjjx , X is0 00a js0 00a 0i 0j s ni s ni s ni s where fx0Ž. Ýis0 axi , fx1Ž. Ýis0 bxi , fx2Ž. Ýis0 cxi , fx3Ž. ni s m X j s m X j s m X j Ýis0 dxi , gx0Ž. Ý js0 axj , gx1Ž. Ý js0 bxj , gx2Ž. Ý js0 cxj , and s m X j XXXXs s s gx3Ž. Ý js0 dxjijijijij. Then we obtain that aa 0, ab 0, ba 0, ac s X s X s X s X s 0, bdij0, ca ij0, ad ij 0, and daij 0 for all i, j by the preced- ing results, the condition that R is reduced, andwx 2, Lemma 1 . Conse- quently X X X a b c abciii jjj XXs 0 adii0 adjj 0 X 00a 00a 0i 0j for all i, j and therefore S is an Armendariz ring. 480 KIM AND LEE

Let S be a reduced ring and let ¡ иии ¦ aa12 a 13 a1n иии 0 aa23 a2 n sN~¥иии g Rnij00 a a3n a, a S ...... ¢§...... 000 0 иии a

Based on Proposition 2, one may suspect that Rn may be also an Armen- dariz ring for n G 4. But the following example erases the possibility.

EXAMPLE 3. Let S be a ring and let ¡ ¦ aa12 a 13 a 14 ~¥0 aa a R sN23 24 a, a g S . 4 00 aa ij ¢§34 000 0 a Let 0100 01y10 0000 00 00 fxŽ.sq x 0000 00 00 000000 00 00 and 0000 0000 0000 0001 gxŽ.sqx 0001 0001 000000 0000 wx s be polynomials in Rx4 . Then fxgxŽ.Ž. 0, but 0100 0000 0000 0001 / 0. 0000 0001 000000 0000 G So R4 is not Armendariz. Similarly, for the case of n 5, we have the same result. ¨ Given a ring R and a bimodule RRM , the tri ial extension of R by M is the ring TRŽ., M s R [ M with the usual addition and the multiplication s q Ž.Ž.Žr112, mr, m 2rr 1212, rm mr 12.. ARMENDARIZ RINGS AND REDUCED RINGS 481

Ž.rm g This is isomorphic to the ring of all matrices0 r , where r R and m g M and the usual matrix operations are used.

COROLLARY 4wx 7, Proposition 2.5 . Let R be a reduced ring. Then the tri¨ial extension TŽ. R, R is an Armendariz ring. Proof. Notice that TRŽ., R is isomorphic to

ab0 U sN0 a 0 a, b g R ½5ž/00a and that each subring of an Armendariz ring is also Armendariz. Thus TRŽ., R is an Armendariz ring by Proposition 2. From Corollary 4, one may suspect that if R is Armendariz then the trivial extension TRŽ., R is Armendariz. But the following example elimi- nates the possibility.

sNÄ4Ž.ab g EXAMPLE 5. Let T be a reduced ring. Then R 0 a a, b T is an sNÄ4Ž.AB g Armendariz ring by Corollary 4. Let S 0 A A, B R and let

01 00 01 y10 ž/ž/ž/ž00 00 00 0y1 / fxŽ.sq x 00 01 00 01 00ž/ž/ž/ž/00 00 00 00 and 01 00 01 10 ž/ž/ž/ž/00 00 00 01 gxŽ.sqx 00 01 00 01 00ž/ž/ž/ž/00 00 00 00 be polynomials in Sxwx. Then fxgxŽ.Ž.s 0, but

