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Let $L/K$ be a Galois extension with Galois group $G$ and let $H \leq G$ be a subgroup. If one finds an element $\alpha \in L$ so that $\sigma \alpha = \alpha$ for all $\sigma \in H$ and $\sigma \alpha \neq \alpha$ for all $\sigma \not\in H$, then is the fixed field $L^H$ necessarily generated as $K(\alpha)$?

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Yes. By Galois correspondence $K(\alpha)=L^{H'}$ for a uniquely determined subgroup $H'\le G$. Because $\sigma$ fixes $\alpha$ if and only if $\sigma\in Gal(L/K(\alpha))=H'$ the conclusion follows.

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  • $\begingroup$ Of course -- thanks. I had $K(\alpha) \subset L^H$, but I should have just used the other direction of the correspondence to get $\supset$. $\endgroup$
    – MT_
    Commented Oct 17, 2017 at 3:40

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