Lemma 10.58.2. A graded ring $S$ is Noetherian if and only if $S_0$ is Noetherian and $S_{+}$ is finitely generated as an ideal of $S$.
Proof. It is clear that if $S$ is Noetherian then $S_0 = S/S_{+}$ is Noetherian and $S_{+}$ is finitely generated. Conversely, assume $S_0$ is Noetherian and $S_{+}$ finitely generated as an ideal of $S$. Pick generators $S_{+} = (f_1, \ldots , f_ n)$. By decomposing the $f_ i$ into homogeneous pieces we may assume each $f_ i$ is homogeneous. By Lemma 10.58.1 we see that $S_0[X_1, \ldots X_ n] \to S$ sending $X_ i$ to $f_ i$ is surjective. Thus $S$ is Noetherian by Lemma 10.31.1. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: