On the direct sum of dual-square-free modules

Abstract

A module \(M\) is called square-free if it contains no non-zero isomorphic submodules \(A\) and \(B\) with \(A \cap B =0\). Dually, \(M\) is called dual-square-free if \(M\) has no proper submodules \(A\) and \(B\) with \(M =A +B\) and \(M/A \cong M/B\). In this paper we show that if \(M = \oplus _{i \in I}M_{i}\), then \(M\) is square-free iff each \(M_{i}\) is square-free and \(M_{j}\) and \( \oplus _{j \neq i \in I}M_{i}\) are orthogonal. Dually, if \(M  = \oplus _{i =1}^{n}M_{i}\), then \(M\) is dual-square-free iff each \(M_{i}\) is dual-square-free, \(1 \leqslant i \leqslant n\), and \(M_{j}\) and \( \oplus _{i \neq j}^{n}M_{i}\) are factor-orthogonal. Moreover, in the infinite case, we show that if \(M = \oplus _{i \in I}S_{i}\) is a direct sum of non-isomorphic simple modules, then \(M\) is a dual-square-free. In particular, if \(M =A \oplus B\) where \(A\) is dual-square-free and \(B = \oplus _{i \in I}S_{i}\) is a direct sum of non-isomorphic simple modules, then \(M\) is dual-square-free iff \(A\) and \(B\) are factor-orthogonal; this extends an earlier result by the authors in [2, Proposition 2.8].