Let $k$ be a finite field of order $q$ in characteristic $p$, let $n$ be a positive integer not divisible by $p$, and let $K$ be the splitting field of $X^n-1$ over $k$. Prove that $[K:k]$ equals the smallest positive integer $d$ such that $n\mid q^d-1$.
My approach: It's easy to show that if $k$ is a finite field of order $q$ with $\text{char} \ k=p$ then $k=\mathbb{F}_{q}$ with $q=p^m$ with $m\geq 1$. If $K$ is the splitting field of $X^n-1$ over $k$ then $K=\mathbb{F}_{q^r}$ for some suitable $r\geq 1$.
Let $\alpha \in K=\mathbb{F}_{q^r}$ be a root of $X^n-1$ then $\alpha^n-1=0$ and $\alpha\neq 0$.
Thus, $\alpha^n=1$ and $\alpha \in \mathbb{F}^{\times}_{q^r}$ where $\mathbb{F}^{\times}_{q^r}$ is the cyclic group of order $q^r-1$.
But I have some difficulties:
1) From $\alpha^n=1$ it follows that $o(\alpha)\mid n$. How to show that $o(\alpha)=n$? I was trying to show it using that $(n,p)=1$ but I failed.
Remark: Because if we can show that $o(\alpha)=n$ then the desired result follows immediately from cyclicity of multiplicative group of field $\mathbb{F}_{q^r}$.
I have spent one day in order to come up with this approach. So please do not duplicate this question and can anyone show how to answer my question, please.
Detailed explanation would be great!
Best Answer
It is not true that $o(\alpha)=n$ for every root $\alpha\in K$ of $X^n-1$; after all $1\in K$ is also a root of $X^n-1$. But because $K$ is the splitting field of $X^n-1$ there exists a root $\alpha\in K$ of $X^n-1$ with $o(\alpha)=n$:
Because $\gcd(n,p)=1$ the polynomial $X^n-1$ is separable over $k$, meaning that it has precisely $n$ roots in the splitting field $K$. These roots form a subgroup of $K^{\times}$ of order $n$, and because $K^{\times}$ is cyclic so is this subgroup. Then for any generator $\alpha$ of this subgroup you have $o(\alpha)=n$.