EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 4, 2023, 2156-2168 ISSN 1307-5543 – ejpam.com Published by New York Business Global Generalized Reflexive Structures Properties of Crossed Products Type Eltiyeb Ali1,2 1 Department of Mathematics, College of Science and Arts, Najran University, KSA 2 Department of Mathematics, Faculty of Education, University of Khartoum, Sudan Abstract. Let R be a ring and M be a monoid with a twisting map f : M ×M → U(R) and an action map ω : M → Aut(R). The objective of our work is to extend the reflexive properties of rings by focusing on the crossed product R ∗M over R. In order to achieve this, we introduce and examine the concept of strongly CM -reflexive. Although a monoid M and any ring R with an idempotent are not strongly CM -reflexive in general, we prove that R is strongly CM -reflexive under some additional conditions. Moreover, we prove that if R is a left p.q.-Baer (semiprime, left APP -ring, respectively), then R is strongly CM -reflexive. Additionally, for a right Ore ring R with a classical right quotient ring Q, we prove R is strongly CM -reflexive if and only if Q is strongly CM -reflexive. Finally, we discuss some relevant results on crossed products. 2020 Mathematics Subject Classifications: 16S36, 16N60, 16U99. Key Words and Phrases: Left p.q.-Baer-ring, crossed product monoid R ∗M , CM -quasi Ar- mendariz ring, strongly CM -reflexive ring. 1. Introduction Unless otherwise stated, we assume that R is an associative ring with identity and M is a monoid. The concept of reflexive properties of rings was first studied by Mason [1]. In particular, a right ideal I of R is said to be reflexive if xRy ⊆ I implies yRx ⊆ I for any x, y ∈ R. This concept is also specialized to the zero ideal of a ring, where a ring R is said to be reflexive if its zero ideal is reflexive. Moreover, a ring R is called completely reflexive if xy = 0 implies yx = 0 for any x, y ∈ R. It is worth noting that reduced rings are completely reflexive, and every completely reflexive ring is semicommutative, as shown in the literature [1]. Several authors have discussed extensions of reflexive rings, including strongly reflexive rings, strongly M -reflexive rings, Armendariz rings, reversible rings, and reflexive on skew monoid rings, in numerous publications (see, for example, [2], [3], [4] and [5]). According to [6], a ring R is said to be anM -Armendariz ring of crossed product type relative to the given DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4918 Email addresses: eltiyeb76@gmail.com, emali@nu.edu.sa (E. Ali) https://www.ejpam.com 2156 © 2023 EJPAM All rights reserved. E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2157 twisting f and action ω, or an M -quasi Armendariz ring (or simply a CM -Armendariz ring or CM -quasi Armendariz ring, respectively), if for any ϕ = ∑n i=1 aigi, ψ = ∑m j=1 bjhj ∈ R ∗M such that ϕψ = 0 (resp., ϕ(R ∗M)ψ = 0), it follows that aiωgi(bj) = 0 (resp., aiRωgil(bj) = 0) for all i, j and all gi, hj , l ∈M . The focus of this paper is on investigating strongly CM -reflexive rings, which are a reflexive-like property defined for the monoid crossed product R ∗M with respect to the given twisting map f and action map ω. This concept is a generalization of several other reflexive properties, including reflexive rings, strongly reflexive rings, strongly M -reflexive rings, and skew monoid rings. The paper is devoted to presents several results, including, if R is a semiprime, then R is strongly CM -reflexive for a u.p.-monoid M . Also, if R is a left p.q.