Microsoft Word - 110-131   110   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 T-Abso and T-Abso Quasi Primary Fuzzy Submodules Wafaa H. Hanoon Department of Computer Science, College of Education for Girls, University of Kufa, Iraq. wafaah.hannon@uokufa.edu.iq Khalaf m Y.Hate Department of Mathematics, College of Education for Pure Science Ibn- Haitham University of Baghdad, Baghdad, Iraq. dr.hatamyahya@yahoo.com Article history: Received 4 November 2018, Accepted 24 November 2018, Publish January 2019 Abstract Let Ḿ be a unitary R-module and R is a commutative ring with identity. Our aim in this paper to study the concepts T-ABSO fuzzy ideals, T-ABSO fuzzy submodules and T-ABSO quasi primary fuzzy submodules, also we discuss these concepts in the class of multiplication fuzzy modules and relationships between these concepts. Many new basic properties and characterizations on these concepts are given. Keywords: T-ABSO fuzzy ideal, T-ABSO fuzzy submodule, Quasi- prime fuzzy submodule, T-ABSO primary fuzzy submodule, T-ABSO quasi primary fuzzy submodule, Multiplication fuzzy module. 1. Introduction In this paper all ring is commutative with identity and all modules are unitary. Deniz S. et al in [1] presented the concept of 2-absorbing fuzzy ideal which is a generalization of prime fuzzy ideal. Prime submodule which play an important turn in the module theory over a commutative ring. A prime submodule N of an R-module Ḿ, N≠Ḿ, with property 𝑎∈𝑅, 𝑥∈Ḿ, 𝑎𝑥∈𝑁 implies that ∈𝑁 or 𝑎 ∈ (𝑁: Ḿ) [2]. This concept was generalized to concept of prime fuzzy submodule which was presented by Rabi [3]. In 1999, Abdul-Razakm, presented and studied quasi-prime submodule let N < Ḿ, N be called a quasi-prime if for 𝑎, 𝑏 ∈𝑅, 𝑚∈Ḿ, 𝑎𝑏𝑚∈𝑁, implies either 𝑎𝑚∈𝑁 or b𝑚∈𝑁 [4]. In 2001, Hatam generalized it to fuzzy quasi-prime submodules [5]. Darani, et al in [6] presented the definition of 2-absorbing submodule. Let N < Ḿ, N be called 2-absorbing submodule of Ḿ if whenever r, b ∈ R, x ∈ Ḿ and rbx ∈ N, then rx ∈ N or bx ∈ N or rb ∈ (N: Ḿ). Hatam and wafaa expanded this concept that is: if X be a fuzzy module of an R- module Ḿ. A proper fuzzy submodule A of X is called T-ABSO fuzzy submodule if whenever 𝑎 , 𝑏 be fuzzy singletons of R, and 𝑥 ⊆ 𝑋, ∀ 𝑠, 𝑙, 𝑣 ∈ 𝐿, such that 𝑎 𝑏 𝑥 ⊆ 𝐴, then either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝐴 or 𝑏 𝑥 ⊆ 𝐴 [7]. McCasland and Moore presented the concept of Ḿ-radical of N such: Let N be a proper module of a nonzero R-module Ḿ, then the Ḿ-radical of N, denoted by Ḿ-rad N is defined to be the intersection of all prime module including N, see [8]. Mostafanasab et al, were presented the connotation of 2-absorbing primary submodule. So, A proper submodule N of an R-module Ḿ is called 2-absorbing primary submodule of Ḿ if whenever a, b∈ R and m∈ Ḿ and abm∈ N, then am∈ Ḿ-rad N or bm∈ Ḿ-rad N or ab∈ 𝑁: Ḿ ,   111   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 [9]. Rabi and Hassan in 2008 were presented the concept of quasi primary fuzzy submodule. A proper fuzzy submodule A of fuzzy module X is said to be quasi primary fuzzy submodule if (A:B) is a primary fuzzy ideal of R for each fuzzy submodule B of X such that A ⊂ B [10]. Suat K. et al, studied and presented the connotation of 2-absorbing quasi primary submodule, i.e., A proper submodule N of Ḿ is said to be 2-absorbing quasi primary submodule if the condition abq∈ N implies either ab∈ 𝑁: Ḿ or aq∈ Ḿ-rad(N) or bq∈ Ḿ-rad(N) for every a, b∈ R and q∈ Ḿ [11]. This paper is composed of two sections. In section (1) we present the definition of T-ABSO fuzzy ideals and we give some characterizations of this definition for ideals. Also many properties and outcomes of this concept are given. In section (2) we present the definition of T-ABSO fuzzy submodules, many basic properties and outcomes are studied. In section (3) we present the concept of T-ABSO quasi primary fuzzy submodules and we study the relationships this concept with among T- ABSO fuzzy submodules and T-ABSO primary fuzzy submodules. Several important results have been demonstrated. Note that we denote to fuzzy module, submodule. 2. T-ABSO F. Ideals In this section, we introduce the concepts of T-ABSO and T-ABSO primary ideals. Some concepts and propositions which are needed in the next section. Definition 1. [1] Let Ĥ be a non-constant F. ideal of R. Then Ĥ is called T-ABSO F. ideal if for any F. points 𝑎 , 𝑏 , 𝑟 of R, 𝑎 𝑏 𝑟 ∈ Ĥ implies that either 𝑎 𝑏 ∈ Ĥ or 𝑎 𝑟 ∈ Ĥ or 𝑏 𝑟 ∈ Ĥ. The following proposition characterize T-ABSO F. ideal in terms of its level ideal. Lemma 2. [1] Let A be F. ideal of R. If A is T-ABSO F. ideal, then 𝐴 is T-ABSO ideal of R, ∀ v ∈ L, Recall that Let Ĥ be any F. ideal of R. Then the radical F. of Ĥ, denoted by Ĥ, is defined by: Ĥ =∩{Ṷ:Ṷ is a prime F. ideal of R containing Ĥ} [12]. Now, we give these propositions which are used in the next section. Proposition 3. Suppose that R be a ring and Ĥ is T-ABSO F. ideal of R. Then Ĥ is T-ABSO F. ideal of R and 𝑎 ⊆Ĥ for each F. singleton 𝑎 ⊆ Ĥ, ∀ v ∈ L. Proof. Let Ĥ be T-ABSO F. ideal and 𝑎 ⊆ Ĥ, hence 𝑎 ∈ Ĥ . Then 𝑎 ∈ Ĥ . So that Ĥ 𝑎 𝑣. Thus 𝑎 ⊆ Ĥ. Since 𝑎 𝑎 , so that 𝑎 ⊆ Ĥ. Now, let 𝑎 , 𝑏 , 𝑟 be F. singletons of R such that 𝑎 𝑏 𝑟 ⊆ Ĥ. Then 𝑎 𝑏 , 𝑟 𝑎 𝑏 𝑟 ⊆ Ĥ. Since Ĥ is T- ABSO F. ideal, then either 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ, since 𝑎 𝑏 𝑎 𝑏 , 𝑎 𝑟 𝑎 𝑟 , 𝑏 𝑟 𝑏 𝑟 hence either 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ. So that either 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ. Thus Ĥis T-ABSO F. ideal of R.   112   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Lemma 4. Let Ĥ⊆ Ƥ be F. ideal of a ring R, where Ƥ is a prime F. ideal. Then the following expressions are equivalent: 1- Ƥ is a minimal prime F. ideal of Ĥ; 2- For each F. singleton 𝑎 ⊆ Ƥ, there exists F. singleton 𝑏 of R\Ƥ and a non-negative integer n such that 𝑏 𝑎 ⊆ Ĥ , ∀ v, l ∈ L. Proof. (1) ⇒ (2) Let Ƥ be a minimal prime F. ideal of Ĥ and 𝑎 ⊆ Ƥ, suppose that for every F. singleton 𝑏 of R\Ƥ, 𝑏 𝑎 ⊈ Ĥ, ∀n∈ N. Inparticular, 𝑎 ⊈ Ĥ, ∀n ∈N. Let A={1, 𝑎 , 𝑎 ,...} and B={K: K is F. ideal of R such that K∩A=∅, Ĥ⊆ K⊆ Ƥ}. Then B≠∅, since Ĥ⊆B, it is obvious B is partially ordered by inclusion. By [13], B has a maximal F. ideal say Ṷ. Then Ṷ is a prime F. ideal by [12], such that Ĥ⊆ Ṷ⊆ Ƥ. Since Ƥ is a minimal prime F. ideal of Ĥ, so Ṷ=Ƥ this is a contradiction to 𝑎 ⊆ Ƥ=Ṷ, hence 𝑏 𝑎 ⊆ Ĥ. (2) ⇒(1) Suppose that for each F. singleton 𝑎 ⊆ Ƥ, there exists F. singleton 𝑏 of R\ Ƥ and n∈N such that 𝑏 𝑎 ⊆ Ĥ. Let K be a prime F. ideal of R such that Ĥ⊆ K⊆ Ƥ. We claim that Ƥ⊆ K. Since 𝑎 ⊆ Ƥ , then there exists F. singleton 𝑏 ⊆ R\ Ƥ and n∈N such that 𝑏 𝑎 ⊆ Ĥ⊆ K. Since K is a prime F. ideal, then either 𝑏 ⊆ K or 𝑎 ⊆ K. Hence 𝑎 ⊆ K as 𝑏 ⊆ R\ Ƥ. So that Ƥ⊆ K, then Ƥ=K; that is Ƥ is a minimal prime F. ideal of Ĥ. Proposition 5. Suppose that Ĥ is T-ABSO F. ideal of a ring R. Then there are at most two prime F ideals of R that are minimal over Ĥ.  