3.3 The fundamental theorem of Galois theory

Let $E$ be an extension of $F$ . A subextension of $E/F$ is an extension $M/F$ with $M\subset E$ , i.e., a field $M$ with $F\subset M\subset E$ . When $E$ is Galois over $F$ , the subextensions of $E/F$ are in one-to-one correspondence with the subgroups of $\operatorname{Gal}(E/F)$ . More precisely, there is the following statement.

Theorem 3.16 (Fundamental theorem of Galois theory)

Let $E$ be a Galois extension of $F$ with Galois group $G$ . The map $H\mapsto E^{H}$ is a bijection from the set of subgroups of $G$ to the set of subextensions of $E/F$ ,

\{\text{subgroups }H\text{ of }G\}\overset{1\colon 1}{\leftrightarrow}\{\text{% subextensions }F\subset M\subset E\},

with inverse $M\mapsto\operatorname{Gal}(E/M)$ . Moreover,

(a)

the correspondence is inclusion-reversing: $H_{1}\supset H_{2}\iff E^{H_{1}}\subset E^{H_{2}};$
(b)

indexes equal degrees: $(H_{1}\colon H_{2})=[E^{H_{2}}\colon E^{H_{1}}]$ ;
(c)

$\sigma H\sigma^{-1}\leftrightarrow\sigma M$ , i.e., $E^{\sigma H\sigma^{-1}}=\sigma(E^{H})$ ; $\operatorname{Gal}(E/\sigma M)=\sigma\operatorname{Gal}(E/M)\sigma^{-1}.$
(d)

$H$ is normal in $G\iff E^{H}$ is normal (hence Galois) over $F$ , in which case

$\operatorname{Gal}(E^{H}/F)\simeq G/H.$

Proof.

For the first statement, we have to show that $H\mapsto E^{H}$ and $M\mapsto\operatorname{Gal}(E/M)$ are inverse maps. Let $H$ be a subgroup of $G$ . Then, Corollary 3.5 shows that $\operatorname{Gal}(E/E^{H})=H$ . Let $M/F$ be a subextension. Then $E$ is Galois over $M$ by (3.13), which means that $E^{\operatorname{Gal}(E/M)}=M\,$ .

(a) We have the obvious implications,

H_{1}\supset H_{2}\implies E^{H_{1}}\subset E^{H_{2}}\implies\operatorname{Gal% }(E/E^{H_{1}})\supset\operatorname{Gal}(E/E^{H_{2}}).

As $\operatorname{Gal}(E/E^{H_{i}})=H_{i}$ , this proves (a).

(b) Let $H$ be a subgroup of $G$ . According to 3.11b,

(\operatorname{Gal}(E/E^{H})\colon 1)=[E\colon E^{H}].

This proves (b) in the case $H_{2}=1$ , and the general case follows, using that

\displaystyle(H_{1}\colon 1)

\displaystyle\overset{\phantom{1.20}}{=}(H_{1}\colon H_{2})(H_{2}\colon 1)\quad

\displaystyle[E\colon E^{H_{1}}]

\displaystyle\overset{\text{\ref{ef10}}}{=}[E\colon E^{H_{2}}][E^{H_{2}}\colon E% ^{H_{1}}].

\tau\alpha=\alpha\iff\sigma\tau\sigma^{-1}(\sigma\alpha)=\sigma\alpha.

Therefore, $\tau$ fixes $M$ if and only if $\sigma\tau\sigma^{-1}$ fixes $\sigma M\,$ , and so $\sigma\operatorname{Gal}(E/M)\sigma^{-1}=\operatorname{Gal}(E/\sigma M)$ . This shows that $\sigma\operatorname{Gal}(E/M)\sigma^{-1}$ corresponds to $\sigma M.$

(d) Let $H$ be a normal subgroup of $G$ . Because $\sigma H\sigma^{-1}=H$ for all $\sigma\in G$ , we must have $\sigma E^{H}=E^{H}$ for all $\sigma\in G$ , i.e., the action of $G$ on $E$ stabilizes $E^{H}$ . We therefore have a homomorphism

\sigma\mapsto\sigma|E^{H}\colon G\rightarrow\operatorname{Aut}(E^{H}/F)

whose kernel is $H$ . As $(E^{H})^{G/H}=F$ , we see that $E^{H}$ is Galois over $F$ (by Theorem 3.9) and that $G/H\simeq\operatorname{Gal}(E^{H}/F)$ (by 3.11b).

Conversely, suppose that $M$ is normal over $F$ , and let $\alpha_{1},\ldots,\alpha_{m}$ generate $M$ over $F$ . For $\sigma\in G$ , $\sigma\alpha_{i}$ is a root of the minimal polynomial of $\alpha_{i}$ over $F$ , and so lies in $M$ . Hence $\sigma M=M$ , and this implies that $\sigma H\sigma^{-1}=H$ (by (c)). _□

Remark 3.17

Let $E/F$ be a Galois extension, so that there is an order reversing bijection between the subextensions of $E/F$ and the subgroups of $G$ . From this, we can read off the following results.

(a) Let $M_{1},M_{2},\ldots,M_{r}$ be subextensions of $E/F$ , and let $H_{i}$ be the subgroup corresponding to $M_{i}$ (i.e., $H_{i}=\operatorname{Gal}(E/M_{i})$ ). Then (by definition) $M_{1}M_{2}\cdots M_{r}$ is the smallest field containing all $M_{i}$ ; hence it must correspond to the largest subgroup contained in all $H_{i}$ , which is $\bigcap H_{i}$ . Therefore

\operatorname{Gal}(E/M_{1}\cdots M_{r})=H_{1}\cap...\cap H_{r}.

