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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 May 2026

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.