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help with this interpretation..

Choose a prime $l$ such that $l\neq p$ and $l$ is a quadratic residue modulo $p$. Choose an integer $m\geq 1$ such that $l^m-1$ is divisible by $p$. Let $\theta$ be a primitive element of $F_{l^m}$ and pun $\alpha=\theta^{(l^m-1)/p}$. Then, the order of $\alpha$ is $p$; i.e., $1=\alpha^0=\alpha^p,\alpha^1=\alpha,\alpha^2,\alpha^3,...,\alpha^{p-1}$ are pairwise distinct and $x^p-1=\prod_{i=0}^{p-1}(x-\alpha^i).$

I know that $(\alpha)^p=(\theta^{(l^m-1)/p})^p=\theta^{l^m-1}=1$ but why $1=\alpha^0=\alpha^p,\alpha^1=\alpha,\alpha^2,\alpha^3,...,\alpha^{p-1}$ are pairwise distinct and $x^p-1=\prod_{i=0}^{p-1}(x-\alpha^i).$ ??

mathnew2
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  • If we are in characteristic $l$, $x^p-1$ has distinct factors. – Angina Seng Dec 03 '17 at 06:50
  • In $\mathbb{C}[x]$ : $x^p -1 = \prod_{j=0}^{p-1} (x-\zeta_p^j)$ where $\zeta_p = e^{2i \pi /p}$. This stays true in $k[x]$ for any field $k$ of characteristic $\ne p$ provided $\zeta_p$ is a primitive $p$-th root of unity, ie. its order is $p$. – reuns Dec 03 '17 at 07:00

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