Let $R$ be a commutative ring with unity , $M$ be a finitely generated $R$-module with a submoduke $N$ such that $M/N$ is a free module , then off-course $M/N$ is a free module of finite rank (as $M/N$ is finitely generated) i.e. $\exists$ unique $ n\in \mathbb N$ such that $R^n \cong M/N$ ,; my question is :
Is it true that $N$ is finitely generated ?
Yep. Consider the short exact sequence
$$ 0 \rightarrow N \rightarrow M \rightarrow M/N \rightarrow 0. $$
Since $M/N$ is free, we can find $\hat{p} \colon M/N \rightarrow M$ such that $p \circ \hat{p} = \operatorname{id}_{M/N}$ (if $M/N$ is generated freely by $[m_i]$ then define $\hat{p}([m_i]) = m_i$). By the splitting lemma, we can find also $\hat{i} \colon M \rightarrow N$ such that $\hat{i} \circ i = \operatorname{id}_N$ where $i \colon N \rightarrow M$ is the inclusion. This implies that $\hat{i}$ is onto and so $N$ is a quotient of a finitely-generated module, hence finitely-generated.