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Some properties of graded generalized 2-absorbing submodules

  • Shatha Alghueiri and Khaldoun Al-Zoubi EMAIL logo
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Demonstratio Mathematica
From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

Let G be an abelian 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 generalized 2-absorbing submodules and their homogeneous components. Special attention has been paid, when graded rings are graded gr-Noetherian, to find extra properties of these graded submodules.

Keywords: graded generalized 2-absorbing submodule; graded 2-absorbing submodule; graded 2-absorbing primary submodule
MSC 2010: 13A02; 16W50

1 Introduction and preliminaries

Throughout this paper all rings are commutative with identity and all modules are unitary.

Badawi in [1] introduced the concept of 2-absorbing ideals of commutative rings. The notion of 2-absorbing ideals was extended to 2-absorbing submodules in [2] and [3]. The concept of generalized 2-absorbing submodules, as a generalization of 2-absorbing submodules, was introduced in [4].

In [5], Refai and Al-Zoubi introduced the concept of graded primary ideal. The concept of graded 2-absorbing ideals, as a generalization of graded prime ideals, was introduced and studied by Al-Zoubi et al. in [6]. In [7], Al-Zoubi and Sharafat introduced the concept of graded 2-absorbing primary ideal, which is a generalization of graded primary ideal. The concept of graded 2-absorbing submodules was introduced by Al-Zoubi and Abu-Dawwas in [8] and studied in [9,10].

Recently, the authors in [11] introduced the concept of graded generalized 2-absorbing submodules as a generalization of graded 2-absorbing submodules.

Here, we introduce several results concerning graded generalized 2-absorbing submodules and their homogeneous components. First, we customized the definition of “graded generalized 2-absorbing submodule” of proper graded submodule to direct summand. Graded generalized 2-absorbing submodule of the graded R -module over a graded ring R is classified with the help of the annihilator and the graded ideal. So Theorems 2.5 and 2.6 are important. Then the concept of graded G 2 -absorbing submodule was moved to localization. It has been proved in Theorem 3.2 that direct summand of graded submodule of graded R-module over G-graded g r -Noetherian ring R is g - G 2 -absorbing submodule by means of the definitions of g -idempotent submodule and fully g -idempotent submodule. It is also exemplified in Example 3.3 that the theorem does not work if the conditions in the hypothesis of Theorem 3.2 are not met. According to this module, it has been proven that the e-primary ideal of the annihilator coincides with g-primary submodule of a g-multiplication module in Theorem 3.7. Characterization with finite cartesian product of g-G2-absorbing submodule. In Theorems 3.8 and 3.9 are given using Theorems 3.6 and 3.7. It has been proved by the application of the Zorn lemma that the graded ideal set is the maximal element with the help of multiplicatively closed subset in Theorem 3.10. Theorem 3.12 is given as a dual notion of Theorem 3.10. In terms of the direct sum of graded R -module over G -graded g r -Noetherian ring R in Theorems 3.14, 3.15 and 3.16, graded submodule being g − G 2 -absorbing submodule is classified.

Throughout the paper, we apply the notations introduced in [8]. For more information about graded rings and graded modules, see [12,13, 14,15].

2 Graded G 2 -absorbing submodules

Definition 2.1

[11, Definition 2.2] Let R be a G -graded ring, M a graded R -module and N a proper graded submodule of M . Then N is said to be a graded generalized 2-absorbing or graded G 2 -absorbing submodule of M if whenever r h , s α ∈ h ( R ) and m λ ∈ h ( M ) with r h s α m λ ∈ N , which implies either r h k 1 ∈ ( N : R m λ ) or s α k 2 ∈ ( N : R m λ ) or r h s α ∈ ( N : R M ) , for some k 1 , k 2 ∈ Z + .

Definition 2.2

Let R be a G -graded ring, M a graded R -module, N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G . We say that N g is a g -generalized 2-absorbing or g - G 2 -absorbing submodule of the R e -module M g if N g ≠ M g ; and whenever r e , s e ∈ R e and m g ∈ M g with r e s e m g ∈ N g , then either r e k 1 ∈ ( N g : R e m g ) or s e k 2 ∈ ( N g : R e m g ) or r e s e ∈ ( N g : R e M g ) , for some k 1 , k 2 ∈ Z + .

Let R be a G -graded ring and M a graded R -module. The graded radical of a graded ideal I , denoted by Gr ( I ) , is the set of all x = ∑ g ∈ G x g ∈ R such that for each g ∈ G there exists n g > 0 with x g n g ∈ I . Note that if r is a homogeneous element, then r ∈ Gr ( I ) if and only if r n ∈ I for some n ∈ N (see [5]). The graded radical of a graded submodule N of M , denoted by Gr M ( N ) , is defined to be the intersection of all graded prime submodules of M containing N . If N is not contained in any graded prime submodule of M , then Gr M ( N ) = M (see [16]).

Lemma 2.3

Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R and N a graded G 2 -absorbing submodule of M . If r g ∈ h ( R ) , m λ ∈ h ( M ) , and h ∈ G such that r g I h m λ ⊆ N , then either r g ∈ Gr ( ( N : R m λ ) ) or I h ⊆ Gr ( ( N : R m λ ) ) or r g I h ⊆ ( N : R M ) .

Proof

Suppose that r g ∉ Gr ( ( N : R m λ ) ) and r g I h ⊈ ( N : R M ) . Hence, there exists i h ′ ∈ I h such that r g i h ′ ∉ ( N : R M ) . Now, r g i h ′ m λ ∈ N implies that i h ′ ∈ Gr ( ( N : R m λ ) ) as N is a graded G 2 -absorbing submodule of M . Now, let i h ∈ I h , then r g ( i h + i h ′ ) m λ ∈ N . Hence, either i h + i h ′ ∈ Gr ( ( N : R m λ ) ) or r g ( i h + i h ′ ) ∈ ( N : R M ) . If i h + i h ′ ∈ Gr ( ( N : R m λ ) ) , then i h ∈ Gr ( ( N : R m λ ) ) since i h ′ ∈ Gr ( ( N : R m λ ) ) . If r g ( i h + i h ′ ) ∈ ( N : R M ) , then r g i h ∉ ( N : R M ) . But r g i h m λ ∈ N which yields that i h ∈ Gr ( ( N : R m λ ) ) as N is a graded G 2 -absorbing submodule of M . Therefore, we obtain I h ⊆ Gr ( ( N : R m λ ) ) .□

Lemma 2.4

Let R be a G -graded ring, M a graded R-module, I = ⊕ g ∈ G I g , J = ⊕ g ∈ G J g be two graded ideals of R and N a graded G 2 -absorbing submodule of M . If m λ ∈ h ( M ) and g , h ∈ G such that J g I h m λ ⊆ N , then either J g ⊆ Gr ( ( N : R m λ ) ) or I h ⊆ Gr ( ( N : R m λ ) ) or J g I h ⊆ ( N : R M ) .

