EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 11, No. 1, 2018, 244-259 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Weak PS-rings over Skew Hurwitz Series 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 2 Mathematics Department, Faculty of Science, Al-Azhar University, Cairo, Egypt Abstract. The notion of PS-rings is extended to the class of weak PS-rings. We explore the algebraic properties of such class and study its relation with some other rings such as a local ring and a semisimple NI-ring. Also, we show the following result concerning, the ring of skew Hurwitz series, A = (HR, σ): Let R be a σ-compatible NI-ring with nil(R) nilpotent, σ(e) = e for every idempotent e ∈ R and R a torsion free as a Z-module. If R is a weak right PS-ring, then A = (HR, σ) is a weak right PS-ring. 2010 Mathematics Subject Classifications: 06F05, 16W60, 16D25, 16P60 Key Words and Phrases: PS-modules, PS-rings, skew Hurwitz series rings 1. Introduction All rings are assumed to be an associative ring with identity and modules are nonzero unitary right modules, unless otherwise stated. For a nonempty subset X of R, rR(X) (or `R(X)) denote the right (or left) annihilator of X over R. Also, for a ring R, nil (R) denotes the set of all nilpotent elements of R and Id (R) denotes the set of all idempotent elements of R. Furthermore, we use J(R) for the Jacobson radicals of a ring R. Recall that: A ring R is called an NI-ring if nil (R) is a two sided ideal in R and R is called a reduced ring if nil (R) = (0). According to Nicholson and Watters, in ([15], 1988), a right R-module MR is called a right PS-module if every simple submodule is projective, equivalently if its right socle, Soc(MR) = ∑ {B : B is a simple submodule of MR} . is projective. The class of PS-modules is closed under direct sums and submodules. A left PS-module RM is defined similarly. The study of PS-modules was initiated by Gordon, in ([6], 1969). A ring R is said to be a right (left) PS-ring if RR (RR) is a right (left) PS-module. The notion of PS-rings is not left-right symmetric (see Example 11). ∗Corresponding author. Email addresses: m farahat79@yahoo.com (Mohamed A. Farahat), SalhaAlbogamy@hotmail.com (S. T. Al-Bogamy) http://www.ejpam.com 244 c© 2018 EJPAM All rights reserved. M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 245 Example 1 ([15]). (a) If Soc(MR) = 0, then MR is a right PS-module. (b) Any projective semisimple module is a PS-module. (c) Every regular module is a PS-module. (d) Every nonsingular right R-module is a PS-module. For any subset X of R, the left annihilator of X in a right R-module MR is denoted by `M (X) = {m ∈M | mX = 0} . Similarly, one can consider the right annihilator of X in a left R-module RM, where X is any subset of R. The following result is due to Weimin in ([22], 1992), which gives an equivalent condi- tion for a right R-module MR to be a right PS-module. Theorem 1 ([22]). The following statements are equivalent for a right R-module MR : (1) MR (RM) is a right (left) PS-module. (2) If L is a maximal right (left) ideal of R, then `M (L) = Re (rM (L) = eR), where e ∈ Id (R) . (3) If L is a maximal right (left) ideal of R, then either L = Re (L = eR), or `M (L) = (0) (rM (L) = (0)), where e ∈ Id (R) . Example 2 ([15]). (a) Every right (left) PP-ring is a right (left) PS-ring. In particular, every right (left) Baer ring is a right (left) PS-ring. (b) Every semiprime ring is a right (left) PS-ring. But the converse is not true. (c) If `R(J(R)) = 0, then R is a PS-ring, since we have J(R) ⊆ L for every maximal right (left) ideal L in R and hence `R(L) ⊆ `R(J(R)) = 0 implies that for every