Microsoft Word - 452-462 IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|452    Coclosed Rickart Modules Ghaleb Ahmed galeb.ahmud3@gmail.com Dept.of Mathematics/College of Education for Pure Science( Ibn-Al- Haitham)/ University of Baghdad Abstract Let 𝑀 be a right module over an arbitrary ring 𝑅 with identity and S End 𝑀 . In this work, the coclosed rickart modules as a generalization of rickart modules is given. We say a module 𝑀 over 𝑅 coclosed rickart if for each 𝑓 ∈ End 𝑀 , Ker 𝑓 is a coclosed submodule of 𝑀. Basic results over this paper are introduced and connections between these modules and otherwise notions are investigated. Keywords: coclosed rickart modules, relatively coclosed rickart modules, rickart modules, cosemisimple modules, modules with CCIP. IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|453    1. Introduction 𝑅 is denoted a ring has unity and 𝑀 is studied as a left S- right 𝑅-bimodule where S End 𝑀 is the endomorphism ring of 𝑀. Also 𝑟 𝑆𝑓 𝑟 𝑓 Ker 𝑓 𝑚 ∈ 𝑀 | 𝑓 𝑚 0 is the right annihilator of each element 𝑓 ∈ 𝑆. In [2] is presented a generalization of rickart modules by using the concept of purity. Further, dual coclosed rickart modules is introduced in [3]. We say a submodule 𝐿 of a module 𝑀 is coclosed in 𝑀 when ≪ then 𝐿 𝐾 for each 𝐾 ⊆ 𝐿 [1]. Equivalently, for each proper submodule 𝐾 ⊂ 𝐿, there is a submodule 𝑁 of 𝑀 where 𝐿 𝑁 𝑀 while 𝐾 𝑁 𝑀 iff 𝐿 is coclosed of 𝑀. Our goal of this research is to present and discuss another generalization of rickart modules namely coclosed rickart modules. A module 𝑀 over 𝑅 is said coclosed rickart when each 𝑓 ∈ End 𝑀 , Ker 𝑓 is a coclosed of 𝑀. Other studies in [5],[6],[7],[8],[11] and [12] are related topics. The paper is divided into four sections. Section two, includes introducing the concept of coclosed rickart modules and providing facts of this argument.While direct summands of coclosed rickart modules are shown to inherit the property (Proposition 2.6), this is not so for direct sums (Remark 2.7).We obtain a condition which allows direct sums of coclosed rickart modules is coclosed rickart  (Proposition 2.8). Section three is devoted to look for any connection between coclosed rickart modules and other modules. We see that coclosed rickart modules  and rickart modules  coincide in lifiting modules (Proposition 3.4). In section four, we present and study the concept of relatively coclosed rickart modules. By using the CCIP, we will provide a condition for modules to be relatively coclosed rickart ( for example, Theorem 4.7,  Proposition 4.11) where an 𝑅-module 𝑀 has the coclosed intersection property (in brief CCIP) if, the intersection of any two coclosed submodules of 𝑀 is coclosed [4]. Many results are investigated, we see that the rings 𝑅 for which each right module over 𝑅 relatively rickart are precisely right cosemisimple rings (Theorem 4.13). Coclosed Rickart Modules A comprehensive study of coclosed rickart modules is given in this section. We provide several characterizations and some properties of this of modules are investigated. We begin with the next. Definition 2.1. We call a module 𝑀 over 𝑅 is coclosed rickart when each 𝑓 ∈ End 𝑀 , Ker 𝑓 is a coclosed submodule of 𝑀. Remarks and