Proof

Suppose that J g ⊈ Gr ( ( N : R m λ ) ) and I h ⊈ Gr ( ( N : R m λ ) ) , then there exist j g ′ ∈ J g and i h ′ ∈ I h such that j g ′ ∉ Gr ( ( N : R m λ ) ) and i h ′ ∉ Gr ( ( N : R m λ ) ) . Now, by Lemma 2.3, j g ′ I h m λ ⊆ N yields that j g ′ I h ⊆ ( N : R M ) and so ( J g \ Gr ( ( N : R m λ ) ) ) I h ⊆ ( N : R M ) . Also, since i h ′ ∉ Gr ( ( N : R m λ ) ) , J g ( I h \ Gr ( ( N : R m λ ) ) ) ⊆ ( N : R M ) . Hence, we obtain j g ′ i h ′ ∈ ( N : R M ) . Now, let j g ∈ J g and i h ∈ I h , then j g ′ i h ∈ ( N : R M ) and j g i h ′ ∈ ( N : R M ) . Thus, ( j g + j g ′ ) ( i h + i h ′ ) m λ ∈ N yields that either j g + j g ′ ∈ Gr ( ( N : R m λ ) ) or i h + i h ′ ∈ Gr ( ( N : R m λ ) ) or ( j g + j g ′ ) ( i h + i h ′ ) ∈ ( N : R M ) . If j g + j g ′ ∈ Gr ( ( N : R m λ ) ) , then j g ∉ Gr ( ( N : R m λ ) ) , which follows that j g ∈ J g \ Gr ( ( N : R m λ ) ) and then j g i h ∈ ( N : R M ) . Similarly, if i h + i h ′ ∈ Gr ( ( N : R m λ ) ) , then j g i h ∈ ( N : R M ) . Now, if ( j g + j g ′ ) ( i h + i h ′ ) = j g i h + j g i h ′ + j g ′ i h + j g ′ i h ′ ∈ ( N : R M ) , then we obtain j g i h ∈ ( N : R M ) . Therefore, J g I h ⊆ ( N : R M ) .□

The following theorem gives us a characterization of a graded G 2 -absorbing submodule.

Theorem 2.5

Let R be a G -graded ring, M a graded R -module and N a proper graded submodule of M. Let I = ⊕ g ∈ G I g , J = ⊕ g ∈ G J g be two graded ideals of R and K = ⊕ g ∈ G K g a graded submodule of M. Then the following statements are equivalent:

  1. N is a graded G 2 -absorbing submodule of M .

  2. If g , h , λ ∈ G with J g I h K λ ⊆ N , then either J g ⊆ Gr ( ( N : R K λ ) ) or I h ⊆ Gr ( ( N : R K λ ) ) or J g I h ⊆ ( N : R M ) .

Proof

(i) ⇒ (ii) Suppose that N is a graded G 2 -absorbing submodule of M and let g , h , λ ∈ G with J g I h K λ ⊆ N and J g I h ⊈ ( N : R M ) . Now, let k λ ∈ K λ , then since J g I h k λ ⊆ N , either J g ⊆ Gr ( ( N : R k λ ) ) or I h ⊆ Gr ( ( N : R k λ ) ) by Lemma 2.4. Assume that there exist k 1 λ , k 2 λ ∈ K λ such that J g ⊈ Gr ( ( N : R k 1 λ ) ) and I h ⊈ Gr ( ( N : R k 2 λ ) ) . Hence, I h ⊆ Gr ( ( N : R k 1 λ ) ) and J g ⊆ Gr ( ( N : R k 2 λ ) ) . Now, since J g I h ( k 1 λ + k 2 λ ) ⊆ N , either J g ⊆ Gr ( ( N : R k 1 λ + k 2 λ ) ) or I h ⊆ Gr ( ( N : R k 1 λ + k 2 λ ) ) , which yields that either J g ⊆ Gr ( ( N : R k 1 λ ) ) or I h ⊆ Gr ( ( N : R k 2 λ ) ) , a contradiction. Therefore, either J g ⊆ Gr ( ( N : R K λ ) ) or I h ⊆ Gr ( ( N : R K λ ) ) .

( i i ) ⇒ ( i ) Assume that ( i i ) holds. Let r g , s h ∈ h ( R ) and m λ ∈ h ( M ) such that r g s h m λ ∈ N . Now, let J = ⟨ r g ⟩ and I = ⟨ s h ⟩ be two graded ideals of R generated by r g and s h , respectively, and let K = ⟨ m λ ⟩ be a graded submodule of M generated by m λ . Thus, J g I h K λ ⊆ N , then we obtain either J g ⊆ Gr ( ( N : R K λ ) ) or I h ⊆ Gr ( ( N : R K λ ) ) or J g I h ⊆ ( N : R M ) . Hence, either r g ⊆ Gr ( ( N : R m λ ) ) or s h ⊆ Gr ( ( N : R m λ ) ) or r g s h ⊆ ( N : R M ) . Therefore, N is a graded G 2 -absorbing submodule of M .□

Recall from [7] that if I = ⊕ g ∈ G I g is a graded ideal of a G -graded ring R , then I e is said to be an e -2-absorbing primary ideal of R e if I e ≠ R e ; and whenever r e , s e , t e ∈ R e with r e s e t e ∈ I e , then either r e s e ∈ I e or r e t e ∈ Gr ( I e ) or s e t e ∈ Gr ( I e ) .

Theorem 2.6

Let R be a G -graded ring, M a graded R-module, N = ⊕ g ∈ G N g , K = ⊕ g ∈ G K g be two graded submodules of M and g ∈ G . If N g is a g - G 2 -absorbing submodule of an R e -module M g , then the following statements hold:

  1. If K g ⊈ N g , then ( N g : R e K g ) is an e -2-absorbing primary ideal of R e .

  2. ( N g : R e M g ) is an e -2-absorbing primary ideal of R e .

Proof

( i ) Let r e , s e , t e ∈ R e such that r e s e t e ∈ ( N g : R e K g ) . Then either r e n 1 t e K g ⊆ N g for some n 1 ∈ Z + or s e n 2 t e K g ⊆ N g for some n 2 ∈ Z + or r e s e M g ⊆ N g as N g is a g - G 2 -absorbing submodule of M g . Therefore, ( r e t e ) n 1 K g ⊆ N g or ( s e t e ) n 2 K g ⊆ N g or r e s e K g ⊆ N g . Therefore, ( N g : R e K g ) is an e -2-absorbing primary ideal of R e .

( i i ) The proof follows from M g ⊈ N g and part ( i ) .□

Recall from [6] that if I = ⊕ g ∈ G I g is a graded ideal of a G -graded ring R , then I e is said to be an e -2 -absorbing ideal of R e if I e ≠ R e ; and whenever r e , s e , t e ∈ R e with r e s e t e ∈ I e , then either r e s e ∈ I e or r e t e ∈ I e or s e t e ∈ I e .

Theorem 2.7

Let R be a G -graded ring and I = ⊕ g ∈ G I g a graded ideal of R . If I e is an e -2-absorbing primary ideal of R e , then Gr ( I e ) is an e -2-absorbing ideal of R e .

