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If we have $F \leq E \leq \mathbb{C}$, I know that for any $\alpha \in E$ and $\sigma \in \operatorname{Gal}(E/F)$, we have that $\sigma$ maps roots of $m_{\alpha,F}(x)$ to roots of $m_{\alpha,F}(x)$. I am wondering if there is any way to know, given another $\beta$ such that $m_{\alpha,F}(\beta) = 0$, if there exists some $\tau \in \operatorname{Gal}(E/F)$ such that $\tau(\alpha) = \beta$. It is my understanding that this is sometimes the case and sometimes not.

For a more specific example, I have a practice problem that says:

Let $F \leq E \leq \mathbb{C}$. Suppose that $E$ is a finite Galois extension of $F$ and there exist $\alpha \in E$ and $\sigma \in \operatorname{Gal}(E/F)$ such that $\sigma(\alpha)= - \alpha$. Prove that if $\beta \in E$ is a root of $m_{\alpha,F}(x)$, then there exists $\tau \in \operatorname{Gal}(E/F)$ such that $\tau(\beta) = - \beta$.

Clearly if we have a $\gamma \in \operatorname{Gal}(E/F)$ such that $\gamma(\alpha) = \beta$, then the composition of $\gamma$ and $\sigma$ is the element we desire, but I am unsure of how to show that $\gamma \in \operatorname{Gal}(E/F)$, or if it even is for that matter.

Thanks!

Ben A
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    the galois group acts transitively on roots of irreducible polynomials see: https://math.stackexchange.com/questions/140927/when-the-group-of-automorphisms-of-an-extension-of-fields-acts-transitively – wormram Apr 16 '21 at 01:57
  • Note that in your first paragraph you're not assuming $E/F$ is Galois, whereas in your specific example you are making that assumption. That's important. – Gerry Myerson Apr 16 '21 at 04:03

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