The object of this paper is to introducea class of module which is a fully stable Banach Algebra module.

ALI, MANAL

MOHAMMED ALI, MUNA JASIM

English

International Institute for Science, Technology and Education (IISTE): E-Journals

Mathematical Theory and Modeling                                                                                                                                                  www.iiste.org ISSN 2224-5804 (Paper)    ISSN 2225-0522 (Online) Vol.6, No.1, 2016   136 Fully Stable Banach Algebra Module MUNA JASIM MOHAMMED ALI, MANAL ALI Department of  -College of Science for women  -University of Baghdad  Baghdad-Iraq  -Mathematics  Abstract The object of this paper is to introducea class of module which is a fully stable Banach Algebra module. Introduction Given a Banachspace 𝐸 is aBanach left 𝐴-module if 𝐸 is a left 𝐴 −module, and  ‖𝑎. 𝑥‖ ≤ ‖𝑎‖‖𝑥‖(𝑎 ∈ 𝐴, 𝑥 ∈𝐸)[1] .Recall that a submoduleN of an 𝑅 −module 𝑀 is said to be stable, if 𝑓(𝑁) ⊆ 𝑁 for each 𝑅 −homomorphism𝑓: 𝑁 → 𝑀. In case each submodule of it is stable ,𝑀 is called a fully stable module [2],throughout this paper we introduce the concept of  full stability for modules . A Banachalgebra module 𝑀 is called fully stable Banach 𝐴-module if for every submodule 𝑁 of 𝑀 and for each multiplier 𝜃: 𝑁 ⟶ 𝑀  such that 𝜃(𝑁) ⊆ 𝑁.Structure of fully stableBanach A- module in term of their elements is considered see(2.5) Studying Baer criterion gives another characterization  of fully stable Banach𝐴-modulein proposition (2. 8)  1. Preliminaries In this section the fundamental basic concepts and primitive results are given. Definition (1.1) [2]  A submoduleN of an 𝑅 −module 𝑀 is said to be stable, if 𝑓(𝑁) ⊆ 𝑁 for each 𝑅 −homomorphism 𝑓: 𝑁 ⟶ 𝑀,In case each submodule of it is stable , 𝑀 is called a fully stable module. Examples (1.2) [2] a) The 𝑍 −module 𝑍 of all integers is not fully stable .For, define 𝜃: 2𝑍 ⟶ 𝑍by𝜃(2𝑛) = 3𝑛 for each 𝑛 ∈ 𝑍 Clearly ,𝜃  is a 𝑍 −homomomrphism .But 𝜃(2𝑍) ⊈2Z  b) Let 𝑀 be an𝑅 −module .For any ideal 𝐼 of 𝑅  , 𝑎𝑛𝑛𝑀(𝐼) = {𝑚 ∈ 𝑀| 𝐼𝑚 = (0)}  . Is astable submodule of𝑀   In fact, for any 𝑅 −homomorphism 𝑓: 𝑎𝑛𝑛𝑀(𝐼) ⟶ 𝑀 And each 𝑚 ∈ 𝑎𝑛𝑛𝑀(𝐼). 𝐼𝑚 = (0), 𝐼𝑓(𝑚) = 𝑓(𝐼𝑚) = 𝑓((0)) = (0) Hence  𝑓(𝑚) ∈ 𝑎𝑛𝑛𝑀(𝐼).  Thus 𝑎𝑛𝑛𝑀(𝐼)  is stable submodule. Recall that , a submodule𝑁 of an 𝑅 −module 𝑀 is said to be annihilator if   𝑁 = 𝑎𝑛𝑛𝑀(𝐼)   for some ideal 𝐼of 𝑅. As in (𝑏) by the a bove an 𝑅 −module , in which all its submodules are annihilators is fully stable.  