Rings in which every nilpotent is central
Burcu Ungor, Sait Halicioglu, Handan Kose, Abdullah Harmanci
aa r X i v : . [ m a t h . R A ] D ec RINGS IN WHICH EVERYNILPOTENT IS CENTRAL
Burcu Ungor and Sait Halicioglu
Department of Mathematics, Ankara University, Turkey [email protected] [email protected]
Handan Kose
Department of Mathematics, Ahi Evran University, Turkey [email protected] and
Abdullah Harmanci
Department of Mathematics, Hacettepe University, Turkey [email protected]
Abstract
In this paper, we introduce a class of rings in which every nilpo-tent element is central. This class of rings generalizes so-calledreduced rings. A ring R is called central reduced if every nilpo-tent element of R is central. For a ring R , we prove that R iscentral reduced if and only if R [ x , x , . . . , x n ] is central reducedif and only if R [[ x , x , . . . , x n ]] is central reduced if and only if R [ x , x − , x , x − , . . . , x n , x − n ] is central reduced. Moreover, if R isa central reduced ring, then the trivial extension T ( R, R ) is centralArmendariz.
Key words: reduced rings, central reduced rings, central semicom-mutative rings, central Armendariz rings, 2-primal rings. Throughout this paper all rings are associative with identity unless other-wise stated. A ring is reduced if it has no nonzero nilpotent elements. A ring R is called semicommutative if for any a, b ∈ R , ab = 0 implies aRb = 0.Recently a generalization of semicommutative rings is given in [3]. A ring R is called central semicommutative if for any a, b ∈ R , ab = 0 implies arb is a central element of R for each r ∈ R . A ring R is said to be abelian ifevery idempotent in R is central. A ring R is called right (left) principallyquasi-Baer [8] if the right (left) annihilator of a principal right (left) idealof R is generated by an idempotent. Finally, a ring R is called right (left)principally projective if the right (left) annihilator of an element of R isgenerated by an idempotent [7].In this paper, we introduce central reduced rings as a generalization ofreduced rings. Clearly, reduced rings are central reduced. We supply someexamples to show that all central reduced rings need not be reduced. Amongothers we prove that central reduced rings are abelian and there exists anabelian ring but not central reduced. Therefore the class of central reducedrings lies strictly between classes of reduced rings and abelian rings. Weprove that every central reduced ring is weakly semicommutative, centralsemicommutative, 2-primal, abelian and so directly finite, and a ring R iscentral reduced if and only if the Dorroh extension of R is central reduced.Moreover, it is proven that if R is a right principally projective ring, then R is central reduced if and only if R [ x ] / ( x n ) is central Armendariz, where n ≥ x n ) is the ideal generated by x n . Itis shown that a ring R is central reduced if and only if the polynomialring R [ x , x , . . . , x n ] is central reduced if and only if the power series ring R [[ x , x , . . . , x n ]] is central reduced if and only if the Laurent polynomialring R [ x , x − , x , x − , . . . , x n , x − n ] is central reduced. Finally we prove thatif R is a central reduced ring, then the trivial extension T ( R, R ) is centralArmendariz.Throughout this paper, Z denotes the ring of integers and for a positiveinteger n , Z n is the ring of integers modulo n . In this section we introduce a class of rings in which every nilpotent elementis central. We now give our main definition.
Definition 2.1
A ring R is called central reduced if every nilpotent elementof R is central. Commutative rings and reduced rings are central reduced. Every unit-central ring (i.e., every unit element of R is central [13]) is central reduced.One may suspect that central reduced rings are reduced. We now give anexample to show that central reduced rings need not be reduced. Example 2.2
Let S be a commutative ring and R = S [ x ] / ( x ). Then R isa commutative ring and so it is central reduced. If a = x + ( x ) ∈ R , then a = 0. Therefore R is not a reduced ring.Recall that a ring R is semiprime if aRa = 0 implies a = 0 for a ∈ R .Our next aim is to find conditions under which a central reduced ring isreduced. Proposition 2.3 If R is a reduced ring, then R is central reduced. Theconverse holds if R satisfies any of the following conditions. (1) R is a semiprime ring. (2) R is a right (left) principally projective ring. (3) R is a right (left) principally quasi-Baer ring. Proof.