01 00 01 10 ž/ž/ž/ž/00 00 00 01 / 0. 00 01 00 01 00ž/ž/ž/ž/00 00 00 00 Thus S is not Armendariz. By Anderson and Camillowx 1, Theorem 2 , polynomial rings over Armen- dariz rings are also Armendariz. So we may conjecture that skew polyno- 482 KIM AND LEE mial rings over Armendariz rings are also Armendariz. Recall that for a ring R with an endomorphism ␣ of R and an ␣-derivation ␦ of R, the Ore extension of R, denoted by Rxwx; ␣, ␦ , is the ring obtained by giving the polynomial ring over R with the new multiplication xr s ␣Ž.r q ␦ Ž.r for all r g R. When ␦ s 0, we write Rxwx; ␣ for Rx w; ␣, 0 x and call it a skew polynomial ring Ž.also called an Ore extension of endomorphism type . However there exists an Armendariz ring R over which the skew polyno- mial ring is not an Armendariz ring as in the following. ޚ EXAMPLE 6. Let 2 be the ring of integers modulo 2 and consider the ޚ [ ޚ s ޚ [ ޚ ring 22with the usual addition and multiplication. Let R 22. Then R is a commutative reduced ring; hence R is Armendariz byw 2, Lemma 1x . Now let ␣: R ª R be defined by ␣ŽŽa, b ..s Žb, a .. Then ␣ is an automorphism of R. We claim that Rxwx; ␣ is not Armendariz. Let fyŽ.s Ž1, 0 .q wŽ.1, 0 xyx and gyŽ.s Ž0, 1 .q wŽ.1, 0 xyx be elements in Rxwxwx; ␣ y . Then fygyŽ.Ž.s 0, but Ž 1, 0 .wŽ. 1, 0 xxwx/ 0. Therefore Rx; ␣ is not an Armendariz ring. Armendariz obtained some results on reduced rings inwx 2 . We here obtain the same results on Armendariz rings. A ring is called abelian if every idempotent of it is central.

LEMMA 7. Armendariz rings are abelian. Proof. By the method in the proof ofwx 1, Theorem 6 . Given a ring R, the formal power series ring over R is denoted by Rxww xx .

LEMMA 8. Suppose that a ring R is abelian. Then we ha¨e the following: Ž.1 E¨ery idempotent of Rwx x is in R and R wx x is abelian. Ž.2 E¨ery idempotent of Rww x xx is in R and R ww x xx is abelian. Proof. Rw x x is a subring of Rxww xx and so it is enough to proveŽ. 2 . For g ww xx 2 s s q q иии q n q иии f Rx, assume that f f, where f e01ex exn . Then we have the system of equations

2 s Ž.0 e00e ; q s Ž.1 ee01ee 10e 1; q q s Ž.2 ee02ee 11ee 20e 2; ...; q q иии q s Ž.nee0 n ee1 ny1 een 0 en ; ... . ARMENDARIZ RINGS AND REDUCED RINGS 483

Observing that Eq.Ž. 0 yields that e0 is an idempotent, so it is central. If we q s multiply Eq.Ž. 1 on the left side by e00, then ee10eee100ee1. But s s s eee010ee 01because e0is central. So ee01 0 and so e1 0; hence Eq. q s Ž.2 becomes ee02ee 20e 2. If we multiply Eq.Ž. 2 on the left side by e0, q s s s then ee02eee 020ee 02. But eee 020ee 02. Hence ee02 0 and so s s e2 0. Now assume that k is a positive integer such that ei 0 for all F F q q s 1 i k. Then equation Ž.k 1 becomes ee0 kq1 eekq10 ekq1, and s s g ww xx so ekq100 by the same method. Therefore f e R and also Rx is abelian. By Kaplanskywx 3 , a ring R is called Baer if the right annihilator of every nonempty subset of R is generated by an idempotent. Closely related to these rings are p.p.-rings; a ring R is called a right p.p.-ring if each principal right ideal of R is projective, or equivalently, if the right annihilator of each element of R is generated by an idempotent. A ring R is called a p.p.-ring if it is both a right and a left p.p.-ring. Baer rings are clearly right p.p.-rings. We denote the right annihilator over a ring R by ᎐ rRŽ.. Abelian right p.p.-rings are reduced Ž for, letting R be an abelian right p.p.-ring and r 2 s 0 for r g R, then there exists e2 s e g R such s g s s s that rrRRŽ.eR; hence r rr Ž.implies r er re 0 . and so they are also left p.p.-rings. Thus we may obtain the following two theorems with the help ofwx 2, Theorems A and B . However we prove them a little independently in this paper.

