# Power-Serieswise McCoy Modules

Authors: Xinquan Zhang
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

Extending the notion of McCoy modules and power-serieswise Armendariz modules, we introduce the concept of power-serieswise McCoy modules. Various basic properties of these modules are discussed and many illustrative examples are provided. Moreover, power-serieswise McCoy modules are used to obtain results on some polynomial extensions and matrix extensions of a module.


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## Body

## 1. Introduction

Throughout the paper, all rings considered are associative with identity and modules are unitary right modules. LetN be the set of natural integers. The symbol Eij stands for the matrix unit. For a set X of a module MR, rR(X) stands for the right annihilator of X in the ring R. We write Rx,Rx,Rx-1;x, and Rx-1;x for the polynomial ring, the power-serieswise ring, the Laurent polynomial ring, and Laurent power-serieswise polynomial ring over R, respectively. For a module MR, we consider (1)Mx=∑i=0smixi∣s≥0,mi∈M,Mx=∑i=0∞mixi∣mi∈M,Mx-1;x=∑i=-stmixi∣s≥0,t≥0,mi∈M,Mx-1;x=∑i=-s∞mixi∣s≥0,mi∈M.Each of these is an abelian group under an obvious addition operation. Moreover, Mx becomes an Rx module under the following product operation: for m(x)=∑i=0∞mixi∈Mx,f(x)=∑j=0∞ajxj∈Rx, (2)mxfx=∑k=0∞∑i+j=kmiajxk. Similarly, M[x],Mx,x-1 and Mx,x-1 are R[x]-module, R[x,x-1]-module, and Rx,x-1-module, respectively.Chhawchharia, Rege [1], and Nielsen [2] independently called a ring Rright McCoy (respectively,left McCoy) if whenever f(x)g(x)=0 for f(x),g(x)∈R[x]∖{0}, there exists r∈R∖{0} (respectively, s∈R∖{0}) such that f(x)r=0 (respectively, sg(x)=0). A ring isMcCoy if it is both left and right McCoy. The term “McCoy ring" was coined because McCoy [3] had shown that every commutative ring satisfies the above-mentioned condition. Armendariz rings (i.e., if f(x)=∑i=0saixi and g(x)=∑j=0tbjxj in R[x] satisfy f(x)g(x)=0, then aibj=0 for each i and j) are well known example of McCoy ring. For more details of rings with McCoy conditions, one may refer to [4–6], etc. The McCoy properties of rings were extended to power-serieswise rings in [7]. A ring R is called aright power-serieswise McCoy ring if whenever f(x),g(x)∈R[[x]]∖{0} satisfy f(x)g(x)=0, then there exists a nonzero element r∈R such that f(x)r=0. A left power-serieswise McCoy can be defined similarly. A left and right power-serieswise McCoy is calledpower-serieswise McCoy. Recall that a ring R isreduced [8] if it has no nonzero nilpotent elements. In [7], it was claimed that all reduced rings were power-serieswise McCoy and so are McCoy.Due to [9], a module MR isArmendariz if nibj=0 for all i,j whenever n(x)=∑i=0snixi∈M[x] and g(x)=∑j=0tbjxj∈R[x] satisfy n(x)g(x)=0. As both generalization of McCoy rings and Armendariz modules, McCoy modules were introduced in [10]. A module MR is calledMcCoy if m(x)g(x)=0 where m(x)=∑i=0smixi∈M[x] and g(x)=∑j=0tbjxj∈R[x]∖{0}, then there exists r∈R∖{0} such that m(x)r=0. In [11], Lee and Zhou extended the concept of Armendariz modules to power-serieswise modules and introduced the concept of power-serieswise Armendariz modules. A module MR is said to bepower-serieswise Armendariz if m(x)=∑i=0∞mixi∈Mx,g(x)=∑j=0∞bjxj∈Rx with m(x)g(x)=0, then mibj=0 for all i and j.Motivated by the above, we introduce the notion of power-serieswise McCoy modules as a straightforward extensions of McCoy modules and power-serieswise McCoy rings. We show that the class of power-serieswise Armendariz modules are properly contained in the class of power-serieswise McCoy modules. Basic properties of this class of modules are investigated, and examples are provided. In addition, we study the relationship between a module and its polynomial module (respectively, matrix module) by means of power-serieswise McCoy properties.


