proof of fundamental theorem of Galois theory
The theorem is a consequence of the following lemmas, roughly corresponding to the various assertions in the theorem. We assume $L/F$ to be a finitedimensional Galois extension^{} of fields with Galois 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 $L$ is Galois 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 Galois extension; it remains to prove that $L/K$ is also normal and separable. Since $L$ is normal and finite over $F$, it is the splitting field^{} of a polynomial^{} $f\in F[X]$ over $F$. Now $L$ is also the 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 the minimal polynomial ${f}_{K}^{\alpha}$ of $\alpha $ over $K$ divides ${f}_{F}^{\alpha}$, which has no double roots in its splitting field by the separability of $L/F$. Therefore ${f}_{K}^{\alpha}$ has no double roots in its 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 the fact that $K\supset F$. ∎
Lemma 2.
The function $\varphi $ from the set of extension fields of $F$ contained in $L$ to the set of subgroups of $G$ defined by
$$\varphi (K)=\mathrm{Gal}(L/K)$$ 
is an inclusionreversing 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 fixed field 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 $ is inclusionreversing. ∎
The following lemmas show that normal subextensions of $L/F$ are Galois 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 normal subgroup^{} 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 minimal polynomial of $\alpha $ over $F$. Since $L/F$ is normal, $f$ splits into linear factors in $L[X]$. Suppose ${\alpha}^{\prime}\in L$ is another zero of $f$, 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 extension of $F$, 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 a subextension of a separable extension is separable, ${L}^{H}/F$ is a Galois extension.
The map $r$ is welldefined by the implication^{} $1\Rightarrow 2$ from Lemma 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 we can 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. ∎
Title  proof of fundamental theorem of Galois theory 

Canonical name  ProofOfFundamentalTheoremOfGaloisTheory 
Date of creation  20130322 14:26:38 
Last modified on  20130322 14:26:38 
Owner  pbruin (1001) 
Last modified by  pbruin (1001) 
Numerical id  5 
Author  pbruin (1001) 
Entry type  Proof 
Classification  msc 12F10 
Classification  msc 11R32 
Classification  msc 11S20 
Classification  msc 13B05 