Let R be a ring, MR a right R-module, G a totally ordered group, σ a map from G into the group of automorphisms of R which assigns to each x ∈ G an automorphism σ_x ∈ Aut(R), τ a map from G × G to U(R) (the group of unit elements of R) and M((G; σ ; τ)) the Malcev-Neumann series module. Then, under some certain conditions, we show that MR is a right zip R-module if and only if M((G; σ ; τ))_{R((G;σ ;τ))} is a right zip R((G; σ ; τ))-module, where R((G; σ ; τ)) is the Malcev-Neumann series ring

Abd-Elmalk, Hanan

Radwan, Abdelaziz E.

Salem, Refaat

Le Matematiche (Dipartimento di Matematica e Informatica, Università degli Studi di Catania)

LE MATEMATICHEVol. LXX (2015) – Fasc. I, pp. 115–123doi: 10.4418/2015.70.1.9ZIP PROPERTY ON MALCEV-NEUMANN SERIES MODULESR. SALEM - A. E. RADWAN - H. ABD-ELMALKLet R be a ring, MR a right R-module, G a totally ordered group, σa map from G into the group of automorphisms of R which assigns toeach x ∈ G an automorphism σx ∈ Aut(R), τ a map from G×G to U(R)(the group of unit elements of R) and M((G;σ ;τ)) the Malcev-Neumannseries module. Then, under some certain conditions, we show that MRis a right zip R-module if and only if M((G;σ ;τ))R((G;σ ;τ)) is a right zipR((G;σ ;τ))-module, where R((G;σ ;τ)) is the Malcev-Neumann seriesring.1. IntroductionThroughout this paper R denotes an associative ring with identity. Recall from[3] that R is a right zip ring if the right annihilator of a subset X ⊆ R is zero,then rR(X0) = 0 for a finite subset X0 of X , equivalently for a left ideal L of R ifrR(L)= 0, then there exists a finitely generated ideal L0⊆ L such that rR(L0)= 0.The concept of zip rings was initiated by Zelmanowitz [8] where it wasnot so called zip at that time, however he showed that any ring satisfying thedescending chain condition on right annihilator ideals is a right zip ring but theconverse is not true.Extensions of zip rings were studied by several authors. In [1] Beachy andBlair showed that if R is a commutative zip ring, then R[x] is a zip ring.Entrato in redazione: 4 maggio 2014AMS 2010 Subject Classification: 06F05, 16W60, 13E10.Keywords: Zip ring, zip module, Malcev-Neumann series ring, Malcev-Neumann series module.116 REFAAT SALEM - ABDELAZIZ E. RADWAN - HANAN ABD-ELMALKIn ([4], Theorem 1) Hong et al showed that if R is an Armendariz ring, thenR is a right zip ring if and only if R[x] is a right zip ring.In ([2], Theorem 2.8) Cortes studied skew polynomial extension over ziprings and he showed that, if σ is an automorphism of R and R is σ -Armendariz,then R is a right zip ring if and only if R[x;σ ] is a right zip ring.Recall from [9] that a right R-module MR is called a right zip module pro-vided that if the right annihilator of a subset X of MR is zero, then there exists afinite subset X0 ⊆ X such that rR(X0) = 0.In the following section we introduce results concerned with the transferof a right zip property of MR and a twisted Malcev-Neumann series moduleextension M((G;σ ;τ)).2. Zip Modules over Twisted Malcev-Neumann Series RingsLet R be a ring, G a totally ordered group, σ a map from G into the group ofautomorphisms of R which assigns to each x ∈G an automorphism σx ∈Aut(R)where σ1 = idR with 1 the identity of group G, and τ a map