Example. [2106B]

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 $\{ \pm \sqrt{ 2 } \}$.