## 2. Power-Serieswise McCoy Modules

Definition 1.
LetM be a module over a ring R and Mx the corresponding polynomial module over Rx. MR is called power-serieswise McCoy if whenever m(x)g(x)=0 where m(x)=∑i=0∞mixi∈M[[x]] and g(x)=∑j=0∞bjxj∈R[[x]]∖{0}, there exists a nonzero element r∈R such that m(x)r=0.It is clear that power-serieswise McCoy modules are McCoy andRR is power-serieswise McCoy iff R is a right power-serieswise McCoy ring. Every submodule of a power-serieswise McCoy module is power-serieswise McCoy. In particular, if I is a right ideal of a power-serieswise McCoy ring R, then IR is a power-serieswise McCoy module.The following result is motivated by [10, Proposition 2.3].Proposition 2.
LetI be any nonzero ideal of a ring R, then (R/I)R is a power-serieswise McCoy module.Proof.
Give anyf(x)¯∈(R/I)[[x]]. Take a nonzero element r∈I(⊆R). Since I is an ideal of R, f(x)r∈I[[x]]. That is, f(x)¯r=0.The following example reveals that McCoy module need not be power-serieswise McCoy.Example 3 (see [7, Example 2.6]).
LetF be a field, A=Fa,b,c the free algebra of polynomials in noncommuting indeterminates a,b,c over F, and I the ideal of F+A generated by cc,ac,crc with r∈A. Let R=(F+A)/I. Identify f with f+I for simplicity. Consider the equality (3)a-abxc+bcx+b2cx2+⋯+bncxn+⋯=ac=0with a-abx,c+bcx+b2cx2+⋯+bncxn+⋯∈R[[x]]∖{0}. For any nonzero r∈R, if (a-abx)r=0, then ar=0 and abr=0. The first equality implies r∈Fc, but abc≠0, a contradiction. Thus, RR is not power-serieswise McCoy. However, the ring R is McCoy since it is Armendariz by [12, Example 2.1].Theorem 4.
Power-serieswise Armendariz modules are power-serieswise McCoy.Proof.
LetMR be a power-serieswise Armendariz module and m(x)g(x)=0, where m(x)=∑i=0∞mixi∈M[[x]]; there exists j0 such that bj0∈R∖{0}. Hence, mibj0=0 for each i which deduces that m(x)bj0=0. Therefore, MR is a power-serieswise McCoy module.Lete2=e∈R. A module MR isp.p.-module[13] if for any m∈M,rR(m)=eR. A module M is calledabelian[13] if for any m∈M and any a∈R, any idempotent e∈R, mae=mea. In view of [13, Corollary 2.9] and Theorem 4, we have the following result.Corollary 5.
g ( x ) = ∑ j = 0 ∞ b j x j ∈ R [ [ x ] ] ∖ { 0 }. Then mibj=0 for all i,j. Since g(x)≠0, if MR is an abelian and p.p.-module, then M is power-serieswise McCoy.The converse of Theorem4 does not hold generally.Example 6.
LetS be a reduced ring and let (4)R4=aa12a13a140aa23a2400aa34000a∣a,aij∈S. Then R4 is not an Armendariz ring by [8, Example 3], so is not power-serieswise Armendariz. Since S is reduced, R4 is a power-serieswise McCoy ring by [7, Corollary 2.9].A moduleMR is calledreduced [11] if for any m∈M and any a∈R, ma=0 implies that mR∩Ma=0. All reduced modules are Armendariz and hence McCoy.Remark 7.
In [13], a ring R is called strong regular if for any x∈R, there exists y∈R such that x=x2y. By [14, Theorem 2.16], all modules over strong regular rings are reduced, so is power-serieswise Armendariz by [11, Lemma 1.5], and hence power-serieswise McCoy.A moduleMR issemicommutative [9] if whenever m∈M and a∈R satisfy ma=0, then mra=0 for every r∈R. A module MR is said to befinitely cogenerated [15] if for every set Ai∣i∈I of submodules Ai of M with ⋂i∈IAi=0, there is a finite subset Ai∣i∈I0 (i.e., I0⊂I and I0 is finite) with ⋂i∈I0Ai=0.Cui and Chen [10] proved that a semicommutative module over a reduced ring is McCoy; we have the following result.Proposition 8.
LetR be a reduced ring and RR a finitely cogenerated module. Then a semicommutative module MR is power-serieswise McCoy.Proof.
Letm(x)=∑i=0∞mixi∈M[[x]],g(x)=∑j=0∞bjxj∈R[[x]]∖{0} with m(x)g(x)=0. Assume that b0≠0. (If not, set g(x)=g′(x)xk with a minimal k such that bk≠0, and we have m(x)g′(x)=0.) This implies the following system of equations: 0m0b0=0;1m0b1+m1b0=0;2m0b2+m1b1+m2b0=0;⋯⋯km0bk+m1bk-1+⋯+mkb0=0;⋯⋯ Since MR is semicommutative, we have m0b1b0=0 from Eq.0. Multiplying 1 on the right by b0 yields m1b02=0. Similarly, multiplying 2 on the right by b02 gives m2b03=0 because m0b2b02=m1b1b02=0. Continuing this procedure, one has mib0i+1=0 for all i. Since R is reduced, b0≠0 implies b0i≠0 for each i. Thus b0iR≠0 and b0R⊇b02R⊇⋯ imply ⋂i∈Nb0iR≠0. Since RR is a finitely cogenerated module, ⋂i∈I0b0iR≠0 for some finite I0⊆N. Take r∈⋂i∈I0b0iR. Then m(x)r=0. Hence MR is power-serieswise McCoy.A module isuniform[15] if any two nonzero submodules have a nonzero intersection. We have a similar result as in [10].Theorem 9.
Let{Mi}i∈Λ be a family of McCoy R-modules for an index set Λ. Then
(1) ifRR is uniform, then a direct sum M=∐i∈ΛMi is power-serieswise McCoy;
(2) ifΛ is an infinite set and RR is uniform and finitely cogenerated, then a direct product M=∏i∈ΛMi is power-serieswise McCoy.Proof.
Ifm(x)g(x)=0 with m(x)=∑k=0∞miki∈Λxk∈M[[x]] and g(x)∈R[[x]]∖{0}, let mi(x)=∑k=0∞mikxk∈Mi[[x]]. Since Mi is power-serieswise McCoy and mi(x)g(x)=0, there exists ri∈R∖{0} such that mi(x)ri=0 for each i∈Λ.
We now prove (1). Note that the setΛ1=i∈Λ∣mix≠0 is finite. Put U=⋂i∈Λ1riR. Since RR is uniform, we have U≠0. Take any r∈U∖{0}. Then mi(x)r=0 for each i, whence m(x)r=0. Thus, M=∐i∈ΛMi is power-serieswise McCoy.
To show (2), the assumptionRR is uniform implies that ⋂i∈Λ0riR≠0 for any finite subset Λ0⊆Λ. As RR is finite cogenerated, it follows that ⋂i∈ΛriR≠0. Similar to the proof of (1), we are done.LetMR be a module. Put (5)TM=x∈M∣rRx≠0. If R is a commutative domain, then T(M) is a submodule of M called thetorsion submodule of M (see [15]).LetMi be a module over a ring Ri where i∈Λ. Set m(x)=∑p=0∞(mip)i∈Λxp∈(∏i∈ΛMi)[[x]],g(x)=∑q=0∞(biqi∈Λxq∈(∏i∈ΛRi)[[x]]. It is clear that (∏i∈ΛMi)[[x]]→∏i∈Λ(Mi[[x]]) is an abelian group isomorphism via the usual map. So m(x)∈(∏i∈ΛMi)[[x]] can be viewed as (∑p=0∞mipxp)i∈Λ∈∏i∈Λ(Mi[[x]]). Similarly, (∏i∈ΛRi)[[x]]≅∏i∈Λ(Ri[[x]]). Then g(x)∈(∏i∈ΛRi)[[x]] can be viewed as (∑q=0∞biqxq)i∈Λ∈∏i∈Λ(Ri[[x]]). So we can define (6)mxgx=∑p=0∞mipxpi∈Λ∑q=0∞biqxqi∈Λ.Theorem 10.
