EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 2, 2019, 332-347 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On weak graded rings II Najla Al-Subaie1, M. M. Al-Shomrani2,∗ 1 Department of Mathematics, Taif University, Taif, Saudi Arabia 2 King Abdulaziz University, Faculty of Science, P.O.Box 80203, Jeddah 21589, Saudi Arabia Abstract. The G-weak graded rings are rings graded by a set G of left coset representatives for the left action of a subgroup H of a finite group X. The main aim of this article is to study the concept of G-weak graded rings and continue the investigation of their properties. Moreover, some results concerning G-weak graded rings of fractions are derived. Finally, some additional examples of G-weak graded rings are provided. 2010 Mathematics Subject Classifications: 16W50, 13A02, 16D25 Key Words and Phrases: Weak graded rings, fully weak graded rings, left coset representatives, graded rings of fractions, homogeneous elements. 1. Introduction Recall that, for a group X and a ring R, R is called X-graded if, for each element g in the group X, there is an additive subgroup Rg of R, such that R = ⊕ g∈X Rg and, for all g, h ∈ X, we have RgRh ⊆ Rgh. Group graded rings as well as Clifford theory for group graded rings were studied and their properties were investigated by many mathematicians, see for exampl [8], [9],[10], [12], [16], [21] and [22]. Nevertheless, rings and modules can be graded by using semigroups instead of groups leading to more general results as we can see in [1], [11], [13], [14], [15] and [19]. Many ways have been used to investigate the properties of these rings. In [7], Cohen and Montgomery introduced an interesting way using duality theorems, see also [5]. Another useful way is the associated graded ring construction which states that for a valuation ring R, we can associate a ring RG graded by the valuation group G. This ring seems to be easier to be studied and the properties can be lifted back from RG to R. This way is one of the motivations for studying graded rings, see [16] for more details. Moreover, some mathematicians introduced categorical methods to study these graded rings such as the study of separable functors introduced in [17] and [20]. Most of these methods have been ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i2.3380 Email addresses: njlalsubaie@gmail.com (N. Al-Subaie), malshomrani@hotmail.com (M. Al-Shomrani) http://www.ejpam.com 332 c© 2019 EJPAM All rights reserved. Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 333 introduced for the finite group-grading. However, more additional investigations have been done considering the infinite case, see for example [2]. In [6], a fixed set G of left coset representatives for the left action of a subgroup H on a group X was constructed and a binary operation on G, which has a left identity and the right division property, was defined. This binary operation is not associative in a travail way. However, the associativity was considered by using a ”cocycle” f : G×G −→ H. The dependence