### On the direct sum of dual-square-free modules

Y. Ibrahim, M. Yousif

#### 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].

#### Keywords

square-free and dual-square-free modules, abelian and quasi-duo rings

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