Automorphisms of splitting field of $x^p-x-a$ Let $p$ be a prime and consider the splitting field of $f(x) = x^{p} - x - a$ over $\mathbb{F}_{p}$.  I have worked out that the splitting field is $\mathbb{F}_{p}(\beta)$, where $\beta$ is a root of $f$.  Then the maps $\tau^{n}$, where $\tau: \beta \mapsto \beta + 1$ permute the roots, but I am having trouble seeing that they are actually automorphisms.  For example,
$$\tau(\beta + (\beta + k)) = \beta + (\beta + k) + 1,\\
\tau(\beta) + \tau(\beta + k) = \beta + 1 + (\beta + k) + 1.$$
What am I missing?

Since the map
$$ u\to u^p $$
is an automorphism (the Frobenius' automorphism) of $\mathbb{F}_p(\beta)\simeq\mathbb{F}_{p^p}$, the roots of your polynomial are:
$$\xi,\xi^p,\xi^{p^2},\ldots,\xi^{p^{p-1}}.$$
But since $\xi^p = \xi+a$ this is the same as saying:
$$\xi, \xi+a,\xi^p+a,\ldots,\xi^{p^{p-2}}+a,$$
or just:
$$\xi, \xi+a,\xi+2a,\ldots,\xi+(p-1)a,$$
hence:
$$\xi,\xi+1,\ldots,\xi+(p-1).$$
This also follows from the rather simple fact that:
$$ x^p-x = \prod_{\nu\in\mathbb{F}_p}(x-\nu). $$