Following remark  show us ,it issufficies to consider stability over a very restricted class of submodules.   Mathematical Theory and Modeling                                                                                                                                                  www.iiste.org ISSN 2224-5804 (Paper)    ISSN 2225-0522 (Online) Vol.6, No.1, 2016   137 [2]  Remark (1.3) Let 𝑀 be an𝑅 −module. If every cyclic submodule of 𝑀 is stable then 𝑀 is fully stable module. Proposition (1.4) [2]  An 𝑅 −module 𝑀 is fully stable if and only if for each 𝑥, 𝑦in 𝑀, 𝑦∉(x)  implies 𝑎𝑛𝑛𝑅(𝑥)  ⊈    𝑎𝑛𝑛𝑅(𝑦). Corollary (1.5) [2] Let 𝑀 be a fully stable 𝑅 −module .Then for each 𝑥, 𝑦in 𝑀,          𝑎𝑛𝑛𝑅(𝑦) = 𝑎𝑛𝑛𝑅(𝑥) implies (𝑥) = (𝑦). Definition (1.6) [2] Let 𝑀  be an 𝑅 −module, and  𝑁be  anysubmodule of 𝑀. We say that 𝑁satisfies Baer criterion if for every 𝑅 −homomorphism 𝑓: 𝑁 ⟶ 𝑀, there exists an element 𝑟 ∈ 𝑅 such that 𝑓(𝑛) = 𝑟𝑛  for each 𝑛 ∈ 𝑁An 𝑅 −module 𝑀 is said to be satisfy Baer criterion if each submodule of 𝑀 satisfies Baer criterion ,that is for every submodule 𝑁 of 𝑀 and 𝑅 −homomorphism 𝑓: 𝑁 ⟶ 𝑀, there exists an element 𝑟 in 𝑅 such that  𝑓(𝑛) = 𝑟𝑛  for each 𝑛 ∈ 𝑁   .   Note :-From above definition notice that every  module which satisfies Baer criterion is fully stable. In thefollowing proposition and its corollary obtain another characterization of fully stable modules. Proposition (1.7) [2] Let 𝑀 be anR −module .Then Baer criterion holds for cyclic submodules of M if and only if 𝑎𝑛𝑛𝑀(𝑎𝑛𝑛𝑅(𝑥)) =(𝑥)  𝑓𝑜𝑟 𝑒𝑎𝑐ℎ  𝑥 ∈ 𝑀. Corollary (1.8) [2] An𝑅 −module𝑀is fully stable if and onlyif  𝑎𝑛𝑛𝑀(𝑎𝑛𝑛𝑅(𝑥)) = (𝑥)  𝑓𝑜𝑟 𝑒𝑎𝑐ℎ  𝑥 𝑖𝑛 𝑀. 2. Main results We see that we need first to define follows Definition (2.1) [3] A map from a left Banach𝐴 −module 𝑋 in to a left Banach 𝐴 −module 𝑌 (A is not necessarily commutative ) is said a multiplier if it satisfies 𝑇(𝑎. 𝑥) = 𝑎. 𝑇𝑥for all 𝑎 ∈ 𝐴, 𝑥 ∈ 𝑋 . Definition (2.2) [4]  For a nonempty subset 𝑀 in a left Banach 𝐴 −module 𝑋, the annihilater  𝑎𝑛𝑛𝐴(𝑀) of 𝑀 is 𝑎𝑛𝑛𝐴(𝑀) = {𝑎 ∈𝐴 ; 𝑎. 𝑥 = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝑀} .   Definition (2.3) [5] Aleft Banach𝐴 −module 𝑋 is called 𝑛 −generated for 𝑛 ∈ 𝑁  if there exists𝑥1, … , 𝑥𝑛 ∈ 𝑋such  thateach 𝑥 ∈ 𝑋  can represented as𝑥 = ∑ 𝑎𝑘 . 