maximal right (left) ideal L in R, hence `R(L) = 0. In [15], the class of PS-rings is closed under the formulation of polynomials and power series extensions, by other words, if a ring R is a right PS-ring, then so is R [x] and R [[x]] . The converse of this result is false by the following example: Example 3 ([15], Example 3.2). If R = Z4, then R[x] and R[[x]] are PS-rings, but R is not a PS-ring. Nicholson and Watters, in [15], proved that: A ring R is a right PS-ring if and only if the full matrix ring Mn(R), where n is a positive integer, is a right PS-ring. The authors in ([14], 1998), showed that the commutative PS-ring condition is preserved by the generalized power series rings under certain conditions. Recently, Salem, Farahat and Abd- ELmalk, in ([20], 2015), investigated PS-modules over Ore extensions and skew generalized power series extensions. Also, Farahat and Al-Harthy, in ([3], 2017), investigated PS- modules over generalized Mal’cev-Neumann series rings. In ([18], 2017), Paykan proved that, under suitable conditions, if R is a right PS-ring, then so the skew inverse power series rings. As a generalization of the annihilator concept, Ouyang, in ([16], 2009), introduced the weak annihilator (or nilpotent annihilator), for a nonempty subset X of a ring R, the weak M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 246 annihilator of X in R is defined as follows: NR (X) = {a ∈ R |xa ∈ nil (R) for all x ∈ X } . It can be easily shown that ab ∈ nil (R)⇔ ba ∈ nil (R) for all a, b ∈ R. Therefore there is no way to distinguish between the right weak annihilator and the left weak annihilator. Obviously, rR (X) ⊆ NR (X) for any subset X of R and if R is reduced, then rR (X) = NR (X) . For example, let D be an integral domain and R = T2 (D) = {( a b 0 c ) | a, b, c ∈ D } = ( D D 0 D ) . We can check that nil (R) = {( 0 b 0 0 ) | b ∈ D } = ( 0 D 0 0 ) is a two sided ideal in R, hence R is an NI-ring. Consider the subset X = {( x 0 0 x ) | x ∈ D } of R. Then rR (X) = 0 but NR (X) = {( 0 y 0 0 ) | y ∈ D } = ( 0 D 0 0 ) = nil (R) . Thus rR (X) $ NR (X) . Hence the weak annihilator is a nontrivial generalization of the annihilator. Proposition 1 ([16]). Let R be a ring. Then: 1) If X ⊆ Y ⊂ R, then NR(Y ) ⊆ NR(X), 2) X ⊆ NR(NR(X)), 3) NR(NR(NR(X))) = NR(X), 4) If S ≤ R and X ⊆ S, then NS(X) = NR(X) ∩ S. Lemma 1. Let R be an NI-ring. If ab ∈ nil (R) , then arb ∈ nil (R) for all a, b, r ∈ R. Proof. Let a, b ∈ R such that ab ∈ nil (R) . Hence ba ∈ nil (R). We thus get bar ∈ nil (R) for each r ∈ R. It follows that arb ∈ nil (R) for each r ∈ R. Remark 1. If R is an NI-ring, then NR (X) is an ideal of R for any subset X of R. In particular, NR (a) is an ideal of R for any a ∈ R. In this paper, we introduce the class of weak PS-rings and investigate some of its characterization and study the transfer of weak PS-condition between a base ring R and the extension ring of skew Hurwitz series A = (HR, σ). M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 247 2. Rings Satisfy The Weak PS-Condition Motivated by the definition of the PS-condition and the definition of the weak annihi- lator, we introduced the notion of weak PS-condition as follows: Definition 1. A ring R satisfies the right weak PS-condition if, for every maximal right ideal L of R, either NR (L) ⊆ nil (R) or NR (L) = Re, (principal left ideal generated by e), where e ∈ Id(R). Similarly, we can define the left weak PS-condition. A ring R satisfies the weak PS-condition if it satisfies both the right and the left weak PS-conditions. Remark 2. 1) In the definition of the weak PS-condition, we use ”or” to mean that one of the two options only holds, but not both, unless that NR (L) = (0). Since nil (R) ⋂ Id (R) = (0). 