Proof. Assume that K={Ƥ : Ƥ is a prime F. ideal of R which is minimal over Ĥ}. Let K have at least three prime F. ideals. Let Ƥ , Ƥ ∈ 𝐾 be two different prime F. ideals. Then there exists F. singleton 𝑎 ⊆ Ƥ \Ƥ and there exists F singleton 𝑏 ⊆ Ƥ \Ƥ . We show that 𝑎 𝑏 ⊆ Ĥ. By lemma (4), there exist F. singletons 𝑥 ⊈ Ƥ and 𝑦 ⊈ Ƥ , such that 𝑥 𝑎 ⊆ Ĥ and 𝑦 𝑏 ⊆ Ĥ for some n, m ≥1. Since Ĥ is T-ABSO F. ideal of R, we have 𝑥 𝑎 ⊆ Ĥ and 𝑦 𝑏 ⊆ Ĥ. Since 𝑎 , 𝑏 ⊈ Ƥ ∩ Ƥ and 𝑥 𝑎 , 𝑦 𝑏 ⊆ Ĥ ⊆ Ƥ ∩ Ƥ , we get 𝑥 ⊆ Ƥ \Ƥ and 𝑦 ⊆ Ƥ \Ƥ , thus 𝑥 , 𝑦 ⊈ Ƥ ∩ Ƥ . Since 𝑥 𝑎 ⊆ Ĥ and 𝑦 𝑏 ⊆ Ĥ, have 𝑥 𝑦 𝑎 𝑏 ⊆ Ĥ. Observe that 𝑥 𝑦 ⊈ Ƥ and 𝑥 𝑦 ⊈ Ƥ . Since 𝑥 𝑦 𝑎 ⊈ Ƥ and 𝑥 𝑦 𝑏 ⊈ Ƥ , we conclude that neither 𝑥 𝑦 𝑎 ⊆ Ĥ nor 𝑥 𝑦 𝑏 ⊆ Ĥ and hence 𝑎 𝑏 ⊆ Ĥ. Now, suppose there exists Ƥ ∈ K such that Ƥ is neither Ƥ nor Ƥ .Then we can choose 𝑟 ⊆ Ƥ \ Ƥ ∪ Ƥ , 𝑐 ⊆ Ƥ \ Ƥ ∪ Ƥ and 𝑑 ⊆ Ƥ \ Ƥ ∪ Ƥ . By the same way we show that 𝑟 𝑐 ⊆ Ĥ. Since Ĥ⊆Ƥ ∩ Ƥ ∩ Ƥ and 𝑟 𝑐 ⊆ Ĥ, we get either 𝑟 ⊆ Ƥ or 𝑐 ⊆ Ƥ this is a discrepancy. Hence K has at most two prime F. ideals of R. Proposition 6 Let Ĥ be T-ABSO F. ideal of R. Then one of the following expressions must hold 1- Ĥ = Ƥ is a prime F. ideal of R such that Ƥ ⊆ Ĥ 2- Ĥ= Ƥ ∩ Ƥ , Ƥ Ƥ ⊆ Ĥ, and Ĥ ⊆ Ĥ where Ƥ , Ƥ are the only distinct prime F. ideals of R that are minimal over Ĥ.   113   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Proof. By proposition (5), we get either Ĥ=Ƥ is a prime F. ideal of R or Ĥ = Ƥ ∩ Ƥ , where Ƥ , Ƥ are the only distinct prime F. ideals of R that are minimal over Ĥ. Assume that Ĥ = Ƥ is prime F. ideal of R. Let F. singletons 𝑎 , 𝑏 ⊆ Ƥ . By proposition (3), we have 𝑎 , 𝑏 ⊆ Ĥ. So that 𝑎 𝑎 𝑏 𝑏 ⊆ Ĥ. Since Ĥ is T-ABSO F. ideal, we get 𝑎 𝑎 𝑏 𝑎 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑏 𝑏 𝑎 𝑏 𝑏 ⊆ Ĥ or 𝑎 𝑏 ⊆ Ĥ. From each case implies that 𝑎 𝑏 ⊆ Ĥ, and so Ƥ ⊆ Ĥ. Suppose that Ĥ = Ƥ ∩ Ƥ , where Ƥ , Ƥ are the only distinct prime F. ideals of R that are minimal over Ĥ. Let F singletons 𝑎 , 𝑏 ⊆ Ĥ. By the same way of above, we have 𝑎 𝑏 ⊆ Ĥ and hence Ĥ ⊆ Ĥ. Now, we show that, Ƥ Ƥ ⊆ Ĥ. By proposition (3), we have 𝑥 ⊆ Ĥ for each F singleton 𝑥 ⊆ Ĥ . Let be F singleton 𝑦 ⊆ Ƥ \Ƥ and 𝑟 ⊆ Ƥ \Ƥ . By the proof of proposition (5), we have 𝑦 𝑟 ⊆ Ĥ. Let F singletons 𝑐 ⊆ Ĥ and 𝑑 ⊆ Ƥ \Ƥ , choose F singleton 𝑓 ⊆ Ƥ \Ƥ . Then 𝑓 𝑑 ⊆ Ĥ by the proof of proposition (5) and 𝑐 𝑓 ⊆ Ƥ \Ƥ . Thus 𝑐 𝑑 𝑓 𝑑 𝑐 𝑓 𝑑 ⊆ Ĥ. So that 𝑐 𝑑 ⊆ Ĥ. By the same method we show that if 𝑐 ⊆ Ĥ and 𝑑 ⊆ Ƥ \Ƥ .Thus Ƥ Ƥ ⊆ Ĥ. Proposition 7 Let Ĥ be T-ABSO F. ideal of R such that Ĥ = Ƥ is a prime F. ideal, of Rand suppose that Ĥ≠Ƥ. For each F. singleton 𝑎 ⊆ P\Ĥ , let 𝐴 𝑏 ⊆ 𝑅: 𝑏 𝑎 ⊆ Ĥ , ∀ v, l ∈L. Then 𝐴 is a prime F. ideal of R included Ƥ. Futhermore, either 𝐴 ⊆ 𝐴 or 𝐴 ⊆ 𝐴 for each F. singletons 𝑎 , 𝑏 ⊆ Ƥ\Ĥ . Proof. Let 𝑎 ⊆ Ƥ\Ĥ. Since Ƥ ⊆ Ĥ by proposition (6), we have Ƥ⊆ 𝐴 . Assume that Ƥ≠𝐴 and 𝑏 𝑟 ⊆ 𝐴 for some F. singleton 𝑏 , 𝑟 of R. Since Ƥ⊆ 𝐴 , we may suppose that 𝑏 ⊈ Ƥ and 𝑟 ⊈ Ƥ, hence 𝑏 𝑟 ⊈ Ĥ. Since 𝑏 𝑟 ⊆ 𝐴 we have 𝑏 𝑟 𝑎 ⊆ Ĥ. Since Ĥ is T-ABSO F. ideal of R and 𝑏 𝑟 ⊈ Ĥ, we have either 𝑏 𝑎 ⊆ Ĥ or 𝑟 𝑎 ⊆ Ĥ, thus either 𝑏 ⊆ 𝐴 or 𝑟 ⊆ 𝐴 . Hence 𝐴 is a prime F. ideal of R included Ƥ. Now, let 𝑎 , 𝑏 ⊆ Ƥ\Ĥ for F. singletons 𝑎 , 𝑏 of R and assume that F singleton 𝑟 ⊆ 𝐴 \ 𝐴 . Since Ƥ ⊆ 𝐴 , so 𝑟 ⊆ 𝐴 \Ƥ . We show that 𝐴 ⊆ 𝐴 . Let F singleton 𝑥 of R such that 𝑥 ⊆ 𝐴 . Since Ƥ⊆ 𝐴 , we may suppose that 𝑥 ⊆ 𝐴 Ƥ. Since 𝑟 ⊈ Ƥ and 𝑥 ⊈ Ƥ, we have 𝑟 𝑥 ⊈ Ĥ. Since 𝑟 𝑎 𝑏 𝑥 ⊆ Ĥ and 𝑟 𝑥 , 𝑟 𝑏 ⊈ Ĥ, we have 𝑎 𝑏 𝑥 ⊆ Ĥ. Hence 𝑎 𝑥 ⊆ Ĥ since 𝑎 𝑏 𝑥 ⊆ Ĥ and 𝑥 𝑏 ⊆ Ĥ. Hence 𝑥 ⊆ 𝐴 . So that 𝐴 ⊆ 𝐴 . Proposition 8. Assume that Ĥ is F. ideal of R such that Ĥ≠ Ĥ and Ĥ is a prime F. ideal of R. Then the following expressions are equivalent: 1- Ĥ is T-ABSO F. ideal of R; 2- 𝐴 𝑏 ⊆ 𝑅: 𝑏 𝑎 ⊆ Ĥ , ∀ v,l ∈L, is a prime F. ideal of R for each F. singleton 𝑎 ⊆ Ĥ \ Ĥ.   114   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Proof. (1) ⇒ (2) This is obvious by proposition (7). (2)⇒(1) Assume that 𝑎 𝑏 𝑟 ⊆ Ĥ for F. singletons 𝑎 , 𝑏 , 𝑟 of R. Since Ĥ is a prime F. ideal of R, we may suppose that 𝑎 ⊆ Ĥ . If 𝑎 ⊆ Ĥ, then 𝑎 𝑏 ⊆ Ĥ. Thus suppose that 𝑎 ⊆ Ĥ \Ĥ. Hence 𝑏 𝑟 ⊆ 𝐴 . But 𝐴 is a prime F. ideal of R , then by proposition (7), either 𝑏 𝑎 ⊆ Ĥ or 𝑟 𝑎 ⊆ Ĥ. Thus Ĥ is T- ABSO F. ideal of R. Proposition 9. Assume that Ĥ is a non-constant proper F. ideal of a ring R. Then the following expressions are equivalent: 1- Ĥ is T-ABSO F. ideal of R; 2- If Ṷ𝐾𝑇 ⊆ Ĥ for F. ideals Ṷ, 𝐾, 𝑇 of R, Ṷ𝐾 ⊆ Ĥ or 𝐾𝑇 ⊆ Ĥ or Ṷ𝑇 ⊆ Ĥ. Proof. (1)⇒(2) Assume that Ṷ𝐾𝑇 ⊆ Ĥ for F. ideals Ṷ, 𝐾, 𝑇 of R. By proposition (5), we have Ĥ is a prime F. ideal of R or Ĥ = Ƥ ∩ Ƥ where Ƥ , Ƥ are non-constant distinct prime F. ideals of R that are minimal over Ĥ. If Ĥ= Ĥ, then it is readily showed that, Ṷ𝐾 ⊆ Ĥ or 𝐾𝑇 ⊆ Ĥ or Ṷ𝑇 ⊆ Ĥ. Thus suppose that Ĥ≠ Ĥ. We see the following: (1) Assume that Ĥ is a prime F. ideal of R. Then we perhaps suppose that Ṷ ⊆ Ĥ andṶ ⊈ Ĥ. Let F. singleton 𝑎 of R such that 𝑎 ⊆ Ṷ\Ĥ. Since 𝑎 𝐾𝑇 ⊆ Ĥ, we have 𝐾𝑇 ⊆ 𝐴 where 𝐴 𝑏 ⊆ 𝑅: 𝑏 𝑎 ⊆ Ĥ . Since 𝐴 is a prime F. ideal of R by proposition (8), we have either 𝐾 ⊆ 𝐴 or 𝑇 ⊆ 𝐴 . If 𝐾 ⊆ 𝐴 and 𝑇 ⊆ 𝐴 for each F. singleton 𝑥 ⊆ Ṷ\Ĥ, then Ṷ𝐾 ⊆ Ĥ (and Ṷ𝑇 ⊆ Ĥ) and we are finished. Hence suppose that 𝐾 ⊆ 𝐴 and 𝑇 ⊈ 𝐴 for some F. singleton 𝑟 ⊆ Ṷ\Ĥ. Since {𝐴 : 𝑤 ⊆ Ṷ\Ĥ}, is a set of prime F. ideals of R that are linearly ordered by proposition (7), since 𝐾 ⊆ 𝐴 and 𝑇 ⊈ 𝐴 , we have 𝐾 ⊆ 𝐴 for some F. singleton 𝑧 ⊆ Ṷ\Ĥ. ThusṶ𝐾 ⊆ Ĥ. (2) Assume that Ĥ = Ƥ ∩ Ƥ where Ƥ , Ƥ are non-constant distinct prime F. ideals of R that are minimal over Ĥ. We suppose that Ṷ ⊆ Ƥ. If either 𝐾 ⊆ Ƥ or 𝑇 ⊆ Ƥ , then either Ṷ𝐾 ⊆ Ĥ or Ṷ𝑇 ⊆ Ĥ because Ƥ Ƥ ⊆ Ĥ by proposition (6). Hence suppose that Ṷ ⊆ Ĥ andṶ ⊈ Ĥ . By the same way in (1) and by proposition (7), we are finished from this proof. (2) ⇒ (1) it is trivial. Now, we give the concept of T-ABSO quasi primary F. ideal as follows: Definition 10. A proper F. ideal Ĥ of R is called T-ABSO quasi primary F. ideal of R if Ĥ is T-ABSO F. ideal of R. Proposition 11. A proper F. ideal Ĥ of R is T-ABSO quasi primary F. of R iff whenever for each F. singleton 𝑎 , 𝑏 , 𝑟 of R, ∀s, l, h ∈ L, such that 𝑎 𝑏 𝑟 ⊆ Ĥ, then 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ .   115   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Proof. (⇐) Suppose that Ĥ is a proper F. ideal of R and whenever for each F. singleton 𝑎 , 𝑏 , 𝑟 of R, such that 𝑎 𝑏 𝑟 ⊆ Ĥ, then 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ . Let 𝑎 𝑏 𝑟 ⊆ Ĥ , 𝑎 𝑟 ⊈ Ĥ and 𝑏 𝑟 ⊈ Ĥ . Since 𝑎 𝑏 𝑟 ⊆ Ĥ , then there exists n∈ 𝑍 such tha 𝑎 𝑏 𝑟 𝑎 𝑏 𝑟 ⊆ Ĥ. Since 𝑎 𝑟 ⊈ Ĥ and 𝑏 𝑟 ⊈ Ĥ, then we have 𝑎 𝑏 𝑎 𝑏 ⊆ Ĥ. So that 𝑎 𝑏 ⊆ Ĥ . Thus Ĥ is T-ABSO F. ideal of R and so that Ĥ is T- ABSO quasi primary F. of R. (⇒) Let Ĥ be T-ABSO quasi primary F. ideal of R and for each F. singleton 𝑎 , 𝑏 , 𝑟 of R, such that 𝑎 𝑏 𝑟 ⊆ Ĥ. Since Ĥ ⊆ Ĥ and Ĥ is T-ABSO F. ideal of R. So that 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ . The proposition specificities T-ABSO quasi primary F. ideal in terms of its level ideal is given as follow Proposition 12. A F. ideal Ĥ of R is T-ABSO quasi primary F. iff the level ideal Ĥ is T-ABSO quasi primary ideal of R, ∀ v ∈ L. Proof. ( ⟹) Let abr∈ Ĥ for each a, b, r ∈ R then Ĥ(abr)≥ v hence 𝑎𝑏𝑟 ⊆ Ĥ. So that 𝑎 𝑏 𝑟 ⊆ Ĥ where v = min{s,l, k}. Since Ĥ is T-ABSO quasi primary F., then either 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ hence either 𝑎𝑏 ⊆ Ĥ or 𝑎𝑟 ⊆ Ĥ or 𝑏𝑟 ⊆ Ĥ and so ab∈ Ĥ or ar ∈ Ĥ or br ∈ Ĥ . Thus Ĥ is T-ABSO quasi primary ideal of R. (⟸) Let 𝑎 𝑏 𝑟 ⊆ Ĥ for F. singletons 𝑎 , 𝑏 , 𝑟 of R, ∀ s, l, k ∈ L. Hence 𝑎𝑏𝑟 ⊆ 𝐴, where v = min{s,l, k}, so that Ĥ(abr)≥ v and abr∈ Ĥ . But Ĥ is T-ABSO quasi primary ideal then either ab∈ Ĥ or ar ∈ Ĥ or br ∈ Ĥ , hence either 𝑎𝑏 ⊆ Ĥ or 𝑎𝑟 ⊆ Ĥ or 𝑏𝑟 ⊆ Ĥ. So that either 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ . Thus Ĥ is T-ABSO quasi primary F. ideal of R. The following theorem gives a characterization of T-ABSO quasi primary F. ideal. Theorem 13. Let Ĥ be a proper F. ideal of R. Then Ĥ is T-ABSO quasi primary F. ideal iff whenever Ṷ𝐾𝑇 ⊆ Ĥ for some F. ideals Ṷ, 𝐾, 𝑇 of R, then Ṷ𝐾 ⊆ Ĥ or Ṷ𝑇 ⊆ Ĥ or 𝐾𝑇 ⊆ Ĥ .Proof. (⇐) Assume that Ṷ𝐾𝑇 ⊆ Ĥ for some F. ideals Ṷ, 𝐾, 𝑇 of R, then Ṷ𝐾 ⊆ Ĥ or Ṷ𝑇 ⊆ Ĥ or 𝐾𝑇 ⊆ Ĥ and let 𝑎 𝑏 𝑟 ⊆ Ĥ for F. singleton 𝑎 , 𝑏 , 𝑟 of R . Hence 𝑎 𝑏 𝑟 ⊆ Ĥ and so that 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ . Then 𝑎 𝑏 ⊆ Ĥ or 𝑎 𝑟 ⊆ Ĥ or 𝑏 𝑟 ⊆ Ĥ. By proposition (11), then Ĥ is T-ABSO quasi primary F. ideal of R. (⇒) Assume that Ĥ is T-ABSO quasi primary F. ideal of R and Ṷ𝐾𝑇 ⊆ Ĥ for some F. ideals Ṷ, 𝐾, 𝑇 of R, then Ṷ𝐾𝑇 ⊆ Ĥ . Since Ĥ is T-ABSO F. ideal of R, then Ṷ𝐾 ⊆ Ĥ or Ṷ𝑇 ⊆ Ĥ or 𝐾𝑇 ⊆ Ĥ by proposition (9).   116   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 3. T-ABSO F. Subm. In this section we present the concept of T-ABSO F. subm. and we introduce many basic properties and results about this concept. Definition 14. Let X be F. M. of an R-M. Ḿ. A proper F. subm. A of X is called T-ABSO F. subm. if whenever 𝑎 , 𝑏 be F. singletons of R, and 𝑥 ⊆ 𝑋 , ∀ 𝑠, 𝑙, 𝑣 ∈ 𝐿 such that 𝑎 𝑏 𝑥 ⊆ 𝐴, then either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝐴 or 𝑏 𝑥 ⊆ 𝐴 , see [7]. The proposition specificities T-ABSO F. subm. in terms of its level subm. is given as follow: Proposition 15. Let A be T-ABSO F. subm. of F. M. X of an R-M. Ḿ., iff the level subm. 𝐴 is T-ABSO subm. of 𝑋 , for all v ∈ L, see[7]. Remarks and Examples 1. The intersection of two distinct prime F. subms. of F. M. X of an R-M, Ḿ is T-ABSO F. subm. Proof. Let A and B be two distinct prime F. subms. of X. Suppose that F. singletons 𝑎 , 𝑏 of R, 𝑥 ⊆ 𝑋 such that 𝑎 𝑏 𝑥 ⊆ 𝐴 ∩ 𝐵, but 𝑎 𝑥 ⊈ 𝐴 ∩ 𝐵 and 𝑏 𝑥 ⊈ 𝐴 ∩ 𝐵. Then 𝑎 𝑥 ⊈ 𝐴, 𝑏 𝑥 ⊈ 𝐴, 𝑎 𝑥 ⊈ 𝐵 and 𝑏 𝑥 ⊈ 𝐵 these are impossible since A and B are prime F. subms. So suppose that 𝑎 𝑥 ⊈ 𝐴 and 𝑏 𝑥 ⊈ 𝐵. Since 𝑎 𝑏 𝑥 ⊆ 𝐴 and 𝑎 𝑏 𝑥 ⊆ 𝐵, then 𝑏 ⊆ 𝐴: 𝑋 and 𝑎 ⊆ 𝐵: 𝑋 . So that 𝑎 𝑏 ⊆ 𝐴: 𝑋 ∩ 𝐵: 𝑋 𝐴 ∩ 𝐵: 𝑋 . Thus 𝐴 ∩ 𝐵 is T-ABSO F. subm. of X. (2). Every prime F. subm. is T-ABSO F. subm. Proof. Let A be a prime F. subm. of F. M. X of an R-M. Ḿ. Let𝑎 𝑏 𝑥 ⊆ 𝐴 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋. Then 𝑎𝑏𝑥 ⊆ 𝐴 where v = min{s, l, k }. Since A is a proper subm. of X then 𝐴 is a proper subm. of 𝑋 , hence 𝐴 is prime subm. of 𝑋 . So that 𝐴 is T-ABSO subm. (see [14]), hence 𝑎𝑏 ∈ 𝐴: 𝑋 , then either 𝑎𝑏 ⊆ 𝐴: 𝑋 or 𝑎𝑥 ⊆ 𝐴 or 𝑏𝑥 ⊆ 𝐴. So either ab ∈ 𝐴 : 𝑋 or ax ∈ 𝐴 or bx∈𝐴 . Since 𝐴 : 𝑋 𝐴: 𝑋 by [5]. So that Then either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝐴 or 𝑏 𝑥 ⊆ 𝐴. Thus A is T-ABSO F. subm. of X. However, the converse incorrect in general, for example: Let X: 𝑍 → 𝐿 such that X(y) = 1 𝑖𝑓 𝑦 ∈ 𝑍 0 𝑜. 𝑤. It is obvious that X is F. M. of 𝑍 as Z-M. Let A: 𝑍 → 𝐿 such that A(y) = 𝑣 𝑖𝑓 𝑦 ∈ 6 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that A is F. subm. of X. Now 𝐴 = 6 is not prime subm. of 𝑍 ,since 2. 3 ∈ 6 but 3 ∉ 6 and 2∉( 6 : 𝑍 6𝑍 . But 6 2 ∩ 3 is T-ABSO subm.of 𝑍 as Z-M. by [14]. So 𝐴 is T-ABSO subm., but not prime subm.,implies that A is T-ABSO F. subm., but not prime F. subm. (3) It obvious every quasi-prime F. subm. is T-ABSO F. subm. However T-ABSO F. subm. may not be quasi-prime F. subm. for example: Let X:Z→L such that X(y)= 1 𝑖𝑓 𝑦 ∈ 𝑍 0 𝑜. 𝑤.   117   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 It is obvious that X is F. M. of Z-M. Z. Let A: Z→L such that A(y) = 𝑣 𝑖𝑓 𝑦 ∈ 6𝑍 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that A is F. subm. of X. 𝐴 =6Z is T-ABSO subm. of Z, since if x, y, z∈ Z and xyz∈ 6Z= 𝐴 then at least one of x, y and z is even or one of them is 6. Then either xy∈ 𝐴 or xz∈ 𝐴 or yz∈ 𝐴 . But 6Z= 𝐴 is not quasi-prime, since 2.3.1∈6Z, but 2.1∉6Z and 3.1∉6Z. So that A is T-ABSO F. subm., but A is not quasi-prime F. subm. (4) Let A, B be two F. subms. of F. M. X of an R-M. Ḿ, and B⊂A. If A is T-ABSO F. subm. of X, then it is not necessary that B is a T-ABSO F. subm., for example: Let X: 𝑍 → 𝐿 such that X(y) = 1 𝑖𝑓 𝑦 ∈ 𝑍 0 𝑜. 𝑤. It is obvious that X is F. M. of Z-M. 𝑍 . Let A: 𝑍 → 𝐿 such that A(y) = 𝑣 𝑖𝑓 𝑦 ∈ 2 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 And B: 𝑍 → 𝐿 such that B(y) = 𝑣 𝑖𝑓 𝑦 ∈ 12 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that A and B are F. subms. of X. Now, 𝐴 = 2 and 𝐵 = 12 where 𝐵 ⊂ 𝐴 , since 𝐴 = 2 is maximal subm. of 𝑍 as Z- M., then 𝐴 is prime subm. by[15]. Implies that 𝐴 is T-ABSO subm. by [14]. But 2.2. 3 ∈ 𝐵 , 2. 3 ∉ 𝐵 and 2.2=4∉(𝐵 : 𝑍 12𝑍. Thus 𝐵 is not T-A BSO subm. Of 𝑍 as Z-M. hence B is not T-ABSO F. subm. (5) Let A and B be F. subms. of F. M. X of an R-M . Ḿ and A⊂B. If A is T-ABSO F . subm. of X, then A is T-ABSO F. subm. of B. Proof. If B =X, then don´t need to prove. Let 𝑎 𝑏 𝑥 ⊆ 𝐴 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝐵, implies 𝑎𝑏𝑥 ⊆ 𝐴 hence 𝑣 min 𝑠, 𝑙, 𝑘 abx∈𝐴 , where a,b∈R, x∈𝐵 . Since A⊂B implies where 𝐴 ⊂ 𝐵 . Since A is T-ABS O F. subm. of X, then 𝐴 is T-ABSO subm. Of 𝑋 . Hence 𝐴 is T-ABSO subm. Of 𝐵 by [14], so that either ab ∈ 𝐴 : 𝐵 → 𝑎𝑏 ∈ 𝐴: 𝐵 or ax ∈ 𝐴 or bx ∈ 𝐴 , then 𝑎𝑏 ⊆ 𝐴: 𝐵 or 𝑎𝑥 ⊆ 𝐴 or 𝑏𝑥 ⊆ 𝐴, implies either 𝑎 𝑏 ⊆ 𝐴: 𝐵 of T-A or 𝑎 𝑥 ⊆ 𝐴 or 𝑏 𝑥 ⊆ 𝐴. Thus A is T-ABSO F. subm. of B. (6) The sum BSO F. subm. is not necessary T- ABSO F. subm., for example: Let X: Z→L such that X(y)= 1 𝑖𝑓 𝑦 ∈ 𝑍 0 𝑜. 