Examples Clearly each cosemisimple module is coclosed rickart, but not conversely, where An 𝑅-module 𝑀 is called cosemisimple if each submodule of 𝑀 is coclosed in 𝑀 [1]. For example, the ℤ-module ℤ is coclosed rickart because each 𝑓 ∈ End 𝑀 , Ker 𝑓 0 is a coclosed submodule in 𝑀 but it is not cosemisimple since the only proper coclosed submodule in the ℤ-module ℤ is the zero submodule. (1) It is obvious that rickart module is coclosed rickart while the reverse is not hold as follows. Consider the ring ∏ F , F F is a field and 𝑅 is not semisimple then by [9, Theorem 2.25], there exists a right module 𝑀 over 𝑅, it is not rickart. And, 𝑅 is commutative regular (in sense Von Neumann) so by [1], 𝑅 is cosemisimple (or V-ring) implies that Rad 0 for each submodule 𝑁 of 𝑀. So each submodule of 𝑀 is coclosed in 𝑀 implies it is a coclosed rickart module. IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|454    (2) When 𝑀 is a coclosed simple 𝑅-modul yeild 𝑀 need not be coclosed rickart where an 𝑅-module 𝑀 is called coclosed simple if 𝑀  0 and it has no coclosed submodules except {0} and 𝑀 4 . Study ℤ as ℤ-module is coclosed simple, while it is not coclosed rickart since there exists an endomorphism 𝑓 ∶ ℤ ⟶ ℤ defined by 𝑓 𝑚 𝑚2 for each 𝑚 ∈ ℤ , implies Ker 𝑓 0, 2 is not a coclosed submodule in ℤ . (3) When 𝑀 is a quasi-Dedekind over 𝑅 implies 𝑀 is coclosed rickart, where 𝑀 is quasi-Dedekind if every 0 𝑓 ∈ End 𝑀 , Ker 𝑓 0 [10 . The reverse is not hold. For example, ℤ as ℤ-module is coclosed rickart while it is not quasi- Dedekind. (4) If 𝑀 is a coclosed rickart coclosed simple module, implies 𝑀 is quasi-Dedekind. Proof. When 𝑀 is a coclosed rickart over 𝑅 and 0 𝑓 ∈ End 𝑀 , implies Ker 𝑓 is coclosed in 𝑀. But 𝑀 is coclosed simple with 𝑓 0, thus Ker 𝑓 0 , as required. (5) A homomrphic picture of a coclosed rickart module may not be coclosed rickart. Consider the natural epimorphism 𝑓 ∶ ℤ ⟶ ℤ . The ℤ -module ℤ is coclosed rickart, while Im 𝑓 ℤ4 is not a coclosed rickart over ℤ. (6) Every integral domain is coclosed rickart. Proof. When 𝑅 as in assumption, implies 𝑅 is commutative implies that for each 𝑎 ∈ 𝑅 and 𝑓 ∈ End 𝑅 ≅ 𝑅, we can define 𝑓 ∶ 𝑅 ⟶ 𝑅 by 𝑓 𝑟 𝑟𝑎 for each 𝑟 ∈ 𝑅. It follows that Ker 𝑓 𝑟 ∈ 𝑅| 𝑓 𝑟 0 𝑟 ∈ 𝑅| 𝑟𝑎 0 𝑎𝑛𝑛 𝑎 0 is coclosed in 𝑅. (7) The reverse of Remark (7) need not be true generally. For example, the ring ℤ is coclosed rickart but it is not integral domain. Proposition Under an isomorphism, the coclosed rickart property is moved. Proof. Let 𝑀 and 𝑀 be modules over 𝑅, where 𝑀 is coclosed rickart and 𝑓 ∶ 𝑀 → 𝑀 an isomorphism. Let 𝑔  End 𝑀 , we prove that Ker 𝑔 is a coclosed submodule in 𝑀 . Let 𝐾 be any proper submodule such that 𝐾 ⊂ Ker 𝑔 implies that 𝑓 (𝐾 ⊂ 𝑓 Ker 𝑔 . But 𝑓 ( Ker 𝑔 Ker 𝑓 𝑔𝑓 , to show this. Let x  𝑓 ( Ker 𝑔 ), 𝑥 𝑓 𝑦 and 𝑦  Ker 𝑔 implies that 𝑓 𝑥 𝑦 and 𝑔 𝑦 0. Then 𝑓 𝑔 𝑓 𝑥 𝑓 𝑔 𝑦 0, implies 𝑥  Ker 𝑓 𝑔𝑓 ). For the other, let 𝑥  Ker 𝑓 𝑔𝑓 ), 𝑓 𝑔 f ( x ) 0 and so 𝑓 𝑥  Ker 𝑓 𝑔). One can easily see that Ker 𝑓 𝑔 Ker 𝑔, hence 𝑓 𝑥  Ker 𝑔, 𝑥  𝑓 Ker 𝑔), it follows that 𝑓 ( Ker 𝑔 ) = Ker 𝑓 𝑔𝑓 ). This means that 𝑓 ( K ) ⊂ ker ( 𝑓 𝑔𝑓 ), but 𝑓 𝑔𝑓  𝑓 ∈ End (𝑀 ) and 𝑀 coclosed rickart then Ker ( 𝑓 𝑔𝑓 ) is a coclosed , there exists a submodule 𝑁 of 𝑀 , 𝑓 ( Ker 𝑔 ) + 𝑁 = 𝑀 but 𝑓 ( 𝐾 ) + 𝑁 𝑀 . This means that Ker 𝑔 + 𝑓 𝑁 = 𝑀 but 𝐾 + 𝑓 𝑁 𝑀 , it follows that Ker 𝑔 is a coclosed submodule in 𝑀 . From a module to any of its submodules or inversely, coclosed rickart property does not always passed in general as follows Examples If 𝑁 is a submodule of a coclosed rickart module 𝑀 over 𝑅 , implies 𝑁 may not be coclosed rickart. Study 𝑀 ℚ  ℤ as ℤ-module. By [9, Ex 2.5], 𝑀 is rickart, yield 𝑀 is coclosed rickart. Now consider 𝑁 ℤ  ℤ a submodule of 𝑀. See that 𝑁 may not be a coclosed rickart since there is 𝑓 ∶ 𝑁 → 𝑁 defined by 𝑓 𝑚, 𝑛 0, 𝑚 where 𝑚, 𝑛  ℤ. Thus Ker 𝑓 𝑚, 𝑛 ∈ ℤ  ℤ | 𝑓 𝑚, 𝑛 = 0, 0 𝑚, 𝑛 | 𝑚 0 } = 2ℤ  ℤ is not a coclosed submodule in ℤ  ℤ , because 2ℤ ≅ ℤ ≅ ℤ  ℤ ℤ  ℤ ≪ ℤ  ℤ ℤ  ℤ ≅ ℤ . IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|455    (1) If each proper submodule of 𝑀 is a coclosed rickart, implies 𝑀 may not be coclosed rickart. as the next, ℤ4 as ℤ-module in which every proper submodule is semisimple module and hence they are coclosed rickart modules. But ℤ4 is not coclosed rickart. We record the next from [1]. Lemma Let 𝑁, 𝐻 be submodules of an 𝑅-module 𝑀 and 𝐾 ⊂ 𝑁. If 𝐾 is coclosed in 𝑀, implies 𝐾 is coclosed in 𝑁 and the reverse is hold if 𝑁 is coclosed in 𝑀. Proposition. Each direct summand of a coclosed rickart module is coclosed rickart. Proof. Let 𝑀 be a coclosed rickart module over 𝑅 and 𝐴 a direct summand of 𝑀, then 𝑀 𝐴𝐵 for some submodule 𝐵 of 𝑀. Let 𝑓 ∈ End 𝐴 , then we have the following 𝑀 → 𝐴 → 𝐴 → 𝑀. Say 𝑔 𝑖 𝑓 , 𝑔 ∈ EndR 𝑀 . Therefore Ker 𝑔 is a coclosed in 𝑀. See Ker 𝑔 Ker 𝑓 𝐵. To show this, let 𝑚 ∈ Ker 𝑓 𝐵, 𝑚 𝑥 𝑦 where 𝑥 ∈ Ker 𝑓 and 𝑦 ∈ 𝐵. Hence 𝑔 𝑚 𝑔 𝑥 𝑦 𝑖 𝑓  𝑥 𝑦 𝑖 𝑓 𝑥 𝑓 𝑥 0. Therefore 𝑚 ∈ Ker 𝑔 implies that Ker 𝑓 𝐵  Ker 𝑔. Conversely, let 𝑚 ∈ Ker 𝑔  𝑀 𝐴𝐵. let 𝑚 𝑥 𝑦 , 𝑥  𝐴 and 𝑦  𝐵, then 𝑔 𝑚 𝑔 𝑥 𝑦 𝑖 𝑓  𝑥 𝑦 0. Thus 𝑖 𝑓 𝑥 0 and hence 𝑓 𝑥 0. That is 𝑥 ∈ Ker 𝑓 and since 𝑦 ∈ 𝐵. Thus 𝑚 𝑥 𝑦 ∈ Ker 𝑓 𝐵 , Ker 𝑔  Ker 𝑓 𝐵. That is, Ker 𝑔 Ker 𝑓 𝐵. Clearly Ker 𝑓 ⋂ 𝐵 0. Therefore Ker 𝑔 Ker 𝑓  𝐵 and hence Ker 𝑓 is a coclosed submodule in Ker 𝑔. But Ker 𝑔 is coclosed in 𝑀, then Ker 𝑓 is coclosed in 𝑀. But 𝐴 is containing Ker 𝑓, so by lemma 2.5, Ker 𝑓 is coclosed in 𝐴. Therefore 𝐴 is a coclosed rickart 𝑅-module. Remark The direct sum of coclosed rickart modules need not be coclosed rickart, as follows. ℤ  ℤ2 as ℤ-module is not coclosed rickart while each of ℤ and ℤ2 is a coclosed rickart module. Recall that a submodule 𝑁 of an 𝑅-module 𝑀 is said to be fully invariant if 𝑓 𝑁 is included in 𝑁 for each 𝑓 ∈ End 𝑀 [13]. We call 𝑀 is duo module if every submodule in 𝑀 is fully invariant [13]. Proposition Let 𝑀 ⊕ ∈ 𝑀 be a duo module over 𝑅. 𝑀 is a coclosed rickart module if and only if 𝑀 is a coclosed rickart for each 𝑖 ∈  . Proof. First side is already from Proposition 2.6. The other side, assume 𝑓 𝑓 ∈ End 𝑀 where 𝑓 ∈ Hom 𝑀 , 𝑀 . Since 𝑀 is fully invariant in 𝑀 ⊕ ∈ 𝑀 , implies Hom 𝑀 , 𝑀 ) 0 for each 𝑖 𝑗 [13, Lemma 1.9]. Further 𝑓 𝑀 ⊆ 𝑀 for all 𝑖 ∈ , this implies that Ker𝑓 ⊕ ∈ Ker 𝑓 . We claim that Ker𝑓 is a coclosed submodule in 𝑀. To show this, let 𝐾 ⊂ ⊕ ∈ Ker 𝑓 be a submodule of 𝑀, then 𝐾 is a fully invariant of 𝑀. By [7, Result 2.1], 𝐾 ⊕ ∈ 𝐾 ∩ 𝑀 , let 𝐾 𝐾 ∩ 𝑀 for each 𝑖 ∈ .Easily to check 𝐾 ⊆ Ker 𝑓 . Since 𝑀 is coclosed