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?

  • $\begingroup$ Yes, you can say that. $\endgroup$ – user26857 Jan 2 at 9:31

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.