on the choice of representatives was shown as follows: For a given subgroup H of a group X, different sets of representatives G and G for the left cosets can be chosen. These cosets are related by an arbitrary function γ : X/H −→ H, so that if s ∈ G then γ([s])s ∈ G, where [s] denotes the coset Hs. In [3], the concept of group graded rings was extended by using a set G of left coset representatives with specific binary operation. This new concept was called G-weak graded rings. In [4], some properties of weak graded rings were investigated. Moreover, graded rings by using the product H ×G were also discussed. In this article, we consider the G-weak graded rings and continue the investigation of their properties. More specifically, for a finite group X, a subgroup H, a fixed set of left coset representatives (G, ∗) and a G-weak graded ring R with unity, the following results are proved: (i) If K is a subring of R containing all of its G-homogeneous elements, then K is a G-weak graded subring. (ii) If x is a unit element such that x ∈ Rs for some s ∈ G, then x−1 ∈ RsL where sL is the left inverse of s. (iii) WGrU (R), the set of all weak graded units of R, is a subgroup of U(R). Moreover, some results considering the G-weak graded rings of fraction are proved. Finally, some additional examples of G-weak graded rings are introduced. Throughout this article, we shall assume that all groups are finite, all rings are com- mutative with unities and all vector spaces are finite dimensional. 2. Preliminaries In this section, we list some important definitions and results that will be used later in this article. Definition 1. [6] Let X be a group and H be a subgroup of X. We call G ⊂ X a set of left coset representatives if, for every x ∈ X, there is a unique s ∈ G such that x ∈ Hs. In addition, we call the decomposition x = us, for u ∈ H and s ∈ G, the unique factorization of x. In what follows, G stands for a fixed set of left coset representatives for the action of the subgroup H of X on the group X and e is the identity element in X. Definition 2. [6] For elements s, t ∈ G we define f(s, t) ∈ H and s ∗ t ∈ G by the unique factorization st = f(s, t)(s ∗ t) in X, where f is the cocycle map. Moreover, the functions . : G×H → H and / : G×H → G are defined by the unique factorization su = (s.u)(s/u) for s, s / u ∈ G and u, s . u ∈ H. Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 334 The binary operation ∗ on G has a unique left identity eG ∈ G and satisfying the right division property, i.e., for all s, t ∈ G there is a unique solution p ∈ G satisfying the equation p ∗ s = t [6]. If e ∈ G, then eG = e is also a right identity. Also, there is a unique left inverse sL for every s ∈ G satisfying the equation sL ∗ s = eG. Proposition 1. [6] The following identities between (G, ∗) and f are satisfied for all s, t, p ∈ G and all u, v ∈ H: s . (t . u) = f(s, t) ( (s ∗ t ) . u)f ( s / (t . u), t / u )−1 , (s ∗ t) / u = ( s / (t . u) ) ∗ (t / u) , s . uv = (s . u) ( (s / u) . v ) , s / uv = (s / u) / v , f(p, s)f(p ∗ s, t) = ( p . f(s, t) ) f ( p / f(s, t), s ∗ t ) and ( p / f(s, t) ) ∗ (s ∗ t) = (p ∗ s) ∗ t. Proposition 2. [6] The following identities between (G, ∗) and f are satisfied for all t ∈ G and all v ∈ H: eG / v = eG, eG . v = eGve −1 G , t . e = e, t / e = t, f(eG, t) = eG, t . e−1 G = f ( t / e−1 G , eG )−1 and ( t / e−1 G ) ∗ eG = t. Now, we include the definition of G-weak graded rings and some related results form [3] where the binary operation ∗ is as in Definition 2. Definition 3. [3], [4] Let X be a group, H be a subgroup of X and (G, ∗) be a fixed set of left coset representatives for the left action of H on X. A ring R is called a G-weak graded ring if R = ⊕ s∈G Rs (1) and RsRt ⊆ Rs∗t for all s, t ∈ G, (2) where Rs is an additive subgroup for each s ∈ G. If (2) is replaced by RsRt = Rs∗t for all s, t ∈ G, (3) then, R is called a fully (or strongly) G-weak graded ring. Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 335 It can be noted that any ring R can be put into a G-weak graded ring by placing R = ReG and Rs = 0 for all s ∈ G with s 6= eG. This is called the trivial G-weak graded ring. Proposition 3. [3] Let G be a fixed set of left coset representatives for a subgroup H of a group X and R be a G-weak graded ring with identity. Then, 1R ∈ ReG , where 1R is the multiplicative identity of R. Proposition 4. [3] Let R be a G-weak graded ring. Then the eG-component ReG is a subring of R. 3. G-weak graded subrings and related results In [3] and [4], the definition of weak graded rings was given and some of their properties were derived. In this section, we continue the investigation of the properties of weak graded rings. We start by giving a definition for a G-homogeneous element of a G-weak graded ring R. Definition 4. Let R be a G-weak graded ring. Then, a non-zero element rs ∈ R is said to be a weak graded or G-homogeneous element of grade s if there exists an s-component Rs of R such that rs ∈ Rs. The grade of rs is denoted by 〈rs〉 = s. The set of all G-homogeneous elements of R is defined and written as h(R) = ∪s∈GRs. Remark 1. By Definition 3, every element r ∈ R has a unique decomposition written as r = ∑ s∈G rs with rs ∈ Rs for all s ∈ G. These {rs}s∈G are called the G-homogeneous components of r. However, the sum ∑ s∈G rs is finite (in other words almost all rs are zero). Definition 5. Let K be a subring of a G-weak graded ring R. Then, K is said to be a G-weak graded subring of R if K itself is a G-weak graded ring. Theorem 1. Let K be a subring of a G-weak graded ring R. If K contains all the G- homogeneous components for each k ∈ K, then K is a G-weak graded subring of R. Proof. Clearly, we can write K as K = ∑ s∈GKs as additive subgroups. Then for every k ∈ K, we have k = ∑ s∈G ks where ks ∈ Ks for all s ∈ G. We need to prove that K = ⊕ s∈GKs and KsKt ⊆ Ks∗t which we do as follows: (i) K = ⊕ s∈GKs. Since K is a subring of R and R = ⊕ s∈GRs, we have Ks = K ∩ Rs for all s ∈ G. Also, as R is a G-weak graded ring, the direct sum condition implies that Rs ⋂ ( ∑ t∈GRt) = {0} for all s ∈ G with s 6= t, which