LetMi be a module over a ring Ri where i∈Λ. If Mi(i∈Λ) is a power-serieswise McCoy Ri-module, then ∏i∈ΛMi is a power-serieswise McCoy ∏i∈ΛRi-module. The converse holds if T(Mi)≠Mi for all i∈Λ.Proof.
Suppose thatMi is a power-serieswise McCoy Ri-module for any i∈Λ. Let m(x)g(x)=0 where m(x)=∑p=0∞mipi∈Λxp∈(∏i∈ΛMi)[[x]] and g(x)=∑q=0∞biqi∈Λxq∈(∏i∈ΛRi)[[x]]∖{0}. Set mi(x)=∑p=0∞mipxp, gi(x)=∑q=0∞biqxq. Then mi(x)∈Mi[[x]],gi(x)∈Ri[[x]]. By the discussion above, we have (7)0=mxgx=mixi∈Λgixi∈Λ=mixgixi∈Λ.So mi(x)gi(x)=0 for all i∈Λ. Since g(x)≠0, there exists k∈Λ such that gk(x)≠0. The assumption that Mk is a power-serieswise McCoy Rk-module implies that there exists rk∈Rk∖{0} such that mk(x)rk=0. Take (8)r=rjj∈Λ=rk,j=k;0,elsewise.Then r∈∏i∈ΛRi∖{0} and m(x)r=mixrii∈Λ=0.
Conversely, fix anyk∈Λ. Let mk(x)=∑p=0∞mkpxp∈Mk[[x]],gk(x)=∑q=0∞bkqxq∈Rk[[x]]∖{0} with mk(x)gk(x)=0 where mkp∈Mk,bkq∈Rk. Write (9)mx=msxs∈Λ=mkx,s=k;ms,s≠kandms∈Ms∖TMs,and (10)gx=gtxt∈Λ=gkx,t=k;0,t≠k.Then (11)mxgx=msxgsxs∈Λ=0.Since ∏i∈ΛMi is power-serieswise McCoy ∏i∈ΛRi-module, there exists r=rii∈Λ∈∏i∈ΓRi∖{0} such that m(x)r=0. That is, (12)mxr=msrs,s≠k;mkxrk,s=k. If s≠k, then msrs=0 and ms∈Ms∖T(Ms) imply rs=0. However, r≠0 yields rk≠0. Hence, Mk is a right power-serieswise McCoy Rk-module. Since k∈Λ is arbitrary, we have done.Proposition 11.
LetR be a commutative domain and M an R-module. The module MR is power-serieswise McCoy if and only if its torsion submodule T(M) is power-serieswise McCoy.Proof.
Letm(x)=∑i=0∞mixi∈M[[x]] and g(x)=∑j=0∞bjxj∈R[[x]] satisfy m(x)g(x)=0. Then (13)m0b0=0,m0b1+m1b0=0,…,m0bn+m1bn-1+⋯+mnb0=0,…. We may assume b0≠0 since g(x)≠0. Now multiplying m0b1+m1b0=0 by b0 on the right yields m1b02=0; thus b02 annihilates both m0 and m1. Similarly, multiplying m0b2+m1b1+m2b0=0 on the right by b02 yields m2b03=0. Continuing this procedure, one has mib0i+1=0 for all i. So mi∈T(M), that is m(x)∈T(M)[[x]]. As T(M) is power-serieswise McCoy, there exists r∈R∖{0} such that m(x)r=0. So MR is power-serieswise McCoy. The other implication is trivial.Recall that aclassical right quotient ring for a ring R is a ring Q which contains R as a subring in such a way that every regular element (that is, nonzero-divisor) of R is invertible in Q and Q=aμ-1∣a,μ∈R,μregular.Proposition 12.
