$R \subset \mathbb{Q}[x]$ 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 $a_1\in A$, $<a_1x> \subset<a_1x,x^2>\subset….$
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.
Best Answer
It's unclear what your chain is, or how you expect it to continue, but note that as a graded ideal, your first ideal is $\newcommand\ZZ{\Bbb{Z}}0\oplus a_1\ZZ \oplus \newcommand\QQ{\Bbb{Q}}\QQ\oplus \QQ\oplus \cdots$ already, since we can multiply by $\frac{q}{a_1}x^i$ for $q\in\Bbb{Q}$, $i\ge 1$. Thus your first ideal equals your second ideal.
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.