2) It is clear that in reduced rings weak PS-condition and PS-condition are equivalent. Example 4. Every domain (which is not a field) is a reduced weak PS-ring, since the left (right) annihilator of every right (left) maximal ideal is zero and hence generated by idempotent. Also, every field is a reduced weak PS-ring, since the zero ideal is the only maximal ideal in a field F, clearly the left (right) annihilator of the zero ideal is F which is generated by 1 which is an idempotent in F. Example 5. If R is a commutative local ring with a unique maximal ideal L, then: 1) R is an NI-ring and L = nil(R), 2) NR (L) = R = 〈1〉 , the principal ideal generated by 1, which proves that a commutative local ring is a weak PS-ring. 3) By above we get many examples of weak PS-rings, e.g., R = Zpn , for any prime number p and for any positive integer n. The unique maximal ideal of R = Zpn is L = 〈p〉 = J(R). Also, a local domain R = F [[x]], where F is any field with unique maximal ideal L = 〈x〉 = J(R). Example 6. Consider the commutative reduced ring R = Z6, since nil(R) = {0}. R has two maximal ideals which are L1 = {0, 2, 4} and L2 = {0, 3}. Note that Id(R) = {0, 1, 3, 4}. Now we compute the weak annihilator for these ideals as follows: NR (L1) = `R(L1) = rR(L1) = L2 = {0, 3} = 〈 3 〉 , and NR (L2) = `R(L2) = rR(L2) = L1 = {0, 2, 4} = 〈 4 〉 . Hence Z6 is a (weak) PS-ring. Example 7. Consider the commutative reduced ring R = Z10 with nil(R) = {0} and Id(R) = {0, 5, 6}. We can check that, R has two maximal ideals only which are L1 = {0, 2, 4, 6, 8} and L2 = {0, 5}. Now we compute the weak annihilator for these two ideals as follows: NR (L1) = L2 = {0, 5} = 〈 5 〉 , M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 248 and NR (L2) = L1 = {0, 2, 4, 6, 8} = 〈 6 〉 . Hence Z10 is a (weak) PS-ring. Example 8. Consider the commutative nonreduced ring R = Z12 with nil(R) = {0, 6} and Id(R) = {0, 1, 9}. We can check that, R has two maximal ideals only which are L1 = {0, 2, 4, 6, 8, 10} and L2 = {0, 3, 6, 9}. Now we compute the weak annihilator for these two ideals as follows: NR (L1) = L2 = {0, 3, 6, 9} = 〈 9 〉 , and NR (L2) = L1 = {0, 2, 4, 6, 8, 10} = 〈 2 〉 = 〈 10 〉 . Clearly, NR (L2) " nil(R) and NR (L2) not generated by an idempotent in R. Hence Z12 is not a weak PS-ring. Now we turn to compute the left (right) annihilator for these maximal two-sided ideals as follows: `R (L1) = rR(L1) = {0, 6} = nil(R). Which is not generated by an idempotent element in R, and `R (L2) = rR(L2) = 〈 0 〉 . Hence Z12 is not a PS-ring. Example 9. Let F be a field and R = ( F F F F ) . The only maximal right ideals (which are not left ideals) of R are: L1 = ( F F 0 0 ) & L2 = ( 0 0 F F ) , also, J1 = ( F 0 F 0 ) & J2 = ( 0 F 0 F ) , are the only maximal left ideals (which are not right ideals) of R. nil (R) = ( 0 F 0 0 )⋃( 0 0 F 0 ) , which is neither a left nor a right ideal of R. Hence R is not an NI-ring. We can check the following facts: 1) NR(L1) = J2 = Re where e = ( 0 1 0 1 ) ∈ Id(R), 2) NR(L2) = J1 = Rf where f = ( 1 0 1 0 ) ∈ Id(R). Using the above discussion, we conclude that R is a weak right PS-ring. By the same way we can check the following: M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 249 3) NR(J1) = L2 = hR where h = ( 0 0 1 1 ) ∈ Id(R), 4) NR(J2) = L1 = kR where k = ( 1 1 0 0 ) ∈ Id(R). By the above discussion, we conclude that R is a weak left PS-ring. Therefore R is a weak PS-ring. Example 10. Let F be a filed and R = T2 (F ) = ( F F 0 F ) = {( a b 0 c ) | a, b, c ∈ F } . We can check that nil (R) = {( 0 b 0 0 ) | b ∈ F } = ( 0 F 0 0 ) is a two