TITLE: Is every (left) graded-Noetherian graded ring (left) Noetherian?
QUESTION [6 upvotes]: I call a $\mathbb{Z}$-graded (non-commutative, associative, unital) ring $A$ (left) graded-Noetherian if every homogeneous (left) ideal is finitely generated, and (left) Noetherian if it is (left) Noetherian as a ring.
In the commutative setting, I think I can prove that a graded-Noetherian ring is Noetherian. This basically follows from Hilbert's basis theorem (graded-Noetherianity suffices to show that $A_0$ is Noetherian and that $A$ is finitely generated over $A_0$).
Since Hilbert's basis theorem fails noncommutatively the same line of reasoning will not work in the noncommutative setting.
Q: Is every (left) graded-Noetherian graded ring (left) Noetherian?
I asked this question in MSE a few days ago and got no replies at all, not sure at all where it belongs.

REPLY [6 votes]: The answer is yes, by Corollary 2.2 in C. Nastasescu, F. Van Oystaeyen, Graded rings with finiteness conditions II, Comm. Algebra 13 (1985), 605-618.
More generally, we have the following.
Let $G$ be a commutative group. Every epimorphism $\psi\colon G\twoheadrightarrow H$ of commutative groups gives rise to the $\psi$-coarsening functor $\bullet_{[\psi]}$ from the category of $G$-graded rings to the category of $H$-graded rings. We say that $\psi$-coarsening preserves noetherianness if whenever $R$ is a $G$-graded ring that is noetherian as a $G$-graded ring, then the $H$-graded ring $R_{[\psi]}$ is noetherian as an $H$-graded ring. Now, it follows from the aforementioned result by Nastasescu and Van Oystaeyen (as well as by a result independently proven two years earlier by Goto and Yamagishi):

$\psi$-coarsening preserves noetherianness for every epimorphism $\psi$ whose source is $G$ if and only if $G$ is of finite type.

For details on the yoga of coarsening cf. this article.