consequently implies that Ks ⋂ ( ∑ t∈GKt) = {0}. Thus, K = ⊕ s∈GKs. (ii) KsKt ⊆ Ks∗t. Let Ks = K ∩Rs and Kt = K ∩Rt for some s and t in G. Hence, KsKt = (K ∩Rs)(K ∩Rt) ⊆ K ∩ (RsRt) ⊆ K ∩ (Rs∗t) = Ks∗t, Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 336 which completes the proof. It can be noted that the last two lines in part (i) of the proof can be, equivalently, written as follows: Ks ∩ ( ∑ t∈G Kt) = (K ∩Rs) ∩ ( ∑ t∈G K ∩Rt) = K ∩ (Rs ∩ ∑ t∈G Rt) = K ∩ {0} = {0}, for all s ∈ G with s 6= t. Corollary 1. A subring K of a G-weak graded ring R is a G-weak graded subring if and only if K = ⊕ s∈G(K ∩Rs). Theorem 2. Let X be a group, H be a subgroup of X and (G, ∗) be a fixed set of left coset representatives. Suppose that G has a right inverse sR for each s ∈ G and that sR = sL = s−1. Then, the G-weak graded ring R is fully if and only if 1R ∈ RsLRs for all s ∈ G. Proof. Let R be a fully G-weak graded ring. Then, we have 1R ∈ ReG = RsL∗s = RsLRs. On the other hand, let 1R ∈ RsLRs. Since R is G-weak graded ring, we have RtRs ⊆ Rt∗s. So, we only need to show that Rt∗s ⊆ RtRs which we do as follows: Rt∗s = Rt∗s1R ⊆ Rt∗sRsLRs ⊆ R(t∗s)∗sLRs = R( t/f(s,sL) ) ∗(s∗sL) Rs = R( t/f(s,sL) ) ∗eG Rs = R(t/e−1 G )∗eGRs = RtRs, as required. Theorem 3. If R is a G-weak graded ring and I is an ideal of ReG, then IR ∩ReG = I. Proof. We have to show that I ⊆ IR ∩ ReG and IR ∩ ReG ⊆ I. It is obvious that I ⊆ IR∩ReG as, for any i ∈ I, we can write i = i1R ∈ IR as well as i ∈ ReG since I ⊆ ReG . So, it is enough to show that IR ∩ ReG ⊆ I which we do as follows: Suppose that x ∈ IR ∩ReG which implies that x ∈ IR and x ∈ ReG . Since x ∈ IR, we can write x as x = ∑ finite ir, (4) for i ∈ I and r ∈ R. As R is a G-weak graded ring, equation (4) can be rewritten as x = ∑ finite irs , for s ∈ G, where rs ∈ Rs. Also, as x ∈ ReG , hence x can be written as x = ireG . Thus, x ∈ I since I is an ideal of ReG . Therefore, IR ∩ReG ⊆ I which completes the proof. Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 337 Definition 6. For a G-weak graded ring R, a unit x ∈ U(R) is said to be weak graded unit or a G-homogeneous unit if x ∈ Rs for some s ∈ G, where U(R) is the group of all units in R. The set of all weak graded units in R is denoted by WGrU (R). Theorem 4. Let R be a G-weak graded ring with unity and let x be an element in U(R). If x ∈ Rs, for some s ∈ G, then x−1 ∈ RsL. Proof. Let x ∈ Rs for some s ∈ G and let x−1 = ∑ t∈G rt where rt ∈ Rt such that all but a finite number of them are zero. Hence, rtx ∈ RtRs ⊆ Rt∗s for any t ∈ G. Thus, x−1x = ∑ t∈G rtx is the unique expansion for x−1x in the direct sumR = ⊕ t∈GRt∗s which is equivalent to R = ⊕ s∈GRs since G is closed under the binary operation ∗. But, on the other side, we have x−1x = 1R ∈ ReG = RsL∗s. Consequently, ∑ t∈G rtx = 1R which implies that rt = 0 for all t 6= sL. So, if we put t = sL we get rsLx = 1R. Therefore, rsL = x−1 ∈ RsL as required. Theorem 5. If R is a G-weak graded ring, then the set of all its weak graded units, WGrU (R), is a subgroup of U(R) and the map 〈−〉 : WGrU (R) −→ G, satisfies the homomorphism property of groups with 〈−〉−1(eG) = WGrU(ReG), where 〈−〉 is the G- grade. Proof. First, to show that the set WGrU (R) is a subgroup of U(R), let x, y ∈ WGrU (R). Then x ∈ Rs and y ∈ Rt for some s, t ∈ G. Now, since R is a G-weak graded ring, we have xy ∈ RsRt ⊆ Rs∗t. Also, as G is closed under the operation ∗, we can put s ∗ t = p for some p ∈ G. Hence, xy is weak graded for all x, y ∈WGrU (R). In addition, as x, y ∈WGrU (R) ⊆ U(R) and that U(R) is a group under the ring multiplication, then xy ∈ U(R). Thus xy is a weak graded unit for any x, y ∈WGrU (R), i.e., xy ∈WGrU (R). Moreover, if x ∈WGrU (R), then, by Theorem 4, x−1 ∈WGrU (R). Next, to show that the map 〈−〉 satisfies the homomorphism property of groups, let x, y ∈WGrU (R) and suppose that x ∈ Rs and y ∈ Rt for some s, t ∈ G, i.e. 〈x〉 = s and 〈y〉 = t. Thus, xy ∈ RsRt ⊆ Rs∗t = Rp, for some p ∈ G. Consequently, 〈xy〉 = p = s ∗ t = 〈x〉 ∗ 〈y〉. Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 338 Finally, 〈−〉−1(eG) = {x ∈WGrU (R) : 〈x〉 = eG} = {x ∈WGrU (R) : x ∈ ReG} = WGrU(ReG), as required. It should be noted that G is, in general, neither a group nor even a monoid. Proposition 5. If a ring R is a G-weak graded ring, then the conjugation in R defines an action given by ϕR : reG , x −→ rxeG = x−1reGx of the group WGrU (R) as automorphisms of the subring ReG, where reG ∈ ReG and x ∈WGrU (R). Proof. The conjugation by a weak graded unit x in Rs for some s ∈ G is an automor- phism of the ring R satisfying: Rx eG = x−1ReGx ⊆ RsLReGRs ⊆ RsLReG∗s = RsLRs ⊆ ReG . Since WGrU (R) is a group, the proof is completed. 4. G-weak graded rings of fractions We start this section by recalling the following well known results [18]: (a) Let R be a ring and K be a multiplicatively closed subset of R such that 1R ∈ K, 0 /∈ K. Then the left ring of fractions with respect to K, K−1R, exists if and only if R satisfies the left Ore conditions with respect to K, i.e.: (i) If rk = 0, for some k ∈ K and r ∈ R, then there is an element k′ ∈ K such that k′r = 0. (ii) For r ∈ R and k ∈ K, there are elements r′ ∈ R and k′ ∈ K such that k′r = r′k. If Ore conditions with respect to K are satisfied, then K−1R = RK = { r k : r ∈ R, k ∈ K } . The addition and multiplication operations on RK are defined, respectively, by r k + r′ k′ = k′r+kr′ kk′ and r k · r′ k′ = r1r′ k′1k for k′1 ∈ K, r1 ∈ R with k′1r = r1k ′. (b) For every M in R-Mod, we can construct a fraction K−1M which is a left K−1R- module. Moreover, K−1M ∼= K−1R⊗R M . Now, we prove the following lemma which will be used to prove the the next theorem. Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 339 Lemma 1. Let X be a group, H be a subgroup of X and G ⊂ X be a set of left coset representatives. Then, for all s, t ∈ G, the multiplication s∗ t in G has a left inverse given by: (s ∗ t)L = ( tL / f ( sL / f(s, t), s ∗ t )−1 ) ∗ ( sL / f(s, t) ) . Proof. We have to show that(( tL / f ( sL / f(s, t), s ∗ t )−1 ) ∗ ( sL / f(s, t) )) ∗ (s ∗ t) = eG. (5) To do so, we start with the left hand side of equation (5) as follows:(( tL / f ( sL / f(s, t), s ∗ t )−1 ) ∗ ( sL / f(s, t) )) ∗ (s ∗ t) =(( tL / f ( sL / f(s, t), s ∗ t )−1) / f ( sL / f(s, t), s ∗ t )) ∗ ( (sL / f(s, t)) ∗ (s ∗ t) ) =( tL / f ( sL / f(s, t), s ∗ t )−1 f ( sL / f(s, t), s ∗ t )) ∗ ( (sL ∗ s) ∗ t ) = tL ∗ t = eG. The right division property yields ( tL / f ( sL / f(s, t), s ∗ t )−1 ) ∗ ( sL / f(s, t) ) = (s ∗ t)L as required. Theorem 6. Let R be a G-weak graded ring and K be a multiplicatively closed set of G-homogeneous elements not containing 0. Then the localization RK can be written as the direct sum of its G-components as follows: RK = ⊕ s∈G (RK)s, (as additive subgroups) with, RK = { r k : r ∈ R, k ∈ K } and (RK)s = { r k ∈ RK : r and k are G-homogeneous and 〈k〉L ∗ 〈r〉 = s } such that f(s, t) = f(s, p) for all s, t, p ∈ G, where 〈k〉, 〈r〉 are the G-grades of k and r respectively. Proof. We begin the proof by showing that (RK)s is an additive subgroup of RK as follows: If r k , r′ k′ ∈ (RK)s, then 〈k〉L ∗ 〈r〉 = s = 〈k′〉L ∗ 〈r′〉. Hence, r k + r′ k′ = k′r+kr′ kk′ . Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 340 Consequently, 〈k ′r + kr′ kk′ 〉 = 〈kk′〉L ∗ 〈kr′〉 = ( 〈k〉 ∗ 〈k′〉 )L ∗ (〈k〉 ∗ 〈r′〉) = (( 〈k′〉L / f ( 〈k〉L / f(〈k〉, 〈k′〉), 〈k〉 ∗ 〈k′〉 )−1) ∗ (〈k〉L / f(〈k〉, 〈k′〉) )) ∗ ( 〈k〉 ∗ 〈r′〉 ) = (( 〈k′〉L / f ( 〈k〉L / f(〈k〉, 〈k′〉), 〈k〉 ∗ 〈k′〉 )−1) / f ( 〈k〉L / f(〈k〉, 〈k′〉), 〈k〉 ∗ 〈r′〉 )) ∗ ( 〈k〉L / f(〈k〉, 〈k′〉) ∗ ( 〈k〉 ∗ 〈r′〉 )) = ( 〈k′〉L / f ( 〈k〉L / f(〈k〉, 〈k′〉), 〈k〉 ∗ 〈k′〉 )−1 f ( 〈k〉L / f(〈k〉, 〈k′〉), 〈k〉 ∗ 〈r′〉 ) ∗ (( 〈k〉L ∗ 〈k〉 ) ∗ 〈r′〉 )) = 〈k′〉L ∗ (eG ∗ 〈r′〉) = 〈k′〉L ∗ 〈r′〉 = s. Also, if r k ∈ (RK)s, then −( r k ) = −r k ∈ (RK)s . Indeed, if r k ∈ (RK)s, then 〈k〉L ∗ 〈r〉 = s = 〈k〉L ∗ 〈−r〉 as r is a G-homogeneous element, i.e. r is contained in fixed component and since each component are additive subgroup of R yields −r has the same G-grade. Hence, (RK)s is an additive subgroup of RK for all s ∈ G. Next, we have to show that RK = ⊕ s∈G(RK)s. It is obvious that RK = ∑ s∈G(RK)s. So, let r k ∈ (RK)sj ⋂ {(RK)s1 + . . . + (RK)sj−1 + (RK)sj+1 + . . . + (RK)sn} for all sj 6= s1, . . . , sj−1, sj+1, . . . , sn. Thus, 〈k〉L ∗ 〈r〉 = sj and 〈k〉L ∗ 〈r〉 = si , where i = 1, ..., j−1, j+1, ..., n. This means that, either (RK)sj = (RK)si which is a contradiction or r k = 0RK which implies RK = ⊕ s∈G(RK)s as required. 5. Additional examples of G-weak graded rings In this section, we give additional examples of G-weak and fully G-weak graded rings that are not trivially constructed. Example 1. Consider a ring R to be the ring of all 2× 2 matrices over the field R, i.e., R = M2(R) = {( a b c d ) : a, b, c, d ∈ R } . Let X be the dihedral group D6 = < x, y : x6 = y2 = 1, xy = yx5 > and H be the non-normal subgroup {1, x3, y, x3y}. Choose G = {1, x, x5} to be the set of left coset representatives. Then the ∗ and f operations as well as the actions / and . are given by the following tables: Thus, R = R1 ⊕Rx ⊕Rx5, where R1 = {( a 0 0 d ) : a, d ∈ R } , Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 341 Table 1: ∗ and f operations. ∗ 1 x x5 1 1 