First statement is clear. Conversely, assume that R is a centralreduced ring and a ∈ R with a = 0. Then a is central. Now consider thefollowing cases.(1) Let R be a semiprime ring. Since axa = 0 for all x ∈ R , it followsthat a = 0. Therefore R is reduced.(2) Let R be a right principally projective ring. Then there exists anidempotent e ∈ R such that r R ( a ) = eR . Thus a = ea = ae = 0, and so R is reduced. A similar proof may be given for left principally projectiverings.(3) Same as the proof of (2). (cid:3) Corollary 2.4 If R is a central reduced ring, then the following conditionsare equivalent. (1) R is a right principally projective ring. (2) R is a left principally projective ring. (3) R is a right principally quasi-Baer ring. (4) R is a left principally quasi-Baer ring. Proof.
It follows from Proposition 2.3 since in either case R is reduced. (cid:3) Note that the homomorphic image of a central reduced ring need not becentral reduced. Consider the following example.
Example 2.5
Let D be a division ring, R = D [ x, y ] and I = < x > where xy = yx . Since R is a domain, R is central reduced. On the other hand, x + I is a nilpotent element of R/I but not central. Hence
R/I is not centralreduced.We now determine under what conditions the homomorphic image of acentral reduced ring is also central reduced.
Proposition 2.6
Let R be a central reduced ring. If I is a nil ideal of R ,then R/I is central reduced.
Proof.
Let a + I ∈ R/I with ( a + I ) n = 0 for some positive integer n .Then a n ∈ I and there exists a positive integer m such that ( a n ) m = 0.Since R is central reduced, a is central. Hence ab − ba ∈ I for all b ∈ R andso a + I is central. Therefore R/I is central reduced. (cid:3)
It is well known that a ring is a domain if and only if it is prime andreduced. In addition to this fact, we have the following proposition whenwe deal with central case.
Proposition 2.7
Let R be a ring. Then R is a domain if and only if R isa prime and central reduced ring. Proof.
First assume R is a domain. It is clear that R is prime andreduced and so central reduced. Conversely, assume R is a prime and centralreduced ring. Let a, b ∈ R with ab = 0. Then rab = 0 for all r ∈ R . Since( bra ) = 0, bra and therefore bRa is contained in the center of R . Let s ∈ R and asbrasb ∈ ( asb ) R ( asb ) for any r ∈ R . Hence asbrasb = abras b = 0,since bra is central and ab = 0. It proves ( asb ) R ( asb ) = 0. Being R prime,we have asb = 0 for all s ∈ R and so aRb = 0. By invoking the primenessof R again we get a = 0 or b = 0. Therefore R is a domain. (cid:3) It is well known that every reduced ring is semicommutative. In ourcase we have the following.
Proposition 2.8
Every central reduced ring is central semicommutative.
Proof.
Let R be a central reduced ring and a, b ∈ R with ab = 0 and r ∈ R . Since ab = 0, ba is central. So barb = rbab = 0. Then ( arb ) = 0.By hypothesis arb is central. (cid:3) For any positive integer n and a ring R , let T n ( R ) denote the n × n upper triangular matrix ring over the ring R and R n ( R ) denote the subring { ( a ij ) ∈ T n ( R ) | all a ii ’s are equal for i = 1 , , ..., n } of T n ( R ).The following example shows that the converse of Proposition 2.8 maynot be true in general. Example 2.9
Let F be a field. By [3, Lemma 1.10] R ( F ) is central semi-commutative. We prove that it is not central reduced. Consider the nilpo-tent element A = . For B = ∈ R ( F ), AB is notequal to BA and so R ( F ) is not central reduced.It is clear that prime semicommutative rings are reduced. For centralcase we have the following result. Proposition 2.10
Let R be a prime central semicommutative ring. Then R is reduced. Proof.