-Baer (semiprime, left APP -ring, respectively), then R is strongly CM -reflexive for a strictly totally ordered monoid. Additionally, if R is an M -compatible ring and M is a monoid with twisting f and action ω as above, then for any reduced ideal I of R such that R/I is strongly CM - reflexive, then R is strongly CM -reflexive. Moreover, for a right Ore ring R with classical right quotient ring Q, we show that R is strongly CM -reflexive if and only if Q is strongly CM -reflexive. Finally, we discuss example and some results in the subject. To begin with, we introduce some notions and notations relevant to this paper. Let ω : M → Aut(R) be a monoid homomorphism. For h ∈M , we denote by ωh the automorphism ω(h). The crossed product R ∗M over R is defined as the set of all finite sums R ∗M = {xhh|xh ∈ R, h ∈ M}, where addition is defined component-wise and multiplication is defined using the distributive law and two rules known as action and twisting. Specifically, for l, h ∈ M and x ∈ R, we have hx = ωh(x)h and l h = f(l, h)l h, where f : M ×M → U(R) is a twisted function and U(R) denotes the set of units of R. Here, the twisted function f and the action ω of M on R satisfy the following conditions: ωl(ωh(x)) = f(l, h)ωl(ωh(x)f(l, h) −1), ωl(f(h, k))f(l, hk) = f(l, h)f(l h, k), f(1, l) = f(l, 1) = 1 for all l, h, k ∈ M . It is worth noting that the monoid crossed product is a general ring construction. Given a monoid crossed product R ∗M with twisting f and action ω, if the twisting f is trivial, (i.e., f(a, b) = 1) for all a, b ∈M , then R ∗M is the skew monoid ring R ∗M . If both the twisting f and the action ω are trivial, then R ∗M is a monoid ring denoted by R[M ] (see [7] and [8]). A monoid M is said to be a u.p.-monoid (unique product monoid) if, for any two nonempty finite subsets X and Y of M , there exists a unique element h ∈M that can be written in the form h = uv with u ∈ X and v ∈ Y . An ordered monoid (M,⪯) is said to be strictly ordered if the following condition holds: whenever g, k, h ∈ M with g ≺ k, it follows that gh ≺ kh and hg ≺ hk. 2. Generalized Reflexive rings of crossed product type In this section, we will discuss the concept of strongly reflexive properties in the context of a monoid of crossed product R∗M , where R is a ring and M is a monoid with a twisting map f :M ×M → U(R) and an action map ω :M → Aut(R). E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2158 Definition 1. A ring R is said to be strongly M -reflexive of crossed product type with respect to the given twisting map f and action map ω (or simply, strongly CM -reflexive) if for any ϕ = c1l1+ c2l2+ · · ·+ cnln and ψ = a1h1+a2h2+ · · ·+amhm ∈ R ∗M satisfying that ϕ(R ∗M)ψ = 0 implies that ciωli(ωg(Raj)) = 0, then ψ(R ∗M)ϕ = 0 for each i, j and for all g, li, hj ∈M. Remark 1. (1) If a ring R is strongly CM -reflexive with a trivial twisting map f , then we refer to the monoid M as a skew strongly M -reflexive ring. If R is strongly CM - reflexive with a trivial action map ω, then we call R a strongly TM -reflexive (i.e., twisted strongly M -reflexive) ring. Note that when both f and ω are trivial, R is simply strongly M -reflexive. In particular, if M = (N ∪ {0},+) and both f and ω are trivial, then R is strongly CM -reflexive if and only if R is strongly reflexive. (2) If R is a strongly CM -reflexive ring with a trivial twisting map f , then any M - invariant subring S (i.e., ωg(S) ⊆ S for all g ∈M) of R is strongly CM -reflexive. An ideal I of a ring R is considered to be right s-unital if there exists an element e ∈ I for every t ∈ I such that te = t. A ring is referred to as a left APP -ring if the left annihilator lR(Rt) is right s-unital as an ideal of R for any element t ∈ R. In their work [9], Nasr-Isfahani and Moussavi introduced a ring R with an endomor- phism ω and defined it as ω-weakly rigid if the condition cRt = 0 holds if and only if c ω(Rt) = 0 for any c, t ∈ R. It is worth noting that the category of ω-rigid rings and ω-compatible rings is a limited one, and it is evident that every ω-compatible ring falls un- der the category of ω-weakly rigid rings. However, there exist several classes of ω-weakly rigid rings that do not belong to the category of ω-compatible rings. By [10], R is α- rigid if and only if R is α-compatible and reduced. According to [9], any prime ring that has an automorphism ω is considered to be ω-weakly rigid. If a monoid homomorphism ω : M → Aut(R) is weakly-rigid (compatible), it means that the ring R is also weakly rigid (compatible) with respect to each g ∈M under the automorphism ωg. Lemma 1. [11, Lemma 1.1]. If M is a u.p.