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.

I asked this question in MSE a few days ago and got no replies at all, not sure at all where it belongs.

• Graded in what? an arbitrary magma? a monoid? an abelian group? The group of integers $\mathbf{Z}$? The monoid $\mathbf{N}$ of nonnegative integers? I also guess that rings are assumed associative and unital. – YCor Jun 17 '18 at 12:27
• $\mathbb{Z}$-graded, and ofcourse associative and unital. – Anonymous Coward Jun 17 '18 at 12:43

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.
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