# Generalized Malcev-Neumann Series Modules with the Beachy-Blair Condition

Authors: Mohamed A. Farahat
Journal: Algebra
 (2015)

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## Abstract

We introduce a new class of extension rings called thegeneralized Malcev-Neumann series ring R((S;σ;τ)) with
coefficients in a ring R and exponents in a strictly ordered monoid S
which extends the usual construction of Malcev-Neumann series rings. Ouyang et al. in 2014 introduced the modules with the Beachy-Blair condition
as follows: A right R-module satisfies the right Beachy-Blair condition
if each of its faithful submodules is cofaithful. In this paper, we study
the relationship between the right Beachy-Blair condition of a right R-module MR and its Malcev-Neumann series module extension MSR((S;σ;τ)).


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## Body

## 1. Introduction

Throughout this paperR denotes an associative ring with identity; S,·,⩽ is a strictly ordered monoid (i.e., S,⩽ is an ordered monoid satisfying the conditions that if s<s′, then st<s′t and ts<ts′ for s,s′,t∈S). Recall that a subset X of S,⩽ is said to beartinian if every strictly decreasing sequence of elements of X is finite and that X isnarrow if every subset of pairwise order-incomparable elements of X is finite. Suppose the two maps σ:S→EndR and τ:S×S→UR (the group of invertible elements of R). Let A=RS;σ;τ denote the set of all formal sums f=∑x∈Saxx¯ such that suppf={x∈S∣ax≠0} is an artinian and narrow subset of S, with componentwise addition and the multiplication rule is given by (1)∑x∈Saxx¯∑y∈Sbyy¯=∑z∈S∑{(x,y)∣xy=z}axσxbyτx,yz¯,for each ∑x∈Saxx¯ and ∑y∈Sbyy¯∈A. In order to ensure the associativity, it is necessary to impose two additional conditions on σ and τ: namely, for all x,y,z∈S,(i)
σx(τ(y,z))τ(x,yz)=τ(x,y)τ(xy,z),(ii)
σxσy=η(x,y)σxy, where η(x,y) denotes the automorphism of R defined by (2)η(x,y)(r)=τ(x,y)rτx,y-1∀r∈R.It is now routine to check thatA=R((S;σ;τ)) is a ring which is calledthe ring of generalized Malcev-Neumann series. We can assume that the identity element of A is 1¯; this means that (3)σ1=IdR,τ(x,1)=τ(1,x)=1.In this case r↦r1¯ is an embedding of R as a subring into A.For eachf∈A∖{0} we denote by πf the set of minimal elements of suppf. If S,≤ is a strictly totally ordered monoid, then suppf is a nonempty well-ordered subset of S and πf consists of only one element.Clearly, the above construction generalizes the construction of Malcev-Neumann series rings, in case ofS=G (an ordered group), which was introduced independently by Malcev and Neumann (see [1, 2]).If the order≤ is the trivial order, then A=R((S;σ;τ)) is the usual crossed product ring RS;σ;τ. Also, if the monoid S has the trivial order and τ is trivial, then A=R((S;σ;τ)) is the usual skew monoid ring RS;σ. However if the monoid S has the trivial order and σ is trivial, then A=R((S;σ;τ)) is the usual twisted monoid ring RS;τ. Finally, if the monoid S has the trivial order and σ and τ are trivial, then A=R((S;σ;τ)) is the usual monoid ring RS (see Sections 3.2 and  3.3 in [3]).Moreover, ifα is a ring endomorphism of R, set S=Z≥0 endowed with the trivial order. Define σ:S→EndR via σx=αx for every x∈Z≥0 and τx,y=1 for any x,y∈Z. We have A=R((S;σ;τ)) is the usual skew polynomial ring Rx,α. However if ≤ is the usual order, then A=R((S;σ;τ)) is the usual skew power series ring Rx,σ. If α is a ring automorphism of R,S=Z and ≤ is the usual order, then A=R((S;σ;τ)) is the usual ring of skew Laurent power series Rx,x-1,α.At the same time, if we set alsoσs=IdR∈End(R) for all s∈S, then it is easy to check that polynomial rings, Laurent polynomial rings, formal power series rings, and Laurent power series rings are special cases of A=R((S;σ;τ)).IfMR is a unitary right R-module, thenthe Malcev-Neumann series moduleB=MS is the set of all formal sums ∑x∈Smxx¯ with coefficients in M and artinian and narrow supports, with pointwise addition and scalar multiplication rule is defined by (4)∑x∈Smxx¯∑y∈Sayy¯=∑z∈S∑{(x,y)∣xy=z}mxσxayτx,yz¯,where ∑x∈Smxx¯∈B and ∑y∈Sayy¯∈A. One can easily check that (i) and (ii) ensure that MS is a unitary right A-module. For each φ∈B∖{0} we denote by πφ the set of minimal elements of suppφ. If S,≤ is a strictly totally ordered monoid, then suppφ is a nonempty well-ordered subset of S and πφ consists of only one element.Recall from Faith [4] that a ring R is called aright zip ring and if the right annihilator rR(X) of a subset X⊆R is zero, then rR(X0)=0 for a finite subset X0 of X. Although the concept of zip rings was initiated by Zelmanowitz [5] it was not called so at that time.Recall from [6] that a right R-module MR is called aright zipmodule provided that if the right annihilator of a subset X of MR is zero, then there exists a finite subset X0⊆X such that rRX0=0.According to Rodríguez-Jorge [7], a ring R satisfiesthe right Beachy-Blair condition if its faithful right ideals are cofaithful; that is, if I is a right ideal of R such that rR(I) vanishes, then rR(I0)=0 for a finite subset I0 of I. Clearly, a right zip ring is a right Beachy-Blair ring.Ouyang et al. in [8] generalized the right Beachy-Blair condition from rings into modules as follows: A right R-module MR is calledmodule with the Beachy-Blair condition provided that if the right annihilator of a submodule NR of MR is zero, then there exists a finite subset N0⊆N such that rRN0=0.The main aim of the present paper is to investigate conditions for the Malcev-Neumann series modulesMSR((S;σ;τ)) to satisfy the right Beachy-Blair condition. The proofs of our results obtained here are very similar to those obtained by Ouyang et al. in [8] and by Salem et al. in [9].