sided ideal in R, hence R is an NI-ring. The only maximal (two sided) ideals of R are: L = ( F F 0 0 ) and J = ( 0 F 0 F ) We can check that NR(L) = nil (R) and NR(J) = nil (R) . Therefore we conclude that R is a weak PS-ring. Example 11. Consider the semigroup ring V = Z2[S] = {αa+ βb : α, β ∈ Z2} = {0, a, b, a+ b = c}, where S = {a, b} is the semigroup with the following Cayley multiplication table × a b a a b b a b Clearly, Id(V ) = {0, a, b} & nil (V ) = {0, c}. We can check the following: 1) I = {0, a} is a left maximal ideal in V which is not a right ideal, 2) J = {0, b} is a left maximal ideal in V which is not a right ideal, 3) K = {0, c} = nil (V ) is a maximal 2-sided ideal in V, hence V is an NI-ring, 4) We can check that NV (I) = K = nil (V ) , NV (J) = K = nil (V ) and NV (K) = V = a× V = b× V. Therefore V is a weak left PS-ring. Since the only right maximal ideal in V, is K = {0, c} = nil (V ) and NV (K) = V which is not generated by an idempotent as a left ideal, so V is not a right weak PS-ring. Also, we can check the following: rV (I) = 0 = 0 × V, rV (J) = 0 = 0 × V, and rV (K) = V = a × V = b × V. Hence V is a left PS-ring. For the unique maximal right ideal K in V, we have `V (K) = V which is not generated by an idempotent as a left ideal. Hence V is not a right PS-ring. M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 250 Lemma 2. Let R be an NI-ring, L a maximal right ideal in R and R = R /nil (R) . Then: 1) L contains N = nil (R) , 2) L = L /N is a maximal right ideal in R, and 3) If e ∈ Id (R) , then e ∈ Id(R). Proof. 1) Let the contrary, i.e., N = nil (R) " L, then there is 0 6= a ∈ N, such that a /∈ L. So L+ aR = R, thus b+ ar = 1, for some b ∈ L and r ∈ R. Since every nilpotent element lies in J(R), we have b = 1− ar is a right invertible element in R, which impossible since b ∈ L and L is a maximal right ideal in R. 2) Let the contrary, i.e., L = L /N is not a maximal right ideal in R, then there is a proper right ideal K = K /N in R, for some right ideal K in R contain N, such that L ⊂ K in R, which implies clearly that L ⊂ K in R, which contradicts for the maximality of L. 3) Let e ∈ Id(R), then (e)2 = (e+N) (e+N) = e2 +N = e+N = e, hence e ∈ Id(R). Theorem 2. Suppose that R is an NI-ring. If R is a weak right PS-ring, then R = R /nil (R) is a right PS-ring. Proof. Let R be a weak right PS-ring and L be a maximal right ideal in R. From Lemma 2, we have L contains N and L = L /N is a maximal right ideal in R. Since R is a weak right PS-ring, we have either NR(L) ⊆ nil (R) or NR(L) = Re, for some e ∈ Id (R) . Case (1): Assume first that NR(L) ⊆ nil (R) . Let a ∈ `R ( L ) . Then aL = 0 = nil (R) , which implies that aL ⊆ nil (R) . Hence a ∈ NR(L) ⊆ nil (R) . Thus a = 0 which implies that `R ( L ) = (0) and we conclude that R = R /nil (R) is a right PS-ring. Case (2): Assume that NR(L) = Re, for some e ∈ Id (R) . Let a ∈ `R ( L ) . Then aL = 0 = nil (R) , which implies that aL ⊆ nil (R) . Hence a ∈ NR(L) = Re and so a = re, for some r ∈ R. Therefore a = re. Thus `R ( L ) = Re and we conclude that R = R /nil (R) is a right PS-ring. Example 12. In Example 8 we conclude that the ring R = Z12 is not a weak PS-ring. We can check that R = R /nil (R) ∼= Z6, which, by Example 6, is a weak PS-ring. So we get here an example to show that the converse direction of Theorem 2 does not hold. Example 13. In Example 11 we conclude that the ring R = V is a weak left PS-ring. We turn now to check the condition of weak PS-ring on the direct sum A = V ⊕ V. In A, we have: nil (A) = nil (V )⊕ nil (V ) = {(0, 0), (0, c), (c, 0), (c, c)}, and Id (A) = Id (V )⊕ Id (V ) = {(0, 0), (0, a), (a, 