Rickart implies that Ker 𝑓 is a coclosed submodule of 𝑀 for all 𝑖 ∈ . Thus there is a submodule 𝑁 of 𝑀 , Ker 𝑓 𝑁 𝑀 but 𝐾 𝑁 𝑀 . This implies that ⊕ ∈ Ker 𝑓 ∑ 𝑁∈ ⊕ ∈ 𝑀 but ⊕ ∈ 𝐾 ∑ 𝑁∈ ⊕ ∈ 𝑀 . Put 𝑁 ∑ 𝑁∈ . So we have Ker𝑓 𝑁 𝑀 but 𝐾 𝑁 𝑀 and hence Ker𝑓 is a coclosed submodule of 𝑀. Therefore 𝑀 is coclosed rickart. Proposition 2.9. Let 𝑅 be a ring. The next statements are equivalently (1) 𝑅 is a coclosed rickart 𝑅-module for any index set . IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|456    (2) All projective 𝑅-modules are coclosed rickart modules. (3) All free 𝑅-modules are coclosed rickart modules. Proof. 1  2 Assume 𝑀 is projective over 𝑅, we get a free module 𝐹 over 𝑅 with an epimorphism 𝑓 ∶ 𝐹  𝑀. We have the short exact sequence 0 → Ker 𝑓 → 𝑅 → 𝑀 → 0 where 𝐹  𝑅 for some index set . But 𝑀 is projective then the sequence splits. Thus 𝑅  Ker 𝑓  𝑀. Because 𝑅 is a coclosed rickart module, therefore from Proposition 2.3, Ker 𝑓  𝑀 is also coclosed rickart. Hence from Lemma 2.5, 𝑀 is coclosed rickart. It is clear 2  1 and 2  3 . Similarly, we can prove 1  3 . Proposition Let 𝑀 be a coclosed simple coclosed rickart module which has a nonzero maximal submodule 𝑁, then 𝑀 ⊕ is not a coclosed rickart module. Proof. Suppose 𝑀 ⊕ is coclosed rickart, 𝑓 ∈ End 𝑀 ⊕ defined by 𝑓 𝑚, 𝑛 0, 𝑚 for all 𝑚 ∈ 𝑀, 𝑛 ∈ . Then Ker 𝑓 𝑚, 𝑛 ∈ 𝑀 ⊕ | 𝑓 𝑚, 𝑛 0, 0 𝑚, 𝑛 ∈ 𝑀 ⊕ 𝑚 𝑁 𝑁 𝑁 ⊕ is a coclosed of 𝑀 ⊕ . But 𝑁 is a coclosed of 𝑁 ⊕ , then by Lemma 2.5, 𝑁 is a coclosed of 𝑀 ⊕ . Again by result 2.5, 𝑁 is a coclosed of 𝑀 , this implies a false because 𝑀 is coclosed simple. Yeild 𝑀 ⊕ is not coclosed rickart. Coclosed Rickart Modules and Other Topics This part looks for any relationship between coclosed rickart modules and other concepts. We first study the next condition (*) for an 𝑅-module : For any submodule 𝑁 of 𝑀 for which ≅ 𝐻 where 𝐻 is a summand of 𝑀, yeild 𝑁 is a coclosed submodule of 𝑀. Proposition. Every coclosed rickart module satisfies the condition (*). Proof. When 𝑁 is a submodule of a coclosed rickart 𝑀 over 𝑅 with ≅ 𝐻 where 𝐻 is a summand in 𝑀. Hence 𝑓 ∶ → 𝐻 is an isomorphism, implies 𝑀 → → 𝐻 → 𝑀. Then 𝑖𝑓𝜋 ∈ End 𝑀 Ker 𝑖𝑓𝜋 𝜋 Ker 𝑖𝑓 𝜋 0 𝑁. Because 𝑀 is coclosed rickart, then 𝑁 = Ker 𝑖𝑓𝜋 is coclosed in 𝑀. Corollary When 𝑀 is a coclosed rickart module over 𝑅 with the condition (*) for each submodule 𝑁 of 𝑀, therefore 𝑀 is a cosemisimple. Proof. Obvious by Proposition 3.1. The next result is to find a certain situation under which the reverse of Proposition 3.1 is hold. Proposition If 𝑀 a module has the condition (*) with Im𝑓 isomorphic to a summand of 𝑀 for all 𝑓 ∈ End 𝑀 , implies 𝑀 is coclosed rickart. Proof. When 𝑓 ∈ End 𝑀 , from assumption, Im𝑓 ≅ 𝐻 where 𝐻 is a direct summand of 𝑀, implies that ≅ 𝐻. So by the condition (*), Ker 𝑓 is coclosed of 𝑀, as asserted. IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|457    Let 𝑀 be a module over 𝑅. We record that, 𝑀 lifiting when each submodule 𝑁 of 𝑀, there is a direct summand 𝐾 of 𝑀 such that 𝐾 ⊆ 𝑁 and ≪ [1]. Proposition Every lifiting coclosed rickart module is rickart module. Proof. When 𝑀 be a lifiting coclosed rickart over 𝑅 and 𝑓 ∈ End 𝑀 . Because 𝑀 is lifting, there exists a summand 𝐾 of 𝑀, 𝐾 ⊆ Ker 𝑓 with ≪ . But 𝑀 is coclosed