𝑥𝑘𝑘=1 for some     𝑎1, … , 𝑎𝑛 ∈ 𝐴   .A cyclic module is just a 1-generated one. Now,we start by introducing the concept of stability forBanach 𝐴 −module. Definition (2. 4) Let 𝑋be Banach𝐴 − module ,  𝑋 is called fully stableBanach𝐴 −module if for every submodule 𝑁 of 𝑋 and for each multiplier    𝜃: 𝑁 ⟶ 𝑋   such that 𝜃(𝑁)  ⊆   𝑁. Mathematical Theory and Modeling                                                                                                                                                  www.iiste.org ISSN 2224-5804 (Paper)    ISSN 2225-0522 (Online) Vol.6, No.1, 2016   138 In the following proposition  wediscuss another characterization of fully stable modules .  Notations:- Let 𝑋 a Banach 𝐴 −module  1)𝑁𝑥 =  {𝑛𝑥|  𝑛 ∈ 𝑁,   𝑥 ∈ 𝑋} 𝐾𝑦  = {𝑘𝑦| 𝑘 ∈ 𝐾,    𝑦 ∈ 𝑋} 2)  𝑎𝑛𝑛𝐴𝑁𝑥  = {𝑎 ∈ 𝐴 , 𝑎. 𝑛𝑥 = 0,   ∀  𝑛𝑥 ∈ 𝑁𝑥} 𝑎𝑛𝑛𝐴𝐾𝑦  = {𝑎 ∈ 𝐴,   𝑎. 𝑘𝑦 = 0,   ∀𝑘𝑦 ∈ 𝐾𝑦}  Proposition (2.5) 𝑋is fully stable Banach 𝐴 −module if and only if for each 𝑥, 𝑦 ∈ 𝑋 And𝑁𝑥, 𝐾𝑦subsetsof 𝑋, 𝑦 ∉  𝑁𝑥 implies 𝑎𝑛𝑛𝐴(𝑁𝑥) ⊈ 𝑎𝑛𝑛𝐴(𝐾𝑦). Proof :- Suppose that 𝑋 is fully stable Banach 𝐴 −module there exists 𝑥, 𝑦 ∈ 𝑋 such that 𝑦 ∉ 𝑁𝑥  and 𝑎𝑛𝑛𝐴(𝑁𝑥) ⊆𝑎𝑛𝑛𝐴(𝐾𝑦) Define   𝜃: < 𝑁𝑥 >⟶ 𝑋     by   𝜃(𝑎. 𝑛𝑥) = 𝑎. 𝑘𝑦,for all 𝑎 ∈ 𝐴 if𝑎. 𝑛𝑥 = 0     then  𝑎 ∈ 𝑎𝑛𝑛𝐴(𝑁𝑥) ⊆ 𝑎𝑛𝑛𝐴(𝐾𝑦) This implies that  𝑎. 𝑘𝑦 = 0,hence𝜃  is well define , clear 𝜃 is a multiplier ,because 𝑋 is fully stable , there exsits an element  𝑡 ∈ 𝐴such that𝜃(𝑚𝑥) = 𝑡𝑚𝑥      𝑓𝑜𝑟 𝑒𝑎𝑐ℎ  𝑚𝑥 ∈ 𝑁𝑥 In particular,𝑘𝑦 = 𝜃(𝑛𝑥) = 𝑡𝑛𝑥 ∈ 𝑁𝑥 Which is a contradiction Conversely, assume that there is a subset𝑁𝑥  of 𝑋  and a multiplier 𝜃: < 𝑁𝑥 >⟶ 𝑋   such that 𝜃(𝑁𝑥) ⊈ 𝑁𝑥  then there exists an element 𝑚𝑥 ∈ 𝑁𝑥    such that 𝜃(𝑚𝑥) ∉ 𝑁𝑥  . Let  𝑠 ∈ 𝑎𝑛𝑛𝐴(𝑁𝑥)therefor  𝑠𝑛𝑥 = 0  , 𝑠𝜃(𝑚𝑥) =𝜃(𝑠𝑚𝑥) = 𝜃(𝑠𝑡𝑛𝑥) = 𝜃(𝑡𝑠𝑛𝑥)   = 𝜃(0) = 0. Hence 𝑎𝑛𝑛𝐴(𝑁𝑥) ⊆ 𝑎𝑛𝑛𝐴(𝜃(𝑚𝑥))   , which is a contradiction.∎  Corollary (2. 6)  Let 𝑋 be a fully stable Banach 𝐴 −module .Then for each 𝑥, 𝑦  in 𝑋, 𝑎𝑛𝑛𝐴(𝐾𝑦) = 𝑎𝑛𝑛𝐴(𝑁𝑥)      implies 𝑁𝑥 = 𝐾𝑦      Proof:- Assume that there are two elements 𝑥, 𝑦 in 𝑋  such that 𝑎𝑛𝑛𝐴(𝑁𝑥) = 𝑎𝑛𝑛𝐴(𝐾𝑦)   and 𝑁𝑥 ≠ 𝐾𝑦 Then without loss of generality there is an element 𝑧𝑥  in 𝑁𝑥  not in 𝐾𝑦  .By proposition (2.5)we have 𝑎𝑛𝑛𝐴(𝐾𝑦) ⊈ 𝑎𝑛𝑛𝐴(𝑍𝑥)  but 𝑎𝑛𝑛𝐴(𝑁𝑥) ⊆ 𝑎𝑛𝑛𝐴(𝑍𝑥),hence,𝑎𝑛𝑛𝐴(𝐾𝑦) ⊈ 𝑎𝑛𝑛𝐴(𝑁𝑥)   which is a contradiction ∎ Mathematical Theory and Modeling                                                                                                                                                  www.iiste.org ISSN 2224-5804 (Paper)    ISSN 2225-0522 (Online) Vol.6, No.1, 2016   139 Definition (2. 