Microsoft Word - 92-101 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 92          Approximaitly Quasi-primary Submodules   Ali Sh. Ajeel Omar A. Abdulla Haibat K. Mohammadali     Ali.shebl@st.tu.edu.iq omar.aldoori87@gmail.com H.mohammadali@tu.edu.iq Abstract In this paper, we introduce and study the notation of approximaitly quasi-primary submodules of a unitary left 𝑅-module 𝑄 over a commutative ring 𝑅 with identity. This concept is a generalization of prime and primary submodules, where a proper submodule 𝐸 of an 𝑅-module 𝑄 is called an approximaitly quasi-primary (for short App-qp) submodule of 𝑄, if 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, implies that either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 , for some 𝑛 ∈ 𝑍 . Many basic properties, examples and characterizations of this concept are introduced. Keywords: Prime submodules, Primary submodules, Socle of modules, Radical of submodules, Multiplication modules, Nonsingular modules. 1. Introduction In this article all rings are commutative with identity, and all modules are left unitary 𝑅- modules. Dauns, J. in 1978 introduced and studied the concept of prime submodule, where a proper submodule 𝐸 of an 𝑅- module 𝑄 was prime if 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, implying that either 𝑞 ∈ 𝐸 or 𝑟𝑄 ⊆ 𝐸 [1]. Recently many generalizations of prime submodule have been introduced for example, see [2-5]. Primary submodules as a generalization of prime submodules was first introduced in [6], where a proper submodule 𝐸 of 𝑄 was called primary submodule if whenever 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, implying that either 𝑞 ∈ 𝐸 or 𝑟 𝑄 ⊆ 𝐸, for some 𝑛 ∈ 𝑍 . The concept of quasi-primary ideal which was introduced and studied by Fuchs, L. [7], where a proper ideal 𝐼 of a ring 𝑅 was called quasi-primary ideal if 𝑟𝑠 ∈ 𝐼, for 𝑟, 𝑠 ∈ 𝑅, implying that 𝑟 ∈ √𝐼 or 𝑠 ∈ √𝐼, where √𝐼 𝑟 ∈ 𝑅: 𝑟 ∈ 𝐼 for some 𝑛 ∈ 𝑍 . In Ibn Al Haitham Journal for Pure and Applied Science Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Directorate General of education Salahaddin, the ministry of education, Tikrit, Iraq.  Directorate General of education Salahaddin, the ministry of education, Tikrit, Iraq.  Department of Mathematics, College of Computer Sciences and Mathematics, Tikrit University, Tikrit, Iraq. Doi: 10.30526/33.4.2513 Article history: Received 7 January 2020, Accepted 12 February 2020, Published in October 2020   93  Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 particular 𝐼 is quasi-primary ideal of 𝑅 if and only if √𝐼 is a prime ideal of 𝑅 [7, p. 176]. In 2016 Hosein, F. et. Extended the notation of quasi-primary ideal to submodules, where a proper submodule 𝐸 of an 𝑅-module 𝑄 was called quasi-primary if 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, implying that either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 or 𝑟 ∈ 𝐸: 𝑄 , “where 𝑟𝑎𝑑 𝐸 define the intersection of all prime submodules of 𝑄 contining 𝐸 [8]”. Those two concepts led us to introduce the notation of approximaitly quasi-primary submodule as generalization of prime and primary submodules, where a proper submodule 𝐸 of an 𝑅-module 𝑄 is called an approximaitly quasi- primary (for short App-qp) submodule of 𝑄, if 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, implies that either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 , for some 𝑛 ∈ 𝑍 . The socle of a module 𝑄 denoted by 𝑠𝑜𝑐 𝑄 is the intersection of all essential submodules of 𝑄 [9]. Several results of approximaitly quasi-primary are introduced. 2. Approximaitly Quasi-primary Submodules In this part of the paper, we introduce the definition of approximaitly quasi-primary submodule and give it some basic properties and characterizations. Definition (1) A proper submodule 𝐸 of an 𝑅-module 𝑄 is called an approximaitly quasi-primary (for short App-qp) submodule of 𝑄, if 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, implies that either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 , for some 𝑛 ∈ 𝑍 . And an ideal 𝐴 of a ring 𝑅 is called App-qp ideal of 𝑅 if 𝐴 is an App-qp submodule of an 𝑅-module 𝑅. Remarks and examples (2) 1) It is clear that every primary submodule is an App-qp, but not conversely. The following example explains that: Consider the 𝑍-module 𝑍 , the submodule 𝐸 〈0〉 is not primary submodule of 𝑍-module 𝑍 , since 4. 3 ∈ 〈0〉, for 4 ∈ 𝑍, 3 ∈ 𝑍 , but 3 ∉ 〈0〉 and 4 ∉ 〈0〉: 𝑍 √12𝑍 6𝑍. But 𝐸 〈0〉 is an App-qp submodule of the 𝑍-module 𝑍 , since for all 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑍 such that 𝑟𝑞 ∈ 𝐸, implies that either 𝑞 ∈ 𝑟𝑎𝑑 〈0〉 𝑠𝑜𝑐 𝑍 〈6〉 〈2〉 〈2〉 or 𝑟 ∈ 〈0〉 𝑠𝑜𝑐 𝑍 : 𝑍 〈2〉: 𝑍 √2𝑍 2𝑍. That is if 4. 3 ∈ 𝐸, for 4 ∈ 𝑍, 3 ∈ 𝑍 , and 3 ∉ 𝑟𝑎𝑑 〈0〉 𝑠𝑜𝑐 𝑍 〈2〉 but 4 ∈ 〈0〉 𝑠𝑜𝑐 𝑍 : 𝑍 2𝑍. 2) It is clear that every prime submodule is an App-qp submodule, but not conversely. The following example shows that: Consider the 𝑍-module 𝑍 , the submodule 𝐸 〈0〉 is not prime submodule of the 𝑍-module 𝑍 , since 2. 2 ∈ 𝐸, for 2 ∈ 𝑍, 2 ∈ 𝑍 , but 2 ∉ 𝐸 and 2 ∉ 〈0〉: 𝑍 4𝑍. While 𝐸 is an App-qp submodule of the 𝑍-module 𝑍 , since 𝑠𝑜𝑐 𝑍 〈2〉 and for all 𝑟 ∈ 𝑍, 𝑞 ∈ 𝑍 such that 𝑟𝑞 ∈ 𝐸, implies that either 𝑞 ∈ 𝑟𝑎𝑑 〈0〉 𝑠𝑜𝑐 𝑍 〈2〉 〈2〉 〈2〉 or 𝑟 ∈ 〈0〉 𝑠𝑜𝑐 𝑍 : 𝑍 √2𝑍 2𝑍. That is if 2. 2 ∈ 𝐸, for 2 ∈ 𝑍, 2 ∈ 𝑍 implies that 2 ∈ 𝑟𝑎𝑑 〈0〉 𝑠𝑜𝑐 𝑍 〈2〉 and 2 ∈ 〈0〉 𝑠𝑜𝑐 𝑍 : 𝑍 2𝑍. 3) It is clear that every quasi-prime submodule is an App-qp submodule, but not conversely, where a proper submodule 𝐸 of 𝑄 is called quasi-prime if 𝑟𝑠𝑞 ∈ 𝐸. For 𝑟, 𝑠 ∈ 𝑅, 𝑞 ∈ 𝑄, implies that either 𝑟𝑞 ∈ 𝐸 or 𝑠𝑞 ∈ 𝐸 [10]. The following example explains that:   94  Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 Consider the 𝑍-module 𝑍, and the submodule 4𝑍 is not quasi-prime submodule of 𝑍, since 2.2.1 4 ∈ 4𝑍,, but 2.1 ∉ 4𝑍. While 4𝑍 is an App-qp submodule of the 𝑍-module 𝑍, since for all 𝑟 ∈ 𝑍, 𝑞 ∈ 𝑍 such that 𝑟𝑞 ∈ 4𝑍, implies that either 𝑞 ∈ 𝑟𝑎𝑑 4𝑍 𝑠𝑜𝑐 𝑍 〈2〉 0 〈2〉 or 𝑟 ∈ 4𝑍 𝑠𝑜𝑐 𝑍 : 𝑍 √4𝑍 2𝑍. That is, if 2.2 ∈ 4𝑍, implies that 2 ∈ 𝑟𝑎𝑑 4𝑍 