Let $F$ be a field and $p(x)$ be an irreducible polynomial in $F[x]$. Let $K$ be a splitting field of $f(x)$.
Let $R:=\{\alpha_1,\alpha_2, \alpha_3,\dots,\alpha_k\}$ be the set of all roots of $p(x)$ in $K$. Then, it is easy to check that \begin{align*} \varphi_1:F(\alpha_1) &\to F(\alpha_2)\subseteq K \\ \alpha_1 & \to \alpha_2 \end{align*} is an $F$-isomorphism i.e. bijective ring homomorphism with $\varphi_1(x)=x$ for all $x\in F$.

Now, I want to extend the map $\varphi_1$ to an $F$-homomorphism \begin{align*} \varphi_2: F(\alpha_1)(\alpha_2)\to K \end{align*} Clearly, if such an extension $\varphi_2$ exists then it is determined by $\varphi_2(\alpha_2)$, and $\varphi_2(\alpha_2)$ must belong to $R$.

Question. How do we guarantee that such an extension $\varphi_2$ exist? Thanks!


Let $h(t) \in F(\alpha_1)[t]$ be the minimal polynomial of $\alpha_2$ over $F(\alpha_1)$, apply $\varphi_1$ to the coefficients of $h$ to obtain $h^{\varphi_1}(t) \in \varphi_1(F(\alpha_1))[t]$, set $\varphi_2(\alpha_2) = \beta_2$ for any root $\beta_2 \in K$ of $h^{\varphi_1}(t)$.

$\beta_2 $ always exists because $K/F$ is normal. Continuing this way you'll have $\varphi_k \in Aut(K/F)$.


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