Artin's theorem on fixed fields

Remarks

For $\alpha \in L$, let $\alpha_1, \ldots, \alpha_r$ be the distinct elements of the orbit $G\alpha$. The polynomial $\prod_{i=1}^r (x - \alpha_i)$ is fixed by $G$, so it has coefficients in $K$; it is separable and splits in $L$. Hence every element of $L$ is separable over $K$ of degree at most $\abs{G}$, and $L/K$ is normal and separable. By the primitive element theorem, every finite subextension of $L/K$ is then generated by one element, so it has degree at most $\abs{G}$. Hence $[L:K] \le \abs{G}$. Conversely, $G \subseteq \Gal(L/K)$ and $\abs{\Gal(L/K)} \le [L:K]$ give the reverse inequality. Another proof uses Dedekind's lemma on independence of characters.

Example: let $L = \C(t)$ and let $G = \set{\id, \tau}$ with $\tau(f)(t) = f(-t)$. The fixed field is $\C(t^2)$, and indeed $[\C(t) : \C(t^2)] = 2$.

The hypothesis that $G$ is finite is necessary. The automorphism of $\C(t)$ with $t \mapsto t + 1$ generates an infinite group whose fixed field is $\C$, and $\C(t)/\C$ is not even algebraic.

Going deeper

For a student

This theorem is one half of the fundamental theorem of Galois theory. It shows that every subgroup $H$ of a finite Galois group $\Gal(L/K)$ is the Galois group of $L$ over the fixed field $L^H$, so the map $H \mapsto L^H$ is injective. It also produces Galois extensions directly: the symmetric group $S_n$ acting on $k(x_1, \ldots, x_n)$ shows that $k(x_1, \ldots, x_n)$ is Galois over the field of symmetric rational functions with group $S_n$.