Let a ∈ R with a n = 0 for some positive integer n . Since R iscentral semicommutative, a n − Ra is contained in the center of R . Hence( axa n − ) R ( axa n − ) = 0 for all x ∈ R . By hypothesis axa n − = 0 for all x ∈ R , and so aRa n − = 0. Thus a = 0 or a n − = 0. If a = 0, then theproof is completed. If a n − = 0, then we also have a = 0, by using thesimilar technique as above. (cid:3) The next example shows that for a ring R and an ideal I , if R/I iscentral reduced, then R need not be central reduced. Example 2.11
Let R = (cid:20) F F F (cid:21) , where F is any field. Since (cid:20) (cid:21) is a nilpotent but not central element of R , R is not central reduced. Nowconsider the ideal I = (cid:20) F F (cid:21) of R . Then R/I is central reduced becauseof the commutativity property of
R/I . Lemma 2.12
Let R be a prime ring. If R/I is a central reduced ring witha reduced ideal I , then R is a reduced ring. Proof.
Let
R/I be a central reduced ring. By Proposition 2.8,
R/I iscentral semicommutative. To complete the proof we show that R is centralsemicommutative. Let a, b ∈ R with ab = 0. Since bIa ⊆ I and ( bIa ) = 0, bIa = 0. Therefore (( aRb ) I ) = 0 and so ( aRb ) I = 0. Since R/I iscentral semicommutative and ( a + I )( b + I ) = I , aRb + I ∈ C ( R/I ), that is, arbr − r arb ∈ I for all r, r ∈ R and so ( arbr − r arb ) ∈ ( arbr − r arb ) I =0 by ( aRb ) I = 0. Then for all r, r ∈ R we have arbr = r arb and so aRb is contained in the center of R . Thus R is central semicommutative. ByProposition 2.10, R is reduced. (cid:3) Recall that a ring R is called weakly semicommutative [15], if for any a, b ∈ R , ab = 0 implies arb is a nilpotent element for each r ∈ R . Proposition 2.13
Let R be a central reduced ring. Then R is weakly semi-commutative. Proof.
Let a, b ∈ R with ab = 0. Since R is central reduced, ba is centralin R . Hence for each r ∈ R , ( arb ) = arbarb = ar bab = 0. Therefore R isa weakly semicommutative ring. (cid:3) The following example shows that there is a weakly semicommutativering which is not central reduced.
Example 2.14
In [15], it is proved that R ( R ) is a weakly semicommuta-tive ring for a ring R . Now consider the nilpotent element a = of R ( R ). Since ab = ba for b = ∈ R ( R ), R ( R ) is notcentral reduced.Let P ( R ) denote the prime radical and N ( R ) the set of all nilpotentelements of the ring R . The ring R is called 2 -primal if P ( R ) = N ( R )(See namely [10] and [12]). In [18, Theorem 1.5] it is proved that everysemicommutative ring is 2-primal. In this direction we prove Theorem 2.15
Every central reduced ring is 2-primal. The converse holdsfor semiprime rings.
Proof.
