Automorphisms of splitting field of $x^p-x-a$

Question

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?

Answer 1

Your map sends $\beta \mapsto \beta + 1$ but not $x \mapsto x+1$ for all $x\in F_p(\beta)$ this is your mistake:

$$\tau(\beta+(\beta+k))\neq \beta+(\beta+k)+1.$$

For example if $k\in F_p$, $\tau(k)=k$ but $\tau(k)\neq k+1$.

Answer 2

I would cheat and use the fact $u \to u^p$ is an automorphism. In fact, this always generates the automorphism group of a finite field extension of $\mathbb{F}_p$.

Another plan would be to use the characterization of homomorphisms with domain $\mathbf{F}_p(\beta)$ to show that there is a unique homomorphism $\mathbf{F}_p(\beta) \to \mathbf{F}_p(\beta)$ that sends $\beta \to \beta + 1$

You could also piece something together using the fact $f(x) = f(x+1)$.

Another starting point is that the automorphism group has to be order $p$, but it also has to act transitively on the set of $p$ roots.

Answer 3

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: $$ x^p-x = \prod_{\nu\in\mathbb{F}_p}(x-\nu). $$