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I have to prove that given $F \subset E$, where $F$ is a field of characteristic p and supposing that $p(x)=x^p-x-a$ is irreducible in $F[X]$, with $\alpha \in E$ a root of $p(x)$. Then is $F(\alpha)/F$ a Galois extension.

I'm trying to see that this is a normal and separable extension.

So far I know that, given the characteristic of the field, the polynomial $f(x)=x^p−x$ has the property:

$f(x_1+x_2)=f(x_1)+f(x_2)$

With $x_1$ and $x_2$ being two elements of an extension field of $F$. And because of little Fermat we know $f(k)=k^p−k=0$ for all $k∈F$. So, if $\alpha$ is a root of $p(x)=x^p−x-a$, then:

$p(r+k)=f(r+k)-a=f(r)+f(k)-a=p(r)+f(k)=0,$

so all the elements $r+k$ with $k∈F$ are roots of $p(x)$, and as there are $p$ of them. So we've shown that there exists an irreducible polynomial whose roots, together with the elements of $F$, generate $F(\alpha)$ (clearly), and therefore it is a normal extension. But how can I prove it is separable? Or is there a better way to prove this is a Galois extension? Thank you, for any help.

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    $\begingroup$ I think you can only apply little Fermat to elements in $\mathbb{F}_p$, not all elements in $F$. $\endgroup$ – Alex Vong May 26 '17 at 15:04
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    $\begingroup$ What about looking at the GCD of $f(x)$ and $f'(x)$ to see that $f$ has no repeated roots? $\endgroup$ – sharding4 May 26 '17 at 15:08
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Hints for you to understand and justify: (doing arithmetic modulo $\;p\;$ all along)

$$\begin{align*} &\bullet f'(x)=px^{p-1}-1=-1\neq0\\{}\\ &\bullet f(\alpha)=0\implies\;\forall\,m\in\Bbb F_p\cong\Bbb Z/p\Bbb Z\,,\;\;\; f(\alpha+m)=0\end{align*}$$

...and thus $\;F(\alpha)\;$ contains all the roots of $\;f(x)\;$ ...

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