Question about algebraic extensions. 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 $P_{\alpha}$.
The question is:
why we can take an automorphism $\tau:K(\alpha)\longrightarrow \overline{K}$ such as $\tau(\alpha)=\beta$ and $\tau_{|_K}=Id$?
I see that in https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Extension
 A: It's a pretty known theorem that $K(\alpha)\cong K[x]/(P_{\alpha})$. Why is it true? Define $\varphi: K[x]\to K(\alpha)$ by $f\to f(\alpha)$. The kernel is exactly the ideal generated by $P_{\alpha}$, so by the first isomorphism theorem $K[x]/(P_{\alpha})\cong Im(\varphi)$. It is not hard to check that the image is all $K(\alpha)$. 
Now to your question. $P_{\alpha}$ is irreducible over $K$ (because it is a minimal polynomial of an element) and hence it must be the minimal polynomial of $\beta$ over $K$. So applying the previous theorem we get:
$K(\alpha)\cong K[x]/(P_{\alpha})\cong K(\beta)\subseteq\overline{K}$ 
The polynomial $f(x)=x$ is a polynomial in $K[x]$ which satisfies $f(\alpha)=\alpha$. So if we remember how the isomorphisms are defined we can see that it sends $\alpha\to x+(P_{\alpha})\to\beta$. Similarly, if $k\in K$ then the constant polynomial $f(x)=k$ satisfies $f(\alpha)=k$. So the isomorphism sends $k\to k+(P_{\alpha})\to k$. So indeed this isomorphism works as the identity on $K$ and sends $\alpha$ to $\beta$. 
