# 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 finite-dimensional Galois extension of fields with Galois group

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

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 $\operatorname{Gal}(L/K)$ 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 $f^{\alpha}_{F}\in F[X]$ be its minimal polynomial over $F$. Then the minimal polynomial $f^{\alpha}_{K}$ of $\alpha$ over $K$ divides $f^{\alpha}_{F}$, which has no double roots in its splitting field by the separability of $L/F$. Therefore $f^{\alpha}_{K}$ has no double roots in its splitting field for any $\alpha\in L$, so $L$ is separable over $K$.

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

###### Lemma 2.

The function $\phi$ from the set of extension fields of $F$ contained in $L$ to the set of subgroups of $G$ defined by

 $\phi(K)=\operatorname{Gal}(L/K)$

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

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

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

###### Proof.

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

 $\phi^{-1}\circ\phi(K)=K\quad\hbox{and}\quad\phi\circ\phi^{-1}(H)=H$

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 $\operatorname{Gal}(L/K)$ 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

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

so $\phi(K)\supset\phi(K^{\prime})$. This shows that $\phi$ is inclusion-reversing. ∎

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. 1.

$L^{H}$ is normal over $F$.

2. 2.

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

3. 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

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

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

 $\displaystyle r\colon G$ $\displaystyle\to$ $\displaystyle\operatorname{Gal}(L^{H}/F)$ $\displaystyle\sigma$ $\displaystyle\mapsto$ $\displaystyle\sigma|_{L^{H}}$

induces a natural identification

 $\operatorname{Gal}(L^{H}/F)\cong G/H.$
###### 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 well-defined 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 $L\neq L^{H}$, 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 $\sigma\in\operatorname{Gal}(L^{H}/F)$ 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 ProofOfFundamentalTheoremOfGaloisTheory 2013-03-22 14:26:38 2013-03-22 14:26:38 pbruin (1001) pbruin (1001) 5 pbruin (1001) Proof msc 12F10 msc 11R32 msc 11S20 msc 13B05