Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 102 This work is licensed under a Creative Commons Attribution 4.0 International License. Fuzzy Soc-Semi-Prime Sub-Modules Abstract In this paper, we study a new concept of fuzzy sub-module, called fuzzy socle semi-prime sub-module that is a generalization the concept of semi-prime fuzzy sub-module and fuzzy of approximately semi-prime sub-module in the ordinary sense. This leads us to introduce level property which studies the relation between the ordinary and fuzzy sense of approximately semi-prime sub-module. Also, some of its characteristics and notions such as the intersection, image and external direct sum of fuzzy socle semi-prime sub-modules are introduced. Furthermore, the relation between the fuzzy socle semi-prime sub-module and other types of fuzzy sub-module presented. Keyword: ℱ-module, ℱ-sub-module, ℱ-prime sub-module, Socle of ℱ-module. 1.Introduction The concept of fuzzy sets was introduced by Zadeh in1965[1]. Many authors indeed presented fuzzy subrings and fuzzy ideals. The concept of fuzzy module was introduced by Negoita and Relescu in 1975 [2]. Since then several authors have studied fuzzy modules. The concept of semi-prime fuzzy sub-module was introduced by Rabi 2004[3]. The concept of approximately semi-prime sub-module was introduced by Ali 2019[4]. The socle of M is a summation of simple sub-modules of an ℛ-module M and denoted by 𝑆𝑜𝑐(𝑀). But, the fuzzy socle of ℱ-module X an ℛ-module M is a summation of simple ℱ-sub-modules of 𝑋 and denoted by 𝐹 − 𝑆𝑜𝑐(𝑋). Ibn Al Haitham Journal for Pure and Applied Science Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Doi: 10.30526/35.1.2804 Article history: Received 1, November, 2021, Accepted,16 , December, 2021, Published in January 2022. Saad S.Merie SaadSaleem@uokirkuk.edu.iq Depatment of Mthmatics, College of Education of Pure Science, Ibn Al- Haitham, University of Baghdad, Baghdad – Iraq. Hatam Yahya Khalf dr.hatamyahya@yahoo.com Depatment of Mthmatics, College of Education of Pure Science, Ibn Al- Haitham, University of Baghdad, Baghdad – Iraq. https://creativecommons.org/licenses/by/4.0/ mailto:SaadSaleem@uokirkuk.edu.iq mailto:dr.hatamyahya@yahoo.com Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 103 Preliminaries " There are various definitions and characteristics in this section of ℱ-sets , ℱ-modules , and prime ℱ-sub-modules. Definition 1.1 [1] Let D be a non- empty set and I is closed interval [0, 1] of real numbers. An ℱ-set B in D (an ℱ-subset of D) is a function from D into I. Definition 1.2 [1] AN ℱ-set B of a set D is said to be ℱ-constant if 𝐵(𝑥) = 𝑡, ∀ 𝑥 ∈ 𝐷 𝑡 ∈ [0, 1] Definition 1.3 [1] Let 𝑥𝑡 : 𝐷 → [0, 1] be an ℱ-set in D, where x ∈D , t ∈ [0, 1] defined by: 𝑥𝑡 (𝑦) = { 𝑡 𝑖𝑓 𝑥 = 𝑦 0 𝑖𝑓 𝑥 ≠ 𝑦 for all 𝑦 ∈ 𝐷. 𝑥𝑡 is said to be an ℱ-singleton or ℱ-point in D. Definition 1.4 [5] Let 𝐵 be an ℱ-set in D, for all t ∈ [0, 1], the set B𝑡 = {𝑥 ∈ 𝐷; B(𝑥) ≥ 𝑡} is said to be a level subset of 𝐵. Remark 1.5 [6] Let Α and Β be two ℱ-sets in S, then: 1- Α = 𝛣 if and only if Α(𝑥) = 𝛣(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ S. 2- Α ⊆ 𝛣 if and only if Α(𝑥) ≤ 𝛣(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ S. 3- Α = 𝛣 if and only if Α𝑡 = 𝛣𝑡 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 ∈ [0,1]. If Α < 𝐵 and there exists x ∈S such that Α(𝑥) < 𝛣(𝑥), then A is a proper ℱ-subset of Β and written as Α < 𝛣. By part (2), we can deduce that 𝑥𝑡 ⊆ Α if and only if Α(𝑥) ≥ 𝑡 . Definition 1.6 [6] If Μ is an ℛ-module. An ℱ-set X of Μ is called ℱ-module of an ℛ-module Μ if : 1- 𝑋(𝑥 − 𝑦) ≥ min{𝑋(𝑥), 𝑋(𝑦)}𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥, 𝑦 ∈ Μ}. 2- 𝑋(𝑟𝑥) ≥ 𝑋(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ Μ 𝑎𝑛𝑑 𝑟 ∈ ℛ . 