1
$\begingroup$

I have been able to show that for a graded subring $S$ of $R[x]$ (where $R$ is a noetherian domain) that in order to show that $S$ is noetherian, it suffices to consider homogeneous ideals of $S$. This proof was very similar to that of Hilbert's basis theorem.

I'm now considering graded subrings in $R[x,y]$ and I would like to prove a similar result to the one above, but this time that it suffices to consider homogeneous ideals in this new $\mathbb{N}^2$ grading. Does anyone have any suggestions? I'm hoping it will follow by an induction but I don't see how at the moment.

I should mention that the grading I'm using on $R[x]$ is the standard one with $R$ in degree $0$ and $x$ in degree $1$. For $R[x,y]$ I would use $R$ in degree $(0,0)$, $x$ in degree $(1,0)$ and $y$ in degree $(0,1)$ with $x \prec y$.

$\endgroup$
0
$\begingroup$

This follows from a general result about coarsening of graded noetherianity.

Let $G$ be an abelian group, and let $R$ be a $G$-graded ring. Then, the following statements are equivalent: (i) $G$ is of finite type; (ii) If $\psi\colon G\rightarrow H$ is an epimorphism of abelian groups and the $G$-graded ring $R$ is noetherian, then the $H$-graded ring $R_{[\psi]}$ is noetherian.

Here, a noetherianity of a graded ring means that every homogeneous ideal is of finite type. Moreover, the coarsening $R_{[\psi]}$ of a $G$-graded ring $R=\bigoplus_{g\in G}R_g$ with respect to an epimorphism of abelian groups $\psi\colon G\rightarrow H$ is the $H$-graded ring whose underlying ring is the ring underlying $R$ and whose graded component of degree $h$ is $\bigoplus_{g\in\psi^{-1}(h)}R_g$ for every $h\in H$.

For a proof, see Theorem 1.1 in this article and Remark 2.14 in this article.

| cite | improve this answer | |
$\endgroup$

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.