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If $R$ is a graded ring then $R$ is Noetherian if and only if $R_0$ is Noetherian and $R$ is finitely generated as an $R_0$-algebra.

If $R$ is Noetherian then $R_0 = R/R_+$ is Noetherian and $R_+ $ is finitely generated, but for the converse I don't know what to do.

Maybe I need to use the following: A subset of homogeneous elements generates $R$ as an algebra if and only if they generate $R_+$ as an ideal of $R$, but I don't know how.

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If $R$ is finitely generated as an $R_0$-algebra, then there exists an $n\geq 0$ and a surjection $$ R_0[x_1,\dots, x_n]\to R. $$ $R_0[x_1,\dots, x_n]$ is a polynomial algebra over a noetherian ring, and hence noetherian, which implies that $R$ is noetherian, as it is a quotient of a noetherian ring.

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  • $\begingroup$ does the surjection exist because $R$ is a f.g $R_0$-algebra? $\endgroup$ – allizdog Dec 7 '16 at 5:18
  • $\begingroup$ @allizdog Yes, $A$ is a finitely generated $R$-algebra (essentially by definition) if there is a surjection $R[x_1,\dots, x_n]\to A$ for some $n$. $\endgroup$ – Stahl Dec 7 '16 at 5:19

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