Galois Theory HT25, Artin's lemma

Suppose:

• $K$ is a field
• $H \le \text{Aut} _ \text{Rings}(K)$ finite subgroup

Then:

• $K^H = K^{\text{Aut} _ {K^H}(K)}$, and so in particular $K/K^H$ is Galois (and hence finite)
• $H \cong \text{Gal}(K/K^H)$

We shall first prove that

\[K^H = K^{\text{Aut} _ {K^H}(K)}\]

We have $K^H \subseteq K^{\text{Aut} _ {K^H}(K)}$ by definition. On the other hand, $H \le \text{Aut} _ {K^H}(K)$ by definition, so that $K^H \supseteq K^{\text{Aut} _ {K^H}(K)}$. Thus $K^H = K^{\text{Aut} _ {K^H}(K)}$.By a previous result, we also know that $[K : K^H] \le \vert H \vert $ and hence it is a finite extension. So then bythe resultthat says $F = K^G$ implies $K/F$ is Galois (where here $F$ is $K^H$), it follows byanother resultthat

\[[K : K^H] = |\text{Aut} _ {K^H}(K)|\]

Therefore we must have

\[|\text{Aut} _ {K^H}(K)| \le |H|\]

Since $H \le \text{Aut} _ {K^H}(K)$, we also have that

\[|H| \le |\text{Aut} _ {K^H}(K)|\]

and hence there must be of equal size, and since one is included in the other the two must actually be equal. Therefore $K/K^H$ is a finite Galois extension with Galois group $H$.

Suppose that $\sigma$ has finite order and let $G = \langle \sigma \rangle$. By Artin’s lemma,

\[[\mathbb Q(x) : \mathbb Q(x)^\sigma] \le |G|\]

and hence $\mathbb Q(x) / \mathbb Q(x)^\sigma$ is a finite extension.

Conversely, suppose that $\mathbb Q(x) / \mathbb Q(x)^\sigma$ is algebraic. Then it is finite, because $\mathbb Q(x)$ is generated by $x$ over $\mathbb Q(x)^\sigma$ and thus $\mathbb Q(x) / \mathbb Q(x)^\sigma$ is simple and algebraic.

Hence

\[G \le \text{Aut} _ {\mathbb Q(x)^G}(\mathbb Q(x)) \le \text{Aut}(\mathbb Q(x))\]

Then $\text{Aut} _ {\mathbb Q(x)^G}(\mathbb Q(x))$ embeds as a subgroup of the group of permutations of the roots of the minimal polynomial of $x$ over $\mathbb Q(x)^\sigma$, and is thus finite. Hence $G$ is finite.