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The Dual Of Summand-square-free Modules

Posted on:2020-06-03Degree:MasterType:Thesis
Country:ChinaCandidate:S Z DingFull Text:PDF
GTID:2370330596474240Subject:Basic mathematics
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Square-free module is derived from the study of automorphism-invariant module which is invariant under automorphisms of its injective envelope.Recent years,more and more scholars have focused on the automorphism-invariant module and its generalizations.In 2013,Singh and Srivastava extended the structure of automorphism-invariant module:if let M be an automorphism-invariant module,then M has a decomposition M = A(?)B,where A is quasi-injective,B is square-free.In 2015,Mazurek and Nielsen promoted the square-free module to the summand-square-free module.In 2017,Tutuncii and Kikumasa introduced a new concept of dual-square-free module.On the basis of these discussions,This paper mainly discusses the following condition of the module M,if whenever its factor module is isomorphic to N2 = N(?)N for N,which is the direct summand of M,then N= 0.we call the module as the dual of summand-square-free module(referred to as DSSF module).This paper has three parts.in the first chapter,we put forward the background,devel-opment and research ideas of this paper.Character 2 mainly discussed the property of the dual of summand-square-free module.Note that any factor module of a DSSF module is not necessary DSSF.In the process of exploring the properties of DSSF module,we proved that the equivalent description of it:under the condition of(D2),M has no proper direct summand A and B with M= A+B and M/A(?)M/B if and only if M is dual-summand-square-free,and showed various examples of DSSF module in this section.And then we proved that being DSSF is a Morita invariant property of modules,and pointed out if R is a commutative ring,right DSSF ring is not a Morita invariant property.In general,a ring R which is square free as a right R-module,need not be square free as a left R-module.Then we proved that for a semilocal ring,R is right DSSF if and only if R is left DSSF.When M is a supplemented DSSF module,if M is dual automorphism-invariant,then M has(C3).Finally we put M ?H1(?)H2(?)…(?)Hn,with H1,H2,...,Hn are hollow modules,if M is DSSF module,then it is H-Supplemented.In the third chapter,we mainly discussed the rings whose DSSF modules are closed under submodules or essential extensions.and found that such a ring is closely related to a semiperfect ring,we showed that every one-sided ideal of a right DSSF semiperfect ring is two-sided.but in general7,a one-sided ideal of a right DSSF semiperfect ring need not be two-sided.(see corollary 3.8)Since any nonzero module over a right perfect ring has a maximal submodule,under the Lemma 3.6,we obtained that a DSSF module over a right perfect ring is cyclic.Next we proved that if R is right perfect ring and E(RR)is DSSF,then E(RR)=RR.In the end we proved two important theorems:Theorem 3.10 Let R be a right(or left)perfect ring and let e1,e2,…,en be a basic set of primitive idempotents.Then the following conditions are equivalent:(a)any right ideal of R is DSSF;(b)R is basic and every submodule of any DSSF R-module is DSSF;(c)(1)R has a ring decomposition R=?in=1 eiR7 where eiR=ei is right Artin right uniserial;(2)for every i?{1,2,...,n?}e eiR/eiJ/eiJ2(?)…(?)eiJki-1/eiJki(?)eiJki-1= Soc(eiR).where ki is a natural number with eiJki-1?0 and eiJki =0.Theorem 3.13 Let R be a right perfect ring and let e1,e2,...,en bea basic set of rimitive idempotents.Then the following conditions are equivalent:(a)the injective hull of any factor module of RR is DSSF;(b)R is basic and every essential extension of any DSSF module is DSSF;(c)R is basic serial QF-ring,for everr i??1,2,…,n?,eiR/eiJ(?)eiJ/eiJ2(?)…(?)eiJki-1/eiJki(?)Soc(eiR).here ki is a natural number with eiJki-1?0 and eiJki=0.
Keywords/Search Tags:automorphism-invariant modules, square-free modules, semiperfect rings, semilocal rings, uniserial modules
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