Some properties of graded comultiplication modules

Khaldoun Al-Zoubi; Amani Al-Qderat

doi:10.1515/math-2017-0016

Article Open Access

Some properties of graded comultiplication modules

Khaldoun Al-Zoubi and Amani Al-Qderat

Published/Copyright: March 4, 2017

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$Open Mathematics$

From the journal Open Mathematics Volume 15 Issue 1

Abstract

Let G be a group with identity e. Let R be a G-graded commutative ring and M a graded R-module. In this paper we will obtain some results concerning the graded comultiplication modules over a commutative graded ring.

Keywords: Graded comultiplication module; Graded multiplication module; Graded submodule

MSC 2010: 13A02; 16W50

1 Introduction and preliminaries

Graded multiplication modules (gr-multiplication modules) over commutative graded ring have been studied by many authors extensively (see [1–7]). As a dual concept of gr-multiplication modules, graded comultiplication modules (gr-comultiplication modules) were introduced and studied by Ansari-Toroghy and Farshadifar [8]. A graded R-module M is said to be graded multiplication module (gr-multiplication module) if for every graded submodule N of M there exists a graded ideal I of R such that N = IM (see [3]). A graded R-module M is said to be graded comultiplication module (gr-comultiplication module) if for every graded submodule N of M there exists a graded ideal I of R such that N = (0 :_M I), where (0 :_M I)= {m ∈ M : mI = 0} (see [8]). Also it was shown that M is a gr-comultiplication module if and only if for each graded submodule N of M, N = (0 :_M Ann_R(N)). Here we will study the class of graded comultiplication modules and obtain some further results which are dual to classical results on graded multiplication modules (see Section 2).

First, we recall some basic properties of graded rings and modules which will be used in the sequel. We refer to [9] and [10] for these basic properties and more information on graded rings and modules. Let G be a group with identity e and R be a commutative ring with identity 1_R. Then R is a G-graded ring if there exist additive subgroups R_g of R such that R = ⊕_g∈G R_g and R_g R_h ⊆ R_gh for all g, h ∈ G. The elements of R_g are called to be homogeneous of degree g where the R_g’s are additive subgroups of R indexed by the elements g ∈ G. If x ∈ R, then x can be written uniquely as ∑_g∈G x_g, where x_g is the component of x in R_g. Also we write, h(R) = ⋃_{g ∈ G} R_g. Moreover, R_e is a subring of R and 1_R ∈ R_e. Let I be an ideal of R. Then I is called a graded ideal of (R, G) if I = ⊕_{g ∈ G}(I ∩ R_g). Thus, if x ∈ I, then x = ∑_{g ∈ G} x_g with x_g∈ I. An ideal of a G-graded ring need not be G-graded.

Let R be a G-graded ring and M an R-module. We say that M is a G-graded R-module (or graded R-module) if there exists a family of subgroups {M_g}_g∈G of M such that M = ⨁ g ∈ G M g (as abelian groups) and R_gM_h⊆ M_gh for all g, h ∈ G. Here, R_gM_h denotes the additive subgroup of M consisting of all finite sums of elements r_gs_h with r_g ∈ R_g and s_h ∈ M_h. Also, we write h(M)=⋃g∈GMg and the elements of h(M) are called to be homogeneous.

Let M=⨁g∈GMg be a graded R-module and N a submodule of M. Then N is called a graded submodule of M if N=⨁g∈GNg where N_g = N ∩ M_g for g ∈ G. In this case, N_g is called the g-component of N.

Let R be a G-graded ring and M a graded R-module. A proper graded ideal I of R is said to be graded maximal ideal (or gr-maximal ideal) of R if J is a graded ideal of R such that I ⊆ J ⊆ R, then I = J or J = R.

A non-zero (resp. a proper) graded ideal I of a G-graded ring R is said to be gr-large (resp. gr-small) if for every non-zero (resp. proper) graded ideal J of R, we have I ∩ J ≠ 0 (resp. I + J ≠ R).

A graded submodule N of a graded R-module M is said to be graded minimal (gr-minimal) if it is minimal in the lattice of graded submodules of M.

A non-zero (resp. a proper) graded submodule N of a graded R-module M is said to be gr-large (resp. gr- small) if for every non-zero (resp. proper) graded submodule L of M, we have N ∩ L ≠ 0 (resp. L + N ≠ M).

A graded R-module M is said to be gr-uniform (resp. gr-hollow) if each of its proper graded submodules is gr-large (resp. gr-small).

A graded R-module M is said to be gr-simple if (0) and M are its only graded submodules.

