EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 1, 2021, 164-172 ISSN 1307-5543 – ejpam.com Published by New York Business Global Ore Extension Rings Satisfy the Weak PS-Rings Mohamed A. Farahat1,2,∗, Salha T. Al-Bogamy1 1 Department of Mathematics and Statistics, Faculty of Science, Taif University, Taif, El-Haweiah, Kingdom of Saudi Arabia (KSA) 2 Mathematics Department, Faculty of Science, Al-Azhar University, Cairo, Egypt Abstract. The main result of this paper is that: If R is a weak right PS-ring, then A = R[x;α, δ], the Ore extension ring, is a weak right PS-ring whenever the following conditions hold on R is an (α, δ)-compatible NI-ring with nil(R) nilpotent, α(e) = e and δ(e) = 0 for every idempotent e ∈ R. 2020 Mathematics Subject Classifications: 16D25, 16P60, 16W60 Key Words and Phrases: PS-ring, weak PS-ring, Ore extensions 1. Introduction Throughout this article, all rings are associative with unity (R denotes such a ring) and all modules are unital R-modules unless explicitly indicated otherwise. According to Nicholson and Watters [6], MR is called a PS-module if every simple submodule is projective, equivalently if its socle, Soc (MR) , is projective. Examples of PS-modules include nonsingular modules and modules with zero socle. A left PS-module RM is defined analogously. A ring R is said to be a left PS-ring if RR is a PS-module. Equivalently, if the left annihilator of every maximal right ideal of R is a principal left ideal generated by an idempotent. Some examples of PS-rings include semiprime and p.p.-rings are PS-rings. In particular every Baer ring is a PS-ring. The notion of PS-rings is not left-right symmetric (cf. [6]). In [6], the authors proved that, if R is a PS-ring so also are R[x] and R[[x]]. The converse of this result is false in general by the following example: Example 1 ([6], Example 3.2). If R = Z4, then R[x] and R[[x]] are PS-rings but R is not PS-ring. Many authors investigated the behavior of PS-rings with respect to their extensions. Salem et. al., in ([9], 2015), characterized PS-modules over Ore extensions and skew gen- eralized power series extensions. Also, Farahat and Al-Harthy, in ([3], 2017), investigated ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i1.3886 Email addresses: m farahat79@yahoo.com,m.farahat @tu.edu.sa (M. A. Farahat), SalhaAlbogamy@hotmail.com (Salha T. Al-Bogamy) http://www.ejpam.com 164 c© 2021 EJPAM All rights reserved. M. A. Farahat, Salha T. Al-Bogamy / Eur. J. Pure Appl. Math, 14 (1) (2021), 164-172 165 PS-modules over generalized Mal’cev-Neumann series rings. In ([8], 2017), Paykan proved that, under suitable conditions, if R is a right PS-ring, then so the skew inverse power series rings. Recently, Farahat and Al-Bogamy, in ([2], 2018), extend the notation of PS-rings to weak PS-rings. Recall the definition of weak PS-rings from [2]: A ring R satisfies the right weak PS-condition if, the weak annihilator of every maximal right ideal of R is a principal left ideal generated by an idempotent. Similarly, the left weak PS-condition was defined. A ring R satisfies the weak PS-condition if it satisfies both the right and the left weak PS-conditions. The following are some examples of rings satisfy the right weak PS-condition. Example 2 ([2]). 