Lemma 9.21.6. Let K be a field. Let G be a finite group acting faithfully on K. Then the extension K/K^ G is Galois, we have [K : K^ G] = |G|, and the Galois group of the extension is G.
Proof. Given \alpha \in K consider the orbit G \cdot \alpha \subset K of \alpha under the group action. Consider the polynomial
The key to the whole lemma is that this polynomial is invariant under the action of G and hence has coefficients in K^ G. Namely, for \tau \in G we have
because the map \beta \mapsto \tau (\beta ) is a permutation of the orbit G \cdot \alpha . Thus P \in K^ G[x]. Since also P(\alpha ) = 0 as \alpha is an element of its orbit we conclude that the extension K/K^ G is algebraic. Moreover, the minimal polynomial Q of \alpha over K^ G divides the polynomial P just constructed. Hence Q is separable (by Lemma 9.12.4 for example) and we conclude that K/K^ G is separable. Thus K/K^ G is Galois. To finish the proof it suffices to show that [K : K^ G] = |G| since then G will be the Galois group by Lemma 9.21.2.
Pick finitely many elements \alpha _ i \in K, i = 1, \ldots , n such that \sigma (\alpha _ i) = \alpha _ i for i = 1, \ldots , n implies \sigma is the neutral element of G. Set
and observe that the action of G on K induces an action of G on L. We will show that L has degree |G| over K^ G. This will finish the proof, since if L \subset K is proper, then we can add an element \alpha \in K, \alpha \not\in L to our list of elements \alpha _1, \ldots , \alpha _ n without increasing L which is absurd. This reduces us to the case that K/K^ G is finite which is treated in the next paragraph.
Assume K/K^ G is finite. By Lemma 9.19.1 we can find \alpha \in K such that K = K^ G(\alpha ). By the construction in the first paragraph of this proof we see that \alpha has degree at most |G| over K. However, the degree cannot be less than |G| as G acts faithfully on K^ G(\alpha ) = L by construction and the inequality of Lemma 9.15.9. \square
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