Example. $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right)$ [22016B]

Let $\mathbb{F}_{p^{n}}$ be the unique (up to isomorphism) field extension of $\mathbb{F}_{p}$ of degree $n$.

Claim: $\mathbb{F}_{p^{n}} / \mathbb{F}_{p}$ is Galois.

By construction in Theorem 1801, $\mathbb{F}_{p^{n}}$ is the splitting field of $x^{p^{n}} - 1$ over $\mathbb{F}_{p}$. Therefore, $\mathbb{F}_{p^{n}} / \mathbb{F}_{p}$ is normal.

By (2) of Proposition 1902, every finite extension of $\mathbb{F}_{p}$ is separable. Hence $\mathbb{F}_{p^{n}} / \mathbb{F}_{p}$ is separable.

Therefore, $\mathbb{F}_{p^{n}} / \mathbb{F}_{p}$ is Galois.

Claim: $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) \cong \mathbb{Z} / n\mathbb{Z}$.

Denote by $\operatorname{Fr} : \mathbb{F}_{p^{n}} \to \mathbb{F}_{p^{n}}$ the Frobenius endomorphism. Since $\operatorname{Fr}$ is injective and $\mathbb{F}_{p^{n}}$ is finite, $\operatorname{Fr}$ is bijective and is an automorphism of $\mathbb{F}_{p^{n}}$. Theorem 1801 also shows that $\alpha^{p} = \alpha$ for all $\alpha \in \mathbb{F}_{p}$. Hence, $\operatorname{Fr}$ fixes $\mathbb{F}_{p}$. Then $$ \mathbb{Z} \to \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) , \qquad m \mapsto \operatorname{Fr}^{m} $$ is a group homomorphism.

It is not injective since the domain is infinite while the codomain is fintie. Its kernel equals $\mathbb{Z} / m\mathbb{Z}$ for some positive integer $m$. Then $\operatorname{Fr}^{m} = \operatorname{id}_{\mathbb{F}_{p^{n}}}$, so every $\alpha \in \mathbb{F}_{p^{n}}$ is a root of $x^{p^{m}} - x$. The polynomial $x^{p^{m}} - x$ has at most $p^{m}$ roots, so $p^{m} \geq p^{n}$. It follows that $m \geq n$. On the other hand, by Theorem 1801, every $\alpha \in \mathbb{F}_{p^{n}}$ satisfies $\alpha^{p^{n}} = \alpha$, so $\operatorname{Fr}^{n} = \operatorname{id}_{\mathbb{F}^{p^{n}}}, n \in m\mathbb{Z}$. Thus, $m = n$ and $\mathbb{Z} / n\mathbb{Z}$ is isomorphism to a subgroup of $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right)$. Then $$ \lvert \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) \rvert = [\mathbb{F}_{p^{n}} : \mathbb{F}_{p}] = n = \lvert \mathbb{Z} / n\mathbb{Z} \rvert $$ implies that $$ \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) \cong \mathbb{Z} / n\mathbb{Z}. $$ More generally, for all $n \mid m$, $\mathbb{F}_{p^{m}} / \mathbb{F}_{p^{n}}$ is a Galois extension with $\mathrm{Gal}\left( \mathbb{F}_{p^{m}} / \mathbb{F}_{p^{n}} \right) \cong \mathbb{Z} / (m / n)\mathbb{Z}$.