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Let $r>0$. Is $\mathbb{F}_{p^r}/\mathbb{F}_{p}$ a Galois extension? If so, why?

I know that it is a finite extension, with $[\mathbb{F}_{p^r}:\mathbb{F}_{p}]=r$. To show that it is a Galois extension, it suffices to show that $|Aut(\mathbb{F}_{p^r}/\mathbb{F}_{p})|=r$.

The notaion $Aut(K/F)$ indicates the group of field automorphisms $K\to K$ such that the automorphism fixes every element of $F$.

But, how do I show $|Aut(\mathbb{F}_{p^r}/\mathbb{F}_{p})|=r$? A simple, easy to grasp proof without holes in it would be ideal.

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  • $\begingroup$ This isn't the easiest way to show it, do you know the theorem that splitting fields are Galois? I would recommend showing that this is a splitting field of a certain polynomial $\endgroup$ – Sheel Stueber May 24 '18 at 17:52
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    $\begingroup$ I know that the splitting field of a seperable polynomial is Galois. Now, which "certain polynomial" are you thinking of? :) $\endgroup$ – Pascal's Wager May 24 '18 at 17:54
  • $\begingroup$ yes sorry separable polynomials. The certain polynomial I was thinking of it in Lord Shark's answer below $\endgroup$ – Sheel Stueber May 24 '18 at 18:14
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Every field $K$ of characteristic $p$ has the Frobenius endomorphism $F:x\mapsto x^p$. This is a homomorphism of fields, and so is injective. If $K$ is finite, then $F$ must be bijective, so an automorphism. On $\Bbb F_p$, $F$ acts trivially.

The fixed points of $F^t$ are the solutions of $x^{p^t}-x$. Every element of $\Bbb F_{p^r}$ is a solution of $x^{p^r}-x$, if $t<r$ then not all elements of $\Bbb F_{p^r}$ is a solution of $x^{p^t}-x$ since that polynomial has fewer than $p^r$ zeros. Thus $F$ has order $r$ on $\Bbb F_{p^r}$. As $|\Bbb F_{p^r}:\Bbb F_p| =r$, then the Galois group must consist of the powers of $F$.

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