confused about Lemma 5.33 (about Galois Theory) in Rotman's textbook

Question

The confusion occurs at very beginning of the proof (highlighted with red rectangle)。

Why are the roots of $x^p-1\in k[x]$ denoted by $1, \omega, \omega^2, \cdots, \omega^{p-1}$? Why they are all distinct? How to use the condition $p=char(k)$?

Answer 1

Hint: $x^p-1 = (x-1)^p$ if the characteristic is $p>0$.

In the case of characteristic $\ne p$, use the derivative: $(x^p-1)' = px^{p-1}\ne 0$. It has no roots in common with $x^p-1$ and so the roots are distinct.

Answer 2

First of all, if $\omega$ is the $p$-th root of unity, then $\omega^i$ is also a root of $x^p-1$, and this is just because $(\omega^i)^p - 1 = 1 - 1 = 0$. Thus $1, \omega, \omega^2, \cdots, \omega^{p-1}$ are all roots of $x^p - 1$.

They are all distinct because if $\omega^i = \omega^j$, then $\omega^{i-j} = 1$, which means $(i-j) | p$, but $p$ is a prime. So, this can only happen if $i-j = 0$, or $p = 0$, i.e. $p = \text{char}(k)$. This is where you need that condition. So $i = j$, and thus all the above roots are distinct.