3- 𝑋(0) = 1. Proposition 1.7 [7] Let 𝐶 be an ℱ-set of an ℛ-module Μ. Then the level subset C𝑡 of Μ , ∀ t ∈ [0, 1] is a sub- module of M if and only if C is an ℱ-sub-module of ℱ-module of an ℛ-module Μ. Definition 1.8 [8] Let X and A be two ℱ-modules of ℛ-module Μ. A is said to be an ℱ-sub-module of X if Α ⊆ 𝑋. Proposition 1.9 [5] Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 104 Let Α be an ℱ-set of an ℛ-module Μ. Then the level subset Α𝑡 , t ∈ [0, 1] is a sub-module of Μ if Α is an ℱ-sub-module of X where X is an ℱ-module of an ℛ-module Μ. Now, we go over various ℱ-sub-module attributes that will be useful in the next section. Lemma 1.10 [6] If 𝑟𝑡 be an ℱ-singleton of ℛ and Α be an ℱ-module of an ℛ-module Μ.Then for any w ∈ Μ (𝑟𝑡 Α)(𝑤) = { sup{inf (𝑡, 𝐴(𝑥))}: 𝑖𝑓 𝑤 = 𝑟𝑥} 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑥 ∈ Μ 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 Where 𝑟𝑡 : ℛ → [0, 1], defined by 𝑟𝑡 (𝑧) = { 𝑡 𝑖𝑓 𝑟 = 𝑧 0 𝑖𝑓 𝑟 ≠ 𝑧 For all 𝑧 ∈ ℛ Definition 1.11 [6] Let Α and 𝛣 be two ℱ-sub-modules of an ℱ-module X of ℛ-module Μ. The residual quotient of Α and 𝛣 denoted by (Α ∶ 𝛣) is the ℱ-subset of ℛ defined by: (Α ∶ 𝛣)(r) = sup {t ∈ [0, 1] ∶ 𝑟𝑡 𝛣 ⊆ Α}, for all 𝑟 ∈ ℛ. That is (Α ∶ 𝛣) = {𝑟𝑡 ∶ 𝑟𝑡 B ⊆ Α; 𝑟𝑡 is an ℱ − singleton of ℛ}. If 𝛣 = 〈𝑥𝑘 〉, then (Α ∶ 〈𝑥𝑘 〉) = {𝑟𝑡 ∶ 𝑟𝑡 𝑥𝑘 ⊆ Α; 𝑟𝑡 is an ℱ − singleton of ℛ }. Lemma 1.12 [9] Let Α be an ℱ-sub-module of ℱ-module X, (Α 𝑡 : 𝑋𝑡 ) ≥ (Α: 𝑋)𝑡 ,For all t ∈ [0, 1]. Also , we can prove that by Lemma 2.3.3.[6]. It follows that if , 𝑋 = Α ⊕ 𝛣,where 𝐴, 𝛣 ≤ 𝑋 then 𝑋𝑡 = (Α ⊕ 𝛣)𝑡 = Α𝑡 ⊕ 𝛣𝑡 . Definition 1.13 [10] Let f be a mapping from a set Μ into a set Ν and let Α be ℱ-set in Μ. The image of Α is denoted by f (Α), where f (Α) is defined by: 𝑓 (Α) (𝑦) = { sup{Α(𝑧): 𝑧 ∈ 𝑓 −1(𝑦) ≠ ∅} 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑦 ∈ Ν 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 Note that, if 𝑓 is a bijective mapping, then 𝑓 (Α)(𝑦) = Α(𝑓 −1(𝑦)) Proposition 1.14 [11] Let f be a mapping from a set Μ into a set Ν. Assume that X and Y are ℱ-modules of M and N respectively, let Α be an ℱ-sub-module of X, then f (Α) is an ℱ-sub-module of Y. Definition 1.15 [12] An ℱ-subset K of a ring ℛ is called ℱ-ideal of ℛ, if ∀ 𝑥, 𝑦 ∈ ℛ : 1- 𝐾(𝑥 − 𝑦) ≥ min {𝐾(𝑥), 𝐾(𝑦)} . 2- 𝐾(𝑥𝑦) ≥ max {𝐾(𝑥), 𝐾(𝑦)} . Definition 1.16 [13] Let X be an ℱ-module of an ℛ-module Μ, let A be an ℱ-sub-module of X and K be an ℱ - ideal of ℛ, the product KA of K and Α is defined by: KΑ(𝑥) = { sup {𝑖𝑛𝑓{𝐾(𝑟1 ), … . , 𝐾(𝑟𝑛 ), Α(𝑥1 ), … , Α(𝑥𝑛 )}} 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑟𝑖 ∈ ℛ, 𝑥𝑖 ∈ Μ, 𝑛 ∈ Ν 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 105 Note that K Α is an ℱ-sub-module of X, and (KΑ)𝑡 = K𝑡 Α𝑡 ,∀ t ∈ [0, 1]. Definition 1.17 [9] Let X be an ℱ-module of an ℛ-module Μ, An ℱ-sub-module U of X is called completely prime if whenever 𝑟𝑏 𝑚𝑡 ⊆ 𝑈,with 𝑟𝑏 ≠ 01 is an ℱ-singleton of ℛ and 𝑚𝑡 is an ℱ- singleton of Ximplies that 𝑚𝑡 ⊆ 𝑈 for each t, b ∈ [0,1]. Definition 1.18 [6] Let Α and Β be two ℱ-sub-modules of an R-module Μ. The addition A + Β is defined by: (Α + Β)(x) = sup{𝑚𝑖𝑛{Α(𝑦), Β(𝑧)} 𝑤𝑖𝑡ℎ 𝑥 = 𝑦 + 𝑧, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥, 𝑦, 𝑧 ∈ Μ }. Furthermore, Α + Β is an ℱ-sub-module of an ℛ-module Μ. Corollary 1.19 [8] If X is an ℱ-module of an ℛ-module Μ and 𝑥𝑡 ⊆ X, then for all ℱ-singleton 𝑟𝑘 of ℛ, 𝑟𝑘 𝑥𝑡 = (𝑟𝑥)𝜆, where 𝜆 = min {𝑡, 𝑘}. Proposition 1.20 [6] Let Α and Β be two ℱ-sub-modules of an ℱ-module X of an ℛ-module Μ. Then the residual quotient of Α and 𝛣 (Α ∶ Β) is an ℱ-ideal of ℛ. Proposition 1.21 [14] Let 𝑓: 𝑀 ⟶ Ν be an ℛ-homomorphisim, then 𝑓(𝑆𝑜𝑐(𝑀)) ⊆ 𝑆𝑜𝑐(Ν). Definition 1.22 [15] Let X be an ℱ-module of an R-module Μ, X is called ℱ-simple if and only if X has no proper ℱ-sub-modules (in fact X is ℱ-simple if and only if X has only itself and 01 ). Definition 1.23 [16] 𝐴 ℱ-module 𝑋 is called