Let R be a central reduced ring. It is obvious that P ( R ) ⊆ N ( R ).For the converse inclusion, let a ∈ N ( R ) with a n = 0 for some positiveinteger n . Then ( RaR ) n = 0 ⊆ P ( R ), and so RaR ⊆ P ( R ). Hence we have a ∈ P ( R ). Therefore N ( R ) ⊆ P ( R ). Conversely, let R be a semiprime and2-primal ring. Then P ( R ) = 0 and so N ( R ) = 0. Hence R is reduced andso central reduced. This completes the proof. (cid:3) Theorem 2.16
Let R be a ring. Then the following are equivalent. (1) R is central reduced. (2) R/P ( R ) is central reduced with P ( R ) ⊆ C ( R ) where C ( R ) is thecenter of R . Proof. (1) ⇒ (2) Clear from Theorem 2.15.(2) ⇒ (1) Let x ∈ R with x n = 0 for some positive integer n and R denote the ring R/P ( R ). Since R is central reduced, x = x + P ( R ) iscentral in R . This implies that ( RxR ) n = 0 ⊆ P ( R ), and so RxR ⊆ P ( R ).Hence x ∈ P ( R ) = 0, thus x ∈ P ( R ). By hypothesis, x is central in R . (cid:3) Proposition 2.17
Every central reduced ring is abelian.
Proof.
Let R be a central reduced ring and e = e ∈ R . Then xe − exe and ex − exe are central for all x ∈ R since ( xe − exe ) = 0 and ( ex − exe ) = 0.Hence ( xe − exe ) e = 0 and e ( ex − exe ) = 0 for all x ∈ R . So we have xe = ex for all x ∈ R . Therefore R is abelian. (cid:3) Every abelian ring need not be central reduced, as the following exampleshows.
Example 2.18
Consider the ring R = (cid:26)(cid:20) a bc d (cid:21) : a ≡ d ( mod , b ≡ c ≡ mod (cid:27) Since (cid:20) (cid:21) and (cid:20) (cid:21) are the only idempotents of R , R is abelian.On the other hand, (cid:20) (cid:21) is a nilpotent element of R but not centralbecause (cid:20) (cid:21) (cid:20) (cid:21) = (cid:20) (cid:21) (cid:20) (cid:21) . Hence R is not centralreduced.Recall that a ring R is called directly finite whenever a, b ∈ R , ab = 1implies ba = 1. Then we have the following. Corollary 2.19 If R is a central reduced ring, then R is directly finite. Proof.
Clear from Proposition 2.17 since every abelian ring is directlyfinite. (cid:3)
A ring R is called nil clean [9] if there exist an idempotent e and anilpotent b in R such that a = e + b . We can give a relation between centralreduced and commutative rings by using nil clean rings. Proposition 2.20
Let R be a central reduced ring. If R is nil clean, thenit is commutative. Proof.
Let a ∈ R . Since R is nil clean, that is, an idempotent e anda nilpotent b exist in R such that a = e + b . By hypothesis, e and b arecentral, and so a is central. (cid:3) Let I be an index set and { R i } i ∈ I be a class of rings. Then R i is centralreduced for all i ∈ I if and only if Q i ∈ I R i is central reduced. Then the nextresult is an immediate consequence of Proposition 2.17. Corollary 2.21
Let R be a ring. Then the following are equivalent.1. R is central reduced.2. R is abelian and for any idempotent e ∈ R , eR and (1 − e ) R arecentral reduced.3. There is a central idempotent e ∈ R with eR and (1 − e ) R are centralreduced. Recall that a ring R is said to be regular if for any a ∈ R there exists b ∈ R with a = aba , while a ring R is called strongly regular if for any a ∈ R there exists b ∈ R such that a = a b .Now we give some relations between reduced, central reduced, regular,strongly regular and abelian rings. Also following theorem provides someconditions for the converses of Proposition 2.3 and Proposition 2.17(1). Theorem 2.22
Let R be a ring. Then the following are equivalent. R is strongly regular. (2) All R -modules are flat and R is central reduced. (3) All cyclic R -modules are flat and R is central reduced. (4) R is regular and central reduced. (5) R is regular and reduced. (6) R is regular and abelian. Proof. (1) ⇒ (2) Since R is strongly regular, R is central reduced. Onthe other hand, R is regular and so every R -module is flat.(2) ⇒ (3) and (3) ⇒ (4) Obvious.(4) ⇒ (5) Let a ∈ R with a = 0. By hypothesis there exists b ∈ R such that a = aba . Since ab is an idempotent, we have a = a b = 0 byProposition 2.17(1). Hence R is reduced.(5) ⇒ (6) Clear.(6) ⇒ (1) Let a ∈ R . By hypothesis, there exists b ∈ R such that a = aba . Since ab is an idempotent, ab is central. Hence a = a b andtherefore R is strongly regular. (cid:3) Let S denote a multiplicatively closed subset of a ring R consisting ofcentral regular elements. Let S − R be the localization of R at S . Then wehave the following lemma. Lemma 2.23
A ring R is central reduced if and only if S − R is centralreduced. Proof.
