Number of field homomorphism extending a given one Given a number fields extension $F\subseteq K$ I'm trying to prove that the characteristic polynomial of an element $\alpha\in K$ - which is defined to be the characteristic polynomial of the $F$-endomorphism of $K$ given by the multiplication $x\mapsto \alpha x$ - is a power of its minimal polynomial.
I know that if $F\subseteq K \subseteq E$ is the Galois closure of $K/F$ then the minimal polynomial of $\alpha$ is
$$ f^{min}_\alpha (x)=\prod_{\sigma\in Gal(E/F)}(x-\sigma(\alpha))$$
while if $\text{Hom}_F(K,\mathbb{C})=\{\sigma_1, \dots, \sigma_n\}$ its characteristic polynomial is given by
$$ f^{char}_\alpha (x)=\prod_{i=1}^n (x-\sigma_i(\alpha))$$
The result is the following
$$ f^{char}_\alpha (x)=f^{min}_\alpha(x)^{[K/F(\alpha)]}$$
and the textbook proves it as follows. The multiplicity of the factor $(x-\sigma_i(\alpha))$ in the characteristic polynomial is given by the number of $\sigma_j$ s.t. $\sigma_j(\alpha)=\sigma_i(\alpha)$ which is the number of $F$-homomorphism $K\to E$ extending $\sigma_i\big|_{F(\alpha)}$.
How do I prove that number is given by $[K/F(\alpha)]$? If $K/F$ was Galois I would use the fundamental theorem, but in this general case?

Let $\beta$ be a primitive element for $K/F(\alpha)$. Then with $p_\beta(x)\in F(\alpha)[x]$ the minimal polynomial for $\beta$ with $\deg p_{\beta}(x)=k$, we have

$$K\cong F(\alpha)[x]/(p_\beta(x))$$

And each choice of root gives at most one different embedding of $F(\alpha)(\beta)$ into the Galois closure, so the number of embeddings is bounded by $[K:F(\alpha)]$, i.e. there are at most that many. However, each root gives a different embedding as--if we number the roots as $\beta=\beta_1,\ldots,\beta_k$ then writing the isomorphisms

$$\phi_i: F(\alpha)(\beta_i)=K_i\to F(\alpha)[x]/(p_{\beta}(x))$$

where the equivalence class $[x]\in F(\alpha)[x]/(p_{\beta}(x))$ is mapped to $\beta_i$ as usual, then $\phi_i\circ \phi_j^{-1}$, $i\ne j$, is a non-identity--$\beta_i\mapsto\beta_j$ and $p_\beta(x)$ is irreducible--isomorphism between $K_i$ and $K_j$ as subfields of the Galois closure, hence the number of embeddings is at least $[K:F(\alpha)]$ proving equality.

Note the first inequality holds in general, the second one relies on the fact that an irreducible polynomial in characteristic $0$ is separable, i.e. all roots are distinct. This is critical for the equality, as the general case only has that the number of embeddings is at most $[K:F(\alpha)]$.