THEOREM 9. Let R be an Armendariz ring. Then R is a p. p.-ring if and only if Rwx x is a p. p.-ring. s q q иии q m g Proof. Assume that R is a p.p.-ring. Let p a01ax axm wx 2 s g s s Rx. There exists eiie R such that raRiiŽ. eR, for i 0, 1, . . . , m. s иии 2 s g s Fm Let e ee01 emi. Then by Lemma 7, e e R and eR s0 raRiŽ.. s q q иии q m s wx: s So pe ae01aex aexmR0. Hence eR x rpw xxŽ.. Let q q q иии q n g s b01bx bxnRrpw xxŽ.. Since pq 0 and R is Armendariz, abij s F F F F g иии s 0 for all 0 i m,0 j n. Then bj ee01 eRm eR for all s g wx wxs j 0, 1, . . . , n. Hence q eR x . Consequently eR x rpRw xxŽ.and thus Rxwxis a p.p.-ring. Conversely, assume that Rxwxis a p.p.-ring. Let a g R. By Lemma 8, g s wx there exists an idempotent e R such that raRw xxŽ. eR x . Hence raRŽ. s l s raRw xxŽ. R eR and therefore R is a p.p.-ring.

THEOREM 10. Let R be an Armendariz ring. Then R is a Baer ring if and only if Rwx x is a Baer ring. Proof. Assume that R is Baer. Let A be a nonempty subset of Rxwx and let A* be the set of all coefficients of elements of A. Then A*isa s g nonempty subset of R and so rARŽ.* eR for some idempotent e R. 484 KIM AND LEE

g wx: s q Since e rARw xxŽ., we get eR x rARw xxŽ.. Now, let g b01bx q иии q t g s s g bxtRrAw xxŽ.. Then Ag 0 and hence fg 0 for any f A. g s Thus b01, b ,...,btRrAŽ.* eR since R is Armendariz. Hence there g s q q иии q t s q exist c01, c ,...,ct R such that g ec01ec x ect x ecŽ 01cx q иии q t g wx wx cxt . eR x . Therefore Rx is Baer. Conversely, assume that Rxwxis a Baer ring. Let B be a nonempty s wx g subset of R. Then rBRw xxŽ. eR x for some idempotent e R by Lemma s 8. Hence rBRŽ. eR and therefore R is a Baer ring.

In the following text we obtain similar results for the formal power series rings.

PROPOSITION 11. Suppose that a ring R is abelian. Then we ha¨e the following:

Ž.1 If Rww x xx is a p. p.-ring, then R is a p. p.-ring. Ž.2 If Rww x xx is a Baer ring, then R is a Baer ring.

Proof. By the same methods in the proofs of Theorems 9 and 10.

COROLLARY 12. Suppose that a ring R is an Armendariz ring. Then we ha¨e the following:

Ž.1 If Rww x xx is a p. p.-ring, then R is a p. p.-ring. Ž.2 If Rww x xx is a Baer ring, then R is a Baer ring.

Proof. Combining Lemma 7 and Proposition 11.

Remark. The converse of Corollary 12Ž. 1 is not true in general by the following argument. Take the ring R inw 5, Example 1Ž. 1x . Notice that R is a Boolean ring and hence it is a p.p.-ring. R is also an Armendariz ring because it is reduced. However Rxww xx is not a p.p.-ring by the argument in wx4, Example 4 .

