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Research Article

On nil-McCoy rings relative to a monoid

ORCID Icon & | (Reviewing Editor)
Article: 1426184 | Received 24 Oct 2017, Accepted 29 Dec 2017, Published online: 13 Feb 2018

Abstract

The concept of nil-M-McCoy (nil-McCoy ring relative to monoid M), which are generalizations of McCoy ring and nil-M-Armendariz rings have been introduced, and we investigate their properties. It is shown that every NI ring is nil-M-McCoy for any unique product monoid M, it has also been shown that every semicommutative rings is nil-M-McCoy for any unique product monoid and any strictly totally ordered monoid M. Moreover, it is proved that for an ideal I of R, if I is semicommutative and R / I is nil-M-McCoy then R is nil-M-McCoy for any strictly totally ordered monoid. We extend and unify many known results related to McCoy rings and nil-Armendariz ring.

AMS Subject Classifications:

Public Interest Statement

The current paper attempts to generalize Nil-macCoy rings with Monoid to present a novel work with the title of Nil-M-MacCoy. The survey of literature demonstrates that studies on Nill-Armendariz and Nil-MacCoy and Nil-M-Armendariz and have already been researched. The theorems, lemmas and corollaries which have been already studied concerning this topic to be compared with the new work to generalize or complete it. It paves the way for the willing researchers to continue the trend with the novel areas that have been investigated.

1. Introduction

All rings considered here are associative with identity. For a ring R, nil(R) denotes the set of nilpotent elements. Rege and Chhawchharia (Citation1997) introduced the notion of an Armendariz ring Recall that a ring R is called reduced if a2=0 implies that a=0, for all aR; R is symmetric if abc=0 implies acb=0, for a,b,cR; R is reversible if ab=0 implies ba=0, for all a,bR; R is semicommutative if ab=0 implies aRb=0, for all a,bR.

Recall that a monoid M is called an u.p.-monoid (unique product monoid) if for any two nonempty finite subsets A,BM there exists an element gM that can be written uniquely in the form ab where aA and bB. The class of u.p.monoids is quite large and important (see Birkenmeier & Park, Citation2003).

According to Nielsen (Citation2006), a ring R is called right McCoy whenever polynomials f(x),g(x)R[x]-{0} satisfy f(x)g(x)=0, there exists a nonzero rR such that f(x)r=0. We define left McCoy ring similarly. If a ring is both left and right McCoy, we say that the ring is a McCoy ring. Liu (Citation2005), studied a generalization of Armendariz ring, which is called M-Armendariz ring if whenever α=i=1naigi, β=j=1mbjhjR[M], with gi,hjM satisfy αβ=0, then ajbj=0 for each i,j.

Zhao, Zhu, and Gu (Citation2012) introduced the notion of a nil-McCoy ring. A ring R is said to be right nil-McCoy if f(x)g(x)nil(R)[x], where f(x)=i=0maixi, g(x)=j=0nbjxjR[x]-{0}, implies that there exists rR-{0} such that airnil(R) for 0im. Left nil-McCoy ring are defined similarly. A ring is nil-McCoy if it is both left and right nil-McCoy. According to Hashemi (Citation2010), a ring R is called right M-McCoy if whenever α=i=1naigi, β=j=1mbjhjR[M]-{0}, with gi,hjM, ai,bjR satisfy αβ=0, then αr=0 for some nonzero rR. Left M-McCoy ring are defined similarly. Also, if R is both left and right M-McCoy, then we say R is M-McCoy.

Our results also generalized and unify the above-mentioned concepts by introducing the notion of nil-M-McCoy ring.

2. Nil-McCoy rings relative to a monoid

We begin this section by the following definition and also we study properties of nil-M-McCoy rings.