0), (0, b), (b, 0), (a, b), (b, a)}. Consider now the following maximal two sided ideal in A, L = K ⊕ V = {(0, 0), (0, c), (a, 0), (a, c), (b, 0), (b, c), (c, 0), (c, c)}. M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 251 By direct computations we get NA (K ⊕ V ) = NV (K)⊕NV (V ) = V ⊕K = {(0, 0), (0, a), (0, b), (0, c), (c, 0), (c, a), (c, b), (c, c)}. Clearly, NA (K ⊕ V ) is not contained in nil (A) and not generated by idempotent in A. Therefore A = V ⊕ V is not a weak left (right) PS-ring, i.e., the direct sum of weak left (right) PS-rings not necessary be a weak left (right) PS-ring. Recall that a nonzero right ideal I of R is a right essential ideal if I has nonzero intersection with every nonzero right ideal of R. A singular right ideal Sr (R) of R is defined by Sr (R) = {a ∈ R | rR (a) is an essential right ideal of R} . Similarly, we can define a singular left ideal S` (R) of R is defined by S` (R) = {a ∈ R | `R (a) is an essential left ideal of R} . A ring R is called right (left) nonsingular if Sr (R) = 0 (S` (R) = 0) and right (left) singular if Sr (R) = R (S` (R) = R). We define a weak singular ideal of an NI-ring R as follows: NS (R) = {a ∈ R | NR (a) is an essential ideal of R} . We extend the definitions of a right (left) nonsingular ring and a right (left) singular ring to a weak nonsingular ring and a weak singular ring, respectively, as follows: Definition 2. An NI-ring R is called a weak singular ring if NS (R) = R and R is called a weak nonsingular ring if NS (R) = 0. Example 14. 1) In example 5, R = Z4, we can check that NS (R) = R, hence R is a weak singular ring. But Sr (R) = S` (R) = {0, 2} = nil (R) , hence R is neither a nonsingular ring nor a singular ring. 2) In example 6, R = Z6, we can check that NS (R) = Sr (R) = S` (R) = (0), hence R is a (weak) nonsingular ring. 3) In example 7, R = Z10, we can check that NS (R) = Sr (R) = S` (R) = (0), hence R is a (weak) nonsingular ring. 4) In example 8, R = Z12, we can check that NS (R) = {0, 6} = nil (R) , hence R is neither a weak nonsingular ring nor a weak singular ring. But Sr (R) = S` (R) = {0, 6} = nil (R) , hence R is neither a nonsingular ring nor a singular ring. 5) In example 11, R = V, we can check that NS (V ) = {0, c} = K = nil (V ) , hence V is neither a weak nonsingular ring nor a weak singular ring. But Sr(V ) = {0, c} = K and S` (V ) = (0), hence V is a left nonsingular ring which is not a right nonsingular ring. Lemma 3. For any ring R, we have Sr (R) ⊆ NS (R) . M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 252 Proof. If a ∈ Sr (R) , then rR (a) is an essential right ideal. Since rR (a) ⊆ NR (a) , we conclude that NR (a) is also essential ideal. Thus a ∈ NS (R) . Therefore Sr (R) ⊆ NS (R) . Corollary 1. 1) Every weak nonsingular ring is a right (left) nonsingular ring. 2) Every right (left) singular ring is a weak singular ring. Remark 3. The converse of the last Corollary need not be true by the following examples: 1) In example 14, (1), R = Z4, is a weak singular ring. But R is not a singular ring. 2) In example 14, (5), R = V, is a left nonsingular ring. But R is not a weak singular ring. Corollary 2. Every weak nonsingular ring is a PS-ring. Proof. Since every nonsingular ring is a PS-ring, the result follows. Theorem 3. A semisimple NI-ring is a weak right (left) PS-ring. Proof. It is well-known that any right (resp. left) ideal I in a semisimple ring R has the form I = eR (resp. I = Re), where e ∈ Id (R) . Since R is an NI-ring, we have NR (L) is an ideal of R and the proof is complete. Remark 4. A ring R, in Example 9, is a simple ring, hence R is a semisimple ring which is not an NI-ring. But R