The Stacks project

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. $\square$