Galois Theory HT25, Artin's lemma

Created: May 16, 2025 | Updated: May 29, 2025 | Read markdown | About these notes | View in context | Study these flashcards |

Flashcards

Statement

@State Artin’s lemma.

artins-lemma

Suppose:

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

Then:

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

Builds on ∆fixed-field-definition, ∆galois-group-definition, ∆galois-extension-definition

Proof

@Prove Artin’s lemma (∆artins-lemma), i.e. that if

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

then:

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

We shall first prove that

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

We have $K^H \subseteq K^{\mathrm{Aut} _ {K^H}(K)}$ by definition. On the other hand, $H \le \mathrm{Aut} _ {K^H}(K)$ by definition, so that $K^H \supseteq K^{\mathrm{Aut} _ {K^H}(K)}$. Thus $K^H = K^{\mathrm{Aut} _ {K^H}(K)}$.

By a previous result (∆artins-lemma-precursor-proof), we also know that $[K : K^H] \le \vert H \vert $ and hence it is a finite extension. So then by the result (∆f-is-fixed-field-implies-k-f-is-galois-proof) that says $F = K^G$ implies $K/F$ is Galois (where here $F$ is $K^H$), it follows by another result (∆galois-implies-g-equals-degree-proof) that

\[[K : K^H] = \vert \mathrm{Aut} _ {K^H}(K) \vert \]

Therefore we must have

\[ \vert \mathrm{Aut} _ {K^H}(K) \vert \le \vert H \vert \]

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

\[ \vert H \vert \le \vert \mathrm{Aut} _ {K^H}(K) \vert \]

and hence they 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 $H$.

Builds on ∆artins-lemma, ∆artins-lemma-precursor-proof, ∆f-is-fixed-field-implies-k-f-is-galois-proof, ∆galois-implies-g-equals-degree-proof

Examples

Suppose that

\[\sigma : \mathbb Q(x) \to \mathbb Q(x)\]

is an automorphism of the field of fractions of $\mathbb Q[x]$ (or equivalently, the field of rational functions over $\mathbb Q$).

@Prove that a field extension $\mathbb Q(x) / \mathbb Q(x)^\sigma$ is algebraic iff $\sigma$ is of finite order in $\mathrm{Aut} _ {\mathbb Q}(\mathbb Q(x))$.

qx-extension-algebraic-iff-sigma-finite-order

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

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

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

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

Then $\mathrm{Aut} _ {\mathbb Q(x)^G}(\mathbb Q(x))$ 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.



Related posts