is a weak right (left) PS-ring. Example 15. In Example 11, R = V, we have V is a left nonsingular ring which is not a right nonsingular ring. Therefore V is a left PS-ring which is not a reduced ring. Given a ring R and RMR an R-R-bimodule, the trivial extension of R by M is the ring T (R,M) = R⊕M with the usual addition and the multiplication (r1,m1) (r2,m2) = (r1r2, r1m2 +m1r2) , where r1, r2 ∈ R and m1,m2 ∈M. This is isomorphic to the ring of all matrices ( r m 0 r ) , where r ∈ R and m ∈ M with the usual matrix operations. Trivial extensions attracted attention when people searched for nonreduced rings which are Armendraiz [2]. The article of Rege and Chhawchharia ([19], 1997) seems to be the first to consider the Armendraiz property of trivial extensions. Proposition 2. Let R be a reduced ring. If T (R,M) is a weak right PS-ring, then R is a right PS-ring. Proof. Set A = T (R,M) . Since R is a reduced ring, we can easily conclude that nil (A) ∼= {( 0 m 0 0 ) |m ∈M } and hence A = A /nil (A) ∼= R, which completes the proof. M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 253 Proposition 3. Let R be a reduced ring. If T (R,R) is a weak right PS-ring, then R is a right PS-ring. Observe that T (R,R) ∼= R[x] /〈 x2 〉 , so we get the following Corollary: Corollary 3. Let R be a reduced ring. If R[x] /〈 x2 〉 is a weak right PS-ring, then R is a right PS-ring. 3. Skew Hurwitz Series Rings Satisfy The Weak PS-Condition Rings of formal power series have been of interest and have had important applications in many areas, one of which has been differential algebra. In an earlier paper by Keigher [10], the ring of Hurwitz series, a variant of the ring of formal power series was considered, and some of its properties, especially its categorical properties, were studied. In the papers ([11, 12]) Keigher demonstrated that the ring of Hurwitz series has many interesting applications in differential algebra and in the discussion about weak normalization. The product of series using the binomial coefficients, was studied also in papers by Fliess [5] and Taft [21]. Ring-theoretical properties of Hurwitz series rings and its skew have been investigated by many authors ([10, 11, 12, 7, 8, 4]). In this section we study the skew Hurwitz series rings over a noncommutative rings and examine their structures and properties. The elements of HR are ordinary formal series, i.e., HR = { ∞∑ i=0 aix i |ai ∈ R } . The operation of addition in HR is a componentwise addition and the operation of multiplication is defined by the following: For each two Hurwitz series f(x) = ∞∑ i=0 aix i and g(x) = ∞∑ j=0 bjx j ∈ HR, ( ∞∑ i=0 aix i ) ∞∑ j=0 bjx j  = ∞∑ n=0 ( n∑ i=0 ( n i ) aibn−i ) xn, where ( n i ) = Cn i = n! i!×(n−i)! , is the binomial coefficient. From the above discussion we can say that: The multiplication is subject to the relation (Hurwitz multiplication rule) ( aix i ) ( bjx j ) = ( i+ j i ) aibjx i+j = ( i+ j j ) aibjx i+j . It is now routine to check that HR is a ring with identity 1R which is called the Hurwitz series rings over R. Clearly, R is a subring of HR. Despite the apparent similarity between the formal power series rings R[[x]] and the Hurwitz series rings HR but there are substantial difference between them for example, M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 254 the ideal 〈 x, x2, x3, ... 〉 is principal in R[[x]] which is the same as the ideal 〈x〉 but the ideal〈 x, x2, x3, ... 〉 in HR is not a finitely generated. Here x2 /∈ 〈x〉 since x.x = 2x2. Clearly, if R has Q as a subring, then the ideal 〈 x, x2, x3, ... 