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[parent] proof of fundamental theorem of Galois theory (Proof)

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 $$ G=\Gal(L/F). $$ 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 $\Gal(L/K)$ 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 $\Gal(L/K)$ is a subgroup of $G$ is clear from the fact that $K\supset F$ . $ \qedsymbol$

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 $$ \phi(K)=\Gal(L/K) $$ 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 identities $$ \phi^{-1}\circ\phi(K)=K\quad\hbox{and}\quad \phi\circ\phi^{-1}(H)=H $$ 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 $\Gal(L/K)$ 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 $K'$ of $F$ with $F\subset K\subset K'\subset L$ , we have $$ \sigma\in\Gal(L/K')\iff\sigma\in\Gal(L/K), $$ so $\phi(K)\supset\phi(K')$ . This shows that $\phi$ is inclusion-reversing. $ \qedsymbol$

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. $1\Rightarrow2$ : 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 normality of $L^H/F$ .

$2\Rightarrow3$ : 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 $$ \sigma\tau\sigma^{-1}\in\Gal(L/L^H)=H $$ for all $\sigma\in G,\tau\in H$ .

$3\Rightarrow1$ : 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 $\alpha'\in L$ is another zero of $f$ , and let $\sigma\in G$ be such that $\sigma(\alpha')=\alpha$ (such a $\sigma$ always exists). By assumption, for all $\tau\in H$ we have $\tau':=\sigma\tau\sigma^{-1}\in H$ , so that $$ \tau(\alpha')=\sigma^{-1}\tau'\sigma(\alpha') =\sigma^{-1}\tau'(\alpha)=\sigma^{-1}(\alpha)=\alpha'. $$ This shows that $\alpha'$ lies in $L^H$ as well, so $f$ splits in $L^H[X]$ . We conclude that $L^H$ is normal over $F$ . $ \qedsymbol$

Lemma 4   Let $H$ be a normal subgroup of $G$ . Then $L^H$ is a Galois extension of $F$ , and the homomorphism \begin{eqnarray*} r\colon G&\to&\Gal(L^H/F) \\ \sigma&\mapsto&\sigma\vert_{L^H} \end{eqnarray*}induces a natural identification $$ \Gal(L^H/F)\cong G/H. $$
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 $1\Rightarrow2$ 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 $\sigma\in\Gal(L^H/F)$ 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. $ \qedsymbol$




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Cross-references: first isomorphism theorem, kernel, primitive element theorem, automorphism, surjective, implication, well-defined, map, separable extension, induces, homomorphism, implies, fixed, equality, normal subgroup, the following are equivalent, quotient groups, properties, fixed field, inverse, bijection, function, clear, roots, divides, minimal polynomial, polynomial, splitting field, separable, normal, contained, extension fields, subgroups, Galois group, fields, Galois extension, finite-dimensional, consequence, theorem

This is version 2 of proof of fundamental theorem of Galois theory, born on 2004-06-23, modified 2004-06-29.
Object id is 5958, canonical name is ProofOfFundamentalTheoremOfGaloisTheory.
Accessed 3702 times total.

Classification:
AMS MSC12F10 (Field theory and polynomials :: Field extensions :: Separable extensions, Galois theory)
 11R32 (Number theory :: Algebraic number theory: global fields :: Galois theory)
 11S20 (Number theory :: Algebraic number theory: local and $p$-adic fields :: Galois theory)
 13B05 (Commutative rings and algebras :: Ring extensions and related topics :: Galois theory)

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