from G×G to U(R)(the group of invertible elements of R). Let A= R((G;σ ;τ)) denote the set of allformal sums f = ∑x∈Gaxx such that supp( f ) = {x ∈ G| ax 6= 0} is a well orderedsubset of G, with componentwise addition and the multiplication rule is givenby(∑x∈Gaxx)(∑y∈Gbyy) = ∑z∈G( ∑{(x,y)|xy=z}axσx(by)τ(x,y))z,for each ∑x∈Gaxx and ∑y∈Gbyy ∈ A. In order to ensure the associativity it is neces-sary that(i) σx(τ(y,z))τ(x,yz) = τ(x,y)τ(xy,z) and(ii) σxσy = η(x,y)σxy,where η(x,y) denotes the automorphism of R induced by the unit τ(x,y), for allx,y,z∈G, see ([5], Lemma 1.1). It is now routine to check that A is a ring whichis called the ring of Malcev-Neumann series.The Malcev-Neumann construction appeared for the first time in the latterpart of the 1940 (the Laurent series ring, a particular case of Malcev-Neumannring, was used before by Hilbert). Using them, Malcev and Neumann inde-pendently showed (in 1948 and 1949, respectively.) that the group ring of anordered group over a division ring can be embedded in a division ring. Sincethen, the construction has appeared in many papers, mainly in the study of vari-ous properties of division rings and related topics.ZIP PROPERTY ON MALCEV-NEUMANN SERIES MODULES 117In [7] Sonin generalized the construction to obtain Malcev-Neumann mod-ules over Malcev-Neumann rings as follows:If MR is a right R-module, then the Malcev-Neumann series module B =M((G;σ ;τ)) is the set of all formal sums ∑x∈Gmxx with coefficients in M andwell-ordered supports, with pointwise addition and scalar multiplication ruledefined by(∑x∈Gmxx)(∑y∈Gayy) = ∑z∈G( ∑{(x,y)|xy=z}mxσx(ay)τ(x,y))z,where ∑x∈Gmxx ∈ B and ∑y∈Gayy ∈ A. One can easily check that (i) and (ii) ensurethat M((G;σ ;τ)) is a right A-module.Let V be a subset of MR, then V ((G;σ ;τ)) is defined as follows:V ((G;σ ;τ)) = {ϕ = ∑x∈Gmxx ∈ B |0 6= mx ∈V and x ∈ supp(ϕ)}.For ϕ = ∑x∈Gmxx ∈ B, let Cϕ = {mx| x ∈ supp(ϕ)} and for a subset V ⊆ B, wehave CV = ∪ϕ∈VCϕ .As usual we shall identify R with the subring R1G ⊆ A, identify G with thesubgroup 1RG of invertible elements in A, and identify MR with the submoduleM1G ⊆ B.In this section, we generalize the results of [6] to the Malcev-Neumann se-ries modules. We start with the following definitions, see [10] and the literaturetherein for more details.Definition 2.1. A ring R is called σ -compatible if, for all a,b ∈ R and x ∈ G,ab = 0 if and only if aσx(b) = 0.Definition 2.2 ([10]). A right R-module MR is called σ -compatible if, for eachm ∈M, a ∈ R and x ∈ G, ma = 0 if and only if mσx(a) = 0.Definition 2.3. A ring R is called (G,σ)-Armendariz if whenever f g = 0 im-plies axσx(by) = 0 for each x ∈ supp( f ) and y ∈ supp(g), where f = ∑x∈Gaxx andg = ∑y∈Gbyy be elements of A.We extend the (G,σ)-Armendariz concept to modules as follows:Definition 2.4. A right R-module MR is called (G,σ)-Armendariz if wheneverϕ f = 0 implies mxσx(ay) = 0 for each x ∈ supp(ϕ) and y ∈ supp( f ), whereϕ = ∑x∈Gmxx ∈ B and f = ∑y∈Gayy ∈ A.118 REFAAT SALEM - ABDELAZIZ E. RADWAN - HANAN ABD-ELMALKIt is clear that, R is a (G,σ)-Armendariz and σ -compatible ring if and onlyif RR is a (G,σ)-Armendariz and σ -compatible module.For a subset X of MR, we define rA(X) as the set:rA(X) = { f ∈ A| (x1) f = 0 for each