1) Any local ring is a right weak PS-ring. 2) The ring Zpq of integers modulo pq, where p and q are distinct prime numbers, is a reduced (weak) PS-ring. 3) Let F be a field and R = ( F F F F ) is a right weak PS-ring. 4) A semisimple NI ring is a right weak PS-ring. In this paper, we study the transfer of right weak PS-condition between a base ring R and its Ore extension A = R[x;α, δ]. 2. Notations (1) Id (R) denotes idempotents of R. (2) nil (R) denotes nilpotents of R. (3) For a nonempty subset X of R, NR(X) denotes the weak annihilator of X over R, i.e., NR(X) = {a ∈ R |ax ∈ nil (R) for all x ∈ X } . It can be easily shown that ab ∈ nil (R)⇔ ba ∈ nil (R) for all a, b ∈ R. (4) R is NI if nil (R) is a two sided ideal in R. 3. Ore extension rings satisfy the weak PS-condition Ore extensions, named after Øystein Ore (1899–1968), are special types of ring ex- tensions whose properties are relatively well understood. These extensions cover a large class of noncommutative polynomial extensions. are special types of ring extensions whose properties are relatively well understood. The definition of noncommutative polynomial rings with identity was first introduced by Øystein Ore [7]. Ever since the appearance of Ore’s fundamental paper [7], Ore extensions have played an important role in noncom- mutative. Ore extensions have wide applications. Not only do they provide interesting M. A. Farahat, Salha T. Al-Bogamy / Eur. J. Pure Appl. Math, 14 (1) (2021), 164-172 166 examples in noncommutative algebra, they have also been a valuable tool used first by David Hilbert (1862–1943) in the study of the independence of geometry axioms. Let R be a ring with identity 1 and α an endomorphism of R. Then a map δ : R −→ R is called an α-derivation of R if δ (a+ b) = δ (a) + δ (b) and δ (ab) = δ (a) b+ α (a) δ (b) , for all a, b ∈ R. We denote by A = R [x;α, δ] , the Ore extension of R whose elements are polynomials over R, the addition is defined as usual and the multiplication is subject to the relation (Ore commutation rule) xa = α (a)x+ δ (a) , for each a ∈ R. We assume that 1 is the identity element of A = R [x;α, δ] . This means that α(1) = 1 and δ (1) = 0, since x = x1 = α (1)x+ δ (1)⇒ α(1) = 1 and δ (1) = 0. Notation ([5]). For integers i, j with j ≥ i ≥ 0, λji ∈ End(R,+) denotes the map which is the sum of all possible ”words” in α and δ built with i letters of α and j − i letters of δ. For instance λ00 = IdR, λ j j = αj , λj0 = δj and λjj−1 = αj−1δ + αj−2δα+ ...+ δαj−1. Lemma 1 ([5]). For any positive integer n and r ∈ R, we have xnr = n∑ i=0 λni (r)xi. This formula uniquely determines a general product of (left) polynomials in R [x;α, δ] and will be used freely in what follows. The ring-theoretical properties of Ore extension have been investigated by many au- thors (see [9], [8], [5], [1], [4], for instance). There are many other papers addressed δ = 0 and α an automorphism or the case where α is the identity. However the recent surge of interest in quantum groups and quantized algebras