Definition 2.1

Let M be a monoid and R a ring. We say that R is right nil-M-McCoy (right nil-McCoy ring relative to M), if whenever α=a1g1++angn, β=b1h1++bmhmR[M]-{0}, with gi,hjM, ai,bjR satisfy αβnil(R)[M], then airnil(R) for some nonzero rR and for each 1in. Left nil-M-McCoy rings are defined similarly. If R is both left and right nil-M-McCoy, then we say R is nil-M-McCoy.

Let M=(N{0},+). Then a ring R is nil-M-McCoy (resp., nil-M-Armendariz) if and only if R is nil-McCoy (resp., nil-Armendariz). If M={e}, then for any ring R, R is nil-M-McCoy and nil-M-Armendariz.

Example 2.1

Let Z2 be the ring of integers modulo 2 and R be a nil-Armendariz ring andR=Z2a,b,c,d,e/I,

where I is the ideal generated by the relations: ac=0,ad+bc=0,bd=0,ea=eb=ec=ed=ee=de=ce=be=0. Get S be subring generated by elements not involving e. Let α=a+bg and β=c+dh such that g,hR[M]-{0} and let g=h then αβ=0. It is straightforward to see that the element as is not nilpotent for any nonzero sS. However, R is nil-M-McCoy as αe,eβnil(R)[M] for any elements α,βR[M].

For any αR[M], we denote by Cα the set of all coefficients of α.

Proposition 2.1

For any "u.p."-monoid M, every NI ring is nil-M-McCoy.

Proof

Let M be an "u.p."-monoid and R be a NI ring. Suppose that α=a1g1++amgm,β=b1h1++bnhnR[M]-{0} satisfy αβnil(R)[M]. Since nil(R) is an ideal of R, R / nil(R) is M-McCoy by Liu (Citation2005, Proposition 2.6). For any α=i=1maigiR[M], denote α¯=i=1m(ai+nil(R))gi(R/nil(R))[M]. It is easy to see that the mapping φ:R[M](R/nil(R))[M]. Defined by φ(αi)=α¯i is a ring homomorphism. Since αβnil(R)[M], Cαβnil(R). Hence, we have α¯β¯=αβ¯nil(R¯)[M]. So a¯ib¯jnil(R¯)=0 for ij, since R / nil(R) is M-McCoy. Thus, aibjnil(R) for ij. Choosing r=bn0 and s=am0, we have airnil(R) and sbjnil(R), for ij. Therefore, R is nil-M-McCoy.

Example 2.2

Let M be "u.p."-monoid and K be field. Suppose S=K<a|a2=0>. Then the ring R=sasass is nil-M-McCoy, since R is a NI ring.

To prove Theorem 2.1, we state the following lemmas.

Lemma 2.1

Let M be cyclic group of order n2 and R a ring with 10. Then R is not nil-M-McCoy.

Proof

Suppose that M={e,h,,hn-1}. Let α=1e+1h++1hn-1 and β=1e+(-1)h then αβnil(R)[M], but aic=1cnil(R) for each nonzero cR. Therefore, R is not right nil-M-McCoy. Similarly R is not left nil-M-McCoy. Therefore, R is not nil-M-McCoy.

Lemma 2.2

Let M be a monoid and N a submonoid of M. If R is nil-M-McCoy, then R is nil-N-McCoy.

Proof

It is clear by Liu (Citation2005, lemma 1.12) and Hashemi (Citation2010, lemma 1.12).

Lemma 2.3

Liu (Citation2005, Lemma 1.13). Let M and N be "u.p."-monoid. Then so is the monoid M×N.

Example 2.3

Let M,NM2(R) and let MN be “u.p.”-monoids. LetA=aibicidi,eifigihiM×Ni=1,2,...,p

andB=ajbjcjdj,ejfjgjhjM×Nj=1,2,,q. Since M is a "u.p."-monoid, there exist ij with 1ip and 1jq such that aibicidiajbjcjdj is uniquely presented by considering two subsetsa1b1c1d1,...,apbpcpdpanda1b1c1d1,...,aqbqcqdqofM.

