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1 International Journal of Pure and Applied Mathematics Volume 91 No , ISSN: (printed version); ISSN: (on-line version) url: doi: PAijpameu NIL 3-ARMENDARIZ RINGS RELATIVE TO A MONOID Eltiyeb Ali 1, Ayoub Elshokry 2 1,2 Department of Mathematics Northwest Normal University Lanzhou, , PR CHINA 1,2 Department of Mathematics University of Khartoum Omdurman, SUDAN Abstract: For a monoid M, we introduce nil 3-M-Armendariz, which are a common generalization of nil 3-Armendariz and 3-M-Armendariz rings, and investigates their properties We show that a ring R is nil 3-M-Armendariz ring if and only if for any n N,T n (R) is nil 3-M-Armendariz, where M is a monoid Also we show that if a ring R is semicommutative which is also nil 3-M-Armendariz, then R is nil 3-(M N)-Armendariz, where N is a unique product monoid AMS Subject Classification: 16S36, 16U20, 16N60, 16U99 Key Words: unique product monoid, 3-Armendariz ring, nil 3-Armendariz ring, 3-M-Armendariz ring, nil 3-M-Armendariz 1 Introduction Throughout this article, R and M denote an associative ring, not necessary with identity and a monoid, respectively Given a ring R, the polynomial ring over R is denoted by R[x] Rege and Chhawchharia [12], introduced the notion of an Armendarizring AringRiscalledArmendarizifwheneverpolynomialsf(x) = Received: November 30, 2013 Correspondence author c 2014 Academic Publications, Ltd url: wwwacadpubleu

2 88 E Ali, A Elshokry a 0 +a 1 x+ +a n x n, g(x) = b 0 +b 1 x+ +b m x m R[x], satisfy f(x)g(x) = 0, then a i b j = 0 for each i and j The name Armendariz ring was chosen because Armendariz [8, Lemma 1], has shown that a reduced ring (ie, a ring without nonzero nilpotent elements) satisfies this condition Some properties of Armendariz rings have been studied in Hong and et al [3], Anderson and Camillo [4], Kim and Lee [15], Huh et al [2] and Lee and Wong [17] In [20], Suiyi, introduced the notion of 3-Armendariz rings A ring R is called a 3- Armendariz ring if whenever polynomials f(x) = a 0 +a 1 x+ +a n x n,g(x) = b 0 +b 1 x+ +b m x m,h(x) = c 0 +c 1 x+ +c r x r R[x],satisfyf(x)g(x)h(x) = 0, then a i b j c k = 0, for all i,j and k Armendariz rings are thus a generalization of reduced rings, and therefore, nilpotent elements play an important role in this class of rings In fact, in [4], Anderson and Camillo proved that if n 2, then R[x]/(x n ) is an Armendariz ring if and only if R is reduced Zhongkui [10], studied a generalization of Armendariz rings, which are called M-Armendariz rings, where M is a monoid A ring R is called M-Armendariz if whenever elements α = a 1 g 1 + +a n g n,β = b 1 h 1 + +b m h m R[M], satisfy αβ = 0, then a i b j = 0 for each i,j, where g i,h j M Elshokry and et al [1], studied a generalization of M-Armendariz rings, which are called 3-M-Armendariz rings, where M is a monoid A ring R is called 3-M-Armendariz if whenever elements α = a 1 g 1 + +a n g n,β = b 1 h 1 + +b m h m,γ = c 1 l 1 + +c r l r R[M], satisfy αβγ = 0, then a i b j c k = 0 for each i,j,k, where g i,h j,l k M Armendariz ring are abelian by Kim and Lee [15] Subrings of M-Armendariz rings are also M-Armendariz by Zhongkui [10] Subrings of 3-Armendariz rings are also 3-Armendariz by Suiyi [20] Subrings of 3-M-Armendariz rings are also 3- M-Armendariz by Elshokry and et al [1] According to Antoine [16] A ring R is called nil-armendariz if whenever two polynomials f(x), g(x) R[x], satisfy f(x)g(x) nil(r)[x] then ab nil(r) for all a coef(f(x)) and b coef(g(x)), where coef(f(x)) denote the subset of R of the coefficients of f(x) Mohammed and et al [14], introduced the notion of nil M-Armendariz rings A ring R is said to be nil M-Armendariz rings, if whenever elements α = a 1 g 1 + +a n g n,β = b 1 h 1 + +b m h m R[M], satisfy αβ nil(r)[m], when a i b j nil(r) for each i,j Zhang Cuiping and Chen Jianlong [21], introduced the notion of weak M-Armendarizrings A ringris said to beweak M-Armendariz if whenever elements α = a 1 g 1 + +a n g n and β = b 1 h 1 + +b m h m R[M], satisfy αβ = 0, then a i b j nil(r) for each i,j In [11], Liu and Zhao introduced weak Armendariz rings which a generalization of Armendariz rings A ring R is called weak Armendariz if whenever the product of two polynomials is zero then the product of their coefficients is nilpotent Weak Armendariz rings have the property that motivates the study of the nilpotent elements in this class of

