Galois Group of Finite Fields
Definition/Theorem (Frobenius Automorphism).
Let $\mathbb{F}_p$ be a finite field. We define the Frobenius automorphism:
Then $\sigma_p$ is a field automorphism of $\overline{\mathbb{F}_p}$.
We also define its powers by: $\sigma_p^n(\alpha) = \alpha^{p^n}$, and denote this as $\sigma_{p^n}$. That is, $\sigma_{p^n} := \sigma_p^n$.
Proof.
\[(\alpha + \beta)^p = \alpha^p + \beta^p\]
Since $\overline{\mathbb{F}_p}$ has characteristic $p$, the binomial expansion yields:because all intermediate binomial coefficients $\binom{p}{k}$ are divisible by $p$ for $1 \leq k \leq p-1$.
Thus:
\[\sigma_p(\alpha + \beta) = \alpha^p + \beta^p = \sigma_p(\alpha) + \sigma_p(\beta)\]and:
\[\sigma_p(\alpha \beta) = (\alpha \beta)^p = \alpha^p \beta^p = \sigma_p(\alpha) \sigma_p(\beta)\]so $\sigma_p$ is a ring homomorphism.
Since $\overline{\mathbb{F}_p}$ is a field, a nonzero element $\alpha$ satisfies $\alpha^p = 0$ only if $\alpha = 0$, so the map is injective. Moreover, as $\overline{\mathbb{F}_p}$ is algebraically closed, the polynomial $x^p - \alpha$ has a root for all $\alpha\in\overline{\mathbb{F}_p}$, so the map is surjective.
Therefore, $\sigma_p$ is a field automorphism of $\overline{\mathbb{F}_p}$.
Theorem. (Galois Group of Finite Field Extension)
Let $\mathbb{F}_{q^n}/\mathbb{F}_q$ be a finite field extension, where $q = p^r$ is a prime power. Then:
Proof.
As proved in Separable Extensions, every finite field is perfect. Therefore, the extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ is separable. Moreover, $\mathbb{F}_{q^n}$ is the splitting field of the polynomial $x^{q^n} - x$ over $\mathbb{F}_q$, so the extension is normal. Hence, $\mathbb{F}_{q^n}/\mathbb{F}_q$ is Galois.Define the Frobenius automorphism $\sigma_q : \alpha \mapsto \alpha^q$. Since this fixes all elements in $\mathbb{F}_q$, and maps $\mathbb{F}_{q^n}$ to itself, it is an element of $\text{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$.
To compute the order of $\sigma_q$, note that:
- Since every element $\alpha$ in $\mathbb{F}_{q^n}$ satisfies $\alpha^{q^n} = \alpha$, we have $\sigma_q^n = \text{id}$ on $\mathbb{F}_{q^n}$.
- Suppose $\sigma_q^k = \text{id}$ for some $0 < k < n$. Then for all $\alpha \in \mathbb{F}_{q^n}$, $\alpha = \sigma_q^k(\alpha) = \alpha^{q^k}$. Hence every element satisfies $\alpha^{q^k} = \alpha$, so $\alpha$ lies in $\mathbb{F}_{q^k}$. This forces $\mathbb{F}_{q^n} \subset \mathbb{F}_{q^k}$, which contradicts $k < n$.
Therefore, the order of $\sigma_q$ is exactly $n$.
Since $\text{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$ has order $[\mathbb{F}_{q^n} : \mathbb{F}_q] = n$, and $\langle \sigma_q \rangle$ is a subgroup of order $n$, it must be the whole Galois group, and we have
\[\text{Gal}(\mathbb{F}_{q^n} / \mathbb{F}_q) = \langle \sigma_q \rangle \cong \mathbb{Z}/n\mathbb{Z}.\]