It is routine. (cid:3)
Theorem 2.24
Let R be a ring. Then the following are equivalent. (1) R is central reduced. (2) R [ x , x , . . . , x n ] is central reduced. (3) R [[ x , x , . . . , x n ]] is central reduced. (4) R [ x , x − , x , x − , . . . , x n , x − n ] is central reduced. Proof.
The equivalencies of (1), (2) and (3) are clear by showing that R is central reduced if and only if R [ x ] is central reduced. One way is clear. Soassume that R is central reduced. Let f ( x ) = a + a x + . . . + a n x n ∈ R [ x ]be nilpotent. To complete the proof it is enough to show a , a , . . . , a n arecentral. If f ( x ) = 0, then we have a = 0 (1) a a + a a = 0 (2) a a + a + a a = 0 (3) · · · Then a is central and by (2) we have 2 a a = 0. (3) implies 2 a a + a = 0.The latter gives 2 a a = − a . Hence a = 0. So a is central. We maycontinue in this way to obtain a , . . . , a n central. Now assume f ( x ) = 0.Then a = 0 (1) a a + a a a + a a = 0 (2) a a + a a + a a a + a a a + a a + a a = 0 (3) a a + a a a + a a a + a a a + a a a + a + a a a + a a a + a a a + a a = 0 (4) · · · (1) implies a is central. Hence from (4) we have 3 a a +3 a a a +3 a a a + a = 0. Since a = 0 and a is central, a = 0 and so a is central.Continuing on this way we get a , . . . , a n central. Similarly if f ( x ) m = 0, wemay prove that all coefficients of f ( x ) are central.(2) ⇒ (4) Let S = { , x, x , . . . } be a multiplicatively closed subset of R [ x ].By Lemma 2.23, R [ x, x − ] = S − R [ x ] is central reduced.(4) ⇒ (2) Clear. (cid:3) The following result can be easily obtain from Theorem 2.24.
Corollary 2.25
Let R be a ring and G a finitely generated free abeliangroup. Then the following are equivalent. (1) R is central reduced. RG is central reduced. It is known that reduced rings are nonsingular and commutative non-singular rings are reduced. But in our case there is no relation betweennonsingular and central reduced rings, as the following examples show.
Examples 2.26 (1) The ring Z is central reduced but not nonsingular.(2) The ring of 2 × (cid:20) (cid:21) is a nilpotent element but not central. Thusthis ring is not central reduced.The Dorroh extension D ( R, Z ) = { ( r, n ) : r ∈ R, n ∈ Z } of a ring R is a ring with operations ( r , n ) + ( r , n ) = ( r + r , n + n ) and( r , n )( r , n ) = ( r r + n r + n r , n n ). Obviously R is isomorphicto the ideal { ( r,
0) : r ∈ R } of D ( R, Z ). Then we obtain the following. Proposition 2.27
A ring R is central reduced if and only if the Dorrohextension D ( R, Z ) of R is central reduced. Proof.
