On nil-mccoy rings relative to a monoid

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1 PURE MATHEMATICS RESEARCH ARTICLE On ni-mccoy rings reative to a monoid Vahid Aghapouramin 1 * and Mohammad Javad Nikmehr 2 Received: 24 October 2017 Accepted: 29 December 2017 First Pubished: 25 January 2018 *Corresponding author: Vahid Aghapouramin Department of Mathematics Karaj Branch Isamic AzadUniversity Karaj Iran E-mai: nikmehr@kntu.ac.ir Reviewing editor: Lishan Liu Qufu Norma University China Additiona information is avaiabe at the end of the artice Abstract: The concept of ni-m-mccoy (ni-mccoy ring reative to monoid M) which are generaizations of McCoy ring and ni-m-armendariz rings have been introduced and we investigate their properties. It is shown that every NI ring is ni-m-mccoy for any unique product monoid M it has aso been shown that every semicommutative rings is ni-m-mccoy for any unique product monoid and any stricty totay ordered monoid M. Moreover it is proved that for an idea I of R if I is semicommutative and R / I is ni-m-mccoy then R is ni-m-mccoy for any stricty totay ordered monoid. We extend and unify many known resuts reated to McCoy rings and ni-armendariz ring. Subjects: Agebra; Pure Mathematics; Mathematics & Statistics for Engineers Keywords: M-McCoy rings; Ni-M-Armendariz rings; Ni-M-McCoy rings; u.p.-monoid AMS subject cassifications: 16U99; 16S15 1. Introduction A rings considered here are associative with identity. For a ring R ni(r) denotes the set of nipotent eements. Rege and Chhawchharia (1997) introduced the notion of an Armendariz ring. Reca that a ring R is caed reduced if a 2 = 0 impies that a = 0 for a a R; R is symmetric if abc = 0 impies acb = 0 for a b c R; R is reversibe if ab = 0 impies ba = 0 for a a b R; R is semicommutative if ab = 0 impies arb = 0 for a a b R. Reca that a monoid M is caed an u.p.-monoid (unique product monoid) if for any two nonempty finite subsets A B M there exists an eement g M that can be written uniquey in the form ab where a A and b B. The cass of u.p.monoids is quite arge and important (see Birkenmeier & Park 2003). According to Niesen (2006) a ring R is caed right McCoy whenever poynomias f (x) g(x) R[x] {0} satisfy f (x)g(x) =0 there exists a nonzero r R such that f (x)r = 0. We define eft McCoy ring simiary. If a ring is both eft and right McCoy we say that the ring is a McCoy ring. Liu (2005) studied a generaization of Armendariz ring which is caed M-Armendariz ring if ABOUT THE AUTHORS Vahid Aghapouramin is currenty a PhD student at the department of mathematics Karaj Branch Isamic Azad university Karaj Iran. His research interests are pure mathematics Commutative Agebra Noncommutative Agebra. Mohammad Javad Nikmehr is a associate professor Facuty of Mathematics K.N. Toosi University of Technoogy P.O. Box Tehran Iran. His research interest are Commutative Agebra Graph Theory Noncommutative Agebra Agebraic Combinatorics. PUBLIC INTEREST STATEMENT The current paper attempts to generaize NimacCoy rings with Monoid to present a nove work with the tite of Ni-M-MacCoy. The survey of iterature demonstrates that studies on Ni- Armendariz and Ni-MacCoy and Ni-M-Armendariz and have aready been researched. The theorems emmas and coroaries which have been aready studied concerning this topic to be compared with the new work to generaize or compete it. It paves the way for the wiing researchers to continue the trend with the nove areas that have been investigated The Author(s). This open access artice is distributed under a Creative Commons Attribution (CC-BY) 4.0 icense. Page 1 of 12

