3.3 Galois Extension

Common notations

In this section:

• we always assume that $K\subseteq E$ and $K\subseteq F$;
• If $K\subseteq E\subseteq F$, then we say $E$ is intermediate extension;
• for a field homomorphism: $f:E\to F$, it always have $f(0)=0$ and $f(1)=1$, that is, $f$ is injective.

Definition

We say a field homomorphism $f:E\to F$ is a $K$-homomorphism if it is also a homomorphism of $K$-vector spaces.

In fact, $f$ is a $K$-homomorphism iff $f{|}_K=\mathrm{id}$.

\begin{proof} Note that $f(k)=kf(1)=k$ and $f(ke)=kf(e)$. \end{proof}

Definition

Denote ${\mathrm{Aut}}_K(F)=\{\sigma :F\to F\text{mbox}whichisK\text{mbox}-homomorphismandafieldautomorphism\}$.

Remark. ${\mathrm{Aut}}_K(F)$ is a group; if $[F:K]<\infty$, then ${\mathrm{Aut}}_K(F)=\{\sigma :F\to F\text{mbox}whichisK\text{mbox}-homomorphism\}$ by $\sigma$ injective and $[F:K]$ finite.

Lemma

For any $\sigma \in {\mathrm{Aut}}_K(F)$ and $f\in K[T]$, if $f(u)=0$, then $f(\sigma (u))=0$.

\begin{proof} Easy. \end{proof}

Remark. This lemma tells us ${\mathrm{Aut}}_K(F)$ acts on $\{F\text{mbox}-rootsoff(T)\in K[T]\}$.

Examples.

• ${\mathrm{Aut}}_K(K)={\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{Q}[2^{1/3}])=\{1\}$.
• ${\mathrm{Aut}}_{\mathbb{R}}(\mathbb{C})\simeq {\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{Q}[\sqrt{2}])\simeq \mathbb{Z}/2\mathbb{Z}$.
• ${\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{Q}(\sqrt{2},\sqrt{3}))={\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{Q}(\sqrt{2}+\sqrt{3}))$ and so $\mathrm{Irr}(\sqrt{2}+\sqrt{3},\mathbb{Q})=(x^2-1{)}^2-8x^2$, then we have ${\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{Q}(\sqrt{2}+\sqrt{3}))\simeq \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. See here.

Lemma

Let $K\subseteq F$, and let $H<{\mathrm{Aut}}_{K}F$. Let $H^{\prime }=F^H$, then $K\subseteq H\subseteq F$ is an intermediate extension.

Let $K\subseteq E\subseteq F$, and let $E^{\prime }={\mathrm{Aut}}_{E}F$. Then $E^{\prime }<{\mathrm{Aut}}_{K}F$ is a subgroup.

Example. Consider the field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and the corresponding Galois group ${\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{Q}(\sqrt{2},\sqrt{3}))=\{{\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_4\}$, where ${\sigma }_1=\mathrm{id}$, ${\sigma }_2((\sqrt{2},\sqrt{3}))=(\sqrt{2},-\sqrt{3})$, ${\sigma }_3((\sqrt{2},\sqrt{3}))=(-\sqrt{2},\sqrt{3})$ and ${\sigma }_4((\sqrt{2},\sqrt{3}))=(-\sqrt{2},-\sqrt{3})$.

• There are $5$ subgroups.
  • When $H=⟨{\sigma }_4⟩$, $F^H=\mathbb{Q}(\sqrt{6})$.
  • When $H=⟨{\sigma }_2⟩$, $F^H=\mathbb{Q}(\sqrt{2})$.
  • When $H=⟨{\sigma }_3⟩$, $F^H=\mathbb{Q}(\sqrt{3})$.
  • When $H=\{\mathrm{id}\}$, $F^H=\mathbb{Q}(\sqrt{2},\sqrt{3})$.
  • When $H=⟨{\sigma }_2,{\sigma }_3⟩$, $F^H=\mathbb{Q}$.
• Conversely, there are $5$ intermediate fields.
  • When $E=\mathbb{Q}(\sqrt{2})$, $E^{\prime }=⟨{\sigma }_2⟩$.
  • When $E=\mathbb{Q}(\sqrt{3})$, $E^{\prime }=⟨{\sigma }_3⟩$.
  • When $E=\mathbb{Q}(\sqrt{6})$, $E^{\prime }=⟨{\sigma }_4⟩$.
  • When $E=\mathbb{Q}$, $E^{\prime }=\{\mathrm{id}\}$
  • When $E=\mathbb{Q}(\sqrt{2},\sqrt{3})$, $E^{\prime }=⟨{\sigma }_2,{\sigma }_3⟩$.

Remark. There seems to be a nice correspondence between subgroups and subfields. However, how many intermediate fields exactly are there? By theorem of primitive element, there exists $\alpha$ such that $\mathbb{Q}\subseteq \mathbb{Q}(\alpha )\subseteq \mathbb{Q}(\sqrt{2},\sqrt{3})$ for any intermediate field. Recall an exercise in HW, it is easy to prove that there are $5$ intermediate fields exactly.

Another examples.

• ${\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{Q}(\sqrt{5},2^{1/3}))\simeq \mathbb{Z}/2\mathbb{Z}$ has $2$ subgroups but has more than $2$ intermediate extensions.
• ${\mathrm{Aut}}_{{\mathbb{F}}_p(x)}({\mathbb{F}}_p(x^{1/p}))=\{1\}$, but ${\mathbb{F}}_p(x^{1/p})$ has $2$ intermediate extensions.

Definition

Let $K\subseteq F$. If $({\mathrm{Aut}}_{K}F{)}^{\prime }=F^{{\mathrm{Aut}}_{K}F}=K$, then say $F$ is a Galois extension of $K$. For short, we say $F/K$ is a Galois extension or say $F$ is Galois over $K$. In this case, write $\mathrm{Gal}(F/K)={\mathrm{Aut}}_{K}F$ and call it the Galois group of the Galois extension.

fundamental theorem of Galois theory: the finite case

Suppose $F/K$ is a finite Galois extension. Then there is a bijection between

$$\{E:K\subseteq E\subseteq F\}⟷\{H:H<\mathrm{Gal}(F/K)\}$$

given by $E\to E^{\prime }$, $H^{\prime }\leftarrow H$. Furthermore,

• If $K\subseteq E_1\subseteq E_2\subseteq F$, then $E_2^{\prime }<E_1^{\prime }$ and $[E_2:E_1]=[E_1^{\prime }:E_2^{\prime }]$.
• For any $K\subseteq E\subseteq F$, $F/E$ is also Galois and $E/K$ is Galois iff $E^{\prime }$ is a normal subgroup. In this case, $\mathrm{Gal}(E/K)=\mathrm{Gal}(F/K)/E^{\prime }=\mathrm{Gal}(F/K)/\mathrm{Gal}(F/E)$.

\begin{proof} See 3.5 Proof of Fundamental Theorem of Galois Theory. \end{proof}

Remark. Galois extension is not transitive, that is, it is possible that $A/B$ and $B/C$ are Galois, but $A/C$ is not Galois. The reason is normality is not transitive, see here.