Let char $F = p >0$ and let $f (t) :=t^p-t-a \in F [t] $. Suppose also that $a \neq b^p-b $ for any $b \in F $. Find $G (K/F) $ where $K $ is a splitting field of $f $.
Here is my work so far:
I've shown that $f $ has no multiple roots by derivative test. Also, I've shown that if $\alpha $ is a root of $f $ then so is $\alpha + k $ for all $0 \leq k \leq p-1$. Then, I've shown that $f $ is irresucible iff $f $ has no roots in $F $. Most, if not all, of these can be found online if you are interested.
Let $\alpha $ be a root of $f $. Then we know that the set $A:=\{\alpha + k | k =0,1,...,p-1\}$ is set of roots of $f $. From the above we know that (since $f $ can have at most $p $ roots) $A$ is the set of all roots of $f $. As any $\sigma \in G (K/F) $ sends a root of $f $ to a root of $f $, we know that any $\sigma$ satisfy $\sigma (\alpha) = \alpha + k $ for some $k =0,1,...$ or $p-1$. It can easily be seen that $F (\alpha) $ is a splitting field of $f $ so that we only need to specify the act of $\sigma$ on $\alpha $. Of course $1 \in G (K/F) $. How can we know that any other map (i.e. $k \neq 0$) is possible, i.e. doesn't lead to any contradictions? Also, what has $a \neq b^p-b $ for any $b \in F $ to do with anything?
Thanks for your help
As you say $K=F(\alpha)$ is a splitting field. The derivative of $f(X)=X^p-X-a$ is $-1$, which is certainly coprime to $f(X)$, so $K/F$ is a separable extension, and so a Galois extension. Then the Galois group acts transitively on the zeros of $f$, since $f$ is irreducible over $F$, so for each $k\in\Bbb F_p$ there is $\sigma_k$ in the Galois group with $\sigma_k(\alpha)=\alpha+k$. This is clearly unique. Moreover $\sigma_k\circ\sigma_l=\sigma_{k+l}$ etc.