3 NIL 3-ARMENDARIZ RINGS RELATIVE TO A MONOID 89 rings Recall that a ring R is called semicommutative if for all a,b R,ab = 0 implies arb = 0 Yang Suiyi [20], introduced the notion of Condition (P), for all a,b,c R, if (abc) 2 = 0, then abc = 0 In [19], Wu Hui-feng introduced the concept of weak 3-Armendariz ring which a generalization of 3-Armendariz ring and weak Armendariz ring and investigate their properties A ring R is called weak 3-Armendariz if whenever the product of three polynomials is zero then the product of their coefficients is nilpotent If M = {e}, then every ring is 3-M-Armendariz, so it is nil 3-M-Armendariz Thus nil 3-M-Armendariz rings need not be nil 3-Armendariz Hence nil 3-M-Armendariz rings are a common generalization of 3-M-Armendariz rings and nil 3-Armendariz rings If M = (N {0},+), then a ring R is nil 3-M-Armendariz if and only if R is nil-3-armendariz Recall that a monoid M is called a up-monoid (unique product monoid) if for any two nonempty finite subsets A,B M there exists an element g M uniquely presented in the form ab where a A and b B The class of up- monoids is quite large and important (see Birkenmeier and Park [9], Passman [5]) For example, this class includes the right or left ordered monoids, submonoids of a free group, and torsion-free nilpotent groups Every up-monoid M has non unity element of finite order Motivated by results in Elshokry and et al [1], Suiyi [20], Zhongkui [10], Habibi and Moussavi [13], Antoine [16], Ebrahim Hashemi [7] and Mohammed and et al [14], we will investigate a generalization of 3-M-Armendariz which are called nil 3-M-Armendariz For a ring R, we denote by R[M] the monoid ring over R, and by nil(r) the set of nilpotent elements in R If α R[M],coef(α) denote the subset of R of the coefficients of α 2 Nil 3-Armendariz Rings Relative to a Monoid For a monid M, e will always stand for the identity of M We denote by T n (R) the n n upper triangular matrix over a ring R Definition 21 [6, Definition 25] A ring R is said to be nil 3-Armendariz if whenever polynomials f(x), g(x), h(x) R[x], satisfy f(x)g(x)h(x) nil(r)[x] then abc nil(r) for all a coef(f(x)),b coef(g(x)) and c coef(h(x)) Definition 22 Let M be a monoid A ring R is called nil 3-Armendariz relative to M (nil 3-M-Armendariz) if whenever elements α = a 1 g a n g n,β = b 1 h b m h m and γ = c 1 l 1 + +c r l r R[M], satisfy αβγ nil(r)[m], then a i b j c k nil(r) for each i,j,k

4 90 E Ali, A Elshokry Lemma 23 [18, Proposition 1] If R a reduced rings then R satisfies condition (P), but the converse is not true Clearly, any subrings of nil 3-M-Armendariz rings are nil 3-M-Armendariz and any 3-M-Armendariz ring is nil 3-M-Armendariz In the following, we will see that the converse is not true Proposition 24 Let R be a ring and M a monoid Then R is a nil 3-M-Armendariz ring if and only if, for any n,t n (R) is a nil 3-M-Armendariz ring Proof We note that any subring of nil 3-M-Armendariz ring is a nil 3- M-Armendariz ring Thus if T n (R) is a nil 3-M-Armendariz ring, then so is R Conversely, let α = A 1 g A n g n,β = B 1 h B m h m and γ = C 1 l 1 + +C r l r be elements of T n (R)[M] It is easy to see that there exists an isomorphism of rings T n (R)[M] T n (R[M]) define by: p i=1 a i 11 a i 12 a i 13 a i 1n 0 a i 22 a i 23 a i 2n 0 0 a i 33 a i 3n g i a i nn p i=1 ai 11 g p i i=1 ai 12 g p i i=1 ai 13 g i p i=1 ai 1n g i 0 p i=1 ai 22 g p i i=1 ai 23 g i p i=1 ai 2n g i 0 0 p i=1 ai 33 g i p i=1 ai 3n g i p i=1 ai nng i Assume that αβγ nil(t n (R))[M] Let A i = a i 11 a i 12 a i 13 a i 1n 0 a i 22 a i 23 a i 2n 0 0 a i 33 a i 3n a i nn,b j = b j 11 b j 12 b j 13 b j 1n 0 b j 22 b j 23 b j 2n 0 0 b j 33 b j 3n b j nn