x ∈ X} .Lemma 2.5. Let MR be a right R-module. Then rA(X) = rR(X)((G;σ ;τ)), forany subset X of MR.Proof. Let f = ∑g∈Gagg ∈ rA(X). Then for each x ∈ X we have (x1) f = 0. Thus0 = (x1)(∑g∈Gagg) = ∑g∈Gxσ1(ag)τ(1,g)g = ∑g∈Gxagτ(1,g)g,which implies that xagτ(1,g) = 0 for each g ∈ supp( f ). Since τ(1,g) is invert-ible, xag = 0. Hence ag ∈ rR(X) for each g ∈ supp( f ). So f ∈ rR(X)((G;σ ;τ))and rA(X)⊆ rR(X)((G;σ ;τ)).On the other hand, suppose that f = ∑g∈Gagg ∈ rR(X)((G;σ ;τ)), then ag ∈rR(X) for each g ∈ supp( f ). Thus xag = 0 for each x ∈ X and g ∈ supp( f ).We have xσ1(ag) = 0 and we have that xσ1(ag)τ(1,g) = 0 for each x ∈ X andg ∈ supp( f ). Hence (x1) f = 0 for each x ∈ X , and it follows that f ∈ rA(X). SorR(X)((G;σ ;τ))⊆ rA(X). Therefore rA(X) = rR(X)((G;σ ;τ)).For a right R-module MR, we definerR(2M) = {rR(U)|U ⊆M} ,rA(2B) = {rA(V )|V ⊆ B} .The above Lemma gives us the map ψ : rR(2M) −→ rA(2B) defined by ψ(I) =I((G;σ ;τ)) for every I ∈ rR(2M). Obviously ψ is an injective map.In the following Lemma we show that ψ is a bijective map if and only if MRis (G,σ)-Armendariz.Lemma 2.6. Let MR be a σ -compatible module. The following conditions areequivalent:(1) MR is a (G,σ)-Armendariz module.(2) ψ : rR(2M)−→ rA(2B) defined by ψ(I) = I((G;σ ;τ)) is a bijective map.Proof. (1)⇒(2)It is only necessary to show that ψ is surjective. Let V ⊆ B and T = CV =∪ϕ∈VCϕ = ∪ϕ∈V {mx| x ∈ supp(ϕ)}. We show thatrA(V ) = ψ(rR(T )) = rR(T )((G;σ ;τ))ZIP PROPERTY ON MALCEV-NEUMANN SERIES MODULES 119and it is enough to show that rA(ϕ) = rR(Cϕ)((G;σ ;τ)) for each ϕ = ∑x∈Gmxx ∈V . In fact, let f = ∑y∈Gayy ∈ rA(ϕ). Then ϕ f = 0. Since MR is a (G,σ)-Armendariz and σ -compatible module, mxay = 0 for each x ∈ supp(ϕ) andy∈ supp( f ). Then ay ∈ rR(Cϕ) for each y∈ supp( f ). Thus f ∈ rR(Cϕ)((G;σ ;τ))and rA(ϕ)⊆ rR(Cϕ)((G;σ ;τ)). Now, let f = ∑y∈Gayy ∈ rR(Cϕ)((G;σ ;τ)). Thenay ∈ rR(Cϕ) for each y ∈ supp( f ). Hence mxay = 0 for each x ∈ supp(ϕ) andy ∈ supp( f ). Since MR is σ -compatible, it follows that mxσx(ay) = 0, whichimplies that mxσx(ay)τ(x,y) = 0 for each x ∈ supp(ϕ) and y ∈ supp( f ). Hence0 = ∑z∈G( ∑{(x,y)|xy=z }mxσx(ay)τ(x,y))z = ϕ f .So f ∈ rA(ϕ) and it follows that rR(Cϕ)((G;σ ;τ))⊆ rA(ϕ). Consequently,rA(V ) = ∩ϕ∈V rA(ϕ) = ∩ϕ∈V rR(Cϕ)((G;σ ;τ))= (∩ϕ∈V rR(Cϕ))((G;σ ;τ))= rR(T )((G;σ ;τ)) = ψ(rR(T )).(2)⇒(1)Let f = ∑y∈Gayy ∈ A and ϕ = ∑x∈Gmxx ∈ B such that ϕ f = 0. Then f ∈ rA(ϕ).By assumption rA(ϕ) = T ((G;σ ;τ)) for some right ideal T of R. Hence f ∈T ((G;σ ;τ)) which implies that ay ∈ T ⊆ rA(ϕ) for each y ∈ supp( f ). So,ϕ(ay1) = 0 and we have that0 = (∑x∈Gmxx)(ay1) = ∑x∈Gmxσx(ay)τ(x,1)xfor each x ∈ supp(ϕ) and y ∈ supp( f ). Since τ(x,1) is an invertible element, itfollows that mxσx(ay) = 0 for each x ∈ supp(ϕ) and y ∈ supp( f ). Therefore MRis a (G,σ)-Armendariz module.Theorem 2.7. Let MR be σ -compatible and a (G,σ)-Armendariz module. ThenMR is a right zip R-module if and only if BA is a right zip A-module.Proof. Suppose that BA is a right zip A-module and X ⊆MR such that rR(X) = 0.Let Y = {m1| m ∈ X} be the embedding of X in BA. Then, by Lemma 2.5, wehaverA(Y ) = { f ∈ A| (m1) f = 0, for all m ∈ X}= rA(X) = rR(X)((G;σ ;τ)) = 0.120 REFAAT SALEM - ABDELAZIZ E. RADWAN - HANAN ABD-ELMALKSince BA is a right