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

$P = \prod \nolimits _{\beta \in G \cdot \alpha } (x - \beta ) \in K[x]$

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

$P^\tau = \prod \nolimits _{\beta \in G \cdot \alpha } (x - \tau (\beta )) = \prod \nolimits _{\beta \in G \cdot \alpha } (x - \beta ) = P$

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

$L = K^ G(\{ \sigma (\alpha _ i); 1 \leq i \leq n, \sigma \in G\} ) \subset K$

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. $\square$

Comment #5818 by Zhang on

In the second displayed equation, $P^{\sigma}$ should probably be $P^{\tau}$.

Comment #8509 by Et on

I dont get the second to last paragraph, how do you know that adding an element to L won't increase it?

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