-monoid, then M is cancellative (i.e., for ℓ, h, λ ∈M, if ℓλ = hλ or λℓ = λh, then ℓ = h). Lemma 2. Suppose R is a ring and M is a u.p.-monoid with a twisting map f :M×M → U(R) and an action map ω :M → Aut(R). If R is an M -rigid ring, then the monoid ring R ∗M is reduced. Proof. Assume that ϕ = c1h1 + · · · + cnhn ∈ R ∗M satisfies ϕ2 = 0. According to Proposition 2.2 [6], R is CM -Armendariz, this implies ciωhi(bj)f(li, hj))(lihj) = 0 for all i and j, by Lemma 1, M is a cancellative so ciωhi(bj) = 0 . As R is an M -rigid, then R is a reduced, we can conclude that ci = 0 for all 1 ≤ i ≤ n. Consequently, ϕ = 0, and hence R ∗M is a reduced. Theorem 1. Let R be a semiprime ring and M be a u.p.-monoid with a twisting map f : M ×M → U(R) and an action map ω : M → Aut(R). If R is an M -compatible ring, then R is strongly CM -reflexive. E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2159 Proof. The evidence has been modified from the Theorem 1.1 of [12]. Let ϕ = c1l1 + c2l2 + · · ·+ cnln, ψ = a1h1 + a2h2 + · · ·+ amhm ∈ R ∗M satisfy ϕ(R ∗M)ψ = 0. Then for any r ∈ R and g ∈M, we have (c1l1 + c2l2 + · · ·+ cnln)gr(a1h1 + a2h2 + · · ·+ amhm) = 0. (2.1) We will employ mathematical induction on n to demonstrate that ciRωli(ωg(aj)) = 0 for all 1 ≤ i ≤ n, 1 ≤ j ≤ m, and for any g ∈M . This can be achieved by utilizing the fact thatM is a compatible monoid. If we take n = 1, then we have (c1l1)gr(a1h1+a2h2+· · ·+amhm) = 0. Therefore, for each 1 ≤ j ≤ m, we have c1Rωl1(ωg(aj))f(li, hj))(lihj) = 0. By Lemma 1, M is a cancellative, this means l1hi ̸= l1hj for any i and j with 1 ≤ i ̸= j ≤ m. Thus, c1Rωl1(ωg(aj)) = 0. For the case where n ≥ 2, we can use the assumption that M is a uniquely presented monoid to find s and t with 1 ≤ s ≤ n and 1 ≤ t ≤ m such that lsght is uniquely represented by considering two subsets K = {l1g, l2g, . . . , lng} and H = {h1, h2, . . . , hm} of the monoid M . Without loss of generality, we may assume that s = 1 and t = 1. From Eq. (2.1), we can deduce that c1ωl1(ωg(Ra1))f(l1, h1)(l1h1) = 0, which implies that c1Rωl1(ωg(a1)) = 0. Since ωg and ωl1 are automorphisms of R, we have c1Rωl1(ωg(a1)) = 0. As a result, for every z ∈ R, we have c1Rωl1(ωg(a1za1))f(l1, h1) = 0, which implies that 0 = (c1l1 + c2l2 + · · ·+ cnln)gra1z(gra1z(a1h1 + a2h2 + · · ·+ amhm) = (c2l2 + · · ·+ cnln)gr(a1za1h1 + a1za2h2 + · · ·+ a1zamhm). By applying the induction hypothesis, it follows that ciωli(ωg(ra1zaj)) = 0 for all 2 ≤ i ≤ n and 1 ≤ j ≤ m. Thus, we have ciRωli(ωg(a1))Rωli(ωg(a1)) = 0, which implies that ciRωli(ωg(a1)) = 0 for all 2 ≤ i ≤ n, as R is a semiprime ring. Therefore, we have ciRωli(ωg(a1)) = 0 for all 1 ≤ i ≤ n. As a result, the Eq. (2.1) becomes (c1l1 + c2l2 + · · · + cnln)gr(a2h2 + · · · + amhm) = 0. We can repeat this process to show that ciωli(ωg(raj)) = 0 for all g ∈ M and all i, j. This shows that ciRωli(ωg(aj)) = 0. Consequently, we can see that ajRωhj (ωg(ci)) = 0 for all g ∈M , 1 ≤ j ≤ m, and 1 ≤ i ≤ n. Therefore, R is strongly CM -reflexive. The following example demonstrates the existence of a ring R over a field F that is not strongly CM -reflexive. Example 1. Let M be a monoid with at least two elements, and let S = M2(F ) be the matrix ring over a field F with a twisting map f :M ×M → U(R), then S is not strongly CM -reflexive. Solution. Take e ̸= h ∈M, we define ω :M → Aut(S) by ωh (( a d 0 c )) = ( a −d 0 c ) . If the twisting map f is trivial (i.e., f(x, y) = 1 for all x, y ∈ M), then the ring S is not strongly CM -reflexive. To see this, consider ϕ = E12e+E11h and ψ = (E11+E12)h ∈ S∗M . For φ = (E11 + E22)h ∈ S ∗M , we can easily verify that ϕφψ = 0. However, we have ψφϕ ̸= 0, which implies that S is not strongly CM -reflexive. E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2160 A ring R is categorized as a right PP -ring or left PP -ring if the right or left annihilator of an element in R, respectively, is generated by an idempotent. A (quasi-) Baer ring is one where the right annihilator of every nonempty subset or every right ideal of R is generated by an idempotent. Principally quasi-Baer rings, introduced by Birkenmeier et al. [13], extend the concept of quasi-Baer rings. A ring R is referred to as left principally quasi- Baer or simply left p.q.