5 NIL 3-ARMENDARIZ RINGS RELATIVE TO A MONOID 91 and C k = c k 11 c k 12 c k 13 c k 1n 0 c k 22 c k 23 c k 2n 0 0 c k 33 c k 3n c k nn Then we have p i=1 ai 11 g p i i=1 ai 12 g p i i=1 ai 13 g i p i=1 ai 1n g i 0 p i=1 ai 22 g p i i=1 ai 23 g i p i=1 ai 2n g i 0 0 p i=1 ai 33 g i p i=1 ai 3n g i p i=1 ai nn g i q j=1 bj 11 h q j j=1 bj 12 h q j j=1 bj 13 h j q j=1 bj 1n h j 0 q j=1 bj 22 h q j j=1 bj 23 h j q j=1 bj 2n h j 0 0 q j=1 bj 33 h j q j=1 bj 3n h j q j=1 bj nnh j d k=1 ck 11 l d k k=1 ck 12 l d k k=1 ck 13 l k d k=1 ck 1n l k 0 d k=1 ck 22 l d k k=1 ck 23 l k d k=1 ck 2n l k 0 0 d k=1 ck 33 l k d k=1 ck 3n l k nil(t n (R[M])) d k=1 ck nnl k Because T n (R)[M] = T n (R[M]) Also nil(r) R R R 0 nil(r) R R nil(t n (R)) = 0 0 nil(r) It follows that ( i) p a i ss g i=1 for s = 1,2,,n ( q d ) b j ss h j c k ss l k nil(r)[m], j=1 k=1

6 92 E Ali, A Elshokry Since R is nil 3-M-Armendariz, there exists m ijks N such that (a i ss bj ss ck ss )m ijks = 0 for any s and any i,j,k Let m ijk = max{m ijk1,m ijk2,,m ijkn } Then (a i 11 bj 11 ck 11 )m ijk (A i B j C k ) m ijk 0 (a i 22 = bj 22 ck 22 )m ijk 0 0 (a i nn bj nnc k nn )m ijk = Thus ((A i B j C k ) m ijk) n = 0 and so A i B j C k nil(t n (R)), for each i,j,k This shows that T n (R) is nil 3-M-Armendariz ring Corollary 25 Let M be a monoid If a ring R is 3-M-Armendariz, then, for any n, T n (R) is nil 3-M-Armendariz Now we can give the example of nil 3-M-Armendariz ring which is not 3-M-Armendariz Example 26 Let M be a monoid Let S be a nil 3-M-Armendariz ring Then the ring a a 12 a 13 a 1n 0 a a 23 a 2n R n = 0 0 a a 3n a,a ij S a is not 3-M-Armendariz by Elshokry and et al [1, Example 214] when n 4, but R n is nil 3-M-Armendariz by Proposition 24, since any subring of nil 3-M-Armendariz ring is nil 3-M-Armendariz From Proposition 24, one may suspect that if R is nil 3-M-Armendariz then every n-by-n full matrix ring M n (R) over R is nil 3-M-Armendariz, where n 2 But the following example erases the possibility