## 2. Generalized Malcev-Neumann Series Modules with the Beachy-Blair Condition

We start this section with the following notions and definitions.LetV be a subset of MR; then (5)VS=φ=∑x∈Smxx¯∈B∣0≠mx∈V,x∈suppφ.Definition 1.
A ringR is called S-compatible if, for all a,b∈R and x∈S, ab=0 if and only if aσxb=0.Definition 2.
A rightR-module MR is called S-compatible if, for each m∈M, a∈R, and x∈S, ma=0 if and only if mσxa=0.Definition 3.
A ringR is called S-Armendariz if whenever fg=0 implies axσx(by)=0 for each x∈suppf and y∈suppg, where f=∑x∈Saxx¯ and g=∑y∈Sbyy¯ are elements of A.We extend theS-Armendariz concept to modules as follows.Definition 4.
A rightR-module MR is called S-Armendariz if whenever φf=0 implies mxσx(ay)=0 for each x∈suppφ and y∈suppf, where φ=∑x∈Smxx¯∈B and f=∑y∈Sayy¯∈A.It is clear thatR is an S-Armendariz (S-compatible) ring if and only if RR is an S-Armendariz (S-compatible) module.For a subsetU of MR, we define rAU as the set (6)rAU={f∈A∣u1¯f=0foreachu∈U}.Lemma 5.
LetMR be a right R-module. Then rAU=rRUS;σ;τ, for any subset U of MR.Proof.
Letf=∑s∈Sass¯∈rAU. Then for each u∈U we have u1¯f=0. Thus (7)0=u1¯∑s∈Sass¯=∑s∈Suσ1asτ1,ss¯=∑s∈Suass¯,which implies that uas=0 for each s∈suppf. Hence as∈rRU for each s∈suppf. So f∈rRUS;σ;τ and rAU⊆rRUS;σ;τ.
On the other hand, suppose thatf=∑s∈Sass¯∈rRUS;σ;τ; then as∈rRU for each s∈suppf. Thus uas=0 for each u∈U, which implies that uσ1asτ1,s=0 for each u∈U and s∈suppf. Hence u1¯f=0 and f∈rAU. So rRUS;σ;τ⊆rAU. Therefore rAU=rRUS;σ;τ.WhenMR=RR we have the following consequence of Lemma 5.Corollary 6.
ConsiderrAU=rRUS;σ;τ, for any subset U of R.Note the following: forφ=∑x∈Smxx¯∈B, let Cφ={mx∣x∈S and for a subset V⊆MS, we have CV=∪φ∈VCφ.Lemma 7.
LetMR be an S-compatible and S-Armendariz R-module. Then (8)rAV=rRCVS;σ;τfor any V⊆B.Proof.
LetV⊆B and T=CV=∪φ∈VCφ=∪φ∈V{mx∣x∈S}. We show that rAV=rRTS;σ;τ and it is enough to show that rAφ=rRCφS;σ;τ for each φ=∑x∈Smxx¯∈V. In fact, let f=∑y∈Sayy¯∈rAφ. Then φf=0. Since MR is an S -Armendariz module, mxay=0 for each x∈suppφ and y∈suppf. Then ay∈rRCφ for each y∈suppf. Thus f∈rRCφS;σ;τ and rAφ⊆rRCφS;σ;τ. Now, let f=∑y∈Sayy¯∈rRCφS;σ;τ. Then ay∈rRCφ for each y∈suppf. Hence mxay=0 for each x∈suppφ and y∈suppf. Since MR is S-compatible, it follows that mxσxay=0, which implies that mxσxayτx,y=0 for each x∈suppφ and y∈suppf. Consequently (9)0=∑z∈S∑{(x,y)∣xy=z}mxσxayτx,yz¯=φf.So f∈rAφ and it follows that rRCφS;σ;τ⊆rAφ. So (10)rAV=⋂φ∈VrAφ=⋂φ∈VrRCφS;σ;τ=⋂φ∈VrRCφS;σ;τ=rRTS;σ;τ=rRCVS;σ;τ.For a rightR-module MR, we define (11)rR2M=rRU∣U⊆M,rA2B=rAV∣V⊆B.Lemma 5 gives us the map Π:rR2M→rA2B defined by ΠI=IS;σ;τ for every I∈rR2M. Obviously Π is an injective map.In the following lemma we show thatΠ is a bijective map if and only if MR is S-Armendariz.Lemma 8.
LetMR be an S-compatible R-module. The following conditions are equivalent.(1)
MR is an S-Armendariz R-module.(2)
