Let $M$ be $R$-module, where $R$ is commutative ring with $1,$ and $N$ be submodule of $M.$ If $M$ is free of finite rank, so is $N \ ?$
Answer: False. Let $R=M=\mathbb{Z}_{6}$ and $N=2\mathbb{Z}_6.$ Clearly, $M$ is free of finite rank. Since $|N|\neq |M^k|, $ we have $\ N \not\cong M^k, \ \forall k \in \mathbb{Z}_{>0},$
May I know if my proof is correct? Thank you.