Let $F = \mathbb{Q}(\zeta_{p})$ where $\zeta_{p} = e^{ 2\pi 1 / p }$ is a primitive $p$-th root of unity.

Claim: $F / \mathbb{Q}$ is Galois.

$F$ is the splitting field of $x^{p} - 1$ over $\mathbb{Q}$ by Example 1505, so $F / \mathbb{Q}$ is normal by (4) of Proposition 2205. $\operatorname{ch}(\mathbb{Q}) = 0$, so $F / \mathbb{Q}$ is separable by (5) of Proposition 1704. Hence, $F / \mathbb{Q}$ is Galois.

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

We showed that $m_{\zeta_{p}, \mathbb{Q}}(x) = \Phi_{p}(x) = \frac{x^{p} - 1}{x - 1} = \prod_{1 \leq j \leq p-1} (x - \zeta_{p}^{j})$ in Example 1505. For each integer $j$ coprime to $p$, $\zeta_{p}^{j}$ is a root of $\Phi_{p}(x)$. By Proposition 1709 and its proof, we know that $$ \mathbb{Q}(\zeta_{p}) \to \mathbb{Q}(\zeta_{p}^{j}) = \mathbb{Q}(\zeta_{p}), \qquad a(\zeta_{p}) \mapsto a(\zeta_{p}^{j}) \quad \forall a(x) \in \mathbb{Q}[x] $$ is a field isomorphism. By definition, we can check that $\sigma^{i} \circ \sigma^{j} = \sigma^{ij}$ and $\sigma^{j} = \operatorname{id}_{F}$ if and only if $j \equiv 1 \pmod{p}$. It follows that $$ \begin{align*} \left( \mathbb{Z} / p\mathbb{Z} \right)^{\times} &\to \mathrm{Gal}\left( \mathbb{Q}(\zeta_{p}) / \mathbb{Q} \right) \\ \overline{j} &\mapsto \sigma_{j} \end{align*} $$ is well defined and is an injective group homomorphism. On the other hand, $$ \lvert \mathrm{Gal}\left( \mathbb{Q}(\zeta_{p}) / \mathbb{Q} \right) \rvert = [\mathbb{Q}(\zeta_{p}) : \mathbb{Q}] = \deg \Phi_{p}(x) = p - 1 = \lvert \left( \mathbb{Z} / n\mathbb{Z} \right) ^{\times} \rvert . $$ Therefore, the group homomorphism is bijective and is a group isomorphism.

For general positive integer $n$, we will show that $\mathrm{Gal}\left( \mathbb{Q}(\zeta_{n}) / \mathbb{Q} \right) \cong \left( \mathbb{Z} / n\mathbb{Z} \right)^{\times}$.

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

1. (D&F Example (2) at p.563) More generally, any quadratic extension $K$ of any field $F$ of characteristic different from $2$ is Galois.
2. (D&F Example (4) at p.563) The extension $\mathbb{Q}(\sqrt{ 2 }, \sqrt{ 3 })$ is Galois over $\mathbb{Q}$ because it is the splitting field of $(x^{2} - 2)(x^{2} - 3)$.
3. (D&F Example (6) at p.566) The field $\mathbb{Q}(\sqrt[4]{ 2 })$ is not Galois over $\mathbb{Q}$ while we have $\mathbb{Q} \subset \mathbb{Q}(\sqrt{ 2 }) \subset \mathbb{Q}(\sqrt[4]{ 2 })$ and both $\mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q}$ and $\mathbb{Q}(\sqrt[4]{ 2 }) / \mathbb{Q}(\sqrt{ 2 })$ are Galois.

(D&F Example (8) at p.566) The inseparable extension $\mathbb{F}_{2}(x)$ over $\mathbb{F}_{2}(t)$ where $x^{2} - t = 0$ is not Galois.