Π:rR2M→rA2B defined by ΠI=IS;σ;τ is a bijective map.Proof.
(1)⇒(2).
It is only necessary to show thatΠ is surjective. Let V⊆B and T=CV. Since ΠrRT=rRTS;σ;τ, the proof of this direction follows directly from Lemma 7.
(2)⇒(1).
Letf=∑y∈Sayy¯∈A and φ=∑x∈Smxx¯∈B such that φf=0. Then f∈rAφ. By assumption rAφ=TS;σ;τ for some right ideal T of R. Hence f∈TS;σ;τ which implies that ay∈T⊆rAφ for each y∈suppf. So, φay1¯=0 and we have that (12)0=∑x∈Smxx¯ay1¯=∑x∈Smxσxayτx,1x¯for each x∈suppφ and y∈suppf. Thus mxσxay=0 for each x∈suppφ and y∈suppf. So, MR is an S-Armendariz module.Recall that a ring is reduced if it has no nonzero nilpotent elements. Reduced rings have been studied for over forty-eight years (see [10]). In 2004, the reduced ring concept was extended to modules by Lee and Zhou [11] as follows: a right R-module MR is reduced if, for any m∈MR and any a∈R, ma=0 implies mR∩Ma=0. Clearly, if MR is reduced, then, for all m∈MR and a∈R, ma=0 implies mRa=0. It is clear that R is a reduced ring if and only if RR is a reduced module.Now, we are able to prove the main result.Theorem 9.
LetMR be a reduced, S-compatible, and S-Armendariz right R-module. If MR satisfies the right Beachy-Blair condition, then BA satisfies the right Beachy-Blair condition.Proof.
Suppose that a rightR-module MR satisfies the right Beachy-Blair condition and J is a right A-submodule of B such that rA(J)=0.
From Lemma8, we conclude that rRCJS;σ;τ=ΠrRCJ=rAJ=0. Thus rRCJ=0.
LetCJR denote the right R-submodule of MR generated by CJ. Since CJ⊂CJR, we have rR(CJR)⊂rR(CJ)=0. Since MR satisfies the right Beachy-Blair condition, there exists a finite subset (13)X=∑i=1ntqitrit∣qit∈CJ,rit∈R,1≤t≤k⊂CJR,such that rR(X) = 0. Let (14)X0=q11,q21,…,qn11,q12,q22,…,qn22,q1k,q2k,…,qnkk.Then X0 is a finite subset of CJ. Now we will see that rR(X0)=0. Let a∈rRX0; then qita=0 for 1≤i≤nt and1≤t≤k. Since MR is a reduced R-module, then qitrita=0 for 1≤i≤nt and1≤t≤k. Then for each ∑i=1ntqitrit∈X, we have ∑i=1ntqitrita=0. Therefore a∈rR(X) = 0, and so rR(X0) = 0 is proved.
For eachqit∈X0, there exists an element φit∈J such that qit∈Cφit. Let V be a minimal subset of J such that φit∈V for each qit∈X0; then V is a finite subset of J and X0⊂CV. Thus rR(CV)⊂rR(X0)=0. Now we show that rA(V)=0. Let the contrary; that is, rA(V)≠0, and suppose that f=∑y∈Sbyy¯∈rA(V)∖{0}; then φf=0 for each φ=∑x∈Saxx¯∈V. Let y∈supp(f); since MR is an S-Armendariz and S-compatible module, we have axby=0 for all ax∈Cφ and each φ∈V. Hence by∈rR(CV)=0, a contradiction. Hence rA(V)=0 is proved. Thus BA satisfies the right Beachy-Blair condition.WhenMR=RR we have the following consequence of Theorem 9.Corollary 10.
Suppose thatR is a reduced, S-compatible, and S-Armendariz ring. If R satisfies the right Beachy-Blair condition, then A satisfies the right Beachy-Blair condition.



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