2 whenever α = n i=1 g i β = m j=1 b j R[M] with g i M satisfy αβ = 0 then a j b j = 0 for each i j. Zhao Zhu and Gu (2012) introduced the notion of a ni-mccoy ring. A ring R is said to be right ni- McCoy if f (x)g(x) ni(r)[x] where f (x) = m =0 i xi g(x) = n b j=0 j xj R[x] {0} impies that there exists r R {0} such that r ni(r) for 0 i m. Left ni-mccoy ring are defined simiary. A ring is ni-mccoy if it is both eft and right ni-mccoy. According to Hashemi (2010) a ring R is caed right M-McCoy if whenever α = n a g i=1 i i β = m b =1 j j R[M] {0} with g i M b j R satisfy αβ = 0 then αr = 0 for some nonzero r R. Left M-McCoy ring are defined simiary. Aso if R is both eft and right M-McCoy then we say R is M-McCoy. Our resuts aso generaized and unify the above-mentioned concepts by introducing the notion of ni-m-mccoy ring. 2. Ni-McCoy rings reative to a monoid We begin this section by the foowing definition and aso we study properties of ni-m-mccoy rings. Definition 2.1 Let M be a monoid and R a ring. We say that R is right ni-m-mccoy (right ni-mccoy ring reative to M) if whenever α = a a n g n β = b b m h m R[M] {0} with g i M b j R satisfy αβ ni(r)[m] then r ni(r) for some nonzero r R and for eac i n. Left ni-m-mccoy rings are defined simiary. If R is both eft and right ni-m-mccoy then we say R is ni- M-McCoy. Let M =( {0} +). Then a ring R is ni-m-mccoy (resp. ni-m-armendariz) if and ony if R is ni- McCoy (resp. ni-armendariz). If M ={e} then for any ring R R is ni-m-mccoy and ni-m-armendariz. Exampe 2.1 Let 2 be the ring of integers moduo 2 and R be a ni-armendariz ring and R = 2 a b c d e I where I is the idea generated by the reations: ac = 0 ad + bc = 0 bd = 0 ea = eb = ec = ed = ee = de = ce = be = 0. Get S be subring generated by eements not invoving e. Let α = a + bg and β = c + dh such that g h R[M] {0} and et g = h then αβ = 0. It is straightforward to see that the eement as is not nipotent for any nonzero s S. However R is ni-m-mccoy as αe eβ ni(r)[m] for any eements α β R[M]. For any α R[M] we denote by C α the set of a coefficients of α. Proposition 2.1 For any "u.p."-monoid M every NI ring is ni-m-mccoy. Proof Let M be an "u.p."-monoid and R be a NI ring. Suppose that α = a a m g m β = b b n h n R[M] {0} satisfy αβ ni(r)[m]. Since ni(r) is an idea of R R / ni(r) is M-McCoy by Liu (2005 Proposition 2.6). For any α = m i=1 g i R[M] denote α = m i=1 ( + ni(r))g i (R ni(r))[m]. It is easy to see that the mapping φ:r[m] (R ni(r))[m]. Defined by φ(α i )=α i is a ring homomorphism. Since αβ ni(r)[m] C αβ ni(r). Hence we have αβ = αβ ni(r)[m]. So b j ni(r) =0 for i j since R / ni(r) is M-McCoy. Thus b j ni(r) for i j. Choosing r = b n 0 and s = a m 0 we have r ni(r) and sb j ni(r) for i j. Therefore R is ni-m- McCoy. Page 2 of 12

