A problem about normal extensions and automorphisms

Question

this is my problem:

Suppose $K|F$ is a normal extension. Prove that for every $\alpha ,\beta \in K$ that have the same minimal polynomial over $F$ there is a $F$-algebra automorphism of $K$ (automorphism which is identity over $F$) $\phi:K\rightarrow K$ such that $\phi(\alpha)=\beta$.

we know that if $\alpha ,\beta \in K$ have the same minimal polynomial over $F$ then $F(\alpha)\simeq_F F(\beta)$,but how can i continue?

any hint is welcomed!

thank you very much!

Answer 1

if the minimal polynomial is $f$ perhaps one could find an $F$-isomorphism between $F(\alpha)$ and $F[X]/\langle f\rangle$ and similarly for $F(\beta)$ so that you could thereby construct an equivalence $F(\alpha) \cong F(\beta)$?

Answer 2

It could be proved that there exists a $F$-isomorphism $$\sigma:F(\alpha)=F[\alpha]\longrightarrow f[\beta]=F(\beta)\subseteq K$$ such that $\sigma(\alpha)=\beta$.

Now as $\bar{K}/F(\alpha)$ is algebraic and $\bar{K}$ is algebraically closed over $K$, by Theorem 3 of Lorenz , there is a field monomorphism extension $\displaystyle\tilde{\sigma}:\bar{K}\longrightarrow\bar{K}$ where $\tilde{\sigma}|_{F(\alpha)}=\sigma$.

Then, again by $F3$ of Lorenz, as $\tilde{\sigma}$ is an endomorphsim & monomorphism over $\bar{K}/F$ which is algebraic, hence $\tilde{\sigma}$ is an isomorphsim over $\bar{K}/F$.

Also by below theorem, we can infer that $\tilde{\sigma}(K)=K$.

So it suffices to define $\phi$ as follows: $$\phi:=\tilde{\sigma}|_{K}:K\longrightarrow K$$

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Theorem. Let $K/F$ and $L/K$ be normal and algebraic extensions, respectively. Then for every $F$*-isomorphsim* $\psi$ over $L$ : $$\psi(K)=K$$