Normal Basis for Galois extension $ \Bbb K / \Bbb F$ is defined as looking at $\Bbb K$ as a vector space over $\Bbb F $ with the basis of $(\sigma(a)| \sigma \in Gal(\Bbb K / \Bbb F))$ for $a \in \Bbb K$

I need to proof that $ \Bbb K=\Bbb F(a)$

I know that Galois group is acting transitively on the roots of the minimal separable and normal polynomial over $\Bbb K$: $f(x)=\prod_{i=1}^n (x-\sigma_i(a))$

How I proof that for every i, $\sigma_i(a)=a^k $ for some $k \in \Bbb N$?

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    $\begingroup$ It is not true that $\sigma_i(a) = a^k$ for some $k$; nearly everything is a counterexample to that. $\endgroup$ – Magdiragdag Jun 9 '18 at 8:59
  • $\begingroup$ Any thoughts on my answer, Daniel? $\endgroup$ – Gerry Myerson Jun 10 '18 at 12:36
  • $\begingroup$ Earth to Daniel, come in, please. $\endgroup$ – Gerry Myerson Jun 12 '18 at 8:37
  • $\begingroup$ I agree it may be any structure with the binary action on $\alpha$ such that $\sigma_i(a)=a^3-a$ $\endgroup$ – Daniel Vainshtein Jun 12 '18 at 16:59

If the conjugates of $\alpha$ form a basis for the extension, then the number of conjugates must equal the degree of the extension, so the degree of $\alpha$ must equal the degree of the extension, so then degree of $F(\alpha)$ over $F$ must equal the degree of the extension, so $F(\alpha)=K$.


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