7 NIL 3-ARMENDARIZ RINGS RELATIVE TO A MONOID 93 Example 27 Let M be a monoid with M 2 and R a ring with identity Take e g M Let S = M 2 (F) Let α = ( γ = ) ( 1 0 e+ 0 0 ( be elements in S[M] Then αβγ = 0 But ( )( ) ( 1 0 g,β = 0 1 ) ( 0 0 e+ 1 1 )( ) = is not nilpotent Thus S is not nil 3-M-Armendariz ) g ( ) g, Corollary 28 Let M be a monoid and R a ring {( Then R) is nil 3-M- } a b Armendariz,IfandonlyifthetrivialextensionT(R,R) = a,b R 0 a is a nil 3-M-Armendariz Proof It follows from Proposition 24 Proposition 29 The class of nil 3-M-Armendariz rings is closed under finite direct products Proof Let R = s β R s be the finite direct product of R s where β = {1,2,,p}, R s is nil 3-M-Armendariz Suppose αβγ nil(r)[m] for some elements α = a 1 g 1 +a 2 g 2 + +a n g n, β = b 1 h 1 +b 2 h 2 + +b m h m andγ = c 1 l 1 + c 2 l 2 + +c r l r R[M], wherea i = (a i1,a i2,,a ip ),b j = (b j1,b j2,,b jp ),c k = (c k1,c k2,,c kp ), are elements of the product ring R Set α s = Σ n i=1 a isg i, β s = Σ m j=1 b jsh j and γ s = Σ r k=1 c ksl k R[M] Since αβγ nil(r)[m] then Σ i+j+k=u a i b j c k nil(r),1 u n+m+rsoσ i+j+k=u (a i1 b j1 c k1,,a ip b jp c kp ) = 0,andsoΣ i+j+k=u (a is b js c ks ) nil(r),1 s pthusα s β s γ s nil(r s )[M],1 s p Since R s is nil 3-M-Armendariz, then we have a is b js c ks nil(r s ) Now, for each i,j,k, there exist positive integers m ijks such that (a is b js c ks ) m ijks = 0, in the ring R s,1 s p If we take m ijk = max{m ijks : 1 s p}, then it is clear that (a is b js c ks ) m ijk = 0 Therefore a i b j c k nil(r) This means that R is nil 3-M-Armendariz Theorem 210 Let M be a up-monoid and nil(r) an ideal of R Then R is nil 3-M-Armendariz )

8 94 E Ali, A Elshokry Proof Let α = n i=1 a ig i,β = m j=1 b jh j and γ = r k=1 c kl k in R[M], satisfy αβγ nil(r)[m] Since nil(r) is an ideal of R, the ring R = R/nil(R) is reduced By Lemma 23, R = R/nil(R) satisfies condition (P) and so 3- M-Armendariz, by [1, Theorem 26] Also, αβγ nil(r)[m] implies that ᾱ β γ = 0 So ā i bj c k = 0, for each i,j and k, since R is 3-M-Armendariz Thus a i b j c k nil(r), for each i,j and k, and the result follows Thus, Theorem 210 implies that this class involves nil(r) R Moreover, if we take M = (N {0},+), in Theorem 210, it follows that nil(r) R are in fact nil 3-Armendariz [6, Proposition 23] Proposition 211 Let M be a up-monoid and R a reduced ring Then R is nil 3-M-Armendariz Proof Since R is reduced, hence nil(r) = 0 is an ideal of R Thus, the result follows from Theorem 210 Corollary 212 Let M be a up-monoid and R satisfies condition (P) Then R is nil 3-M-Armendariz Proof Let M be a up-monoid and R satisfies condition (P) Then by [1, Theorem 26], R is 3-M-Armendariz Thus, R is nil 3-M-Armendariz Corollary 213 Let M be a up-monoid and R a semicommutative ring Then R is nil 3-M-Armendariz Proof Since R is a semicommutative ring, by [11, Lemma 31], nil(r) is an ideal of R Hence the result follows from Theorem 210 Let (M, ) be an ordered monoid If for any g,g,h M,g < g implies that gh < g h and hg < hg, then (M, ) is called a strictly ordered monoid A monoid M is said to be totally orderable if (M, ) is an ordered monoid for some total order Since each strictly totally ordered monoid is up-monoid, hence we have the following results Corollary 214 Let M be a strictly totally ordered monoid and nil(r) an ideal of R Then R is nil 3-M-Armendariz Corollary 215 Let M be a strictly totally ordered monoid and R a reduced ring Then R is nil 3-M-Armendariz