3 Exampe 2.2 Let M be "u.p."-monoid and K be fied. Suppose S = K < a a 2 = 0 >. Then the ring ( ) s as R = is ni-m-mccoy since R is a NI ring. as s To prove Theorem 2.1 we state the foowing emmas. Lemma 2.1 Let M be cycic group of order n 2 and R a ring wit 0. Then R is not ni-m-mccoy. Proof Suppose that M ={e h h n 1 }. Let α = 1e + 1h + + 1h n 1 and β = 1e + ( 1)h then αβ ni(r)[m] but c = 1c ni(r) for each nonzero c R. Therefore R is not right ni-m-mccoy. Simiary R is not eft ni-m-mccoy. Therefore R is not ni-m-mccoy. Lemma 2.2 Let M be a monoid and N a submonoid of M. If R is ni-m-mccoy then R is ni-n-mccoy. Proof It is cear by Liu (2005 emma 1.12) and Hashemi (2010 emma 1.12). Lemma 2.3 Liu (2005 Lemma 1.13). Let M and N be "u.p."-monoid. Then so is the monoid M N. Exampe 2.3 Let M N M 2 ( ) and et M N be u.p. -monoids. Let A = {(( ai b i c i d i ) ( ei f i g i h i )) } M N i = p and B = {(( a j c j b j d j ) ( e j f j g j h j )) } M N j = 1 2 q (. Since )( M is a "u.p."-monoid ) there exist i j wit i p and 1 j q such that ai b i a j b j is uniquey presented by considering two subsets c i d i c j d j { ( a1 b 1 c 1 d 1 ) (... a p c p b p d p )} { ( a 1 b 1 and c 1 d 1 ) (... a q c q b q d q )} of M. Consider two subsets {( ek f k ) (( ai b i ) ( ek f k )) } N A and g k h k c i d i g k h k {( ) (( ) ( )) } e f a j b j e f N B ofn. g h c j d j g h ( ek f Since N is a u.p. -monoid there exist k such that k g k h presented. k (( ) ( ))(( ) ai b Now it is easy to see that i ek f k a j b j presented. Thus M N is a u.p. -monoid. c i d i g k h k c j d j )( e f g h ( e f g h ) is uniquey )) is uniquey Page 3 of 12

4 Reca that a monoid M is caed torsion-free if the foowing property hods: if g h M and k 1 are such that g k = h k then g = h. Let T(H) be the set of eements of finite order in an abeian group H. H is said to be torsien-free If T(H)={e}. Theorem 2.1 Let H be finitey generated abeian group. Then the foowing conditions on H are equivaent: (1) H is torsion-free. (2) There exists a ring R with R 2 such that ni-h-mccoy. Proof (1) (2) If H is finitey generated abeian group with T(H) ={e} then H a finite direct product of the group by Lemma 2.3 H is u.p. -monoid. Let R be a NI ring. Then by Proposition 2.1 R is ni-h-mccoy. (2) (1) If h T(H) and h e then N = h is a cycic group of finite order. If a ring R {0} is ni-h- McCoy then by Lemma 2.2 R is ni-n-mccoy a contradiction with Lemma 2.1. Therefore every ring R {0} is not ni-h-mccoy. A ring R is caed right Ore if given a b R with b reguar there exist a 1 b 1 R with b 1 reguar such that ab 1 = ba 1. eft Ore rings can be defined simiary. Proposition 2.2 Let M be a monoid and R a right Ore ring with its cassica right quotient ring Q. If R is right ni-m-mccoy then Q is right ni-m-mccoy. Proof Let α = m α g β = n β h i=0 i i j=0 j j be nonzero poynomias of Q[M] such that αβ ni(q)[m]. Since Q is a cassica right quotient ring we assume that α i = u 1 and β j = b j v 1 for b j R for a i j and reguar eements u v R. For eac there exists c j R and a reguar eement w R such that u 1 b j = c j w 1. Denote α = m a g and i=0 i i β = n b =0 j j R[M] {0}. We have αβ = m n α β g h i=0 j=0 i j i j = m n =0 j=0 i u 1 b j v 1 g i = m n a c i=0 j=0 i j w 1 v 1 g i = m n a c i=0 j=0 i j (vw) 1 g i = α β (vw) 1. So α β ni(r)[m]. Since R is right ni-m-mccoy there exists a nonzero eement c R such that c ni(r). So c = α i (uc) ni(r) for each i and uc is a nonzero eement in Q. Hence Q is a right ni-m-mccoy ring. Theorem 2.2 Let M be a monoid with M 2. Then the finite direct sum of ni-m-mccoy rings is ni- M-McCoy. Proof It suffices to show