Suppose that there exists the classical right quotient ringQ of a ring R and M is a Q-module. Then MR is power-serieswise McCoy if and only if MQ is power-serieswise McCoy.Proof.
Suppose thatMR is a power-serieswise McCoy module. Let m(x)=∑i=0∞mixi∈M[[x]] and g(x)=∑j=0∞bjxj∈Q[[x]]∖{0} with m(x)g(x)=0. Since Q is the classical right quotient ring of R, by [16, Proposition 2.1.16] we may assume that bj=bj′μ-1 with bj′∈R and some regular element μ∈R. Set g′(x)=∑j=0∞bj′xj, then g′(x)∈R[[x]]∖{0} and 0=m(x)g(x)=∑i=0∞∑j=0∞mibj′xi+jμ-1=m(x)g′(x)μ-1. So m(x)g′(x)=0. Since MR is power-serieswise McCoy, there exists r∈R∖{0}(⊆Q∖{0}) such that m(x)r=0. Therefore, MQ is power-serieswise McCoy.
Conversely, note thatR is the subring of Q, so MR is power-serieswise McCoy.


## 3. Polynomial Modules

In this section, we study the extensions of power-serieswise McCoy moduleM, such as M[x] and M[x;x-1]. And the properties of special matrix extensions of a power-serieswise McCoy module are also considered.LetM be a right R-module. If S is a multiplicatively closed subset of central regular elements of R, set (14)S-1M=ms∣m∈M,s∈S,S-1R=rt∣r∈R,t∈S,then S-1M has an S-1R-module structure. M is called S-torsion free if whenever m∈M∖{0} and s∈S, we have ms≠0 (see [17]).Proposition 13.
LetM be an S-torsion free right R-module. Then MR is power-serieswise McCoy if and only if S-1M is power-serieswise McCoy as an S-1R-module.Proof.
Suppose thatMR is a power-serieswise McCoy module. Let m′(x)=∑i=0∞mi/sxi∈S-1M[[x]] and g(x)=∑j=0∞bj/txj∈S-1R[[x]]∖{0} satisfy m′(x)g(x)=0. Thus m(x)g(x)=0 where m(x)=∑i=0∞mixi∈M[[x]],g(x)=∑j=0∞bjxj∈R[[x]]∖{0}. Since M is power-serieswise McCoy, there exists r∈R∖{0}(⊆S-1R∖{0}) such that m(x)r=0, so m′(x)r=0. Hence S-1M is power-serieswise McCoy.
Conversely, letn(x)=∑i=0∞nixi∈M[[x]],h(x)=∑j=0∞cjxj∈R[[x]]∖{0} satisfy n(x)h(x)=0. Since S-1M is power-serieswise McCoy, there exists r/t∈S-1R∖{0} such that n(x)r/t=0, thus n(x)r=0. So MR is power-serieswise McCoy.Corollary 14.
A ringR is power-serieswise McCoy if and only if so is S-1R.Theorem 15.
LetMR be a module. Then the following statements are equivalent:(1)
M is a power-serieswise McCoy R-module.(2)
M[x] is a power-serieswise McCoy R[x]-module.(3)
M[[x]] is a McCoy R[[x]]-module.(4)
M[x;x-1] is a power-serieswise McCoy R[x;x-1]-module.(5)
M[[x;x-1]] is a McCoy R[[x;x-1]]-module.Proof.
Some of the proof is similar to that of [10, Theorem 3.2].
( 1 ) ⇒ ( 2 ) Suppose that MR is power-serieswise McCoy. Let m(y)=∑i=0∞mixyi∈M[x][[y]] and g(y)=∑j=0∞bjxyj∈R[x][[y]]∖{0} with m(y)g(y)=0, where mi(x)∈M[x],bj(x)∈R[x]. The degree of zero polynomial is taken to be 0. Let (15)kn=degm0x+degm1x+⋯+degmnx+degb0x+degb1x+⋯+degbnx+1. If we set m(x)=∑i=0∞mi(x)xiki,b(x)=∑j=0∞bj(x)xjkj, then b(x)≠0, and the set of coefficients of mix′s (respectively, bjx′s) equals the set of coefficients of m(x) (respectively, b(x)). Since the equation m(y)g(y)=0 deduced that m(x)b(x)=0, there exists r∈R∖{0}(⊆R[x]∖{0}) such that m(x)r=0. Thus m(y)r=0. Hence, M[x] is power-serieswise McCoy.
