3.5 Proof of Fundamental Theorem of Galois Theory

In this section, we prove the fundamental theorem of Galois theory. Firstly we prove the following theorem.

Theorem

Let $F/K$ be a finite extension. Then TFAE:

• $F/K$ is Galois;
• $F/K$ is separable and normal;
• $F$ is a splitting field of a separable polynomial $f(T)\in K[T]$.

\begin{proof} i)→ii) Let $u\in F$. Suppose $\mathrm{Irr}(u,K)=f(T)$ factors as ${\prod }_{i=1}^n(T-u_i)$ in some $\overline{F}[T]$. We aim to show $u_i\in F$ and $u_i$ are distinct. Now suppose (after reordering) $u_1,\cdots ,u_m$ are all the distinct $F$-roots with $u_1=u$ and write $g(T)={\prod }_{j=1}^m(T-u_j)\in F[T]$. Note that ${\mathrm{Aut}}_{K}F$ acts on the set $\{u_1,\cdots ,u_m\}$ because given any $\sigma \in {\mathrm{Aut}}_{K}F$, $\sigma (g(T))={\prod }_{j=1}^m(T-\sigma (u_j))=g(T)$. Hence, $g(T)\in F^{{\mathrm{Aut}}_{K}F}[T]=K[T]$ and so $g(T)=f(T)$. Therefore, $u_1,\cdots ,u_n$ are all contained in $F$ and they are distinct.

ii)→iii) Since $F/K$ is normal, $F$ is a splitting field of some $f(T)\in K[T]$. Each irreducible factor of $f(T)$ is the minimal polynomial of some $u\in F$ and so it is separable.

iii)→i) By ^m1looz, we need to find $[F:K]$-many elements in ${\mathrm{Aut}}_{K}F$. By induction on $[F:K]$. Pick any root $\alpha$ of $f(T)$ and let $p(T)=\mathrm{Irr}(\alpha ,K)\mid f(T)$.

We will construct elements in ${\mathrm{Aut}}_{K}F$ using $K\subseteq K(\alpha )\subseteq F$. Let $p(T)={\prod }_{i=1}^s(T-{\alpha }_i)$ with $\alpha ={\alpha }_1$. Recall that for any $1⩽i⩽s$ we have ${\theta }_i:K(\alpha )\simeq K[T]/(p(T))\simeq K({\alpha }_i)$. Then by ^28f933, ${\theta }_i$ extends to ${\theta }_i:F\to F$, and these are distinct elements in ${\mathrm{Aut}}_{K}F$ by ${\theta }_i(\alpha )={\alpha }_i$.

Since $f(T)$ is separable over $K(\alpha )[T]$ and $F$ is a splitting field of $f$, we have $F/K(\alpha )$ is Galois by induction hypothesis. By ^m1looz ${\mathrm{Aut}}_{K(\alpha )}F$ has exactly $n/s$ elements, denoted as $\{{\beta }_1,\cdots ,{\beta }_{n/s}\}$. It is easy to show ${\theta }_i{\beta }_j$ are distinct elements in ${\mathrm{Aut}}_{K}F$. Otherwise, suppose $Q_i{\beta }_j=Q_a{\beta }_b$, then $Q_i(\alpha )=Q_a(\alpha )$ and $i=a$. Thus $j=b$.

Now we finish the proof by induction. \end{proof}

Corollary

If $F/K$ is Galois and $F\supseteq E\supseteq K$, then $F/E$ is Galois.

\begin{proof} By ^232bdc ii). \end{proof}

Now we are ready to show the fundamental theorem of Galois theory.

• 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, $\mathrm{Gal}(E/K)=\mathrm{Gal}(F/K)/E^{\prime }=\mathrm{Gal}(F/K)/\mathrm{Gal}(F/E)$.

Link to original

\begin{proof} Step 0. Show the bijection. That is:

• $E\mapsto E^{\prime }\mapsto (E^{\prime }{)}^{\prime }=E^{\prime \prime }$, aiming to show $E^{\prime \prime }=E$;
• $H\mapsto H^{\prime }\mapsto H^{\prime \prime }$, aiming to show $H^{\prime \prime }=H$.

Note that $E^{\prime \prime }=F^{E^{\prime }}=F^{{\mathrm{Aut}}_{E}F}=E$ because $F/E$ is Galois. Also, note that $H^{\prime \prime }={\mathrm{Aut}}_{H^{\prime }}F={\mathrm{Aut}}_{F^H}F$ and $H⩽H^{\prime \prime }$. By ^l2077i, $F/F^H$ is Galois and $[F:F^H]=|\mathrm{Gal}(F/F^H)|=|H^{\prime \prime }|$ by ^m1looz and $[F:F^H]=|H|$ by ^ehefrs. Hence, $|H|=|H^{\prime \prime }|$ and so $H=H^{\prime \prime }$. Now we finish the proof.