that if R 1 R 2 are ni-m-mccoy rings then so is R 1 R 2. Let α =(a 1 1 b1)g (a1 m b1 )g and β m m =(a2 1 b2)h (a2 n b2)h (R R n n 1 2 )[M] {0} be such that αβ ni(r 1 R 2 )[M]. Therefore we show that there exists r s R 1 R 2 such that r ni(r 1 R 2 ) and sb j ni(r 1 R 2 ) for each =(a 1 i b1) b i j =(a 2 j b2) and for a r =(r r ) s =(s s j ). Now et f 1 = a 1 g a1 g g = m m 1 b1g b1 g f = m m 2 a2h a2h g = n n 2 b2h b2h. Then n n f 1 f 2 ni(r 1 )[M] and g 2 ni(r 2 )[M]. Since by hyopothesis R 1 and R 2 are ni-m-mccoy a 1 r ni(r ) i 1 1 b 1 r ni(r ) for eac i m. Thus for eac i j 2 2 m(a1 i b1)(r r ) ni(r R i ) and aso s 1 a 2 j ni(r 1 ) s 2 b 2 j ni(r 2 ) for eac j n. Thus for eac j n(s 1 s 2 )(a 2 j b2) ni(r R ). j 1 2 Therefore R 1 R 2 is ni-m-mccoy. Exampe 2.4 Let R be an M-McCoy reduced ring and M a monoid with M 2 and et a 11 a 12 a 1n 0 a T n (R) ={ 22 a 2n j R}. 0 0 a nn Page 4 of 12

5 Then T n (R) is not M-McCoy for n 4 by Ying et a. (2008 Theorem 2.1). But T n (R) is ni-m-mccoy by Theorem 2.3 beow. Hence a ni-m-mccoy ring is not a trivia extension of an M-McCoy. Theorem 2.3 Let M be a monoid with M 2. Then the foowing conditions are equivaent: (1) R is ni-m-mccoy. (2) T n (R) is ni-m-mccoy. Proof (1) (2) In a simiar way proposition 2.12 (Nikmehr Fatahi & Amraei 2011). It is easy to see that there exists an isomorphism of rings T n (R)[M] T n (R[M]) defined n 1n i=1 ai g n 11 i i=1 ai 2 i n i=1 ai g 1n i n 0 22 n 2n 0 i=1 g i ai g 22 i n i=1 ai g 2n i. i=1 0 0 nn 0 0 n i=1 ai g nn i Suppose that α = A A n g n and β = B B m h m T n (R)[M] {0} are such that αβ ni(t n (R))[M] where A i B j T n (R). We caim there exists B T n (R) such that A i B ni(t n (R)) for each i. Assume that A i = n n 0 0 nn B = c j 11 c j 12 c j 1n 0 c j 22 c j 2n 0 0 c j nn and et α s = n i=1 ai ss g i R[M] and β s = m j=1 cj ss R. Aso by observing that ni(t n (R)) = ni(r) R R 0 ni(r) R 0 0 ni(r) thus we have α s β s ni(r)[m] for eac s n. Since R is a ni-m-mccoy ring there exists some positive integer m is such that ( ss bj ss )m i s = 0 for any s and any i. Let m i = max{m is 1 s n}. Then ((A i B) m i ) n = 0 and so A i B ni(t n (R)). Therefore T n (R) is ni-m-mccoy. (2) (1) Suppose that T n (R) is ni-m-mccoy. Note that R is isomorphic to the subring a a a a R of T n (R). Thus R is ni-m-mccoy since each sub ring of a ni-m-mccoy ring is aso ni-m-mccoy. Let R be a ring and et S n (R) = a a 12 a 1n 0 a a 2n 0 0 a a a ij R Page 5 of 12

6 and T(R n) ={ a 1 a 2 a n 0 a 1 a n a 1 a ij R} and T(R R) be the trivia extension of R by R. Using the same method in the proof of Theorem 2.3 we have the foowing resuts: Coroary 2.1 Let M be a monoid with M 2. Then the foowing conditions are equivaent: (1) R is ni-m-mccoy; (2) S n (R) is ni-m-mccoy; (3) T(R n) is ni-m-mccoy; (4) T(R R) is ni-m-mccoy; (5) R[x] x n is ni-m-mccoy for each n 2. Proof Using the same method in the proof of Theorem 2.3 we have (1) (2) (1) (3) and (1) (4). It is easy to see that T(R n) is a sub ring of the trianguar matrix rings with matrix addition and mutipication. We can denote eements of T(R n) by (a 0 a 1 a n 1 ) then T(R n) is a ring with addition point-wise and mutipication given by (a 0 a 1 a n 1 )(b 0 b 1 b n 1 )=(a 0 b 0 a 0 b 1 + a 1 b 0 a 0 b n 1 + +a n 1 b 0 ) for each b j R. On the other hand there is a ring isomorphism φ: R[x] T(R