Let $R$ be an integral domain,$M$ is a finitely generated $R$ module. Prove that rank$(M)=r$ iff $M$ has a free submodule $N \equiv R^r$ , such that $M/N$ torsion . If $R$ is a PID then $N$ may be chosen so that $0\to N\to M\to M/N\to 0$ splits.
If rank$(M)=r $ then there exists a maximal independent set of elements of $M$ say $S=\{m_1,m_2,...,m_r\}$ take the free module $R^r$ and an isomorphism $\phi: R^r \to \tilde S$ defined by $$\phi(e_i)=m_i,$$ where $\tilde S$ denote the submodule generated by $S$. Take $N=\tilde S$ if $M/N$ is not torsion there will be contradiction in maximality of $S$. So , one side is done .
How to show the converse part and the case where $R$ is a PID?