A question about algebraic extension and how to extend an automorphism Let be $L/K$ an algebraic extension, $\alpha\in L$ and $P_\alpha$ the minimal polynomial of $\alpha$ over $K$. We denote $\beta\in L-\{\alpha\}$ another root of $P_{\alpha}$.
I can take an automorphism $\tau:K(\alpha)\longrightarrow \overline{K}$ such as $\tau(\alpha)=\beta$ and $\tau_{|_K}=Id$.
The question is:
why i can extend $\tau$ to an automorphism $\sigma:L\longrightarrow \overline{K}\hspace{0.15cm}$, i.e, $\hspace{0.15cm}\sigma_{|_{K(\alpha)}} = \tau$ ?
I see that in https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Extension
 A: The main claim here is that if $L/K$ is an algebraic field extension and $\Omega$ is any algebraically closed field (for example $\overline{K}$) then any homomorphism $\varphi: K\to \Omega$ has an extension to a homomorphism $\hat{\varphi}:L\to\Omega$. I'll assume here that $L/K$ is finite, for infinite algebraic extensions it can be done similarly using Zorn's lemma. 
First assume that $L=K(\alpha)$, a simple algebraic extension. Let $\varphi: K\to\Omega$ be a homomorphism, and we want to extend it to $L$. Let $\hat{K}=\varphi(K)$. Since any field homomorphism is injective we know $\varphi: K\to\hat{K}$ is an isomorphism. Also, if $P_{\alpha}=\sum_{i=0}^n c_ix^i$ is the minimal polynomial of $\alpha$ over $K$ then let $\hat{P_{\alpha}}=\sum_{i=0}^n \varphi(c_i)x^i\in\hat{K}[x]$. Since $\varphi$ is a bijection and preserves all the field operations it is easy to see that $\hat{P_{\alpha}}$ is irreducible over $\hat{K}$. So if we let $\beta\in\Omega$ be a root of $\hat{P_{\alpha}}$ (which exists, since $\Omega$ is algebraically closed) then $\hat{P_{\alpha}}$ is its minimal polynomial over $\hat{K}$. So just like we did in your previous question, there is a chain of isomorphisms:
$L=K(\alpha)\cong K[x]/(P_{\alpha})\cong \hat{K}[x]/(\hat{P_{\alpha}})\cong \hat{K}(\beta)\subseteq\Omega$
And if $k\in K$ then this chain sends $k\to k+(P_{\alpha})\to \varphi(k)+\hat{P_{\alpha}}\to \varphi(k)$. So this homomorphism $L\to\Omega$ is indeed an extension of $\varphi$. 
General case: Now assume $L/K$ is any finite extension and let $\varphi:L\to\Omega$ be a homomorphism, and we want to prove it has an extension to $L$. We use induction on $[L:K]$. If $[L:K]=1$ there is nothing to prove. Assume the statement is true for extensions of degree up to $n-1$ and suppose $[L:K]=n$. We split into two cases:
Case 1: $L/K$ has no intermediate fields. In that case for any $\alpha\in L\setminus K$ we have $L=K(\alpha)$. In that case we already know an extension exists.
Case 2: There is an intermediate field $K\subset M\subset L$. Then $[M:K],[L:M]<n$. By induction hypothesis $\varphi$ has an extension to $\varphi_0:M\to\Omega$. And again by induction hypothesis $\varphi_0$ has an extension to $\hat{\varphi}:L\to\Omega$. 
Conclusion:  Finally we can answer your question. $\tau:K(\alpha)\to\overline{K}$ is a homomorphism, so it can be extended to $L$. (since $L$ is an extension of $K(\alpha)$). 