rickart, implies Ker 𝑓 is a coclosed in 𝑀, hence Ker 𝑓 𝐾, as desired. Examples Investigate ℤ  ℤ is lifiting as ℤ - module. But ℤ  ℤ is not coclosed rickart as ℤ - module, if otherwise, then by Proposition 2.5, each summand of ℤ  ℤ is coclosed rickart, whereas this contradicts that ℤ  0 ≅ ℤ is not coclosed rickart as ℤ - module. (1) ℤ as ℤ -module is coclosed rickart while not lifiting. A module 𝑀 is called Hopfian if every epimorphism 𝑓 ∈ End 𝑀 is an isomorphism [1]. Proposition When 𝑀 is a coclosed rickart coclosed simple module over 𝑅 implies 𝑀 is Hopfian. Proof. When 0 𝑓 ∈ End 𝑀 is an epimorphism. Because 𝑀 is coclosed rickart, implies Ker 𝑓 is a coclosed of 𝑀. While 𝑀 is coclosed simple and 𝑓 0, it follows that Ker 𝑓 0, and hence 𝑀 is Hopfian. Record that, let 𝑀 be a module over 𝑅, we call it scalar when each 𝑓 ∈ End 𝑀 , there is 𝑎 ∈ 𝑅 with 𝑓 𝑚 𝑚𝑎 for each 𝑚 ∈ 𝑀 [14]. Proposition When 𝑀 is a scalar coclosed rickart module, yield for each 0 𝑓 ∈ End 𝑀 , there is 0 𝑎 ∈ 𝑅 with 𝑟 𝑎 is coclosed of 𝑀. Proof. When 0 𝑓 ∈ End 𝑀 and 𝑀 a scalar 𝑅-module yield there is 𝑎 ∈ 𝑅, 𝑓 𝑚 𝑚𝑎 for every 𝑚 ∈ 𝑀. But 𝑀 is coclosed Rickart, therefore 𝑟 𝑎 is a coclosed of 𝑀. Corollary. If 𝑅 is a commutative via unity coclosed rickart over 𝑅, it follows for any 0 𝑓 ∈ End 𝑅 , there is 0 𝑎 ∈ 𝑅 and 𝑟 𝑎 is a coclosed submodule of 𝑅. Proof. Due to every commutative ring 𝑅 is a scalar, then result is already obtained by Proposition 3.7. Proposition Assume 𝑀 is a scalar faithful over 𝑅, implies 𝑅 is coclosed rickart if only if End 𝑅 is coclosed rickart. Proof. Because 𝑀 is a scalar module over 𝑅, then by [14, Lemma 6.2], End 𝑅 ≅ . But 𝑀 is faithful implies that 𝑅 ≅ End 𝑅 , as required. IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|458    Remarks and Examples (1) The factor module of a coclosed rickart module may not be coclosed rickart. Study ℤ as ℤ -module is coclosed rickart, whereas ≅ 𝑍 is not a coclosed rickart. (2) When 𝑀 is a coclosed rickart module, yeild is coclosed rickart for each summand 𝑁 of 𝑀. (3) If 𝑁 is a submodule of 𝑀. When 𝑁 and are coclosed rickart modules implies 𝑀 need not be coclosed rickart. Study 𝑀 = ℤ ℤ2 over ℤ and 𝑁 0  ℤ2. Thus 𝑁 and ≅ ℤ are coclosed rickart ℤ-modules while 𝑀 is not coclosed rickart. Relatively Coclosed Rickart Modules In this part, we study relatively coclosed rickart modules. Main results of this type of modules are investigated. The family of rings 𝑅 for which each right 𝑅-module is relatively coclosed rickart is right cosemisimple rings. Our concern is: When do modules have the relatively coclosed rickart property. Definition. Let 𝑀 and 𝑁 be 𝑅-modules. 𝑀 is called relatively coclosed rickart to 𝑁 if for every 𝑓 ∈ Hom 𝑀, 𝑁 , Ker 𝑓 is a coclosed submodule of 𝑀. As special case, 𝑀 is coclosed rickart if and only if 𝑀 is relatively coclosed rickart to 𝑀. Remarks with Examples (1) Obviously every cosemisimple 𝑅-module 𝑀 is relatively coclosed rickart to any 𝑅- module 𝑁. (2) Let 𝑀 and 𝑁 be 𝑅-modules. If 𝑀 is relatively coclosed rickart to 𝑁, then 𝑁 need not be relatively coclosed rickart to 𝑀. For example, let ℤn and ℤ as ℤ-modules. Then ℤn is relatively coclosed rickart to ℤ for each positive integer 𝑛 greater than one, in fact Homℤ ℤ , ℤ 0. On the other