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?

  • $\begingroup$ 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? $\endgroup$ – 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 $\sigma(\alpha) = \alpha+1$."

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    $\begingroup$ Thank you so much. I've gotten so used to proving that something is an automorphism by showing the homomorphism condition and bijectivity that it didn't occur to me one of the steps was implicit in the statement. But the way you describe it makes sense. The reference to Theorem 8 was also particularly helpful. $\endgroup$ – Struggling Student Apr 28 '15 at 20:31

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