Question: I want to show that $S_0$ being Noetherian does not imply that $S$ is Noetherian.
Thoughts: Here, a ring $S$ is graded if $S=S_0\oplus S_1\oplus\dots\oplus S_n\cdots$ of the additive abelian subgroups $S_n$ and if multiplication satisfies $S_iS_j\subseteq S_{i+j}$ for all $i,j\geq0$. We let $S_+=S_1\oplus\cdots$. Now, I know that I can make the above question true if we also assume that $S_+$ is a finitely generated ideal of $S$. So for the proof of my question, I am sort of stuck on whether or not it is best to just try and prove directly, or if I should start with assuming that $S_+$ is not finitely generated and show that it is true then. Any help would be greatly appreciated! Thank you.
Let $S=k[x_1,\ldots, x_n,\ldots]$ denote the infinite polynomial ring over a field $k$ (infinitely many transcendental variables $x_i$). Then there is a grading by degree, and we see that $S_0=k$, which is of course Noetherian while the ring is not. Indeed, take the ascending chain $$ (x_1)\subsetneq (x_1,x_2)\subsetneq \cdots \subsetneq (x_1,\ldots, x_n)\subsetneq \cdots $$ which does not stabilize.
Edit: Undeleted since Atticus did not write up his answer in the meanwhile.