Let R be a reduced ring. Then the trivial extension T s TRŽ., R is

Ž. Ž.0 r Armendariz by Corollary 4; notice that the prime radical PT of T is 00 with r g R Žhence it is Armendariz by applying the definition of Armen- dariz rings to rings without identity.Ž. and that TrPT ( R is reduced Ž.hence it is also Armendariz . So one may suspect that if a ring R is an abelian ring such that RrPRŽ.and PR Ž.are Armendariz, then R is Armendariz, where PRŽ.is the prime radical of R. However, the following example erases the possibility. ARMENDARIZ RINGS AND REDUCED RINGS 485

EXAMPLE 13. Let ޚ be the ring of integers and let

ac R sNa y b ' c ' 0mod2.Ž. ½5ž/0 b

Ž.sNÄŽ.0 c ' Ž.4 Ž. Then PR 00 c 0 mod 2 and so PR is Armendariz. Also the

Ž.00 Ž. 10 only idempotents of R are00 and 01 . So R is abelian. Next note that RrPRŽ.is reduced and so it is Armendariz. In fact,

a 0 RrPRŽ.sNa y b ' 0Ž. mod 2 ½5ž/0 b ( Ä4Ž.a, b N a y b ' 0mod2. Ž .

If Ž.a, b 2 s Ža22, b .Ž.s 0, 0 , then a s 0 and b s 0.

Ž.sqŽ.22 Ž. 02 Now we claim that R is not Armendariz. Let fx 00 00 x and

Ž.sqŽ0 2 . Ž. 02Ž.Ž.s Ž.Ž. 22 02 / gx 0 y200x. Then fxgx 0, but 00000. Therefore R is not an Armendariz ring. Moreover we conjecture that R is an Armendariz ring if for any nonzero proper ideal I of R, RrI and I are Armendariz. However, we also have a counterexample to this situation as in the following.

EXAMPLE 14. Let F be a field and consider the ring

FF R s . ž/0 F

Then by Example 1, R is not Armendariz. Now we show that RrI and I are Armendariz for any nonzero ideal I of ŽFF .Ž.0 F Ž.0 F R. Note that the only nonzero proper ideals of R are00 , 0F , and00 . s Ž.FF r ( r First, let I 00. Then R I F and so R I is Armendariz. So we claim that I is Armendariz. For fxŽ., gx Ž.g Ixwx, suppose that fxgxŽ.Ž.s 0, s ␣ q ␣ q иии q␣ n s ␤ q ␤ q иии q␤ m where fxŽ. 01x n x and gxŽ. 01x m x . ␣ s Ž.abii ␤ s Ž.cdjj F F F F Let ij00 and 00, where 0 i n and 0 j m. Assume ␣ / ␤ / s s / s that 000 and 0. Then ac 0000 ad000.If a 0, then c 0 s s / and d000, which is a contradiction. So a 0 and hence b00. This ␣␤s F F implies that 0 j 0 for all j,0 j m. Hence the coefficient of x in ␣␤s s s / s fxgxŽ.Ž.is 10 0. Then ac100 ad 10.If a 1 0, then c0 0 and s ␣␤s F F d010, which is also a contradiction. So j 0 for all j,0 j m. ␣␤s F F F F Continuing this process, we have ij 0 for all i, j,0 i n 0 j m. Therefore I is Armendariz. 486 KIM AND LEE

s Ž.0 F r ( r Next let J 0 F . Then R K F and so R J is Armendariz. By the same method, we have that J is Armendariz.

s Ž.0 F r ( [ r Finally, let K 00. Then R K F F and so R K is Armendariz. Also K 2 s 0 and so K is Armendariz. Notice that for any nontrivial idempotent e g R,if R is an Armendariz ring then so is eRe; but Mat nŽ.R is not Armendariz and so ‘‘Armendariz’’ is not a Morita invariant property. But one may suspect that if eRe is an Armendariz ring for any nontrivial nonidentity idempotent e of R then R is an Armendariz ring. However, it is not true in general, by the following example. ޚ EXAMPLE 15. Let 2 be the ring of integers modulo 2 and consider the ޚޚ ring R s Ž.22ޚ . Then by Example 1, R is not Armendariz. Notice that the 0 2 only nontrivial nonidentity idempotents of R are

10 00 11 01 , , , and , ž/ž/ž/00 01 00 ž/ 01

( ޚ and that eRe 2 is an Armendariz ring for any nontrivial nonidentity idempotent e in R.