hand, ℤ is not relatively coclosed rickart to ℤn, since there exists the natural homomorphism 𝑓 ∈ Homℤ ℤ , ℤ defined by 𝑓 𝑚 𝑚 for each ∈ ℤ, for some 𝑛 1, Ker 𝑓 𝑛ℤ is not coclosed in ℤ . (3) When 𝑀 is a coclosed rickart module over 𝑅, implies 𝑀 needs not be relatively coclosed rickart to an 𝑅-module 𝑁. For example, ℤ as ℤ-module is coclosed rickart. Whereas ℤ is not relatively coclosed rickart to ℤn as ℤ-module for any 𝑛 1. (4) If 𝑀 is relatively coclosed rickart to an 𝑅-module 𝑁, then 𝑀 may not be coclosed rickart. For example, consider the ℤ-module ℤ is relatively coclosed rickart to the ℤ-module ℤ , because Homℤ ℤ , ℤ 0. But ℤ is not coclosed rickart. (5) When 𝑀 is a coclosed simple or quasi-Dedekind over 𝑅, yeild 𝑀 need not be relatively coclosed rickart to a module 𝑁 over 𝑅. Discuss ℤ over ℤ is coclosed simple and quasi-Dedekind while not relatively coclosed rickart to the ℤ-module ℤn, for each 𝑛 1. Theorem Let 𝑀 and 𝑁 be 𝑅-modules. The next statements are equivalent (1) 𝑀 is relatively coclosed rickart to 𝑁. (2) Each submodule 𝐵 of 𝑁 and each summand 𝐴 of 𝑀 is relatively coclosed rickart to 𝐵. (3) Each direct 𝐴 of 𝑀 and for each coclosed 𝐵 of 𝑁 and each 𝑓 ∈ Hom 𝑀, 𝐵 , Ker 𝑓| is a coclosed in 𝐴. IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|459    Proof. 1  2 Assume 𝑀 is relatively coclosed rickart to 𝑁. When 𝐴 is a direct summand of 𝑀 and 𝐵 of 𝑁. Let 𝑓 ∈ Hom 𝐴, 𝐵 . Study the next 𝑀 𝐴 ⊕ 𝐻 → 𝐴 → 𝐵 → 𝑁 where 𝐻 is a submodule of 𝑀. Say 𝑔 𝑖 𝑓  ∈ Hom 𝑀, 𝑁 . This implies that Ker 𝑔 is a coclosed in 𝑀. By similar steps of result 2.6, we gain Ker 𝑓 is a coclosed submodule in 𝐴. Therefore 𝐴 is relatively coclosed Rickart to 𝐵. 2  3 Let 𝐵 be a coclosed submodule in 𝑁 and 𝐴 a direct summand of 𝑀. Let 𝑓 ∈ Hom 𝑀, 𝐵 , then 𝑓| ∈ Hom 𝐴, 𝐵 . Since 𝐴 is relatively coclosed Rickart to 𝐵, implies Ker𝑓| is a closed of 𝐴. 3  1 by taking 𝐴 𝑀 and 𝐵 𝑁. The next two lemmas are proved in [4]. Lemma A module 𝑀 has the CCIP iff every coclosed submodule in 𝑀 gets the CCIP . Lemma When 𝑀 be a module over 𝑅 with the CCIP, yeild each decomposition 𝑀 𝐴 𝐵 and for every 𝑓 ∈ Hom 𝐴, 𝐵 , ker 𝑓 is a coclosed submodule of 𝑀. We give the following results Proposition Under an isomorphism, CCIP is transformed . Proof. When 𝑀 and 𝑀 are modules over 𝑅 with 𝑀 has The CCIP and 𝑓 ∶ 𝑀 → 𝑀 an isomorphism. Assume 𝐴, 𝐵 are coclosed of 𝑀 . To show 𝐴 ∩ 𝐵 is again coclosed of 𝑀 . Then there is submodules 𝐶, 𝐷 of 𝑀 and 𝑓 𝐶 𝐴, 𝑓 𝐷 𝐵 implies 𝐶 𝑓 𝐴 , 𝐷 𝑓 𝐵 . It is easy to check that 𝐶 𝑓 𝐴 , 𝐷 𝑓 𝐵 are coclosed submodules of 𝑀 . It follows that 𝐶 ∩ 𝐷 𝑓 𝐴 ∩ 𝑓 𝐵 is a coclosed of 𝑀 . It is not hard to see that 𝑓 𝐶 ∩ 𝐷 is a coclosed submodule of 𝑀 . On the other hand, 𝑓 𝐶 ∩ 𝐷 𝐴 ∩ 𝐵 implies the result is gained. Theorem If 𝑀 is a module over 𝑅 with the CCIP and 𝐴 𝐵 is a coclosed of 𝑀. Yeild 𝐴 is relatively coclosed rickart module to 𝐵. Proof. Assume that 𝑀 has the CCIP. Then by lemma 4.4, every coclosed submodule of 𝑀 has the CCIP implies that 𝐴 𝐵 has the CCIP. By lemma 4.5, for every 𝑓 ∈ Hom 𝐴, 𝐵 , Ker 𝑓 is a coclosed submodule in 𝐴𝐵. But Ker 𝑓  𝐴, then by lemma 2.5, Ker 𝑓 is a coclosed in 𝐴. Therefore 𝐴 is relatively coclosed Rickart to 𝐵. As immediate consequences we have Corollary When 𝑀 and 𝑁 be modules over 𝑅. If 𝑀 