(b) Let $H$ be a subgroup of $G$ and let $M=E^{H}$ . The largest normal subgroup contained in $H$ is $N=\bigcap\nolimits_{\sigma\in G}\sigma H\sigma^{-1}$ (see GT, 4.1), and so $E^{N}$ is the smallest normal extension of $F$ containing $M$ . Note that, by (a), $E^{N}$ is the composite of the fields $\sigma M$ . It is called the normal, or Galois, closure of $M$ in $E$ .

Proposition 3.18

Let $E$ and $L$ be extensions of $F$ contained in some common field. If $E/F$ is Galois, then $EL/L$ and $E/E\cap L$ are Galois, and the map

\sigma\mapsto\sigma|E\colon\operatorname{Gal}(EL/L)\rightarrow\operatorname{% Gal}(E/E\cap L)

is an isomorphism.

Proof.

Because $E$ is Galois over $F$ , it is the splitting field of a separable polynomial

$f\in F[X]$ . Then $EL$ is the splitting field of $f$ over $L$ , and $E$ is the splitting field of $f$ over $E\cap L$ . Hence $EL/L$ and $E/E\cap L$ are Galois. Every automorphism $\sigma$ of $EL$ fixing the elements of $L$ maps roots of $f$ to roots of $f$ , and so $\sigma E=E$ . There is therefore a homomorphism

\sigma\mapsto\sigma|E\colon\operatorname{Gal}(EL/L)\rightarrow\operatorname{% Gal}(E/E\cap L)\text{.}

If $\sigma\in\operatorname{Gal}(EL/L)$ fixes the elements of $E$ , then it fixes the elements of $EL$ , and hence is the identity map. Thus, $\sigma\mapsto\sigma|E$ is injective. If $\alpha\in$ $E$ is fixed by all $\sigma\in\operatorname{Gal}(EL/L)$ , then $\alpha\in E\cap L$ . By Corollary 3.5,

this implies that the image of $\sigma\mapsto\sigma|E$ is $\operatorname{Gal}(E/E\cap L)$ . _□

Corollary 3.19

Suppose, in the proposition, that $L$ is finite over $F$ . Then

[EL\colon F]=\frac{[E\colon F][L\colon F]}{[E\cap L\colon F]}\text{.}

Proof.

According to Proposition 1.20,

[EL\colon F]=[EL\colon L][L\colon F],

but

[EL\colon L]\overset{\ref{ft18f}}{=}[E\colon E\cap L]\overset{\ref{ef10}}{=}% \frac{[E\colon F]}{[E\cap L\colon F]}\text{.}

_□

Proposition 3.20

Let $E_{1}$ and $E_{2}$ be extensions of $F$ contained in some common field. If $E_{1}$ and $E_{2}$ are Galois over $F$ , then $E_{1}E_{2}$ and $E_{1}\cap E_{2}$ are Galois over $F$ , and the map

\sigma\mapsto(\sigma|E_{1},\sigma|E_{2})\colon\operatorname{Gal}(E_{1}E_{2}/F)% \rightarrow\operatorname{Gal}(E_{1}/F)\times\!\operatorname{Gal}(E_{2}/F)

is an isomorphism of $\operatorname{Gal}(E_{1}E_{2}/F)$ onto the subgroup

H=\{(\sigma_{1},\sigma_{2})\mid\sigma_{1}|E_{1}\cap E_{2}=\sigma_{2}|E_{1}\cap E% _{2}\}

of $\operatorname{Gal}(E_{1}/F)\times\!\operatorname{Gal}(E_{2}/F)$ .

Proof: Let $a\in E_{1}\cap E_{2}$ , and let $f$ be its minimal polynomial over $F$ . Then $f$ has

$\deg f$ distinct roots in $E_{1}$ and $\deg f$ distinct roots in $E_{2}$ . Since $f$ can have at most $\deg f$ roots in $E_{1}E_{2}$ , it follows that it has $\deg f$ distinct roots in $E_{1}\cap E_{2}$ . This shows that $E_{1}\cap E_{2}$ is normal and separable over $F$ , and hence Galois (3.9). As $E_{1}$ and $E_{2}$ are Galois over $F$ , they are splitting fields for separable polynomials $f_{1},f_{2}\in F[X]$ . Now $E_{1}E_{2}$ is a splitting field for $\textrm{lcm}(f_{1},f_{2})$ , and hence it also is Galois over $F$ . The map $\sigma\mapsto(\sigma|E_{1},\sigma|E_{2})$ is clearly an injective homomorphism, and its image is contained in $H$ . We’ll prove that the image is the whole of $H$ by counting.

From the fundamental theorem,

\frac{\operatorname{Gal}(E_{2}/F)}{\operatorname{Gal}(E_{2}/E_{1}\cap E_{2})}% \simeq\operatorname{Gal}(E_{1}\cap E_{2}/F)\text{,}

and so, for each $\sigma_{1}\in\operatorname{Gal}(E_{1}/F)$ , $\sigma_{1}|E_{1}\cap E_{2}$ has exactly $[E_{2}\colon E_{1}\cap E_{2}]$ extensions to an element of $\operatorname{Gal}(E_{2}/F)$ . Therefore,

(H\colon 1)=[E_{1}\colon F][E_{2}\colon E_{1}\cap E_{2}]=\frac{[E_{1}\colon F]% \cdot[E_{2}\colon F]}{[E_{1}\cap E_{2}\colon F]},

which equals $[E_{1}E_{2}\colon F]$ by (3.19) $.\hfill\square$