𝑠𝑜𝑐 𝑍 〈2〉 and 2 ∈ 4𝑍 𝑠𝑜𝑐 𝑍 : 𝑍 2𝑍. The following results are characterizations of App-qp submodules. Proposition (3) Let 𝑄 be an 𝑅-module, and 𝐸 be a proper submodule of 𝑄. Then 𝐸 is an App-qp submodule of 𝑄 if and only if 𝐼𝐹 ⊆ 𝐸, for 𝐼 is an ideal of 𝑅 and 𝐹 is a submodule of 𝑄, implies that either 𝐹 ⊆ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝐼 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Proof Suppose 𝐼𝐹 ⊆ 𝐸, for 𝐼 is an ideal of 𝑅 and 𝐹 is a submodule of 𝑄 with 𝐹 ⊈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 , then there exists 𝑘 ∈ 𝐹 such that 𝑘 ∉ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 . Now we have 𝐼𝐹 ⊆ 𝐸, then for any 𝑎 ∈ 𝐼, 𝑎𝑘 ∈ 𝐸. Since 𝐸 is an App-qp submodule of 𝑄 and 𝑘 ∉ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 , it follows that 𝑎 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 , that is 𝐼 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Assume that 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, then 𝑟𝑞 〈𝑟〉〈𝑞〉, that is 𝐼𝐹 ⊆ 𝐸 where 𝐼 〈𝑟〉, 𝐹 〈𝑞〉, then by hypothesis, either 𝐹 ⊆ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝐼 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Hence either𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Thus 𝐸 is an App-qp submodule of 𝑄. The following Corollary is a direct consequence Proposition (3). Corollary (4) Let 𝑄 be an 𝑅-module, and 𝐸 be a proper submodule of 𝑄.Then, 𝐸 is an App-qp submodule of 𝑄 if and only if for every submodule 𝐹 of 𝑄 and every 𝑟 ∈ 𝑅 with 𝑟𝐹 ⊆ 𝐸, implies that either 𝐹 ⊆ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Proposition (5) A zero submodule of a non-zero 𝑅-module 𝑄 is an App-qp submodule of 𝑄 if and only if 𝑎𝑛𝑛 𝐹 ⊆ 𝑠𝑜𝑐 𝑄 : 𝑄 for all non-zero submodule 𝐹 of 𝑄, with 𝐹 ⊈ 𝑟𝑎𝑑 0 𝑠𝑜𝑐 𝑄 . Proof Let 𝐹 be a non-zero submodule of 𝑄, such that 𝐹 ⊈ 𝑟𝑎𝑑 0 𝑠𝑜𝑐 𝑄 , and let 𝑥 ∈ 𝑎𝑛𝑛 𝐹 , implies that 𝑥𝐹 0 but 0 is an App-qp submodule of 𝑄 and 𝐹 ⊈ 𝑟𝑎𝑑 0 𝑠𝑜𝑐 𝑄 , it follows by Corollary (4) that 𝑥 𝑄 ⊆ 0 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 , that is 𝑥 ∈ 𝑠𝑜𝑐 𝑄 : 𝑄 . Hence 𝑎𝑛𝑛 𝐹 ⊆ 𝑠𝑜𝑐 𝑄 : 𝑄 . Suppose that 𝑥𝐹 ⊆ 0 , for 𝑟 ∈ 𝑅 and 𝐹 is a non-zero submodule of 𝑄, with 𝐹 ⊈ 𝑟𝑎𝑑 0 𝑠𝑜𝑐 𝑄 . Since 𝑥𝐹 ⊆ 0 it follows that 𝑥 ∈ 𝑎𝑛𝑛 𝐹 , by hypothesis 𝑥 ∈ 𝑠𝑜𝑐 𝑄 : 𝑄 , that is 𝑥 ∈ 0 𝑠𝑜𝑐 𝑄 : 𝑄 . Hence 𝑥 𝑄 ⊆ 0 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Thus by Corollary (4) a zero submodule of an 𝑅-module 𝑄 is an app-primary submodule of 𝑄.   95  Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 Proposition (6) Let 𝑄 be an 𝑅-module, and 𝐸 be a proper submodule of 𝑄. Then, 𝐸 is an App-qp submodule of 𝑄 if and only if for every 𝑞 ∈ 𝑄, 𝐸: 𝑞 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑄 with 𝑞 ∉ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 . Proof Suppose that 𝐸 is an App-qp submodule of 𝑄, and 𝑟 ∈ 𝐸: 𝑞 , implies that 𝑟𝑞 ∈ 𝐸. Since 𝐸 is an App-qp submodule of 𝑄. and 𝑞 ∉ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 , then 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 , that is, 𝑟 ∈ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑄 . Thus 𝐸: 𝑞 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑄 . Let 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, and suppose that 𝑞 ∉ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 . Since 𝑟𝑞 ∈ 𝐸 it follows that 𝑟 ∈ 𝐸: 𝑞 by hypothesis 𝑟 ∈ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑄 . Hence, 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Thus 𝐸 is an App-qp submodule of 𝑄. Proposition (7) Let 𝑄 be an 𝑅-module, and 𝐸 be a proper submodule of 𝑄. Then, 𝐸 is an App-qp submodule of 𝑄 if and only if 𝐸: 𝑟 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑟 for 𝑟 ∈ 𝑅, 𝑛 ∈ 𝑍 . Proof Suppose that 𝐸 is an App-qp submodule of 𝑄, and let 𝑞 ∈ 𝐸: 𝑟 , such that 𝑞 ∉ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 . Since 𝑞 ∈ 𝐸: 𝑟 it follows that 𝑟𝑞 ∈ 𝐸. But 𝐸 is an App-qp submodule of 𝑄. and 𝑞 ∉ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 , then 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑄 for some 𝑛 ∈ 𝑍 . That is 𝑟 𝑞 ∈ 𝐸 𝑠𝑜𝑐 𝑄 for all 𝑞 ∈ 𝑄, it follows that 𝑞 ∈ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑟 . Thus 𝐸: 𝑟 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑟 . Let 𝑟𝑞 ∈ 𝐸, for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, and suppose that 𝑞 ∉ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 . Since 𝑟𝑞 ∈ 𝐸 it follows that 𝑞 ∈ 𝐸: 𝑟 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑟 , implies that 𝑞 ∈ 𝐸 𝑠𝑜𝑐 𝑄 : 𝑟 , that is 𝑟 𝑞 ∈ 𝐸 𝑠𝑜𝑐 𝑄 for all 𝑞 ∈ 𝑄, hence 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 . Thus 𝐸 is an App-qp submodule of 𝑄. Before we give the next result we need to recall the following Lemma. Lemma (8) [11, Coro. (9.9)] Let 𝐸 be a submodule of an 𝑅-module 𝑄, then 𝑠𝑜𝑐 𝐸 𝐸 ∩ 𝑠𝑜𝑐 𝑄 . Proposition (9) Let 𝐸 and 𝐹 are proper submodules of an 𝑅-module 𝑄 with 𝐸 ⊂ 𝐹 and 𝑠𝑜𝑐 𝑄 ⊆ 𝐹. If 𝐸 is an App-qp submodule of 𝑄, then 𝐸 is an App-qp submodule of 𝐹. Proof Let 𝑟𝑞 ∈ 𝐸, with 𝑟 ∈ 𝑅, 𝑞 ∈ 𝐹 ⊆ 𝑄. Since 𝐸 is an App-qp submodule of 𝑄, then either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 , for some 𝑛 ∈ 𝑍 . That is either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 ∩ 𝐹 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 ∩ 𝐹. But since 𝑠𝑜𝑐 𝑄 ⊆ 𝐹, then by modular law we have either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 ∩ 𝐹 𝑠𝑜𝑐 𝑄 ∩ 𝐹 or 𝑟 𝑄 ⊆ 𝐸 ∩ 𝐹 𝑠𝑜𝑐 𝑄 ∩ 𝐹 . Now by Lemma (8) 𝑠𝑜𝑐 𝑄 ∩ 𝐹 𝑠𝑜𝑐 𝐹 , so either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 ∩ 𝐹 𝑠𝑜𝑐 𝐹 ⊆ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝐹 or 𝑟 𝑄 ⊆ 𝐸 ∩ 𝐹 𝑠𝑜𝑐 𝐹 ⊆ 𝐸 𝑠𝑜𝑐 𝐹 . Hence 𝐸 is an App-qp submodule of 𝐹.   96  Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 Remark (10) If 𝐸 is an App-qp submodule of an 𝑅-module 𝑄, then 𝐸: 𝑄 need not to be an App-qp ideal of 𝑅. The following example explains that: Consider the 𝑍-module 𝑍 , the submodule 𝐸 〈0〉 is an App-qp submodule of the 𝑍- module 𝑍 [see Remarks and Examples (2) (1)]. But 𝐸: 𝑍 12𝑍 