has brought renewed interest in general Ore extensions, due to the fact that many of these quantized algebras and their repre- sentations can be expressed in terms of Ore extension rings. When we move from these “unmixed” polynomials to the general case with an endomorphism α and an α-derivation δ, we face a much greater challenge. Annin [1] introduced the notion of (α, δ)-compatibility as follows: Definition 1 ([1]). Given a module MR, an endomorphism α : R −→ R and an α- derivation δ : R −→ R. We say that MR is α-compatible if for each m ∈M and r ∈ R, we have mr = 0⇔ mα(r) = 0. Moreover, we say that MR is δ-compatible if for each m ∈M and r ∈ R, we have mr = 0 =⇒ mδ(r) = 0. If MR is both α-compatible and δ-compatible, we say that MR is (α, δ)-compatible. M. A. Farahat, Salha T. Al-Bogamy / Eur. J. Pure Appl. Math, 14 (1) (2021), 164-172 167 A ring R is called (α, δ)-compatible if RR is an (α, δ)-compatible module. The (α, δ)- compatible condition on the module MR is a natural, independently interesting condition from which we can derive a number of interesting properties, and it will be of invaluable service in the proof of our main results. Remark ([1]). (1) If MR is α-compatible, then MR is αi-compatible for all i ≥ 1, (2) If MR is δ-compatible, then MR is δi-compatible for all i ≥ 1, (3) If MR is (α, δ)-compatible, then for each m ∈ M and r ∈ R, we have mr = 0 =⇒ mλji (r) = 0 for all j ≥ i ≥ 0. In what follows, we characterize Ore extension rings that satisfy the weak PS-condition. We need first the following Lemmas which will help us in our target. Lemma 2. Let R be an (α, δ)-compatible NI ring. If a ∈ nil (R) , then λji (a) ∈ nil (R) for all j ≥ i ≥ 0. Proof. Clearly for any R-endomorphism α, we have αk (nil (R)) ⊆ nil (R) , for any positive integer k. Since R is an (α, δ)-compatible ring, we have also that δk (nil (R)) ⊆ nil (R) . Since R is an NI ring, we conclude that λji (a) ∈ nil (R) for any a ∈ nil (R) and for all j ≥ i ≥ 0. Lemma 3. Let R be an (α, δ)-compatible NI ring with nil (R) nilpotent and f(x) = n∑ i=0 aix i ∈ A = R [x;α, δ] . Then f(x) ∈ nil (A) if and only if ai ∈ nil (R) for all inte- gers 0 ≤ i ≤ n. Proof. (=⇒) Suppose that f(x) ∈ nil (A) . Then there exists some positive integer k such that 0 = (f(x))k = ( a0 + a1x+ a2x 2 + ...+ anx n )k . Then 0 = (f(x))k = “lower terms” + anα n(an)α2n(an)...α(k−1)n(an)xkn. Hence anα n(an)α2n(an)...α(k−1)n(an) = 0, and α-compatiblility of R, gives an ∈ nil (R) . So λji (an) ∈ nil (R) for all j ≥ i ≥ 0. Let Q = a0 + a1x+ a2x 2 + ...+ an−1x n−1. Then we have 0 = (Q+ anx n)k = (Q+ anx n) (Q+ anx n) ... (Q+ anx n)︸ ︷︷ ︸ k-factor = ( Q2 +Qanx n + anx nQ+ anx nanx n ) ... (Q+ anx n) = Qk + ∆, where ∆ ∈ A. Note that the coefficients of ∆ can be written as sums of monomials in ai and λvu(aj), where ai, aj ∈ {a0, a1, a2, ..., an} and v ≥ u ≥ 0, and each monomial has an and λvu(an) as a factor. Since nil (R) is an ideal, we obtain that each monomial of ∆ is in nil (R) and so ∆ ∈ nil (R) [x;α, δ] . Thus we obtain Qk = ( a0 + a1x+ a2x 2 + ...+ an−1x n−1 )k M. A. Farahat, Salha T. Al-Bogamy / Eur. J. Pure Appl. Math, 14 (1) (2021), 164-172 168 = “lower terms” + an−1α n−1(an−1)...