Let R be a central reduced ring and ( r, n ) ∈ D ( R, Z ) with( r, n ) m = 0 for some positive integer m . Since n m = 0, it follows that n = 0and so r m = 0. By hypothesis r is central. Then ( r, n )( s, a ) = ( s, a )( r, n ) forany ( s, a ) ∈ D ( R, Z ). Therefore D ( R, Z ) is central reduced. The converseis clear. (cid:3) Let R be a ring and M an ( R, R )-bimodule. Recall that the trivialextension of R by M is defined to be ring T ( R, M ) = R ⊕ M with theusual addition and the multiplication ( r , m )( r , m ) = ( r r , r m + m r ).This ring is isomorphic to the ring (cid:26)(cid:20) r m r (cid:21) : r ∈ R, m ∈ M (cid:27) with theusual matrix operations and isomorphic to R [ x ] / ( x ), where ( x ) is the idealgenerated by x . The trivial extension of R by M need not be a centralreduced ring, as the following example shows.3 Example 2.28
Let H be the division ring of quaternions over the real num-bers. Then H is a reduced ring but not commutative. Consider the nilpotentelement (cid:20) i (cid:21) of T ( H , H ). Since (cid:20) i (cid:21) (cid:20) j j (cid:21) = (cid:20) j j (cid:21) (cid:20) i (cid:21) , T ( H , H ) is not central reduced.It can be easily shown that for a positive integer n ≥ M n ( R ) and T n ( R ) can not be central reduced even if R is commutative. But we havethe following result when we deal with T ( R, R ). Proposition 2.29
Let R be a ring. Then the following are equivalent. (1) R is commutative. (2) T ( R, R ) is central reduced. Proof. (1) ⇒ (2) By hypothesis, T ( R, R ) is commutative, and so it iscentral reduced.(2) ⇒ (1) Let x, y ∈ R . Since (cid:20) x (cid:21) ∈ T ( R, R ) is nilpotent, itcommutes with (cid:20) y y (cid:21) ∈ T ( R, R ). Hence we have xy = yx . (cid:3) Let R be a ring and f ( x ) = n P i =0 a i x i , g ( x ) = s P j =0 b j x j ∈ R [ x ]. Regeand Chhawchharia [17] introduce the notion of an Armendariz ring, thatis, f ( x ) g ( x ) = 0 implies a i b j = 0 for all i and j . The name of the ringwas given due to Armendariz who proved that reduced rings satisfied thiscondition [6]. The interest of this notion lies in its natural and useful rolein understanding the relation between the annihilators of the ring R andthe annihilators of the polynomial ring R [ x ]. So far, Armendariz ringsare generalized in different ways (see namely, [11], [16]). In particular, aring R is called linear Armendariz [14], if the product of two linear poly-nomials in R [ x ] is zero, then each product of their coefficients is zero. Aring R is called central linear Armendariz [1], if the product of two lin-ear polynomials in R [ x ] is zero, then each product of their coefficients is4central. According to Harmanci et al. [2], a ring R is called central Ar-mendariz if f ( x ) g ( x ) = 0 implies that a i b j is central element of R for all i and j . A ring R is called weak Armendariz [16] if f ( x ) g ( x ) = 0, then a i b j is a nilpotent element of R for each i and j , while a ring R is called nil-Armendariz [5] if f ( x ) g ( x ) has nilpotent coefficients, then a i b j is nilpotentfor 0 ≤ i ≤ n , 0 ≤ j ≤ s . Clearly every nil-Armendariz ring is weak Armen-dariz. In [4, Theorem 5], Anderson and Camillo proved that for a ring R and n ≥ R [ x ] / ( x n ) is Armendariz if and only if R is reduced.For central reduced rings, we have Theorem 2.30
Let R be a right principally projective ring and n ≥ anatural number. Then R is central reduced if and only if R [ x ] / ( x n ) is centralArmendariz. Proof.
Suppose R is a central reduced ring. By Proposition 2.3, R is areduced ring. From [4, Theorem 5], R [ x ] / ( x n ) is Armendariz and so centralArmendariz. Conversely, assume that R [ x ] / ( x n ) is central Armendariz. Byhypothesis and [2, Theorem 2.5], R [ x ] / ( x n ) is Armendariz. It follows from[4, Theorem 5] that R is reduced and so central reduced. (cid:3) Central reduced rings allow us to get the following result.