9 NIL 3-ARMENDARIZ RINGS RELATIVE TO A MONOID 95 Clearly (Z,+) is a strictly totally ordered monoid So a ring R is nil 3-Z- Armendariz, if whenever α = Σ p i= n a ix i,β = Σ q i= m b jx j and γ = Σ s k= r c kx k R[x;x 1 ], satisfy αβγ nil(r)[x;x 1 ], then a i b j c k nil(r) for each i,j,k Corollary 216 Let R be a ring satisfies condition (P) Then R is nil 3-Z-Armendariz Proof Since R be a ring satisfies condition (P) Then by [1, Corollary 28], R is 3-Z-Armendariz, hence R is nil 3-Z-Armendariz Corollary 217 Let R be a semicommutative ring Then R is nil 3-Z- Armendariz Proposition 218 Let M be a monoid, R be a nil 3-M-Armendariz ring and α i nil(r)[m], for 1 i n If α 1 α 2 α n nil(r)[m], then a 1 a 2 a n nil(r), where a i coef(α i ) In particular, nil(r[m]) nil(r)[m] Proof Let a i be a coefficient of α i, for each i We have α 1 (α 2 α n ) nil(r)[m] Thus a 1 b nil(r), for each b coef(α 2 α 3 α n ), since R is nil 3-M-Armendariz So a 1 α 2 α 3 α n nil(r)[m] Hence (a 1 α 2 )(α 3 α n ) nil(r)[m] Thus, a 1 a 2 b nil(r), for each b coef(α 3 α 4 α n ) By continuing in this way, we have a 1 a 2 a n nil(r) and the proof is complete Theorem 219 Let R be a ring and M be a monoid If nil(r[m]) = nil(r)[m], then R is nil 3-M-Armendariz Proof Suppose α = n i=1 a ig i,β = m j=1 b jh j and γ = r k=1 c kl k be elements of R[M] be such that αβγ nil(r)[m] = nil(r[m]) So there exists a positive integer s such that (αβγ) s = 0 Therefore, we have (a i b j c k ) s = (a i b j c k ) (a i b j c k ) = 0, by Proposition 218 and thus a i b j c k nil(r), for each i,j and k Hence R is a nil 3-M-Armendariz ring Observe that if nil(r) is an ideal, then by [6, Proposition 26], R is nil 3-Armendariz More generally we obtain the following Proposition 220 Let M be a monoid and I R be a nil ideal of R Then R is nil 3-M-Armendariz if and only if so is R/I Proof SinceI nil(r),wehavenil(r/i) = nil(r)/isoαβγ nil(r)[m] if and only if ᾱ β γ nil(r/i)[m] Also, abc nil(r) if and only if ā b c nil(r/i) Therefore R is nil 3-M-Amendariz if and only if R/I is nil 3-M- Armendariz

10 96 E Ali, A Elshokry TakingM = (N {0},+), inproposition 220, it follows that foranynilideal I R, we have R is nil 3-Armendariz if and only if so is R/I [6, Proposition 26] Recall that an element u of a ring R is right regular if ur = 0 implies r = 0 for r R Similarly, left regular elements can be defined An element is regular if it is both left and right regular (and hence not a zero divisor) Proposition 221 Let R be a ring and be a multiplicative monoid in R consisting of central regular elements Then R is nil 3-M-Armendariz if and only if so is 1 R Proof Let R be nil 3-M-Armendariz ring, and S = 1 R Put αβγ = 0, where α = n i=1 a ig i,β = m j=1 b jh j and γ = r k=1 c kl k S[M] We may assume that a i = ε i u 1,b j = η j v 1 and c k = µ k w 1 with ε i,η j, µ k are in R for all i,j and k, and u,v,w We will show that a i b j c k nil(s) Now we have Hence nil(s)[m] αβγ = n m r i=1 j=1 k=1 a ib j c k g i h j l k = n m r i=1 j=1 k=1 ε iη j µ k u 1 v 1 w 1 g i h j l k = ( n m r i=1 j=1 k=1 ε iη j µ k g i h j l k )(uvw) 1 n m i=1 j=1 k=1 r ε i η j µ k g i h j l k nil(r)[m] Since R is nil 3-M-Armendariz, ε i η j µ k nil(r) for all i,j and k and so a i b j c k = ε i u 1 η j v 1 µ k w 1 = ε i η j µ k (uvw) 1 nil(s), for all i,j,k Thus, S is nil 3-M-Armendariz The converse it is obvious that subring of nil 3-M-Armendariz is nil 3-M-Armendariz Proposition 222 Let M be a monoid, R be a ring and e an idempotent of R If e is central in R, then the following statements are equivalent: 1 R is nil 3-M-Armendariz; 2 er and (1 e)r are nil 3-M-Armendariz Proof We only need to prove (2) (1) Let α = n i=1 a ig i,β = m j=1 b jh j and γ = r k=1 c kl k be elements in R[M] be such that αβγ nil(r)[m] Let α 1 = Σ n i=1 (ea i)g i,α 2 = Σ n i=1 (1 e)a ig i,β 1 = Σ m j=1 (eb j)h j,β 2 = Σ m j=1 (1 e)b jh j and γ 1 = Σ r k=1 (ec k)l k,γ 2 = Σ r k=1 (1 e)c kl k So α 1 β 1 γ 1 nil(er)[m] and