is not App-qp ideal of 𝑍 because 4.3 ∈ 12𝑍, for 4,3 ∈ 𝑍, but 3 ∉ 𝑟𝑎𝑑 12𝑍 𝑠𝑜𝑐 𝑍 〈6〉 0 〈6〉 and 4 ∉ 12𝑍 𝑠𝑜𝑐 𝑍 : 𝑍 √12𝑍 6𝑍. Now before we offer under certain condition the residual of App-qp submodule is an App-qp ideal we need to revise the following Lemma: Recall that an 𝑅-module 𝑄 is called multiplication if every submodule 𝐸 of 𝑄 is of the form 𝐸 𝐼𝑄 for some ideal 𝐼 of 𝑄 [12]. Lemma (11) [12, Coro. 14(i)] Let 𝑄 be a faithful multiplication 𝑅-module, then 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑅 𝑄. Proposition (12) Let 𝑄 be a faithful multiplication 𝑅-module and 𝐸 be a proper submodule of 𝑄. Then 𝐸 is an App-qp submodule of 𝑄 if and only if 𝐸: 𝑄 is an App-qp ideal of 𝑅. Proof ⟹ Let 𝑟𝑠 ∈ 𝐸: 𝑄 , for 𝑟, 𝑠 ∈ 𝑅, so 𝑟𝑠𝑄 ⊆ 𝐸. But 𝐸 is an App-qp submodule of 𝑄 then by Corollary (4) either 𝑠𝑄 ⊆ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 , for some 𝑛 ∈ 𝑍 . Since 𝑄 is multiplication then 𝑟𝑎𝑑 𝐸 𝐸: 𝑄 𝑄 , and since 𝑄 is faithful multiplication then by Lemma (11) 𝑠𝑜𝑐 𝑅 𝑄 𝑠𝑜𝑐 𝑄 , we get either 𝑠𝑄 ⊆ 𝐸: 𝑄 𝑄 𝑠𝑜𝑐 𝑅 𝑄 or 𝑟 𝑄 ⊆ 𝐸: 𝑄 𝑄 𝑠𝑜𝑐 𝑅 𝑄, that is either 𝑠 ∈ 𝐸: 𝑄 𝑠𝑜𝑐 𝑅 or 𝑟 ⊆ 𝐸: 𝑄 𝑠𝑜𝑐 𝑅 ⊆ 𝐸: 𝑄 𝑠𝑜𝑐 𝑅 : 𝑅 . Hence 𝐸: 𝑄 is an App-qp ideal of 𝑅. ⟸ Suppose that 𝐸: 𝑄 is an App-qp ideal of 𝑅, and 𝐼𝐹 ⊆ 𝐸, for 𝐼 is an ideal of 𝑅 and 𝐹 is a submodule of 𝑄. Since 𝑄 is multiplication then 𝐹 𝐽𝑄 for some ideal 𝐽 of 𝑅, that is 𝐼𝐽𝑄 ⊆ 𝐸, implies that 𝐼𝐽 ⊆ 𝐸: 𝑄 . But 𝐸: 𝑄 is an App-qp ideal of 𝑅 then either 𝐽 ⊆ 𝐸: 𝑄 𝑠𝑜𝑐 𝑅 or 𝐼 ⊆ 𝐸: 𝑄 𝑠𝑜𝑐 𝑅 : 𝑅 𝐸: 𝑄 𝑠𝑜𝑐 𝑅 for some 𝑛 ∈ 𝑍 . It follows that either 𝐽𝑄 ⊆ 𝐸: 𝑄 𝑄 𝑠𝑜𝑐 𝑅 𝑄 or 𝐼 𝑄 ⊆ 𝐸: 𝑄 𝑄 𝑠𝑜𝑐 𝑅 𝑄. Since 𝑄 is faithful multiplication then by Lemma (11) 𝑠𝑜𝑐 𝑅 𝑄 𝑠𝑜𝑐 𝑄 , and since 𝑄 is multiplication then 𝐸: 𝑄 𝑄 𝐸 and 𝑟𝑎𝑑 𝐸 𝐸: 𝑄 𝑄. Hence either 𝐽𝑄 ⊆ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝐼 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 , that is either 𝐹 ⊆ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝐼 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 . Hence, by Proposition (3) 𝐸 is an App-qp submodule of 𝑄. Recall that an 𝑅-module 𝑄 is called non-singular if 𝑍 𝑄 𝑄, where 𝑍 𝑄 𝑞 ∈ 𝑄: 𝑞𝐽 0 for some essentail ideal 𝐽 of 𝑅 [9]. We need to recall the following Lemma: Lemma (13) [9, Coro. (1.26)] If 𝑄 is a non-singular 𝑅-module, then 𝑠𝑜𝑐 𝑅 𝑄 𝑠𝑜𝑐 𝑄 .   97  Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 Proposition (14) Let 𝐸 be a propoer submodule of a non-singular multiplication 𝑅-module 𝑇. Then, 𝐸 is an App-qp submodule of 𝑄 if and only if 𝐸: 𝑄 is an App-qp ideal of 𝑅. Proof Follow as in Proposition (12) by using Lemma (13). We need to recall the following Lemma: Lemma (15) [13, Coro. of Theo. 9] Let 𝐼 and 𝐽 are ideals of a ring 𝑅, and 𝑄 be a finitely generated multiplication 𝑅-module. Then 𝐼𝑄 ⊆ 𝐽𝑄 if and only if 𝐼 ⊆ 𝐽 𝑎𝑛𝑛 𝑄 . Proposition (16) Let 𝑄 be a faithful finitely generated multiplication 𝑅-module and 𝐼 is an App-qp ideal of 𝑅. Then 𝐼𝑄 is an App-qp submodule of 𝑄. Proof Let 𝑟𝐹 ⊆ 𝐼𝑄 for 𝑟 ∈ 𝑅, and 𝐹 is a submodule of 𝑄 with 𝑟 𝑄 ⊈ 𝐼𝑄 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Since 