3. ARMENDARIZ RINGS AND REDUCED RINGS

Inwx 1 , Anderson and Camillo assert that it does seem possible for R to be reduced but the classical quotient ring QRŽ.of R is not reduced. But we have the following affirmative result.

THEOREM 16. Suppose that there exists the classical right quotient ring QŽ. R of a ring R. Then R is reduced if and only if Q Ž. R is reduced. Proof. It is enough to show that if R is reduced then QRŽ.is reduced. Let q be a nonzero element of QRŽ.with q 2 s 0. Then there exist a, b g R with b regular such that q s aby1.Soaby1aby1 s 0. Clearly bay1 g QRŽ.and so there exist c, d g R with d regular such that bay1 s cdy1. Then acŽ. bd y1 s acdy1 by1 s aby1 aby1 s 0 and so ac s 0. Hence Ž.ca 2 s 0 and so ca s 0 since R is reduced. Now from bay1 s cdy1,we have ad s bc in QRŽ..Soada s bca s 0. Thus ad s 0 and so a s 0 since d is regular, which is a contradiction. Therefore QRŽ.is reduced. Anderson and Camillo also assert that for a semiprime left and right Noetherian ring R, R is Armendariz if and only if QRŽ.is reduced in the argument afterwx 1, Theorem 6 . In the following corollary we add another condition. ARMENDARIZ RINGS AND REDUCED RINGS 487

COROLLARY 17. Let R be a ¨on Neumann regular ring and suppose that there exists the classical right quotient ring of R, QRŽ.. Then the following statements are equi¨alent: Ž.1 R is Armendariz. Ž.2 R is reduced. Ž.3 Q Ž R . is reduced. Ž.4 Q Ž R . is Armendariz. Proof. Ž.3 « Ž.2 and Ž. 4 « Ž.1 are straightforward. Ž. 2 « Ž.1 : Byw 2, Lemma 1x .Ž. 1 « Ž.3 : Assume that R is Armendariz. Then by Lemma 7, R is abelian von Neumann regular and so it is reduced; hence QRŽ.is reduced by Theorem 16.Ž. 3 « Ž.4 : Bywx 2, Lemma 1 . Anderson and Camillowx 1, Theorem 7 proved that if R is a prime ring which is left and right Noetherian, then R is Armendariz if and only if R is reduced. We obtain this result under a weaker condition.

PROPOSITION 18. Suppose that R is a semiprime right and left Goldie ring. Then R is Armendariz if and only if R is reduced. Proof. Since reduced rings are Armendariz bywx 2, Lemma 1 , it is enough to show that if R is Armendariz then R is reduced. By hypothesis, R has the right and left classical quotient ring QRŽ., which is semisimple ArtinianŽ. up to isomorphism by the Goldie theorem. Since we have right and left common denominators for finite sets of elements in QRŽ.byw 6, Lemma 2.1.8x , it follows that for fxŽ., gx Ž.g QR Ž .wx x there exist regular elements a and b in R such that afŽ. x , gxb Ž.g Rxwx. Assume that fxgxŽ.Ž.s 0. Recall that R is Armendariz and that a, b are regular; hence each coefficient of fxŽ.annihilates each coefficient of gxŽ.. Thus QRŽ.is Armendariz. On the other hand, QRŽ.is von Neumann regular; hence QRŽ.is reduced bywx 1, Theorem 6 and so R is reduced.

COROLLARY 19. Suppose that R is a semisimple Artinian ring. Then R is an Armendariz ring if and only if R is a finite direct sum of di¨ision rings. Proof. By Proposition 18.

ACKNOWLEDGMENTS

The authors express their gratitude to Professor Jae Keol Park for valuable suggestions and helpful comments. The first named author was supported in part by the Post-Doctoral Fellowship of KOSEF in 1998 during his stay at Pusan National University, while the second named author was supported by the Research Grant, Pusan National University in 1998 and 1999, and in part by the Korea Research Foundation, Project No. 1998-001-D00003. 488 KIM AND LEE

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