The automorphism of splitting field of x^p-x+a over Z_p. I'd like to solve the question.
Let $L$ the splitting field of $f(x)=x^p-x+a$ ($a\neq 0$) over $\mathbb Z_p$.


*

*$g : L \rightarrow L$, $g(\alpha)=\alpha+1$ where $\alpha$ is a root of $f(x)$ is automorphism.

*$\text{Aut}(L/\mathbb Z_p)=\mathbb Z_p$?
I solved the follwings fact


*

*$f$ has no root in $\mathbb Z_p$

*If $\alpha$ is root of $\mathbb Z_p$, then $\alpha+1$ is root of $\mathbb Z_p$

*$f$ is irreducible over $\mathbb Z_p$

*The number of irreducible of which order is $p$ over $\mathbb Z_p$ is $p^{p-1} -1$
I tried to use the frobenius automorphism, but I failed.
How to solve this problem? Please help.
 A: This is a part of Artin–Schreier theorem.
Since $x^p-x-a$ is irreducible (using your remark n.3), then $\mathbb F_p(\alpha)$ (with $\alpha$ a root of $f(x)$) is the splitting field of $f(x)$. Also, if $\alpha$ is a root, then the set of roots of $f(x)$ is
$$
\{\alpha, \alpha+1,...,\alpha+p-1\}
$$
that are distinct and exactly $p$ (I used your second remark). Hence the field extension $L=\mathbb F_p(\alpha)$ over $\mathbb F_p$ is a Galois extension of degree $p$ (because $\alpha$ has degree $p$ over $\mathbb F_p$ and it's minimal polynomial is separable).
The map $g:L\rightarrow L$ such that $g(\alpha)=\alpha+1$ is well defined: infact, to define a $g:L\rightarrow L$ it's sufficient to define the image of $\alpha$ and this image has to be a root of the minimal polynomial of $\alpha$. Since $\alpha+1$ is a root of $f(x)$ that is the minimal polynomial of $\alpha$, $g$ is well defined. Also, since $L$ over $\mathbb F_p$ is normal, then $g(L) = L$ and $g$ is an automorphism.
To compute $\text{Gal}(L/\mathbb F_p)$ is sufficient to notice that is a group of cardinality equal to $p$, so it is cyclic and isomorphic to $\mathbb Z_p$. Moreover a generator of this group is the automorphism $g$ that has order $p$.