𝑁 has the CCIP, yeild 𝑀 is relatively coclosed rickart to 𝑁. Corollary If 𝑀 be a module over . If 𝑀  𝑀 gets CCIP, implies 𝑀 is coclosed rickart module. Remark The reverse of Corollary 4.8 need not be hold generally. Study ℤ as ℤ -module is cosemisimple. Then it is relatively coclosed rickart for each 𝑅-module 𝑁. Let ℤ , 𝑀 = ℤ ℤ be modules over ℤ. We show that 𝑀 does not have the CCIP. Let 𝐴 1, 0 ℤ, and 𝐵 1, 1 ℤ be the submodules generated by (1, 0 and (1, 1) respectively. Obviously 𝐴 and 𝐵 are summands of 𝑀. Then , 𝐵 are coclosed in 𝑀. But 𝐴⋂𝐵 2ℤ  0 is not a coclosed submodule in 𝑀, since 2  0 ≅ ℤ  ℤ  ≪ ℤ ℤ ℤ  ≅ ℤ4 ℤ2. IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|460    Proposition. Let 𝑀 ∈ be a family modules over 𝑅 and 𝑁 an 𝑅-module where  1,2, … , 𝑛 . The next are equivalent (1) If 𝑁 has the CCIP, then 𝑁 is relatively coclosed Rickart to  𝑀 . (2) 𝑁 is relatively coclosed rickart to 𝑀 for all 𝑖 1,2, … , 𝑛. Proof. 1  2 Via Theorem 4.3. 2  1 When 𝑁 is relatively coclosed rickart to 𝑀 for all 𝑖 1,2, … , 𝑛 and 𝑁 has the CCIP. To show that 𝑁 is relatively coclosed rickart to  𝑀 . Let 𝑓 ∈ Hom 𝑁,  𝑀 . Let us consider the following sequence 𝑁 →  𝑀 → 𝑀 , it is evident that Im𝑓 ∑ Im 𝜌 𝑓 where 𝜌 is the natural projection map of  𝑀 onto 𝑀 for 𝑖 1,2, … , 𝑛. Yield 𝑓 𝜌 𝑓, 𝜌 𝑓, … , 𝜌 𝑓 and Ker 𝑓 ∩ Ker 𝜌 𝑓 . But 𝜌 𝑓 Hom 𝑁, 𝑀 and 𝑁 is relatively coclosed rickart to 𝑀 implies that Ker 𝜌 𝑓 is a coclosed submodule in 𝑁. Because 𝑁 has the CCIP, therefore Ker 𝑓 ∩ Ker 𝜌 𝑓 is a coclosed submodule in 𝑁, and hence 𝑁 is relatively coclosed rickart to  𝑀 . As an immediate result Corollary. When 𝑀 ∈ is a family modules over 𝑅 where  1,2, … , 𝑛 . The next are equivalent (1) If 𝑀 has the CCIP for all 𝑗 1,2, … , 𝑛, then 𝑀 is relatively coclosed rickart to  𝑀 . (2) 𝑀 is relatively coclosed rickart to 𝑀 for all 𝑖 1,2, … , 𝑛. A characterization of cosemisimple rings via relatively coclosed rickart modules is given as follows. Theorem. The next are equivalent (1) 𝑅 is a cosemisimple right 𝑅-module. (2) All 𝑅-modules are cosemisimple. (3) All 𝑅-modules are relatively coclosed rickart to any 𝑅-module. (4) All 𝑅-modules have the CCIP. (5) All injective over 𝑅 have the CCIP. (6) All injective over 𝑅are cosemisimple. (7) All quasi injective over 𝑅 have the CCIP. (8) All quasi injective over 𝑅 are cosemisimple. Proof. 1  2 follows by [1, Theorem 1.12]. 1  3 , 2  4  5 , 6  5 and 8  7 are obvious.. 3  1 When 𝐼 is an ideal of 𝑅. Because all 𝑅-modules are relatively coclosed rickart to any 𝑅-module. Then the 𝑅-module 𝑅 is relatively coclosed rickart to the module as 𝑅-module. Because there exists the natural epimorphism 𝜋 ∶ 𝑅 → , hece Ker 𝜋 𝐼 is a coclosed right ideal of 𝑅. This implies that 𝑅 is cosemisimple right 𝑅-module. 4  2 When 𝑀 is an 𝑅-module and 𝑁 a submodule of 𝑀, implies 𝑀  has CCIP. Let 𝑓 ∶ 𝑀 → be the natural epimorphisim. By Lemma 4.5, Ker 𝑓 is a coclosed submodule in 𝑀. Hence 𝑀 is coclosed rickart. 