𝑄 is faithful multiplication then by Lemma (11) 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑅 𝑄, that is 𝑟 𝑄 ⊈ 𝐼𝑄 𝑠𝑜𝑐 𝑅 𝑄 for some 𝑛 ∈ 𝑍 , it follows that 𝑟 ∉ 𝐼 𝑠𝑜𝑐 𝑅 𝐼 𝑠𝑜𝑐 𝑅 : 𝑅 implies that 𝑟 𝑅 ⊈ 𝐼 𝑠𝑜𝑐 𝑅 , Now, since 𝑟𝐹 ⊆ 𝐼𝑄 and 𝑄 is a multiplication then 𝐹 𝐽𝑄 for some ideal 𝐽 of 𝑅, thus 𝑟𝐽𝑄 ⊆ 𝐼𝑄. Hence by Lemma (15) 𝑟𝐽 ⊆ 𝐼 𝑎𝑛𝑛 𝑄 , but 𝑄 is a faithful, then 𝑟𝐽 ⊆ 𝐼 0 𝐼. Since 𝐼 is an App-qp ideal of 𝑅 and 𝑟 𝑅 ⊈ 𝐼 𝑠𝑜𝑐 𝑅 then by Corollary (4) either ⊆ √𝐼 𝑠𝑜𝑐 𝑅 , hence 𝐽𝑄 ⊆ √𝐼𝑄 𝑠𝑜𝑐 𝑅 𝑄. It follows by Lemma (11) 𝐽𝑄 ⊆ 𝑟𝑎𝑑 𝐼𝑄 𝑠𝑜𝑐 𝑄 . That is 𝐹 ⊆ 𝑟𝑎𝑑 𝐼𝑄 𝑠𝑜𝑐 𝑄 . Hence by Corollary (4) 𝐼𝑄 is an App-qp submodule of 𝑄. Proposition (17) Let 𝑄 be a finitely generated multiplication non-singular 𝑅-module and 𝐼 is an App-qp ideal of 𝑅 with 𝑎𝑛𝑛 𝑄 ⊆ 𝐼. Then 𝐼𝑄 is an App-qp submodule of 𝑄. Proof Follows similar as in Proposition (16) and using Lemma (13). Proposition (18) Let 𝑄 be a faithful finitely generated multiplication 𝑅-module and 𝐸 be a proper submodule of 𝑄. Then the following statements are equivalent. 1) 𝐸 is an App-qp submodule of 𝑄. 2) 𝐸: 𝑄 is an App-qp ideal of 𝑅. 3) 𝐸 𝐼𝑄 for some an App-qp ideal 𝐼 of 𝑅. Proof (1) 2) It follows by Proposition (12). (2) (3) It is clear. (3) (2) Suppose that 𝐸 𝐼𝑄 for some App-qp ideal 𝐼 of 𝑅. Since 𝑄 is a multiplication, then 𝐸 𝐸: 𝑄 𝑄 𝐼𝑄. But 𝑄 is faithful finitely generated multiplication, then 𝐼 𝐸: 𝑄 , it follows that 𝐸: 𝑄 an App-qp ideal of 𝑅.   98  Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 Proposition (19) Let 𝑄 be a finitely generated multiplication non-singular 𝑅-module and 𝐸 be a proper submodule of 𝑄. Then the following statements are equivalent. 1) 𝐸 is an App-qp submodule of 𝑄. 2) 𝐸: 𝑄 is an App-qp ideal of 𝑅. 3) 𝐸 𝐽𝑄 for some an App-qp ideal 𝐽 of 𝑅 with 𝑎𝑛𝑛 𝑄 ⊆ 𝐽. Proof It follows similar as Proposition (18) by using Proposition (14) and Lemma (15). We need the following Lemma. Lemma (20) [14. Coro. (1.3)] Let 𝑓: 𝑄 ⟶ 𝑄 be an 𝑅-epimorphism and 𝐸 is a submodule of 𝑄 with 𝑘𝑒𝑟 𝑓 ⊆ 𝐸, then 𝑓 𝑟𝑎𝑑 𝐸 𝑟𝑎𝑑 𝑓 𝐸 . Proposition (21) Let 𝑓: 𝑄 ⟶ 𝑄 be an 𝑅-epimorphism and 𝐸 is an App-qp submodule of 𝑄 . Then 𝑓 𝐸 is an App-qp submodule of 𝑄. Proof It is clear that 𝑓 𝐸 is a proper submodule of 𝑄. Now, suppose that 𝑟𝑞 ∈ 𝑓 𝐸 , for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄, implies that 𝑟𝑓 𝑞 ∈ 𝐸 . But 𝐸 is an App-qp submodule of 𝑄 , it follows that either 𝑓 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . It follows that by Lemma (20), either 𝑞 ∈ 𝑓 𝑟𝑎𝑑 𝐸 𝑓 𝑠𝑜𝑐 𝑄 ⊆ 𝑟𝑎𝑑 𝑓 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑓 𝑓 𝑄 ⊆ 𝑓 𝐸 𝑓 𝑠𝑜𝑐 𝑄 ⊆ 𝑓 𝐸 𝑠𝑜𝑐 𝑄 .That is either 𝑞 ∈ 𝑟𝑎𝑑 𝑓 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝑓 𝐸 𝑠𝑜𝑐 𝑄 . Hence 𝑓 𝐸 be an App-qp submodule of 𝑄. Proposition (22) Let 𝑓: 𝑄 ⟶ 𝑄 be an 𝑅-epimorphism and 𝐸 is an App-qp submodule of 𝑄 with ker 𝑓 ⊆ 𝐸 . Then 𝑓 𝐸 is an App-qp submodule of 𝑄 . Proof 𝑓 𝐸 is a proper submodule of 𝑄 . If not, that is 𝑓 𝐸 𝑄 . Let 𝑞 ∈ 𝑄, then 𝑓 𝑞 ∈ 𝑄 𝑓 𝐸 , so there exists 𝑥 ∈ 𝐸 such that 𝑓 𝑞 𝑓 𝑥 , implies that 𝑓 𝑞 𝑥 0, that is 𝑞 𝑥 ∈ 𝐹𝑒𝑟 𝑓 ⊆ 𝐸, it follows that 𝑞 ∈ 𝐸. Thus,  𝐸 𝑄 contradiction. Now suppose that 𝑟𝑞 ∈ 𝑓 𝐸 , for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄 , 𝑓 𝑞 𝑞 for some 𝑞 ∈ 𝑄 (since 𝑓 is onto), that is 𝑟𝑞 𝑟𝑓 𝑞 𝑓 𝑟𝑞 ∈ 𝑓 𝐸 , it follows that there exists 𝑒 ∈ 𝐸 such that 𝑓 𝑟𝑞 𝑓 𝑒 , that is 𝑓 𝑒 𝑟𝑞 0, so 𝑒 𝑟𝑞 ∈ 𝑘𝑒𝑟 𝑓 ⊆ 𝐸, implies that 𝑟𝑞 ∈ 𝐸. But 𝐸 is an App-qp submodule of 𝑄, then either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Hence, by using Lemma (20) either 𝑞 𝑓 𝑞 ∈ 𝑓 𝑟𝑎𝑑 𝐸 𝑓 𝑠𝑜𝑐 𝑄 ⊆ 𝑟𝑎𝑑 𝑓 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 𝑟 𝑓 𝑄 ⊆ 𝑓 𝐸 𝑓 𝑠𝑜𝑐 𝑄 ⊆ 𝑓 𝐸 𝑠𝑜𝑐 𝑄 . Thus 𝑓 𝐸 is an App-qp submodule of 𝑄 . Remark (23) The intersection of two App-qp submodules of an 𝑅-module 𝑄 need not to be an App-qp submodule of 𝑄.The following example explains that:   99  Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 Consider the 𝑍-module 𝑍 and the submodules 2𝑍, 3𝑍 are App-qp submodules of 𝑍-modules 𝑍 (because they are prime) but 2𝑍 ∩ 3𝑍 6𝑍 is not App-qp submodule of 𝑍-module 𝑍, since 2.3 ∈ 6𝑍, but 3 ∉ 𝑟𝑎𝑑 6𝑍 𝑠𝑜𝑐 𝑍 6𝑍 0 6𝑍 and 2 ∉ 6𝑍 𝑠𝑜𝑐 𝑍 : 𝑍 6𝑍: 𝑍 √6𝑍 6𝑍. We need the following Lemma: Lemma (24) [15, Theo. 15(3)] Let 𝑄 be a multiplication 𝑅-module and 𝐸, 𝐹 be a submodules of 𝑄. Then 𝑟𝑎𝑑 𝐸 ∩ 𝐹 𝑟𝑎𝑑 𝐸 ∩ 𝑟𝑎𝑑 𝐹 . Proposition (25) Let 𝐸 and 𝐹 be a proper submodules of multiplication 𝑅-module 𝑄 with 𝑠𝑜𝑐 𝑄 ⊆ 𝐸 or 𝑠𝑜𝑐 𝑄 ⊆ 𝐹. If 𝐸 and 𝐹 are App-qp submodules of 𝑄, then 𝐸 ∩ 𝐹 is an App-qp submodule of 𝑄. Proof Suppose 𝑟𝑞 ∈ 𝐸 ∩ 𝐹 for 𝑟,∈ 𝑅, 𝑞 ∈ 𝑄, then 𝑟𝑞 ∈ 𝐸 and 𝑟𝑞 ∈ 𝐹. But both 𝐸 and 𝐹 are App-qp submodules of 𝑄, then either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 and either 𝑞 ∈ 𝑟𝑎𝑑 𝐹 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐹 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Hence either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 ∩ 𝑟𝑎𝑑 𝐹 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 ∩ 𝐹 𝑠𝑜𝑐 𝑄 . If 𝑠𝑜𝑐 𝑄 ⊆ 𝐹 ⊆ 𝑟𝑎𝑑 𝐸 , then 𝐹 𝑠𝑜𝑐 𝑄 𝐹 and 𝑟𝑎𝑑 𝐹 𝑠𝑜𝑐 𝑄 𝑟𝑎𝑑 𝐹 . Thus either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑇 ∩ 𝑟𝑎𝑑 𝐹 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 ∩ 𝐹. It follows that by modular law either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 ∩ 𝑟𝑎𝑑 𝐹 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 ∩ 𝐹 𝑠𝑜𝑐 𝑄 . Hence by Lemma (24) either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 ∩ 𝐹 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 ∩ 𝐹 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Thus 𝐸 ∩ 𝐹 is an App-qp submodule of 𝑄. Similarly if 𝑠𝑜𝑐 𝑄 ⊆ 𝐸, we got 𝐸 ∩ 𝐹 is an App-qp submodule of 𝑄. Proposition (26) Let 𝑄 𝑄 ⊕ 𝑄 be an 𝑅-module, where 𝑄 , 𝑄 are 𝑅-modules, and 𝐸 𝐸 ⊕ 𝐸 be a submodule of 𝑄, with 𝐸 , 𝐸 are submodules of 𝑄 , 𝑄 respectively with 𝑟𝑎𝑑 𝐸 ⊆ 𝑠𝑜𝑐 𝑄 . If 𝐸 is an App-qp submodule of 𝑄, then 𝐸 is an App-qp submodule of 𝑄 and 𝐸 is an App- qp submodule of 𝑄 . Proof Let 𝑟𝑞 ∈ 𝐸 , for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄 , then 𝑟 𝑞 , 0 ∈ 𝐸. Since 𝐸 is an App-qp submodule of 𝑄, then 