About the structure of the unit group of a Galois ring.

Question

Let $\tau=\{0,1,\xi,\ldots,\xi^{p^m-2}\}$ the Teichmüller set of $R=GR(p^s,p^{sm})$ with maximal ideal $(p)$, $c=a_0+a_1p+\cdots+a_{s-1}p^{s-1}$ the $p$-adic representation of $c\in R$ where $a_j\in\tau$,for $j\in\{0,1,\ldots,s-1\}$. Furthermore, consider the abelian group $\varepsilon=\{1+\pi:\pi\in(p)\}$ with $o(\varepsilon)=p^{m(s-1)}$ and the cyclic group $\langle{\xi}\rangle$ with $o(\langle{\xi}\rangle)=p^{m}-1$.

I've already proof the next statements:

1. $R^{*}=\langle{\xi}\rangle\otimes\varepsilon$ (Direct product)

2. $\forall{(1+\pi\in\varepsilon)}:{ord(1+\pi)\mid p^{s-1}}$

3. $c\in(p)\iff a_0=0$

Now, I need proof the next statement "If $ord(c)\mid (p^{m}-1)$ then $c=\xi^{i}$ with $i\in\{0,1,\ldots,p^m-2\}$.

My attempt of proof:

Let $c\in R-\{0\}$ with $c=a_0+a_1p+\cdots+a_{s-1}p^{s-1}$ and $d=ord(c)$. As $d\mid p^m-1$ thus $p^m-1=dt$, so $p^m=dt+1$, using this we have:

\begin{eqnarray} c^{p^m}&=&(a_0+a_1p+\cdots+a_{s-1}p^{s-1})^{p^m}\\ c^{dt+1}&=&(a_0+a_1p+\cdots+a_{s-1}p^{s-1})^{p^m}\\ c&=&(a_0+a_1p+\cdots+a_{s-1}p^{s-1})^{p^m} \end{eqnarray}

Since $a_0\in\tau$, $a_0=0$ or $a_0=\xi^k$ with $k\in\{0,1,\ldots,p^{s-2}\}$.

If $a_0=0$, $c\in(p)$ and $1=c^d=p^dc'^d=p(p^{d-1}c´^d)$ thus $p$ is a unit !, then $a_0\ne 0$ and $c$ can be rewrited as:

\begin{eqnarray} c&=&\left(a_0\left(1+\frac{a_1}{a_0}p+\cdots+\frac{a_s-1}{a_0}p^{s-1}\right)\right)^{p^m}\\ &=&a_o^{p^m}(1+pr)^{p^m} \end{eqnarray}

I just wanna know a hint or correction in my attempt because I don't see how to go ahead.

Answer 1

By the direct product representation, $c = ab = ba$ with $a\in\langle\xi\rangle$ and $b\in \epsilon$. As $a$ and $b$ commute, $\operatorname{ord}(c) = \operatorname{lcm}(\operatorname{ord}(a),\operatorname{ord}(b))$ which is assumed to be a divisor of $p^m - 1$. So $\operatorname{ord}(b) \mid p^m-1$. On the other hand, from $b\in\epsilon$, $\operatorname{ord}(b) \mid \#\epsilon = p^{m(s-1)}$. Hence $\operatorname{ord}(b) \mid \gcd(p^m - 1, p^{m(s-1)}) = 1$. Therefore, $\operatorname{ord}(b) = 1$, so $b = 1$ and $c = a\in \langle \xi\rangle$.