Solution: Let $m \in M$. Consider the set $Rm = {rm \mid r \in R}$. This is a submodule of $M$, and $M$ is a direct sum of these submodules.
Exercise 6.5: Let $A$ be an algebra over a field $F$. Show that $A$ is a simple algebra if and only if $A$ has no nontrivial ideals. herstein topics in algebra solutions chapter 6 pdf
Exercise 6.1: Let $M$ be a module over a ring $R$. Show that $M$ is a direct sum of cyclic modules. Solution: Let $m \in M$
You can download the PDF solution manual for Chapter 6 of "Topics in Algebra" by Herstein from the following link: [insert link] herstein topics in algebra solutions chapter 6 pdf