Suppose $S$ is a positively graded ring, and $S$ is generated by $S_1$ (the degree $1$ part of $S$). If we know $\operatorname{Proj}(S)$ is Noetherian, can we deduce $S$ is a Noetherian ring? Could you give a proof or give a counterexample?
Any help is appreciated. Thanks!
No, this is false. Let $S$ be the graded ring given by $S_0=\Bbb Z$ and $S_d=\Bbb C[x]_d$ for $d>0$. Then $\operatorname{Proj} S\cong \operatorname{Proj} \Bbb C[x]$ (ref), so $\operatorname{Proj} S\cong \operatorname{Spec} \Bbb C$, a point. But $S_+$ is not finitely generated for cardinality reasons: any finite collection of generators of $S_+$ give at most countably many elements of $S_+\cap S_1$, and $S_1$ is uncountable.
