When is a hypersurface in a quasi-polynomial ring finite dimensional?

Question

$\newcommand\la{\langle}\newcommand\ra{\rangle}\newcommand\laa{\langle\!\langle}\newcommand\raa{\rangle\!\rangle}$ Assume in the following that every polynomials is a sum of monomials of degree at least two.

For $K$ an algebraically closed(this is probably needed?) field and $f_i$ polynomials, the algebra $A=K[x_i]/(f_i)$ has finite vector space dimension iff the $f_i$ have finitely many solutions and then the vector space dimension of $A$ is equal to the number of solutions.

In particular, because of this result algebras of the form $K[x_i]/(f)$ for a single polynomial $f$ should never be finite dimensional when we have at least two variables.

Let $K\la x_i\ra$ be the non-commutative polynomial ring in $n$ variables $x_i$ and define $K\laa x_i\raa:=K\la x_i\ra/ \la x_i x_j+x_j x_i \ra$ for $i \neq j$ to be the quasi-polynomial ring (or is there already a (better) name for this in the literature?).

So instead of the commutativity relations $x_i x_j=x_j x_i$ we have $x_i x_j =- x_j x_i$.

Interestingly we can have polynomials $f$ such that $K\laa x_i\raa/(f)$ is finite dimensional. Here is an example class of such polynomials. Let $f_{n,a_i}:=\sum\limits_{r=1}^{n-1}{x_r^{a_r}}-x_n^{a_n}$ where $a_i \geq 2$. So for $n=3$ those polynomials are the Fermat polynomials of the form $x_1^{a_1}+x_2^{a_2}-x_3^{a_3}$. Let $A_{n,a_i}:=K\laa x_i\raa/(f_{n,a_i})$.

Question 1: What is the $K$-dimension of $A_{n,a_i}$ depending on the parameters $a_i$ and $n$? When is this algebra finite dimensional?

For $n=3$ the algebra is for example finite dimensional for $a_1=3,a_2=3,a_3=2$ with dimension 26 and infinite dimensional for $a_1=2,a_2=2,a_3=3$. Does the vector space dimension have a number theoretic meaning in case it is finite?

Question 2: For which polynomials $f$ is a quotient of the quasi-polynomial ring $K\laa x_i\raa/(f)$ finite dimensional and what is the $K$-dimension?

Is there a nice interpretation in terms of properties or even certain zeros of $f$?

Answer 1

Let me assume that $\newcommand\la{\langle}\newcommand\ra{\rangle}\newcommand\laa{\langle\!\langle}\newcommand\raa{\rangle\!\rangle} K$ is algebraically closed (this is not necessary, but it means I can work instead with the more symmetric polynomial $f_{n,a_i}=\displaystyle\sum_{r=1}^n(x_r)^{a_r}.$) If $K$ is of characteristic two then the ring $K\laa x_i\raa$ coincides with the usual polynomial algebra, and so $K\laa x_i\raa/(f_{n,a_i})$ is never trivial unless $n\leq 1.$ So let's assume also that $K$ is of characteristic not equal to two.

Then I claim that the algebra $K\laa x_i\raa/(f_{n,a_i})$ is finite-dimensional if and only if at most one of the $a_i$ is even.

First let me show that if two or more of the $a_i$ are even then $K\laa x_i\raa/(f_{n,a_i})$ is infinite-dimensional. WLOG, assume $a_1$ and $a_2$ are even. Then we have a surjective map $K\laa x_i\raa/(f_{n,a_i})\rightarrow K\laa x_1,x_2\raa/(x_1^{a_1}+x_2^{a_2})$ sending all the other $x_i$ to zero. So it suffices to show that this latter algebra is infinite-dimensional.

Now note that $K[x_1^2,x_2^2]$ lies inside the center of $K\laa x_i\raa$. Therefore, $K\laa x_1,x_2\raa/(x_1^{a_1}+x_2^{a_2})$ contains $K[x_1^2,x_2^2]/(x_1^{a_1}+x_2^{a_2})$ as a subalgebra and hence is infinite-dimensional, as desired.

The other direction is a little more involved; we split into two cases. First we assume that none of the $a_i$ are even.

Then, consider the anticommutators $\{x_i,f_{n,a_i}\}:=x_if_{n,a_i}+f_{n,a_i}x_i.$ These are necessarily zero in $K\laa x_i\raa/(f_{n,a_i})$. We compute $\{x_i,f_{n,a_i}\}=2x_i^{a_i+1},$ so as $K$ is not of characteristic two, we see that $x_i^{a_i+1}$ lies in the kernel of $K\laa x_i\raa\rightarrow K\laa x_i\raa/(f_{n,a_i})$. This is sufficient to see that $K\laa x_i\raa/(f_{n,a_i})$ is finite-dimensional.

Now, assume that $a_1$ is even, but the other $a_i$ are odd. We proceed similarly to the previous case. The anticommutator $\{x_1,f_{n,a_i}\}$ is still equal to $2x_1^{a_1+1}$, so we conclude that $x_1^{a_1+1}$ is trivial inside $K\laa x_i\raa/(f_{n,a_i}).$ But the other anticommutators are a little more complicated now - we have $\{x_i,f_{n,a_i}\}=2x_i^{a_i+1}+2x_ix_1^{a_1}$. To deal with this we use a trick: Note that this divides $x_i^{2(a_i+1)}-x_i^2x_1^{2a_i}.$ As we deduced earlier that $x_1^{a_1+1}$ is sent to zero, this means that $x_i^{2(a_i+1)}$ is also sent to zero. Thus, $K\laa x_i\raa/(f_{n,a_i})$ is finite-dimensional, as desired.