Is this generalization of Hilbert's basis theorem already known (and is it even true)?

Question

Hilbert's basis theorem states that if $R$ is a commutative noetherian ring (i.e. every ideal $I\subseteq R$ is finitely generated) then $R[x]$ is noetherian as well.

I immediately thought that this could be proven using the fact that $R[x]/(x)\cong R$, where $(x)$ is the ideal generated by the polynomial $x$. Indeed I am pretty sure that I came up with a proof for it. The main part is just showing that for any (commutative) ring $S$ and ideals $I\subseteq S$ and $J\subseteq S$, the quotient ideal $I/J\subseteq S/J$ is not finitely generated when $I$ is not finitely generated and $J$ is finitely generated (where $I/J$ is the image of $I$ under the quotient map $q:S\rightarrow S/J$).

Then it follows easily that a ring $R$ is noetherian if there is a finitely generated ideal $J\subseteq R$ s.t. $R/J$ is noetherian. (If $R$ had a non-finitely generated ideal $I\subseteq R$ then $I/J$ would also not be finitely generated which is a contradiction).

This generalizes Hilbert's basis theorem, since $R[x]/(x)\cong R$ and $(x)$ is obviously finitely generated.

My question is if this generalization can even be true or if I have made a bad mistake and if it is true wether this is already known. I can not imagine that it is both true and not known, since the proof was pretty easy, but I couldn't find anything on the internet and you never know.

Answer 1

Let $R \subset \mathbb{Q}[x]$ be the subring made with the $P \in \mathbb{Q}[x]$ such that $P(0) \in \mathbb{Z}$.

It’s easy to see that $2R=\{P \in \mathbb{Q}[x],\, P(0) \in 2\mathbb{Z}\}$. So $R/2R \cong \mathbb{F}_2$ is a Noetherian ring, but (for instance) the sequence of ideals (of $R$) $(2^{-n}x)$ is increasing and not stationary.

Answer 2

the problem with your proof is, that I/J only makes sense if J is a subset of I.

So you can only prove that every ideal in R[x] which is contained in (x) is finitely generated.