Consider two subsetsekfkgkhkNaibicidi,ekfkgkhkAandelflglhlNajbjcjdj,elflglhlBofN.

Since N is a “u.p.”-monoid, there exist k,l such that ekfkgkhkelflglhl is uniquely presented.

Now it is easy to see that aibicidi,ekfkgkhkajbjcjdj,elflglhl is uniquely presented. Thus M×N is a “u.p.”-monoid.

Recall that a monoid M is called torsion-free if the following property holds: if g,hM and k1 are such that gk=hk, then g=h.

Let T(H) be the set of elements of finite order in an abelian group H. H is said to be torsien-free If T(H)={e}.

Theorem 2.1

Let H be finitely generated abelian group. Then the following conditions on H are equivalent:

(1)

H is torsion-free.

(2)

There exists a ring R with |R|2 such that nil-H-McCoy.

Proof

(1)(2) If H is finitely generated abelian group with T(H)={e}, then HZ×Z××Z, a finite direct product of the group Z, by Lemma 2.3, H is “u.p.”-monoid. Let R be a NI ring. Then, by Proposition 2.1, R is nil-H-McCoy.

(2)(1) If hT(H) and he, then N=h is a cyclic group of finite order. If a ring R{0} is nil-H-McCoy, then by Lemma 2.2, R is nil-N-McCoy, a contradiction with Lemma 2.1. Therefore, every ring R{0} is not nil-H-McCoy.

A ring R is called right Ore if given a,bR with b regular, there exist a1,b1R with b1 regular such that ab1=ba1. left Ore rings can be defined similarly.

Proposition 2.2

Let M be a monoid and R a right Ore ring with its classical right quotient ring Q. If R is right nil-M-McCoy, then Q is right nil-M-McCoy.

Proof

Let α=i=0mαigi, β=j=0nβjhj be nonzero polynomials of Q[M] such that αβnil(Q)[M]. Since Q is a classical right quotient ring, we assume that αi=aiu-1 and βj=bjv-1 for ai,bjR for all ij and regular elements u,vR. For each j, there exists cjR and a regular element wR such that u-1bj=cjw-1. Denote α=i=0maigi and β=j=0nbjhjR[M]-{0}. We have αβ=i=0mj=0nαiβjgihj=i=0mj=0naiu-1bjv-1gihj=i=0mj=0naicjw-1v-1gihj=i=0mj=0naicj(vw)-1gihj=αβ(vw)-1. So αβnil(R)[M]. Since R is right nil-M-McCoy, there exists a nonzero element cR such that aicnil(R). So aic=αi(uc)nil(R), for each i, and uc is a nonzero element in Q. Hence, Q is a right nil-M-McCoy ring.

Theorem 2.2

Let M be a monoid with |M|2. Then the finite direct sum of nil-M-McCoy rings is nil-M-McCoy.

Proof

It suffices to show that if R1,R2 are nil-M-McCoy rings, then so is R1R2. Let α=(a11,b11)g1++(am1,bm1)gm and β=(a12,b12)h1++(an2,bn2)hn(R1R2)[M]-{0} be such that αβnil(R1R2)[M]. Therefore we show that there exists r,sR1R2 such that airnil(R1R2) and sbjnil(R1R2) for each ai=(ai1,bi1), bj=(aj2,bj2) and for all r=(r1,r2),s=(s1,s2). Now let f1=a11g1++am1gm, g1=b11g1++bm1gm, f2=a12h1++an2hn, g2=b12h1++bn2hn. Then f1f2nil(R1)[M] and g1g2nil(R2)[M]. Since by hyopothesis R1 and R2 are nil-M-McCoy ai1r1nil(R1), bj1r2nil(R2) for each 1im. Thus for each 1im(ai1,bi1)(r1,r2)nil(R1R2), and also s1aj2nil(R1), s2bj2nil(R2) for each 1jn. Thus for each 1jn(s1,s2)(aj2,bj2)nil(R1R2). Therefore, R1R2 is nil-M-McCoy.

