Let $R$ be a commutative ring with unity, and $S$ be a multiplicatively closed set. Consider a $R$-module $M$ and a submodule $N\subset S^{-1}M$ (where $S^{-1}M$ is considered as a module over $S^{-1}R$).
I managed to show that there exists a submodule $N'$ of $M$ such that $S^{-1}N'=N$. Taking the set of numerators of $N$ satisfies these conditions.
I got stuck trying to prove that, if $N$ is finitely generated, then we can choose $N'$ to be finitely generated. Is this claim even true?
We may assume without loss of generality that $N$ has a generating set of the form $\{m_1/1,\ldots,m_n/1\}$, but I can't seem to prove that the set $\{m_1,\ldots,m_n\}$ generates a submodule $N'$ such that $S^{-1}N'=N$.
Let $N$ be generated by elements $m_i/1$. Let $N'$ be the submodule of $M$ generated by the $m_i$. Then $S^{-1} N'$ is a submodule of $S^{-1} M$ generated by the $m_i/1$ and hence is equal to $N$.