Torsion elements of $SL_3(\mathbb{F}_p[x])$? (Quick question)

Question

Is every element of $SL_3(\mathbb{F}_p[x])$ a torsion element?

Here are my thoughts:

First of all, the group is noncommutative, so a torsion element is an element of finite order.

I'm thinking of examples of torsion elements: Any generator (since a generator p times is the identity.) Any element that has a rotation matrix as one of its factors. And there are a bunch more examples.

Then I can't think of any element that does not have torsion in this group.

Could someone please refer me to a result that makes this more precise? Your help will be very much appreciated!

Answer 1

There are elements of infinite order in this group.

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Consider the matrix $$ A=\left(\begin{array}{cc}1&x\\ x&1+x^2\end{array}\right)\in SL_2(\Bbb{F}_p[x]). $$ I claim that the order of $A$ is infinite.

Many ways to see this. See the comments by KCd below this post for a very compact argument. Another explicit proof goes as follows.

Let $K$ be an algebraic closure of the prime field, $K=\overline{\Bbb{F}_p}$. The characteristic polynomial of $A$ is $$ \chi_A(T)=T^2-(2+x^2)T+1. $$ This has no zeros in $K$. For if $\lambda$ is a root of $\chi_A(T)$, then $1/\lambda$ is the other. But their sum $\lambda+1/\lambda=2+x^2$ is transcendental over the prime field. Hence the eigenvalues of $A$ cannot be roots of unity (those are all contained in $K$). Consequently $A^M\neq I_2$ for all integers $M>0$. QED

It is trivial to extend $A$ to a $3\times3$ matrix in $SL_3(\Bbb{F}_p[x])$ by adding a $1$ along the diagonal. The resulting matrix must have infinite order as well.

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My first solution (see the edit history) used the fact that if $g$ is any element of $K$, then $$ \left(\begin{array}{cc}1&g\\ g&1+g^2\end{array}\right)\in SL_2(K) $$ is a homomorphic image of $A$. Surely the orders of such matrices are unconstrained, and hence $A$ must have infinite order. However, that only lead to a rather kludgy argument, so I removed it (some comments below may refer to it).

Answer 2

In $\mathrm{GL}_n(\mathbf{F}_q(t))$, the characteristic polynomial $\det(B-x\mathrm{Id})$ of every element of finite order lies in $\mathbf{F}_q[x]$. Indeed, all its roots being roots of unity, they lie in $\overline{\mathbf{F}_q}$, hence so do their symmetric polynomials. Since the only elements in $\mathbf{F}_q(t)$ that are algebraic over over $\mathbf{F}_q$ are the constants, this observation follows.

Hence, consider a companion matrix $M$, relative to some monic polynomial $P$ of degree $n$ (here $n=3$) with coefficients in $\mathbf{F}_q(t)$, i.e., $P\in\mathbf{F}_q(t)[x]$. If $P\notin\mathbf{F}_q[x]$ (and $P(0)\neq 0$ to ensure that $M$ is invertible), then $M$ has infinite order. The determinant of $M$ is $P(0)=(-1)^{n-1}$.

For instance, for $n=2,3$ the following (companion, for the polynomials $x^2-tx+1$, $x^3-tx^2-1$ respectively) matrices are in $\mathrm{SL}_n(\mathbf{F}_p[t])$ and have infinite order and determinant 1: $$\begin{pmatrix}0 & -1\\1 & t\end{pmatrix},\quad \begin{pmatrix}0 & 0 & 1\\ 1 & 0 & 0\\0 & 1 & t\end{pmatrix}.$$

Of course this applies to non-companion matrices. For instance the one in Jyrki's answer: the trace is not constant, hence it has infinite order.

This is an iff condition: if all coefficients of the characteristic polynomial are constant with respect to $t$, then these are roots of unity, so some power of the matrix is unipotent (i.e. char. polynomial $(x-1)^n$ and hence has finite order (since the characteristic is finite). This yields a immediate algorithm to check whether a matrix in $\mathrm{GL}_n(\mathbf{F}_q(t))$ has finite order: iff its char. pol. is constant w.r.t $t$.