-Baer if the left annihilator of a principal left ideal in R is generated by an idempotent. It is important to note that biregular rings and quasi-Baer rings are examples of left p.q.-Baer rings. For more information and examples of left p.q.-Baer rings, see Birkenmeier et al. ([13], [14]) and Liu [15]. Since right PP -rings and left p.q.-Baer rings both fall under the category of left APP [16], the following results can be deduced. Theorem 2. Suppose R is a reduced ring, M is a strictly totally ordered monoid with a twisting map f : M ×M → U(R) and an action map ω : M → Aut(R) that is compatible with the multiplication in M . If R is a left p.q.-Baer ring, then R is strongly CM -reflexive. Proof. The proof is a variant of the proof given in Proposition 2.9 [17]. Let ϕ = c1l1+c2l2+ · · ·+cnln, ψ = a1h1+a2h2+ · · ·+amhm ∈ R∗M satisfy ϕ(R∗M)ψ = 0. Since M is a strictly totally ordered monoid, we can assume that li ⪯ lj and hi ⪯ hj whenever i < j. Now, we claim ciωli(ωg(Raj)) = 0 for all i, j. Let r be an element of R. Then, we have ϕ(re)ψ = 0 since ϕ(R ∗M)ψ = 0. Thus, we have 0 = ϕ(re)ψ = c1rf(l1, e)a1f(l1, h1)l1h1 + · · ·+ [cnrf(ln, e)am−2f(ln, hm−2)lnhm−2 + cn−1rf(ln−1, e)am−1f(ln−1, hm−1)ln−1hm−1 + cn−2rf(ln−2, e)lmf(ln−2, hm)ln−2hm] + [cnrf(ln, e)am−1f(ln, hm−1)lnhm−1 + an−1rf(ln−1, e)amf(ln−1, hm)ln−1hm] + cnrf(ln, e)amf(ln, hm)lnhm. (2.2) It follows that cnrf(ln, e)amf(ln, hm) = 0 since lnhm is of highest order in the lih ′ js. Hence cnrf(ln, e)am = 0. This shows that cn ∈ ℓR(Rf(ln, e)am) = ℓR(Ram). Hence, ℓR(Ram) = Rem for some idempotent em by hypothesis. Replacing r by rem in Eq. (2.2) we obtain 0 = c1remf(l1, e)a1f(l1, h1)l1h1 + · · ·+ [cnremf(ln, e)am−2f(ln, hm−2)lnhm−2 +cn−1remf(ln−1, e)am−1f(ln−1, hm−1)ln−1hm−1]+cnremf(ln, e)am−1f(ln, hm−1)lnhm−1(2.3) So cnremf(ln, e)am−1f(ln, hm−1) = 0, because lnhm−1 is of highest order in {lihj |1 ≤ i ≤ n, 1 ≤ j ≤ m} {ln−1hm, lnhm}. Hence cnremf(ln, e)am−1 = 0. Since Rem is an ideal of R and em ∈ Rem, we have emr ∈ Rem and thus emr = emrem for all r ∈ R. On the other hand, we also have cn = cnem since cn ∈ ℓR(Ram) = Rem. Hence cnrf(ln, e)am−1 = cnemrf(ln, e)am−1 = cnemremf(ln, e)am−1 = cnremf(ln, e)am−1 = 0. This implies that cn ∈ ℓR(Ram + Ram−1), and hence ℓR(Ram + Ram−1) = Rem−1 for some idempotent em−1 ∈ R since R is a left p.q.-Baer ring. Replacing r by rem−1 in equation (2.3) we obtain cnrem−1f(ln, e)am−2f(ln, hm−2) = 0 in the same way as above. This shows that cn ∈ ℓR(Ram+Ram−1 +Ram−2). Continuing this process we obtain cnRat = 0 for all t = 1, 2, . . . ,m. So, we have (c1l1+c2l2+ · · ·+cn−1ln−1)(R∗M)(a1h1+a2h2+ · · ·+amhm) = 0. Using induction on m+n, we obtain ciωli(ωg(Raj)) = 0 for all i, j. So it is easy to see that ajωhj (ωg(Rci)) = 0 by a reduced ness. Therefore, R is strongly CM -reflexive. If N is an ideal of the monoid M with twisting f : M × M → U(R) and action E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2161 ω : M → Aut(R), then the restrictions f |N×N : N ×N → U(R) and ω|N : N → Aut(R) are induced twisting and action. Proposition 1. Let R be an M -compatible ring and M be a commutative, cancellative monoid and N be an ideal of M with a center element λ. If R is strongly CN -reflexive, then R is strongly CM -reflexive. Proof. Let ϕ = ∑n i=1 cili, ψ = ∑m j=1 ajhj ∈ R ∗M satisfying ϕφψ = 0 for any φ =∑v r=1 ℓrgr ∈ R ∗M. Since λ ∈ N is a center element, this implies that λl1, λl2, . . . , λln, λg1λ, λg2λ, . . . , λgvλ, h1λ, h2λ, . . . , hmλ ∈ N, such that λli ̸= λlj , λgiλ ̸= λgjλ and hiλ ̸= hjλ for all i ̸= j. Then, we have ϕ1φ1ψ1 = n∑ i=1 m∑ j=1 v∑ r=1 (ciωli(ℓrωλ(aj)))f(liλ, hj)(λ 2ligrhjλ 2) = 0. Since φ, ϕ and ψ are nonzero in R∗M, so ϕ1 and ψ1 are nonzero elements in (R∗M)[N ]. Moreover, from ϕφψ = 0 and ω compatible automorphism, λ a center