𝑤. It is obvious that X is F. M. of Z-M. Z. Let A: Z→L such that A(y) = 𝑣 𝑖𝑓 𝑦 ∈ 2𝑍 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that A is F. subm. of X. Let B: Z→L such that B(y) = 𝑣 𝑖𝑓 𝑦 ∈ 3𝑍 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that B is F. subm. of X. Now, 𝐴 =2Z , , 𝐵 =3Z where 𝐴 and 𝐵 be T- ABSO subms. of Z-M. Z, but 𝐴 𝐵 𝑍 𝑋 is not T-ABSO subm., implies that A+B=X is not T-ABSO F. subm. (7) Let A and B be two F. subms. of F. M. X of an R-M. Ḿ. If A is T-ABSO F. subm. then it is not necessary that B is T-ABSO F. subm., for example: Let X: Z→L such that X(y)= 1 𝑖𝑓 𝑦 ∈ 𝑍 0 𝑜. 𝑤.   118   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 It is obvious that X is F. M. of Z-M. Z. Let A: Z→L such that A(y) = 𝑣 𝑖𝑓 𝑦 ∈ 12𝑍 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 Let B: Z→L such that B(y) = 𝑣 𝑖𝑓 𝑦 ∈ 10𝑍 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that A and B are F. subms. of X . Now, 𝐴 =2Z , 𝐵 =20Z where 𝐴 is T- ABSO subm. of Z as Z-M., but 2Z≅20Z and 𝐵 =20Z is not T-ABSO subm. of Z as Z-M. since 2.2.5∈𝐵 =20Z, but 2.5∉ 𝐵 =20Z and 2.2∉ 𝐵 =20Z. Thus A≅B where A is T- ABSO F. subm. of X and B is not T-ABSO F. subm. of X. (8) The intersection of two T- ABSO F. subms. need not be T-ABSO F. subm., for example: Let X: Z→L such that X(y)= 1 𝑖𝑓 𝑦 ∈ 𝑍 0 𝑜. 𝑤. It is obvious that X is F. M. of Z-M. Z. Let A: Z→L such that A(y) = 𝑣 𝑖𝑓 𝑦 ∈ 12𝑍 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 Let B: Z→L such that B(y) = 𝑣 𝑖𝑓 𝑦 ∈ 10𝑍 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that A and B are F. subms. of X . 𝐴 =12Z , 𝐵 =10Z are T-ABSO subms. in the Z as Z-M. But 𝐴 ∩ 𝐵 =12Z∩10Z=120Z which is not T-ABSO since 2.6.10 ∈120Z, but 2.10∉120Z and 6.10∉120Z and 2.6∉120Z. Hence A and B subms., but A∩B is not T- ABSO F. subm (9) Let A be T-ABSO are two T-ABSO F. subm. of F. M. X of an R-M. Ḿ. Then for each B⊆X, either B⊆A or B∩A is T-ABSO F. subm. of B. Proof. Assume that B⊈A then B∩A⊊B Let 𝑎 , 𝑏 be F. singletons of R and 𝑥 ⊆ 𝐵, such that 𝑎 𝑏 𝑥 ⊆ 𝐵 ∩ 𝐴 , implies 𝑎 𝑏 𝑥 ⊆ 𝐴. Since A is T-ABSO F. subm., thus either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝐴 or 𝑏 𝑥 ⊆ 𝐴. Then either 𝑎 𝑏 ⊆ 𝐵 ∩ 𝐴: 𝐵 or 𝑎 𝑥 ⊆ 𝐵 ∩ 𝐴 or 𝑏 𝑥 ⊆ 𝐵 ∩ 𝐴. Thus B∩A is T-ABSO F. subm. of B. Proposition 17. Let 𝑓: Ḿ ⟶ Ḿ be an epimorphism, where 𝑋 , 𝑋 are F. M. of R- M. Ḿ and Ḿ resp. If B is T-ABSO F. subm. of 𝑋 , then 𝑓 𝐵 is T-ABSO F. subm. of 𝑋 . Proof. Since B is F. subm. of 𝑋 , then 𝑓 𝐵 is F. subm. of 𝑋 , since f is epimorphism. Let 𝑎 𝑏 𝑥 ⊆ 𝑓 𝐵 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋 . Then 𝑎 𝑏 𝑓 𝑥 ⊆ 𝐵 and since B is TABSO F. subm., then either 𝑎 𝑓 𝑥 ⊆ 𝐵 or 𝑏 𝑓 𝑥 ⊆ 𝐵 or 𝑎 𝑏 ⊆ 𝐵: 𝑋 . Hence either 𝑎 𝑥 ⊆ 𝑓 𝐵 or 𝑏 𝑓 𝑥 ⊆ 𝑓 𝐵 or 𝑎 𝑏 𝑋 ⊆ 𝐵. But 𝑓 𝑋 ⊆ 𝑋 , so that 𝑎 𝑏 𝑓 𝑋 ⊆ 𝐵, hence 𝑎 𝑏 𝑋 ⊆ 𝑓 𝐵 , implies 𝑎 𝑏 ⊆ 𝑓 𝐵 : 𝑋 Thus 𝑓 𝐵 is T-ABSO F. subm. of 𝑋 Proposition 18. Let 𝑓: Ḿ ⟶ Ḿ be an epimorphism, and 𝑋 , 𝑋 are F. M. of R-M. Ḿ and Ḿ resp. Let 𝐴 ⊆ 𝑋 such that F-ker f ⊆ A . Then A is T-ABSO F. subm. of 𝑋 iff f (A) is T-ABSO F. subm. of 𝑋 . Proof. ⇒ Let 𝑎 , 𝑏 be F. singletons of R and 𝑦 ⊆ 𝑋 where 𝑦 𝑓 𝑥 for some F. singleton 𝑥 ⊆ 𝑋 , such that 𝑎 𝑏 𝑦 ⊆ 𝑓 𝐴 . Hence 𝑎 𝑏 𝑓 𝑥 ⊆ 𝑓 𝐴 𝑎 𝑏 𝑓 𝑥 ⊆ 𝑓 𝐴 since f is onto. Then 𝑎 𝑏 𝑓 𝑥 𝑓 𝑧 for some F. singleton 𝑧 ⊆ 𝐴. So that 𝑓 𝑎 𝑏 𝑥   119   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 𝑓 𝑧 , hence 𝑓 𝑎 𝑏 𝑥 𝑓 𝑧 0 ; that is 𝑓 𝑎 𝑏 𝑥 𝑧 0 , implies 𝑎 𝑏 𝑥 𝑧 ⊆ 𝐹 𝑘𝑒𝑟𝑓 ⊆ 𝐴. So that 𝑎 𝑏 𝑥 ⊆ 𝐴. Since A is T-ABSO F. subm., then either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝐴 or 𝑏 𝑥 ⊆ 𝐴. Hence either 𝑎 𝑏 𝑋 ⊆ 𝐴 → 𝑓 𝑎 𝑏 𝑋 ⊆ 𝑓 𝐴 or 𝑓 𝑎 𝑥 ⊆ 𝑓 𝐴 or 𝑓 𝑏 𝑥 ⊆ 𝑓 𝐴 , implies either 𝑎 𝑏 𝑓 𝑋 ⊆ 𝑓 𝐴 → 𝑎 𝑏 𝑋 ⊆ 𝑓 𝐴 or 𝑎 𝑓 𝑥 ⊆ 𝑓 𝐴 or 𝑏 𝑓 𝑥 ⊆ 𝑓 𝐴 . Then either 𝑎 𝑏 ⊆ 𝑓 𝐴 : 𝑋 or 𝑎 𝑦 ⊆ 𝑓 𝐴 or 𝑏 𝑦 ⊆ 𝑓 𝐴 . Thus f (A) is T-ABSO F. subm. of 𝑋 . (⇐) Let 𝑎 𝑏 𝑥 ⊆ 𝐴 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋 . Hence 𝑓 𝑎 𝑏 𝑥 ⊆ 𝑓 𝐴 , implies 𝑎 𝑏 𝑓 𝑥 ⊆ 𝑓 𝐴 . But f (A) is T-ABSO F. subm., then either 𝑎 𝑏 ⊆ 𝑓 𝐴 : 𝑋 or 𝑎 𝑓 𝑥 ⊆ 𝑓 𝐴 or 𝑏 𝑓 𝑥 ⊆ 𝑓 𝐴 . If 𝑎 𝑏 ⊆ 𝑓 𝐴 : 𝑋 , then 𝑎 𝑏 𝑋 ⊆ 𝑓 𝐴 , implies 𝑎 𝑏 𝑓 𝑋 ⊆ 𝑓 𝐴 since f is onto. Hence 𝑓 𝑎 𝑏 𝑋 ⊆ 𝑓 𝐴 , so that 𝑎 𝑏 𝑋 ⊆ 𝐴 ; that is 𝑎 𝑏 ⊆ 𝐴: 𝑋 . If 𝑎 𝑓 𝑥 ⊆ 𝑓 𝐴 then 𝑓 𝑎 𝑥 𝑓 𝑧 for some F. singleton 𝑧 ⊆ 𝐴 ,∀ n∈ L. Hence 𝑓 𝑎 𝑥 𝑓 𝑧 0 , implies 𝑎 𝑥 𝑧 ⊆ 𝐹 𝑘𝑒𝑟𝑓 ⊆ 𝐴. So that 𝑎 𝑥 ⊆ 𝐴 . If 𝑏 𝑓 𝑥 ⊆ 𝑓 𝐴 , then by the same way above, we have 𝑏 𝑥 ⊆ 𝐴. Therefore, A is T-ABSO F. subm. of 𝑋 . Proposition 19. Let A be a proper F. subm. of F. M. X of an R-M Ḿ. Then A is T-ABSO F. subm. of X iff 𝑎 𝑏 𝐵 ⊆ 𝐴 for F. singletons 𝑎 , 𝑏 of R and B is F . subm. of X implies 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝐵 ⊆ 𝐴 or 𝑏 𝐵 ⊆ 𝐴. Proof. ⇒ Let A be T-ABSO F. subm. and 𝑎 𝑏 𝐵 ⊆ 𝐴. Assume that 𝑎 𝑏 ⊈ 𝐴: 𝑋 , 𝑎 𝐵 ⊈ 𝐴 and 𝑏 𝐵 ⊈ 𝐴. Then there exist F. singletons 𝑥 , 𝑦 ⊆ 𝐵, such that 𝑎 𝑥 ⊈ 𝐴 and 𝑏 𝑦 ⊈ 𝐴. Since 𝑎 𝑏 𝑥 ⊆ 𝐴 and 𝑎 𝑏 ⊈ 𝐴: 𝑋 , 𝑎 𝑥 ⊈ 𝐴, we have 𝑏 𝑥 ⊆ 𝐴. Also since 𝑎 𝑏 𝑦 ⊆ 𝐴 and 𝑎 𝑏 ⊈ 𝐴: 𝑋 , 𝑏 𝑦 ⊈ 𝐴, we have 𝑎 𝑦 ⊆ 𝐴. Now, since 𝑎 𝑏 𝑥 𝑦 ⊆ 𝐴 and 𝑎 𝑏 ⊈ 𝐴: 𝑋 , we have 𝑎 𝑥 𝑦 ⊆ 𝐴 or 𝑏 𝑥 𝑦 ⊆ 𝐴. If 𝑎 𝑥 𝑦 ⊆ 𝐴, then 𝑎 𝑥 𝑎 𝑦 ⊆ 𝐴 and since 𝑎 𝑦 ⊆ 𝐴, we get 𝑎 𝑥 ⊆ 𝐴, this is a discrepancy. If 𝑏 𝑥 𝑦 ⊆ 𝐴, then 𝑏 𝑥 𝑏 𝑦 ⊆ 𝐴 and since 𝑏 𝑥 ⊆ 𝐴, we get 𝑏 𝑦 ⊆ 𝐴 this is a discrepancy. Thus either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝐵 ⊆ 𝐴 or 𝑏 𝐵 ⊆ 𝐴. (⇐) It is obvious. The next theorem gives a characterization of T-ABSO F. subm. Theorem 20. Let A be a proper F. subm. of F. M. X of an R-M. Ḿ. Then the following expressions are equivalent: 1- A is T-ABSO F. subm. of X; 2- If ĤṶB⊆A, for some F. ideals Ĥ, Ṷ of R and F. subm. B of X, then either ĤB⊆A or ṶB⊆A or ĤṶ⊆(A: X) . Proof. (1)⇒(2) Suppose that A is T-ABSO F. subm. of X and ĤṶB⊆A for some F. ideals Ĥ, Ṷ of R and some F. subm. B of X. Let ĤṶ ⊈ (A: X), to prove ĤB⊆A or ṶB⊆A. Assume that ĤB⊈A and ṶB⊈A, then there exist F. singletons 𝑎 ⊆ Ĥ and 𝑏 ⊆ Ṷ, such that 𝑎 𝐵 ⊈ 𝐴 and 𝑏 𝐵 ⊈ 𝐴. But 𝑎 𝑏 𝐵 ⊆ 𝐴 and neither 𝑎 𝐵 ⊈ 𝐴 nor 𝑏 𝐵 ⊈ 𝐴 and A is T-ABSO F. subm., so that 𝑎 𝑏 ⊆ 𝐴: 𝑋 . Since ĤṶ⊈ (A: X), then there exist F. singletons 𝑥 ⊆ Ĥ and 𝑦 ⊆ Ṷ, such that 𝑥 𝑦 ⊈(A: X). But 𝑥 𝑦 𝐵 ⊆ 𝐴 , so that 𝑥 𝐵 ⊆ 𝐴 or 𝑦 𝐵 ⊆ 𝐴 by proposition (19).   