# Showing explicitly that $\alpha \mapsto \alpha+1$ is an automorphism in the Galois group of $x^p-x+a$?

I have already shown that $x^p-x+a$ is irreducible and separable over $\mathbb{F}_p$ and I have shown that if $\alpha$ is a root, so is $\alpha+1$, so $\alpha$ generates the extension, i.e. the splitting field and Galois closure is $\mathbb{F}_p(\alpha)$. I know the Galois group is cyclic because it is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ (because it has order $p$, the degree of the extension being the degree of the minimal polynomial). But I have been asked to explicitly'' show that the group is cyclic by showing that $\alpha \mapsto \alpha+1$ is an automorphism which generates the group. However, I cannot see how to compute the homomorphism conditions (additive and multiplicative)? I know that such a map would have to fix $\mathbb{F}_p$, the ground field, but I don't know how to compute, say, $\sigma(\alpha + (\alpha+k))$ to show that this is $\sigma(\alpha) + \sigma(\alpha+k)$, for example. Am I not seeing something? Should I be expressing $\alpha+k$ as a linear combination of powers of $\alpha$ and going from there?

• Note: we cannot assume the sum of two roots is also a root. In fact, if it were, then $\alpha + \alpha + k = \alpha + l$ which contradicts $\alpha \notin \mathbb{F}_p$. So how do I compute the map applied to the sum of two roots? – Struggling Student Apr 27 '15 at 22:03

I think your confusion stems from what it means to "define" $\sigma$ by $\alpha \mapsto \alpha+1$. Let $F = \mathbb{F}_p$ and $K = \mathbb{F}_p(\alpha)$. If the only conditions you put on $\sigma$ are that $\sigma : K \to K$ is a function fixing $F$ with $\sigma(\alpha) = \alpha+1$, then $\sigma$ need not be a homomorphism. For instance, I could define $$\sigma(\beta) = \begin{cases} \beta & \text{if } \beta \in F\\ \alpha + 1 & \text{if } \beta \in K \setminus F \end{cases}$$ and then $\sigma$ satisfies both the properties above, but is certainly not a homomorphism.
The point is that there is a unique homomorphism $\sigma : K \to K$ fixing $F$ such that $\sigma(\alpha) = \alpha+1$. One way to prove this is just by brute force computation: any element of $K$ is a polynomial in $\alpha$, i.e., of the form $\sum_{i=1}^n c_i \alpha^i$ for some $c_i \in F$. So define $\sigma : K \to K$ by $\sigma\left(\sum_{i=1}^n c_i \alpha^i\right) = \sum_{i=1}^n c_i (\alpha+1)^i$ and check that this is a homomorphism.
A slicker and more instructive proof uses the universal property of the polynomial ring. Let $\varphi : F[X] \to F(\alpha)$ be the evaluation map at $\alpha+1$, i.e., $\varphi(f(X)) = f(\alpha+1)$ for any polynomial $f \in F[X]$. One can show that this is a homomorphism. (Indeed, this is the universal property of $F[X]$; see p. $5$ here.) Moreover, since $X^p - X + a$ is the minimal polynomial for $\alpha + 1$, then $\ker(\varphi) = (X^p - X + a)$, so $\varphi$ descends to the quotient, inducing a homomorphism $\overline{\varphi}: F[X]/(X^p - X + a) \to F(\alpha)$. But $F[X]/(X^p - X + a) \cong F(\alpha)$ by definition, so composing with this isomorphism yields our desired map $\sigma : F(\alpha) \to F(\alpha)$ with $\sigma(\alpha) = \varphi(X) = \alpha + 1$. (A more general theorem is stated in Dummit and Foote, Theorem $8$, $\S 13.1$, p. $519$.)
So you don't have to show that $\sigma$ is a homomorphism. By saying, "define $\sigma$ by $\alpha \mapsto \alpha + 1$," you are implicitly saying "let $\sigma$ be the unique homomorphism $K \to K$ fixing $F$ with $\sigma(\alpha) = \alpha+1$."