7) A Banach𝐴 −module 𝑋 is said to satisfy Baer criterion if eachsubmoduleof  𝑋 satisfies Baer criterion ,that is for every submodule 𝑁 of 𝑋 and 𝐴 − multiplier𝜃: 𝑁 ⟶ 𝑋     ,there exists an element 𝑎 in 𝐴 such that 𝜃(𝑛) = 𝑎𝑛for all 𝑛 ∈ 𝑁. In the following proposition and its corollary another characterization of fully stable Banach𝐴 −module is given. Proposition (2.8)  Let 𝑋 be a Banach 𝐴 −module .Then Baer criterion holds for cyclic submodules of 𝑋 if and only if 𝑎𝑛𝑛𝑋(𝑎𝑛𝑛𝐴(𝑁𝑥)) =  𝑁𝑥   𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑥 ∈ 𝑋. Proof :- Assume that Baer criterion holds. Let  𝑦 ∈ 𝑎𝑛𝑛𝑋(𝑎𝑛𝑛𝐴(𝑁𝑥)) Define   𝜃: < 𝑁𝑥 >⟶ 𝑋     by   𝜃(𝑎. 𝑛𝑥) = 𝑎. 𝑘𝑦         ,for all 𝑎 ∈ 𝐴 Let 𝑎1. 𝑛𝑥 = 𝑎2. 𝑛𝑥 (𝑎1 − 𝑎2)𝑛𝑥 = 0  ,       𝑎1 − 𝑎2 ∈ 𝑎𝑛𝑛𝐴(𝑁𝑥) ( 𝑎1 − 𝑎2) ∈ 𝑎𝑛𝑛𝐴(𝐾𝑦) ⟶       (𝑎1 − 𝑎2)𝑘𝑦 = 0       𝑎1𝑘𝑦 = 𝑎2𝑘𝑦 hence ,𝜃  is well define. It is clear that clear 𝜃   is an𝐴 − multiplier. By the assumption, there exists an element 𝑡 ∈ 𝐴   such that  𝜃(𝑚𝑥) = 𝑡𝑚𝑥      𝑓𝑜𝑟 𝑒𝑎𝑐ℎ  𝑚𝑥 ∈ 𝑁𝑥 This implies that   .In particular,𝑘𝑦 = 𝜃(𝑛𝑥) = 𝑡𝑛𝑥 ∈ 𝑁𝑥𝑎𝑛𝑛𝑋(𝑎𝑛𝑛𝐴(𝑁𝑥)) ⊆  𝑁𝑥  ;   ℎ𝑒𝑛𝑐𝑒 𝑎𝑛𝑛𝑋(𝑎𝑛𝑛𝐴(𝑁𝑥)) = 𝑁𝑥 Conversely, assume that  𝑎𝑛𝑛𝑋(𝑎𝑛𝑛𝐴(𝑁𝑥)) =  𝑁𝑥 For each  𝑁𝑥 ⊆ 𝑋.Then for each 𝐴 − multiplier𝜃: 𝑁𝑥 → 𝑋 And s ∈ 𝑎𝑛𝑛𝐴(𝑁𝑥),we have 𝑠𝜃(𝑛𝑥) = 𝜃(𝑠𝑛𝑥) = 0 Thus  𝜃(𝑛𝑥) ∈ 𝑎𝑛𝑛𝑋(𝑎𝑛𝑛𝐴(𝑁𝑥)) =  𝑁𝑥    ,  then 𝜃(𝑛𝑥) =  𝑡 𝑛𝑥  for some 𝑡 ∈ 𝐴,thus Baer criterion is holds. Corollary (2.9) 𝑋isfully stable Banach𝐴  −module if and only if    𝑎𝑛𝑛𝑋(𝑎𝑛𝑛𝐴(𝑁𝑥)) =  𝑁𝑥   𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑥 ∈ 𝑋. References [1]  Matthew David Peter Daws , Banach algebras of operators  The University of Leeds, School of Mathematics, Department of Pure Mathematics , (2004). [2]M.S.ABBAS,On Fully Stable Modules , Ph. D ,Thesis , University of Baghdad, Iraq,(1990). [3]JANKO  BRAˇCIˇC,Simple Multipliers onBanachModules .University of Ljubljana,Slovenia,GlasgowMathematical Journal Trust ,(2003). [4]J. BRAˇCIˇC ,Local Operators onBanach Modules, University of Ljubljana, Slovenia,Mathematical Proceedings of the Royal Irish Academy,(2004). [5]Antonio M.Cegarra, Projective Covers of Finitely  Generated Banach Modules and the Structure of Some Banach Algebras, O.Yu.Aristov,Russia,(2006). 

Fully Stable Banach Algebra Module

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Fully Stable Banach Algebra Module

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