Theorem 2.31
Let R be a central reduced ring. Then the followings hold. (1) R is nil-Armendariz. (2) R is weak Armendariz. (3) R is central Armendariz. Proof. If R is central reduced, then it is 2-primal by Theorem 2.15and so N ( R ) is an ideal of R . Proposition 2.1 in [5] states that in a ringin which the set of all nilpotent elements forms an ideal, then the ring isnil-Armendariz. Therefore R is weak Armendariz and central Armendariz. (cid:3) Corollary 2.32 If R is a central reduced ring, then R [ x ] / ( x n ) is nil-Armendariz. Proof. If R is central reduced, then it is nil-Armendariz by Theorem2.31. From [5, Proposition 4.1], R [ x ] / ( x n ) is nil-Armendariz. (cid:3) We now give a useful lemma without proof to show that the trivialextension of a central reduced ring is central Armendariz.
Lemma 2.33
The following hold for a ring R with a, b ∈ R . (1) The sum of central nilpotent elements of R is nilpotent. (2) If b is central nilpotent, then ba and ab are nilpotent. (3) If aba is central nilpotent, then ab and ba are nilpotent. Note that if R is a reduced ring, by [17, Proposition 2.5] trivial extension T ( R, R ) is Armendariz and so it is central Armendariz. In [1, Lemma 2.18],it is proved that for a central reduced ring R , the trivial extension T ( R, R ) iscentral linear Armendariz. Here we extend this result to central Armendarizrings. Note that in proving Theorem 2.34 we use the results in Lemma 2.33without mention.
Theorem 2.34 If R is a central reduced ring, then the trivial extension T ( R, R ) is central Armendariz. Proof.
Let f ( x ) = (cid:20) a a ′ a (cid:21) + (cid:20) a a ′ a (cid:21) x + . . . + (cid:20) a n a ′ n a n (cid:21) x n = (cid:20) f ( x ) f ( x )0 f ( x ) (cid:21) , g ( x ) = (cid:20) b b ′ b (cid:21) + (cid:20) b b ′ b (cid:21) x + . . . + (cid:20) b t b ′ t b t (cid:21) x t = (cid:20) g ( x ) g ( x )0 g ( x ) (cid:21) be elements of T ( R, R )[ x ], where f ( x ) = a + a x + ... + a n x n , f ( x ) = a ′ + a ′ x + ... + a ′ n x n , g ( x ) = b + b x + ... + b t x t , g ( x ) = b ′ + b ′ x + ... + b ′ t x t .Assume that f ( x ) g ( x ) = 0. We have f ( x ) g ( x ) = (cid:20) f ( x ) g ( x ) f ( x ) g ( x ) + f ( x ) g ( x )0 f ( x ) g ( x ) (cid:21) = 0 . Hence f ( x ) g ( x ) = 0 and f ( x ) g ( x ) + f ( x ) g ( x ) = 0. By Theorem 2.31, R is both nil-Armendariz and central Armendariz. Hence a i b j and b j a i are6central nilpotent in R for all i, j . As for the coefficients of f ( x ) g ( x ) + f ( x ) g ( x ) = 0, we have a b ′ + a ′ b = 0 (1) a b ′ + a b ′ + a ′ b + a ′ b = 0 (2) a b ′ + a b ′ + a b ′ + a ′ b + a ′ b + a ′ b = 0 (3) · · · Multiplying (1) from the right by a , we have a b ′ a + a ′ b a = 0. Since b a is central nilpotent, a b ′ a is central nilpotent, so a b ′ and b ′ a arecentral nilpotent. From (1) a ′ b and b a ′ are central nilpotent.Multiplying (2) from the right by a , we have a b ′ a + a b ′ a + a ′ b a + a ′ b a = 0. Since b ′ a , b a and b a are central nilpotent, a b ′ a and so a b ′ and b ′ a are central nilpotent. Now multiplying (2) from the left by b and using central nilpotency of b a , b a and b