$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt{ 2 })$. One can check that $$ \tau : \mathbb{Q}(\sqrt{ 2 }) \to \mathbb{Q}(\sqrt{ 2 }), \qquad \tau(a + b\sqrt{ 2 }) = a - b\sqrt{ 2 }, \, a, b \in \mathbb{Q} $$ is an automorphism in $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$. If $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$, then $\sigma(\sqrt{ 2 })^{2} = \sigma(\sqrt{ 2 }^{2}) = \sigma(2) = 2$. Hence, $\sigma(\sqrt{ 2 }) = \pm \sqrt{ 2 }$. If $\sigma(\sqrt{ 2 }) = \sqrt{ 2 }$, then $\sigma = \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}$, then $\sigma = \tau$. Thus, $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right) = \{ \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}, \tau \} \cong \mathbb{Z} / 2\mathbb{Z}$ ($\tau \circ \tau = \operatorname{id}_{\mathbb{Q}(\sqrt{ 2 })}$).

Let $K / F$ be an algebraic field extension. If $\alpha \in K$ with $m_{\alpha, F}(x)$ and $\sigma \in \mathrm{Aut}(K / F)$, then $\varphi(\alpha)$ is a root of $m_{\alpha, F}(x)$. Therefore, we have the map $$ \mathrm{Aut}\left( F(\alpha) / F \right) \to \{ \text{root of }m_{\alpha, F}(x) \text{ in }F(\alpha) \}, \qquad \sigma \mapsto \sigma(\alpha). $$ The map is injective because an automorphism $\sigma \in \mathrm{Aut}\left( F(\alpha) / F \right)$ is completely determined by $\sigma(\alpha)$. Next, we show that it is also surjective. If $\beta \in F(\alpha)$ is a root of $m_{\alpha, F}(x)$, then $F(\beta) = F(\alpha)$ ($[F(\alpha) : F(\beta)] = \frac{[F(\alpha) : F]}{F(\beta) : F} = \frac{\deg m_{\alpha, F}(x)}{\deg m_{\beta, F}(x)} = 1$), and $$ \sigma: F(\alpha) \to F(\beta) = F(\alpha), \qquad a(\alpha) \mapsto a(\beta) \, \forall a(x) \in F[x] $$ defines an automorphism of $F(\alpha)$ fixing $F$ by Theorem 1408. Thus, $\beta = \sigma(\alpha)$ belongs to the image of the map. This shows that it is surjective.

Example 2106 (1) can be viewed as a special case of this. The minimal polynomial of $\sqrt{ 2 }$ over $\mathbb{Q}$ is $x^{2} - 2$, and the set $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$ is in bijection with the set of roots of $x^{2} - 2$ in $\mathbb{Q}(\sqrt{ 2 })$ which is $\{ \pm \sqrt{ 2 } \}$.

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt[3]{ 2 })$. The minimal polynomial of $\sqrt[3]{ 2 }$ over $\mathbb{Q}$ is $x^{3} - 2$. $x^{3} - 2$ has three roots $$ \theta_{1} = \sqrt[3]{ 2 }, \quad \theta_{2} = \sqrt[3]{ 2 }e^{ 2 \pi i / 3 } = \sqrt[3]{ 2 } \frac{-1 + \sqrt{ 3 }}{2}, \quad \theta_{3} = \sqrt[3]{ 2 } e^{ 4 \pi i / 3 } = -\frac{\sqrt[3]{ 2 }(1 + \sqrt{ 3 })}{2} $$ in $\overline{\mathbb{Q}}$, but only $\sqrt[3]{ 2 }$ belongs to $\mathbb{Q}(\sqrt[3]{ 2 })$. $x^{3} - 2$ has only one root in $\mathbb{Q}(\sqrt[3]{ 2 })$. By Example 2106 (2), $$ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q} \right) = \{ 1 \}, \qquad \text{the trivial group}. $$

$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$, which is the splitting field of $x^{3} - 2$ over $\mathbb{Q}$ with $[\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}] = 6$ (part 2 of Example 1505). We show that $\mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \cong S_{3}$.

Keep the notation in Example 2106 (4). We have $$ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) = \left< \sigma_{1}, \sigma_{2} \right> \cong S_{3}, $$ with $$ \begin{align*} \sigma_1 : && \sqrt[3]{2} &\longmapsto \sqrt[3]{2}\frac{-1 + \sqrt{-3}}{2}, & \sigma_2 : && \sqrt[3]{2} &\longmapsto \sqrt[3]{2}, \\ && \frac{-1 + \sqrt{-3}}{2} &\longmapsto \frac{-1 + \sqrt{-3}}{2}, & && \frac{-1 + \sqrt{-3}}{2} &\longmapsto \frac{-1 - \sqrt{-3}}{2}. \end{align*} $$ One can check that $$ \begin{align*} \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}(\sqrt[3]{ 2 }) \right) &= \left< \sigma_{2} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2} \right) \right) & = \left< \sigma_{2} \sigma_{1} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 - \sqrt{ -3 }}{2} \right) \right) & = \left< \sigma_{2} \sigma_{1}^{2} \right> \quad (\cong \mathbb{Z} / 2\mathbb{Z}), \\ \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}(\sqrt{ -3 }) \right) & = \left< \sigma_{1} \right> \quad (\cong \mathbb{Z} / 3\mathbb{Z}), \end{align*} $$ and $$ \begin{align*} \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{2} \right> } &= \mathbb{Q}(\sqrt[3]{ 2 }), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{2} \sigma_{1} \right> } &= \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2} \right), \\ \mathbb{Q}(\sqrt[3]{ 2 }. \sqrt{ -3 })^{\left< \sigma_{2}\sigma_{1}^{2} \right> } &= \mathbb{Q}\left( \sqrt[3]{ 2 } \frac{-1 - \sqrt{ -3 }}{2} \right), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{1} \right> } &= \mathbb{Q}(\sqrt{ -3 }), \\ \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })^{\left< \sigma_{1}, \sigma_{2} \right> } &= \mathbb{Q}. \end{align*} $$