α (k−1)(n−1)(an−1)x k(n−1) ∈ nil (R) [x;α, δ] . Therefore an−1α n−1(an−1)...α (k−1)(n−1)(an−1) ∈ nil (R) and so an−1 ∈ nil (R) . By using induction on n we obtain ai ∈ nil (R) for all 0 ≤ i ≤ n. (⇐=) Consider the finite subset S = {a0, a1, a2, ..., an} ⊆ nil (R) . Since R is an NI ring with nil (R) nilpotent, there exist integers ki such that (aiR)ki = 0, 0 ≤ i ≤ n. Let k = k0 + k1 + ...+ kn + 1. Then we have (aiR)k = 0, 0 ≤ i ≤ n. We have (f(x))k = ( a0 + a1x+ a2x 2 + ...+ anx n )k = n∑ i=0 aiλ i 0(a0) + ( n∑ i=0 a0λ i 0(a1) + n∑ i=1 aiλ i 1(a0) ) x + ( n∑ i=0 aiλ i 0(a2) + n∑ i=1 aiλ i 1(a1) + n∑ i=2 aiλ i 2(a0) ) x2 +...+ ( k∑ s=0 ( n∑ i=s aiλ i s(ak−s) )) xk + ...+ anα n(an)xn. We show that the coefficients of (f(x))k can be written as sums of monomials of length k in ai and λvu(aj), where ai, aj ∈ {a0, a1, a2, ..., an} and v ≥ u ≥ 0 are integers. By using (α, δ)-compatiblility of R and (aiR)k = 0, 0 ≤ i ≤ n, we have ai1λ vi2 ui2 (ai2)λ vi3 ui3 (ai3)...λ vik uik (aik) = 0, where {ai1 , ai2 , ..., aik} ⊆ S. Thus (f(x))k = 0. Hence f(x) is a nilpotent of A = R [x;α, δ] . Corollary 1. If f(x) = a0 + a1x + a2x 2 + ... + anx n ∈ A = R [x;α, δ] , R is an (α, δ)- compatible ring and satisfies any one of the following conditions: 1) R is a Noetherian ring, 2) R has either the ACC or DCC on left annihilators, then f(x) ∈ nil (A) if and only if ai ∈ nil (R) for all 0 ≤ i ≤ n. Proof. If R satisfies any one of the conditions (1) and (2), then R is an NI ring with nil (R) nilpotent. Hence the result follows directly from Lemma 3. Lemma 4. Let R be an (α, δ)-compatible NI ring with nil (R) nilpotent and a, b ∈ R. Then ab ∈ nil (R) if and only if aλvu(b) ∈ nil (R) , where v ≥ u ≥ 0 are integers. Proof. (=⇒) Suppose that ab ∈ nil (R) , so ba ∈ nil (R) . Assume that f(x) = b and g(x) = ax ∈ A = R [x;α, δ] . Then f(x)g(x) ∈ nil (A) , so g(x)f(x) = aδ(b) + aα(b)x ∈ nil (R) [x;α, δ] . Thus aδ(b), aα(b) ∈ nil (R) . Now suppose that h(x) = α(b) and k(x) = M. A. Farahat, Salha T. Al-Bogamy / Eur. J. Pure Appl. Math, 14 (1) (2021), 164-172 169 ax ∈ A = R [x;α, δ] . Then h(x)k(x) ∈ nil (A) , so k(x)h(x) = aδ(α(b)) + aα2(b)x ∈ nil (R) [x;α, δ] . Thus aδ(α(b)), aα2(b) ∈ nil (R) . Since aδ(b) ∈ nil (R) , for p(x) = δ(b) and q(x) = ax ∈ A = R [x;α, δ] , we have p(x)q(x) ∈ nil (A) , so q(x)p(x) = aδ2(b)+aα(δ(b))x ∈ nil (R) [x;α, δ] . Thus aδ2(b), aα(δ(b)) ∈ nil (R) . Continuing in this process we get aαn1(δm1(αn2(δm2 ...αni(δmj (b))))) ∈ nil (R) , where ni,mj are nonnegative integers. Thus aλvu(b) ∈ nil (R) , where v ≥ u ≥ 0 are integers. (⇐=) Suppose that aλvu(b) ∈ nil (R) , where v ≥ u ≥ 0 are integers. By using (α, δ)- compatiblility of R, we can conclude that ab ∈ nil (R) . Proposition 1. Let R be an (α, δ)-compatible NI ring with nil (R) nilpotent, f(x) = n∑ i=0 aix i and g(x) = m∑ j=0 bjx j ∈ A = R [x;α, δ] . Then f(x)g(x) ∈ nil (A) if and only if aibj ∈ nil (R) for all integers 0 ≤ i ≤ n and 0 ≤ j ≤ m. Proof. Suppose that f(x) = n∑ i=0 aix i and g(x) = m∑ j=0 bjx j ∈ A such that f(x)g(x) ∈ nil (A) . Since R is an (α, δ)-compatible NI ring with nil (R) nilpotent, we get, from Lemma 3, the following: ∆n+m = anα n(bm) ∈ nil (R) , (1) ∆n+m−1 = anα n(bm−1) + an−1α n−1(bm) + anλ n n−1(bm) ∈ nil (R) , (2) ∆n+m−2 = anα n(bm−2) + n∑ i=n−1 aiλ i n−1(bm−1) + n∑ i=n−2 aiλ i n−2(bm) ∈ nil (R) . (3) From Eq.