I am surprised that I didn't find a reference for the following question.

Q: Is there any characterization of the finite subgroups of $GL_n( \mathbb{F}_p [T_1, \dots, T_n])$? Can we do so more generally for $GL_n(\mathbb{Z}/ p^n \mathbb{Z} \ [T_1, \dots, T_n] )$.

Edit: I had previously state an incorrect statement. Thank you for the clarifying answers.

  • 3
    $\begingroup$ Are you sure that the Jordan canonical form is applicable in this setting? The ring $\mathbb F_p[T_1, \ldots, T_n]$ is not a field, so I am not sure that the proof of the J.c.f. goes through. As a specific example, the matrix $\begin{bmatrix} 1 & 1 \\ T & T+1 \end{bmatrix}$ is invertible since it has determinant one, but its eigenvalues are not in your ring... so what would its Jordan canonical form be like? $\endgroup$
    – user94041
    Dec 20, 2016 at 1:17
  • 3
    $\begingroup$ @user94041 Indeed, it is is easy to check that this element has infinite order. It satisfies a polynomial equation $X^2 - (T+2) X +1 =0$, and the ring $\mathbb F_p[X,T]/(X^2-(T+2) X +1)$ is isomorphic to $\mathbb F_p[X,X^{-1}]$ with $T = X + X^{-1}- 2$, so $X$ has infinite order. $\endgroup$
    – Will Sawin
    Dec 20, 2016 at 1:41
  • 7
    $\begingroup$ The kernel of $GL_n(Z/p^kZ[T_1,\dots,T_n])\to GL_n(Z/pZ[T_1,\dots,T_n])$ is a locally finite nilpotent $p$-group of finite exponent, so $GL_n(Z/pZ[T_1,\dots,T_n])$ is really the interesting part. This is a finitely generated group, and has a finite index subgroup in which there is no element of prime order $\neq p$. It might be true that all its finite subgroups of order coprime to $p$ (which are thus of bounded order) are conjugate to subgroups of $GL_n(Z/pZ)$ (at least within $GL_n(Z/pZ(T_1,\dots,T_n))$). $\endgroup$
    – YCor
    Dec 20, 2016 at 4:05
  • 1
    $\begingroup$ @YCor Your description is correct. I just want to note that $GL_2(\mathbb{F}_q[T_1,\dots,T_n])$ is not finitely generated. Already for $n=1$, one needs elementary matrices $e_{12}(T^k)$ for all $k$ and there is no commutator formula to generate higher powers from $e_{12}(T)$ (by Nagao's theorem). Finite generation is ok for $GL_n$ with $n\geq 3$ (but $GL_3$ is not finitely presented, by Behr's theorem). $\endgroup$ Dec 20, 2016 at 10:23
  • $\begingroup$ @MatthiasWendt yes thanks, I forgot to assume $n\ge 3$ for finite generation. $\endgroup$
    – YCor
    Dec 20, 2016 at 13:56

1 Answer 1


Well, polynomial rings are very complicated. I assume the question is more generally about $GL_n(\mathbb{F}_q[T_1,\dots,T_m])$.

The case of one variable can be deduced from the following paper:

  • C. Soulé: Chevalley groups over polynomial rings. In: Homological group theory (Proc. Sympos. Durham) London Math. Soc. Lecture Notes Series 36 (1979), pp. 359-367.

More details of proof can be found in

  • B. Margaux. The structure of the group G(k[t]): variations on a theme of Soulé. Algebra Number Theory 3 (2009), 393-409.

These results compute the fundamental domain for the action of $GL_n(k[T])$ on the associated Bruhat-Tits building. Any finite subgroup will have a fixed point on the building, so it will appear as subgroup of a stabilizer of the $GL_n(\mathbb{F}_q[T])$-action on the building. The stabilizers are explicitly determined, they are semidirect products $U\rtimes L$ where $L$ is a Levi subgroup of a parabolic in $GL_n(\mathbb{F}_q)$ and $U$ is a subgroup of the upper-triangular matrices with $\mathbb{F}_q[T]$-coefficients (in particular the latter is a $p$-group). Probably everything you want to know about finite subgroups and their conjugacy classification can be deduced from that.

In the case $m>1$, several variables, a lot less is known. One can still use Soulé's theorem, applied to $GL_n(\mathbb{F}_q(T_1,\dots,T_{m-1})[T_m])$. Any finite subgroup will have a fixed point on the building and therefore be conjugate to a subgroup of one of the standard stabilizers described above (but of course for the base field $k=\mathbb{F}_q(T_1,\dots,T_{m-1})$). However, the conjugacy involved will be by matrices in $GL_n(\mathbb{F}_q(T_1,\dots,T_{m-1})[T_m])$ which may not be what you are interested in. Also, there may be some finite subgroups which only appear after some denominators are introduced. Anyway, this confirms the description in YCor's comment.

For $m>1$, I don't know of a comprehensible description of a space on which $GL_n(\mathbb{F}_q[T_1,\dots,T_m])$ would act with finite stabilizers and with identifiable fundamental domain. This is already terribly complicated for $GL_2(\mathbb{F}_q[T_1,T_2])$, see

  • S. Krstić and J. McCool. Free quotients of $SL_2(R[X])$. Proc. Amer. Math. Soc. 125 (1997), 1585-1588. (paper can be found here)

A consequence of their result is that $SL_2(\mathbb{F}_q[T_1,T_2])$ surjects onto a free group of infinite rank. This provides lots of elements of infinite order and probably also tells us that the conjugacy classification of finite subgroups is very complicated. On the other hand, these results also tell us that finite subgroups don't say that much about the structure of $GL_n(\mathbb{F}_q[T_1,\dots,T_m])$ when $m>1$.


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