Proof

Let r e , s e , t e ∈ R e such that r e s e t e ∈ Gr ( I e ) and neither r e t e ∈ Gr ( I e ) nor s e t e ∈ Gr ( I e ) . Hence, there exists n ∈ Z + such that ( r e s e t e ) n = r e n s e n t e n ∈ I e . Since I e is an e -2-absorbing primary and r e t e ∉ Gr ( I e ) and s e t e ∉ Gr ( I e ) , we obtain r e n s e n = ( r e s e ) n ∈ I e . Thus, r e s e ∈ Gr ( I e ) . Therefore, Gr ( I e ) is an e -2-absorbing ideal of R e .□

Corollary 2.8

Let R be a G -graded ring, M a graded R -module, N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G . If N g is a g - G 2 -absorbing submodule of an R e -module M g , then Gr ( ( N g : R e M g ) ) is an e -2-absorbing ideal of R e .

Proof

We have ( N g : R e M g ) is an e -2-absorbing primary ideal of R e by Theorem 2.6. Therefore, Gr ( ( N g : R e M g ) ) is an e -2-absorbing ideal of R e by Theorem 2.7.□

Let R be a G -graded ring, M a graded R -module, N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G . An R e -module M g is said to be a g -multiplication module if for every submodule N g of M g there exists an ideal I e of R e such that N g = I e M g (see [17]).

Corollary 2.9

Let R be a G -graded ring, M a graded R-module, N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G . If M g is a g -multiplication R e -module and N g is a g - G 2 -absorbing submodule of M g such that Gr ( ( N g : R e M g ) ) = ( N g : R e M g ) , then N g is a g -2-absorbing submodule of M g .

Proof

By Theorem 2.6 we have ( N g : R e M g ) is an e -2-absorbing primary ideal of R e . Hence, ( N g : R e M g ) = Gr ( ( N g : R e M g ) ) is an e -2-absorbing ideal of R e by Corollary 2.8. Now, let r e , s e ∈ R e and m g ∈ M g such that r e s e m g ∈ N g . Hence, r e s e R e m g ⊆ N g and as M g is a g -multiplication module we obtain r e s e ( R e m g : R e M g ) M g ⊆ N g . Thus, r e s e ( R e m g : R e M g ) ⊆ ( N g : R e M g ) , which yields that either r e s e ∈ ( N g : R e M g ) or r e ( R e m g : R e M g ) ⊆ ( N g : R e M g ) or s e ( R e m g : R e M g ) ⊆ ( N g : R e M g ) . So, either r e s e ∈ ( N g : R e M g ) or r e R e m g = r e ( R e m g : R e M g ) M g ⊆ N g or s e R e m g = s e ( R e m g : R e M g ) M g ⊆ N g , then either r e s e ∈ ( N g : R e M g ) or r e m g ∈ N g or s e m g ∈ N g . Therefore, N g is a g -2-absorbing submodule of M g .□

Recall that from [18] that a graded zero-divisor on a graded R -module M is an element r g ∈ h ( R ) for which there exists m h ∈ h ( M ) such that m h ≠ 0 but r g m h = 0 . The set of all graded zero-divisors on M is denoted by G - Z d v R ( M ) .

The following result studies the behavior of graded G 2 -absorbing submodules under localization.

Theorem 2.10

Let R be a G-graded ring, M a graded R-module and S ⊆ h ( R ) be a multiplicatively closed subset of R.

  1. If N is a graded G 2 -absorbing submodule of M such that ( N : R M ) ∩ S = ∅ , then S − 1 N is a graded G 2 -absorbing submodule of S − 1 M .

  2. If S − 1 N is a graded G 2 -absorbing submodule of S − 1 M such that G - Z d v R ( M / N ) ∩ S = ∅ , then N is a graded G 2 -absorbing submodule of M .

Proof

( i ) Since ( N : R M ) ∩ S = ∅ , S − 1 N is a proper graded submodule of S − 1 M . Now, let r g s 1 , s h s 2 ∈ h ( S − 1 R ) and m λ s 3 ∈ h ( S − 1 M ) such that r g s 1 s h s 2 m λ s 3 = r g s h m λ s 1 s 2 s 3 ∈ S − 1 N . Hence, there exists s 4 ∈ S such that s 4 r g s h m λ ∈ N which yields that either r g k 1 s 4 m λ ∈ N for some k 1 ∈ Z + or s h k 2 s 4 m λ ∈ N for some k 2 ∈ Z + or r g s h ∈ ( N : R M ) as N is a graded G 2 -absorbing submodule of M . Thus, either r g s 1 k 1 m λ s 3 = r g k 1 s 4 m λ s 1 k 1 s 4 s 3 ∈ S − 1 N or s h s 2 k 2 m λ s 3 = s h k 2 s 4 m λ s 2 k 2 s 4 s 3 ∈ S − 1 N or r g s 1 s h s 2 = r g s h s 1 s 2 ∈ S − 1 ( N : R M ) ⊆ ( S − 1 N : S − 1 R S − 1 M ) . Therefore, S − 1 N is a graded G 2 -absorbing submodule of S − 1 M .

( i i ) Let r g , s h ∈ h ( R ) and m λ ∈ h ( M ) such that r g s h m λ ∈ N , so r g 1 s h 1 m λ 1 = r g s h m λ 1 ∈ S − 1 N . Hence, either r g 1 k 1 m λ 1 ∈ S − 1 N for some k 1 ∈ Z + or s h 1 k 2 m λ 1 ∈ S − 1 N for some k 2 ∈ Z + or r g s h 1 ∈ ( S − 1 N : S − 1 R S − 1 M ) as S − 1 N is a graded G 2 -absorbing submodule of S − 1 M . Since G - Z d v R ( M / N ) ∩ S = ∅ , we have S − 1 ( N : R M ) = ( S − 1 N : S − 1 R S − 1 M ) . Since G - Z d v R ( M / N ) ∩ S = ∅ , we have S − 1 ( N : R M ) = ( S − 1 N : S − 1 R S − 1 M ) . If r g s h 1 ∈ ( S − 1 N : S − 1 R S − 1 M ) = S − 1 ( N : R M ) , then r g s h ∈ ( N : R M ) and we obtain the result. Otherwise, if r g k 1 m λ 1 ∈ S − 1 N there exists t ∈ S such that t r g k 1 m λ ∈ N , which yields that r g k 1 m λ ∈ N since G - Z d v R ( M / N ) ∩ S = ∅ . Similarly, if s h k 2 m λ 1 ∈ S − 1 N , we obtain s h k 2 m λ ∈ N . Therefore, N is a graded G 2 -absorbing submodule of M .□

3 Graded G 2 -absorbing submodules over Gr-Noetherian ring

Let R be a G -graded ring, M a graded R -module, N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G . We say that N g is a g -idempotent submodule of an R e -module M g if N g = ( N g : R e M g ) 2 M g . Also, M g is called a fully g -idempotent if every submodule is a g -idempotent. It is easy to see that every fully g -idempotent module is a g -multiplication.