A graded R-module M is said to be gr-faithful if aM = 0 implies a = 0 for a ∈ h(R).

A graded R-module M is said to be graded finitely generated if there exist x_g1, x_g2, …, x_gn ∈ h(M) such that M = Rx_g1 + … + Rx_gn.

A graded R-module M is said to be gr-Artinian if satisfies the descending chain condition for graded submodules.

A proper graded ideal P of a G-graded ring R is said to be graded prime ideal (gr-prime ideal) if whenever r, s ∈ h(R) with rs ∈ P, then either r ∈ P or s ∈ P (see [11]).

A non-zero graded submodule N of a graded R-module M is said to be a graded second (gr-second) if for each homogeneous element a of R, the endomorphism of M given by multiplication by a is either surjective or zero (see [8]).

2 Results

The following lemma is known (see [12] and [6]), but we write it here for the sake of references.

Lemma 2.1

Let R be a G-graded ring and M a graded R-module. Then the following hold:

If I and J are graded ideals of R, then I + J and I ∩ J are graded ideals.
If N is a graded submodule of M, r ∈ h(R), x ∈ h(M) and I is a graded ideal of R, then Rx, IN and rN are graded submodules of M.
If N and K are graded submodules of M, then N + K and N ∩ K are also graded submodules of M and (N :_R M) = {r ∈ R : rM ⊆ N} is a graded ideal of R.
Let {N_λ} be a collection of graded submodules of M. Then ∑λNλand∩λ⁡Nλ are graded submodules of M.

Recall that a non-zero graded R-module M over a G-graded ring R is said to be gr-prime if Ann_R(M) = Ann_R(K) for every non-zero graded submodule K of M (see[10]).

Theorem 2.2

Let R be a G-graded ring and M a graded R-module. If M is a gr-comultiplication gr-prime R-module, then M is a gr-simple module.

Proof

Let K be a non-zero graded submodule of M. Since M is gr-prime, we have Ann_R(K) = Ann_R(M). By[8, Lemma 3.5], we have K = M. Therefore M is a gr-simple module. □

Recall that a graded R-module M is called finitely gr-cogenerated if for every non-empty family of graded submodules K_r(r ∈ L) of M such that ∩_{r∈ L} K_r = 0, there exists a finite subset F ⊆ L verifying ∩_r∈F K_r = 0 (see [5]).

Theorem 2.3

Let R be a G-graded ring and M a graded R-module. If M is finitely gr-cogenerated gr-comultiplication R-module such that for each graded ideal J of R and for each collection {K_α}_α∊Λ of graded submodules of M, we have (∩_α∊ΛK_α)J = ∩_α∊Λ(K_αJ), then M is gr-Aritinian module.

Proof

Let K₁⊇ K₂⊇ K₃⊇⋯ be a descending chain of graded submodules of M. Set I=AnnR(∩i=1∞Ki). By assumption, J(∩i=1∞Ki)=(∩i=1∞JKi). Hence (∩i=1∞JKi)=0. Since M is finitely gr-cogenerated R-module, there exists a positive integer j such that JK_j = 0. Since M is a gr-comultiplication R-module, Kj⊆(∩i=1∞Ki). This completes the proof because the reverse inclusion is clear. □

Recall that a G-graded ring R is said to be a gr-comultiplication ring if it is a gr-comultiplication R-module (see [8]).

Theorem 2.4

Let R be a gr-comultiplication ring and M a graded R-module.

If M is a gr-faithful R-module, then for each proper graded ideal J of R, (0:_M J)≠ 0 and JM≠ M.
If M is a gr-faithful gr-comultiplication R-module, then for each collection {J_α}_α∊Λ of graded ideals of R, (0:M∩α∈ΛJα)=∑α∈Λ(0:MJα).

Proof

Let J be a proper graded ideal of R. Suppose that (0 :_M J) = 0. If Ann_R (J)M≠ 0, there exists r_g ∊ h(Ann_R(J)) and m_λ∊ h(M) such that r_gm_λ≠ 0. Since Jr_gm_λ = 0 m_λ = 0, r_gm_λ∊(0 :_M J) = 0, which is impossible. Consequently, Ann_R(J)M = 0 and so Ann_R(J)⊆ Ann_R(M) . Thus Ann_R(J) = 0. Since R is gr-comultiplication ring, we conclude that J = Ann_R(Ann_R(J)) = R, which is a contradiction. Therefore (0 :_M J)≠ 0. Now assume that J M = M. Then Ann_R(J) = Ann_R(M) = 0. A similar argument yields a similar contradiction and thus completes the proof.
For each α ∊ Λ since M is gr-faithful it follows that Ann_R(Ann_R(J_α)M) = Ann_R(Ann_R(J_α)) . Since R is a gr-comultiplication ring, we have Ann_R(Ann_R(J_α)) = J_α. By [8, Theorem 3.8(a)], we conclude that (0 :_M ∩_α∊ΛJ_α) = (0 :_M∩_α∊Λ Ann_R(Ann_R(J_α)) = (0 :_M∩_α∊Λ Ann_R(Ann_R(J_α)M)) = Σ_α∊Λ(0 :_M Ann_R(Ann_R (J_α)M)) = Σ_α∊Λ(0 :_M Ann_R (Ann_R (J_α))) = Σ_α∊Λ (0 :_M J_α)). □

