Extension of a field homomorphism $K(\alpha) \to \overline{K}$ to normal extension $L/K$

Question

I am trying to understand the equivalence of the two definitions of normal extension which can be found here https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Extension. In the proof that the second definition implies the first, one needs to show that given a generic finite extension $L/K$, $\alpha \in L$ and $\beta$ another root of the minimum polynomial of $\alpha$ over $K$, I can always extend the injective homomorphism $\tau: K(\alpha) \to \overline{K}$ such that $\tau(\alpha) = \beta$ to an injective homomorphism $\sigma: L \to \overline{K}$ such that $\sigma|_{K(\alpha)}=\tau$. Can someone sketch a proof for this? This part is missing on Wikipedia...

Answer 1

Let $F_0 = K(a), F_{n+1} =F_n(c_{n+1})$ such that $L = F_N$, at each step you need the minimal polynomial $f_{n+1} \in F_n[x]$ of $c_{n+1}$, given an homomorphism $\phi_n : F_n \to \overline{L}$, apply $\phi_n$ to the coefficients of $f_{n+1}$ to obtain $f_{n+1}^{\phi_n} \in \phi_n(F_n)[x]$, send $c_{n+1}$ to one of its roots to obtain an homomorphism $\phi_{n+1}:F_{n+1} \to \overline{L}$.

Doing so with your $K$-homomorphism $\phi_0(a) = b$ yields a $K$-homomorphism $\phi_N : L \to \overline{L}$ such that $\phi_N(a) = b$. If $L/K$ is normal then $\phi_N(L) = L$.