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](https://math.stackexchange.com/questions/2819286/is-every-left-graded-noetherian-graded-ring-left-noetherian) and got no replies at all, not sure at all where it belongs.



Is every (left) graded-Noetherian graded ring (left) Noetherian?
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](https://doi.org/10.1007/s13366-013-0138-6).


