---
tag: X7R15C00ET
kind: statement
role: theorem
name: Artin's theorem on fixed fields
msc: [12F10]
uses: [0KMJ6919WS, 0K1SCPBNKJ, 0GS74FGFEK, 0DS2F7GK05, 0XVR72M51M, 0SVX98HYV1]
about: [0J1E5ZRZRM, 0QMZAQ6QQ4]
sources:
  - url: https://en.wikipedia.org/wiki/Galois_extension
---

## Statement

Let $L$ be a field and let $G$ be a finite subgroup of the [[automorphism group|0K1SCPBNKJ]] of $L$.
Let $K = L^G$ be the [[fixed field|0KMJ6919WS]] of $G$.
Then:

1. the [[degree|0GS74FGFEK]] $[L:K]$ equals the order $\abs{G}$;
2. $L/K$ is a [[Galois extension|0DS2F7GK05]];
3. the [[Galois group|0XVR72M51M]] $\Gal(L/K)$ equals $G$.

## 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|0SVX98HYV1]], 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|N43G50YHG2]].

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.

## Explanations

### L1 human

Start with a field and a finite group of its symmetries.
The elements that no symmetry moves form a smaller field.
Artin's theorem says that the big field is a Galois extension of the small one.
Its degree is the number of symmetries, and its Galois group consists exactly of these symmetries.

### L2 human

This theorem is one half of the [[fundamental theorem of Galois theory|0T1Q8D0FW8]].
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 $S_n$.