a ′ , we have b a ′ b is centralnilpotent and so b a ′ and a ′ b are central nilpotent. Multiplying (2) fromthe right by a and using a b ′ a , a ′ b a and a ′ b a are central nilpotent wehave a b ′ a and so a b ′ and b ′ a are central nilpotent. The remaining termof (2) a ′ b and b a ′ are central nilpotent.Similarly, multiplying (3) from the right by a and using b ′ a , b ′ a , b a , b a and b a are central nilpotent, we have a b ′ a , therefore b ′ a and a b ′ are central nilpotent. Multiplying (3) from the left by b and using b a , b a , b a , b a ′ and b a ′ are central nilpotent, we have b a ′ b , therefore b a ′ and a ′ b are central nilpotent. Multiplying (3) by a from right we have a b ′ a , a b ′ a , a ′ b a , a ′ b a , a ′ b a are central nilpotent. Hence a b ′ a and therefore a b ′ and b ′ a are central nilpotent. Similarly, multiplying(3) by b from the left we have b a b ′ , b a b ′ , b a b ′ , b a ′ b and b a ′ b arecentral nilpotent, therefore b a ′ b and so b a ′ and a ′ b are central nilpotent.Multiplying (3) from the right by a and using a b ′ a , a b ′ a , a ′ b a , a ′ b a , a ′ b a are central nilpotent, then a b ′ a and so a b ′ and b ′ a are centralnilpotent. a ′ b in (3) is central nilpotent since it is a sum of central nilpotentelements, so is b a ′ . Thus all terms of (3) are central nilpotent.To complete the proof, we use induction on i + j for i + j ≤ n + t .Assume that the claim is true for all i + j − i + j ≤ n + t , that is forall k and l with k + l ≤ i + j − a k b ′ l , b ′ l a k , a ′ k b l , b l a ′ k are central nilpotent.Consider the ( i + j )-th equation7 a b ′ i + j + a b ′ i + j − + a b ′ i + j − + ... + a i + j − b ′ + a i + j b ′ + a ′ b i + j + a ′ b i + j − + ... + a ′ i + j − b + a ′ i + j − b + a ′ i + j b = 0.In order to complete the proof by induction, we have to show that all termsare central nilpotent in ( i + j ). We proceed as preceding. Multiplying( i + j ) by a from the right we have a b ′ i + j a + a b ′ i + j − a + a b ′ i + j − a + ... + a i + j − b ′ a + a i + j b ′ a + a ′ b i + j a + a ′ b i + j − a + ... + a ′ i + j − b a + a ′ i + j − b a + a ′ i + j b a . Since b ′ i + j − a , b ′ i + j − a ,..., b ′ a , b ′ a are central nilpotent by induc-tion hypothesis and b i + j a , b i + j − a ,..., b a , b a , b a are central nilpotent, a b ′ i + j a and so a b ′ i + j and b ′ i + j a are central nilpotent. Similarly, multiply-ing ( i + j ) from the left by b , we have b a b ′ i + j + b a b ′ i + j − + b a b ′ i + j − + . . . + b a i + j − b ′ + b a i + j b ′ + b a ′ b i + j + b a ′ b i + j − + . . . + b a ′ i + j − b + b a ′ i + j − b + b a ′ i + j b = 0. Since b a , b a , b a , ..., b a i + j − , b a i + j , b a ′ , b a ′ , ..., b a ′ i + j − , b a ′ i + j − are central nilpotent, b a ′ i + j b is central nilpotent andtherefore b a ′ i + j and a ′ i + j b are central nilpotent. Proceeding in this man-ner, we finally obtain a k b ′ l , b ′ l a k , a ′ k b l , b l a ′ k are central nilpotent for all k and l with k + l ≤ i + j . This completes the induction hypothesis. 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