120   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Now we have the following: (1) If 𝑥 𝐵 ⊆ 𝐴 and 𝑦 𝐵 ⊈ 𝐴, since 𝑎 𝑦 𝐵 ⊆ 𝐴 and 𝑦 𝐵 ⊈ 𝐴 , 𝑎 𝐵 ⊈ 𝐴 , so that 𝑎 𝑦 ⊆ 𝐴: 𝑋 by proposition (19). Since 𝑥 𝐵 ⊆ 𝐴 and 𝑎 𝐵 ⊈ 𝐴 , hence 𝑎 𝑥 𝐵 ⊈ 𝐴. On the other hand, 𝑎 𝑥 𝑦 𝐵 ⊆ 𝐴 and neither 𝑎 𝑥 𝐵 ⊆ 𝐴 nor 𝑦 𝐵 ⊆ 𝐴, we get 𝑎 𝑥 𝑦 ⊆ 𝐴: 𝑋 by proposition (19). But 𝑎 𝑥 𝑦 𝑎 𝑦 𝑥 𝑦 ⊆ 𝐴: 𝑋 and 𝑎 𝑦 ⊆ 𝐴: 𝑋 , we get 𝑥 𝑦 ⊆ 𝐴: 𝑋 this is a discrepancy. (2) If 𝑦 𝐵 ⊆ 𝐴 and 𝑥 𝐵 ⊈ 𝐴 . By the same way of (1), we get a discrepancy. (3) If 𝑥 𝐵 ⊆ 𝐴 and 𝑦 𝐵 ⊆ 𝐴 . Since 𝑦 𝐵 ⊆ 𝐴 and 𝑏 𝐵 ⊈ 𝐴, we have 𝑏 𝑦 𝐵 ⊈ 𝐴. But 𝑎 𝑏 𝑦 𝐵 ⊆ 𝐴 and neither 𝑎 𝐵 ⊆ 𝐴 nor 𝑏 𝑦 𝐵 ⊆ 𝐴. Thus 𝑎 𝑏 𝑦 ⊆ 𝐴: 𝑋 by proposition (19). Since 𝑎 𝑏 ⊆ 𝐴: 𝑋 and 𝑎 𝑏 𝑎 𝑦 ⊆ 𝐴: 𝑋 , we get 𝑎 𝑦 ⊆ 𝐴: 𝑋 . Since 𝑎 𝑥 𝑏 𝐵 ⊆ 𝐴 and neither 𝑏 𝐵 ⊆ 𝐴 nor 𝑎 𝑥 𝐵 ⊆ 𝐴, we have 𝑎 𝑥 𝑏 ⊆ 𝐴: 𝑋 by proposition (19). But 𝑎 𝑥 𝑏 𝑎 𝑏 𝑥 𝑏 ⊆ 𝐴: 𝑋 and since 𝑎 𝑏 ⊆ 𝐴: 𝑋 , we have 𝑥 𝑏 ⊆ 𝐴: 𝑋 . Now, since 𝑎 𝑥 𝑏 𝑦 𝐵 ⊆ 𝐴 and neither 𝑎 𝑥 𝐵 ⊆ 𝐴 nor 𝑏 𝑦 𝐵 ⊆ 𝐴, we get 𝑎 𝑥 𝑏 𝑦 ⊆ 𝐴: 𝑋 by proposition (19), where 𝑎 𝑥 𝑏 𝑦 𝑎 𝑏 𝑎 𝑦 𝑥 𝑏 𝑥 𝑦 ⊆ 𝐴: 𝑋 . But 𝑎 𝑏 𝑎 𝑦 𝑥 𝑏 ⊆ 𝐴: 𝑋 , so that 𝑥 𝑦 ⊆ 𝐴: 𝑋 this is a discrepancy. Thus ĤB⊆A or ṶB⊆A (2) ⇒ (1) It is obvious. Theorem 21. If A is T-ABSO F. subm. of F. M. X of an R-M. Ḿ, then 𝐴: 𝑋 is T-ABSO F. ideal of R. Proof. Let 𝑎 𝑏 𝑟 ⊆ 𝐴: 𝑋 for F. singletons 𝑎 , 𝑏 , 𝑟 of R. If 𝑎 𝑟 ⊈ 𝐴: 𝑋 and 𝑏 𝑟 ⊈ 𝐴: 𝑋 , then there exist F. singletons 𝑥 , 𝑦 ⊆ 𝑋\𝐴, such that 𝑎 𝑟 𝑥 ⊈ 𝐴 and 𝑏 𝑟 𝑦 ⊈ 𝐴 . Since 𝑎 𝑏 𝑟 𝑥 𝑦 ⊆ 𝐴 and A is T-ABSO F. subm., then either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑟 𝑥 𝑦 ⊆ 𝐴 or 𝑏 𝑟 𝑥 𝑦 ⊆ 𝐴. If 𝑎 𝑟 𝑥 𝑦 ⊆ 𝐴 and since 𝑎 𝑟 𝑥 ⊈ 𝐴, then we have 𝑎 𝑟 𝑦 ⊈ 𝐴. So that 𝑎 𝑏 𝑟 𝑦 ⊆ 𝐴 and 𝑏 𝑟 𝑦 ⊈ 𝐴, hence 𝑎 𝑏 ⊆ 𝐴: 𝑋 . By the same method if 𝑏 𝑟 𝑥 𝑦 ⊆ 𝐴 , we get 𝑎 𝑏 ⊆ 𝐴: 𝑋 . Thus 𝐴: 𝑋 is T-ABSO F. ideal of R. Theorem 22. Let X be multiplication F. M. of an R-M. Ḿ, and A is a proper F. subm. of X. If 𝐴: 𝑋 is T-ABSO F. ideal of R, then A is T-ABSO F. subm. of X. Proof. Let 𝑎 𝑏 𝑥 ⊆ 𝐴 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋, then 𝑎 𝑏 𝑥 ⊆ 𝐴. But 𝑥 Ĥ𝑋 for some F. ideal Ĥ of R since X is multiplication F. M., so that 𝑎 𝑏 Ĥ𝑋 ⊆ 𝐴. Thus 𝑎 𝑏 Ĥ ⊆ 𝐴: 𝑋 , so we have that 𝑎 𝑏 Ĥ ⊆ 𝐴: 𝑋 . Since 𝐴: 𝑋 is T- ABSO F. ideal of R, we get either 𝑎 Ĥ ⊆ 𝐴: 𝑋 or 𝑏 Ĥ ⊆ 𝐴: 𝑋 or 𝑎 𝑏 ⊆ 𝐴: 𝑋 by Proposition (9). 1) If 𝑎 Ĥ ⊆ 𝐴: 𝑋 , then 𝑎 Ĥ𝑋 ⊆ 𝐴 and so 𝑎 𝑥 ⊆ 𝐴 . Hence 𝑎 𝑥 ⊆ 𝐴 2) If 𝑏 Ĥ ⊆ 𝐴: 𝑋 , then by the same method 𝑏 𝑥 ⊆ 𝐴 . 3) If 𝑎 𝑏 ⊆ 𝐴: 𝑋 , then 𝑎 𝑏 ⊆ 𝐴: 𝑋 . By combining theorem (21) and theorem (22), we have the following corollary:   121   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Corollary 23. Let A be a proper F. subm. of a multiplication F. M. X of an R-M. Ḿ. Then A is T-ABSO F. subm. of X iff 𝐴: 𝑋 is T-ABSO F. ideal of R. Remark 24. The condition X is multiplication F. M. can't be deleted from theorem (22). See the following example: Let X: 𝑍 → 𝐿 such that X(y) = 1 𝑖𝑓 𝑦 ∈ 𝑍 0 𝑜. 𝑤. It is obvious that X is F. M. of Z-M. 𝑍 . Let A: 𝑍 → 𝐿 such that A(y) = 𝑣 𝑖𝑓 𝑦 ∈ 0 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that A is F. subm. of X. Now, 𝐴 0 is not T-ABSO subm. of 𝑋 𝑍 , since 𝑝 𝑍 0 but 𝑝 𝑍 0 and 𝑝 ∉ 0 : 𝑍 0. Note (0) is a prime ideal in Z , so that 0 : 𝑍 0 is T-ABSO ideal in Z ; that is 𝐴 : 𝑋 is T-ABSO ideal in Z, then 𝐴: 𝑋 is T-ABSO F. ideal in Z. Thus A is not T-ABSO F. subm. of X, but 𝐴: 𝑋 is T-ABSO F. ideal in Z. Now, we gave the following theorem is a characterization of T-ABSO F. subm. Theoerm 25. Let A be a proper F. subm. of a multiplication F. M. X of Ḿ. Then A is T-ABSO F. subm. of X iff 𝐴 𝐴 𝐴 ⊆ 𝐴 implies that 𝐴 𝐴 ⊆ 𝐴 or 𝐴 𝐴 ⊆ 𝐴 or 𝐴 𝐴 ⊆ 𝐴 , where 𝐴 , 𝐴 , 𝐴 are F. subm. of X. Proof. (⇒)Since X is a multiplication F., then 𝐴 Ĥ𝑋 , 𝐴 Ṷ𝑋 and 𝐴 𝐾𝑋 for some F. ideals Ĥ, Ṷ and 𝐾 of R . So that the product of 𝐴 , 𝐴 and 𝐴 as follows: 𝐴 𝐴 𝐴 ĤṶ𝐾𝑋 ⊆ 𝐴. by [16]. Hence ĤṶ𝐾 ⊆ 𝐴: 𝑋 . Since A is T-ABSO F. subm. of X, then 𝐴: 𝑋 is T-ABSO F. ideal by theorem (21). So by proposition (9), either ĤṶ ⊆ 𝐴: 𝑋 or Ĥ𝐾 ⊆ 𝐴: 𝑋 or Ṷ𝐾 ⊆ 𝐴: 𝑋 . Hence either ĤṶ𝑋 ⊆ 𝐴 or Ĥ𝐾𝑋 ⊆ 𝐴 or Ṷ𝐾𝑋 ⊆ 𝐴, then 𝐴 𝐴 ⊆ 𝐴 or 𝐴 𝐴 ⊆ 𝐴 or 𝐴 𝐴 ⊆ 𝐴 . (⇐) Let ĤṶ𝐵 ⊆ 𝐴 for some F. ideals Ĥ, Ṷ of R and B is F. subm. of X. Since X is a multiplication F. M., then B = EX for some F. ideal E of R. Then ĤṶ𝐸𝑋 ⊆ 𝐴 . Let 𝐴 Ĥ𝑋 and 𝐴 Ṷ𝑋 , so that 𝐴 𝐴 𝐵 ĤṶ𝐸𝑋 ⊆ 𝐴. So by hypotheses either 𝐴 𝐵 ⊆ 𝐴 or 𝐴 𝐵 ⊆ 𝐴 or 𝐴 𝐴 ⊆ 𝐴, hence Ĥ𝐸𝑋 ⊆ 𝐴 or Ṷ𝐸𝑋 ⊆ 𝐴 or ĤṶ𝑋 ⊆ 𝐴. Thus Ĥ𝐵 ⊆ 𝐴 or Ṷ𝐵 ⊆ 𝐴 or ĤṶ ⊆ 𝐴: 𝑋 . Therefore, A is T-ABSO F. subm. of X by theorem (20). Now, the definitions of finitely generated F. M. see [17, Definition (2.11)] and faithful F. M. see [3, Definition (3.2.6)]. We give the following proposition. Proposition 26. Let X be a finitely generated multiplication F. M. of an R-M. Ḿ. If Ĥ is T-ABSO F. ideal of R such that F-annX⊆Ĥ, then ĤX is T-ABSO F. subm. of X. Proof. Let 𝑎 𝑏 𝑥 ⊆ Ĥ𝑋 , where 𝑎 , 𝑏 be F. singletons of R and 𝑥 ⊆ 𝑋, hence 𝑎 𝑏 𝑥 ⊆ Ĥ𝑋. But X is a multiplication F. M., then 𝑥 Ṷ𝑋 for some F. ideal Ṷ of R. Thus   122   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 𝑎 𝑏 Ṷ𝑋 ⊆ Ĥ𝑋. So that 𝑎 𝑏 Ṷ ⊆ Ĥ 𝐹 𝑎𝑛𝑛𝑋 Ĥ since F-annX⊆Ĥ. But Ĥ is T-ABSO F. ideal of R, so that either 𝑎 𝑏 ⊆ Ĥ or 𝑎 Ṷ ⊆ Ĥ 𝑏 Ṷ ⊆ Ĥ. Then we have 𝑎 𝑏 𝑋 ⊆ Ĥ𝑋 or 𝑎 Ṷ𝑋 ⊆ Ĥ𝑋 or 𝑏 Ṷ𝑋 ⊆ Ĥ𝑋, so that 𝑎 𝑏 ⊆ Ĥ𝑋: 𝑋 or 𝑎 𝑥 ⊆ Ĥ𝑋 or 𝑏 𝑥 ⊆ Ĥ𝑋, hence 𝑎 𝑏 ⊆ Ĥ𝑋: 𝑋 or 𝑎 𝑥 ⊆ Ĥ𝑋 or 𝑏 𝑥 ⊆ Ĥ𝑋. So that ĤX is T-ABSO F. subm. of X. Corollary 27. Let X be a faithful finitely generated multiplication F. M. of Ḿ. If Ĥ is T-ABSO F. ideal of R, then ĤX is T-ABSO F. subm. of X. Proof. By proposition (26), it follows immediately. Corollary 28. Suppose that X be a faithful finitely generated multiplication F. M. of Ḿ. Then every proper F. subm. of X is T-ABSO iff every proper F. ideal of R is T-ABSO. Proof. (⇐) By corollary (27), it follows immediately. (⟹) Let Ĥ be a proper F. ideal of R. Then A=ĤX is a proper subm. of X . Since A is T-ABSO F. subm., so that 𝐴: 𝑋 is T-ABSO F. ideal by theorem (21). But X is a multiplication F. M., hence A= 𝐴: 𝑋 X by [5]. Thus ĤX= 𝐴: 𝑋 X. Since X is a faithful finitely generated multiplication F. M., then 𝑋 is a faithful finitely generated multiplication M. by [16, 17], implies that 𝑋 =Ḿ is cancellation R-M. by [18]. Hence X is a cancellation F. M. by [8]. Therefore Ĥ= 𝐴: 𝑋 ; that is Ĥ is T-ABSO F. ideal of R. Recall that Let X be F. M. of an R-M. Ḿ, and let A be F. subm. of X. A is called a pure F. subm., if for each F. ideal Ĥ of R such that ĤA=ĤX∩A, see [19]. Proposition 29. Let A be a proper pure F. subm. of F. M. X of Ḿ. If 0 is T-ABSO F. subm. of X , then A is T-ABSO F. subm. of X. Proof. Let 𝑎 𝑏 𝑥 ⊆ 𝐴 where 𝑎 , 𝑏 F. singletons of R and 𝑥 ⊆ 𝑋. Put Ĥ 𝑎 𝑏 , hence 𝑎 𝑏 𝑥 ⊆ Ĥ𝑋 ∩ 𝐴, but ĤX∩A=ĤA . So 𝑎 𝑏 𝑥 𝑎 𝑏 𝑦 , for some F. singleton 𝑦 ⊆ 𝐴 , then 𝑎 𝑏 𝑥 𝑦 ⊆ 0 , but 0 is T-ABSO F. subm., hence 𝑎 𝑥 𝑦 ⊆ 0 or 𝑏 𝑥 𝑦 ⊆ 0 or 𝑎 𝑏 ⊆ 𝐹 𝑎𝑛𝑛𝑋 ⊆ 𝐴: 𝑋 . So we have 𝑎 𝑥 𝑎 𝑦 ⊆ 𝐴 or 𝑏 𝑥 𝑏 𝑦 ⊆ 𝐴 or 𝑎 𝑏 ⊆ 𝐴: 𝑋 . Therefore A is T-ABSO F. subm. of X. Now, we give the concept of a cancellative F. M. as follows: Definition 30. A F. M. X of Ḿ is called a cancellative F. if whenever 𝑎 𝑥 𝑎 𝑦 for F. singletons 𝑎 of R and 𝑥 , 𝑦 ⊆ 𝑋, ∀s,v,k ∈ L, then 𝑥 𝑦 Proposition 31. Let X be a cancellative F. M. of Ḿ, and A be a proper F. subm. of X. Then A is a pure F. subm. of X iff A is T-ABSO F. subm.of X with 𝐴: 𝑋 0 .   