Herstein Topics In Algebra Solutions Chapter 6 Pdf [RECOMMENDED]

Exercise 6.1: Let $M$ be a module over a ring $R$. Show that $M$ is a direct sum of cyclic modules.

Solution: Let $m \in M$. Consider the set $Rm = {rm \mid r \in R}$. This is a submodule of $M$, and $M$ is a direct sum of these submodules. herstein topics in algebra solutions chapter 6 pdf

Solution: Suppose $A$ is simple. Let $I$ be an ideal of $A$. Then $I$ is a submodule of $A$, and since $A$ is simple, $I = 0$ or $I = A$. Exercise 6