# Is a quotient of a free algebra noetherian or artinian?

This is a tutorial exercise for my Modules and Representation theory course.

Let $k$ be a field and $k\langle x,y \rangle$ denote the free algebra over $k$ with two generators. Let $q\in k^*$.

Is the ring $$k\langle x, y \rangle / (yx - qxy),$$ noetherian or artinian?

I know that $k\langle x, y \rangle$ itself is neither artinian nor noetherian. I also know that in the case of $q=1$, the ring is isomorphic to the polynomial ring and so is in that case it is noetherian (by Hilbert's basis theorem) but not artinian, since $(*)$ $I_n = (x^n)$ for $n\geq 0$ is a non stabilising descending chain.

However in the case where $q\neq 1$, I'm not sure how to proceed. Can I use the same chain $(*)$ to show that it's not artinian? And as for noetherian, I have no idea how to proceed.

Thanks in advance for any help.

About Artinianity, you may use the same chain of ideals to show that it cannot be Artinian. Let $A=k\langle x,y\rangle/(yx-qxy)$ and $I_n=(x^n)$ as you did, then to say that it stabilizes means that there exists $n$ such that $x^n\in(x^{n+1})$. However, $A$ is a graded ring and $(x^{n+1})\subseteq A^{\geq n+1}$ while $x^n\in A^n$. By comparing the degrees we reach a contradiction.