5  4 when 𝑀 is a module over an 𝑅, then there is an injective 𝑅-module 𝑁 and a monomorphisim 𝑓: 𝑀 → 𝑁 implies that 𝑀 is isomorphic to Im𝑓. When 𝑓 is an epimorphism, then the result follows. Suppose not, so there is an injective 𝑅-module 𝐾 and IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|461    and a monomorphisim 𝑔: → 𝐾. Let ℎ: 𝑁 → be the natural epimorphism. Let us consider 𝑔ℎ: 𝑁 → 𝐾. Because 𝑁  𝐾 is injective then 𝑁  𝐾 has the CCIP. By Lemma 4.4, ker 𝑔ℎ is a coclosed submodule of 𝑁  and due to 𝑔 is a monomorphisim, implies Ker 𝑔ℎ Ker ℎ Im𝑓 is coclosed of 𝑁. Because 𝑁 has the CCIP, then by Lemma 4.4, Im𝑓 has the CCIP. But 𝑀 ≅ Im𝑓 and due to Proposition 4.6, therefore 𝑀 has the CCIP. 5  6 When 𝑀 is an injective with 𝑁 a submodule in 𝑀 with 𝑓: 𝑀 → is the natural epimorphism. When is injective, then 𝑀  is injective with the CCIP. By Lemma 4.5, Ker 𝑓 𝑁 is a coclosed of 𝑀. Suppose not, when 𝐸 is the injective hull of with 𝑖: → 𝐸 is the inclusion function. Consider 𝑖𝑓: 𝑀 → 𝐸 . Since 𝑀𝐸 is injective, implies 𝑀𝐸 has the CCIP. But 𝑖 is a monomorphisim, then Ker 𝑖𝑓 Ker 𝑓 𝑁 is a coclosed in 𝑀. Therefore 𝑀 is cosemisimple. 7  8 By the same way of the statement 5  6 . Theorem Let 𝑀 be a module over 𝑅. The next are equivalent (1) 𝑀 is coclosed rickart with CCIP. (2) The right annihilator in 𝑀 of each finitely generated left ideal 𝐼 𝑓 , 𝑓 , … , 𝑓 of End 𝑀 is a coclosed. submodule of 𝑀. Proof. 1  2 Let 𝐼 be a nonzero left ideal of End 𝑀 where 𝑀 a coclosed rickart module over 𝑅 with a finite number of generators {𝑓 , 𝑓 , … , 𝑓 }. Because 𝑀 is coclosed Rickart, then 𝑟 𝑓 is a coclosed of 𝑀 for every 𝑖 1,2, … , 𝑛. But 𝑀 with CCIP, it follows that ⋂ 𝑟 𝑓 𝑟 𝐼 is a coclosed of 𝑀. (3)  1 When 𝑓 ∈ End 𝑀 , then 𝑓 is a left ideal of End 𝑀 with one generator. By hypothesis, 𝑟 𝑓 is coclosed of 𝑀, yeild 𝑀 is coclosed rickart. Moreover, ⋂ 𝑟 𝑓 𝑟 𝐼 for any finitely generated left ideal 𝐼 𝑓 , 𝑓 , … , 𝑓 of End 𝑀 . Hence ⋂ 𝑟 𝑓 is a coclosed of 𝑀, thus 𝑀 has the CCIP. References [1] J. Clark; CH. Lomp; N. Vanaja; R. Wisbauer, Lifting modules supplements and projectivity in module theory. Birkhauser Verlag, Basel, Boston, Berlin, 2006. [2] Ghaleb Ahmed, Pure Rickart Modules and Their Generalization,  International Journal of Mathematics Trend and Technology 30 (2) 82-95. 2016 [3] Ghaleb Ahmed, Dual Coclosed Rickart Modules, Ibn Al-Haitham Journal for Pure and Applied Science, to appear. [4] Ghawi, Th.Y. Modules with Closed Intersection (Sum) Property, Ph.D. Thesis, University of Al-Mustansiriyah, Iraq, 2015. [5] I.M.A. Hadi, Ghaleb Ahmed, Modules with ⊛(⊛’or ⊛’’) Condition, The Bulletin of Society for Mathematical Services and Standards 4 ,12-21. 2012 [6] I.M.A. Hadi, Ghaleb Ahmed, Strongly (Comletely) Hollow Submodules I, Ibn Al- Haitham Journal for Pure and Applied Science 3 (25), 393-403. 2012 IHSCICONF 2017 Special Issue Ibn Al-Haitham Journal for Pure and Applied science https://doi.org/ 10.30526/2017.IHSCICONF.1818 For more information about the Conference please visit the websites: http://www.ihsciconf.org/conf/  www.ihsciconf.org   Mathematics|462    [7] I.M.A. Hadi, Ghaleb Ahmed, Strongly (Completely) Hollow Submodules II, Ibn Al- Haitham Journal for Pure and Applied Science 1 (26) ,292- 302. 2013 [8] I.M.A. Hadi, Ghaleb Ahmed, t-Regular Modules, International Journal of Advanced Scientific and Technical Research 6 (1) 128-137. 2016 [9] G. Lee, S. T. Rizvi and C. S. Roman, Rickart Modules, Comm. Algebra 38 2010, 4005-4027. 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