# Is a Noetherian graded ring also a finitely generated $R_0$-module?

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?

• Yes, you can say that. – 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.