TITLE: Centralizers of elements in general linear group over Z mod prime power
QUESTION [6 upvotes]: I would like to know for which elements $x$ in $G:=Gl_n(\mathbb{Z}/\ell^e\mathbb{Z})$ their centralizers $C_G(x):=\{ y \in G \mid xy=yx\}$ are abelian groups. 

Here, $n$ is an integer $\geq 2$ and $\ell^e$ is a prime power. I am especially interested in the case $n=2$.
Of course, if $x$ is a multiple of the identity matrix, then $C_G(x)=G$, and thus is not abelian.
In case $e=1$, i.e. $\mathbb{Z}/\ell^e\mathbb{Z}$ is a field, then I have read a few times (but always without proof) that $C_G(x)$ is abelian, if $x$ is not a multiple of the identity matrix and $n=2$. To be precise, $C_G(x)$ will be either isomorphic to $\mathbb{F}_{\ell^2}^*$, $\mathbb{F}_\ell^* \times \mathbb{F}_\ell^*$, or $\mathbb{F}_\ell \times \mathbb{F}_\ell^*$. 
So the question is what happens for arbitrary prime powers (and for arbitrary $n$)?
The question seems to me like a standard fact which should be contained in every text book about general linear groups. Can anybody give a good reference?
Thanks a lot!

REPLY [9 votes]: Let $G_r=\mathrm{GL}_n(\mathbb{Z}/p^r)$. For $x\in G_r$ the centraliser $C_{G_{r}}(x)$ is abelian iff $x$ is regular iff the reduction mod $p$ of $x$ is regular. This is due to G. Hill, Regular elements and regular characters of $\mathrm{GL}_n(\mathcal{O})$, J. Algebra 174 (1995), no. 2, 610–635. The case $r=1$ is easy and was known earlier.
To add some details, note that Hill's Theorem 3.6 holds when the residue field is $\overline{\mathbb{F}}_p$, but I think his proof goes through for any algebraically closed residue field $\overline{k}$. If $k$ is any field and $C_{\mathrm{GL}_n(k)}(A)$ is abelian for $A\in \mathrm{GL}_n(k)$, then using for example the rational canonical form one can see that $A$ must be $\mathrm{GL}_n(k)$-conjugate to a companion matrix, and so  $k^n$ is a cyclic $k[A]$-module. Extending scalars to $\overline{k}$ we get that $A$ is regular as an element of $\mathrm{GL}_n(\overline{k})$.