(1) and Lemma 4, we obtain anbm ∈ nil (R) . So, bman ∈ nil (R) . If we multiply Eq.(2) on the left side by bm, then we get bman−1α n−1(bm) ∈ nil (R) . Thus, by Lemma 4, we obtain an−1bm ∈ nil (R) . Again from Eq.(2) and Lemma 4, we obtain anbm−1 ∈ nil (R) . Applying the preceding method repeatedly, we deduce that aibj ∈ nil (R) for all integers 0 ≤ i ≤ n and 0 ≤ j ≤ m. Conversely, suppose that f(x) = n∑ i=0 aix i and g(x) = m∑ j=0 bjx j ∈ A such that aibj ∈ nil (R) for all integers 0 ≤ i ≤ n and 0 ≤ j ≤ m. We show that f(x)g(x) ∈ nil (A) . From Lemma 3 and Lemma 4, we get the following: aλvu(b) ∈ nil (R) , where v ≥ u ≥ 0 are integers. Hence f(x)g(x) ∈ nil (A) . Now we can turn to our main Theorems in the paper. Theorem 1. Let R be an (α, δ)-compatible NI ring with nil (R) nilpotent, such that α(e) = e and δ(e) = 0 for every e ∈ Id(R). If R is a weak right PS-ring, then A = R [x;α, δ] is a weak right PS-ring. M. A. Farahat, Salha T. Al-Bogamy / Eur. J. Pure Appl. Math, 14 (1) (2021), 164-172 170 Proof. Let L be a maximal right ideal of A = R [x;α, δ] . We will show that either NA (L) ⊆ nil(A) or NA (L) = Aq, where q ∈ Id(A). Let I be the set of all coefficients of all polynomials in L and let J be the right ideal of R generated by I, i.e., J = 〈I〉r = IR. If J = R, then there exist a1, a2, ..., an ∈ I and r1, r2, ..., rn ∈ R, such that 1 = a1r1 + a2r2 + ...+ anrn. Suppose that ϕ(x) = k∑ i=0 bix i ∈ NA (L) , then for every f(x) = n∑ j=0 ajx j ∈ L, we have ϕ(x)f(x) = ( k∑ i=0 bix i ) n∑ j=0 ajx j  ∈ nil(A). Since R is an (α, δ)-compatible NI ring with nil (R) nilpotent, from Lemma 1, we get that biaj ∈ nil(R), for all integers 0 ≤ i ≤ k and 0 ≤ j ≤ n. Consequently, for every a ∈ I, bia ∈ nil(R), for all integers 0 ≤ i ≤ k. Hence bi ∈ NR (J) = NR (R) = nil(R), for all integers 0 ≤ i ≤ k. Therefore ϕ(x) ∈ nil(A). Hence NA (L) ⊆ nil(A). If J 6= R, we show that J is a maximal right ideal of R. Let r ∈ R−J. If r ∈ L, then r ∈ I and so r ∈ J, which is a contradiction. Thus r /∈ L. Since L is a maximal right ideal of A, we have A = L+ rA. It follows that there exist f(x) = n∑ i=0 aix i ∈ L and h(x) = m∑ j=0 bjx j ∈ A, such that 1 = a0 + rb0. If a0 = 0, then 1 = rb0 ∈ rR and so R = J + rR. If a0 6= 0, then a0 ∈ I ⊂ J which implies that R = J + rR. Hence J is a maximal right ideal of R. Since R is a weak right PS-ring, it follows that either NR (J) ⊆ nil(R) or NR (J) = Re, where e ∈ Id(R). Case (1): Assume that NR (J) ⊆ nil(R). We will show that NA (L) ⊆ nil(A). Let ϕ (x) = k∑ i=0 mix i ∈ NA (L) . Then for every g (x) = n∑ j=0 ajx j ∈ L, we have ϕ (x) g (x) = ( k∑ i=0 mix i ) n∑ j=0 ajx j  ∈ nil(A). Since R is an (α, δ)-compatible NI ring with nil (R) nilpotent, from Lemma 1, we get that biaj ∈ nil(R), for all integers 0 ≤ i and 0 ≤ j. Consequently, for every a ∈ I, bia ∈ nil(R), M. A. Farahat, Salha T. Al-Bogamy / Eur. J. Pure Appl. Math, 14 (1) (2021), 164-172 171 for all integers 0 ≤ i. Hence bi ∈ NR (J) = NR (R) = nil(R), for all integers 0 ≤ i. Therefore ϕ(x) ∈ nil(A) and we have NA (L) ⊆ nil(A). Case (2): Assume that NR (J) = Re, where e ∈ Id(R). We will show that NA (L) = Ah, where h ∈ Id(A). Let ϕ(x) = k∑ i=0 bix i ∈ NA (L) and ϕ(x) /∈ nil(A), then for every f(x) = n∑ j=0 ajx j ∈ L, we have ϕ(x)f(x) = ( k∑ i=0 bix i ) n∑ j=0 ajx j  ∈ nil(A). Since R is an (α, δ)-compatible NI ring with nil (R) nilpotent, from Lemma 1, we get that biaj ∈ nil(R), for all integers 0 ≤ i ≤ k and 0 ≤ j ≤ n. Consequently, for every a ∈ I, bia ∈ nil(R), for all integers 0 ≤ i ≤ k. For any m ∈ J, there exist a1, a2, ..., an ∈ I and r1, r2, ..., rn ∈ R, such that q = a1r1 + a2r2 + ...+ anrn, biq = (bia1) r1 + (bia2) r2 + ...+ (bian) rn, hence biq ∈ nil(R), for all integers 0 ≤ i ≤ k, so bi ∈ NR (J) = Re, for all integers 0 ≤ i ≤ k. Therefore there exist ti ∈ R such that bi = tie, for all integers 0 ≤ i ≤ k. Since for any idempotent e ∈ R we have α(e) = e and δ(e) = 0, we can conclude that ϕ(x) = k∑ i=0 bix i = k∑ i=0 tiex i = ( k∑ i=0 tix i ) e ∈ Ah, where h = e = e2 = h2 ∈ A. Therefore NA (L) = Ah, where h ∈ Id(A) and the result is proved. Theorem 2. Let R be an (α, δ)-compatible NI ring with nil (R) nilpotent. If R is a weak left PS-ring, then A = R [x;α, δ] is a weak left PS-ring. Proof. The proof is similar to the previous proof of Theorem 1. The only thing we need to note here is that, If L is a maximal left ideal of A = R [x;α, δ] , then, by analogue manner as above, we get in case (2) that bi ∈ NR (J) = Re, for all integers 0 ≤ i ≤ k. Therefore there exist ti ∈ R such that bi = eti, for all integers 0 ≤ i ≤ k. So ϕ(x) = k∑ i=0 bix i = k∑ i=0 etix i = e ( k∑ i=0 tix i ) ∈ hA, where h = e = e2 = h2 ∈ A. Therefore NA (L) = hA, where h ∈ Id(A) and the result is proved. REFERENCES 172 Assume that δ is the zero map, then A = R [x;α] , the usual skew polynomial ring over R, and we get the following corollaries: Corollary 2. Let R be an α-compatible NI ring with nil (R) nilpotent, such that α(e) = e for every e ∈ Id(R). If R is a weak right PS-ring, then A = R [x;α] is a weak right PS-ring. Corollary 3. Let R be an α-compatible NI ring with nil (R) nilpotent. If R is a weak left PS-ring, then A = R [x;α] is a weak left PS-ring. Assume that α is the identity map, then A = R [x; δ] , the differential polynomial ring over R, and we get the following corollaries: Corollary 4. Let R be a δ-compatible NI ring with nil (R) nilpotent, such that δ(e) = 0 for every e ∈ Id(R). If R is a weak right PS-ring, then A = R [x; δ] is a weak right PS-ring. Corollary 5. Let R be a δ-compatible NI ring with nil (R) nilpotent. If R is a weak left PS-ring, then A = R [x; δ] is a weak left PS-ring. Assume that α is the identity map and δ is the zero map, then A = R [x] , the usual polynomial ring over R, and we get the following corollary: Corollary 6. Let R be an NI ring with nil (R) nilpotent. If R is a weak right (left) PS-ring, then A = R [x] is a weak right (left) PS-ring. References [1] S. Annin. Cassociated primes over ore extension rings. J. Alg. Appl., 3:193–205, 2004. [2] M. Farahat and S. 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