n) since by <x n > φ(a 0 + a 1 x + +a n 1 x n 1 )=(a 0 a 1 a n 1 ) with R 0 i n 1. So T(R n) R[x] where R[x] <x n > is the ring of poynomia in an indeterminate x and < x n > is the idea generated by x n. Therefore (3) (5).Thus by simiary method we have (4) (5). Therefore a conditions are equivaent. a Let R be a ring and M a monoid. Let G 3 (R) ={ a 21 a 22 a 23 a ij εr}. The G 3 (R) is a subring of 0 0 a fu matrix ring M 3 (R) under usua addition and mutipication. In fact G 3 (R) possesses the simiar form of both the ring of a ower trianguar matrices and the ring of a upper trianguar matrices. A natura probem asks if the ni-m-mccoy property of such subring of M n (R) coincides with that of R. This inspires us to consider the ni-m-mccoy property of G 3 (R). Theorem 2.4 Let M be a monoid with M 2. Then the foowing conditions are equivaent: (1) R is ni-m-mccoy; (2) G 3 (R) is ni-m-mccoy. Proof (1) (2) We first show that ni(g 3 (R)) = ni(r) 0 0 R ni(r) R. Suppose that 0 0 ni(r) a ni(r) 0 0 a 21 a 22 a 23 R ni(r) R and m is a positive integer such that 0 0 a ni(r) a m 11 = am 22 = am 33 = 0. Then a a 21 a 22 a a 33 2m = 0. Hence Page 6 of 12

7 ni(r) 0 0 R ni(r) R ni(g 3 (R)). 0 0 ni(r) a Now assume that a 21 a 22 a 23 ni(g 3 (R)). Then there exists some positive integer m such 0 0 a 33 m a that a 21 a 22 a 23 = 0. Hence a m = 11 am = 22 am 33 = 0 and so 0 0 a 33 a ni(r) 0 0 a 21 a 22 a 23 R ni(r) R. 0 0 a ni(r) ni(r) 0 0 Therefore ni(g 3 (R)) = R ni(r) R. Then by anaogy with the proof of Theorem 2.4 we 0 0 ni(r) can show that G 3 (R) is ni-m-mccoy. (2) (1) it is trivia (see Hashemi 2013). In the proof of the next proposition we wi need the foowing emma. Lemma 2.4 Zhao et a. (2012 Lemma 2.7) Let R be a semicommutative ring and f 1 (x) f 2 (x)... f n (x) be in R[x]. If C f1...f n ni(r) then C f1 C f2 C fn ni(r). Proposition 2.3 Semicommutative rings are ni-m-mccoy rings. Proof Let α = m i=0 g i β = n j=0 b j R[M] {0} with αβ ni(r)[m] then we have C αβ ni(r). It foows that C α C β ni(r) by Lemma 2.4. Hence there exists r = b j for some 0 j n such that r ni(r) with 0 i m. This impies that R is a right ni-m-mccoy. Simiary we can show that R is eft ni-m-mccoy. Therefore R is a ni-m-mccoy. We now have the foowing description of the rings which shows one way to give more ni-m-mc- Coy rings from od ones. Proposition 2.4 Let Ω be an index set and {R I I Ω} a famiy of rings. If R = R I Ω I then R is right ni-m-mccoy if and ony if every R I is right ni-m-mccoy for each I Ω. Proof It is straightforward to verify that if R is right ni-m-mccoy then every R I is right ni-m-mccoy for each I. Conversey if αβ ni(r)[m] for α = m a g β = n b h i=0 i i j=0 j j R[M] {0} where =(I ) I Ω b j =(b ji ) I Ω R. For each I Ω et α I = m =0 i g I i β I = n b j=0 j h I j R I [M] then α I β I ni(r I )[M]. Since β 0 there exists some index J with β J 0. In particuar there exists some nonzero r J R J with J r J ni(r J ) for a 0 i m by the ni-m-mccoy property of R J. Fix r J R J {0} and take r to be the sequence with r J in the J th coordinate and zero esewhere. Ceary r ni(r) and r 0. Coroary 2.2 A finite direct product of right ni-m-mccoy rings is right ni-m-mccoy. Theorem 2.5 For any monoid M. Then we have the foowing statements: Page 7 of 12