Example 2.4

Let R be an M-McCoy reduced ring, and M a monoid with |M|2 and letTn(R)={a11a12a1n0a22a2n00ann|aijR}.

Then Tn(R) is not M-McCoy for n4 by Ying et al. (Citation2008, Theorem 2.1). But Tn(R) is nil-M-McCoy by Theorem 2.3 below. Hence, a nil-M-McCoy ring is not a trivial extension of an M-McCoy.

Theorem 2.3

Let M be a monoid with |M|2. Then the following conditions are equivalent:

(1)

R is nil-M-McCoy.

(2)

Tn(R) is nil-M-McCoy.

Proof

(1)(2) In a similar way, proposition 2.12 (Nikmehr, Fatahi, & Amraei, Citation2011). It is easy to see that there exists an isomorphism of rings Tn(R)[M]Tn(R[M]) definedi=1na11ia12ia1ni0a22ia2ni00annigii=1na11igii=1na12igii=1na1nigi0i=1na22igii=1na2nigi00i=1nannigi.

Suppose that α=A1g1++Angn and β=B1h1++BmhmTn(R)[M]-{0} are such that αβnil(Tn(R))[M], where Ai,BjTn(R). We claim there exists BTn(R) such that AiBnil(Tn(R)) for each i. Assume thatAi=a11ia12ia1ni0a22ia2ni00anni,B=c11jc12jc1nj0c22jc2nj00cnnj,

and let αs=i=1nassigiR[M] and βs=j=1mcssjhjR. Also by observing thatnil(Tn(R))=nil(R)RR0nil(R)R00nil(R),

thus, we have αsβsnil(R)[M] for each 1sn. Since R is a nil-M-McCoy ring, there exists some positive integer mis such that (assibssj)mis=0 for any s and any i. Let mi=max{mis|1sn}. Then ((AiB)mi)n=0 and so AiBnil(Tn(R)). Therefore, Tn(R) is nil-M-McCoy.

(2)(1) Suppose that Tn(R) is nil-M-McCoy. Note that R is isomorphic to the subringa000a000a|aR

of Tn(R). Thus, R is nil-M-McCoy since each sub ring of a nil-M-McCoy ring is also nil-M-McCoy.

Let R be a ring and letSn(R)=aa12a1n0aa2n00a|a,aijR,

andT(R,n)={a1a2an0a1an-100a1|aijR},

and T(RR) be the trivial extension of R by R. Using the same method in the proof of Theorem 2.3, we have the following results:

Corollary 2.1

Let M be a monoid with |M|2. Then the following conditions are equivalent:

(1)

R is nil-M-McCoy;

(2)

Sn(R) is nil-M-McCoy;

(3)

T(Rn) is nil-M-McCoy;

(4)

T(RR) is nil-M-McCoy;

(5)

R[x]/xn is nil-M-McCoy for each n2.

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(Rn) is a sub ring of the triangular matrix rings, with matrix addition and multiplication. We can denote elements of T(Rn) by (a0,a1,,an-1), then T(Rn) is a ring with addition point-wise and multiplication given by(a0,a1,,an-1)(b0,b1,,bn-1)=(a0b0,a0b1+a1b0,,a0bn-1++an-1b0),

for each ai,bjR. On the other hand, there is a ring isomorphism φ:R[x]<xn>T(R,n), since by, φ(a0+a1x++an-1xn-1)=(a0,a1,,an-1), with aiR, 0in-1. So T(R,n)R[x]<xn>, where R[x] is the ring of polynomial in an indeterminate x and <xn> is the ideal generated by xn. Therefore (3)(5).Thus by similarly method we have (4)(5). Therefore all conditions are equivalent.

Let R be a ring and M a monoid. Let G3(R)={a1100a21a22a2300a33|aijϵR}. The G3(R) is a subring of 3×3 full matrix ring M3(R) under usual addition and multiplication. In fact, G3(R) possesses the similar form of both the ring of all lower triangular matrices and the ring of all upper triangular matrices. A natural problem asks if the nil-M-McCoy property of such subring of Mn(R) coincides with that of R. This inspires us to consider the nil-M-McCoy property of G3(R).

