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$\newcommand\la{\langle}\newcommand\ra{\rangle}$Let $K$ be a field and $K\la x,y,z\ra$ the non-commutative polynomial ring in 3 variables.

Question 1: Are there three (fewer is probably not possible?!) polynomials $f,g,h \in K\la x,y,z\ra$, which are sums of monomials of degree at least two, such that the algebra $K\la x,y,z\ra/(f,g,h)$ is finite dimensional?

Is there a systematic way to construct such polynomials?

For two variables we can take for example $f=xy+yx$ and $g=x^3+y^3$, but I do not know (or forgot) such examples for more than 2 variables.

(If possible we should also have (this is equivalent to $(f,g,h)$ being an admissible ideal) that there exists an $n \geq 2$ such that $J^n \subseteq (f,g,h)$ where $J=\la x,y,z\ra$ is the ideal generated by $x,y,z$.)

Question 2: Let $f_i$ for $i=1,\dotsc,n-1$ be polynomials spanning an admissible ideal in the non-commutative polynomial ring in $n-1$ variables $x_i$. Can we find $g$ such that $\la f_i,g\ra$ is an admissible ideal in $K\la x_i,y\ra$?

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    $\begingroup$ Shouldn't this be impossible already for the commutative polynomial rings, because of some sort of dimension reasoning? $\endgroup$
    – darij grinberg
    Commented May 31, 2021 at 17:26
  • $\begingroup$ @darijgrinberg Oh, I think you are right. If this ring is finite dimensional(for just two polynomials f,g, instead of three f,g,h) then also the commutative version is finite dimensional but has Krull dimension at least 1 so cant befinite dimensional as being finite dimensional is equivalent to Krull dimension 0. (or do I have a thinking error?) I added one more polynomial, since at the moment I also do not know an example with three polynomials. $\endgroup$
    – Mare
    Commented May 31, 2021 at 17:36
  • $\begingroup$ Maybe I am missing something, but what's wrong with $K[x,y,z]/(x^2, y^2,z^2)$ in the commutative case? $\endgroup$
    – M.G.
    Commented May 31, 2021 at 17:39
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    $\begingroup$ @M.G. Darij's comment is about $K<x,y,z>/(f,g)$ with just two instead of three polynomials. $\endgroup$
    – Mare
    Commented May 31, 2021 at 17:39
  • $\begingroup$ TeX note: \DeclareMathOperator imparts the spacing of, well, an operator to its argument. Note the difference between $\DeclareMathOperator\la{\langle}\DeclareMathOperator\ra{\rangle}K\la x, y, z\ra$ \DeclareMathOperator\la{\langle}\DeclareMathOperator\ra{\rangle}K\la x, y, z\ra and just $K\langle x, y, z\rangle$ K\langle x, y, z\rangle. I have edited accordingly. (But thanks for using \langle\rangle at all!) $\endgroup$
    – LSpice
    Commented May 31, 2021 at 19:36

1 Answer 1

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Q1. Yes, such 3 monomials exist. Moreover, the generic homogeneous quadratic polynomials $f,g,h$ are such that the dimension of the quotient algebra $K⟨x,y,z⟩/(f,g,h)$ is 28. This follows, in particular, from Th.1.3 of [Natalia Iyudu and Stanislav Shkarin. The Golod-Shafarevich inequality for Hilbert series of quadratic algebras and the Anick conjecture].

Q2. In contrast, such an ideal does not exist if $n\ge 5$. If follows from the Vinberg version of the Golod--Shafarevich theorem that if an ideal $I$ is generated by $r$ elements (which are linear combinations of at least quadratic monomials) in a free associative algebra $F$ with $g$ generators, then the quotient algebra is infinite-dimensional provided that the polynomial $1-gz+rz^2$ has a positive root (cf, for example, Prop. 2.7 in [M.Ershov, Golod-Shafarevich groups: a survey]). In your case we have $r=g=n-1$, so that the quotient algebra is infinite if $n\ge 5$.

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