proof of fundamental theorem of Galois theory

The theorem is a consequence of the following lemmas, roughlycorresponding to the various assertions in the theorem. We assume$L/F$ to be a finite-dimensional Galois extension of fields withGalois group

$$G=\mathrm{Gal}(L/F).$$

The first two lemmas establish the correspondence between subgroups of$G$ and extension fields of $F$ contained in $L$.

Lemma 1.

Let $K$ be an extension field of $F$ contained in $L$. Then $L$ isGalois over $K$, and $\mathrm{Gal}\mathit{}\mathrm{(}L\mathrm{/}K\mathrm{)}$ is a subgroup of $G$.

Proof.

Note that $L/F$ is normal and separable because it is a Galoisextension; it remains to prove that $L/K$ is also normal andseparable. Since $L$ is normal and finite over $F$, it is thesplitting field of a polynomial $f\in F[X]$ over $F$. Now $L$ is alsothe splitting field of $f$ over $K$ (because $F\subset K\subset L$),so $L/K$ is normal.

To see that $L/K$ is also separable, suppose $\alpha \in L$, and let$f_F^{\alpha }\in F[X]$ be its minimal polynomial over $F$. Then theminimal polynomial $f_K^{\alpha }$ of $\alpha$ over $K$ divides$f_F^{\alpha }$, which has no double roots in its splitting field by theseparability of $L/F$. Therefore $f_K^{\alpha }$ has no double roots inits splitting field for any $\alpha \in L$, so $L$ is separable over$K$.

The assertion that $\mathrm{Gal}(L/K)$ is a subgroup of $G$ is clear from thefact that $K\supset F$.∎

Lemma 2.

The function $\varphi$ from the set of extension fields of $F$ containedin $L$ to the set of subgroups of $G$ defined by

$$\varphi (K)=\mathrm{Gal}(L/K)$$

is an inclusion-reversing bijection. The inverse is given by

$${\varphi }^{-1}(H)=L^H,$$

where $L^H$ is the fixed field of $H$ in $L$.

Proof.

The definition of $\varphi$ makes sense because of Lemma 1.The

$${\varphi }^{-1}\circ \varphi (K)=K\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\varphi \circ {\varphi }^{-1}(H)=H$$

for all subgroups $H\subset G$ and all fields $K$ with $F\subset K\subset L$ follow from the properties of the Galois group. The fixedfield of $\mathrm{Gal}(L/K)$ is precisely $K$; on the other hand, since $L^H$is the fixed field of $H$ in $L$, $H$ is the Galois group of $L/L^H$.

For extensions $K$ and $K^{\prime }$ of $F$ with $F\subset K\subset K^{\prime }\subset L$, we have

$$\sigma \in \mathrm{Gal}(L/K^{\prime })\iff \sigma \in \mathrm{Gal}(L/K),$$

so $\varphi (K)\supset \varphi (K^{\prime })$. This shows that $\varphi$ isinclusion-reversing.∎

The following lemmas show that normal subextensions of $L/F$ areGalois extensions and that their Galois groups are quotient groups of$G$.

Lemma 3.

Let $H$ be a subgroup of $G$. Then the following are equivalent:

1. $L^H$ is normal over $F$.

2. $\sigma \left(L^H\right)=L^H$ for all $\sigma \in G$.

3. $\sigma H{\sigma }^{-1}=H$ for all $\sigma \in G$.

In particular, $L^H$ is normal over $F$ if and only if $H$ is a normalsubgroup of $G$.

Proof.

$1\Rightarrow 2$: Since for all $\sigma \in G$ and $\alpha \in L^H$,$\sigma (\alpha )$ is a zero of the minimal polynomial of $\alpha$ over$F$, we have $\sigma (\alpha )\in L^H$ by the of $L^H/F$.

$2\Rightarrow 3$: For all $\sigma \in G,\tau \in H$ the equality

$$\sigma \tau {\sigma }^{-1}(x)=\sigma {\sigma }^{-1}(x)=x$$

holds for all $x\in L^H$ (from the assumption it follows that${\sigma }^{-1}(x)\in L^H$, which is fixed by $\tau$). This implies that

$$\sigma \tau {\sigma }^{-1}\in \mathrm{Gal}(L/L^H)=H$$

for all $\sigma \in G,\tau \in H$.

$3\Rightarrow 1$: Let $\alpha \in L^H$, and let $f$ be the minimalpolynomial of $\alpha$ over $F$. Since $L/F$ is normal, $f$ splitsinto linear factors in $L[X]$. Suppose ${\alpha }^{\prime }\in L$ is another zeroof $f$, and let $\sigma \in G$ be such that $\sigma ({\alpha }^{\prime })=\alpha$(such a $\sigma$ always exists). By assumption, for all $\tau \in H$we have ${\tau }^{\prime }:=\sigma \tau {\sigma }^{-1}\in H$, so that

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

This shows that ${\alpha }^{\prime }$ lies in $L^H$ as well, so $f$ splits in$L^H[X]$. We conclude that $L^H$ is normal over $F$.∎

Lemma 4.

Let $H$ be a normal subgroup of $G$. Then $L^H$ is a Galois extensionof $F$, and the homomorphism

$r:G$$\to$$\mathrm{Gal}(L^H/F)$

$\sigma$$\mapsto$${\sigma |}_{L^H}$

induces a natural identification

$$\mathrm{Gal}(L^H/F)\cong G/H.$$

Proof.

By Lemma 3, $L^H$ is normal over $F$, and because asubextension of a separable extension is separable, $L^H/F$ is aGalois extension.

The map $r$ is well-defined by the implication $1\Rightarrow 2$ fromLemma 3. It is surjective since every automorphism of$L^H$ that fixes $F$ can be extended to an automorphism of $L$ (if$L\ne L^H$, for example, we can choose an $\alpha \in L\setminus L^H$such that $L=L^H(\alpha )$ using the primitive element theorem, and wecan extend $\sigma \in \mathrm{Gal}(L^H/F)$ to $L$ by putting$\sigma (\alpha )=\alpha$). The kernel of $r$ is clearly equal to $H$,so the first isomorphism theorem gives the claimed identification.∎