semi-simple if 𝑋 is a summation of simple ℱ-sub-modules of 𝑋 . Moreover, 𝑋 is called semi-simple if 𝑋 = 𝐹 − 𝑆𝑜𝑐(𝑋). Definition 1.24 [9] Let X be an ℱ-module of an ℛ-module Μ, X is said to be faithful if 𝐹 − 𝑎𝑛𝑛𝑋 = 01 . Where 𝐹 − 𝑎𝑛𝑛𝑋 = {𝑟𝑡 ∶ 𝑟𝑡 𝑥𝑙 = 01 ; for all 𝑥𝑙 ⊆ X and 𝑟𝑡 be an ℱ − singleton of ℛ}. Definition 1.25 [17] Let X be an ℱ-module of an ℛ-module Μ, X is said to be cancellative if whenever 𝑟𝑡 𝑥𝑙 = 𝑟𝑡 𝑦𝑑 for all 𝑥𝑙 , 𝑦𝑑 ⊆ X and 𝑟𝑡 be an ℱ − singleton of ℛ then 𝑥𝑙 = 𝑦𝑑 . Definition 1.26 [3] A proper ℱ-sub-module U of an ℱ-module X of an ℛ-module M is called semi-prime ℱ- sub-module of X if whenever 𝑟𝑏 𝑛𝑚𝑡 ⊆ 𝑈,where 𝑟𝑏 is an ℱ-singleton of ℛ , 𝑚𝑡 is an ℱ- singleton of X and n ∈ 𝑍+implies that 𝑟𝑏 𝑚𝑡 ⊆ 𝑈 for each t, b ∈ [0,1]. Definition 1.27 [4] A proper sub-module E of an ℛ-module Μ is called pproximately semi prime (for a short app-semi-prime) sub-module of Μ if whenever 𝑎𝑚 ∈ 𝐸, for 𝑎 ∈ ℛ, 𝑚 ∈ Μ𝑖𝑚𝑝𝑙𝑖𝑒𝑠 that 𝑎𝑚 ∈ 𝐸 + 𝑆𝑜𝑐(Μ) . Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 106 Definition 1.28 [9] An ℱ-sub-module N of an ℱ-module X of an ℛ-module M is called weakly pure ℱ-sub- module of X if for any ℱ-singleton 𝑟𝑏 of ℛ implies that 𝑟𝑏 𝑁 = 𝑟𝑏 𝑋 ∩ 𝑁 with b ∈ [0,1]. Lemma 1.29 [18] Let X be an ℱ-module of an ℛ-module M and let Α , Β and C are ℱ-sub-modules of X such that C ⊆ Β. Then 𝐶 + (Β⋂Α) = (𝐶 + Α)⋂Β. Proposition 1.30 [14] If Μ be a faithful multiplication ℛ-module, then 𝑆𝑜𝑐(ℛ)Μ = 𝑆𝑜𝑐(Μ) Definition 1.31 [15] Let X be an ℱ-module of an ℛ- module Μ. X is called multiplication ℱ-module if and only if for each ℱsub-module Α of X ,there exists an ℱ-ideal K of ℛ such that Α = KX. Proposition 1.32 [15] AN ℱ-module X of an ℛ-module Μ is a multiplication if and only if every non-empty ℱ - sub-module A of X such that Α = (Α:𝑅 𝑋)𝑋 . Definition 1.33 [19] A sub-module 𝑉 of ℛ-module Μ is called essential if 𝐻 ∩ V ≠ 0. For non-trivial sub-module H of Μ . Definition 1.34 [9] Let X be an ℱ-module of an ℛ-module Μ. An ℱ-sub-module A of X is called essential if 𝐴 ∩ 𝐵 ≠ 01 , for nontrivial ℱ-sub-module B of X. Finally, (shortly fuzzy set, fuzzy sub-module, fuzzy ideal, fuzzy module and fuzzy singleton are ℱ-set, ℱ-sub-module, ℱ-ideal , ℱ-module and ℱ-singleton)." 𝓕-Soc-semi-prime sub-modules In this section, we offer the concept of an ℱ-Soc-semi-prime sub-module as a generalization of ordinary concept(approximately semi-prime sub-module). Some characterizations of ℱ- Soc-prime sub-module are introduced. Definition 2.1 Let 𝑟𝑏 be an ℱ-singleton of ℛ and 𝑚𝑡 is an ℱ-singleton of X , then a proper ℱ-sub-module U of an ℱ-module X of an ℛ-module M is called an ℱ-Socle semi-prime ( for short ℱ-Soc- semi-prime) sub-module(ideal) of X if whenever 𝑟𝑏 𝑛𝑚𝑡 ⊆ 𝑈 with n ∈ 𝑍 + implies that 𝑟𝑏 𝑚𝑡 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) for each t, b ∈ [0,1]. Furthermore, if 𝑟𝑏and 𝑠ℎ are ℱ-singletons of ℛ, then a proper ℱ-ideal L of ℛ is called an ℱ-Socle semi-prime ( for short ℱ-Soc-semi-prime) ideal of ℛ if whenever 𝑟𝑏 𝑛𝑠ℎ ⊆ 𝐿 with n ∈ 𝑍+implies that 𝑟𝑏 𝑠ℎ ⊆ 𝐿 + ℱ − 𝑆𝑜𝑐(ℛ) for each h, b ∈ [0,1]. We will adopt the definition of an ℱ-socle of X in this research as follows: Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 107 ℱ − 𝑆𝑜𝑐(𝑋): 𝑀 → [0,1] such that: ℱ − 𝑆𝑜𝑐(𝑋)(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 𝑆𝑜𝑐(𝑀) ℎ 𝑖𝑓 𝑚 ∉ 𝑆𝑜𝑐(𝑀) 𝑤𝑖𝑡ℎ 0 < ℎ < 1 Lemma 2.2 (ℱ − 𝑆𝑜𝑐(𝑋))𝑡 = 𝑆𝑜𝑐(𝑋𝑡 ) for any ℱ-module X for each t ∈ (0,1] with (ℱ − 𝑆𝑜𝑐(𝑋))𝑡 ≠ 𝑋𝑡 Proof: ℱ − 𝑆𝑜𝑐(𝑋): 𝑀 → [0,1] such that: ℱ − 𝑆𝑜𝑐(𝑋)(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 𝑆𝑜𝑐(𝑀) ℎ 𝑖𝑓 𝑚 ∉ 𝑆𝑜𝑐(𝑀) 𝑤𝑖𝑡ℎ 0 < ℎ < 1 Now, (ℱ − 𝑆𝑜𝑐(𝑋))𝑡 = {𝑚 ∈ 𝑀 ∶ (ℱ − 𝑆𝑜𝑐(𝑋))(𝑚) ≥ 𝑡} So, if 𝑡 = 1 then (ℱ − 𝑆𝑜𝑐(𝑋))𝑡 = 𝑆𝑜𝑐(𝑀) = 𝑆𝑜𝑐(𝑋𝑡 ) If 0 < 𝑡 ≤ ℎ then (ℱ − 𝑆𝑜𝑐(𝑋))𝑡 = 𝑀 = 𝑋𝑡 that is a contradiction If ℎ < 𝑡 < 1 then (ℱ − 𝑆𝑜𝑐(𝑋))𝑡 = 𝑆𝑜𝑐(𝑀) = 𝑆𝑜𝑐(𝑋𝑡 ) Lemma 2.3 Let X be an ℱ-module