Step 1. i) is easy to show by

$$[E_2:E_1]=\frac{[F:E_1]}{[F:E_2]}=\frac{|\mathrm{Gal}(F/E_1)|}{|\mathrm{Gal}(F/E_2)|}=\frac{|E_1^{\prime }|}{|E_2^{\prime }|}.$$

Step 2. Show ii).

We first show the followings.

Definition

For $K\subseteq E\subseteq F$, we say $E$ is stable if for any $\sigma \in {\mathrm{Aut}}_{K}F$, $\sigma (E)\subseteq E$.

Remark. If above holds, then ${\sigma }^{-1}(E)\subseteq E$ and $E\subseteq \sigma (E)$. Thus $\sigma (E)=E$. Furthermore, it implies that $\sigma {|}_E\in {\mathrm{Aut}}_{K}E$.

Lemma

Let $F/K$ be any algebraic extension.

• Let $F\supseteq E\supseteq K$ with $E$ stable. Then $E^{\prime }={\mathrm{Aut}}_{E}F<{\mathrm{Aut}}_{K}F$ is normal.
• If $H<{\mathrm{Aut}}_{K}F$ is normal, then $H^{\prime }=F^H$ is stable.

\begin{proof} i) Let $\tau \in E^{\prime }$, and let $\sigma \in {\mathrm{Aut}}_{K}F$. It suffices to show ${\sigma }^{-1}\tau \sigma \in {\mathrm{Aut}}_{E}F$. Let $e\in E$. Then ${\sigma }^{-1}\tau \sigma (e)=e$ as $\sigma (e)\in E$.

ii) Let $\sigma \in {\mathrm{Aut}}_{K}F$. We want to show $\sigma (u)\in H^{\prime }$ for all $u\in H^{\prime }$. That is, for any $h\in H$, $h(\sigma (u))=\sigma (u)$ for all $u\in H^{\prime }$. Then $({\sigma }^{-1}h\sigma )(u)=u$ by ${\sigma }^{-1}h\sigma \in H$ and we finish the proof. \end{proof}

Lemma

Let $K\subseteq E\subseteq F$, $F/K$ is Galois and $E$ is stable. Then $E/K$ is Galois.

\begin{proof} Note that $E/K$ is Galois iff $E^{{\mathrm{Aut}}_{K}E}=K$ iff $E^{{\mathrm{Aut}}_{K}E}\subseteq K$ iff for any $u\in E\setminus K$ there exists $\sigma \in {\mathrm{Aut}}_{K}E$ such that $\sigma (u)\ne u$. Now we construct $\sigma$ for each given $u$.

Since $F/K$ is Galois, for this $u\in F\setminus K$, there exists $\tilde{\sigma }\in {\mathrm{Aut}}_{K}F$ such that $\tilde{\sigma }(u)\ne u$. Since $E$ is stable, $\tilde{\sigma }{|}_E\in {\mathrm{Aut}}_{K}E$ and it is what we desire. \end{proof}

Lemma

Let $K\subseteq E\subseteq F$ be a finite extension. Suppose $E/K$ is Galois, then $E$ is stable.

\begin{proof} For any given $u\in E$, we aim to show $\sigma (u)\in E$ for all $\sigma \in {\mathrm{Aut}}_{K}F$. Let $\mathrm{Irr}(u,K)={\prod }_{i=1}^n(T-u_i)$ with $u_i\in E$ and $u_1=u$ by $E/K$ Galois and ^232bdc. Then $\sigma (u)=u_j$ for some $1⩽j⩽n$ and so $\sigma (u)\in E$. Therefore, $E$ is stable. \end{proof}

summary of lemmas above

Suppose $K\subseteq E\subseteq F$ is a finite extension and $F/K$ is Galois. Then TFAE:

• $E/K$ is Galois;
• ${\mathrm{Aut}}_{E}F⊲{\mathrm{Aut}}_{K}F$;
• $E$ is stable.

Now we get back to here and continue to prove ii) of the fundamental theorem of Galois theory.

By ^049cad, it remains to show $\mathrm{Gal}(E/K)=\mathrm{Gal}(F/K)/E^{\prime }=\mathrm{Gal}(F/K)/\mathrm{Gal}(F/E)$. Since $E$ is stable, there is a map

$$\alpha :{\mathrm{Aut}}_{K}F\to {\mathrm{Aut}}_{K}E,\sigma \mapsto \sigma {|}_E$$

which is a group homomorphism with $\ker \alpha ={\mathrm{Aut}}_{E}F$. Thus there is an embedding ${\mathrm{Aut}}_{K}F/{\mathrm{Aut}}_{E}F\hookrightarrow {\mathrm{Aut}}_{K}E$. But this is bijective because

$$|\text{LHS}|=\frac{[F:K]}{[F:E]}=[E:K]=|\text{RHS}|.$$

Now we finish the proof. \end{proof}