Theorem 2.5

Let R be G-graded ring and M a gr-comultiplication R-module. If every gr-prime ideal of R is contained in a unique gr-maximal ideal of R, then every gr-second submodule of M contains a unique gr-minimal submodule of M.

Proof

Let N be a gr-second submodule of M. By [8, Theorem 3.9(a)], N contains a gr-minimal submodule of M. Now, assume that K₁ and K₂ are two gr-minimal submodules of M, such that K₁ ⊆ N and K₂ ⊆ N. Hence Ann_R(N)⊆ Ann_R(K₁) and Ann_R (N)⊆ Ann_R (K₂) . Since N is a gr-second submodule of M, by [8, Proposition 3.15(a)], we have Ann_R (N) is a gr-prime ideal of R. Now, as Ann_R (K₂) and Ann_R(K₁) are gr- maximal ideals of R, by assumption we must have Ann_R (K₁) = Ann_R (K₂). Thus by [8, Lemma 3.5], K₁ = K₂. □

Theorem 2.6

Let R be a G-graded ring, M a gr-comultiplication R-module and (0 :_M I)⊆(0 :_M J) for some graded ideals I and J of R. If there exists a gr-finitely generated gr-multiplication submodule N of M such that Ann_R(N)⊆ I, then J⊆ I.

Proof

Let N be a gr-finitely generated gr-multiplication submodule of M. Since (0 :_M I)⊆(0 :_M J), (0 :_N I)⊆(0 :_N J). By [8, Theorem 3.7(a)], N is a gr-comultiplication R-module. By [8, Theorem 3.7(c)], JN⊆ IN. Since N is a gr-finitely generated gr-multiplication module, J⊆ I+Ann_R(N) = I by [1, Lemma 3.9]. □

Lemma 2.7

Let R be a G-graded ring, M a gr-comultiplication R-module and K and N a graded submodules of M. Then (0 :_M(K :_R N)) = Ann_R(K)N.

Proof

Note first that (K :_R N)Ann_R (K) N = Ann_R (K)(K :_R N)N ⊆ Ann_R(K)K = 0. Hence Ann_R(K)N ⊆(0 :_M(K :_R N)) . Since M is a gr-comultiplication module, Ann_R(K)N = (0 :_M I) for some graded ideal I of R and so Ann_R (K) IN = 0. It follows that IN ⊆(0 :_M Ann_R (K)). Since M is a gr-comultiplication module, (0 :_M Ann_R (K)) = K. Thus I ⊆ (K :_R N) and hence (0 :_M (K:_R N))⊆(0 :_M I) = Ann_R (K)N. Therefore Ann_R (K)N = (0 :_M (K :_R N)) □

Let M and M′ be two graded R-modules. A homomorphism of graded R-modules φ : M → M′ is ahomomorphism of R-modules verifying φ(Mg)⊆Mg′ for every g ∊ G (see [10]).

Theorem 2.8

Let R be a G-graded ring and M a gr-comultiplication R-module. Then we have the following

For each graded endomorphism, φ of M, we have φ(M) = Ann_R(ker(φ))M.
If N is a gr-minimal submodule of M, and K₁ and K₂ are graded submodules of M with K₁ ∩ N = K₂ ∩ N = 0, then N ∩ (K₁ + K₂) = 0.
If {N_λ}_λ∈Λ is a collection of graded submodules of M such that ∩_λ∈Λ N_λ = 0 and I = ∑_λ∈Λ An n_R(M_λ), then R = Ann_R(K) + I for every gr-finitely generated graded submodule K of M.

