Roots of $x^n - 1$ in an algebraically closed field of prime characteristic

Question

Let $F$ be an algebraically closed field of characteristic $p$ , and let $n$ be a positive integer.

Consider $ g := x^n - 1 \in F[x]$

Is it true that $ g$ has distinct roots in $F$ if and only if $p$ does not divides $n$ ?

Here is my argument, please could you tell me whether it is correct:

Note roots of g are distinct if and only if $\mathrm{hcf}(g,g') = 1 $ .

Suppose $\beta \in F$ is a root of $g$ .

Now $ \beta^n = 1$, so $\beta \neq 0$ .

Then $g'(\beta) = n \beta^{-1} = 0 $ iff $p \mid n $

unless there is something wrong with how i am checking, i can't find any counterexamples for finite fields (where roots exist)

EDIT:

i think the statement is correct, at least i have found something which implies this in a book "Classical Galois Theory"

EDIT 2:

Noting $(a-b)^p = a^p - b^p $ in a field of characteristic p (what Dilip said)

then $$p \mid n \implies n = mp \implies x^n - 1 = (x^m-1)^p$$

and the polynomial on the RHS has only repeated roots. Can someone show me the other way?

Answer 1

Your argument shows that $g$ and $g'$ have a common root exactly when $p | n$, and as you observe, this implies that $g$ has a multiple root exactly when $p | n$. So your argument is correct.

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Aside: This statement is very well-known. A useful adjective in this context is separable (which just means "has distinct roots"). You might have more luck finding this statement in textbooks if you look for discussions of when cyclotomic polynomials are separable. (The polynomial $x^n - 1$ has a factorization as a product of cyclotomic polynomials $\Phi_d(x)$, as $d$ ranges over the divisors of $n$, so $x^n - 1$ is separable iff all $\Phi_d(x)$ are separable for $d | n$. Now (almost?) every Galois theory text should have the statement that $\Phi_d(x)$ is separable iff $p \not\mid d$.)