〉 in HR will be the same as the principal ideal 〈x〉 . Let σ be an endomorphism of the ring R, with σ(1) = 1. The elements of A = (HR, σ), the ring of skew Hurwitz series, are the ordinary formal series ∞∑ i=0 aix i, where ai ∈ R, with component wise addition and the following operation of multiplication: For each two Hurwitz series f(x) = ∞∑ i=0 aix i and g(x) = ∞∑ j=0 bjx j ∈ A = (HR, σ), ( ∞∑ i=0 aix i ) ∞∑ j=0 bjx j  = ∞∑ n=0 ( n∑ i=0 ( n i ) aiσ i(bn−i) ) xn . From the above discussion we can say that: The multiplication is subject to the relation (Hurwitz multiplication rule) ( aix i ) ( bjx j ) = ( i+ j i ) aiσ i(bj)x i+j = ( i+ j j ) aiσ i(bj)x i+j . By hR and P = (hR, σ), we denote the polynomial Hurwitz ring and skew polynomial Hurwitz ring, respectively. The product, in this case will be as follows: For each two polynomial Hurwitz f(x) = v∑ i=0 aix i and g(x) = u∑ j=0 bjx j ∈ P, ( v∑ i=0 aix i ) u∑ j=0 bjx j  = v+u∑ n=0 ( n∑ i=0 ( n i ) aiσ i(bn−i) ) xn. Annin [1] introduced the notion of σ-compatibility of rings as follows. Definition 3 ([1]). A ring R is said to be σ-compatible if ab = 0 ⇔ aσ(b) = 0, where a, b ∈ R. Some of the basic properties of a σ-compatible ring was given in the following lemmas. Lemma 4 ([1]). If R is a σ-compatible ring, then: 1) σ(1) = 1, 2) σ is a monomorphism, 3) ab = 0⇔ σ(a)b = 0, where a, b ∈ R. Lemma 5 ([1]). Let R be a σ-compatible ring. Then ab = 0 ⇔ aσi (b) = 0 ⇔ σi(a)b = 0 for all i ≥ 0 and a, b ∈ R. In [13], the author introduced the concept of σ-rigid rings for rings with an endomor- phism σ, as follows: M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 255 Definition 4 ([13]). Let R be a ring with an endomorphism σ. If aσ(a) = 0⇒ a = 0, for a ∈ R, then R is called a σ-rigid ring. Some of the basic properties of a σ-rigid ring was given in the following lemma. Lemma 6 ([9]). If R is a σ-rigid ring, then: 1) R is a reduced ring, 2) σ is a monomorphism. The relation between a σ-compatible ring and a σ-rigid ring was studied in the following lemma. Lemma 7 ([9]). A ring R is a reduced σ-compatible ring if and only if R is σ-rigid. Hong, et. al., in ([9], Proposition 5), proved that: If R is a σ-rigid ring, then for each e ∈ Id (R) , we have σ(e) = e. In what follows, we characterize skew Hurwitz series rings that satisfy the weak PS- condition. We shall need the following auxiliary result in the proof of our Theorems. Lemma 8 (2.5 and 2.6 [17]). Let R be a σ-compatible NI-ring with nil(R) nilpotent and R a torsion free as a Z-module. Set K = nil (R) . Then we have: (1) nil (A) = (HK,σ). (2) f (x) = ∞∑ i=0 aix i ∈ nil (A) if and only if ai ∈ nil (R) for all i ≥ 0. (3) If f (x) = ∞∑ i=0 aix i and g (x) = ∞∑ j=0 bjx j ∈ A such that f (x) g (x) ∈ nil (A) , then aibj ∈ K = nil (R) for all i, j ≥ 0. As a corollary we get the following result in the finite case. Lemma 9. Let R be a σ-compatible NI-ring with nil(R) nilpotent and R a torsion free as a Z-module. Set K = nil (R) . Then we have: (1) nil (P ) = (hK, σ). (2) f (x) = k∑ i=0 aix i ∈ nil (P ) if and only if ai ∈ nil (R) for all i ≥ 0. (3) If f (x) = k∑ i=0 aix i and g (x) = n∑ j=0 bjx j ∈ P such that f (x) g (x) ∈ nil (P ) , then aibj ∈ K = nil (R) for all 0 ≤ i ≤ k and 0 ≤ j ≤ n. Now, we can prove our result in the infinite case: Theorem 4. Let R be a σ-compatible NI-ring with nil(R) nilpotent, σ(e) = e for every e ∈ Id(R) and R a torsion free as a Z-module. If R is a weak right PS-ring, then A = (HR, σ) is a weak right PS-ring. M. A. Farahat, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 256 Proof. Let L be a maximal right ideal of A = (HR, σ). 