# Normal Basis for Galois extension $\Bbb K / \Bbb F$ implies $\Bbb K=\Bbb F(a)$

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 $k \in \Bbb N$?

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