123   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Proof. (⇒) Assume that A is a pure F. subm. of X and 𝑎 𝑏 𝑥 ⊆ 𝐴 such that 𝑎 𝑏 ⊈ 𝐴: 𝑋 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋 . Then 𝑎 𝑏 𝑥 ⊆ 𝑎 𝑏 𝑋 ∩ 𝐴 𝑎 𝑏 𝐴 , hence 𝑎 𝑏 𝑥 𝑎 𝑏 𝑦 for some F. singleton 𝑦 ⊆ 𝐴. Since X is a cancellative F. M., then 𝑏 𝑥 𝑏 𝑦 ⊆ 𝐴. Thus A is T-ABSO F. subm. of X. Now, assume that F. singleton 𝑟 ⊆ 𝐴: 𝑋 with 𝑟 0 . Since A≠X there exists F. singleton 𝑥 ⊆ 𝑋\𝐴 such that 𝑟 𝑥 ⊆ 𝑟 𝑋 ∩ 𝐴 𝑟 𝐴 , so there exists F. singleton 𝑦 ⊆ 𝐴, such that 𝑟 𝑥 𝑟 𝑦 , hence 𝑥 𝑦 this is a contradication. So that 𝐴: 𝑋 0 . (⇐) Suppose that A is T-ABSO F. subm. of X. Let 𝑎 𝑏 𝑥 ⊆ 𝑎 𝑏 𝑋 ∩ 𝐴 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋 . We may suppose that 𝑎 𝑏 0 . Since A is T-ABSO F. subm. of X, then either 𝑎 𝑥 ⊆ 𝐴 or 𝑏 𝑥 ⊆ 𝐴. If 𝑏 𝑥 ⊆ 𝐴 and 𝑏 be F. singleton of R, 𝑎 𝑏 𝑥 ⊆ 𝑎 𝑏 𝐴. Thus 𝑎 𝑏 𝑋 ∩ 𝐴 ⊆ 𝑎 𝑏 𝐴. By the same method to prove the case if 𝑎 𝑥 ⊆ 𝐴; that is 𝑎 𝑏 𝐴 ⊆ 𝑎 𝑏 𝑋 ∩ 𝐴. Thus 𝑎 𝑏 𝑋 ∩ 𝐴 𝑎 𝑏 𝐴. So that A is a pure F. subm. 4. T-ABSO Quasi Primary F. Subm. In this section we present the concept of T-ABSO quasi primary F. subm. and study the relationships this concept among T-ABSO F. subm. and T-ABSO primary F. subm. Many basic properties and outcomes are given. Now, we give the following definition: Definition 32. Let A be a proper F. subm. of non-empty F. M. X of an R-M. Ḿ. Then the X-F. radical of A, denoted by X-R(A) is defined to the intersection of all prime F. subm. including A. We give the pursue lemma which are needed in the next proposition. Lemma 33. Let X be a multiplication F. M. of Ḿ, let A be a proper F. subm. of X. Then the following expressions are equivalent: 1- A is a prime F. subm. of X. 2- 𝐴: 𝑋 be a prime F. ideal of R . 3- A=ĤX for some a prime F. ideal Ĥ of R with F-annX⊆Ĥ. Proof. (1)→(2) It follows by [20, proposition (2.5)]. (2)→(3) Since X is a multiplication F. M., so that 𝐴 𝐴: 𝑋 𝑋 by[5]. Put Ĥ= 𝐴: 𝑋 be a prime F. ideal of R. Now, since F-annX= 0 : 𝑋 and 0 : 𝑋 ⊆ 𝐴: 𝑋 Ĥ. So that F-annX ⊆ Ĥ . (3)→(1) Let 𝑎 𝑥 ⊆ 𝐴 for F. singleton 𝑎 of R and 𝑥 ⊆ 𝑋, and 𝑥 ⊈ 𝐴 to prove 𝑎 ⊆ 𝐴: 𝑋 . By(3), A=ĤX for some a prime F. ideal Ĥ of R with F-annX ⊆Ĥ, so that F-annX is a prime F. ideal of R, but F-annX= 0 : 𝑋 , hence 0 : 𝑋 is a prime F. ideal of R. Let 𝑎 𝑏 ⊆ 0 : 𝑋 , for F. singleton 𝑏 of R, and 𝑏 ⊈ 0 : 𝑋 , then 𝑎 ⊆ 0 : 𝑋 . Since 0 : 𝑋 ⊆ 𝐴: 𝑋 , so that 𝑎 ⊆ 𝐴: 𝑋 . Thus A is a prime F. subm. of X. Lemma 34. Let X be a finitely generated multiplication F. M. of Ḿ and let A be F. subm. of X. Then 𝑋 𝑅 𝐴 𝐴: 𝑋 . 𝑋.   124   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Proof. If X-R(A)=X , then the result is directly. So that X-R(A)≠X, if B is any prime F. subm. of X which contains A, we get 𝐴: 𝑋 ⊆ 𝐵: 𝑋 . We prove that 𝐵: 𝑋 is a prime F. ideal. Assume that 𝑎 𝑏 ⊆ 𝐵: 𝑋 for F. singleton 𝑎 , 𝑏 of R, so that 𝑎 𝑏 𝑋 ⊆ 𝐵, then either 𝑏 𝑋 ⊆ 𝐵 or 𝑏 𝑥 ⊆ 𝑋/𝐵 for some F. singleton 𝑥 ⊆ 𝑋 . But B is a prime F. subm. and 𝑎 𝑏 𝑥 ⊆ 𝐵, then either 𝑏 𝑥 ⊆ 𝐵 or 𝑎 ⊆ 𝐵: 𝑋 . Thus 𝑎 ⊆ 𝐵: 𝑋 or 𝑏 ⊆ 𝐵: 𝑋 . So that 𝐵: 𝑋 is a prime F. ideal. Hence 𝐴: 𝑋 ⊆ 𝐵: 𝑋 by [13], then 𝐴: 𝑋 . 𝑋 ⊆ 𝐵: 𝑋 𝑋. Since B is an arbitary prime F. subm. containing A , we get 𝐴: 𝑋 . 𝑋 ⊆ 𝑋 𝑅 𝐴 1 . Now, since X is a multiplication F. M., hence 𝑋 𝑅 𝐴 𝑋 𝑅 𝐴 : 𝑋 𝑋. We must prove that 𝑋 𝑅 𝐴 : 𝑋 ⊆ 𝐴: 𝑋 . Let K be any prime F. ideal such that 𝐴: 𝑋 ⊂ 𝐾. Since K is a prime F. ideal containing F-annX= 0 : 𝑋 , then KX is a prime F. subm. of X containing A= 𝐴: 𝑋 𝑋 by lemma (33). Thus 𝑋 𝑅 𝐴 : 𝑋 𝑋 𝑋 𝑅 𝐴 ⊆ 𝐾𝑋 , hence 𝑋 𝑅 𝐴 : 𝑋 ⊆ 𝐾, then 𝑋 𝑅 𝐴 : 𝑋 ⊆ 𝐴: 𝑋 by [13], hence 𝑋 𝑅 𝐴 𝑋 𝑅 𝐴 : 𝑋 𝑋 ⊆ 𝐴: 𝑋 . 𝑋. So that 𝑅 𝐴 ⊆ 𝐴: 𝑋 . 𝑋 2 . From (1) and (2), we get 𝑅 𝐴 𝐴: 𝑋 . 𝑋 . Before the next proposition we give these lemmas and definition which are needed in the proof of the next proposition. We give this definition as follows: Definition 35. Let X be F. M. of an R-M. Ḿ. If Ƥ is a maximal F. ideal of R then we define 𝐹 𝐺Ƥ 𝑋 𝑥 ⊆ 𝑋 ∶ 1 𝑎 𝑥 0 for some F. sigleton 𝑎 ⊆ Ƥ, ∀𝑣, 𝑠 ∈ 𝐿 . It is obvious 𝐹 𝐺Ƥ 𝑋 is F. subm. of X . X is calld Ƥ-cyclic F. M. if there exist F. singleton 𝑏 ⊆ Ƥ 𝑎𝑛𝑑 𝑥 ⊆ 𝑋 such that 1 𝑏 𝑋 ⊆ 𝑥 , ∀𝑙, 𝑣 ∈ 𝐿. Lemma 36. Let R be a commutative ring with unity. Then F. M. X of an R-M. Ḿ is a multiplication F. M. iff for every maximal F. ideal Ƥ of R either 𝑋 𝐹 𝐺Ƥ 𝑋 or X is Ƥ-cyclic F. M. Proof. (⇒) Assume that X is a multiplication F. M. Let Ƥ be maximal F. ideal of R. Suppose that X=ƤX , let F. singleton 𝑥 ⊆ 𝑋, then 𝑥 Ĥ𝑋 for some F. ideal Ĥ of R. Hence 𝑥 Ĥ𝑋 ĤƤ𝑋 ƤĤ𝑋 Ƥ 𝑥 , then 𝑥 𝑎 𝑥 for some F. sigleton 𝑎 ⊆ Ƥ . Thus 1 𝑎 𝑥 0 , so that 𝑥 ⊆ 𝐹 𝐺Ƥ 𝑋 . It follows that 𝑋 𝐹 𝐺Ƥ 𝑋 Now, suppose that X≠ƤX, then there exists F. sigleton 𝑥 ⊆ 𝑋 , 𝑥 ⊈ Ƥ𝑋. So that there exists an ideal Ṷ of R such that 𝑥 Ṷ𝑋. It is obvious that Ṷ⊈ Ƥ and so 1 𝑏 ⊆ Ṷ for some F. singleton 𝑏 ⊆ Ƥ. Hence 1 𝑏 𝑋 ⊆ 𝑥 . Thus X is Ƥ-cyclic F. M.(⇐) Suppose that for each maximal F. ideal Ƥ of R either 𝑋 𝐹 𝐺Ƥ 𝑋 or X is Ƥ-cyclic F. M. Let A be F. subm. of X and Ĥ 𝐴: 𝑋 . It is obvious that ĤX⊆A. Suppose that F. singleton 𝑦 ⊆ 𝐴 and 𝐾 𝑟 ⊆ 𝑅: 𝑟 𝑦 ⊆ Ĥ𝑋 . Assume that K≠R, then there exists a maximal F. ideal E of R such that K⊆ E by [13, proposition(1.3.2.4)]. If X=𝐹 𝐺 𝑋 then 1 𝑎 𝑦 0 for some F. singleton 𝑎 ⊆ 𝐸 , and 1 𝑎 ⊆ 𝐾 ⊆ 𝐸 this is a discrepancy. Thus by   125   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 hypothesis there exist F. singletons 𝑏 ⊆ 𝐸 , 𝑧 ⊆ 𝑋 such that 1 𝑏 𝑋 ⊆ 𝑧 . It follows that 1 𝑏 𝐴 is F. subm. of 𝑧 and so tha 1 𝑏 𝐴 𝐷 𝑧 where D is F. ideal {𝑟 ⊆ 𝑅: 𝑟 𝑧 ⊆ 1 𝑏 𝐴 of R. Note that 1 𝑏 𝐷 𝑋 𝐷 1 𝑏 𝑋 ⊆ 𝐷 𝑧 ⊆ 𝐴. So that 1 𝑏 𝐷 ⊆ Ĥ. Thus for F. singleton 𝑦 ⊆ 𝐴, 1 𝑏 𝑦 ⊆ 1 𝑏 𝐴 1 𝑏 𝐷 𝑧 ⊆ Ĥ𝑋. So that 1 𝑏 ⊆ 𝐾 ⊆ 𝐸 this is a discrepancy. Thus K=R and 𝑦 ⊆ Ĥ𝑋. Therefore A=ĤX and X is a multiplication F. M. Lemma 37. Let X be a multiplication F. M. of an R-M. Ḿ, then ∩ ∈ Ĥ 𝑋 ∩ ∈ Ĥ 𝐹 𝑎𝑛𝑛𝑋 𝑋 for any non-empty collection of F. ideals Ĥ 𝑖 ∈ Λ of R. Proof. Assume that X is a multiplication F. M. Let Ĥ 𝑖 ∈ Λ be any non-empty collection of F. ideals of R, let Ṷ= ∩ ∈ Ĥ 𝐹 𝑎𝑛𝑛𝑋 , then ṶX= ∩ ∈ Ĥ 𝐹 𝑎𝑛𝑛𝑋 𝑋. It is obvious that Ṷ𝑋 ⊆∩ ∈ Ĥ 𝑋 . Now, let be F. singleton 𝑥 ⊆∩ ∈ Ĥ 𝑋 and let 𝐺 𝑎 ⊆ 𝑅: 𝑎 𝑥 ⊆ Ṷ𝑋 , ∀𝑠, 𝑣 ∈ 𝐿 Suppose that G≠R, then there exists a maximal F. ideal Ƥ of R such that 𝐺 ⊆ Ƥ, it is obvious that 𝑥 ⊈ 𝐹 𝐺Ƥ 𝑋 and hence X is Ƥ-cyclic F. M. by lemma (36). Then there exist F. singletons 𝑎 ⊆ Ƥ and 𝑦 ⊆ 𝑋 such that 1 𝑎 𝑋 ⊆ 𝑦 . Hence 1 𝑎 𝑥 ⊆∩ ∈ Ĥ 𝑦 . for each 𝑖 ∈ Λ there exists F. singleton 𝑏 ⊆ Ĥ , ∀𝑙 ∈ 𝐿, such that 1 𝑎 𝑥 𝑏 𝑦 . Choose 𝑗 ∈ Λ, for each 𝑖 ∈ Λ , 𝑏 𝑦 𝑏 𝑦 , so that 𝑏 𝑏 𝑦 0 , implies that: 1 𝑎 𝑏 𝑏 𝑋 𝑏 𝑏 1 𝑎 𝑋 ⊆ 𝑏 𝑏 𝑦 0 , 1 𝑎 𝑏 𝑏 0 . Thus 1 𝑎 𝑏 1 𝑎 𝑏 ⊆ Ĥ 𝑖 ∈ Λ , then 1 𝑎 𝑏 ⊆ Ṷ. Hence 1 𝑎 𝑥 1 𝑎 𝑏 𝑦 ⊆ Ṷ𝑋 . It follows that 1 𝑎 ⊆ 𝐺 ⊆ Ƥ this is a discrepancy. Thus G=R and 𝑥 ⊆ Ṷ𝑋, so that ∩ ∈ Ĥ 𝑋 ⊆ Ṷ𝑋 implies that ∩ ∈ Ĥ 𝑋 Ṷ𝑋 That is ∩ ∈ Ĥ 𝑋 ∩ ∈ Ĥ 𝐹 𝑎𝑛𝑛𝑋 𝑋. Now, we give the proposition as follows: Proposition 38. Let X be a multiplication finitely generated F. M. of an R-M. Ḿ and A be T-ABSO F. subm. of X. Then one of the following satisfy: 1- X-R(A)=Ƥ is a prime F. subm. of X such that Ƥ ⊆ 𝐴. 2- X-R(A)=Ƥ ∩ Ƥ , Ƥ Ƥ ⊆ 𝐴 and 𝑋 𝑅 𝐴 ⊆ 𝐴 where Ƥ , Ƥ are the only distinct minimal prime F. subms. of A. Proof. By theorem (21), 𝐴: 𝑋 is T-ABSO F. ideal of R. So that either 𝑅 𝐴: 𝑋 Ṷ is a prime F. ideal of R such that Ṷ ⊆ 𝐴: 𝑋 or 𝑅 𝐴: 𝑋 Ṷ ∩ Ṷ , Ṷ Ṷ ⊆ 𝐴: 𝑋 and 𝑅 𝐴: 𝑋 ⊆ 𝐴: 𝑋 where Ṷ , Ṷ are the only distinct minimal prime F. ideals of 𝐴: 𝑋 by proposition (6), where 𝑅 𝐴: 𝑋 𝐴: 𝑋 . if the first case satisfies, then since X is F. multiplication, we have X-R(A)=R 𝐴: 𝑋 X=ṶX is a prime F. subm. of X. Put ṶX=Ƥ by lemma (33) and lemma (34), and Ṷ𝑋 Ṷ 𝑋 ⊆ 𝐴: 𝑋 𝑋 𝐴. Now, suppose that the latter case satisfies, then by lemma(33), Ṷ 𝑋 and Ṷ 𝑋 are the only distinct minimal prime F. subms.   