Theorem 2.4

Let M be a monoid with |M|2. Then the following conditions are equivalent:

(1)

R is nil-M-McCoy;

(2)

G3(R) is nil-M-McCoy.

Proof

(1)(2) We first show that nil(G3(R))=nil(R)00Rnil(R)R00nil(R). Suppose that a1100a21a22a2300a33nil(R)00Rnil(R)R00nil(R), and m is a positive integer such that a11m=a22m=a33m=0. Then a1100a21a22a2300a332m=0. Hence,nil(R)00Rnil(R)R00nil(R)nil(G3(R)).

Now assume that a1100a21a22a2300a33nil(G3(R)). Then there exists some positive integer m such that a1100a21a22a2300a33m=0. Hence a11m=a22m=a33m=0, and soa1100a21a22a2300a33nil(R)00Rnil(R)R00nil(R).

Therefore, nil(G3(R))=nil(R)00Rnil(R)R00nil(R). Then by analogy with the proof of Theorem 2.4, we can show that G3(R) is nil-M-McCoy.

(2)(1) it is trivial (see Hashemi, Citation2013).

In the proof of the next proposition we will need the following lemma.

Lemma 2.4

Zhao et al. (Citation2012, Lemma 2.7) Let R be a semicommutative ring, and f1(x),f2(x),...,fn(x) be in R[x]. If Cf1...fnnil(R), then Cf1Cf2Cfnnil(R).

Proposition 2.3

Semicommutative rings are nil-M-McCoy rings.

Proof

Let α=i=0maigi, β=j=0nbjhjR[M]-{0} with αβnil(R)[M], then we have Cαβnil(R). It follows that CαCβnil(R) by Lemma 2.4. Hence, there exists r=bj for some 0jn such that airnil(R) with 0im. This implies that R is a right nil-M-McCoy. Similarly, we can show that R is left nil-M-McCoy. Therefore, R is a nil-M-McCoy.

We now have the following description of the rings which shows one way to give more nil-M-McCoy rings from old ones.

Proposition 2.4

Let Ω be an index set and {RI|IΩ} a family of rings. If R=IΩRI, then R is right nil-M-McCoy if and only if every RI is right nil-M-McCoy for each IΩ.

Proof

It is straightforward to verify that if R is right nil-M-McCoy, then every RI is right nil-M-McCoy for each I. Conversely, if αβnil(R)[M] for α=i=0maigi, β=j=0nbjhjR[M]-{0}, where ai=(aiI)IΩ, bj=(bjI)IΩR. For each IΩ, let αI=i=0maiIgi, βI=j=0nbjIhjRI[M], then αIβInil(RI)[M]. Since β0 there exists some index J with βJ0. In particular, there exists some nonzero rJRJ with aiJrJnil(RJ) for all 0im by the nil-M-McCoy property of RJ. Fix rJRJ-{0}, and take r to be the sequence with rJ in the Jth coordinate and zero elsewhere. Clearly airnil(R) and r0.

Corollary 2.2

A finite direct product of right nil-M-McCoy rings is right nil-M-McCoy.

Theorem 2.5

For any monoid M. Then we have the following statements:

(1)

nil-M-Armendariz rings are nil-M-McCoy;

(2)

M-McCoy rings are nil-M-McCoy;

(3)

M-Armendariz rings are nil-M-McCoy.

Proof

(1) Let R be a nil-M-Armendariz rings and α=a1g1++amgm, β=b1h1++bnhnR[M]-{0} satisfy αβnil(R)[M]. Then aibjnil(R) for each ij. Since α0 and β0, for 1ln, 1km there exists r=bl, s=akR-{0}. Hence, airnil(R) and sbjnil(R). Therefore, R is nil-M-McCoy.

