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I guess the answer is very trivial, but at the moment I don't get it. Let $p$ prime. Consider the extension $\mathbb{F}_{p^n} | \mathbb{F}_p$ with $n \in \mathbb{N}$ arbitrary. We know that this is a galois extension. But why this extension is separable? Especially for $p|n$ it can't be separable, because there is at least one element who's derivative of his minimal polynomial is zero?

Where is the misstake? Ty for help.

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    $\begingroup$ To answer your question about $\mathbb{F}_{q^n}$ over $\mathbb{F}_q$, where $q$ is a prime power, you can just show that the polynomial $x^{q^n}-x$ has no repeated root in the algebraic closure of $\mathbb{F}_q$ and the roots of $x^{q^n}-x$ are precisely the elements of $\mathbb{F}_{q^n}$. $\endgroup$ Mar 7 '17 at 11:38
  • $\begingroup$ But this extension is algebraic since it's normal. $\endgroup$
    – Hamilcar
    Mar 7 '17 at 11:38
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No, there are no elements $\alpha\in\Bbb{F}_{p^n}$ such that the minimal polynomial $m(x)$ of $\alpha$ would have zero derivative. For $m'(x)$ to be zero, we must have the form $$ m(x)=\sum_{i}a_ix^{pi} $$ for some finite set of coefficients $a_i\in\Bbb{F}_p$. By Little Fermat $a_i^p=a_i$ for all $i$, so in fact $$ m(x)=\left(\sum_ia_ix^i\right)^p. $$ But this contradicts the fact that a minimal polynomial is irreducible.

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  • $\begingroup$ The key implication is: $$m'(x)=0\implies \text{$m(x)$ is not irreducible.}$$ This fails for some transcendental extensions of $\Bbb{F}_p$ but holds for all the finite fields $K$, because we then have $K^p=K$. Look up perfect fields. $\endgroup$ Mar 7 '17 at 11:49

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