126   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 of A and X- R(A) = 𝑅 𝐴: 𝑋 𝑋 Ṷ ∩ Ṷ 𝑋 Ṷ 𝑋 ∩ Ṷ 𝑋 by lemma (37). Moreover Ṷ 𝑋 Ṷ 𝑋 Ṷ Ṷ 𝑋 ⊆ 𝐴: 𝑋 𝑋 𝐴 and 𝑋 𝑅 𝐴 R 𝐴: 𝑋 X R 𝐴: 𝑋 𝑋 ⊆ 𝐴: 𝑋 𝑋 𝐴. We give the definition of T-ABSO primary F. subm. as follows: Definition 39. Let A be a proper F. subm. of F. M. X of Ḿ, A is called T-ABSO primary F. subm. of X if whenever F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋 such that 𝑎 𝑏 𝑥 ⊆ 𝐴 , then either 𝑎 𝑥 ⊆ 𝑋 R 𝐴 or 𝑏 𝑥 ⊆ 𝑋 R 𝐴 or 𝑎 𝑏 ⊆ 𝐴: 𝑋 . The following proposition characterize T-ABSO primary F. subm. in terms of its level subm. Proposition 40. Let A be T-ABSO primary F. subm. of F. M. X of Ḿ. for all v∈ L, iff the level subm. 𝐴 is T-ABSO primary subm. of 𝑋 . Proof. ⇒ Let abx∈𝐴 for any a,b∈R and 𝑥 ⊆ 𝑋 , then A(abx)≥ v , so 𝑎𝑏𝑥 ⊆ 𝐴 implies that 𝑎 𝑏 𝑥 ⊆ 𝐴 where v =min{s, l, k}. Since A be T-ABSO primary F. subm., so either 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑎 𝑏 ⊆ 𝐴: 𝑋 . If 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 , then 𝑎𝑥 ⊆ 𝑋 𝑅 𝐴 , so ax ∈𝑋 𝑅 𝐴 . If 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 , then 𝑏𝑥 ⊆ 𝑋 𝑅 𝐴 , so bx∈𝑋 𝑅 𝐴 . If 𝑎 𝑏 ⊆ 𝐴: 𝑋 then 𝑎𝑏 ⊆ 𝐴: 𝑋 , so ab ∈ 𝐴: 𝑋 𝐴 : 𝑋 . Hence ab∈ 𝐴 : 𝑋 . Thus 𝐴 is T-ABSO primary subm. of 𝑋 . (⟸)Let 𝑎 𝑏 𝑥 ⊆ 𝐴 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋,∀s, l, k ∈L, hence 𝑎𝑏𝑥 ⊆ 𝐴 where 𝑣 min 𝑠, 𝑙, 𝑘 so that A(abx)≥ v, implies abx∈𝐴 , but 𝐴 is T- ABSO primary subm. of 𝑋 so either ax∈ 𝑋 R 𝐴 or bx∈ 𝑋 R 𝐴 or ab∈ 𝐴 : 𝑋 . Since 𝐴 : 𝑋 𝐴: 𝑋 , hence ab∈ 𝐴: 𝑋 . Then either 𝑎𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏𝑥 ⊆ 𝑋 R 𝐴 or 𝑎𝑏 ⊆ 𝐴: 𝑋 , implies either 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑎 𝑏 ⊆ 𝐴: 𝑋 . Thus A be T-ABSO primary F. subm. of X. Remark 41. Every T-ABSO F. subm. is T-ABSO primary F. subm., but the converse in general incorrect, for example: Let X: 𝑍 → 𝐿 such that X(y) = 1 𝑖𝑓 𝑦 ∈ 𝑍 0 𝑜. 𝑤. It is obvious that X is F. M. of Z-M. Z. Let A: 𝑍 → 𝐿 such that A(y) = 𝑣 𝑖𝑓 𝑦 ∈ 12𝑍 0 𝑜. 𝑤. ∀ 𝑣 ∈ 𝐿 It is obvious that A is F. subm. of X. Now, 𝐴 12𝑍 and 𝑋 𝑍 as Z-M. Note that 𝐴 12𝑍 is not T-ABSO subm. since 2.2.3 ∈ 12𝑍 𝐴 but 2.2 ∉ 12𝑍 𝐴 and 2.3 ∉ 12𝑍 𝐴 . But 𝑋 R 𝐴 𝑍 R 12𝑍 2𝑍 ∩ 3𝑍 6𝑍 where 2Z and 3Z are prime subms. of 𝑋 containing 𝐴 . So that 𝐴 is T-ABSO primary subm. of 𝑋 since 2.3=6∈6Z. Thus A is not T- ABSO F. subm., but it is T-ABSO primary F. subm. of X . We give the concept of T-ABSO quasi primary F. subm. as follows:   127   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Definition 42. A proper F. subm. A of F. M. X of Ḿ is called T-ABSO quasi primary F. subm. If 𝑎 𝑏 𝑥 ⊆ 𝐴 implies either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 for each F. singleton 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋 , ∀s,l,v ∈ L. The following proposition characterize T-ABSO quasi primary F. subm. in terms of its level subm. Proposition 43. Let A be T-ABSO quasi primary F. subm. of F. M. X of Ḿ iff the level subm. 𝐴 is T- ABSO quasi primary subm. of 𝑋 ∀ v∈L. Proof. ⇒ Let abx∈ 𝐴 for any a, b∈ R and x∈ 𝑋 , then A(abx)≥ v , so 𝑎𝑏𝑥 ⊆ 𝐴 implies that 𝑎 𝑏 𝑥 ⊆ 𝐴 where v =min{s, l, k}. Since A be a T-ABSO quasi primary F. subm., so either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 . If 𝑎 𝑏 ⊆ 𝐴: 𝑋 then 𝑎𝑏 ⊆ 𝐴: 𝑋, so ab ∈ 𝐴: 𝑋 𝐴 : 𝑋 . Henc eab ∈ 𝐴 : 𝑋 . If 𝑎 𝑥 ⊆ 𝑋 R 𝐴 , then 𝑎𝑥 ⊆ 𝑋 𝑅 𝐴 , so ax∈ 𝑋 𝑅 𝐴 . If 𝑏 𝑥 ⊆ 𝑋 R 𝐴 , then 𝑏𝑥 ⊆ 𝑋 𝑅 𝐴 , so bx∈ 𝑋 𝑅 𝐴 . Thus 𝐴 is a T-ABSO quasi primary subm. of 𝑋 . ⇐ Let 𝑎 𝑏 𝑥 ⊆ 𝐴 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋 , hence 𝑎𝑏𝑥 ⊆ 𝐴 where 𝑣 min 𝑠, 𝑙, 𝑘 so that A(abx)≥ v, implies abx∈𝐴 , but 𝐴 is T-ABSO quasi primary subm. of 𝑋 , so either ab∈ 𝐴 : 𝑋 or ax ∈𝑋 𝑅 𝐴 or bx∈ 𝑋 𝑅 𝐴 . Since 𝐴 : 𝑋 𝐴: 𝑋 , hence ab∈ 𝐴: 𝑋 . Then either 𝑎𝑏 ⊆ 𝐴: 𝑋 or 𝑎𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏𝑥 ⊆ 𝑋 R 𝐴 , implies either 𝑎 𝑏 ⊆ 𝐴: 𝑋 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 where 𝑣 min 𝑠, 𝑙, 𝑘 . Thus A be T-ABSO quasi primary F. subm. of X. Theorem 44. Let A be a proper F. subm. of F. M. X of Ḿ. Then the following expressions are equivalent: 1- A is T-ABSO quasi primary F. subm. of X; 2- For every F. singleton 𝑎 , 𝑏 of R, ∀s, 𝐴: 𝑎 𝑏 𝑋 for some n∈ 𝑍 or 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑎 ∪ 𝑋 𝑅 𝐴 : 𝑏 . 3- For every F. singleton 𝑎 , 𝑏 of R, ∀s,l ∈ L, 𝐴: 𝑎 𝑏 𝑋 for some n∈ 𝑍 or 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑎 or 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑏 . Proof. (1)→(2) Assume that A is T-ABSO quasi primary F. subm. of X, let F. singleton 𝑎 , 𝑏 of R. If 𝑎 𝑏 ⊆ 𝐴: 𝑋 , then 𝑎 𝑏 𝑎 𝑏 ⊆ 𝐴: 𝑋 for some n∈ 𝑍 , hence 𝐴: 𝑎 𝑏 𝑋 . Now, suppose that 𝑎 𝑏 ⊈ 𝐴: 𝑋. Let 𝑥 ⊆ 𝐴: 𝑎 𝑏 , then 𝑎 𝑏 𝑥 ⊆ 𝐴. Since A is T-ABSO quasi primary F. subm., then 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 . So that 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑎 ∪ 𝑋 𝑅 𝐴 : 𝑏 . (2)→(3) By (2), we have 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑎 ∪ 𝑋 𝑅 𝐴 : 𝑏 . So that 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑎 or 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑏 .   128   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 (3)→(1) Let 𝑎 𝑏 𝑥 ⊆ 𝐴 and 𝑎 𝑏 ⊈ 𝐴: 𝑋 for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋, hence 𝑎 𝑏 𝑎 𝑏 ⊈ 𝐴: 𝑋 for some n∈ 𝑍 , then 𝐴: 𝑎 𝑏 𝑋. By (3), we have that 𝑥 ⊆ 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑎 or 𝑥 ⊆ 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑏 . Thus 𝑥 𝑎 ⊆ 𝑋 𝑅 𝐴 or 𝑥 𝑏 ⊆ 𝑋 𝑅 𝐴 . So that A is T-ABSO quasi primary F. subm. of X. Lemma 45. Let X be F. M. of Ḿ. Suppose that A is T-ABSO quasi primary F. subm. of X and 𝑎 𝑏 𝐵 ⊆ 𝐴 for F. singleton 𝑎 , 𝑏 of R, ∀ 𝑠, 𝑙∈ L , and F. subm. B of X . If 𝑎 𝑏 ⊈ 𝐴: 𝑋 , then 𝑎 𝐵 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝐵 ⊆ 𝑋 𝑅 𝐴 . Proof. Since 𝐵 ⊆ 𝐴: 𝑎 𝑏 and 𝐴: 𝑎 𝑏 𝑋 for some n∈ 𝑍 , by theorem (44), we get 𝐵 ⊆ 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑎 or 𝐵 ⊆ 𝐴: 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑏 . Then 𝑎 𝐵 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝐵 ⊆ 𝑋 𝑅 𝐴 . Theorem 46. Let A be a proper F. subm. of F. M. X of Ḿ, then the following expressions are equivalent: 1- A is T-ABSO quasi primary F. subm. of X; 2- For F. singleton 𝑎 of R, ∀s∈L, F. ideal Ĥ of R and F. subm. B of X with 𝑎 Ĥ𝐵 ⊆ 𝐴, then either 𝑎 Ĥ ⊆ 𝐴: 𝑋 or 𝑎 𝐵 ⊆ 𝑋 𝑅 𝐴 or Ĥ𝐵 ⊆ 𝑋 𝑅 𝐴 ; 3- For F. ideals Ĥ, Ṷ of R, and F. subm. B of X with ĤṶ𝐵 ⊆ 𝐴, then eithe ĤṶ ⊆ 𝐴: 𝑋 or Ĥ𝐵 ⊆ 𝑋 𝑅 𝐴 or Ṷ𝐵 ⊆ 𝑋 𝑅 𝐴 . Proof. (1)→(2) Assume that 𝑎 Ĥ𝐵 ⊆ 𝐴 with 𝑎 Ĥ ⊈ 𝐴: 𝑋 and Ĥ𝐵 ⊈ 𝑋 𝑅 𝐴 . Then there exist F. singletons 𝑏 , 𝑟 ⊆ Ĥ , such that 𝑎 𝑏 ⊈ 𝐴: 𝑋 and 𝑟 𝐵 ⊈ 𝑋 𝑅 𝐴 . Now, we prove that 𝑎 𝐵 ⊆ 𝑋 𝑅 𝐴 . Suppose that 𝑎 𝐵 ⊈ 𝑋 𝑅 𝐴 . Since 𝑎 𝑏 𝐵 ⊆ 𝐴 , by lemma (45), we have 𝑏 𝐵 ⊆ 𝑋 𝑅 𝐴 , hence 𝑏 𝑟 𝐵 ⊈ 𝑋 𝑅 𝐴 . By using lemma (45), we have 𝑎 𝑏 𝑟 𝑎 𝑏 𝑎 𝑟 ⊆ 𝐴: 𝑋 , because 𝑎 𝑏 𝑟 𝐵 ⊆ 𝐴. Since 𝑎 𝑏 𝑎 𝑟 ⊆ 𝐴: 𝑋 and 𝑎 𝑏 ⊈ 𝐴: 𝑋 , we have 𝑎 𝑟 ⊈ 𝐴: 𝑋 . Since 𝑎 𝑟 𝐵 ⊆ 𝐴, by lemma (45), we have 𝑟 𝐵 ⊆ 𝑋 𝑅 𝐴 or 𝑎 𝐵 ⊆ 𝑋 𝑅 𝐴 this is a discrepancy. So that 𝑎 𝐵 ⊆ 𝑋 𝑅 𝐴 . (2)→(3) Suppose that ĤṶ𝐵 ⊆ 𝐴 with ĤṶ ⊈ 𝐴: 𝑋 for F. ideals Ĥ, Ṷ of R and F. subm. B of X. Hence 𝑎 Ṷ ⊈ 𝐴: 𝑋 for some F. singleton 𝑎 ⊆ Ĥ . Now, we prove that Ĥ𝐵 ⊆ 𝑋 𝑅 𝐴 or Ṷ𝐵 ⊆ 𝑋 𝑅 𝐴 . Assume that Ĥ𝐵 ⊈ 𝑋 𝑅 𝐴 and Ṷ𝐵 ⊈ 𝑋 𝑅 𝐴 . Since 𝑎 Ṷ𝐵 ⊆ 𝐴, by (2), we have 𝑎 𝐵 ⊆ 𝑋 𝑅 𝐴 , then there exists 𝑦 ⊆ Ĥ such that 𝑦 𝐵 ⊈ 𝑋 𝑅 𝐴 since the assumption Ĥ𝐵 ⊈ 𝑋 𝑅 𝐴 . Since 𝑦 Ṷ𝐵 ⊆ 𝐴, we have 𝑦 Ṷ ⊆ 𝐴: 𝑋 , hence 𝑎 𝑦 Ṷ ⊈ 𝐴: 𝑋 . Since 𝑎 𝑦 Ṷ𝐵 ⊆ 𝐴, we get 𝑎 𝑦 𝐵 ⊆ 𝑋 𝑅 𝐴 and so 𝑦 𝐵 ⊆ 𝑋 𝑅 𝐴 this is a discrepancy. Thus Ĥ𝐵 ⊆ 𝑋 𝑅 𝐴 . (3)→(1) Let 𝑎 𝑏 𝑥 ⊆ 𝐴, for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋. Put Ĥ 𝑎 , Ṷ 𝑏 and 𝐵 𝑥 , then ĤṶ𝐵 ⊆ 𝐴. By (3), we have either ĤṶ ⊆ 𝐴: 𝑋 or Ĥ𝐵 ⊆ 𝑋 𝑅 𝐴 or Ṷ𝐵 ⊆ 𝑋 𝑅 𝐴 ; that is either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝑋   129   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 . Hence either 𝑎 𝑏 ⊆ 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 . Thus A is T-ABSO quasi primary F. subm. of X. Theorem 47. Let X be F. M. of Ḿ, and A be F. subm. of X. Then the following are satisfied: 1- If is a multiplication F. M. and 𝐴: 𝑋 is T-ABSO quasi primary F. ideal of R, then A is T- ABSO quasi primary F. subm. of X. 2- If X is a finitely generated multiplication F. M. and A is T-ABSO quasi primary F. subm. of X, then 𝐴: 𝑋 is T-ABSO quasi primary F. ideal of R. Proof. (1) Assume that X is a multiplication F. M., 𝐴: 𝑋 is T-ABSO quasi primary F. ideal of R and ĤṶB⊆A for F. ideals Ĥ, Ṷ of R and F. subm. B of X. Since X is a multiplication F. M., we have B=KX for some F. ideal K of R. So that ĤṶB=ĤṶKX⊆ A, then ĤṶ𝐾 ⊆ 𝐴: 𝑋 . Since 𝐴: 𝑋 is T-ABSO quasi primary F. ideal of R, so by theorem (13), we have ĤṶ ⊆ 𝐴: 𝑋 or Ĥ𝐾 ⊆ 𝐴: 𝑋 ⊆ 𝑋 𝑅 𝐴 : 𝑋 or Ṷ𝐾 ⊆ 𝐴: 𝑋 ⊆ 𝑋 𝑅 𝐴 : 𝑋 . Hence ĤṶ ⊆ 𝐴: 𝑋 or Ĥ𝐵 ⊆ 𝑋 𝑅 𝐴 or Ṷ𝐵 ⊆ 𝑋 𝑅 𝐴 . Then A is T-ABSO quasi primary F. subm. of X by theorem (46). (2) Assume that A is T-ABSO quasi primary F. subm. of a finitely generated multiplication F. M. X. Let F. singletons 𝑎 , 𝑏 , 𝑟 of R , such that 𝑎 𝑏 𝑟 ⊆ 𝐴: 𝑋 with 𝑎 𝑏 ⊈ 𝐴: 𝑋 . Hence 𝑎 𝑏 𝑟 𝑥 ⊆ 𝐴 for evey F. singleton 𝑥 ⊆ 𝑋. Since A is T-ABSO quasi primary F. subm. of X and 𝑎 𝑏 ⊈ 𝐴: 𝑋. Then we have 𝑎 𝑟 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑟 𝑥 ⊆ 𝑋 𝑅 𝐴 for all 𝑥 ⊆ 𝑋. Hence we have 𝑋 𝑅 𝐴 : 𝑎 𝑟 ∪ 𝑋 𝑅 𝐴 : 𝑏 𝑟 𝑋 , so that 𝑋 𝑅 𝐴 : 𝑎 𝑟 𝑋 or 𝑋 𝑅 𝐴 : 𝑏 𝑟 𝑋 . Then we have 𝑎 𝑟 ⊆ 𝑋 𝑅 𝐴 : 𝑋 𝐴: 𝑋 or 𝑏 𝑟 ⊆ 𝑋 𝑅 𝐴 : 𝑋 𝐴: 𝑋. Thus 𝐴: 𝑋 is T-ABSO quasi primary F. ideal of R. Theorem 48. Let X be a finitely generated multiplication F. M. of Ḿ. For any F. subm. A of X, the following expressions are equivalent: 1- A is T-ABSO quasi primary F. subm. of X; 2- X-R(A) is T-ABSO F. subm. of X. Proof. (1)→(2) Assume that A is T-ABSO quasi primary F. subm. of X. By theorm (47) and proposition (6), then we have 𝐴: 𝑋 Ṷ is a prime F. ideal of R or 𝐴: 𝑋 Ṷ ∩ Ṷ where Ṷ , Ṷ are distinct prime F. ideals minimal over 𝐴: 𝑋 . If 𝐴: 𝑋 Ṷ , hence X- R(A)=ṶX is a prime subm. by lemma (43), so that X-R(A) is T-ABSO F. subm. of X. Now, if 𝐴: 𝑋 Ṷ ∩ Ṷ where Ṷ , Ṷ are distinct prime F. ideals minimal over 𝐴: 𝑋 , then we have X-R(A)= Ṷ ∩ Ṷ 𝑋. Since F- annX= 0 : 𝑋 and 0 : 𝑋 ⊆ 𝐴: 𝑋 and Ṷ , Ṷ are distinct prime F. ideals minimal over 𝐴: 𝑋 . So that 𝐹 𝑎𝑛𝑛𝑋 ⊆ Ṷ , Ṷ . Then X-R(A)= Ṷ 𝐹 𝑎𝑛𝑛𝑋 ∩ Ṷ 𝐹 𝑎𝑛𝑛𝑋 𝑋 Ṷ 𝑋 ∩ Ṷ 𝑋 by lemma (47).   130   Ibn Al-Haitham Jour. for Pure & Appl. Sci. IHJPAS https://doi.org/10.30526/32.1.1930 Vol. 32 (1) 2019 Since Ṷ 𝑋 , Ṷ 𝑋 are two distinct prime F. subms., so that X-R(A) is T-ABSO F. subm. of X by remarks and examples(16)part(1). (2)→(1) Assume that X-R(A) is T-ABSO F. subm. of X. Let 𝑎 𝑏 𝑥 ⊆ 𝐴, for F. singletons 𝑎 , 𝑏 of R and 𝑥 ⊆ 𝑋 . Since A⊆ X-R(A), then 𝑎 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 . But X-R(A) is T-ABSO F. subm. of X, so that 𝑎 𝑏 ⊆ 𝑋 𝑅 𝐴 : 𝑋 𝐴: 𝑋 or 𝑎 𝑥 ⊆ 𝑋 𝑅 𝐴 or 𝑏 𝑥 ⊆ 𝑋 𝑅 𝐴 . Thus A is T-ABSO quasi primary F. subm. of X. By combining theorem (47) and theorem (48), we get the following corollary is beneficial to determine T-ABSO quasi primary F. subm. of a finitely generated multiplication F. M. Corollary 49. For any F. subm. A of a finitely generated multiplication F. M. X of Ḿ. Then the following expressions are equivalent: 1- A is T-ABSO quasi primary F. subm. of X; 2- X-R(A) is T-ABSO F. subm. of X; 3- X-R(A) is T-ABSO primary F. subm. of X; 4- X-R(A) is T-ABSO quasi primary F. subm. of X; 5- 𝐴: 𝑋 is T-ABSO F. ideal of R; 6- 𝐴: 𝑋 is T-ABSO primary F. ideal of R, 7- 𝐴: 𝑋 is T-ABSO quasi primary F. ideal of R; 8- 𝐴: 𝑋 is T-ABSO quasi primary F. ideal of R. 4. Conclusions Through our research we concluded to the concepts (prime and quasi-prime) F. subm. lead to the concept T-ABSO F. subm. we reached the concept T-ABSO F. subm.one of the most important conclusions is the theorem (20), and explan the relationship if A is T-ABSO F. subm. with 𝐴: 𝑋 is T-ABSO F. ideal under the class of a multiplication F. M. in corollary (23). Also we concluded the relationship 𝑋 𝑅 𝐴 with 𝐴: 𝑋 under the class of a multiplication F. M. in lemma (45), and explan the relationships A is T-ABSO quasi primary F. subm.with 𝐴: 𝑋 is T-ABSO quasi primary F. ideal and A is T-ABSO quasi primary F. subm.with 𝑋 𝑅 𝐴 is T-ABSO F. subm. under the class of a multiplication F. M. as in theorem (47), and theorem (48). References 1. Deniz, S.; Gürsel, Y.; Serkan, O.; Bayram, A. E.; Bijan, D. On 2-Absorbing Primary Fuzzy Ideals of Commutative Rings. Mathematical Problems in Engineering. 2017, 2017, 1-7. 2. Lu. C. P. Prime Submodules of Modules. 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