Timeline for Finite subgroups of GL_n of polynomial rings over finite fields
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2016 at 19:34 | vote | accept | Jorge António | ||
| Dec 20, 2016 at 19:34 | history | edited | Jorge António | CC BY-SA 3.0 |
deleted 203 characters in body
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| Dec 20, 2016 at 13:56 | comment | added | YCor | @MatthiasWendt yes thanks, I forgot to assume $n\ge 3$ for finite generation. | |
| Dec 20, 2016 at 10:23 | comment | added | Matthias Wendt | @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). | |
| Dec 20, 2016 at 10:19 | answer | added | Matthias Wendt | timeline score: 11 | |
| Dec 20, 2016 at 4:05 | comment | added | YCor | 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 $GL_n(Z/pZ(T_1,\dots,T_n))$). | |
| Dec 20, 2016 at 1:41 | comment | added | Will Sawin | @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. | |
| Dec 20, 2016 at 1:17 | comment | added | user94041 | 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? | |
| Dec 20, 2016 at 1:03 | history | asked | Jorge António | CC BY-SA 3.0 |