of an ℛ-module M with X(m)=1 for each 𝑚 ∈ 𝑀, if U is an ℱ-sub- module of X is defined by 𝑈: 𝑀 → [0,1] such that: 𝑈(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 𝐸 𝑘 𝑖𝑓 𝑚 ∉ 𝐸 𝑤𝑖𝑡ℎ 0 < 𝑘 < 1 Where E is a sub-module of M. Then U is an ℱ-Soc-semi-prime sub-module of X if and only if E is an app-semi-prime sub-module of M. Proof: First of all, we must define 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋). (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋))(𝑚) = sup {min(𝑈(𝑦) , ℱ − 𝑆𝑜𝑐(𝑋)(𝑧)) , 𝑦 + 𝑧 = 𝑚} So, we have (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋))(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 𝐸 + 𝑆𝑜𝑐(𝑀) 𝑠 𝑖𝑓 𝑚 ∉ 𝐸 + 𝑆𝑜𝑐(𝑀) 𝑤𝑖𝑡ℎ 𝑠 = max {𝑘, ℎ} Where ℱ − 𝑆𝑜𝑐(𝑋): 𝑀 → [0,1] such that: ℱ − 𝑆𝑜𝑐(𝑋)(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 𝑆𝑜𝑐(𝑀) ℎ 𝑖𝑓 𝑚 ∉ 𝑆𝑜𝑐(𝑀) 𝑤𝑖𝑡ℎ 0 < ℎ < 1 Now, Suppose E is an app-semi-prime sub-module of M, to prove that U is an ℱ-Soc-semi-prime sub-module of X. Let 𝑟𝑏 ⊆ ℛ and 𝑚𝑡 ⊆ 𝑋 for each t, b ∈ [0,1] such that (𝑟𝑏) 𝑛𝑚𝑡 ⊆ 𝑈,thus (𝑟𝑛)𝑏 𝑚𝑡 ⊆ 𝑈 that is either 𝑟 𝑛𝑚 ∈ 𝐸 or 𝑟𝑛𝑚 ∉ 𝐸. Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 108 1) If 𝑟𝑛𝑚 ∈ 𝐸, then 𝑟𝑚 ∈ 𝐸 + 𝑆𝑜𝑐(𝑀). Hence (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋))(𝑟𝑚) = 1 this implies 𝑟𝑏 𝑚𝑡 = (𝑟𝑚)𝑡 ⊆ (𝑟𝑚)1 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋). 2) If 𝑟𝑛𝑚 ∉ 𝐸 then 𝑈(𝑟𝑛𝑚) = 𝑘 with 𝑚 ∉ 𝐸 thus 𝑈(𝑚) = 𝑘. Since (𝑟𝑏) 𝑛𝑚𝑡 ⊆ 𝑈 then (𝑟𝑛𝑚)ℷ ⊆ 𝑈 where ℷ = min {𝑏, 𝑡}, that is 𝑈(𝑟 𝑛𝑚) ≥ ℷ thus 𝑘 ≥ ℷ . Now, if ℷ = t this implies 𝑚𝑡 ⊆ 𝑚𝑘 ⊆ 𝑈 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋). That is mean 𝑟𝑏 𝑚𝑡 ⊆ 𝑟𝑏 𝑚𝑘 ⊆ 𝑈 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) If ℷ = b, 𝑈(ℎ) ≥ 𝑘 for any ℎ ∈ 𝑀, and: (𝑟𝑛 𝑏 𝑋𝑀)(ℎ) = { 𝑏 𝑖𝑓 ℎ = 𝑟𝑛𝑎 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑎 ∈ 𝑀 Then we get (𝑟𝑛 𝑏 𝑋𝑀 )(ℎ) ≤ 𝑈(ℎ), hence 𝑟 𝑛 𝑏 𝑋𝑀 ⊆ 𝑈 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) So, each case implies that 𝑟𝑏 𝑚𝑡 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) Therefore U is an ℱ-Soc-semi-prime sub-module of X. Conversely Suppose U is an ℱ-Soc-semi-prime of X. Let 𝑎𝑛𝑥 ∈ 𝑈𝑡 , with 𝑎 ∈ ℛ, n ∈ 𝑍 + and 𝑥 ∈ 𝑋𝑡 it follows that (𝑎𝑛𝑥)𝑡 ⊆ 𝑈, that is (𝑎 𝑛)𝑡 𝑥𝑡 = (𝑎𝑡 ) 𝑛𝑥𝑡 ⊆ 𝑈. But U is an ℱ-Soc-semi-prime of X, then we get 𝑎𝑡 𝑥𝑡 = (𝑎𝑥)𝑡 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) . Thus we get (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋))(𝑎𝑥) ≥ 𝑡 , hence, by (Lemma 1.12) and (Lemma 2.2), we have 𝑎𝑥 ∈ (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋)) 𝑡 = 𝑈𝑡 + (ℱ − 𝑆𝑜𝑐(𝑋)) 𝑡 = 𝑈𝑡 + 𝑆𝑜𝑐(𝑋𝑡 ). That is mean 𝑈𝑡 is an app-semi-prime sub-module of 𝑋𝑡 . Hence 𝑈1 = 𝐸 is an app-semi-prime sub-module of M. The following example shows that the definition of an ℱ-socle of X that we adopt in this research is necessary to prove one side of above lemma. Example 2.4 Let 𝑀 = 𝑍12 as a Z-module and 𝑋: 𝑀 → [0,1] , 𝑈: 𝑀 → [0,1] defined by: 𝑋(𝑚) = 1 𝑖𝑓 𝑚 ∈ 𝑍12 𝑈(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 〈0̅〉 1/4 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 And an ℱ-socle of X is defined by ℱ − 𝑆𝑜𝑐(𝑋): 𝑀 → [0,1] such that: ℱ − 𝑆𝑜𝑐(𝑋)(𝑚) = { 1 𝑖𝑓 𝑥 = 0̅ 2/3 𝑖𝑓 𝑚 ∈ 〈2̅〉 − {0̅} 1/3 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 Where 𝑆𝑜𝑐(𝑀) = 〈2̅〉. That’s clear X is an ℱ-module and U be an ℱ-sub-module of X. We have 𝑈𝑡 is an app-semi-prime sub-module of M for every 𝑡 > 0 . Now, Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 109 (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋))(𝑚) = { 1 𝑖𝑓 𝑥 = 0̅ 2/3 𝑖𝑓 𝑚 ∈ 〈2̅〉 − {0̅} 1/3 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 But, U is not an ℱ-Soc-semi-prime sub-module of X, since for an ℱ-singleton 3̅3 4 ⊆ 𝑋 and an ℱ-singleton 23 4 of ℛ such that (22)3 4 3̅3 4 = 0̅3 4 , where 0̅3 4 ⊆ 𝑈 since 𝑈(0̅) = 1 > 3 4 . but 23 4 3̅3 4 = 6̅3 4 ⊈ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) since (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋))(6̅) = 2 3 ≱ 3 4 . Hence, U is not an ℱ-Soc-semi-prime of sub-module of X. Proposition 2.5 Let U and V are ℱ-sub-modules of an ℱ-module X of an ℛ-module M with V is an ℱ- semi- prime sub-module of X. Then [𝑈:ℛ 𝑉] is an ℱ-Soc-semi-prime ideal of ℛ. Proof : Suppose that 𝑟𝑏 𝑛𝑚𝑡 ⊆ [𝑈:ℛ 𝑉] ,for 𝑟𝑏 ⊆ ℛ, 𝑚𝑡 ⊆ 𝑋,thus 𝑟𝑏 𝑛𝑚𝑡 𝑉 ⊆ 𝑈.So we have 𝑟𝑏 𝑛(𝑚𝑡 𝑉) ⊆ 𝑈, but V is an ℱ-semi-prime sub-module of X. that is 𝑟𝑏 (𝑚𝑡 𝑉) ⊆ 𝑈, hence 𝑟𝑏 𝑚𝑡 𝑉 ⊆ 𝑈 that is mean 𝑟𝑏 𝑚𝑡 ⊆ [𝑈:ℛ 𝑉] ⊆ [𝑈:ℛ 𝑉] + ℱ − 𝑆𝑜𝑐(ℛ) . Proposition 2.6 Let U and V are ℱ-Soc-semi-prime sub-modules of an ℱ-module X of an ℛ-module M with ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑈 , Then U ∩V is an ℱ-Soc-semi-prime sub-module of X. Proof : Let 𝑟𝑏 𝑛𝑚𝑡 ⊆ U ∩V ,for 𝑟𝑏 ⊆ ℛ, 𝑚𝑡 ⊆ 𝑋, that is 𝑟𝑏 𝑛𝑚𝑡 ⊆ U and 𝑟𝑏 𝑛𝑚𝑡 ⊆ V. But U and V are ℱ-Soc-semi-prime sub-modules of X, this implies 𝑟𝑏 𝑚𝑡 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) and 𝑟𝑏 𝑚𝑡 ⊆ 𝑉 + ℱ − 𝑆𝑜𝑐(𝑋). That is mean 𝑟𝑏 𝑚𝑡 ⊆ (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋)) ∩ (𝑉 + ℱ − 𝑆𝑜𝑐(𝑋)), by using modular law we get 𝑟𝑏 𝑚𝑡 ⊆ (U ∩ V) + ℱ − 𝑆𝑜𝑐(𝑋). Hence U ∩V is an ℱ-Soc- semi-prime sub-module of X. Remark 2.7 Every ℱ-semi-prime sub-module is an ℱ-Soc-semi-prime sub-module , but the converse is not true . Proof : Suppose U be an ℱ-semi-prime sub-module of an ℱ-module X of an ℛ-module M and 𝑟𝑏 𝑛𝑚𝑡 ⊆ 𝑈,for 𝑟𝑏 ⊆ 𝑅, 𝑚𝑡 ⊆ 𝑋. Since U is an ℱ-semi-prime sub-module, then we get 𝑟𝑏 𝑚𝑡 ⊆ 𝑈 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) , thus 𝑟𝑏 𝑚𝑡 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋). Therefore U is an ℱ-Soc- prime sub-module. The following example show that the converse is not true Example 2.8 Consider 𝑀 = 𝑍12 as a Z-module and 𝑋: 𝑀 → [0,1] , 𝑈: 𝑀 → [0,1] defined by: 𝑋(𝑚) = 1 𝑖𝑓 𝑚 ∈ 𝑍12 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 110 𝑈(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 〈0̅〉 1/5 𝑖𝑓 𝑚 ∉ 〈0̅〉 And an ℱ-socle of X is defined by ℱ − 𝑆𝑜𝑐(𝑋): 𝑀 → [0,1] such that: ℱ − 𝑆𝑜𝑐(𝑋)(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 〈2̅〉 1/3 𝑖𝑓 𝑚 ∉ 〈2̅〉 Where 𝑆𝑜𝑐(𝑀) = 〈2̅〉. That’s clear X is an ℱ-module and U be an ℱ-sub-module of X. From ( [4] Remark 2.3.2 ) 〈0̅〉 is an app-semi-prime sub-module of M, so by (Lemma 2.3) we get U is an ℱ-Soc-semi-prime sub-module of X. But, U is not an ℱ-semi-prime sub-module of X, since for an ℱ-singleton 31 3 ⊆ 𝑋 and an ℱ- singleton 21 3 of ℛ such that (21 3 )2 31 3 = 01 3 where 01 3 ⊆ 𝑈 since 𝑈(0) = 1 > 1 3 . but 21 3 31 3 = 61 3 ⊈ 𝑈 since 𝑈(6) = 1 5 ≯ 1 3 . Hence, U is not an ℱ-semi-prime sub-module of X. Remark 2.9 Every completely ℱ-sub-module of an ℱ-module X of an ℛ-module M is an ℱ-Soc-semi- prime sub-module of X, but the converse is not true . Proof : We take U as a completely ℱ-sub-module of X with 𝑟𝑏 𝑛𝑚𝑡 ⊆ 𝑈,for 𝑟𝑏 ⊆ 𝑅, 𝑚𝑡 ⊆ 𝑋.Now, if 𝑟𝑏 = 01 then 𝑟𝑏 𝑚𝑡 = 0𝑡 ⊆ 01 ⊆ 𝑈.we get U is an ℱ-Soc- semi-prime sub-module of X. If 𝑟𝑏 ≠ 01 ,thus (𝑟𝑏) 𝑛−1 (𝑟𝑏 𝑚𝑡 ) ⊆ 𝑈, we get (𝑟 𝑛−1)𝑏 (𝑟𝑚)𝑑 ⊆ 𝑈 where 𝑑 = min {𝑏, 𝑡}.Now, since U is a completely ℱ-sub-module of X, then we have (𝑟𝑚)𝑑 ⊆ 𝑈 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋), thus 𝑟𝑏 𝑚𝑡 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋). Therefore U is an ℱ-Soc-semi-prime sub- module. The following example show that the converse is not true Example 2.10 Consider 𝑀 = 𝑍 as a Z-module and 𝑋: 𝑀 → [0,1] , 𝑈: 𝑀 → [0,1] defined by: 𝑋(𝑚) = 1 𝑖𝑓 𝑚 ∈ 𝑍 𝑈(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 2𝑍 1/4 𝑖𝑓 𝑚 ∉ 2𝑍 And an ℱ-socle of X is defined by ℱ − 𝑆𝑜𝑐(𝑋): 𝑀 → [0,1] such that: ℱ − 𝑆𝑜𝑐(𝑋)(𝑚) = { 1 𝑖𝑓 𝑚 ∈ {0} 1/3 𝑖𝑓 𝑚 ∉ {0} (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋))(𝑚) = { 1 𝑖𝑓 𝑚 ∈ 2𝑍 1/3 𝑖𝑓 𝑚 ∉ 2𝑍 Where 𝑆𝑜𝑐(𝑀) = {0}. That’s clear X is an ℱ-module and U be an ℱ-sub-module of X. Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 111 2𝑍 is an app-semi-prime sub-module of M, so by (Lemma 2.3) we get U is an ℱ-Soc-semi- prime sub-module of X. But U is not completely ℱ-sub-module of X, since for an ℱ-singleton 51 3 ⊆ 𝑋 and an ℱ- singleton 21 2 of ℛ such that 21 2 51 3 = 101 3 where 101 3 ⊆ 𝑈 since 𝑈(10) = 1 > 1 3 . but 51 3 ⊈ 𝑈 since 𝑈(5) = 1 4 ≯ 1 3 . Hence, U is not completely ℱ-sub-module of X. Proposition 2.11 Let U be an ℱ-Soc-semi-prime sub-module of an ℱ-module X of an ℛ-module M, Then U is an ℱ-Soc-semi-prime sub-module of X if and only if ∀ ℱ-sub-module S of X and an ℱ-ideal J of ℛ with (𝐽)nS ⊆U for 𝑛 ∈ 𝑍+implies that 𝐽S ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) Proof: ( ) Assume that (𝐽)𝑛 S ⊆U , for S is an ℱ-sub-module of X and 𝐽 is an ℱ-ideal of ℛ, let 𝑥𝑡 ⊆ 𝐽𝑆 with 𝑡 ∈ [0,1] then 𝑥𝑡 = (𝑐1)ℎ1(𝑦1)𝑡1 + (𝑐2)ℎ2(𝑦2)𝑡2 + ⋯ + (𝑐𝑛)ℎ𝑛(𝑦𝑛)𝑡𝑛 , for every (𝑐𝑖 )ℎ𝑖 ⊆ 𝐽 and (𝑦𝑖 )𝑡𝑖 ⊆ 𝑈 where ℎ𝑖 , 𝑡𝑖 ∈ [0,1] for every i=1,2,…..,n. Now, we get ((𝑐𝑖)ℎ𝑖 ) 𝑛(𝑦𝑖 )𝑡𝑖 ⊆ (𝐽) 𝑛S ⊆U hence ((𝑐𝑖 )ℎ𝑖 ) 𝑛(𝑦𝑖 )𝑡𝑖 ⊆ U. But U is an ℱ-Soc-semi-prime sub- module of X implies that (𝑐𝑖 )ℎ𝑖 (𝑦𝑖 )𝑡𝑖 ⊆ U+ℱ − 𝑆𝑜𝑐(𝑋) for each i=1,2,…..,n. So we have 𝑥𝑡 ⊆ U+ℱ − 𝑆𝑜𝑐(𝑋). it follows that 𝐽S ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋). ( ) Let (𝑟𝑏) 𝑛𝑥𝑡 ⊆ 𝑈 for 𝑟𝑏 ⊆ ℛ and 𝑛 ∈ 𝑍 + then 〈𝑟𝑏 𝑛〉〈𝑥𝑡 〉 ⊆ 𝑈, that is 〈𝑟𝑏 〉𝑛〈𝑥𝑡 〉 ⊆ 𝑈 then by hypothesis we get 〈𝑟𝑏 〉〈𝑥𝑡 〉 ⊆ 𝑈, hence 𝑟𝑏 𝑥𝑡 ⊆ 𝑈. That is mean U is an ℱ-Soc- semi-prime sub-module of X. Corollary 2.12 Let U be an ℱ-sub-module of an ℱ-module X of an ℛ-module M, Then U is an ℱ-Soc-semi- prime sub-module of X if and only if ∀ ℱ-sub-module S of X and every ℱ-singleton 𝑟𝑏 of ℛ with (𝑟𝑏) 𝑛 S ⊆U implies that 𝑟𝑏S ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) . Proof: It is clear from (proposition 2.11). Corollary 2.13 Let L be an ℱ- ideal of ℛ, Then L is an ℱ-Soc-semi-prime ideal of ℛ if and only if ∀ ℱ-sub- ideal J of ℛ and every ℱ-singleton 𝑟𝑏 of ℛ with (𝑟𝑏) 𝑛 J ⊆L implies that 𝑟𝑏J ⊆ 𝐿 + ℱ − 𝑆𝑜𝑐(ℛ) . Proof : Clearly from (proposition 2.11). Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 112 Proposition 2.14 : If 𝑟𝑏 𝑛 𝑈 ℱ-Soc-semi-prime sub-module of cancellative ℱ-module X. Where U is an ℱ-sub- modules of X and 𝑟𝑏 is an idempotent ℱ-singleton of R. Then 𝑈 ⊆ 𝑟𝑏 𝑛−1 𝑈 + 𝐹 − 𝑆𝑜𝑐(𝑋) Proof : Let 𝑎𝑡 ⊆ 𝑈 this implies 𝑟𝑏 𝑛𝑎𝑡 ⊆ 𝑟𝑏 𝑛𝑈, for 𝑟𝑏 is an ℱ-singleton of R. But, 𝑟𝑏 𝑛𝑈 is an ℱ- Soc-semi-prime sub-module of X with 𝑎𝑡 ⊆ 𝑋,where 𝑡, 𝑏 ∈ [0,1]. Therefore 𝑟𝑏 𝑎𝑡 ⊆ 𝑟𝑏 𝑛𝑈 + 𝐹 − 𝑆𝑜𝑐(𝑋), that is 𝑟𝑏 2 𝑎𝑡 ⊆ 𝑟𝑏 𝑛+1 𝑈 + 𝑟𝑏 𝐹 − 𝑆𝑜𝑐(𝑋), thus 𝑟𝑏 2𝑎𝑡 ⊆ 𝑟𝑏 2𝑟𝑏 𝑛−1𝑈 + 𝑟𝑏 𝐹 − 𝑆𝑜𝑐(𝑋) ) , but 𝑟𝑏 is an idempotent ℱ-singleton of R. So we get 𝑟𝑏 𝑎𝑡 ⊆ 𝑟𝑏 𝑛𝑈 + 𝑟𝑏 𝐹 − 𝑆𝑜𝑐(𝑋). But, X is a cancellative ℱ-module, we have 𝑎𝑡 ⊆ 𝑟𝑏 𝑛−1 𝑈 + 𝐹 − 𝑆𝑜𝑐(𝑋) ,that is mean 𝑈 ⊆ 𝑟𝑏 𝑛−1 𝑈 + 𝐹 − 𝑆𝑜𝑐(𝑋). Remark 2.15 Every ℱ-semi-prime sub-module is an ℱ-Soc-semi-prime sub-module. Proof: It is Clear by definition of ℱ-semi-prime sub-module. Remark 2.16 If U is an ℱ-Soc-semi-prime sub-module of ℱ-module X, with ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑈. Then U is an ℱ-semi-prime sub-module. Proof: Assume that U is an ℱ-Soc-semi-prime sub-module of an ℱ-module X of an ℛ-module M. Let (𝑟𝑛)𝑏𝑚𝑡 = (𝑟𝑏) 𝑛𝑚𝑡 ⊆ 𝑈, for 𝑟𝑏 ⊆ ℛ, 𝑚𝑡 ⊆ 𝑋,where 𝑡, 𝑏 ∈ [0,1]. Since U is an ℱ- Soc-semi-prime sub-module, then 𝑟𝑏 𝑚𝑡 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑈 but ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑈. Hence U is an ℱ-semi-prime sub-module. Corollary 2.17 If U is an ℱ-Soc-semi-prime sub-module of ℱ-module X, with U be an ℱ-essential sub- module of X. Then U is an ℱ-semi-prime sub-module. Proof: Since U be an ℱ-essential sub-module of X, then by definition of ℱ-socle we have ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑈 and by (Remark 2.16) that is complete the proof. Corollary 2.18 If U is an ℱ-sub-module of ℱ-module X, with ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑈. Then U is an ℱ-semi- prime sub-module of X if and only if U is an ℱ-Soc-prime sub-module of X. Proof: Consequently from (Remark 2.7) and (Remark 2.16). Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 113 Remark 2.19 Let U and V are ℱ-sub-modules of ℱ-module X. If U+V is an ℱ-semi-prime sub-module of X with V ⊆ ℱ − 𝑆𝑜𝑐(𝑋), then U is an ℱ-Soc-semi-prime sub-module of X. Proof: Let (𝑟𝑛)𝑏𝑥𝑘 = (𝑟𝑏) 𝑛𝑥𝑘 ⊆ 𝑈, for 𝑟𝑏 ⊆ ℛ, 𝑥𝑘 ⊆ 𝑋,where 𝑘, 𝑏 ∈ [0,1]. this implies (𝑟𝑛)𝑏𝑥𝑘 ⊆ U + V. But U+V is an ℱ-semi-prime sub-module of X, hence 𝑟𝑏 𝑥𝑘 ⊆ 𝑈 + 𝑉 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) since V ⊆ ℱ − 𝑆𝑜𝑐(𝑋). That is U is an ℱ-Soc-semi-prime sub-module of X. Theorem 2.20 Any ℱ-sub-module of semi-simple ℱ-module X is an ℱ-Soc-semi-prime sub-module of X. Proof: If U is an ℱ-sub-module of an ℱ-module X of an ℛ-module M. Let (𝑟𝑛)𝑏𝑥𝑘 = (𝑟𝑏) 𝑛𝑥𝑘 ⊆ 𝑈, for 𝑟𝑏 ⊆ ℛ, 𝑥𝑘 ⊆ 𝑋,where 𝑘, 𝑏 ∈ [0,1]. But, X is a semi-simple ℱ-module, thus 𝑋 = ℱ − 𝑆𝑜𝑐(𝑋).We have 𝑥𝑘 ⊆ 𝑋 = ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋), this implies 𝑟𝑏 𝑥𝑘 ⊆ 𝑟𝑏 𝑋 = 𝑟𝑏 ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑟𝑏 (𝑈 + ℱ − 𝑆𝑜𝑐(𝑋)) ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) that is mean U is an ℱ-Soc- semi-prime sub-module of X. Proposition 2.21 : If U is a weakly pure ℱ-sub-module of ℱ-module X with (𝑟𝑛)𝑏U is an ℱ-Soc-semi-prime sub-module of X for every non-empty ℱ-singleton 𝑟𝑏 of R, then U is an ℱ-Soc-semi-prime sub-module of X. Proof: Suppose that (𝑟𝑛)𝑏𝑥𝑡 ⊆ 𝑈,with 𝑟𝑏 is an ℱ-singleton of R and 𝑥𝑡 ⊆ 𝑋,where 𝑡, 𝑏 ∈ [0,1].Also (𝑟𝑛)𝑏 𝑥𝑡 ⊆ (𝑟 𝑛)𝑏𝑋 this implies (𝑟 𝑛)𝑏𝑥𝑡 ⊆ 𝑈 ∩ (𝑟 𝑛)𝑏𝑋 = (𝑟 𝑛)𝑏𝑈 since U is a weakly pure ℱ-sub-module of X, but (𝑟𝑛)𝑏𝑈 is an ℱ-Soc-semi-prime sub-module of X, hence 𝑟𝑏 𝑥𝑡 ⊆ (𝑟 𝑛)𝑏𝑈 + 𝐹 − 𝑆𝑜𝑐(𝑋) ⊆ 𝑈 + 𝐹 − 𝑆𝑜𝑐(𝑋). Thus U is an ℱ-Soc-semi-prime sub-module of X. Lemma 2.22 : (𝐴⨁𝐵) + 𝐹 − 𝑆𝑜𝑐(𝑋⨁𝑌) = (𝐴 + 𝐹 − 𝑆𝑜𝑐(𝑋))⨁(𝐵 + 𝐹 − 𝑆𝑜𝑐(𝑌)) for every fuzzy sub- modules A and B of fuzzy modules X and Y respectively. Proof: From (Lemma 2.2) we get (𝐹 − 𝑆𝑜𝑐(𝑋⨁𝑌))𝑡 = 𝑆𝑜𝑐((𝑋⨁𝑌)𝑡 ) For each t ∈ (0,1]. But, 𝑆𝑜𝑐((𝑋⨁𝑌)𝑡 ) = 𝑆𝑜𝑐(𝑋𝑡 ⨁𝑌𝑡 ) and we have 𝑆𝑜𝑐(𝑋𝑡 ⨁𝑌𝑡 ) = 𝑆𝑜𝑐(𝑋𝑡 )⨁𝑆𝑜𝑐(𝑌𝑡 ) That is (𝐹 − 𝑆𝑜𝑐(𝑋⨁𝑌))𝑡 = 𝑆𝑜𝑐(𝑋𝑡 )⨁𝑆𝑜𝑐(𝑌𝑡 ) = (𝐹 − 𝑆𝑜𝑐(𝑋))𝑡 ⨁(𝐹 − 𝑆𝑜𝑐(𝑌))𝑡 Thus (𝐹 − 𝑆𝑜𝑐(𝑋⨁𝑌))𝑡 = [(𝐹 − 𝑆𝑜𝑐(𝑋)) ⨁(𝐹 − 𝑆𝑜𝑐(𝑌)]𝑡 𝐹 − 𝑆𝑜𝑐(𝑋⨁𝑌) = 𝐹 − 𝑆𝑜𝑐(𝑋) ⨁ 𝐹 − 𝑆𝑜𝑐(𝑌) Hence from (Remark 1.5) then Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 114 Proposition 2.23 : If U and V are ℱ-sub-modules of ℱ-modules X and Y respectively, then 1) If 𝑈⨁𝑌 is an ℱ-Soc-semi-prime sub-module of 𝑋⨁𝑌 thus U is an ℱ-Soc-semi-prime sub-module of X. 2) if 𝑋⨁𝑉 is an ℱ-Soc-semi-prime sub-module of 𝑋⨁𝑌 thus V is an ℱ-Soc-semi-prime sub-module of X. Proof : 1) Suppose that 𝑈⨁𝑌 is an ℱ-Soc-semi-prime sub-module of 𝑋⨁𝑌 and 𝑟𝑏 is an ℱ-singleton of R and 𝑥𝑡 ⊆ 𝑋 such that (𝑟 𝑛)𝑏𝑥𝑡 ⊆ 𝑈. Then (𝑟 𝑛)𝑏 (𝑥𝑡 , 𝑦𝑝) = ((𝑟 𝑛)𝑏𝑥𝑡 , (𝑟 𝑛)𝑏𝑦𝑝) ⊆ 𝑈⨁𝑌, for any ℱ-singleton 𝑦𝑝 ⊆ 𝑌, but 𝑈⨁𝑌 is an ℱ-Soc-semi-prime sub-module of 𝑋⨁𝑌. Thus (𝑟𝑏𝑥𝑡 , 𝑟𝑏 𝑦𝑝) ⊆ (𝑈⨁𝑌) + 𝐹 − 𝑆𝑜𝑐(𝑋⨁𝑌), by (Lemma 2.22) we get (𝑟𝑏 𝑥𝑡 , 𝑟𝑏 𝑦𝑝) ⊆ (𝑈 + 𝐹 − 𝑆𝑜𝑐(𝑋))⨁(𝑌 + 𝐹 − 𝑆𝑜𝑐(𝑌)). That is 𝑟𝑏 𝑥𝑡 ⊆ 𝑈 + 𝐹 − 𝑆𝑜𝑐(𝑋) , therefore U is an ℱ-Soc-semi-prime sub-module of X. 2) Similarly as the idea in (1). Lemma 2.24 : If X is an ℱ-module of an ℛ-module M, and M be a faithful multiplication ℛ-module, then: ℱ − 𝑆𝑜𝑐(𝑋) = 𝑋 ℱ − 𝑆𝑜𝑐(ℛ). Proposition 2.25 : Let X be a finitely generated multiplication and faithful ℱ-module of an ℛ-module M, if J is an ℱ-Soc-semi-prime ideal of ℛ then JX is an ℱ-Soc-semi-prime sub-module of X. Proof : Assume that 𝑟𝑏 is an ℱ-singleton of ℛ and 𝑥𝑡 ⊆ 𝑋 such that (𝑟 𝑛)𝑏 𝑥𝑘 = (𝑟𝑏 ) 𝑛𝑥𝑘 ⊆ JX,where 𝑘, 𝑏 ∈ [0,1].that is (𝑟𝑛)𝑏〈𝑥𝑡 〉 ⊆ JX. But X is a multiplication ℱ-module, thus there exists an ℱ-ideal L of ℛ with 〈𝑥𝑡 〉 = 