Proving $F(\alpha)/F$ is a Galois extension, where $\alpha \in E \supset F$ is a root of $x^p-x-a=p(x) \in F[X]$, irreducible in $F$

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.

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

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 $\;f(x)\;$ ...