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Let $u,v,w \in \mathbb{F}_p^{\times}$. Consider the $\mathbb{F}_p$-algebra $V$ generated by $ a,b,c$ and the relations $$u a^2 = bc + cb$$ $$v b^2 = ac + ca$$ $$w c^2 = ab + ba$$ Is $V$ generated by the $a^i b^j c^k$ with $i,j,k \geq 0$ as an $\mathbb{F}_p$-module?

It is enough to show that $b a^i b^j$ and $c a^i b^j$ lie in the span. I tried induction on $i+j$, but I am going in circles. Even for $i+j=2$ already. Maybe it is not true actually? The corresponding question for $2$ generators has a positive answer, though.

Also, does $V$ have a name, or does $V$ belong to a more well-known class of algebras?

Background: I want to understand the nilpotent elements of $A \otimes_{\mathbb{F}_p} \mathbb{F}_{p^3}$, where $A$ is a certain non-commutative $\mathbb{F}_p$-algebra. When there is an irreducible binomial $T^3 - \lambda \in \mathbb{F}_p[T]$ (this is the case when $p \equiv 1 \bmod 3$), we get equations of the type above for any element $a+bT+cT^2$ of $A \otimes_{\mathbb{F}_p} \mathbb{F}_{p^3} = A[T]/\langle T^3 - \lambda \rangle$ that squares to $0$.

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    $\begingroup$ In degree 3, the answer is "yes" if and only if $uvw\neq -1$ (if $uvw=-1$ then $bcb$ is not a linear combination as required). I did the computation by hand. $\endgroup$
    – YCor
    Commented Oct 3, 2023 at 22:22
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    $\begingroup$ In degree 4, the answer seems to be "no" for all values of $u,v,w$ (well, only $uvw$ matters; a diagonal change reduces to $u=v=1$). More precisely, in degree 4, the subspace generated by the 15 "ordered monomials" seems to have codimension 3 (for $uvw\neq -1$) and it misses, for example, $b^2ab$ (while it contains $bc^2b$ if $uvw\neq -1$). $\endgroup$
    – YCor
    Commented Oct 4, 2023 at 7:35

2 Answers 2

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I ran a quick test on magma, asking it the dimension of the algebra defined by your relations with (chosen pretty much at random) $p=5$, $u=2$, $v=3$, $w=4$, and with the extra relations $a^3=0$, $b^3=0$, $c^n=0$, and let $n$ run from $3$ to $30$. The answer was consistent in each case with the statement that the dimension of the resulting algebra is $20n-24$, and therefore inconsistent with your statement, which predicts dimension at most $9n$. So the answer is "no".

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  • $\begingroup$ If you let $u=v=1$ and $w=-1$, then using the same code, you get these same dimensions $20n-24$, seemingly independent of $p$. Mathjax insists on interpreting the star symbol, so I can't copy and paste the code here, but it's simple enough to do yourself. $\endgroup$
    – Dave Benson
    Commented Sep 30, 2023 at 14:53
  • $\begingroup$ And then, if you change it to $a^3=0$, $b^4=0$, $c^4=0$, magma doesn't finish in a reasonable time when $p=5$, which suggests it might be infinite dimensional. $\endgroup$
    – Dave Benson
    Commented Sep 30, 2023 at 15:23
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    $\begingroup$ Thanks a lot! Unfortunately, I don't have access to Magma. I checked the website, apparently I cannot just download or pay for it. I tried to work on these questions with SageMath, but apparently it doesn't really work or I am missing something (for example, in SageMath, $\mathbb{Q}[x]/\langle 1 \rangle$ is not the zero ring, and I don't know why). $\endgroup$
    – Martin Brandenburg
    Commented Sep 30, 2023 at 16:40
  • $\begingroup$ Well, Gap is free and does quite a lot of what Magma does. It's worth trying. $\endgroup$
    – Dave Benson
    Commented Sep 30, 2023 at 22:37
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    $\begingroup$ @MartinBrandenburg There is a magma online calculator that can do some simple computations: magma.maths.usyd.edu.au/calc Also the GAP package QPA is good to enter those finite dimensional algebras as in this answer. $\endgroup$
    – Mare
    Commented Oct 4, 2023 at 5:03
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Regarding the second question "Also, does $V$ have a name, or does $V$ belong to a more well-known class of algebras?": If $u=v=w$, then $V$ is a $3$-dimensional Sklyanin algebra. These are the $K$-algebras with generators $x, y, z$ and relations

$$pyz + qzy + rxx = 0,$$ $$pzx + qxz + ryy = 0,$$ $$pxy + qyx + rzz = 0,$$

where $K$ is a field and $p,q,r$ are elements of $K$. As far as I know, it is an open question when such an algebra has a linear basis consisting of the words $x^iy^jz^k$.

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    $\begingroup$ One can rescale $x,y,z$, so whether $V$ can be transformed into a Sklyanin algebra depend on it's orbit in $\mathbb{F}_p^{\times}/\mathbb{F}_p^{\times 3}$ $\endgroup$
    – Peter Wu
    Commented Mar 13 at 5:30
  • $\begingroup$ So over an algebraic closure (and over a cubic extension) it is always a Sklyanin algebra. In general, it's a twisted form of a Sklyanin algebra. $\endgroup$
    – YCor
    Commented Mar 13 at 9:48
  • $\begingroup$ Moreover, $V$ is always a generalized Sklyanin algebra as defined in Section 7 of the paper Natalia Iyudu, Stanislav Shkarin, Three dimensional Sklyanin algebras and Gröbner bases, Journal of Algebra, Volume 470, 2017, Pages 379-419. $\endgroup$
    – Ralle
    Commented Mar 14 at 3:52

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