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Find $\mathrm{Irr}_{\alpha}^{K}(x)$ if $K = \mathbb{F}_{p}$ and $\alpha$ is a generating element of $\mu_{n}\left(\overline{\mathbb{F}}_{p}\right)$.

It is clear that $\mu_{n}$ has $n$ elements (if $x^{n}-1$ has no roots of multiplicity greater than 1, which I believe can be shown by $(x^{n}-1,nx^{n-1})=1$) in $\overline{\mathbb{F}}_{p}$. Also $\deg \mathrm{Irr}_{\alpha}^{K}(x) \leq n-1$ as $\alpha^{n-1}+\dots + 1 = 0$. However, I can't see why there it is minimal and I am not even sure if it is... Can you give a small, just tiny hint (not a solution)?

user26857
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matheg
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    $x^n-1$ is never irreducible if $n> 1$. Not all its roots have order $n$ (some have order $d| n$). If $n=pm$ then $x^n-1=(x^m-1)^p$ is not separable. The Frobenius lets us express $\alpha$'s minimal polynomial in term of $\alpha$ and the order of $p$ modulo $n/p^k$. – reuns Feb 10 '22 at 16:39
  • What does your $\mu_n$ notation mean? – Gerry Myerson Feb 11 '22 at 11:57
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    Note that in general the polynomial will depend on which generator you pick, as cyclotomic polynomials need not be irreducible mod p. – Lukas Heger Feb 11 '22 at 12:43
  • @GerryMyerson it means group of roots of unity $\mu_{n} = {x\in K| x^{n}=1}$. – matheg Feb 11 '22 at 17:18
  • @LukasHeger Yes, so I am more interested in degree of this polynomial. I suppose it must be the same for all of them? – matheg Feb 11 '22 at 17:20
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    @Egor, yes the degree is the order of $p$ in $(\Bbb Z/m\Bbb Z)^\times$, where $n=p^km$ with $m$ coprime to $n$ – Lukas Heger Feb 11 '22 at 19:44
  • Lukas got everything right. I have discussed irreducibility of cyclotomic polynomials over a finite field an embarrassing number of times, so I won't repeat the stuff from here. I think I have linked most other occurrences to that question. The minimal polynomial is buried in there, but you can find its zeros, and the degree is given by the recipe in Lukas' comment. – Jyrki Lahtonen Feb 13 '22 at 11:09

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