Proof

Let φ be a graded endomorphism of a gr-comultiplication R-module M. Since φ(M) ≅ M/Kerφ, we have Ann_R(φ(M)) = (Kerφ :_R M). Since M is a gr-comultiplication module and φ(M) is a graded submodule of M, we have φ(M) = (0 :_M Ann_R(φ(M)). Hence φ(M) = (0 :_M(Kerφ :_R M)) = Ann_R(ker(φ))M, by Lemma 2.7.
Let N be a gr-minimal submodule of M and let K₁, K₂ be two graded submodules of M such that N ∩ K₂ = N ∩ K₁ = 0. Since M is a gr-comultiplication module, K₁ = (0 :_M Ann_R (K₁)) and K₂ = (0 :_M Ann_R(K₂)). Now (0 :_M Ann_R(K₁)Ann_R(K₂)) ∩ N = N or (0:_MAnn_R(K₁)Ann_R(K₂)) ∩ N = 0 because N is a gr-minimal submodule of M. In the first case, we have N = (0:_NAnn_R(K₁)Ann_R(K₂)) = ((0 :_N Ann_R(K₁)):_NAnn_R(K₂)). It follows that N = ((0 :_N Ann_R(K₁)) :_NAnn_R(K₂)) = ((0 :_M Ann_R(K₁)) ∩ N :_NAnn_R(K₂)) = (K₁ ∩ N :_N Ann_R(K₂)) = N ∩ (0 :_M Ann_R(K₂) = N ∩ K₂ = 0 which is a contradiction. In the second case, N ∩(0 :_M Ann_R(K₁)Ann_R(K₂)) = 0 and since (0 :_M Ann_R(K₁)Ann_R(K₂)) ⊇ (0 :_M Ann_R(K₁) ∩ Ann_R(K₂)) = K₁+K₂, we have N ∩ (K₁ + K₂) = 0.
Let K be a gr-finitely generated graded submodule of M and let Ann_R(K) + I ≠ R. Then there exists a gr-maximal ideal J of R such that Ann_R(K) + I ⊆ J. Since I ⊆ J, we have (0 :_M J) ⊆ (0 :_M I) = ∩_{λ ∈ Λ}(0 :_M Ann_R(N_λ)) = ∩_{λ ∈ Λ} N_λ = 0. Thus, (0 :_M J) = 0 and hence (0 :_N J) = 0. By [1, Theorem 3.8(c)], there exists p ∈ J such that 1 − p ∈ Ann_R(K) ⊆ J, a contradiction: Thus Ann_R(K) + I = R. □

Recall that a proper graded ideal I of a G-graded ring R is said to gr-irreducible if whenever I = I₁ ∩ I₂ for graded ideals I₁ and I₂ of R, then I = I₁ or I = I₂ (see [11]).

Theorem 2.9

Let R be a G-graded ring and M a gr-comultiplication R-module such that Ann_R(M) is gr-irreducible ideal of R. Then M is gr-hollow module.

Proof

Let K₁ and K₂ be graded submodules of M with M = K₁+K₂. Then Ann_R(M) = Ann_R(K₁ + K₂) = Ann_R(K₁) ∩ Ann_R(K₂). Since Ann_R(M) is gr-irreducible ideal of R, either Ann_R(M) = Ann_R(K₁) or Ann_R(M) = Ann_R(K₂) without loss of generallity, we can suppose that Ann_R(M) = Ann_R(K₁). Since M is gr-comultiplication module, M = K₁ by [8, Lemma 3.5]. It follows that M is gr-hollow module. □

Theorem 2.10

Let R be a G-graded ring and M a gr-faithful gr-comultiplication module with the property (0 :_M I) + (0 :_M J) = (0 :_M(I ∩ J)) for any two graded ideals I and J of R. Then a graded submodule N of M is a gr-small if and only if there exists a gr-large ideal I of R such that N = (0 :_M I).

Proof

Suppose first that N is a gr-small submodule of M. Since M is a gr-comultiplication module, there exists graded ideal I of R such that N = (0 :_M I). Suppose I ∩ J = 0 for some graded ideal J of R. Then N + (0 :_M J) = (0 :_M I) + (0 :_M J) = (0 :_M(I ∩ J)) = M. Since N is a gr-small submodule of M, (0 :_M J) = M, and hence J ⊆ Ann_R(M) = 0. Thus I is a gr-large ideal of R. Conversely, suppose that I is a gr-large ideal of R such that N = (0 :_M I) and K is a graded submodule of M such that N + K = (0 :_M I) + K = M. Since M is a gr-comultiplication module, there exists a graded ideal J of R such that K = (0 :_M J). Then (0 :_M(I ∩ J)) = (0 :_M I) + (0 :_M J) = (0 :_M I) + K = M and so I ∩ J ⊆ Ann_R(M) = 0. Since I is gr-large ideal of R, J = 0. Hence K = (0 :_M J) = M. Therefore N = (0 :_M I) is a gr-small submodule of M. □

Theorem 2.11

Let R be a G-graded ring and M a gr-comultiplication module with the property I = Ann_R(0:_M I) for each graded ideal I of R. Then a graded submodule N of M is a gr-large if and only if there exists a gr-small ideal I of R such that N = (0:_M I).