element of N one can easily obtain that ϕ1φ1ψ1 = 0 for any φ1 ∈ (R ∗M)[N ]. Since R is strongly CN - reflexive. Then, ciωli(ωλ(r aj))f(li, hj)(lihj) = 0. So ciωli(ωλ(Raj)) = 0. By a compatible automorphism, we have ajωhj (ωλ(Rci)) = 0. Therefore, R is strongly CM -reflexive. Corollary 1. [4, Proposition 3.1] Let M be a cancellative monoid and N an ideal of M. If R is strongly N -reflexive, then R is strongly M -reflexive. Suppose I is an ideal of R and ω :M → Aut(R) is a monoid homomorphism. We define ω̄ : M → Aut(R/I) as ω̄g(d + I) = ωg(d) + I, where d ∈ R and g ∈ M . It can be shown that ω̄ is a monoid homomorphism. Additionally, the twisting map f : M ×M → U(R) induces a twisting map f̄ :M×M → U(R/I) given by f̄(x, y) = f(x, y)+I. Furthermore, for every ϕ = ∑n i=1 cili ∈ R ∗M , we denote ϕ̄ = ∑n i=1 c̄ili ∈ (R/I) ∗M , where c̄i = ci + I for 1 ≤ i ≤ n. It can be easily verified that the mapping θ : R×M → (R/I)×M defined as θ(ϕ) = ϕ̄ is a ring homomorphism. In a proof presented [4], it was shown that when I is a reduced ideal of R and R/I is strongly M -reflexive, then R is strongly M -reflexive. Similarly, we can establish the following result. Theorem 3. Let M be a u.p.-monoid and I an ideal of R with twisting f :M×M → U(R) and action ω : M → Aut(R). If I is a reduced and R/I is strongly CM -reflexive, then R is strongly CM -reflexive. Proof. Let ϕ = Σni=1cili, ψ = Σmj=1ajhj ∈ R ∗M satisfying ϕ(R ∗M)ψ = 0. We will show that ciωli(ωg(Raj)) = 0 for any i and j. E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2162 Note that in (R/I) ∗M, ϕ̄ = Σni=1c̄ili, ψ̄ = Σmj=1ājhj ∈ (R/I) ∗M, we have 0̄ = ϕ̄((R/I) ∗M)ψ̄ = (c̄1l1 + c̄2l2 + · · ·+ c̄nln)r̄gωli(ωg(ā1h1 + ā2h2 + · · ·+ āmhm))f(li, hj)lihj = (c1 + I)r̄gω̄l1(ωg(a1 + I))f(l1, h1)l1h1 + (c2 + I)r̄gω̄l2(ωg(a2 + I))f(l2, h2)l2h2 + · · ·+ (cn + I)r̄gω̄ln(ωg(am + I))f(ln, hm)lnhm. Thus we have ciωli(ωg(Raj))f(li, hj)(lihj) ⊆ I for all i and j with 1 ≤ i ≤ n and 1 ≤ j ≤ m since R/I is strongly CM -reflexive. By induction on both n and m, considering every g in M , and for 1 ≤ i ≤ n and 1 ≤ j ≤ m. If we take n = 1. Then (c1l1)(R ∗M)(a1h1 + a2h2 + · · · + amhm) = 0. Thus, (c1l1)(rg)(a1h1) + (c1l1)rg(a2h2) + · · ·+ (c1l1)rg(amhm) = c1ωl1(ωg(ra1))f(l1, h1)(l1h1) + c1ωl1(ωg(ra2))f(l1, h2)(l1h2) + · · ·+ c1ωl1(ωg(ram))f(l1, hm)(l1hm) = 0 for any r ∈ R, g ∈ M. By Lemma 1, M is cancellative we have l1hi ̸= l1hj for any i and j with 1 ≤ i ̸= j ≤ m. Then c1ωl1(ωg(raj))f(l1, hj)(l1hj) = 0, j = 1, 2, . . . ,m. Thus, c1ωl1(ωg(Raj)) = 0 for any j. If m = 1, then proof is similar. Now suppose that n ≥ 2 and m ≥ 2. Since M is a u.p.-monoid, there exist i, j with 1 ≤ i ≤ n and 1 ≤ j ≤ m such that lighj is uniquely presented by consider- ing two subsets K = {l1g, l2g, . . . , lng} and H = {h1, h2, . . . , hm} of the monoid M . Without loss of generality, we may assume that i = 1 and j = 1. We can deduce that c1ωl1(ωg(Ra1))f(l1g, h1)l1(gh1) = 0, which implies that c1ωl1(ωg(Ra1)) = 0. Since ωg and ωl1 are automorphisms of R, we have c1ωl1(Ra1) = 0. Let b = ckraq, where r ∈ R, 1 ≤ k ≤ n, 1 ≤ q ≤ m. Then b ∈ I. Since (a1bc1) 2 = 0 and I is reduced, we have a1bc1 = 0. Thus, (a1bc2l2 + a1bc3l3 + · · ·+ a1bcnln)(R ∗M)(a1h1 + a2h2 + · · ·+ amhm) = (a1bλ)(c1l1+ c2l2+ · · ·+ cnln)(R ∗M)(a1h1+ a2h2+ · · ·+ amhm) = 0. By induction, we have a1bciωl1(ωg(Raj)) = 0 for 2 ≤ i ≤ n and 1 ≤ j ≤ m. Thus, (a1bciR)2 = 0. Since I is reduced and ω is automorphism, it follows that a1bciωli(R) = 0. Note that bciωli(Ra1) ⊆ I. Thus bciωli(Ra1) = 0 for any i. Now we have (bc1l1+ bc2l2+ · · ·+ bcnln)(R ∗M)(a1h1+a2h2+ · · ·+amhm) = (bλ)(c1l1+ c2l2+ · · ·+ cnln)(R ∗M)(a1h1 + a2h2 + · · ·+ amhm) = 0. By applying the induction hypothesis, it follows that bciωli(ωg(Raj)) = 0 for all 1 ≤ i ≤ n and 2 ≤ j ≤ m. Thus, we have ciωli(ωg(raj)) = 0 for all i, j and all r ∈ R. Particularly, we have bckωlk(ωg(raq)) = 0 and so b2 = 0. Thus b = 0. This shows that ckωlk(ωg(Raq)) = 0 for any 1 ≤ k ≤ n and 1 ≤ q ≤ m. Consequently, we can see that ajωhj (ωg(Rci)) = 0 for all g ∈ M , 1 ≤ j ≤ m, and 1 ≤ i ≤ n. Therefore, R