A G -graded ring R is called g r -Noetherian if it satisfies the ascending chain condition on graded ideals of R . Equivalently, R is g r -Noetherian if and only if every graded ideal of R is finitely generated (see [15]).

Lemma 3.1

Let R be a G -graded ring, I = ⊕ g ∈ G I g and J = ⊕ g ∈ G J g be two graded ideals of R . If I e is an e -2-absorbing primary ideal of R e and r e , s e ∈ R e such that r e s e J e ⊆ I e , then either r e s e ∈ I e or r e J e ⊆ Gr ( I e ) or s e J e ⊆ Gr ( I e ) .

Proof

Assume that neither r e s e ∈ I e nor r e J e ⊆ Gr ( I e ) nor s e J e ⊆ Gr ( I e ) . Then there exist j 1 e , j 2 e ∈ J e with r e j 1 e ∉ Gr ( I e ) and s e j 2 e ∉ Gr ( I e ) . Now, since r e s e j 1 e ∈ I e and r e s e j 2 e ∈ I e , s e j 1 e ∈ Gr ( I e ) and r e j 2 e ∈ Gr ( I e ) as I e is an e -2-absorbing primary ideal of R e . Now, r e s e ( j 1 e + j 2 e ) ∈ I e and r e s e ∉ I e , we have either r e ( j 1 e + j 2 e ) ∈ Gr ( I e ) or s e ( j 1 e + j 2 e ) ∈ Gr ( I e ) and then either r e j 1 e ∈ Gr ( I e ) or s e j 2 e ∈ Gr ( I e ) , a contradiction. Therefore, either r e s e ∈ I e or r e J e ⊆ Gr ( I e ) or s e J e ⊆ Gr ( I e ) .□

Theorem 3.2

Let R be a G -graded g r -Noetherian ring, M a graded R -module, N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G . If M g is a fully g -idempotent R e -module and ( N g : R e M g ) is an e -2-absorbing primary ideal of R e , then N g is a g - G 2 -absorbing submodule of an R e -module M g .

Proof

Let r e , s e ∈ R e and K = ⊕ h ∈ G K h be a graded submodule of M such that r e s e K g ⊆ N g . Now, since M g is a fully g -idempotent, M g is a g -multiplication module, so we obtain r e s e ( K g : R e M g ) M g ⊆ N g and then r e s e ( K g : R e M g ) ⊆ ( N g : R e M g ) . Hence, by Lemma 3.1, either r e ( K g : R e M g ) ⊆ Gr ( ( N g : R e M g ) ) or s e ( K g : R e M g ) ⊆ Gr ( ( N g : R e M g ) ) or r e s e ∈ ( N g : R e M g ) as ( N g : R e M g ) is an e -2-absorbing primary ideal of R e . If r e s e ∈ ( N g : R e M g ) , then we obtain the result. Now, since R is g r -Noetherian, then so R e . Hence, if r e ( K g : R e M g ) ⊆ Gr ( ( N g : R e M g ) ) , then ( r e ( K g : R e M g ) ) n 1 ⊆ ( N g : R e M g ) , for some n 1 ∈ Z + , which follows that r e n 1 K g = r e n 1 ( K g : R e M g ) n 1 M g ⊆ ( N g : R e M g ) M g = N g as M g is a fully g -idempotent. Similarly, if s e ( K g : R e M g ) ⊆ Gr ( ( N g : R e M g ) ) , then s e n 2 K g ⊆ N g , for some n 2 ∈ Z + . Therefore, N g is a g - G 2 -absorbing submodule of M g .□

The following example shows that Theorem 3.2 is not true in general.

Example 3.3

Let G = Z 2 , then R = Z is a G -graded ring with R 0 = Z and R 1 = { 0 } . Let M = Q be a graded R -module with M 0 = Q and M 1 = { 0 } , where M 0 is not a fully 0-idempotent. Now, consider the graded submodule N = Z of M . Then N 0 is not a 0- G 2 -absorbing submodule of M 0 since 2 ⋅ 3 ⋅ 1 6 ∈ Z and neither 2 ⋅ 3 ∈ ( Z : Z Q ) = { 0 } nor 2 ∈ Gr ( ( Z : Z 1 6 ) ) nor 3 ∈ Gr ( ( Z : Z 1 6 ) ) . However, easy computations show that ( Z : Z Q ) = { 0 } is a 0-2-absorbing primary ideal of Z .

Lemma 3.4

Let R i be a G -graded ring, M i a graded R i -module, for i = 1 , 2 and g ∈ G . Let R = R 1 × R 2 and M = M 1 × M 2 . Then M i g is a fully g -idempotent R i e -module, for i = 1 , 2 if and only if M g is a fully g -idempotent R e -module.

Proof

Suppose that M g is a fully g -idempotent R e -module and N 1 g is a submodule of an R 1 e -module M 1 g . Then N g = N 1 g × { 0 } 2 g is a submodule of M g . Hence, N g = ( N g : R e M g ) 2 M g = ( N 1 g : R 1 e M 1 g ) 2 M 1 g × ( { 0 } 2 g : R 2 e M 2 g ) 2 M 2 g . Thus, N 1 g = ( N 1 g : R 1 e M 1 g ) 2 M 1 g . Therefore, M 1 g is a fully g -idempotent R 1 e -module. Similarly, M 2 g is a fully g -idempotent R 2 e -module. Conversely, let N g be a submodule of M g . Then N g = N 1 g × N 2 g for some submodules N 1 g of M 1 g and N 2 g of M 2 g . But N i g = ( N i g : R i e M i g ) 2 M i g , for i = 1 , 2 , so N g = ( N 1 g : R 1 e M 1 g ) 2 M 1 g × ( N 2 g : R 2 e M 2 g ) 2 M 2 g = ( N g : R e M g ) 2 M g . Therefore, M g is a fully g -idempotent R e -module.□

Let R be a G -graded ring and I = ⊕ g ∈ G I g be a graded ideal of R . Then I e is said to be an e -prime ideal of R e if whenever r e , s e ∈ R e with r e s e ∈ I e implies either r e ∈ I e or s e ∈ I e . Also, I e is said to be an e -primary ideal of R e if whenever r e , s e ∈ R e with r e s e ∈ I e implies either r e ∈ I e or s e ∈ Gr ( I e ) (see [5]).

Lemma 3.5

Let R be a G -graded ring and I i = ⊕ g ∈ G I i g is a graded ideal of R , for i = 1 , 2 . If I i e is an e - P i e -primary ideal of R e for some e -prime ideal P i e of R e , for i = 1 , 2 , then I 1 e ∩ I 2 e is an e -2-absorbing primary ideal of R e .

