Let $\mathbb F$ be a field of characteristic $p \gt 0$ and $a \in \mathbb F$ such that $a \ne b^p – b$ for all $b \in \mathbb F$. If $\mathbb L$ is the splitting field of $f(x) = x^p – x – a $ over $\mathbb F$, show that $\mathbb L$ is cyclic extension of $\mathbb F$, and determine the group $Gal(\mathbb L / \mathbb F)$.
I don't know if I understand correctly but this is my attempt:
Let $L/K$ be a cyclic Galois extension of order $p=$char $K$. Let $\sigma$ be a generator of the Galois group. So there exists an element $x \in L^∗$ such that $z=x+\sigma(x)+\sigma^2(x)+\cdots+\sigma^{p-1}(x)\neq0$. Let fix an element $x$, and knowing that $z\in K$, because it is invariant under $\sigma$.
So $y=(p-1)x+(p-2)\sigma(x)+\cdots+2\sigma^{p-3}(x)+\sigma^{p-2}(x)+0\cdot\sigma^{p-1}(x).$
We see that $\sigma(y)=y+z$, so if we have $u=y/z \to \sigma(u)=u+1 $. So $u \notin K$, so $L=K(u)$. From Fermat's little theorem we have $p(T)=T^p-T=\prod_{i=0}^{p-1}(T-i)$ In characteristic $p$ we have $p(a+b)=p(a)+p(b)$ for all $a,b \in L.$ The minimal polynomial of $u$ so
$\prod_{i=0}^{p-1}(T-\sigma^i(u))=\prod_{i=0}^{p-1}(T-(u+i))=p(T-u)=p(T)-p(u)= T^p-T+(-1)^p\prod_{i=0}^{p-1}(u+i)$ so $L / K$.
Galois group is cyclic and is generated by $\alpha \to \alpha +1$ where $\alpha$ is a root.
Is my attempt on the right track?
Thanks.
Note that $a \neq b^p-b$
Hence if $\alpha$ is a root of $x^p-x-a$ in $L$, then $ \alpha \in L \setminus F$
and
we also have $$(\alpha+1)^p-(\alpha+1)-a=\alpha^p-\alpha-a=0$$
In general, if $\alpha$ is a root of $f(x)=x^p-x-a$, then so are $\alpha, \alpha+1, \alpha+2, \alpha+3 \dots \alpha+(p-1)$
Thus $L=F(\alpha)$ and $L/K$ is Galois extension.
Now note that any automorphism $\sigma:L \to L $ permutes the roots...........