Timeline for Order of unipotent matrices over $\mathbb{Z}/q\mathbb{Z}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2018 at 12:49 | comment | added | Will Sawin | This answer is not right even over the field $\mathbb F_p$. You mean $p ^ { \lceil \log_p n \rceil }$. | |
| Mar 2, 2018 at 11:50 | comment | added | David Wehlau | whoops I misread. I was indeed thinking of the field of order q. | |
| Mar 2, 2018 at 11:44 | comment | added | Peter Heinig | FWIW: there is quite some literature on normal forms for matrices over commutative reduced rings, but $\mathbb{Z}/p^k\mathbb{Z}$ for $k>1$ is not reduced. (So, as a suggestion to the OP of how to perhaps tackle a problem of intermediate difficulty and with more literature to rely upon: unipotent upper-triangular matrices over reduced non-field rings, such as e.g. $\mathbb{Z}/p_1p_2\dotsm p_s\mathbb{Z}$, with $p_i$ the $i$-th prime might be worth studying.) | |
| Mar 2, 2018 at 11:34 | comment | added | Neil Strickland | I think that this is a good and correct answer for the field $\mathbb{F}_q$ of order $q$, but the question was about $\mathbb{Z}/q\mathbb{Z}$, which is of course different. | |
| Mar 2, 2018 at 11:29 | comment | added | user05811 | Note that when $n=2$, the group of $2\times 2$ upper-triangular unipotent matrices over $\mathbb{Z}/q\mathbb{Z}$ is cyclic of order $q$. | |
| Mar 2, 2018 at 11:20 | comment | added | Peter Heinig | Re "answer is $p^{\lceil n/p\rceil}$": this proposed upper bound cannot be true, in particular because it is independent of the prime power. An admissible counterexample is given by $p:=2$, $n:=2$, $q:=2^2=4$, while taking the unipotent upper-triangular matrix to be $M=\begin{matrix} 1&1\\ 0 &1\end{matrix}$. Then $M^2=\begin{matrix} 1&2 \\ 0 &1\end{matrix}$ and $M^3=\begin{matrix} 1&4 \\ 0 &1\end{matrix} =_{\mathbb{Z}/4\mathbb{Z}} =\begin{matrix} 1&0 \\ 0 &1\end{matrix}$, hence the order of $M$ is $3 > 2 = 2^{\lceil 2/2\rceil} = p^{\lceil n/p\rceil}$. | |
| Mar 2, 2018 at 11:13 | comment | added | LSpice | To second @PeterHeinig's question, in what sense can you guarantee that you can conjugate a matrix over $\mathbb Z/q\mathbb Z$ so that it not only is upper triangular, but has its entries in $\mathbb Z/p\mathbb Z$? For that matter, what do you mean by $\mathbb Z/p\mathbb Z$? Since it's implicitly inside $\mathbb Z/q\mathbb Z$, I guess you mean $q'\mathbb Z/q\mathbb Z$ (with $q' = q/p$), but that seems to be quite a strong statement. | |
| Mar 2, 2018 at 10:56 | comment | added | Peter Heinig | What is a reference for 'Jordan canonical form' over rings? Please note that $\mathbb{Z}/p^k\mathbb{Z}$ is not a field if $k>1$. | |
| Mar 2, 2018 at 10:52 | history | answered | David Wehlau | CC BY-SA 3.0 |