On the direct sum of dual-square-free modules

IF 0.3 Q4 MATHEMATICS, APPLIED Algebra & Discrete Mathematics Pub Date : 2022-01-01 DOI:10.12958/adm1807
Yasser Ibrahim, M. Yousif
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引用次数: 1

Abstract

A module M is called square-free if it contains nonon-zero is omorphic submodules A and B with A∩B= 0. Dually, Mis called dual-square-free if M has no proper submodules A and B with M=A+B and M/A∼=M/B. In this paper we show that if M=⊕i∈I Mi, then M is square-free iff each Mi is square-free and Mj and ⊕j=i∈I Mi are orthogonal. Dually, if M=⊕ni=1Mi, then M is dual-square-free iff each Mi is dual-square-free, 1⩽i⩽n, and Mj and ⊕ni=jMi are factor-orthogonal. Moreover, in the in finite case, weshow that if M=⊕i∈ISi is a direct sum of non-is omorphic simple modules, then M is a dual-square-free. In particular, if M=A⊕B where A is dual-square-free and B=⊕i∈ISi is a direct sum ofnon-isomorphic simple modules, then M is dual-square-free iff A and B are factor-orthogonal; this extends an earlier result by theauthors in [2, Proposition 2.8].
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关于双无平方模的直接和
如果模块M包含A∩B= 0的非零的同构子模块A和B,则称为无平方模块。如果M没有固有子模块A和B,且M=A+B和M/A ~ =M/B,则称为双平方自由。本文证明了如果M=⊕i∈i Mi,则M是无平方的,如果每个Mi都是无平方的,且Mj与⊕j=i∈i Mi是正交的。对偶地,如果M=⊕ni=1Mi,则M是双平方自由的(如果每个Mi都是双平方自由的,1≥i≤n),并且Mj和⊕ni=jMi是因子正交的。此外,在有限情况下,证明如果M=⊕i∈ISi是非同构简单模的直接和,则M是双平方自由的。特别地,如果M=A⊕B,其中A是双平方自由的,且B=⊕i∈ISi是非同构简单模的直接和,则如果A与B是因子正交的,则M是双平方自由的;这扩展了作者在[2,命题2.8]中的早期结果。
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来源期刊
Algebra & Discrete Mathematics
Algebra & Discrete Mathematics MATHEMATICS, APPLIED-
CiteScore
0.50
自引率
0.00%
发文量
11
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