11 NIL 3-ARMENDARIZ RINGS RELATIVE TO A MONOID 97 α 2 β 2 γ 2 nil((1 e)r)[m] Thus ea i b j c k nil(er) and (1 e)a i b j c k nil((1 e)r), since er and (1 e)r are nil 3-M-Armendariz Therefore e(a i b j c k ) u ijk and (1 e)(a i b j c k ) v ijk = 0, for some positive integer u ijk and v ijk If we take s = max{u ijk,v ijk 1 i n,1 j m,1 k r}, then we have e(a i b j c k ) s = (1 e)(a i b j c k ) s = 0, for each i,j,k Thus, (a i b j c k ) s = 0 and so a i b j c k nil(r), for each i,j,k This implies that R is nil 3-M-Armendariz and the proof is complete Proposition 223 Let M be a monoid If R is a semicommutative ring which is also nil 3-M-Armendariz, then we have nil(r[m]) = nil(r)[m] Proof Since R is nil 3-M-Armendariz, we have nil(r[m]) nil(r)[m], by Proposition 218 Now, let α = a 1 g a n g n nil(r)[m], and let k > 1 such that a k i = 0 for all i = 1,,n We claim that α nk = 0 The coefficients of α nk can be written as sums of monomials of length nk in the a i s Consider one of these monomials, a i1 a i2 a ink where 1 i j n It must contain at least k occurrences of some a j0 for some 1 j 0 n Since a k j 0 = 0 and R is semicommutative, we have a i1 a i2 a ink = 0 Therefore, we have proved that all the monomials appearing in the coefficients of α nk are 0 Hence nil(r)[m] nil(r[m]), and so nil(r[m]) = nil(r)[m] Zhongkui [10, Proposition 21], it was shown that if R is a reduced and M-Armendariz ring, then R[M] is N-Armendariz, where M is a monoid and N a up-monoid Also Elshokry and et al [1, Proposition 31], it was shown that if R satisfies condition (P), and is 3-M-Armendariz, then R[M] is 3-N- Armendariz For nil 3-M-Armendariz, we have the following results Proposition 224 Let M be a monoid and N a up-monoid If R is a semicommutative ring which is also nil 3-M-Armendariz, then R[M] is nil 3-N-Armendariz Proof By Proposition 223, nil(r[m]) = nil(r)[m] is an ideal of R[M] Since N is a up-monoid, hence by Theorem 210, R[M] is nil 3-N-Armendariz Proposition 225 Let M be a monoid and N a up-monoid If R is a semicommutative ring which is also nil 3-M-Armendariz, then R[N] is nil 3-M-Armendariz

12 98 E Ali, A Elshokry Proof It is easy to see that there exists an isomorphism of rings R[N][M] = R[M][N] defined by ( p i a ip n i )m p i ( p a ip m p )n i Nowsupposethatα i,β j,γ k R[N]aresuchthat( i α ig i )( j β jh j )( k γ kl k ) nil(r[n])[m], where g i,h j,l k M We will show that α i β j γ k nil(r[n]) for all i,j and k Assume that α i = p a ipn p,β j = q b jqn q and γ k = s c ksn s, where n p,n q,n s N for all p,q and s Then ( i ( p a ip n p )g i )( j ( q b jq n q )h j)( k ( s c ks n s )l k) nil(r[n])[m] Thus, in R[M][N] we have ( p ( i a ip g i )n p )( q ( j b jq h j )n q)( s ( k c ks l k )n s) nil(r[m])[n] By Proposition 224, R[M] is nil 3-N-Armendariz, ( i a ip g i )( j b jq h j )( k c ks l k ) nil(r[m]) for all p,q and s Since R is nil 3-M-Armendariz, a ip b jq c ks nil(r) for all i,j,k,p,q,s Hence α i β j γ k nil(r[n]) This means that R[N] is nil 3-M- Armendariz Corollary 226 Let M be a monoid and R a semicommutative ring If R is nil 3-M-Armendariz, then R[x] and R[x;x 1 ] are nil 3-M-Armendariz Proof Since R[x] = R[N {0}] and R[x;x 1 ] = R[Z], the result follows from Proposition 225 Theorem 227 Let M be a monoid and N a up-monoid If R a semicommutative ring which is also nil 3-M-Armendariz, then R is nil 3-(M N)- Armendariz Proof By [1, Theoerem 33], R[M N] = R[M][N], and by Proposition 223,nil(R[M]) = nil(r)[m] Now the assertion follows from Proposition 224