𝑞 , 0 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . But 𝑟𝑎𝑑 𝐸 ⊆ 𝑠𝑜𝑐 𝑄 , implies that 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 , and 𝐸 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 [since 𝐸 ⊆ 𝑟𝑎𝑑 𝐸 ⊆ 𝑠𝑜𝑐 𝑄 . It follows that either 𝑞 , 0 ∈ 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 or 𝑟 𝑄 ⊕ 𝑄 ⊆ 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 , that is either 𝑞 , 0 ∈ 𝑠𝑜𝑐 𝑄 ⊕ 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊕ 𝑄 ⊆ 𝑠𝑜𝑐 𝑄 ⊕ 𝑠𝑜𝑐 𝑄 , hence either 𝑞 ∈ 𝑠𝑜𝑐 𝑄 ⊆ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝑠𝑜𝑐 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 . Thus 𝐸 is an App-qp submodule of 𝑄 .Similarly we can prove that 𝐸 is an App-qp submodule of 𝑄 .   100  Ibn Al-Haitham Jour. for Pure & Appl. Sci. 33 (4) 2020 Proposition (27) Let 𝑄 𝑄 ⊕ 𝑄 be an 𝑅-module, where 𝑄 and 𝑄 are 𝑅-modules. Then, the following are held:  1) 𝐸 is an App-qp submodule of 𝑄 such that 𝑟𝑎𝑑 𝐸 ⊆ 𝑠𝑜𝑐 𝑄 and 𝑠𝑜𝑐 𝑄 𝑄 if and only if 𝐸 ⊕ 𝑄 is an App-qp submodule of 𝑄. 2) 𝐸 is an App-qp submodule of 𝑄 such that 𝑟𝑎𝑑 𝐸 ⊆ 𝑠𝑜𝑐 2 and 𝑠𝑜𝑐 𝑄 𝑄 if and only if 𝑄 ⊕ 𝐸 is an App-qp submodule of 𝑄. Proof 1) ⟹ Let 𝑟 𝑞 , 𝑞 ∈ 𝐸 ⊕ 𝑄 , for 𝑟 ∈ 𝑅, 𝑞 , 𝑞 ∈ 𝑄, then 𝑟𝑞 ∈ 𝐸 . But 𝐸 is an App-qp submodule of 𝑄 and 𝑟𝑎𝑑 𝐸 ⊆ 𝑠𝑜𝑐 𝑄 , then either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Since 𝑠𝑜𝑐 𝑄 𝑄 , then either 𝑞 , 𝑞 ∈ 𝑠𝑜𝑐 𝑄 ⊕ 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 ⊆ 𝑟𝑎𝑑 𝐸 ⊕ 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 or 𝑟 𝑄 ⊕ 𝑄 ⊆ 𝑠𝑜𝑐 𝑄 ⊕ 𝑠𝑜𝑐 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 ⊆ 𝐸 ⊕ 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 . Thus 𝐸 ⊕ 𝑄 is an App-qp submodule of 𝑄. ⟸ Suppose 𝑟𝑞 ∈ 𝐸 , for 𝑟 ∈ 𝑅, 𝑞 ∈ 𝑄 . Then for each 𝑞 ∈ 𝑄 , 𝑞 , 𝑞 ∈ 𝐸 ⊕ 𝑄 , but 𝐸 ⊕ 𝑄 is an App-qp submodule of 𝑄, implies that either 𝑞 , 𝑞 ∈ 𝑟𝑎𝑑 𝐸 ⊕ 𝑄 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 ⊕ 𝑄 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 .it follows that either 𝑞 , 𝑞 ∈ 𝑟𝑎𝑑 𝐸 ⊕ 𝑟𝑎𝑑 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 or 𝑟 (𝑄 ⊕ 𝑄 ⊆ 𝐸 ⊕ 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 , that is either 𝑞 , 𝑞 ∈ 𝑟𝑎𝑑 𝐸 ⊕ 𝑟𝑎𝑑 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑠𝑜𝑐 𝑄 or 𝑟 (𝑄 ⊕ 𝑄 ⊆ 𝐸 ⊕ 𝑄 𝑠𝑜𝑐 𝑄 ⊕ 𝑠𝑜𝑐 𝑄 . Since 𝑠𝑜𝑐 𝑄 𝑄 implies that either 𝑞 , 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 ⊕ 𝑟𝑎𝑑 𝑄 𝑄 or 𝑟 (𝑄 ⊕ 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 ⊕ 𝑄 , that is either 𝑞 ∈ 𝑟𝑎𝑑 𝐸 𝑠𝑜𝑐 𝑄 or 𝑟 𝑄 ⊆ 𝐸 𝑠𝑜𝑐 𝑄 for some 𝑛 ∈ 𝑍 . Hence 𝐸 is an App-qp submodule of 𝑄 . 2) Its follows as in part (1). 3. Conclusion In this paper, we introduce a new generalization of prime and primary submodules called an approximaitly quasi-primary submodule. Many characterizations of this generalization are introduced. Relationships of this generalization with other classes of modules are given. References 1. Dauns, J. Prime Modules, J. Reine Angew, Math. 1978, 2, 156-181. 2. Haibat, K.M.; Omar, A.A. 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