Proof of Artin's Theorem on Fixed Fields

Artin's theorem shows that any finite group of field automorphisms gives rise to a Galois extension, providing the foundation for the fundamental theorem.

Statement

Theorem8.1Artin's theorem

Let $G$ be a finite group of automorphisms of a field $K$ and $F = K^G$ the fixed field. Then $K/F$ is a finite Galois extension with $\mathrm{Gal}(K/F) = G$ and $[K:F] = |G|$ .

Proof

Let $|G| = n$ . We prove three claims.

Claim 1: $[K:F] \leq n$ .

We use the Dedekind independence lemma: distinct field automorphisms $\sigma_1, \ldots, \sigma_n$ of $K$ are linearly independent as functions $K \to K$ over $K$ . (Proof by induction: if $\sum c_i \sigma_i = 0$ minimally, pick $a$ with $\sigma_1(a) \neq \sigma_n(a)$ ; applying the relation to $ax$ and subtracting $\sigma_n(a)$ times the original gives a shorter relation, contradiction.)

Now suppose $[K:F] > n$ . Choose $\alpha_1, \ldots, \alpha_{n+1} \in K$ linearly independent over $F$ . Consider the system of $n$ equations in $n+1$ unknowns:

$\sum_{j=1}^{n+1} \sigma_i(\alpha_j) x_j = 0, \quad i = 1, \ldots, n.$

This has a nontrivial solution $(x_1, \ldots, x_{n+1}) \in K^{n+1}$ . Choose such a solution with the minimal number of nonzero $x_j$ . We may assume $x_1 = 1$ . For any $\sigma \in G$ , applying $\sigma$ to the equations:

$\sum_j \sigma(\sigma_i(\alpha_j)) \sigma(x_j) = 0.$

Since $\sigma G = G$ (as $G$ acts on itself by left multiplication), this is the same system with $x_j$ replaced by $\sigma(x_j)$ . Subtracting: $\sum_j \sigma_i(\alpha_j)(\sigma(x_j) - x_j) = 0$ . By minimality, $\sigma(x_j) = x_j$ for all $j$ , so all $x_j \in F$ . But then $\sum x_j \alpha_j = 0$ with $x_1 = 1$ , contradicting linear independence over $F$ .

Claim 2: $[K:F] \geq n$ .

$G \leq \mathrm{Aut}(K/F)$ , so $|\mathrm{Aut}(K/F)| \geq |G| = n$ . But for any extension, $|\mathrm{Aut}(K/F)| \leq [K:F]$ (number of roots of minimal polynomials). Combining with Claim 1: $[K:F] = n$ .

Claim 3: $K/F$ is Galois with $\mathrm{Gal}(K/F) = G$ .

For any $\alpha \in K$ , the orbit $\{\sigma(\alpha) : \sigma \in G\}$ has at most $n$ elements. The polynomial $\prod_{\sigma \in G}(x - \sigma(\alpha))$ has coefficients fixed by $G$ (symmetric in the orbit), hence in $F$ . So $\alpha$ is a root of a separable polynomial over $F$ that splits in $K$ . This gives normality and separability.

Since $\mathrm{Gal}(K/F) \geq G$ and $|\mathrm{Gal}(K/F)| = [K:F] = n = |G|$ : $\mathrm{Gal}(K/F) = G$ . $\blacksquare$

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Key Lemma

RemarkDedekind's independence of characters

The Dedekind independence lemma (used in the proof) states: if $\sigma_1, \ldots, \sigma_n: K^\times \to L^\times$ are distinct group homomorphisms (characters), then they are linearly independent over $L$ . This generalizes to arbitrary monoid homomorphisms and is the key algebraic fact underlying Artin's theorem.