# Algebra with three anti-commutator relations

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 $$0$$.

• 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.
– YCor
Oct 3 at 22:22
• 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 $uvw\neq -1$).
– YCor
Oct 4 at 7:35

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".
• 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. Sep 30 at 14:53
• 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. Sep 30 at 15:23
• 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). Sep 30 at 16:40