𝐿𝑋. Then we get (𝑟 𝑛)𝑏 𝐿𝑋 ⊆ JX, so (𝑟 𝑛)𝑏𝐿 ⊆ J + ℱ − ann(X) = J since X is a faithful ℱ- module. But J is an ℱ-Soc-semi-prime ideal of ℛ, then by ( Corollary 2.13) implies that 𝑟𝑏 𝐿 ⊆ 𝐽 + ℱ − 𝑆𝑜𝑐(ℛ).Now, by multiplying both sides with X and using (Lemma 2.24) we have 𝑟𝑏 𝐿𝑋 ⊆ 𝐽𝑋 + ℱ − 𝑆𝑜𝑐(ℛ)𝑋 = 𝐽𝑋 + ℱ − 𝑆𝑜𝑐(𝑋). Therefore, JX is an ℱ-Soc-semi-prime sub-module of X. Proposition 2.26 Suppose that U is an ℱ-Soc-semi-prime sub-module of an ℱ-module X and V is an ℱ-semi- prime sub-module of X with ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑉. Then the intersection of U and V is an ℱ- Soc-semi-prime of X. Proof : If 𝑟𝑏 is an ℱ-singleton of ℛ and 𝑥𝑡 ⊆ 𝑋 where 𝑏, 𝑡 ∈ [0,1], such that (𝑟 𝑛)𝑏 𝑥𝑘 = (𝑟𝑏 ) 𝑛𝑥𝑘 ⊆ 𝑈 ∩ 𝑉. This implies (𝑟𝑛)𝑏𝑥𝑡 ⊆ 𝑈 and (𝑟 𝑛)𝑏 𝑥𝑡 ⊆ 𝑉, but U is an ℱ-Soc-semi-prime sub- module of X. So, we have 𝑟𝑏 𝑥𝑘 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋).Now, since V is an ℱ-semi-prime sub- module of X then 𝑟𝑏 𝑥𝑘 ⊆ 𝑉.We get 𝑟𝑏 𝑥𝑘 ⊆ [𝑈 + ℱ − 𝑆𝑜𝑐(𝑋)] ∩ 𝑉, but ℱ − 𝑆𝑜𝑐(𝑋) ⊆ 𝑉 Ibn Al-Haitham Jour. for Pure & Appl. Sci. 53 (1)2022 115 then by using (Lemma 1.29) we have 𝑟𝑏 𝑥𝑘 ⊆ (𝑈 ∩ 𝑉) + ℱ − 𝑆𝑜𝑐(𝑋). That is mean 𝑈 ∩ 𝑉 is an ℱ-Soc-semi-prime of X. Proposition 2.27 Let X be a faithful multiplication ℱ-module of an ℛ-module M, then a proper ℱ-sub-module U is an ℱ-Soc-semi-prime sub-module of if and only if [𝑈:𝑅 𝑋] is an ℱ-Soc-semi-prime ideal of ℛ. Proof: Let (𝑟𝑛)𝑏𝑚𝑡 = (𝑟𝑏) 𝑛𝑚𝑡 ⊆ [𝑈:𝑅 𝑋] with 𝑚𝑡 and 𝑟𝑏 are ℱ-singletons of ℛ where 𝑏, 𝑡 ∈ [0,1]implies that (𝑟𝑛)𝑏 (𝑚𝑡 𝑋) ⊆ 𝑈. But, U is an ℱ-Soc-semi-prime sub-module, so by (Corollary 2.13) then 𝑟𝑏 (𝑚𝑡 𝑋) ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋) . Since X is a multiplication ℱ-module, then by (Preposition 1.32) 𝑈 = [𝑈:𝑅 𝑋]𝑋 , and since X is a faithful multiplication, so by (Lemma 2.24) 𝐹 − 𝑆𝑜𝑐(𝑋) = ℱ − 𝑆𝑜𝑐(ℛ)𝑋. Therefore 𝑟𝑏 𝑚𝑡 𝑋 ⊆ [𝑈:𝑅 𝑋]𝑋 + ℱ − 𝑆𝑜𝑐(ℛ)𝑋, this implies 𝑟𝑏 𝑚𝑡 ⊆ [𝑈:𝑅 𝑋] + ℱ − 𝑆𝑜𝑐(ℛ). Thus [𝑈:𝑅 𝑋] is an ℱ-Soc-semi- prime ideal of ℛ . Conversely Let (𝑟𝑛)𝑏𝐷 = (𝑟𝑏) 𝑛𝐷 ⊆ 𝑈 with 𝑟𝑏 be an ℱ-singleton of ℛ and D is an ℱ-sub-module of X. Since X is a multiplication ℱ-module, then 𝐷 = 𝐽𝑋 for some an ℱ-ideal of ℛ, we get (𝑟𝑛)𝑏𝐽𝑋 ⊆ 𝑈 that is mean (𝑟 𝑛)𝑏𝐽 ⊆ [𝑈:𝑅 𝑋],but [𝑈:𝑅 𝑋] is an ℱ-Soc-semi-prime ideal of ℛ, so by (Corollary 2.12) we have 𝑟𝑏 𝐽 ⊆ [𝑈:𝑅 𝑋] + 𝐹 − 𝑆𝑜𝑐(ℛ) , this implies 𝑟𝑏 𝐽𝑋 ⊆ [𝑈:𝑅 𝑋]𝑋 + ℱ − 𝑆𝑜𝑐(ℛ)𝑋 , then by (Lemma 2.25) we get 𝑟𝑏 𝐽𝑋 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋). Lemma 2.28 Let 𝑓: M → �̅� be isomorphism mapping from an ℛ-module M into an ℛ-module �̅� .If X and �̅� are ℱ-modules of an ℛ-modules M and �̅� respectively. Then f (ℱ − 𝑆𝑜𝑐(𝑋)) ⊆ ℱ − 𝑆𝑜𝑐(�̅�). Proposition 2.29 Let 𝑓: 𝑋 → �̅� be an ℱ-isomorphism from ℱ-module 𝑋 into ℱ-module �̅�, with U is an ℱ-Soc- semi-prime sub-module of 𝑋, such that ker (𝑓) ⊆ 𝑈. Then 𝑓(𝑈) is an ℱ-Soc-semi-prime sub-module of �̅�. Proof : 𝑓(𝑈) is a proper ℱ-sub-module of �̅�. If not, then 𝑓(𝑈) = �̅�. Let 𝑥𝑡 ⊆ 𝑋, so 𝑓(𝑥𝑡 ) ⊆ �̅� = 𝑓(𝑈), that is there exists 𝑦𝑠 ⊆ 𝑈 where 𝑠, 𝑡 ∈ [0,1] such that 𝑓(𝑥𝑡 ) = 𝑓(𝑦𝑠)implies that 𝑓(𝑥𝑡 ) − 𝑓(𝑦𝑠) = 01 then 𝑓(𝑥𝑡 − 𝑦𝑠) = 01, thus 𝑥𝑡 − 𝑦𝑠 ⊆ ker (𝑓) ⊆ 𝑈, it follows that 𝑥𝑡 ⊆ 𝑈.Thus 𝑈 = 𝑋 that is a contradiction. Now, Let (𝑟𝑏 ) 𝑛𝑧𝑐 ⊆ 𝑓(𝑈) with 𝑟𝑏 ⊆ ℛ and 𝑧𝑐 ⊆ �̅� with 𝑏, 𝑐 ∈ [0,1],but 𝑓 is onto 𝑓(𝑥𝑡 ) = 𝑧𝑐 for some 𝑥𝑡 ⊆ 𝑋, therefore (𝑟 𝑛)𝑏 𝑧𝑐 = (𝑟𝑛)𝑏𝑓(𝑥𝑡 ) = 𝑓((𝑟 𝑛)𝑏𝑥𝑡 ) ⊆ 𝑓(𝑈), this implies that there exists 𝑘ℎ ⊆ 𝑈 with ℎ ∈ [0,1] such that 𝑓(𝑘ℎ ) = 𝑓((𝑟 𝑛)𝑏𝑥𝑡 ), that is 𝑓(𝑘ℎ − (𝑟 𝑛)𝑏𝑥𝑡 ) = 01, so 𝑘ℎ − (𝑟 𝑛)𝑏𝑥𝑡 ⊆ ker (𝑓) ⊆ 𝑈. It follows that (𝑟𝑏) 𝑛𝑥𝑡 ⊆ 𝑈. But, U is an ℱ-Soc-semi-prime sub-module of 𝑋, thus 𝑟𝑏 𝑥𝑡 ⊆ 𝑈 + ℱ − 𝑆𝑜𝑐(𝑋). Then by (Lemma 2.28) we have 𝑟𝑏𝑧𝑐 = 𝑟𝑏 𝑓(𝑥𝑡 ) ⊆ 𝑓(𝑈) + 𝑓(ℱ − 𝑆𝑜𝑐(𝑋)) ⊆ 𝑓(𝑈) + ℱ − 𝑆𝑜𝑐(�̅�) . Hence 𝑓(𝑈) is an ℱ-Soc-semi-prime sub- module of �̅�. Ibn Al-Haitham Jour. for Pure & Appl. 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