(2) It follows easily form the definition.

(3) It is easy to see that each M-Armendariz rings is M-McCoy and therefore nil-M-McCoy by (2).

Let Ri be a ring, for each iZ, and let R=Ri and iZRi,1R be the subring generated by iZRi and {1R}. Then we have the following result (see Alhevaz & Moussavi, Citation2010).

Proposition 2.5

Let Ri be a ring, R=iZRi and S=iZRi,1R. Then R is right nil-M-McCoy if and only if each Ri is right nil-M-McCoy if and only if S is right nil-M-McCoy.

Proof

Let each Ri be a right nil-M-McCoy ring and α=i=0maigi and β=j=0nbjhjR[M]-{0} such that αβnil(R)[M], where ai=(aik) and bj=(bjk). If there exists tZ such that ait=0, for each 0im, then we have aicnil(R), where c=(0,,0,1Rt,0,). Now, suppose for each kZ, there exists 0ikm such that aikk0. Since β0, there exists tZ and 0jtn such that bjtt0. Consider αt=i=0taitgi and βt=j=0tbjthjRt[M]-{0}. We have αβnil(R)[M]. Thus, there exists some nonzero ctRt such that aitctnil(R), for each 0im. Therefore, aicnil(R), for each 0im, where c=(0,,0,ct,0,), and so R is right nil-M-McCoy.

Conversely, suppose R is right nil-M-McCoy, tZ, α=i=0maigi and β=j=0nbjhjR[M]-{0} such that αβnil(R)[M]. Let α=i=0m(1,...,1,ai,1,...)gi and β=j=0n(0,...,0,bj,0,...)hjR[M]-{0}. So there exists 0c=(ci) and (1,,1,ai,1,)cnil(R) clearly, c=(0,,0,ct,0,). Thus, we have aictnil(Rt) and so Rt is right nil-M-McCoy. It is easy to see that S is right nil-M-McCoy if and only if each Ri is right nil-M-McCoy.

Let (M,) be an ordered monoid. If for any g1,g2,hM, g1<g2 implies that g1h<g2h and g2h<g1h, then (M,) is called a strictly totally ordered monoid.

Corollary 2.3

Let M be a strictly totally ordered monoid and R a reversible ring. Then R is nil-M-McCoy.

It was shown in Liu (Citation2005, Proposition 1.4), that if M is strictly totally ordered monoid and I is a reduced ideal of R such that R / I is an M-Armendariz ring, then R is M-Armendariz. The following result generalizes this.

Theorem 2.6

Let M be a strictly totally ordered monoid and I an ideal of R. If I is semicommutative (as a ring without identity) and R / I is nil-M-McCoy, then R is nil-M-McCoy.

Proof

Let α=a1g1++angn, β=b1h1++bmhmR[M]-{0} be such that αβnil(R)[M] and g1<g2<...<gn, h1<h2<...<hm. We will use transfinite induction on the strictly totally ordered set (M,) to show that airnil(R) for some nonzero rR and for each i. Note that in (R / I)[M], (a¯1g1+a¯2g2++a¯ngn)(b¯1h1+b¯2h2++b¯mhm)nil(R/I)[M]. The fact that R / I is nil-M-McCoy means that there exists r in R such that ai¯rnil(R/I) thus, there exists pN such that (air)pI for each i. If there exists 1in and 1jm such that gihj=g1h1, then g1gi and h1hj. If g1<gi, then g1h1<gih1gihj=g1h1, a contradiction. Thus, gi=g1. Similarly, hj=h1. Therefore airnil(R) for some nonzero rR and for each i.

Now suppose that ωM is such that airnil(R) for each gi and hj with gihj=ω. Set X={(gi,hj)|gihj=ω}. Then X is a finite set. We write X as {(git,hjt)|t=1,,k} such that gi1<gi2<<gik. Since M is cancellative, gi1=gi2 and gi1hj1=gi2hj2=ω imply hj1=hj2. Since gi1<gi2 and gi1hj1=gi2hj2=w, we have hj2<hj1. Thus hjk<...<hj2<hj1.