( 2 ) ⇒ ( 1 ) Assume that M[x] is power-serieswise McCoy and m(x)g(x)=0, where m(x)=∑i=0∞mixi∈M[[x]],g(x)=∑j=0∞bjxj∈R[[x]]∖{0}. Set my=∑i=0∞miyi and g(y)=∑j=0∞bjyj. Then m(y)∈M[x][[y]],g(y)∈R[x][[y]]∖{0} and m(y)g(y)=0. Since M[x]R[x] is power-serieswise McCoy, there exists r(x)∈R[x]∖{0} such that m(y)r(x)=0. Let c be a nonzero coefficient of r(x) since r(x)≠0, then m(y)c=0, that is m(x)c=0. Hence MR is power-serieswise McCoy.
( 2 ) ⇔ ( 3 ) Suppose that M[x] is power-serieswise McCoy. Let m(y)=∑i=0pmi(x)yi∈M[[x]][y] and g(y)=∑j=0qbj(x)yj∈R[[x]][y]∖{0} with m(y)g(y)=0, where mi(x)=∑s=0∞msixs∈M[[x]],bj(x)=∑t=0∞btjxt∈R[[x]] for i=0,1,…,p and j=0,1,…,q. Note that (16)my=m0x+m1xy+m2xy2+⋯+mpxyp=∑s=0∞ms0xs+∑s=0∞ms1xsy+∑s=0∞ms2xsy2+⋯+∑s=0∞mspxsyp=∑k=0pm0kyk+∑k=0pm1kykx+∑k=0pm2kykx2+⋯+∑k=0pmwkykxw+⋯=∑w=0∞∑k=0pmwkykxw. Write m(x)=∑w=0∞nw(y)xw where nw(y)=∑k=0pmwkyk∈M[y]. Then m(x)∈M[y][[x]]. Similarly, (17)gy=b0x+b1xy+b2xy2+⋯+bqxyq=∑t=0∞bt0xt+∑t=0∞bt1xty+∑t=0∞bt2xty2+⋯+∑t=0∞btqxtyq=∑μ=0qb0μyμ+∑μ=0qb1μyμx+∑μ=0qb2μyμx2+⋯+∑μ=0qbvμyμxv+⋯=∑v=0∞∑μ=0qbvμyμxv. Write g(x)=∑v=0∞cv(y)xv where cv(y)=∑μ=0qbvμyμ∈R[y]. Then g(x)∈R[y][[x]]. Obviously, m(x)g(x)=0. By the assumption, M[y] is a power-serieswise McCoy R[y]-module. Thus there exists g1(y)=∑i=0taiyi∈R[y]∖{0} such that m(x)g1(y)=0. Set g2(x)=∑i=0taixi. Then g2(x)∈R[x]∖{0}(⊆R[[x]]∖{0}) and m(y)g2(x)=0. Hence M[[x]] is McCoy. The other complication can be proved similarly.
( 2 ) ⇔ ( 4 ) Let S={1,x,x2,⋯}. Then S is a multiplicatively closed subset of central regular elements of R[x]. Since M[x;x-1]=S-1M[x], R[x;x-1]=S-1R[x], we have done by Proposition 13.
( 4 ) ⇔ ( 5 ) It is similar to the proof of (2)⇔(3).Corollary 16.
LetR be a ring. Then the following statements are equivalent:(1)
R is a power-serieswise McCoy ring.(2)
R[x] is a power-serieswise McCoy ring.(3)
R[[x]] is a McCoy ring.(4)
R[x;x-1] is a power-serieswise McCoy ring.(5)
R[[x;x-1]] is a McCoy ring.
