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.


The ring you are considering is a particular case of a quantum symmetric algebra. It is also known under the name quantum plane and it is always Noetherian (see Brown and Goodearl's Lectures on Algebraic Quantum Groups, from §I.1.14 on. In particular, Theorem I.1.17).

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.


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.