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I have already been able to show that if the irreducible polynomial $ f $ has degree n, then its roots are $ a, a^q \ldots, a^{q^{n-1}}$, where $ a $ is a roots of $ f $ to some extent.

How can I show that they are all different?

user489941
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  • If $f$ of degree $n$ is irreducible then $\mathbf F_q[x]/(f)$ is a field of order $q^n$ so $\alpha^{q^n} = \alpha$ for all $\alpha$ in that field. In particular, $x^{q^n} \equiv x \bmod f$, so $f$ is a factor of $x^{q^n}-x$. Show with derivatives that $x^{q^n}-x$ is separable in $\mathbf F_q[x]$, and every factor of a separable polynomial is separable. – KCd Feb 01 '18 at 13:46
  • But does not you assume it's f separable?

    For for this congruence each root of f can only have multiplicity 1, which only happens if f is separable

     – user489941 Feb 02 '18 at 01:41
  • I made no such assumption: you said $f$ is irreducible, so that makes $\mathbf F_q[x]/(f)$ a field of order $q^n$, where $n = \deg f$. In a field of order $q^n$, every element is equal to its own $q^n$-th power. None of this needs separability of $f$. – KCd Feb 02 '18 at 06:30
  • at first you do not know if there is a factor $ (x-a)^2 $ in $ f $, if it exists, then it is not worth $ x^{q^n} - x $ since $ (x-a)^2$ does not divide $ x^{q^n} - x$ – user489941 Feb 02 '18 at 13:22
  • That is not relevant: nothing in my argument involved assuming or not assuming that $f(x)$ has a double root. So once I show $f(x)$ is a factor of $x^{q^n}-x$ in $\mathbf F_q[x]$ then that proves $f(x)$ has no double root because $x^{q^n}-x$ has no double root. (This is my last comment. If you are unable to understand the argument I presented then go find someone in person, not a stranger on the internet, and ask for their help.) – KCd Feb 02 '18 at 16:46

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