Proof

Let J e = I 1 e ∩ I 2 e , then Gr ( J e ) = P 1 e ∩ P 2 e is an e -2-absorbing ideal of R e . Now, let r e , s e , t e ∈ R e such that r e s e t e ∈ J e and neither r e t e ∈ Gr ( J e ) nor s e t e ∈ Gr ( J e ) . Thus, r e , s e , t e ∉ Gr ( J e ) = P 1 e ∩ P 2 e . Since Gr ( J e ) = P 1 e ∩ P 2 e is an e -2-absorbing ideal of R e and r e t e , s e t e ∉ Gr ( J e ) , r e s e ∈ Gr ( J e ) . Now, suppose that r e ∈ P 1 e , then r e ∉ Gr ( J e ) and r e s e ∈ Gr ( J e ) ⊆ P 2 e implies r e ∉ P 2 e and s e ∈ P 2 e . Thus, s e ∉ P 1 e . If r e ∈ I 1 e and s e ∈ I 2 e , then r e s e ∈ J e and we are done. Now, suppose that r e ∉ I 1 e . Since I 1 e is an e - P 1 e -primary ideal of R e and r e ∉ I 1 e , s e t e ∈ P 1 e . Also, since s e ∈ P 2 e and s e t e ∈ P 1 e , s e t e ∈ Gr ( J e ) , a contradiction. So r e ∈ I 1 e . Similarly, suppose that s e ∉ I 2 e . Since I 2 e is an e - P 2 e -primary ideal of R e and s e ∉ I 2 e , r e t e ∈ P 2 e . Also, since r e t e ∈ P 2 e and r e ∈ P 1 e , r e t e ∈ Gr ( J e ) , a contradiction. So, s e ∈ I 2 e and then r e s e ∈ J e . Therefore, I 1 e ∩ I 2 e is an e -2-absorbing primary ideal of R e .□

Theorem 3.6

Let R 1 , R 2 be two G -graded rings such that R = R 1 × R 2 and J = ⊕ g ∈ G J g be a proper graded ideal of R . Then the following statements are equivalent.

  1. J e is an e -2-absorbing primary ideal of R e .

  2. Either J e = I 1 e × R 2 e for some e -2-absorbing primary ideal I 1 e of R 1 e or J e = R 1 e × I 2 e for some e -2-absorbing primary ideal I 2 e of R 2 e or J e = I 1 e × I 2 e for some e -primary ideal I 1 e of R 1 e and some e -primary ideal I 2 e of R 2 e .

Proof

( i ) ⇒ ( i i ) Suppose that J e is an e -2-absorbing primary ideal of R e . So, J e = I 1 e × I 2 e where I 1 e and I 2 e are two ideals of R 1 e and R 2 e , respectively. Now, assume that I 2 e = R 2 e , then I 1 e ≠ R 1 e since J e is a proper ideal of R e . Let R e ′ = R e / { 0 } 1 e × R 2 e , then J e ′ = J e / { 0 } 1 e × R 2 e is an e -2-absorbing primary ideal of R e ′ . Now, since R e ≅ R e ′ and I 1 e ≅ J e ′ , I 1 e is an e -2-absorbing primary ideal of R 1 e . Similarly, if I 1 e = R 1 e , then I 2 e is an e -2-absorbing primary ideal of R 2 e . Thus, assume that I 1 e ≠ R 1 e , I 2 e ≠ R 2 e and I 1 e is not an e -primary ideal of R 1 e . So, there exist r 1 e , s 1 e ∈ R 1 e such that r 1 e s 1 e ∈ I 1 e and neither r 1 e ∈ I 1 e nor s 1 e ∈ Gr ( I 1 e ) . Now, let x e = ( r 1 e , 1 2 e ) , y e = ( 1 1 e , 0 2 e ) and z e = ( s 1 e , 1 2 e ) , hence x e y e z e = ( r 1 e s 1 e , 0 2 e ) ∈ J e but neither x e y e = ( r 1 e , 0 2 e ) ∈ J e nor x e z e = ( r 1 e s 1 e , 1 2 e ) ∈ J e nor y e z e = ( s 1 e , 0 2 e ) ∈ Gr ( J e ) = Gr ( I 1 e ) × Gr ( I 2 e ) , which is a contradiction. Thus, I 1 e is an e -primary ideal of R 1 e . Similarly, I 2 e is an e -primary ideal of R 2 e .

( i i ) ⇒ ( i ) If J e = I 1 e × R 2 e for some e -2-absorbing primary ideal I 1 e of R 1 e or J e = R 1 e × I 2 e for some e -2-absorbing primary ideal I 2 e of R 2 e , then it is clear that J e is an e -2-absorbing primary ideal of R e . Hence, assume that J e = I 1 e × I 2 e for some e -primary ideal I 1 e of R 1 e and some e -primary ideal I 2 e of R 2 e . Then I 1 e ′ = I 1 e × R 2 e and I 2 e ′ = R 1 e × I 2 e are e -primary ideals of R e . Hence, I 1 e ′ ∩ I 2 e ′ = I 1 e × I 2 e = J e is an e -2-absorbing primary ideal of R e by Lemma 3.5.□

Theorem 3.7

Let R be a G -graded ring, M a graded R-module, N = ⊕ g ∈ G N g a proper graded submodule of M and g ∈ G . Let M g be a g -multiplication. Then the following statements are equivalent:

  1. N g is a g -primary submodule of an R e -module M g .

  2. ( N g : R e M g ) is an e -primary ideal of R e .

Proof

( i ) ⇒ ( i i ) Let r e , s e ∈ R e such that r e s e ∈ ( N g : R e M g ) and r e ∉ Gr ( ( N g : R e M g ) ) . Thus, r e s e M g ⊆ N g yields that s e M g ⊆ N g and then s e ∈ ( N g : R e M g ) . Therefore, ( N g : R e M g ) is an e -primary ideal of R e .

( i i ) ⇒ ( i ) Let r e ∈ R e and m g ∈ M g such that r e m g ∈ N g and r e ∉ Gr ( ( N g : R e M g ) ) . Hence, r e R e m g = r e ( R e m g : R e M g ) M g ⊆ N g , which yields that r e ( R e m g : R e M g ) ⊆ ( N g : R e M g ) . Thus, we obtain ( R e m g : R e M g ) ⊆ ( N g : R e M g ) , so m g ∈ R e m g = ( R e m g : R e M g ) M g ⊆ ( N g : R e M g ) M g = N g . Therefore, N g is a g -primary submodule of an R e -module M g . □

Theorem 3.8

Let R i be a G -graded g r -Noetherian ring, M i a graded R i -module, N i = ⊕ g ∈ G N i g a graded submodule of M i for i = 1 , 2 and g ∈ G . Let R = R 1 × R 2 and M = M 1 × M 2 such that M g is a fully g-idempotent R e -module. Then:

  1. N 1 g is a g - G 2 -absorbing submodule of M 1 g if and only if N 1 g × M 2 g is a g - G 2 -absorbing submodule of M g .

  2. N 2 g is a g - G 2 -absorbing submodule of M 2 g if and only if M 1 g × N 2 g is a g - G 2 -absorbing submodule of M g .

  3. If N 1 g is a g -primary submodule of M 1 g and N 2 g is a g -primary submodule of M 2 g , then N 1 g × N 2 g is a g - G 2 -absorbing submodule of M g .

