# How to show the following ring is not Noetherian

$$R \subset \mathbb{Q}[x]$$ be the subring consisting of polynomials $$f = a_0 + a_1 x+\cdots+a_n x^n$$ s.t. $$a_0\in \mathbb{Z}$$. Show that $$R$$ is not a Noetherian ring.

Given a subgroup $$A\subset \mathbb{Q}$$ (Abelian group under addition), consider the subset $$I\subset R$$ consisting of polynomials $$a_1x+\cdots+a_nx^n$$ s.t. $$a_1\in A$$.

My attempt: I've shown $$I$$ is an ideal and to construct an ascending chain of ideals, does this work?: Fix $$a_1\in A$$, $$ \subset\subset....$$
The main consideration is coefficient of $$x\in A$$ which, I think, the chain above addresses and intuitively this chain does not stabilize but can I give a more rigorous statement why it doesn't? Thanks.

Indeed, $$I$$ can easily be finitely generated. If $$A=\Bbb{Z}$$ for example, then $$I$$ is generated by $$x$$, so $$I$$ is principal, so we wouldn't expect an ascending chain of ideals in $$I$$.
Instead, let $$I_A$$ denote the ideal associated to the subgroup $$A$$.
Then choose $$A_1\subsetneq A_2\subsetneq \cdots \subsetneq A_n \subsetneq \cdots$$ a strictly ascending chain of subgroups of $$\Bbb{Q}$$. Perhaps $$A_n = \frac{1}{2^n}\Bbb{Z}$$. Then note that the ideals $$I_{A_n}$$ give a strictly ascending chain of ideals in $$R$$. Thus $$R$$ is not Noetherian.
• I see, I understand that $A_n = \frac{1}{2^n}\mathbb{Z}$ is a strictly ascending chain of subgroups of $\mathbb{Q}$ but am not sure what the elements of $I_A$ would look like and how we are guaranteed existence of an ideal associated to the subgroup $A$? – manifolded Mar 18 at 20:33
• @manifolded That's what the ideal $I$ is in the question. It's the ideal associated to the subgroup $A$ from the question. – jgon Mar 18 at 20:35