# 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 $R$:

1. $R$ satisfies the ascending condition on ideals
2. every ideal of $R$ is finitely generated
3. 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.

• I didn't know that the 3. statement is equivalent to 1. and 2. I will try to prove this equivalence. Meanwhile, i was tinking: 1) math.stackexchange.com/questions/533873/… 2) So $M$ is Noetherian 3) Noetherian is closed by submodules, so $R[f]$ is Noetherian as $R$-module 4) Noetherian modules are always finitely generated math.stackexchange.com/questions/2331871/… Are they true? Nov 11, 2017 at 19:00
• In this case i have proved the "strong" fact that $R [f]$ is noetherian as $R$ - module, hence also finitelty generated. But actually sems that if R is Noetherian, then for an R module it is the same to be noetherian or finitely generated, so it isn't a really strong fact. Nov 11, 2017 at 19:17
• @Minato Yes: when you have proved that 2 implies 3, you have that $R[f]$ is also noetherian. The hint is to prove the fact first for $R^n$. Nov 11, 2017 at 20:15

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 $R[f]\subseteq M$.