Proof

( i ) Since M g is a fully g -idempotent R e -module, M i g is a fully g -idempotent R i e -module, for i = 1 , 2 , by Lemma 3.4. Now, suppose that N 1 g is a g - G 2 -absorbing submodule of M 1 g , then by Theorem 2.6 we obtain ( N 1 g : R 1 e M 1 g ) is an e -2-absorbing primary ideal of R 1 e . Thus, ( N 1 g × M 2 g : R e M g ) = ( N 1 g : R 1 e M 1 g ) × R 2 e is an e -2-absorbing primary ideal of R e by Theorem 3.6. Hence, by Theorem 3.2 we obtain N 1 g × M 2 g is a g - G 2 -absorbing submodule of M g . Conversely, suppose that N 1 g × M 2 g is a g - G 2 -absorbing submodule of M g , then ( N 1 g × M 2 g : R e M g ) = ( N 1 g : R 1 e M 1 g ) × R 2 e is an e -2-absorbing primary ideal of R e by Theorem 2.6. So, ( N 1 g : R 1 e M 1 g ) is an e -2-absorbing primary ideal of R 1 e by Theorem 3.6. Thus, by Theorem 3.2 we obtain N 1 g is a g - G 2 -absorbing submodule of M 1 g .

( i i ) The proof is similar to that in part ( i ) .

( i i i ) Let N i g be a g -primary submodule of M i g , then ( N i g : R i e M i g ) is an e -primary ideal of R i e , for i = 1 , 2 . Now, since ( N 1 g × N 2 g : R e M g ) = ( N 1 g : R 1 e M 1 g ) × ( N 2 g : R 2 e M 2 g ) , ( N 1 g × N 2 g : R e M g ) is an e -2-absorbing primary ideal of R e by Theorem 3.6. Therefore, N 1 g × N 2 g is a g - G 2 -absorbing submodule of M g .□

Theorem 3.9

Let R i be a G -graded g r -Noetherian ring, M i a graded R i -module, N i = ⊕ g ∈ G N i g a graded submodule of M i for i = 1 , 2 and g ∈ G . Let R = R 1 × R 2 , M = M 1 × M 2 such that M g is a fully g-idempotent R e -module and N = N 1 × N 2 . Then the following statements are equivalent:

  1. N g is a g - G 2 -absorbing submodule of M g .

  2. Either N 1 g = M 1 g and N 2 g is a g - G 2 -absorbing submodule of M 2 g or N 2 g = M 2 g and N 1 g is a g - G 2 -absorbing submodule of M 1 g or N 1 g and N 2 g are g -primary submodules of M 1 g and M 2 g , respectively.

Proof

( i ) ⇒ ( i i ) Assume that N g = N 1 g × N 2 g is a g - G 2 -absorbing submodule of M g . So, by Theorem 2.6 we obtain ( N g : R e M g ) = ( N 1 g : R 1 e M 1 g ) × ( N 2 g : R 2 e M 2 g ) is an e -2-absorbing primary ideal of R e . Thus, we obtain either ( N 1 g : R 1 e M 1 g ) = R 1 e and ( N 2 g : R 2 e M 2 g ) is an e -2-absorbing primary ideal of R 2 e or ( N 2 g : R 2 e M 2 g ) = R 2 e and ( N 1 g : R 1 e M 1 g ) is an e -2-absorbing primary ideal of R 1 e or ( N 1 g : R 1 e M 1 g ) and ( N 2 g : R 2 e M 2 g ) are e -primary ideals of R 1 e and R 2 e , respectively, by Theorem 3.6. Now, if ( N 1 g : R 1 e M 1 g ) = R 1 e and ( N 2 g : R 2 e M 2 g ) is an e -2-absorbing primary ideal of R 2 e , then N 1 g = M 1 g and N 2 g is a g - G 2 -absorbing submodule of M 2 g by Theorem 3.8. Similarly, if ( N 2 g : R 2 e M 2 g ) = R 2 e and ( N 1 g : R 1 e M 1 g ) is an e -2-absorbing primary ideal of R 1 e , then N 2 g = M 2 g and N 1 g is a g - G 2 -absorbing submodule of M 1 g . If ( N i g : R i e M i g ) is an e -primary ideal of R i e , then since M i g is a g -multiplication R i e -module, N i g is a g -primary submodule of M i g , for i = 1 , 2 , by Theorem 3.7.

( i i ) ⇒ ( i ) Clearly, by Theorem 3.8.□

Theorem 3.10

Let R be a G-graded ring, I = ⊕ g ∈ G I g be a graded ideal of R and S e ⊆ R e be a multiplicatively closed subset of R e with I e ∩ S e = ∅ . Then the set Γ = { J e ∣ J e is an ideal of R e , S e ∩ J e = ∅ , I e ⊆ J e } has a maximal element and such maximal elements are e-prime ideals of R e .

Proof

Since I e ∈ Γ , Γ ≠ ∅ . The set Γ is a partially ordered set with respect to set inclusion “ ⊆ .” Now, let Δ be a totally ordered subset of Γ , then J = ∪ J e ∈ Δ J e is an ideal of R e . Now, let P e be a maximal element of Γ and r e , s e ∈ R e such that r 1 e ∉ P e and r 2 e ∉ P e . Thus, P e ⊊ ( P e + ⟨ r 1 e ⟩ ) , which concludes that ( P e + ⟨ r 1 e ⟩ ) ∩ S e ≠ ∅ , so there exists s 1 e ∈ S e such that s 1 e = p 1 e + r 1 e t 1 e , where p 1 e ∈ P e and t 1 e ∈ R e . Similarly, there exists s 2 e ∈ S e such that s 2 e = p 2 e + r 2 e t 2 e where p 2 e ∈ P e and t 2 e ∈ R e . Hence, s 1 e s 2 e = ( p 1 e + r 1 e t 1 e ) ( p 2 e + r 2 e t 2 e ) = p 1 e p 2 e + p 1 e r 2 e t 2 e + p 2 e r 1 e t 1 e + r 1 e r 2 e t 1 e t 2 e ∈ S e \ P e , which yields that r 1 e r 2 e t 1 e t 2 e ∉ P e and then r 1 e r 2 e ∉ P e . Therefore, P e is an e -prime ideal of R e .□

Theorem 3.11

Let R be a G -graded ring and I = ⊕ g ∈ G I g , P = ⊕ g ∈ G P g be two graded ideals of R . Let P e be an e-prime ideal of R e such that I e ⊆ P e . Then the following statements are equivalent:

  1. P e is a minimal e -prime ideal of R e over I e .

  2. R e \ P e is a multiplicatively closed subset of R e that is maximal with respect to missing I e .

  3. For each x e ∈ P e , there exists y e ∈ R e \ P e and nonnegative integer n such that y e x e n ∈ I e .

Proof

( i ) ⇒ ( i i ) Suppose that P e is a minimal e -prime ideal of R e over I e . Now, let S e = R e \ P e , then S e is a multiplicatively closed subset of R e and there exists a maximal element in the set of ideals of R e containing I e and disjoint from S e . Assume that J e is a maximal then J e is an e -prime ideal of R e by Theorem 3.10. Since P e is a minimal, P e = J e and so S e is a maximal with respect to missing I e .

