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Given a hereditary torsion theory $\tau =(𝕋,𝔽)$ in Mod-$R$, a module $M$ is called $\tau$-supplemented if every submodule $A$ of $M$ contains a direct summand $C$ of $M$ with $A/C$ $\tau -$torsion. A submodule $V$ of $M$ is called $\tau$-supplement of $U$ in $M$ if $U+V=M$ and $U\cap V\le \tau \left(V\right)$ and $M$ is $\tau$-weakly supplemented if every submodule of $M$ has a $\tau$-supplement in $M$. Let $M$ be a $\tau$-weakly supplemented module. Then $M$ has a decomposition $M=M_1\oplus M_2$ where $M_1$ is a semisimple module and $M_2$ is a module with $\tau \left(M_2\right){\le }_e{M}_2$. Also, it is shown that; any finite sum of $\tau$-weakly supplemented...

Let $R$ be a ring. A right $R$-module $M$ is said to be retractable if $𝕋{Hom}_R(M,N)\ne 0$ whenever $N$ is a non-zero submodule of $M$. The goal of this article is to investigate a ring $R$ for which every right R-module is retractable. Such a ring will be called right mod-retractable. We proved that $\left(1\right)$ The ring ${\prod }_{i\in ℐ}{R}_i$ is right mod-retractable if and only if each $R_i$ is a right mod-retractable ring for each $i\in ℐ$, where $ℐ$ is an arbitrary finite set. $\left(2\right)$ If $R\left[x\right]$ is a mod-retractable ring then $R$ is a mod-retractable ring.