$\DeclareMathOperator\Comm{Comm}\DeclareMathOperator\Id{Id}$Consider the variety $\Comm$ of commuting matrices $[A,B]=0$ over some field $K$. It is much studied, and interesting for various reasons.

One has an obvious *free* action of $K^2$: $A\to A+a\Id$, $B\to B+b\Id$ on it.
Hence the number of $F_q$ points is divisible by $q^2$. However actually
it is divisible by a higher power of $q$: for $2\times2$ matrices by $q^3$, for $3\times3$ by $q^5$, etc, so:

**Question** Is there some *free* action of $K^3$ on $\Comm$, or any other geometric explanation for the divisibility above?

Similar higher divisibility seems to hold true for triples, $n$-tuples of commutting matrices, so we have similar questions.

**Remark 1**
Number of $F_q$ points of $\Comm$ has been calculated W.Feit, N. Fine, Pairs of commuting matrices over a finite field, 1960.
For each matrix size $n$ it is given by polynomial in $q$ with integer coefficicents.

E.g., for $2\times2$ matrices, it is $ q^3(q^3+q^2-q)$.

For $3\times3$, it is $q^5(q^7+q^2(q^2-1)(q^3-1)/(q-1) + (q^2-1)(q^3-1) ) $

For general $n$, one has a sum over partions of $n$, with the summand corresponding to the partition $1^{b_1}2^{b_2}\cdots$ being $[n]_q!/ \prod_i [b_i]_q! q^{some~power}$. The leading term $q^{n^2+n}$ comes from partion $1^n$, and the "least term" comes from partion $n^1$ (see Feit-Fine for details).

**Remark 2**
In general count equivalence even to $K^n$ does not imply
algebraic equivalence as discussed here: MO300946, MO301249.
Though in that particular case there might exist some geometric reason.

**Remark 3**
If my notes are correct, the commuting-triples count for $n=2$ is $q^4(q^4+q^3+q^2-q-1)$, and for quadruples is $q^5(q^5+q^4+q^3-q-1)$.
There should be nice generating functions for commuting tuples:
MO271752, MO272045.

One may also observe for any $n$, for $n$-tuples of commuting matrices multiplication by $K^*$ acts freely , except of one point - all matrices are zero, and hence number of $F_q$ points $N$ $(N-1)$ is divisible by $(q-1)$ - for all $n$-tuples. There are also some $Z/2Z$ actions comming from $(A,B)->(B,A)$ and similar, which are free on certain easy to describe part of a scheme.

somegeom explanation, even if it's not in the "show it's a torsor for a suitable vector group" direction. $\endgroup$ – EBz Jan 13 at 11:19