( i i ) ⇒ ( i i i ) Let 0 ≠ x e ∈ P e and S e = { y e x e n ∣ y e ∈ R e \ P e , n = 0 , 1 , 2 , … } . Then R e \ P e ⊊ S e . Since R e \ P e is a maximal, there exists y e ∈ R e \ P e and n ∈ Z + such that y e x e n ∈ I e .

( i i i ) ⇒ ( i ) Assume that I e ⊂ J e ⊆ P e , where J e is an e -prime ideal of R e . If there exists x e ∈ P e \ J e , then there exists y e ∈ R e \ P e and n ∈ Z + such that y e x e n ∈ I e ⊆ J e . But y e ∉ J e , so x e n ∈ J e , a contradiction. Therefore, P e is a minimal e -prime ideal of R e over I e .□

Theorem 3.12

Let R be a G -graded ring and I = ⊕ g ∈ G I g a graded ideal of R . If I e is an e -2-absorbing ideal of R e , then there are at most two e-prime ideals of R e that are minimal over I e .

Proof

Let Γ = { P i e ∣ P i e is an e -prime ideal of R e that is minimal over I e } . Suppose that Γ has at least three elements. Let P 1 e , P 2 e ∈ Γ be two distinct e -prime ideals. Hence, there exist x 1 e ∈ P 1 e \ P 2 e and x 2 e ∈ P 2 e \ P 1 e . First we show that x 1 e x 2 e ∈ I e . By Theorem 3.11, there exist c 2 e ∈ R e \ P 1 e and c 1 e ∈ R e \ P 2 e such that c 2 e x 1 e k 1 ∈ I e and c 1 e x 2 e k 2 ∈ I e for some k 1 , k 2 ∈ Z + . Since x 1 e , x 2 e ∉ P 1 e ∩ P 2 e and I e is an e -2-absorbing ideal of R e , we conclude that c 2 e x 1 e ∈ I e and c 1 e x 2 e ∈ I e . Since x 1 e , x 2 e ∉ P 1 e ∩ P 2 e and c 2 e x 1 e , c 1 e x 2 e ∈ I e ⊆ P 1 e ∩ P 2 e , we conclude that c 1 e ∈ P 1 e \ P 2 e and c 2 e ∈ P 2 e \ P 1 e , and thus c 1 e , c 2 e ∉ P 1 e ∩ P 2 e . Since c 2 e x 1 e ∈ I e and c 1 e x 2 e ∈ I e , we have ( c 1 e + c 2 e ) x 1 e x 2 e ∈ I e . Hence, c 1 e + c 2 e ∉ P 1 e and c 1 e + c 2 e ∉ P 2 e . Since ( c 1 e + c 2 e ) x 1 e ∉ P 2 e and ( c 1 e + c 2 e ) x 2 e ∉ P 1 e , neither ( c 1 e + c 2 e ) x 1 e ∈ I e nor ( c 1 e + c 2 e ) x 2 e ∈ I e , and hence x 1 e x 2 e ∈ I e . Now, suppose that there exists P 3 e ∈ Γ such that P 3 e is neither P 1 e nor P 2 e . Then we can choose y 1 e ∈ P 1 e \ ( P 2 e ∪ P 3 e ) , y 2 e ∈ P 2 e \ ( P 1 e ∪ P 3 e ) , and y 3 e ∈ P 3 e \ ( P 1 e ∪ P 2 e ) . By the previous argument y 1 e y 2 e ∈ I e . Since I e ⊆ P 1 e ∩ P 2 e ∩ P 3 e and y 1 e y 2 e ∈ I e , we conclude that either y 1 e ∈ P 3 e or y 2 e ∈ P 3 e , a contradiction. Hence, Γ has at most two elements.□

Theorem 3.13

Let R be a G -graded ring and I = ⊕ g ∈ G I g a graded ideal of R such that I e is an e -2-absorbing primary ideal of R e . Then one of the following statements must hold.

  1. Gr ( I e ) = P e , where P e is an e -prime ideal of R e .

  2. Gr ( I e ) = P 1 e ∩ P 2 e , where P 1 e and P 2 e are the only distinct e-prime ideals of R e that are minimal over I e .

Proof

Suppose that I e is an e -2-absorbing primary ideal of R e , then Gr ( I e ) is an e -2-absorbing ideal of R e by Theorem 2.7. Hence, since Gr ( Gr ( I e ) ) = Gr ( I e ) , the result follows from Theorem 3.12.□

Theorem 3.14

Let R be G -graded g r -Noetherian ring, M a graded R -module, N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G . If N g is a g - G 2 -absorbing submodule of an R e -module M g and m g ∈ M g \ N g , then either Gr ( ( N g : R e m g ) ) is an e -prime ideal of R e or there exist r e ∈ R e \ Gr ( ( N g : R e m g ) ) and n ∈ Z + such that Gr ( ( N g : R e r e n m g ) ) is an e -prime ideal of R e .

Proof

Let m g ∈ M g \ N g and assume that N g is a g - G 2 -absorbing submodule of an R e -module M g , then Gr ( ( N g : R e M g ) ) is an e -2-absorbing ideal of R e by Corollary 2.8. Now, by Theorem 3.13 we obtain either Gr ( ( N g : R e M g ) ) = P 1 e or Gr ( ( N g : R e M g ) ) = P 1 e ∩ P 2 e , where P 1 e and P 2 e are distinct e -prime ideals of R e . First, assume that Gr ( ( N g : R e M g ) ) = P 1 e . Let r e s e ∈ Gr ( ( N g : R e m g ) ) for some r e , s e ∈ R e , then ( r e s e ) n m g ∈ N g for some n ∈ Z + . Thus, we obtain either r e n t m g ∈ N g or s e n k m g ∈ N g for some t , k ∈ Z + or ( r e s e ) n ∈ ( N g : R e M g ) as N g is a g - G 2 -absorbing submodule of M g . If either r e n t m g ∈ N g or s e n k m g ∈ N g , then either r e ∈ Gr ( ( N g : R e m g ) ) or s e ∈ Gr ( ( N g : R e m g ) ) . Now, if ( r e s e ) n ∈ ( N g : R e M g ) , then r e s e ∈ P 1 e . Since P 1 e is an e -prime ideal of R e , then either r e ∈ P 1 e = Gr ( ( N g : R e M g ) ) ⊆ Gr ( ( N g : R e m g ) ) or s e ∈ P 1 e = Gr ( ( N g : R e M g ) ) ⊆ Gr ( ( N g : R e m g ) ) . Therefore, Gr ( ( N g : R e m g ) ) is an e -prime ideal of R e . Now, we can assume that Gr ( ( N g : R e M g ) ) = P 1 e ∩ P 2 e . If P 1 e ⊆ Gr ( ( N g : R e m g ) ) , then we obtain the result by using the previous argument. If P 1 e ⊈ Gr ( ( N g : R e m g ) ) , then there exists r e ∈ P 1 e \ Gr ( ( N g : R e m g ) ) . Also, P 1 e P 2 e ⊆ Gr ( P 1 e P 2 e ) = Gr ( P 1 e ∩ P 2 e ) = Gr ( ( N g : R e M g ) ) ⊆ Gr ( ( N g : R e m g ) ) . Now, R is g r -Noetherian implies that R e is Noetherian, so there exists n ∈ Z + such that ( P 1 e P 2 e ) n m g ⊆ N g and then r e n P 2 e n m g ⊆ N g . Hence, P 2 e ⊆ Gr ( ( N g : R e r e n m g ) ) and we obtain the result by a similar argument.□