13 NIL 3-ARMENDARIZ RINGS RELATIVE TO A MONOID 99 Let M i,i I, be monoids Denote i I M i = {(g i ) i I there exist only finite i s such that g i e i, the identity of M i } Then i I M i is a monoid with the operation (g i ) i I (g i ) i I = (g i g i ) i I Corollary 228 Let M i,i I be up-monoids and R a semicommutative ring If R is nil 3-M i0 -Armendariz for some i 0 I, then R is nil 3- i I M i- Armendariz Proof Let α = Σ i a i g i,β = Σ j b j h j,γ = Σ k c k l k R[ i I M i] such that αβγ nil(r[ i I M i]) Then α,β,γ R[M 1 M 2 M n ], for some finite subset {M 1,M 2,,M n } {M i i I} Thus α,β,γ R[M i0 M 1 M 2 M n ] The ring R, by Theorem 227 and by induction, is nil 3- (M i0 M 1 M 2 M n )-Armendariz, so a i b j c k nil(r) for all i,j and k Hence R is nil 3- i I M i-armendariz Acknowledgments We would like to thank Professor Liu Zhongkui for his valuable comments and we would like to thank the managements of University of Khartoum and Northwest Normal University Also the authors thank the referee for a very careful reading of the paper References [1] A Elskokry, E Ali, L Zhongkui, On the extension of Armendariz rings relative to a monoid, (Accepted) [2] C Huh, Y Lee and A Smoktunowicz, Armendariz rings and semicommutative rings, Comm Algebra, 30, No 2 (2002), [3] CYHong, NKKimandTKKwak, OnSkewArmendarizrings, Comm Algebra, 31, N0 1 (2003), [4] D D Anderson, V Camillo, Armendariz rings and Gaussian rings, Comm Algebra, 26, No 7 (1998), [5] D S Passman, The Algebraic Structure of Group Rings, John Wiley, New York, (1977)

14 100 E Ali, A Elshokry [6] E Ali, A Elshokry, L Zhongkui, Nil 3-Armendariz rings, Advances Pure Math, 3, No 9 (2013), [7] E Hashemi, Nil-Armendariz rings relative to a monoid, Mediter J Math, 10, No 1 (2013), [8] E P Armendariz, A note on extensions of Baer and pp-rings, J Austral Math Soc, 18 (1974), [9] G F Birkenmeier, JK Park, Triangular matrix representations of ring extensions, J Algebra, 265 (2003), [10] L Zhongkui, Armendariz rings relative to a monoid, Comm Algebra, 33, No 3 (2005), [11] L ZhongKui, R Y Zhao, On weak Armendariz rings, Comm Algebra, 34, No 7 (2006), [12] M B Rege, S Chhawchharia, Armendariz rings, Proc Japan Acad Ser A math Sci, 73 (1997), [13] M Habibi, A Moussavi, Nilpotent elements and nil Armendariz property of monoid rings, J Algebra, App, 11, No 4 (2012), [14] M J Nikmehr, F Fatahi and H Amraei, Nil-Armendariz rings with Applications to a monoid, World App Sci J, 13, No 12 (2011), [15] N K Kim, Y Lee, Armendariz rings and reduced rings, J Algebra, 223 (2000), [16] R Antoine, Nilpotent elements and Armendariz rings, J Algebra 319, (2008), [17] TK Lee, TL Wong, On Armendariz rings, Houston J Math 29, No 3 (2003), [18] Wu Hui-feng, Extensions of Reduced Rings, J Hangzhou Normal Uni, 10, No 5 (2011), [19] Wu Hui-feng, On Weak 3-Armendariz rings, J Hangzhou Normal Uni, 11, No 3 (2012), [20] Y Suiyi, On the extension of Armendariz rings, [D] Lanzhou University, (2008)

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