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

$$\newcommand\la{\langle}\newcommand\ra{\rangle}\newcommand\laa{\langle\!\langle}\newcommand\raa{\rangle\!\rangle}$$ 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 $$f$$?

• @YCor thanks for the edit, but $K<x_i>$ and $K\la x_i\ra$ are different, thats why I used different notation.
– Mare
Jun 1, 2021 at 15:12
• Oops, are you sure you don't wan't a less confusing notation for the skew-symmetric quotient?
– YCor
Jun 1, 2021 at 15:18
• @YCor My choice was not the best but I dont know an official notation. If there is a better notation, I dont mind any edit (but of course it should not have the same name as the non-commutative polynomial ring).
– Mare
Jun 1, 2021 at 15:19
• In any case the inequality signs $<,>$ imply an inadequate spacing. Possibility: $K\langle\!\langle x_i\rangle\!\rangle$?
– YCor
Jun 1, 2021 at 15:19
• @YCor Looks nice but my feeling is that I have seen this before for something else (like Laurant non-commutative polynomials?). But I dont mind using this since no other meaning is important for this thread.
– Mare
Jun 1, 2021 at 15:20

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.

• Thanks, surprisingly for the relations $xy+2yx,x^2+y^4$ we get a finite dimensional algebra over the rationals.
– Mare
Jun 1, 2021 at 20:00
• @Mare I suspect that in general, imposing a relation like $xy+2yx$ instead of $xy+yx$ will make it much easier to find finite dimensional algebras. (The reason being that the quotient has trivial center, unlike the case of your post.)
– dhy
Jun 1, 2021 at 20:11