LetMR be a module. For n∈N∖{0}, Rn denotes the following subring of the ring of all matrices over R: (18)Rn=aa12⋯a1n0a⋯a2n⋮⋮⋱⋮00⋯a∣a,aij∈R, and Mn denotes the following additive group: (19)Mn=mm12⋯m1n0m⋯m2n⋮⋮⋱⋮00⋯m∣m,mij∈M. Then Mn is an Rn-module under the usual matrix addition operation and scalar multiplication.Proposition 17.
TheR-module M is power-serieswise McCoy if and only if the Rn-module Mn is power-serieswise McCoy.Proof.
Suppose thatMR is power-serieswise McCoy. Let m(x)=∑i=0∞Aixi∈Mn[[x]],g(x)=∑j=0∞Bjxj∈R[[x]]∖{0} satisfy m(x)g(x)=0, where (20)Ai=aia12i⋯a1ni0ai⋯a2ni⋮⋮⋱⋮00⋯ai∈Mn,Bj=bjb12j⋯b1nj0bj⋯b2nj⋮⋮⋱⋮00⋯bj∈Rn. Under the usual matrix addition operation, we have(21)mx=∑i=0∞Aixi=∑i=0∞aia12i⋯a1ni0ai⋯a2ni⋮⋮⋱⋮00⋯aixi=∑i=0∞aixi∑i=0∞a12ixi⋯∑i=0∞a1nixi0∑i=0∞aixi⋯∑i=0∞a2nixi⋮⋮⋱⋮00⋯∑i=0∞aixi≜axa12x⋯a1nx0ax⋯a2nx⋮⋮⋱⋮00⋯axand (22)gx=∑j=0∞Bjxj=∑j=0∞bjb12j⋯b1nj0bj⋯b2nj⋮⋮⋱⋮00⋯bjxj=∑j=0∞bjxj∑j=0∞b12jxj⋯∑j=0∞b1njxj0∑j=0∞bjxj⋯∑j=0∞b2njxj⋮⋮⋱⋮00⋯∑j=0∞bjxj≜bxb12x⋯b1nx0bx⋯b2nx⋮⋮⋱⋮00⋯bx.
Now we proceed with the following cases.Case 1.
Suppose thata(x)=0. Take D=E1n∈Rn∖{0}. Thus m(x)D=0.Case 2.
Suppose thata(x)≠0 and b(x)≠0. Since M is power-serieswise McCoy, a(x)b(x)=0 implies that there exists r∈R∖{0} such that a(x)r=0. Take D=rE1n, then m(x)D=0.Case 3.
Suppose thata(x)≠0 and b(x)=0. Since g(x)≠0, there must be a minimal index l such that bkl≠0 and bsl=0,k<s<l, where bij(x)=∑t=0∞bijtxt. Consider the (k, l)-entry of matrix m(x)g(x); we have (23)axbklx=0.It follows from MR is power-serieswise McCoy that there exists r∈R∖{0} such that a(x)r=0. Take D=rE1n∈Rn∖{0}, then m(x)D=0.Therefore,Mn is a power-serieswise McCoy Rn-module for all cases.Conversely, letm(x)=∑i=0aixi∈M[[x]],g(x)=∑j=0∞bjxj∈R[[x]]∖{0} satisfy m(x)g(x)=0 where ai∈M,bj∈R. Take α(x)=m(x)E∈Mn[[x]],β(x)=g(x)E∈Rn[[x]], where E is the unit matrix. It is easy to see that α(x)β(x)=0. Since Mn is a power-serieswise McCoy Rn-module, there exists D=dd12⋯d1n0d⋯d2n⋮⋮⋱⋮00⋯d∈Rn∖{0}, such that α(x)D=0. By D≠0, there exists dst≠0 (if s=t, dst=d) such that m(x)dst=0. Hence MR is power-serieswise McCoy.Corollary 18 (see [7, Corollary 2.9]).
R is a power-serieswise McCoy ring if and only if so is Rn where n≥1.



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*Source: 9860597-2018-10-09.xml*