8 (1) ni-m-armendariz rings are ni-m-mccoy; (2) M-McCoy rings are ni-m-mccoy; (3) M-Armendariz rings are ni-m-mccoy. Proof (1) Let R be a ni-m-armendariz rings and α = a a m g m β = b b n h n R[M] {0} satisfy αβ ni(r)[m]. Then b j ni(r) for each i j. Since α 0 and β 0 for 1 n 1 k m there exists r = b s = a k R {0}. Hence r ni(r) and sb j ni(r). Therefore R is ni-m-mccoy. (2) It foows easiy form the definition. (3) It is easy to see that each M-Armendariz rings is M-McCoy and therefore ni-m-mccoy by (2). Let R i be a ring for each i Z and et R = R i and R 1 i Z i R be the subring generated by R i Z i and {1 R }. Then we have the foowing resut (see Ahevaz & Moussavi 2010). Proposition 2.5 Let R i be a ring R = R and S = R 1 i Z i i Z i R. Then R is right ni-m-mccoy if and ony if each R i is right ni-m-mccoy if and ony if S is right ni-m-mccoy. Proof Let each R i be a right ni-m-mccoy ring and α = m a g and β = n b h i=0 i i j=0 j j R[M] {0} such that αβ ni(r)[m] where =(a k ) and b i j =(b k ). If there exists t such that j at i = 0 for each 0 i m then we have c ni(r) where c =(0 0 1 Rt 0 ). Now suppose for each k there exists 0 i k m such that a k i 0. Since β 0 there exists t and 0 j k t n such that b t j 0. Consider t α t = t i=0 atg and β = t i i t j=0 bt j R t [M] {0}. We have αβ ni(r)[m]. Thus there exists some nonzero c t R t such that a t c ni(r) for each 0 i m. Therefore t ic ni(r) for each 0 i m where c =(0 0 c t 0 ) and so R is right ni-m-mccoy. Conversey suppose R is right ni-m-mccoy t Z α = m a g and β = n b h i=0 i i j=0 j j R[M] {0} such that αβ ni(r)[m]. Let α = m ( a 1...)g and i=0 i i β = n ( b 0...)=0 j j R[M] {0}. So there exists 0 c =(c i ) and (1 1 1 )c ni(r) ceary c =(0 0 c t 0 ). Thus we have c t ni(r t ) and so R t is right ni-m-mccoy. It is easy to see that S is right ni-m-mccoy if and ony if each R i is right ni-m-mccoy. Let (M ) be an ordered monoid. If for any g 2 h M < g 2 impies that h < g 2 h and g 2 h < h then (M ) is caed a stricty totay ordered monoid. Coroary 2.3 Let M be a stricty totay ordered monoid and R a reversibe ring. Then R is ni-m-mccoy. It was shown in Liu (2005 Proposition 1.4) that if M is stricty totay ordered monoid and I is a reduced idea of R such that R / I is an M-Armendariz ring then R is M-Armendariz. The foowing resut generaizes this. Theorem 2.6 Let M be a stricty totay ordered monoid and I an idea of R. If I is semicommutative (as a ring without identity) and R/I is ni-m-mccoy then R is ni-m-mccoy. Proof Let α = a a n g n β = b b m h m R[M] {0} be such that αβ ni(r)[m] and < g 2 <... < g n < h 2 <... < h m. We wi use transfinite induction on the stricty totay ordered set (M ) to show that r ni(r) for some nonzero r R and for each i. Note that in (R / I)[M] (a 1 + a 2 g a n g n )(b 1 + b 2 h b m h m ) ni(r I)[M]. The fact that R / I is ni-m-mccoy means that there exists r in R such that r ni(r I) thus there exists p N such that ( r) p I for each i. If there exists 1 i n and 1 j m such that g i = then g i and. If < g i then < g i g i = a contradiction. Thus g i =. Simiary =. Therefore r ni(r) for some nonzero r R and for each i. Page 8 of 12