Now (gi,hj)Xair=t=1kaitrnil(R). For any t2, gi1hjt<githjt=w, and thus by induction hypothesis, we have ai1rnil(R). For t=2, let (ai1r)q=0. Then (rai1)q+1=0. Thus((ai2r)(ai1r)p+1ai2)(rai1)q+1(r(ai1r)p+1)=0.

Since

((ai2r)(ai1r)p+1ai2)rai1)I,

(rai1)q(r(ai1r)p+1)I,

r(ai1r)pai2I,

and I is semicommutative, it follows that((ai2r)(ai1r)p+1ai2)(rai1)(r(ai1r)p)ai2)(rai1)q(r(ai1r)p+1)=0,

that procedure yields that [(ai2r)(ai1r)p+1]q+3=0. Thus (ai2r)(ai1r)p+1nil(I).

Since [(ai1r)p+1(ai2r)]q+4=0 and (ai2r)(ai1r)p+1nil(I), we have (ai1r)p+1(ai2r)nil(I). Similarly one can show that (aitr)(ai1r)p+1nil(I) and (ai1r)p+1(aitr)nil(I), fort=2,3,...,k.

Since 0=[(ai2r)(ai1r)p+1]q+3=[(ai2r)(ai1r)p+1]...[(ai2r)(ai1r)p+1] and (ai1r)p+1(ai3r)I and I is semicommutative, we have0=[(ai2r)(ai1r)p+1][(ai1r)p+1(ai3r)][(ai2r)(ai1r)p+1×...×[(ai2r)(ai1r)p+1][(ai1r)p+1][(ai1r)p+1(ai3r)][(ai2r)(ai1r)p+1].(1)

Multiplying (1) on the right side by (ai1r)p+1 and on the left side by ai3r, then[(ai3r)(ai2r)(ai1r)2s+2]q+3=0.Hence(ai3r)(ai2r)(ai1r)2s+2nil(I).

Similarly one can show that(airsr)(airs-1r)...(air2r)(air1r)sp+snil(I),

for each s2 and {r1,...,rs}1. Since (gi,hj)Xair=t=1kaitrnil(R), there exists sN such that (aitr)s=0. Clearly, (t=1kaitr)s equal to sums of such elements (ai1r)(ai2r)...(airsr), where rt{1,...,k} for each t. Then(ai1r)s=-(ai1r)(ai2r)...(airsr)(2)

where rt{1,...,k} and {r1,...,rs}1. Multiplying Equation (2) on the right side by (ai1r)sp+s, we have(ai1r)sp+2s=-(ai1r)(ai2r)...(airsr)(ai1r)sp+snil(I)

where rt{1,...,k} and {r1,...,rs}1, since nil(I) is an ideal of I by Lunqun and Jinwang (Citation2013, Lemma 3.1). Hence (ai1r)nil(I).

Since (ai2r)pI, by analogy with the above proof we have that (aitr)(ai2r)p+1nil(I) for each 3tk. Since (ai1r) and (t=1kaitr) are nilpotent, there exists sN such that (ai1r)s=0=(t=1kaitr)s. Then (ai2r)p+1(ai1r)s(ai2r)p+1=0. Since (ai2r)p+1I and I is semicommutative, we have ((ai1r)(ai2r)p+1)s+1=0. Thus (ai1r)(ai1r)nil(I). Hence (aitr)(ai2r)p+1nil(I) for each t2. By analogy with the above proof we have (airsr)(airs-1r)...(air2r)(air1r)sp+snil(I), where s2 and {r1,...,rs}2. Since (ai1r)s=0=(t=1kaitr)s, we have(ai2r)s=-(air1r)(air2r)...(airsr)(3)

where rt{1,...,k}, {r1,...,rs}1 and {r1,...,rs}2. Multiplying Equation (3) on the right side by (ai2r)sp+s, then (ai2r)sp+s=-(air1r)(air2r)...(airsr)(ai2r)sp+snil(I) where rt{1,...,k}, {r1,...,rs}1 and {r1,...,rs}2. Therefore ai2rnil(I). Similarly one can show that ai3r,...,aikr are nilpotent. Then airnil(I) for some nonzero rR and for each i. This implies that R is a right nil-M-McCoy. Similarly, we can show that R is left nil-M-McCoy. Therefore, R is a nil-M-McCoy.