x x5 x x x5 1 x5 x5 1 x f 1 x x5 1 1 1 1 x 1 x3 1 x5 1 1 x3 Table 2: . and / actions. s.u 1 x3 y x3y 1 1 x3 y x3y x 1 x3 y x3y x5 1 x3 y x3y s/u 1 x3 y x3y 1 1 1 1 1 x x x x5 x5 x5 x5 x5 x x Rx = {( 0 0 c 0 ) : c ∈ R } and Rx5 = {( 0 b 0 0 ) : b ∈ R } . Moreover, the inclusion property RsRt ⊆ Rs∗t is satisfied for all s, t ∈ G. This can be detailed as follows: (i) R1R1 ⊆ R1∗1 = R1, as for all ( a1 0 0 d1 ) , ( a2 0 0 d2 ) ∈ R1, we have ( a1 0 0 d1 )( a2 0 0 d2 ) = ( a1a2 0 0 d1d2 ) ∈ R1 = R1∗1. (ii) R1Rx ⊆ R1∗x = Rx, as for all ( a 0 0 d ) ∈ R1 and ( 0 0 c 0 ) ∈ Rx, we have ( a 0 0 d )( 0 0 c 0 ) = ( 0 0 dc 0 ) ∈ Rx = R1∗x. (iii) R1Rx5 ⊆ R1∗x5 = Rx5, as for all ( a 0 0 d ) ∈ R1 and ( 0 b 0 0 ) ∈ Rx5, we have ( a 0 0 d )( 0 b 0 0 ) = ( 0 ab 0 0 ) ∈ Rx5 = R1∗x5 . (iv) RxR1 ⊆ Rx∗1 = Rx, as for all ( 0 0 c 0 ) ∈ Rx and ( a 0 0 d ) ∈ R1, we have ( 0 0 c 0 )( a 0 0 d ) = ( 0 0 ca 0 ) ∈ Rx = Rx∗1. Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 342 (v) RxRx ⊆ Rx∗x = Rx5, as for all ( 0 0 c1 0 ) , ( 0 0 c2 0 ) ∈ Rx, we have ( 0 0 c1 0 )( 0 0 c2 0 ) = ( 0 0 0 0 ) ∈ Rx5 = Rx∗x. (vi) RxRx5 ⊆ Rx∗x5 = R1, as for all ( 0 0 c 0 ) ∈ Rx and ( 0 b 0 0 ) ∈ Rx5, we have ( 0 0 c 0 )( 0 b 0 0 ) = ( 0 0 0 cb ) ∈ R1 = Rx∗x5 . (vii) Rx5R1 ⊆ Rx5∗1 = Rx5, as for all ( 0 b 0 0 ) ∈ Rx5 and ( a 0 0 d ) ∈ R1, we have ( 0 b 0 0 )( a 0 0 d ) = ( 0 bd 0 0 ) ∈ Rx5 = Rx5∗1. (viii) Rx5Rx ⊆ Rx5∗x = R1, as for all ( 0 b 0 0 ) ∈ Rx5 and ( 0 0 c 0 ) ∈ Rx, we have ( 0 b 0 0 )( 0 0 c 0 ) = ( bc 0 0 0 ) ∈ R1 = Rx5∗x. (ix) Rx5Rx5 ⊆ Rx5∗x5 = Rx, as for all ( 0 b1 0 0 ) , ( 0 b2 0 0 ) ∈ Rx5, we have ( 0 b1 0 0 )( 0 b2 0 0 ) = ( 0 0 0 0 ) ∈ Rx = Rx5∗x5 . Therefore, R is a G-weak graded ring. However, it is not a fully G-weak graded ring. For instance, Rx5Rx5 6= Rx5∗x5 since Rx = Rx5∗x5 * Rx5Rx5. Example 2. Let X = (Z6,+) and H =< 3 >= {0, 3}. Choose the set of left coset representatives to be G = {1, 3, 5}. Then the ∗ and f operations as well as the actions /, . are given by the following tables: If we consider the Morita ring T = ( R M N S ) , then we have: T = T1 ⊕ T3 ⊕ T5 , where T1 = ( 0 M 0 0 ) , T3 = ( R 0 0 S ) and T5 = ( 0 0 N 0 ) . Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 343 Table 3: ∗ and f operations. ∗ 3 1 5 3 3 1 5 1 1 5 3 5 5 3 1 f 3 1 5 3 3 3 3 1 3 3 3 5 3 3 3 Table 4: . and / actions. s . u 0 3 3 0 3 1 0 3 5 0 3 s / u 0 3 3 3 3 1 1 1 5 5 5 Moreover, the inclusion property TsTt ⊆ Ts∗t is satisfied for all s, t ∈ G which can illustrated as follows: (i) T3T3 ⊆ T3∗3 = T3, as for all ( r1 0 0 s1 ) and ( r2 0 0 s2 ) ∈ T3, we have ( r1 0 0 s1 )( r2 0 0 s2 ) = ( r1r2 0 0 s1s2 ) ∈ T3 = T3∗3. (ii) T3T1 ⊆ T3∗1 = T1, as for all ( r 0 0 s ) ∈ T3 and ( 0 m 0 0 ) ∈ T1, we have ( r 0 0 s )( 0 m 0 0 ) = ( 0 rm 0 0 ) ∈ T1 = T3∗1. (iii) T3T5 ⊆ T3∗5 = T5, as for all ( r 0 0 s ) ∈ T3 and ( 0 0 s 0 ) ∈ T5, we have ( r 0 0 s )( 0 0 s 0 ) = ( 0 0 sn 0 ) ∈ T5 = T3∗5. (iv) T1T3 ⊆ T1∗3 = T1, as for all ( 0 m 0 0 ) ∈ T1 and ( r 0 0 s ) ∈ T3, we have ( 0 m 0 0 )( r 0 0 s ) = ( 0 ms 0 0 ) ∈ T1 = T1∗3. (v) T1T1 ⊆ T1∗1 = T5, as for all ( 0 m1 0 0 ) and ( 0 m2 0 0 ) ∈ T1, we have ( 0 m1 0 0 )( 0 m2 0 0 ) = ( 0 0 0 0 ) ∈ T5 = T1∗1. Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 344 (vi) T1T5 ⊆ T1∗5 = T3, as for all ( 0 m 0 0 ) ∈ T1 and ( 0 0 n 0 ) ∈ T5, we have ( 0 m 0 0 )( 0 0 n 0 ) = ( mn 0 0 0 ) ∈ T3 = T1∗5. (vii) T5T3 ⊆ T5∗3 = T5, as for all ( 0 0 n 0 ) ∈ T5 and ( r 0 0 s ) ∈ T3, we have ( 0 0 n 0 )( r 0 0 s ) = ( 0 0 nr 0 ) ∈ T5 = T5∗3. (viii) T5T1 ⊆ T5∗1 = T3, as for all ( 0 0 n 0 ) ∈ T5 and ( 0 m 0 0 ) ∈ T1, we have ( 0 0 n 0 )( 0 m 0 0 ) = ( 0 0 0 mn ) ∈ T3 = T5∗1. (ix) T5T5 ⊆ T5∗5 = T1, as for all ( 0 0 n1 0 ) and ( 0 0 n2 0 ) ∈ T5, we have ( 0 0 n1 0 )( 0 0 n2 0 ) = ( 0 0 0 0 ) ∈ T1 = T5∗5. Thus, T is a G-weak graded ring. However, it is not a fully G-weak graded ring. For instance, T5T5 6= T5∗5 since T1 = T5∗5 * T5T5. Example 3. Consider the ring of real quaternions (H,+, ·). Let X = D6 = {1, x, x2, x3, x4, x5, y, xy, x2y, x3y, x4y, x5y} and H = {1, x2, x4} be an additive subgroup of the group X. Take the set of left coset representatives to be G = {1, y, x5, xy}. Then the ∗ and f operations as well as the actions /, . are given by the following tables: Table 5: ∗ and f operations. ∗ 1 y x5 xy 1 1 y x5 xy y y 1 xy x5 x5 x5 xy 1 y xy xy x5 y 1 f 1 y x5 xy 1 1 1 1 1 y 1 1 1 1 x5 1 x4 x4 1 xy 1 x2 x2 1 Najla Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 12 (2) (2019), 332-347 345 Table 6: . and / actions. s . u 1 x2 x4 1 1 x2 x4 y 1 x4 x2 x5 1 x2 x4 xy 1 x4 x2 s / u 1 x2 x4 1 1 1 1 y y y y x5 x5 x5 x5 xy xy xy xy Thus, H = R1 ⊕ Ry ⊕ Rx5 ⊕ Rxy where, R1 = R, Ry = Ri, Rx5 = Rj and Rxy = Rk. The inclusion property can be checked as follows for any r, r′ ∈ R: (i) R1R1 ⊆ R1∗1 = R1, as for all r, r′ ∈ R1, we have (r)(r′) = rr′ ∈ R1 = R1∗1. (ii) R1Ry ⊆ R1∗y = Ry, as for all r ∈ R1, r ′i ∈ Ry, we have (r)(r′i) = (rr′)i ∈ Ry = R1∗y. (iii) R1Rx5 ⊆ R1∗x5 = Rx5, as for all r ∈ R1, r ′j ∈ Rx5, we have (r)(r′j) = (rr′)j ∈ Rx5 = R1∗x5 . (iv) R1Rxy ⊆ R1∗xy = Rxy, as for all r ∈ R1, r ′k ∈ Rxy, we have (r)(r′k) = (rr′)k ∈ Rxy = R1∗xy. (v) RyR1 ⊆ Ry∗1 = Ry, as for all ri ∈ Ry, r ′ ∈ R1, we have (ri)(r′) = (rr′)i ∈ Ry = Ry∗1. (vi) RyRy ⊆ Ry∗y = R1, as for all ri, r′i ∈ Ry, we have (ri)(r′i) = (−rr′) ∈ R1 = Ry∗y. (vii) RyRx5 ⊆ Ry∗x5 = Rxy, as for all ri ∈ Ry, r ′j ∈ Rx5, we have (ri)(r′j) = (rr′)k ∈ Rxy = Ry∗x5 . (viii) RyRxy ⊆ Ry∗xy = Rx5, as for all ri ∈ Ry, r ′k ∈ Rxy, we have (ri)(r′k) = (−rr′)j ∈ Rx5 = Ry∗xy. (ix) Rx5R1 ⊆ Rx5∗1 = Rx5, as for all rj ∈ Rx5 , r′ ∈ R1, we have (rj)(r′) = (rr′)j ∈ Rx5 = Rx5∗1. (x) Rx5Ry ⊆ Rx5∗y = Rxy, as for all rj ∈ Rx5 , r′i ∈ Ry, we have (rj)(r′i) = (−rr′)k ∈ Rxy = Rx5∗y. (xi) Rx5Rx5 ⊆ Rx5∗x5 = R1, as for all rj, r′j ∈ Rx5, we have (rj)(r′j) = (−rr′) ∈ R1 = Rx5∗x5 . (xii) Rx5Rxy ⊆ Rx5∗xy = Ry, as for all rj ∈ Rx5 , r′k ∈ Rxy, we have (rj)(r′k) = (rr′)i ∈ Ry = Rx5∗xy. (xiii) RxyR1 ⊆ Rxy∗1 = Rxy, as for all rk ∈ Rxy, r ′ ∈ R1, we have (rk)(r′) = (rr′)k ∈ Rxy = Rxy∗1. REFERENCES 346 (xiv) RxyRy ⊆ Rxy∗y = Rx5, as for all rk ∈ Rxy, r ′i ∈ Ry, we have (rk)(r′i) = (rr′)j ∈ Rx5 = Rxy∗y. 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