Finitely generated R-module Let $R$ be an integral domain and also Noetherian ring. Let $f$ be an element in Quot$R$ (field of fractions of $R$). Let $R[f]$ be the subring of Quot$R$ generated by $R$ and $ \{ f \}$. Let suppose that $R[f] \subseteq M$ where $M$ is an $R$-submodule of Quot$R$ and $M$ is finitely generated as $R$-module. Is it true that $R[f]$ is also finitely generated as $R$-module?

If is it true, can you give me a rigorous and possibly elementary proof?
The following conditions are equivalent for the commutative ring $R$:


$R$ satisfies the ascending condition on ideals
every ideal of $R$ is finitely generated
every submodule of a finitely generated submodule is finitely generated


If $R$ satisfies one (hence all) of these properties, it is said to be Noetherian.
Here is a quick outline at a pretty basic level. First, a finitely generated module over $R$ is also Noetherian (acc on modules). You can do that by induction on the number of generators and the fact (pretty easy to verify) that if $M$ is an $R$-module, $M'$ a submodule of $M$, then $M$ is noetherian if and only if $M'$ and $M''$ are and take $M'$ to be the submodule generated by one element of your generating set for $M$). Now use this same fact applied to $R[f]\subseteq M$.