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 Hurwitz series 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) = ∞∑ i=0 bix i ∈ NA (L) , then for every f(x) = ∞∑ j=0 ajx j ∈ L, we have ϕ(x)f(x) = ( ∞∑ i=0 bix i ) ∞∑ j=0 ajx j  ∈ nil(A). Since R is a σ-compatible NI-ring with nil(R) a nilpotent and R a torsion free as a Z- module from Lemma 8, we get that biaj ∈ nil(R), for all integers 0 ≤ i and 0 ≤ j. Consequently, for every a ∈ I, bia ∈ nil(R), for all integers 0 ≤ i. Hence bi ∈ NR (J) = NR (R) = nil(R), for all integers 0 ≤ i. 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) = ∞∑ i=0 aix i ∈ L and g(x) = ∞∑ j=0 bjx j ∈ A, such that 1 = a0 + rb0. If a0 = 0, then 1 = rb0 ∈ r.R 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) = ∞∑ i=0 mix i ∈ NA (L) . Then for every g (x) = ∞∑ j=0 ajx j ∈ L, we have ϕ (x) g (x) = ( ∞∑ i=0 bix i ) ∞∑ j=0 ajx j  ∈ nil(A). Since R is a σ-compatible NI-ring with nil(R) a nilpotent, from Lemma 8, 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, S. T. Al-Bogamy / Eur. J. Pure Appl. Math, 11 (1) (2018), 244-259 257 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) = Aq, where q ∈ Id(A). Let ϕ(x) = ∞∑ i=0 bix i ∈ NA (L) and ϕ(x) /∈ nil(A), then for every f(x) = ∞∑ j=0 ajx j ∈ L, we have ϕ(x)f(x) = ( ∞∑ i=0 bix i ) ∞∑ j=0 ajx j  ∈ nil(A). Since R is a σ-compatible NI-ring with nil(R) a nilpotent, from Lemma 8, we get that biaj ∈ nil(R), for all integers 0 ≤ i and 0 ≤ j. Consequently, for every a ∈ I, bia ∈ nil(R), for all integers 0 ≤ i. For any m ∈ J, there exist a1, a2, ..., an ∈ I and r1, r2, ..., rn ∈ R, such that m = a1r1 + a2r2 + ...+ anrn, bim = (bia1) r1 + (bia2) r2 + ...+ (bian) rn, hence bim ∈ nil(R), for all integers 0 ≤ i, so bi ∈ NR (J) = Re, for all integers 0 ≤ i. Therefore there exist ti ∈ R such that bi = tie, for all integers 0 ≤ i, and we have σ(e) = e. Hence ϕ(x) = ∞∑ i=0 bix i = ∞∑ i=0 tiex i = ( ∞∑ i=0 tix i ) e ∈ Aq, where q2 = e2 = e = q ∈ A. Therefore NA (L) = Aq, where q ∈ Id(A) and the result is proved. As a corollary we get the following result in the finite case. Corollary 4. Let R be a σ-compatible NI-ring with nil(R) nilpotent, σ(e) = e for every e ∈ Id(R) and R a torsion free as a Z-module. If R is a weak right PS-ring, then P = (hR, σ) is a weak right PS-ring. Theorem 5. Let R be a σ-compatible NI-ring with nil(R) nilpotent and R a torsion free as a Z-module. If R is a weak left PS-ring, then A = (HR, σ) is a weak left PS-ring. Proof. The proof is similar to the previous proof of Theorem 4. The only thing we need to note here is that, if L is a maximal left ideal of A = (HR, σ), then, by the analogue manner as above, we get in case (2) that bi ∈ NR (J) = eR, for all integers 0 ≤ i. Therefore there exist ti ∈ R such that bi = eti, for all integers 0 ≤ i. So ϕ(x) = ∞∑ i=0 bix i = ∞∑ i=0 etix i = e ( ∞∑ i=0 tix i ) ∈ qA, where q2 = e2 = e = q ∈ A. Therefore NA (L) = qA, where q ∈ Id(A) and the result is proved. As a corollary we get the following result in the finite case. REFERENCES 258 Corollary 5. Let R be a σ-compatible NI-ring with nil(R) nilpotent and R a torsion free as a Z-module. If R is a weak left PS-ring, then P = (hR, σ) is a weak left PS-ring. Assume that σ is the identity map, then the skew Hurwitz series ring A = (HR, σ) is HR, the usual Hurwitz series ring over R, and the skew Hurwitz polynomial ring P = (hR, σ) is hR, the usual Hurwitz polynomial ring over R, so we get the following corollaries: Corollary 6. Let R be an NI-ring with nil(R) nilpotent and R a torsion free as a Z- module. If R is a weak right (left) PS-ring, then usual Hurwitz series ring over HR is a weak right (left) PS-ring. Corollary 7. Let R be an NI-ring with nil(R) nilpotent and R a torsion free as a Z- module. 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