Let N = ⊕ g ∈ G N g be a graded submodule of a graded R -module M and g ∈ G . An efficient covering of N g is a covering N g ⊆ N 1 g ∪ N 2 g ∪ … ∪ N n g , where N i g is a submodule of an R e -module M g such that N g ⊈ N i g for i = 1 , … , n . Also, we say that N g = N 1 g ∪ N 2 g ∪ … ∪ N n g is an efficient union if none of the N i g may be excluded (see [19]).

Theorem 3.15

Let R be a G -graded g r -Noetherian ring, M a graded R-module, N = ⊕ g ∈ G N g and N i = ⊕ g ∈ G N i g be graded submodules of M , for i = 1 , … , n and g ∈ G . Let N g ⊆ N 1 g ∪ N 2 g ∪ … ∪ N n g be an efficient covering consisting of submodules of an R e -module M g , where n > 2 . If Gr ( ( N i g : R e M g ) ) ⊈ Gr ( ( N k g : R e m k g ) ) for all m k g ∈ M g \ N k g whenever i ≠ k , then N i g is not a g - G 2 -absorbing submodule of M g , for i = 1 , … , n .

Proof

Suppose that there exists k ∈ { 1 , … , n } such that N k g is a g - G 2 -absorbing submodule of M g . Since N g ⊆ N 1 g ∪ N 2 g ∪ … ∪ N n g is an efficient covering of N g , N g ⊈ N k g , so there exists m k g ∈ N g \ N k g . It is easy to see that N g = ( N 1 g ∩ N g ) ∪ ( N 2 g ∩ N g ) ∪ … ∪ ( N n g ∩ N g ) is an efficient union of N g , so by [19, Lemma 2.2], we have ∩ i ≠ k ( N g ∩ N i g ) = ∩ i = 1 n ( N g ∩ N i g ) ⊆ N g ∩ N k g . Now, by using Theorem 3.14 , we obtain either Gr ( ( N k g : R e m k g ) ) is an e -prime ideal of R e or there exists r e ∈ R e \ Gr ( ( N k g : R e m k g ) ) such that Gr ( ( N k g : R e r e t m k g ) ) is an e -prime ideal of R e , where t ∈ Z + . Hence, if Gr ( ( N k g : R e m k g ) ) is an e -prime ideal of R e , then Gr ( ( N i g : R e M g ) ) ⊈ Gr ( ( N k g : R e m k g ) ) for all i ≠ k . So, there exists s i e ∈ Gr ( ( N i g : R e M g ) ) \ Gr ( ( N k g : R e m k g ) ) , where i ≠ k . Thus, s i e j i ∈ ( N i g : R e M g ) \ ( N k g : R e m k g ) where i ≠ k and j i ∈ Z + . Let s e = Π i ≠ k s i e and j = max { j 1 , j 2 , … , j k − 1 , j k + 1 , … , j n } , then s e j ∈ ( N i g : R e M g ) \ ( N k g : R e m k g ) for all i ≠ k . Thus, s e j m k g ∈ ∩ i ≠ k ( N g ∩ N i g ) \ ( N g ∩ N k g ) = ∅ , a contradiction. Therefore, N k g is not a g - G 2 -absorbing submodule of M g . Now, if Gr ( ( N k g : R e r e t m k g ) ) is an e -prime ideal of R e , where r e ∈ R e \ Gr ( ( N k g : R e m k g ) ) and t ∈ Z + . So, there exists s i e ∈ Gr ( ( N i g : R e M g ) ) \ Gr ( ( N k g : R e r e t m k g ) ) , where i ≠ k . Hence, s i e j i ∈ ( N i g : R e M g ) \ ( N k g : R e r e t m k g ) where i ≠ k and j i ∈ Z + . Let s e = Π i ≠ k s i e and j = max { j 1 , j 2 , … , j k − 1 , j k + 1 , … , j n } , then s e j ∈ ( N i g : R e M g ) \ ( N k g : R e r e t m k g ) for all i ≠ k . Therefore, s e j r e t m k g ∈ ∩ i ≠ k ( N g ∩ N i g ) \ ( N g ∩ N k g ) = ∅ , a contradiction. Therefore, N k g is not a g - G 2 -absorbing submodule of M g .□

Theorem 3.16

Let R be a G -graded g r -Noetherian ring, M a graded R-module, N = ⊕ g ∈ G N g and N i = ⊕ g ∈ G N i g be graded submodules of M , for i = 1 , … , n where n ⩾ 2 and g ∈ G . If N g ⊆ N 1 g ∪ N 2 g ∪ … ∪ N n g such that at most two of N 1 g , N 2 g , … , N n g are not g - G 2 -absorbing submodules of an R e -module M g and Gr ( ( N i g : R e M g ) ) ⊈ Gr ( ( N k g : R e m k g ) ) for all m k g ∈ M g \ N k g whenever i ≠ k , then N g ⊆ N i g for some i = 1 , … , n .

Proof

If n = 2 , then we obtain the result since if a graded submodule is contained in a union of two graded submodules, then it is contained in one of them. Now, take n > 2 and N g ⊈ N i g for all i = 1 , … , n , then N g ⊆ N 1 g ∪ N 2 g ∪ … ∪ N n g is an efficient covering of N g . So, by Theorem 3.15, N i g is not a g - G 2 -absorbing submodule for all i = 1 , … , n , a contradiction. Hence, N g ⊆ N i g for some i = 1 , … , n .□

Acknowledgements

The authors wish to thank sincerely the referees for their valuable comments and suggestions.

  1. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  2. Conflict of interest: The authors state no conflict of interest.

  3. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2021-08-08
Revised: 2022-04-12
Accepted: 2022-04-25
Published Online: 2022-08-05

© 2022 Shatha Alghueiri and Khaldoun Al-Zoubi, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/dema-2022-0017
Keywords for this article
graded generalized 2-absorbing submodule; graded 2-absorbing submodule; graded 2-absorbing primary submodule
Creative Commons
BY 4.0

Articles in the same Issue

  1. Regular Articles
  2. On some summation formulas
  3. A study of a meromorphic perturbation of the sine family
  4. Asymptotic behavior of even-order noncanonical neutral differential equations
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