Proof. sketch [1709A]

    1. The map $$\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \to \overline{F}, \qquad \varphi \mapsto \varphi(\alpha)$$ is injective since the image of $\varphi$ is completely determined by $\varphi(\alpha)$. To associate it with the roots of $m_{\alpha, F}(x)$ in $\overline{F}$, suppose $\varphi(\alpha) = \beta \in \overline{F}$. Then since $\varphi |_{F} = \operatorname{id}_{F}$ and $m_{\alpha, F}(x) \in F[x]$, we have $\varphi(m_{\alpha, F}(\alpha)) = m_{\alpha, F}(\beta) = \varphi(0) = 0$, hence $\beta$ is a root of $m_{\alpha, F}(x)$ in $\overline{F}$. Thus $\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \leq \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \}$.

      Conversely, given a root $\gamma$ of $m_{\alpha, F}(x)$ in $\overline{F}$, then the kernel of the field homomorphism $F[x] \to \overline{F}, a(x) \mapsto a(\gamma)$ is $\left( m_{\alpha, F}(x) \right)$. Thus we obtain a field embedding $$\tau : F[x] / \left( m_{\alpha, F}(x) \right) \to \overline{F}, \qquad \overline{a(x)} \mapsto a(\gamma).$$ Moreover, by Proposition 1407, $$\sigma : F[x] / \left( m_{\alpha, F}(x) \right) \to F(\alpha), \qquad \overline{a(x)} \mapsto a(\alpha)$$ is a field isomorphism. Then we can check that the field homomorphism $\tau \circ \sigma ^{-1} |_{F} = \operatorname{id}_{F}$ and $\tau \circ \sigma ^{-1} (\alpha) = \gamma$, so $\tau \circ \sigma ^{-1}$ is the desired field embedding. Thus $\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \geq \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \}$.

    2. By $\# \{ \text{roots of }m_{\alpha, F}(x) \text{ in } \overline{F} \} \leq \deg m_{\alpha, F}(x) = [F(\alpha) : F]$, and the equality holds if and only if $\alpha$ is separable over $F$.