Proof

Suppose first that N is a gr-large submodule of M. Since M is a gr-comultiplication module, there exists a graded ideal I of R such that N = (0:_M I) . Suppose I + J = R for some graded ideal J of R. Then N ∩ (0:_M J) = (0:_M I) ∩ (0:_M J) = (0:_M I + J) = (0:_MR) = 0. Since N is a gr-large submodule of M, (0:_M J) = 0. Hence J = Ann_R((0:_M J)) = Ann_R(0) = R. So I is a gr-small ideal of R. Conversely, suppose that I is a gr-small ideal of R such that N = (0:_M I) . Assume K is a graded submodule of M such that K ∩ N = K ∩ (0:_M I) = 0. Since M is gr-comultiplication module, there exists a graded ideal J of R such that K = (0:_M J). It follows that (0:_M (I + J)) = (0:_M I) ∩ (0:_M J) = K ∩ (0:_M I) = 0. Then I + J = Ann_R(0:_M (I + J)) = Ann_R(0) = R. Since I is a gr-small ideal of R, J = R. Hence K = (0:_M J) = (0:_MR) = 0 and so N = (0:_M I) is a gr-large submodule of M. □

Theorem 2.12

Let R be a G-graded ring and M a gr-comultiplication module with the property I = Ann_R(0:_M I) for each graded ideal I of R. Then M is gr-uniform if and only if R is gr-hollow

Proof

Suppose first that M is gr-uniform and I a graded proper ideal of R such that I + J = R for some graded ideal J of R. Then (0:_M I) ∩ (0:_M J) = (0:_M (I + J)) = (0:_MR) = 0. Since M is gr-uniform, (0:_M I) = 0 or (0:_M J) = 0. Then I = Ann_R(0:_M I) = Ann_R(0) = R or J = Ann_R(0:_M J) = Ann_R(0) = R. Since I is graded proper ideal of R, J = R. Hence I is a gr-small ideal of R. Therefore R is gr-hollow. Conversely, let 0 ≠ N and K be two graded submodules of M such that N ∩ K = 0. Since M is a gr-comultiplication module, there exist graded ideals I and J of R such that N = (0:_M I) and K = (0:_M J) . Hence (0:_M (I + J)) = (0:_M I) ∩ (0:_M J) = N ∩ K = 0. Thus I + J = Ann_R(0:_M (I + J)) = Ann_R(0) = R. Since R is gr-hollow, I = R or J = R. Since N ≠ 0, I ≠ R hence J = R. It follows that K = (0:_M J) = (0:_MR) = 0 and so N is gr-large submodule of M. Therefore M is gr-uniform. □

Definition 2.13

Let R be a G-graded ring, M a graded R-module and L a graded submodule of M.

A graded homomorphism φ : L → M is called gr-trivial if there exists r ∈ h(R) such that φ(x) = rx(x ∈ L).
A graded module M is said to be a gr-strongly-self cogenerated provided that for each graded submodule N of M there exists a family φ_λ (λ ∈ Λ) of gr-trivial endomorphisms of M, for some index set Λ, such that N = ∩ _{λ ∈ Λ}kerφ_λ.

Theorem 2.14

Let R be a G-graded ring and M a graded R-module. Then M is a gr-comultiplication module if and only if M is gr-strongly self-cogenerated.

Proof

Suppose first that M is gr-comultiplication R-module and N a graded submodule of M. Since M is a gr-comultiplication module, there exists a graded ideal I of R such that N = (0:_M I) . For each i ∈ I ∩ h(R), let φ_i denote the gr-trivial graded endomorphism of M defined by φ_i(m) = im (m ∈ M). Thus N = ∩ _{i ∈ I}kerφ_i. Conversely, let N be a graded submodule of M. By hypothesis there exists an index set Λ and gr-trivial graded homomorphism φ_λ (λ ∈ Λ) such that N = ∩ _{λ ∈ Λ}kerφ_λ. For each λ ∈ Λ there exists r_λ ∈ h(R) such that θ_λ(m) = r_λm (m ∈ M). Let J denote the graded ideal ∑_{λ ∈ Λ} Rr_λ. It is easy to show that N = (0:_M J) . Therefore M is a gr-comultiplication module. □

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Received: 2016-5-25

Accepted: 2016-12-21

Published Online: 2017-3-4

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