zip A-module, for some m1,m2, . . . ,mn ∈ X there exists afinite set Y′= {m11,m21, . . . ,mn1} such that Y ′ ⊆ Y and rA(Y ′) = 0. Let X ′ ={m1,m2, . . . ,mn} which is a nonempty finite subset of X . Then, from Lemma2.5, we have0 = rA(Y′) = { f ∈ A |(m1) f = 0, for all m ∈ X ′}= rA(X′) = rR(X′)((G;σ ;τ))which implies that rR(X′) = 0. Hence MR is a right zip R-module.Conversely, suppose that MR is a right zip R-module and Y ⊆ BA such thatrA(Y ) = 0. LetT =CY = ∪ϕ∈YCϕ = ∪ϕ∈Y {mx| x ∈ supp(ϕ)} .Then, by Lemma 2.6,0 = rA(Y ) = rR(T )((G;σ ;τ))which implies that rR(T ) = 0. Since MR is a right zip R-module, there exists afinite subset T0 ⊆ T such that rR(T0) = 0. For each m ∈ T0 there exists ϕm ∈ Ysuch that for some x ∈ supp(ϕm), mx = m. Let Y0 be a minimal subset of Y withrespect to inclusion such that ϕm ∈ Y0 for each m ∈ T0. Then Y0 is a nonemptyfinite subset of Y . We considerT1 =CY0 = ∪ϕ∈Y0Cϕ = ∪ϕ∈Y0 {mx| x ∈ supp(ϕ)} .Note that T0 ⊆ T1 and we have that rR(T1) ⊆ rR(T0) = 0. Thus, by Lemma 2.6,we haverA(Y0) = rR(T1)((G;σ ;τ)) = 0.So BA is a right zip A-module.When MR = RR we have the following consequence of the last theorem.Corollary 2.8 ([6], Theorem 2.1). Suppose that R is σ -compatible and a (G,σ)-Armendariz ring. Then R is a right zip ring if and only if A is a right zip ring.Let α be a ring automorphism of R and set G = Z endowed with the usualorder. Define σ : G −→ Aut(R) via σ(x) = αx for every x ∈ Z and τ(x,y) = 1for any x,y ∈ Z. Then M((G;σ ;τ))R((G;σ ;τ)) = M[[x,x−1;α]]R[[x,x−1;α]] the usualskew Laurent power series extension of MR.We can introduce the restricted version of (G,σ)-Armendariz condition onskew Laurent power series modules and skew formal power series modules,respectively, as follows:ZIP PROPERTY ON MALCEV-NEUMANN SERIES MODULES 121Definition 2.9. A right R-module MR is called an α-skew Laurent power se-rieswise Armendariz (shortly, α-SLPA) module if ϕ(x) f (x) = 0, where ϕ(x) =∞∑i=smixi ∈M[[x,x−1;α]] and f (x) =∞∑j=ta jx j ∈ R[[x,x−1;α]] for s, t ∈ Z, impliesthat miα i(a j) = 0 for all i≥ s and j ≥ t.Definition 2.10. A right R-module MR is called an α-skew power serieswiseArmendariz (shortly, α-SPA) module if ϕ(x) f (x) = 0, where ϕ(x) =∞∑i=0mixi ∈M[[x;α]] and f (x) =∞∑j=0a jx j ∈ R[[x;α]], implies that miα i(a j) = 0 for all i≥ 0and j ≥ 0.It is clear that R is an α-SLPA (resp. α-SPA) ring if and only if RR is anα-SLPA (resp. α-SPA) module.Proposition 2.11. Let α be a ring automorphism of R. Then a right R-moduleMR is α-SPA if and only if MR is α-SLPA.Proof. Since M[[x;α]]⊆M[[x,x−1;α]] and R[[x;α]]⊆R[[x,x−1;α]], we can eas-ily conclude that: if MR is α-SLPA, then MR is α-SPA.Conversely, assume that MR is α-SPA and let ϕ(x)=∞∑i=−smixi ∈M[[x,x−1;α]],f (x) =∞∑j=−ta jx j ∈ R[[x,x−1;α]], for s, t ∈ Z≥0, be such that ϕ(x) f (x) = 0. Wehave0 = (ϕ(x) f (x))xs+t = (∞∑i=−smixi)(∞∑j=−ta jx j)xs+t= (∞∑i=−smixi)(∞∑j=−txsα−s(a j)x jxt)= (∞∑i=−smixi)(xs∞∑j=−tα−s(a j)x j+t)= (∞∑i=−smixixs)(∞∑j=−tα−s(a j)x j+t)= (∞∑i=−smixi+s)(∞∑j=−tα−s(a j)x j+t).Set i+ s = k and j+ t = l, we get that0 = (ϕ(x) f (x))xs+t = (∞∑k=0mk−sxk)(∞∑l=0α−s(al−t)xl) = (ϕ(x)xs)g(x).122 REFAAT SALEM - ABDELAZIZ E. RADWAN - HANAN ABD-ELMALKHence, ϕ(x)xs =∞∑k=0mk−sxk ∈ M[[x;α]] and g(x) =∞∑l=0α−s(al−t)xl ∈ R[[x;α]].So,0 = mk−sαk(α−s(al−t)) = mk−sαk−s(al−t)for all k ≥ 0 and l ≥ 0. Hence miα i(a j) = 0 for all i ≥ −s and j ≥ −t, asrequired.From Theorem 2.7, we obtain the following result:Corollary 2.12. Let α be a ring automorphism of R, MR an α-compatible andα-SLPA module. Then MR is a right zip R-module if and only ifM[[x,x−1;α]]R[[x,x−1;α]] is a right zip R[[x,x−1;α]]-module.Proof. Take G = Z and τ(x,y) = 1 for any x,y ∈ Z. For any x ∈ Z, let σx = αx.Then the result follows from Theorem 2.7.Set MR = RR in Corollary 2.12, we get:Corollary 2.13. Let α be a ring automorphism of R, R an α-compatible andα-SLPA ring. Then R is a right zip ring if and only if R[[x,x−1;α]] is a right zipring.AcknowledgementsThe authors wish to express their sincere thanks to the referee for his/her helpfulcomments and valuable suggestions. Also, we would like to thank the refereefor drawing our attention to the result appears in Proposition 2.11. The authorswould like to thank Dr. Mohamed Farahat, Math. Dept. Fac. of Sci. Al-AzharUniv., for his inspiring suggestions on this work.REFERENCES[1] J. Beachy - W. Blair, Rings whose faithful left ideals are cofaithful, Pacific J. Math.58 (1) (1975), 1–13.[2] W. Cortes, Skew polynomial extensions over zip rings, Int. J. Math. Sci. 10 (2008),1–8.[3] C. Faith, Annihilator ideals, associated primes and Kasch–McCoy commutativerings, Comm. Algebra 19 (1991), 1967–1982.[4] C. Hong - N. Kim - T. Kwak - Y. Lee, Extensions of zip rings, J. Pure and Appl.Algebra 195 (2005), 231–242.ZIP PROPERTY ON MALCEV-NEUMANN SERIES MODULES 123[5] D. Passman, Infinite Crossed Products, Academic Press, 1989.[6] R. Salem - A. Hassanein - M. Farahat, Malcev-Neumann series over zip andweak zip rings, Asian-European Journal of Mathematics 5 (4) (2012), DOI:10.1142/S1793557112500581.[7] C. Sonin, Krull dimension of Malcev-Neumann rings, Comm. Algebra 26 (9)(1998), 2915–2931.[8] J. Zelmanowitz, The finite intersection property on annihilator right ideals, Proc.Amer. Math. Soc. 57 (2) (1976), 213–216.[9] C. Zhang - J. Chen, Zip modules, Northeast J. Math. 24 (2008), 240–256.[10] R. Zhao - Y. Jiao, Principal quasi-baerness of modules of generalized power se-ries, Taiwanese J. Math. 15 (2) (2011), 711–722.REFAAT SALEMMathematics Department, Faculty of Science,Al-Azhar University, Cairo, Egypt.e-mail: refaat salem@cic-cairo.comABDELAZIZ E. RADWANMathematics Department, Faculty of Science,Ain Shams University, Cairo, Egypt.e-mail: zezoradwan@yahoo.comHANAN ABD-ELMALKMathematics Department, Faculty of Science,Ain Shams University, Cairo, Egypt.e-mail: hanan abdelmalk@yahoo.com

Zip Property on Malcev-Neumann Series Modules

Let R be a ring, MR a right R-module, G a totally ordered group, σ a map from G into the group of automorphisms of R which assigns to each x ∈ G an automorphism σ_x ∈ Aut(R), τ a map from G × G to U(R) (the group of unit elements of R) and M((G; σ ; τ)) the Malcev-Neumann series module. Then, under some certain conditions, we show that MR is a right zip R-module if and only if M((G; σ ; τ))_{R((G;σ ;τ))} is a right zip R((G; σ ; τ))-module, where R((G; σ ; τ)) is the Malcev-Neumann series ring.<br /

Zip Property on Malcev-Neumann Series Modules

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