The following example shows that the condition R / I is a nil-M-McCoy ring in Theorem 2.6 is not superfluous.

Example 2.5

Let M=(N{0},+), and let F be any field and R=Z2M2(F). Then I=Z20 is a semicommutative ideal of R, but R/IM2(F) is not nil-M-McCoy ring. Let α=(1,E11)+(0,E12)g1+(1,E21)g2+(1,E22)g3 and β=(0,-E21)+(0,E11)h1+(0,E22)h2+(0,-E12)h3R[M]-{0}. Then αβ=0, but if rR and (1,Eij)rnil(R), for each i,j then clearly r=0. Therefore, R is not right nil-M-McCoy, and hence it is not nil-M-McCoy.

The following lemma shows that the condition M is an "u.p."-monoid in Theorem 2.7 is not superfluous.

Lemma 2.5

Let M be a cyclic group of order n2 and R any NI ring. Then R is not nil-M-McCoy.

Proof

Suppose that M={e,h,h2,...,hn-1}. Let α=1e+1h+1h2++1hn-1 and β=1e+(-1)h. Then αβ=0. Therefore, R is not nil-M-McCoy, whereas R is NI ring.

A ring R is a subdirect sum of a family of rings {Ri}iI if there is a surjective homomorphism, where πk:iIRiRk is the kth projection.

Now we consider the case of subdirect sum of nil-M-McCoy.

Proposition 2.6

If R is a subdirect sum of nil-M-McCoy then R is nil-M-McCoy.

Proof

Let Ik for each k{1,...,l} be ideals of R and for them ring R/Ik, is nil-M-McCoy and let k=1lIk=0. Let α=i=0maigi and β=j=0nbjhjR[M]-{0} such that αβnil(R)[M]. Since R/Ik, for each k, is nil-M-McCoy then we have (air)rikjIki,j. Now let rij={rikj|k=1,...,l}, therefore (air)rijk=1lIk=0. Since airnil(R), for each i. Hence, R is nil-M-McCoy.

The ring of Laurent polynomials in x with coefficients in a ring R, consists of all formal sums i=knrixi with obvious addition and multiplication, where riR and kn are (possibly negative) integers, denoted by R[x,x-1]. We finish this paper with the following corollary.

Corollary 2.4

Let R be a semicommutative ring. Then R is nil-Z-McCoy, for any α=a-mx-m+a-(m-1)x-(m-1)++apxp, β=a-nx-n+a-(n-1)x-(n-1)++aqxqR[x,x-1]-{0}, if αβnil(R)[x,x-1], then there exist r,sR-{0} such that airnil(R) and sbjnil(R) for each -mip and -njq.

Proof

Note that R[Z]R[x,x-1] (see Alhevaz, Moussavi, & Habibi, Citation2012).

Additional information

Funding

The authors received no direct funding for this research.

Notes on contributors

Vahid Aghapouramin

Vahid Aghapouramin is currently a PhD student at the department of mathematics, Karaj Branch, Islamic Azad university, Karaj, Iran. His research interests are pure mathematics, Commutative Algebra, Noncommutative Algebra.

Mohammad Javad Nikmehr

Mohammad Javad Nikmehr is a associate professor, Faculty of Mathematics, K.N. Toosi, University of Technology, P.O. Box 16315-1618, Tehran, Iran. His research interest are Commutative Algebra, Graph Theory, Non-commutative Algebra, Algebraic Combinatorics.

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