9 Now suppose that ω M is such that r ni(r) for each g i and with g i = ω. Set X = {(g i ) g i = ω}. Then X is a finite set. We write X as {(g it t ) t = 1 k} such that g i1 < g i2 < < g ik. Since M is canceative g i1 = g i2 and g i1 1 = g i2 2 = ω impy 1 = 2. Since g i1 < g i2 and g i1 1 = g i2 2 = w we have 2 < 1. Thus k <... < 2 < 1. Now (g i ) X r = k t=1 t r ni(r). For any t 2 g i1 t < g it t = w and thus by induction hypothesis we have 1 r ni(r). For t = 2 et (1 r) q = 0. Then (r1 ) q+1 = 0. Thus ((2 r)(1 r) p+1 2 )(r1 ) q+1 (r(1 r) p+1 )=0. Since ((2 r)(1 r) p+1 2 )r1 ) I (r1 ) q (r(1 r) p+1 ) I r(1 r) p 2 I and I is semicommutative it foows that ((2 r)(1 r) p+1 2 )(r1 )(r(1 r) p )2 )(r1 ) q (r(1 r) p+1 )=0 that procedure yieds that [(2 r)(1 r) p+1 ] q+3 = 0. Thus (2 r)(1 r) p+1 ni(i). Since [(1 r) p+1 (2 r)] q+4 = 0 and (2 r)(1 r) p+1 ni(i) we have (1 r) p+1 (2 r) ni(i). Simiary one can show that (t r)(1 r) p+1 ni(i) and (1 r) p+1 (t r) ni(i) fort = k. Since 0 = [(2 r)(1 r) p+1 ] q+3 = [(2 r)(1 r) p+1 ]...[(2 r)(1 r) p+1 ] and (1 r) p+1 (3 r) I and I is semicommutative we have 0 = [(2 r)(1 r) p+1 ][(1 r) p+1 (3 r)][(2 r)(1 r) p+1... [(2 r)(1 r) p+1 ][(1 r) p+1 ][(1 r) p+1 (3 r)][(2 r)(1 r) p+1 ]. (1) Mutipying (1) on the right side by (1 r) p+1 and on the eft side by 3 r then [(3 r)(2 r)(1 r) 2s+2 ] q+3 = 0. Hence (3 r)(2 r)(1 r) 2s+2 ni(i). Simiary one can show that (rs r)(rs 1 r)...(r2 r)(r1 r) sp+s ni(i) for each s 2 and {r 1... r s } 1. Since (g i ) X r = k t=1 t r ni(r) there exists s N such that ( t r) s = 0. Ceary ( k t=1 t r) s equa to sums of such eements (1 r)(2 r)...(rs r) where r t {1... k} for each t. Then Page 9 of 12

10 (1 r) s = (1 r)(2 r)...(rs r)(2) where r t {1... k} and {r 1... r s } 1. Mutipying Equation (2) on the right side by (1 r) sp+s we have (1 r) sp+2s = (1 r)(2 r)...(rs r)(1 r) sp+s ni(i) where r t {1... k} and {r 1... r s } 1 since ni(i) is an idea of I by Lunqun and Jinwang (2013) Lemma 3.1). Hence (1 r) ni(i). Since (2 r) p I by anaogy with the above proof we have that (t r)(2 r) p+1 ni(i) for each 3 t k. Since (1 r) and ( k a t=1 i r) are nipotent there exists s N such that (a t i1 r) s = 0 =( k a t=1 i r) s. Then t (2 r) p+1 (1 r) s (2 r) p+1 = 0. Since (2 r) p+1 I and I is semicommutative we have ((1 r)(2 r) p+1 ) s+1 = 0. Thus (1 r)(1 r) ni(i). Hence (t r)(2 r) p+1 ni(i) for each t 2. By anaogy with the above proof we have (rs r)(rs 1 r)...(r2 r)(r1 r) sp+s ni(i) where s 2 and {r 1... r s } 2. Since (1 r) s = 0 =( k a t=1 i r) s t we have (2 r) s = (r1 r)(r2 r)...(rs r)(3) where r t {1... k} {r 1... r s } 1 and {r 1... r s } 2. Mutipying Equation (3) on the right side by (2 r) sp+s then (2 r) sp+s = (r1 r)(r2 r)...(rs r)(2 r) sp+s ni(i) where r t {1... k} {r 1... r s } 1 and {r 1... r s } 2. Therefore 2 r ni(i). Simiary one can show that 3 r... k r are nipotent. Then r ni(i) for some nonzero r R and for each i. This impies that R is a right ni-m-mccoy. Simiary we can show that R is eft ni-m-mccoy. Therefore R is a ni-m-mccoy. The foowing exampe shows that the condition R/I is a ni-m-mccoy ring in Theorem 2.6 is not superfuous. Exampe 2.5 Let M =(N {0} +) and et F be any fied and R = Z 2 M 2 (F). Then I = Z 2 0 is a semicommutative idea of R but R I M 2 (F) is not ni- M-McCoy ring. Let α =(1 E 11 )+(0 E 12 ) +(1 E 21 )g 2 +(1 E 22 )g 3 and β =(0 E 21 )+(0 E 11 ) +(0 E 22 )h 2 +(0 E 12 )h 3 R[M] {0}. Then αβ = 0 but if r R and (1 E ij )r ni(r) for each ij then ceary r = 0. Therefore R is not right ni-m-mccoy and hence it is not ni-m-mccoy. The foowing emma shows that the condition M is an "u.p."