is strongly CM -reflexive. The notion of complete M -compatibility is important in the following result [18]. Corollary 2. Assuming R is a ring that is completely M -compatible, where M is a monoid with twisting f :M ×M → U(R) and action ω :M → Aut(R), and I is an ideal of R such that I is reduced and R/I is CM -quasi-Armendariz, then R is strongly CM -reflexive. Proof. As CM -quasi-Armendariz rings are strongly CM -reflexive, the result can be E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2163 obtained from Theorem 3. Corollary 3. Suppose that R is a completely M -compatible ring, where M is a monoid with twisting f : M ×M → U(R) and action ω : M → Aut(R). Let I be an ideal of R such that I is reduced and R/I is CM -Armendariz. Then, R is strongly CM -reflexive. Proof. Since CM -Armendariz is a CM -quasi-Armendariz, the result can be derived from Corollary 2. Proposition 2. Assuming R is a ring that is both M -compatible and CM -quasi-Armendariz, where M is a monoid with twisting f :M ×M → U(R) and action ω :M → Aut(R), then R is strongly CM -reflexive if and only if R ∗M is strongly CM -reflexive. Proof. To prove a necessary condition is sufficient. Let ϕ = Σni=1cili, ψ = Σmj=1ajhj ∈ R ∗M satisfying ϕ(R ∗M)ψ = 0. Since R is CM -quasi-Armendariz, we have ciωli(ωg(Raj))f(li, hj)(lihj) = 0 for all i, j. This implies that ciωli(ωg(Raj)) = 0 for all i, j since R is M -compatible. Because R is a reflexive ring, ajRci = 0. Then, ajωhj (ωg(Rci)) = 0 for all i, j, and hence for any r ∈ R, g ∈M, we have ψ(R ∗M)ϕ = Σmj=1Σ n i=1ajωhj (ωg(r ci))f(hj , li)(hjli) = 0. Thus, ajωhj (ωg(r ci)) = 0 since R is M -compatible and CM -quasi-Armendariz. Therefore, R is strongly CM -reflexive. Every left APP -ring is quasi-Armendariz, but not conversely [19, 20]. Proposition 3. Let M be a strictly totally ordered monoid with twisting f : M ×M → U(R) and action ω : M → Aut(R). Let R be an M -compatible left APP -ring. Then R is strongly CM -reflexive if and only if R ∗M is strongly CM -reflexive. Proof. If R is a left APP -ring, then it is M -quasi-Armendariz [21]. Therefore, the result follows from Proposition 2. Corollary 4. Let R be a ring, M be a monoid with twisting f : M ×M → U(R) and action ω :M → Aut(R). If R is a reduced, then R is strongly CM -reflexive. Proof. Since R is reduced, it is quasi-Armendariz. Therefore, the result can be derived from Proposition 2. 3. Some results on ring extensions of Crossed product type Let ∆ be a multiplicative monoid consisting of central regular elements of R. Then, the set ∆−1R := {u−1c|u ∈ ∆, c ∈ R} forms a ring. Suppose ω : M → Aut(R) is a monoid homomorphism such that ωh(∆) ⊆ ∆ for every h ∈M . Then, ω can be extended to ω̄ : M → Aut(∆−1R) defined by ω̄h(u−1c) = ωh(u) −1ωh(c). If f : M ×M → U(R) is a twisted function, then it can be viewed as a twisted function from M ×M to U(∆−1R) by noting that U(R) ⊆ U(∆−1R). E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2164 Theorem 4. Assuming R is an M -compatible ring, where M is a cancellative monoid with twisting f :M×M → U(R) and action ω :M → Aut(R), then R is strongly CM -reflexive if and only if ∆−1R is strongly CM -reflexive, where ∆ is the multiplicative subset of R consisting of all elements that are not zero divisors modulo M . Proof. It is enough showing necessary. Assume that R is strongly CM -reflexive. Let ϕ = Σni=1u −1cili, ψ = Σmj=1v −1ajhj be elements in ∆−1R∗M satisfying ϕφψ = 0, where φ = Σqk=1λ −1bkℓk is any nonzero element in ∆−1R∗M. Then, we have α = (unun−1 . . . u1)ϕ, θ = (λqλq−1 . . . λ1)φ, β = (vmvm−1 . . . v1)ψ are in R ∗M. Since R is strongly CM -reflexive and αθβ = 0 we have (unun−1 . . . u1u −1 i ci)ωli(ωg(b(vmvm−1 . . . v1v −1 j )aj))f(li, hj)(lihj)(vjui) −1 = 0 for all i, j and b ∈ R. It follows that ciωli(ωg(Raj))f(li, hj)(lihj) = 0 for any g ∈ M, because ∆ is a multiplicative monoid consisting of central regular elements of R and all ui, vj , λk ∈ ∆. Hence, (u−1 i ci)ωli(ωg(Rv −1 j )aj)) = ciωli(ωg(Raj))(ωli(vj)ui) −1 = 0 for all i, j and ω is automorphism. Therefore, ∆−1R is