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Let $R=R_0\oplus R_1\oplus R_2\oplus\cdots$ be a graded ring. If $R$ is Noetherian, then I can see that $R_0$ is Noetherian and $R_1\oplus R_2\oplus\cdots$ is a finitely generated ideal of $R.$

But, is it also true that, $R$ is also a finitely generated $R_0$-module?

Edit: As mentioned by Alekos, it's not true. Then, can we say that $R$ is a finitely generated $R_0$-algebra?

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  • $\begingroup$ Yes, you can say that. $\endgroup$ – user26857 Jan 2 at 9:31
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What if we take the graded ring $\Bbb{C}[x]=R$? $R$ is certainly Noetherian, and here $\Bbb{C}=R_0$ is indeed Noetherian. But, as we know $\Bbb{C}[x]$ is not a finitely generated $\Bbb{C}-$vector space. You can soup this example up to $k[x_1,\ldots, x_n]$ or you can use more exotic ground rings etc.

As for your second question, the answer is yes and is explained in detail in the first two lemmas of https://stacks.math.columbia.edu/tag/00JV.

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