-monoid in Theorem 2.7 is not superfuous. Lemma 2.5 Let M be a cycic group of order n 2 and R any NI ring. Then R is not ni-m-mccoy. Proof Suppose that M ={e h h 2... h n 1 }. Let α = 1e + 1h + 1h h n 1 and β = 1e + ( 1)h. Then αβ = 0. Therefore R is not ni-m-mccoy whereas R is NI ring. A ring R is a subdirect sum of a famiy of rings {R i } i I if there is a surjective homomorphism where π k : i I R i R k is the k th projection. Now we consider the case of subdirect sum of ni-m-mccoy. Proposition 2.6 If R is a subdirect sum of ni-m-mccoy then R is ni-m-mccoy. Proof Let I k for each k {1... } be ideas of R and for them ring R I k is ni-m-mccoy and et I = 0. k=1 k Let α = m a g and β = n b h i=0 i i j=0 j j R[M] {0} such that αβ ni(r)[m]. Since R I k for each k is ni-m- McCoy then we have ( r) r i k j I k i j. Now et r ij ={r i k j k = 1... } therefore ( r) r ij I k=1 k = 0. Since r ni(r) for each i. Hence R is ni-m-mccoy. Page 10 of 12

11 The ring of Laurent poynomias in x with coefficients in a ring R consists of a forma sums n i=k r i xi with obvious addition and mutipication where r i R and k n are (possiby negative) integers denoted by R[x x 1 ]. We finish this paper with the foowing coroary. Coroary 2.4 Let R be a semicommutative ring. Then R is ni-z-mccoy for any α = a m x m + a (m 1) x (m 1) + + a p x p β = a n x n + a (n 1) x (n 1) + + a q x q R[x x 1 ] {0} if αβ ni(r)[x x 1 ] then there exist r s R {0} such that r ni(r) and sb j ni(r) for each m i p and n j q. Proof Note that R[Z] R[x x 1 ] (see Ahevaz et a. 2012). Funding The authors received no direct funding for this research. Author detais Vahid Aghapouramin 1 E-mai: vah-50@yahoo.com ORCID ID: Mohammad Javad Nikmehr 2 E-mai: nikmehr@kntu.ac.ir 1 Department of Mathematics Karaj Branch Isamic Azad University Karaj Iran. 2 Facuty of Mathematics K.N. Toosi University of Technoogy P.O. Box Tehran Iran. Citation information Cite this artice as: On ni-mccoy rings reative to a monoid Vahid Aghapouramin & Mohammad Javad Nikmehr Cogent Mathematics & Statistics (2018) 5: References Ahevaz A. & Moussavi A. (2010). Weak McCoy rings reative to a monoid. Internationa Mathematica Forum 47(5) Ahevaz A. Moussavi A. & Habibi M. (2012). On rings having McCoy-Like conditions. Communications in Agebra Birkenmeier G. F. & Park J. K. (2003). Trianguar matrix representations of ring extensions. Journa of Pure and Appied Agebra Hashemi E. (2010). McCoy ring reative to a monoid. Communications in Agebra Hashemi E. (2013). Ni-Armendariz rings reative to a monoid. Journa of Mathematics Liu Z. K. (2005). Armendariz rings reative to a monoid. Communications in Agebra 33(3) Lunqun O. & Jinwang L. (2013). Ni-Armendariz ring reative to a monoid. Arabian Journa of Mathematics Niesen P. P. (2006). Semicommutativy and the McCoy condition. Journa of Agebra 298(1) Nikmehr M. J. Fatahi F. & Amraei H. (2011). Ni-Armendariz rings with appications to a monoid. Word Appied Sciences Journa 13(12) Rege M. B. & Chhawchharia S. (1997). Armendariz rings. Proceedings of the Japan Academy Ser. A Mathematica Sciences 73(1) Ying Z. L. Chen J. L. & Lei Z. (2008). Extensions of McCoy rings. Northeastern Mathematica Journa 24(1) Zhao L. Zhu X. & Gu Q. (2012). Nipotent eements and McCoy rings. Mathematica Hungarica 49(3) Page 11 of 12

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