strongly CM -reflexive. The following statement describes how the strongly CM -reflexive property of a ring R is related to the property of its subrings, which are created by a central idempotent. Proposition 4. The following conditions are equivalent for a ring R, a monoid M with twisting f : M ×M → U(R), an action ω : M → Aut(R), and a central idempotent e of R such that ωg(e) = e : (1) R is strongly CM -reflexive. (2) eR and (1− e)R are strongly CM -reflexive. Proof. (1) ⇒ (2). It is easy. (2) ⇒ (1). Assume that both eR and (1 − e)R are strongly CM -reflexive. Let ϕ = Σni=1cili, ψ = Σmj=1ajhj ∈ R ∗M satisfying ϕ(R ∗M)ψ = 0. Let ϕ1 = Σni=1e cili, ψ1 = Σmj=1e ajhj , ϕ2 = Σni=1(1− e)cili, ψ2 = Σmj=1(1− e)ajhj . clear that ϕ1, ψ1 ∈ (eR) ∗M and ϕ2, ψ2 ∈ ((1− e)R) ∗M. Since e is a central idempotent of R such that ωg(e) = e for each g ∈M and for any r ∈ R we have ϕ1((eR) ∗M)ψ1 = ec1(er)ωl1(ωg(ea1))f(l1, h1)l1h1 + · · ·+ ecn(er)ωln(ωg(eam))f(ln, hm)lnhm = ec1(er)ωl1(ωg(e)ωl1(ωg(a1))f(l1, h1)l1h1 + · · · + ecn(er)ωln(ωg(e))ωln(ωg(am))f(ln, hm)lnhm = ec1e(er)ωl1(a1)f(l1, h1)l1h1 + · · ·+ ecne(er)ωln(am)f(ln, hm)lnhm = ec1e 2(r)ωl1(a1)f(l1, h1)l1h1 + · · ·+ e2cn(r)ωln(am)f(ln, hm)lnhm = ec1e(r)ωl1(a1)f(l1, h1)l1h1 + · · ·+ ecn(r)ωln(am)f(ln, hm)lnhm = e2c1rωl1(a1)f(l1, h1)l1h1 + · · ·+ e2cnrωln(am)f(ln, hm)lnhm = ec1rωl1(a1)f(l1, h1)l1h1 + · · ·+ ecnrωln(am)f(ln, hm)lnhm = e[c1rωl1(a1)f(l1, h1)l1h1 + · · ·+ cnrωln(am)f(ln, hm)lnhm] = eϕ(R ∗M)ψ = 0, E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2165 ϕ2((1− e)R ∗M)ψ2 = (1− e)c1((1− e)r)ωl1(ωg((1− e)a1))f(l1, h1)l1h1 + · · · + (1− e)cn((1− e)r)ωln(ωg((1− e)(1− e)am))f(ln, hm)lnhm = (1− e)c1((1− e)r)ωl1((1− e)a1)f(l1, h1)l1h1 + · · ·+ (1− e)cn(1− e)rωln((1− e)am)f(ln, hm)lnhm = (1− e)[c1rωl1(a1)f(l1, h1)l1h1 + · · ·+ cnrωln(am)f(ln, hm)lnhm] = (1− e)ϕ(R ∗M)ψ = 0. Because eR and (1 − e)R are strongly CM -reflexive subrings of R, we conclude that ψ1((eR) ∗M)ϕ1 = 0, ψ2(((1− e)R) ∗M)ϕ2 = 0. Therefore, we have ψ(R ∗M)ϕ = ψ1((eR) ∗M)ϕ1 + ψ2(((1− e)R) ∗M)ϕ2 = eψ(R ∗M)ϕ+ (1− e)ψ(R ∗M)ϕ = 0. Therefore, R is strongly CM -reflexive, which concludes the proof. Proposition 5. Let R be a ring and M is a strictly ordered monoid with a twisting f : M×M → U(R) and an action ω :M → Aut(R). Assume that R is CM -quasi-Armendariz. Let e be a nonzero idempotent in R such that ωg(e) = e for all g ∈ M . Then, the subring eRe is strongly CM -reflexive. Proof. The proof is a variant of the proof given in Proposition 2.9 [17]. Let ϕ = c1l1 + c2l2 + · · ·+ cnln and ψ = a1h1 + a2h2 + · · ·+ amhm ∈ (eRe) ∗M satisfy ϕ((eRe) ∗ M)ψ = 0. Since M is a strictly totally ordered monoid, we can assume that li ⪯ lj and hi ⪯ hj whenever i < j. Since R is CM -quasi-Armendariz, then so is eRe. Thus, we have ciωli(ωg((eRe)aj))f(li, hj)(lihj) = 0 for all i, j. This implies that ciωli(ωg((eRe)aj)) = 0 for all i, j since R is M -compatible and ω is an automorphism. Therefore, by Proposition 2, eRe is strongly CM -reflexive. Corollary 5. [20, Proposition 3.7] Let e ∈ R be an idempotent. If R is a left APP , then eRe is a left APP -ring. Corollary 6. [22, Corollary 3.19] Let M be a strictly totally ordered monoid and ω :M → End(R) a monoid homomorphism. Assume that e be an idempotent. If R is left APP , then eRe is (M,ω)-quasi-Armendariz. Proposition 6. Let M be a strictly totally ordered monoid with twisting f : M ×M → U(R) and action ω : M → Aut(R). Assume that e be an idempotent. If R is a left APP , then eRe is strongly CM -reflexive. Proof. By Corollary 5, eRe is a left APP . So, eRe is (M,ω)-quasi-Armendariz by Corollary 6. Thus, the result follows from Proposition 5. Let I be an index set and Ri be a ring for each i ∈ I. Let M be a strictly ordered monoid and ωi : M → End(Ri) a monoid homomorphism. Then the mapping ω : M → End( ∏ i∈I Ri) is a monoid homomorphism given by ωg({ri}i∈I) = {(ωi)g(ri)}i∈I} for all g ∈M. E. Ali / Eur. J. Pure Appl. Math, 16 (4) (2023), 2156-2168 2166 Proposition 7. Let Ri be a ring for each i in a finite index set I, and let M be a monoid with a twisting f : M ×M → ⋃ i∈I U(Ri) and an action ωi : M → Aut(Ri) on each Ri. Suppose that each Ri is strongly CM -reflexive. Then, the direct product R = ∏ i∈I Ri, equipped with the product action ω = ∏ i∈I ω i, is strongly CM -reflexive. Proof. Let R = ∏ i∈I Ri be the direct product of rings (Ri)i∈I and Ri is is strongly CM - reflexive for each i ∈ I. Denote the projection R→ Ri as Πi. Suppose that ϕ, ψ ∈ R∗M are such that ϕ(R ∗M)ψ = 0. Set ϕi = ∏ i ϕ, ψi = ∏ i ψ and φi = ∏ i φ. Then ϕi, ψi ∈ Ri ∗M. For any u, v ∈ M, assume ϕ(u) = (cui )i∈I , ψ(v) = (avi )i∈I . Now, for any r ∈ R and any g ∈M, ϕ(R ∗M)ψ = ∑ (u,v)∈Xs(ϕ,crψ) ϕ(u)ωu(ωg(rψ(v)))f(um, vn)umvn = ∑ (u,v)∈Xs(ϕ,crψ) (cui )i∈I(( ∏ i∈I ωi)u(ωg(ria v i ))f(u i m, v i n)u i mv i n)i∈I = ∑ (u,v)∈Xs(ϕ,crψ) (cui )i∈I( ∏ i∈I ωiu)(ωg(ria v i )f(u i m, v i n)u i mv i n)i∈I = ∑ (u,v)∈Xs(ϕ,crψ) (cui ω i u(ωg(ria v i ))f(ϕi, ψi)u i mv i n)i∈I = ∑ (u,v)∈Xs(ϕ,crψ) (ϕi(u)ω i u(riψi(v)))f(ϕi, ψi)u i mv i n)i∈I = ( ∑ (u,v)∈Xs(ϕ,crψ) ϕi(u)ω i u(ωg(riψi(v)) ) f(ϕi, ψi)u i mv i n)i∈I = ( ∑ (u,v)∈Xs(ϕi,criψi) ϕi(u)ω i u(ωg(riψi(v))) ) f(ϕi, ψi)u i mv i n)i∈I = (ϕi(Ri ∗M)ψi)i∈I . Since ϕ(R ∗M)ψ = 0, we have ϕi(Ri ∗M)ψi = 0. Now it follows ϕi(u)ωiu(ωg(riψi(v))) = 0 for any r ∈ R, any u, v, g ∈M and any i ∈ I, since Ri is strongly CM -reflexive. Hence, for any u, v ∈M, ψ(v)ωv(ωg(rϕ(u))) = (ψi(v)ω i v(ωg(riϕi(u))))i∈I = 0 since I is finite. Thus, ψ(v)ωv(ωg(rϕ(u))) = 0 by the compatibility of ω. Therefore, ψ(R ∗M)ϕ = 0. This means that R is strongly CM -reflexive. Theorem 5. Assuming that R is an M -compatible ring and M is a cancellative monoid with a twisting map f : M × M → U(R) and an action map ω : M → Aut(R), and considering R as a right Ore ring with the classical right quotient ring Q, the R is strongly CM -reflexive if and only if Q is strongly CM -reflexive. Proof. It is enough showing necessary. Assume that R is strongly CM -reflexive. Let ϕ = Σmi=1αili, ψ = Σpk=1γkhk be elements in Q∗M satisfying ϕφψ = 0, where φ = Σnj=1βjgj is any nonzero element in Q ∗M. By Proposition 2.1.16 [23], we may assume that αi = aiu −1, βj = bjv −1 and γk = ckw −1 with regular u, v, w ∈ R. Also, Proposition 2.1.16 [23], for each j and k, there exist dj , ek ∈ R and regular s, t ∈ R such that u−1bj = djs −1 REFERENCES 2167 and (vs)−1ck = ekt −1. Suppose ϕ1 = Σmi=1aili, φ1 = Σnj=1bjgj , φ2 = Σnj=1djgj , ψ1 = Σpk=1ckhk, ψ2 = Σpk=1ekhk ∈ R ∗M. Since M is a cancellative monoid by Lemma 1. Thus, gish1 ̸= gjsh1 for gi ̸= gj . Then, we have 0 = ϕφψ = Σmi=1Σ p k=1(aiu −1)ωli(ωg(Rckw −1)) f(li, hk)(lihk) = Σmi=1Σ p k=1aiωli(ωg(Rek))f(li, hk)(lihk)(ωli(t)w) −1 = 0 = ϕ1φ2ψ2(wt) −1. Therefore, ϕ1φ2ψ2 = 0. Since R is strongly CM -reflexive, then ψ2φ2ϕ1 = 0. This implies that ϕ1uφ2ψ2 = ϕ1φ1sψ2 = 0 since u−1bj = djs −1, then sψ2φ1ϕ1 = 0 and (vs)ψ2φ1ϕ1 = 0, so ψ1φ1ϕ1 = 0 since (vs)−1ck = ekt −1. Using Proposition 2.1.16 [23] again, for each i, j there exist φi, ϕj ∈ R ∗M and regular element q, p ∈ R such that w−1bj = ϕjq −1 and (vq)−1ai = φip −1. Let ϕ2 = Σmi=1φili, φ3 = Σnj=1ϕjgj . Then, qψ1φ1ϕ1 = Σmi=1Σ p k=1q(ckw −1)ωhk(ωg(Raiu −1))f(hk, li)(hkli) = Σmi=1Σ p k=1q(ck)ωhk(ωg(Rai))× (ωhk(u)w) −1 = 0 since ψ1φ1ϕ1 = 0. Thus, for all k, i we have ckωhk(ωg(Rai)) = 0, and it follows that ϕ1wφ3qψ2 = Σmi=1Σ p k=1waiωli(ωg(Rek)) = 0 since w−1bj = ϕjq −1. Then, ψ1φ3ϕ1w = 0 since R is strongly CM -reflexive, and so ψ1φ3ϕ1 = 0. Therefore, ψ1φ3ϕ1p = Σpk=1Σ m i=1ckωhk(ωg(Rai))f(hk, li)(hkli)p = ψ1φ3ϕ2(vq) = Σpk=1Σ n j=1ckωhk(ωg(Rdj))(pv) = 0, and thus ψ1φ3ϕ2 = 0. Therefore, ψφϕ = Σpk=1Σ m i=1(ckw −1)ωhk(ωg(Raiu −1)) = Σpk=1Σ m i=1ckωhk(ωg(Rai))(ωhk(u)w) −1 = 0. Thus, ckωhk(ωg(Rai))(up) −1 = 0.Therefore,QisstronglyCM−reflexive. References [1] L. Zhao, X. Zhu, and Q. Gu. Reflexive rings and their extensions. Math. Slovaca, 63(3):417–430, 2013. [2] E. Ali. The reflexive condition on skew monoid rings. Eur. J. Pure Appl. Math, 16(3):1878–1893, 2023. [3] E. Ali, A. Elshokry, and Z. Kui. Strongly α-reversible rings relative to a monoid. Int. J. of Algebra, 8:375–387, 2014. [4] Z. 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