The Galois Correspondence

The identity map is an $F$-automorphism, the composite of two $F$-automorphisms is an $F$-automorphism, and the inverse of an $F$-isomorphism is again an $F$-isomorphism fixing $F$; associativity is inherited from composition. Thus $\mathrm{Aut}(K\mathrm{/}F)$ is a group. Every such map fixes $0$ and $1$, so no zero case is excluded.

For finite $K\mathrm{/}F$, choose $\tau :K\to \mathrm{\Omega }$ as in [A1]. The map $\sigma \mapsto \tau \circ \sigma$ sends $\mathrm{Aut}(K\mathrm{/}F)$ into the set of $F$-embeddings $K\to \mathrm{\Omega }$ and is injective, because $\tau \circ {\sigma }_1=\tau \circ {\sigma }_2$ and injectivity of $\tau$ imply ${\sigma }_1={\sigma }_2$.

Step 1.2 and the definition of separable degree give $\mathrm{\mid }\mathrm{Aut}(K\mathrm{/}F)\mathrm{\mid }\le [K:F{]}_s$, while [L1] gives $[K:F{]}_s\le [K:F]$. If $K=F$, all three numbers are $1$, so this includes the degree-one endpoint.

The empty family is independent vacuously, and a singleton is independent because its character never vanishes. Suppose, for contradiction, that some finite distinct family is dependent; among all nonzero relations choose one with the least support, relabel its supported characters as ${\chi }_1,\dots ,{\chi }_r$, and divide by the first nonzero coefficient to write ${\sum }_{i=1}^r{a}_i{\chi }_i(x)=0$ for every $x\in G$, where $r\ge 2$ and $a_1=1$.

Since ${\chi }_1\mathrm{\ne }{\chi }_r$, choose $g\in G$ with ${\chi }_1(g)\mathrm{\ne }{\chi }_r(g)$. Evaluate the relation of step 1.1 at $gx$ and subtract ${\chi }_r(g)$ times its value at $x$ to obtain ${\sum }_{i=1}^{r-1}{a}_i({\chi }_i(g)-{\chi }_r(g)){\chi }_i(x)=0$ for every $x\in G$.

The new relation has support smaller than $r$ but is nonzero because its ${\chi }_1$-coefficient is ${\chi }_1(g)-{\chi }_r(g)\mathrm{\ne }0$. This contradicts the minimality in step 1.1, so no nontrivial relation exists and the characters are linearly independent.

Restricted to $K^{\times }$, the distinct automorphisms ${\sigma }_1,\dots ,{\sigma }_m$ are distinct characters $K^{\times }\to K^{\times }$, so [L1] makes them linearly independent as functions. Hence their evaluation vectors span $K^m$: otherwise a nonzero linear functional on their span would give a nontrivial $K$-linear relation among the ${\sigma }_i$. Choose nonzero $x_1,\dots ,x_m\in K$ such that the evaluation matrix $A=({\sigma }_i(x_j){)}_{i,j}$ is invertible. For $m=1$, one may take $x_1=1$.

Suppose $c_1,\dots ,c_m\in K^G$ satisfy ${\sum }_j{c}_j{x}_j=0$. Applying each ${\sigma }_i$ and using ${\sigma }_i(c_j)=c_j$ gives $A(c_1,\dots ,c_m{)}^{\mathsf{T}}=0$, so invertibility of $A$ forces every $c_j=0$. Thus $x_1,\dots ,x_m$ are linearly independent over $K^G$.

A $K^G$-linearly independent family of $m=\mathrm{\mid }G\mathrm{\mid }$ elements of $K$ gives $[K:K^G]\ge m=\mathrm{\mid }G\mathrm{\mid }$.

Define the $K$-linear map $T:K^{m+1}\to K^m$ by $T(c_1,\dots ,c_{m+1})=({\sum }_j{c}_j{\sigma }_i(x_j){)}_{i=1}^m$. Since the domain has dimension $m+1$ and the codomain dimension $m$, [L1] gives a nonzero vector in $\ker T$.

Among nonzero vectors in $\ker T$, choose $c=(c_j)$ with least support and scale it so its first supported coordinate is $1$. For $\tau \in G$, applying $\tau$ to all equations $T(c)=0$ and reindexing the rows by ${\sigma }_i\mapsto \tau {\sigma }_i$ shows that $\tau (c)=(\tau (c_j))$ also lies in $\ker T$. The vector $\tau (c)-c$ has a zero in the normalized coordinate and support strictly smaller than that of $c$ unless it vanishes; minimality therefore gives $\tau (c_j)=c_j$ for every $j$ and every $\tau \in G$, so all $c_j$ lie in $K^G$.

The identity row of $T(c)=0$ is ${\sum }_j{c}_j{x}_j=0$, a nontrivial $K^G$-linear dependence among the arbitrary $m+1$ elements. Thus no $m+1$ elements of $K$ are linearly independent over $K^G$, and $[K:K^G]\le m=\mathrm{\mid }G\mathrm{\mid }$. This includes $m=1$ and also covers repeated or zero $x_j$.

The opposing bounds [L1] and [L2] give $[K:K^G]=\mathrm{\mid }G\mathrm{\mid }$. In particular $K\mathrm{/}{K}^G$ is finite. For $G=\{1\}$ this says $K^G=K$ and the degree is $1$.

Every $\sigma \in G$ fixes $K^G$, so $G\subseteq \mathrm{Aut}(K\mathrm{/}{K}^G)$. By [L3] and step 1.1, $\mathrm{\mid }\mathrm{Aut}(K\mathrm{/}{K}^G)\mathrm{\mid }\le [K:K^G]=\mathrm{\mid }G\mathrm{\mid }$; a finite set containing $G$ and having at most its cardinality equals $G$.

If $K^G=K^H=E$, then [L1] applied to each group gives $G=\mathrm{Aut}(K\mathrm{/}E)=H$. Thus equal fixed fields force equal groups, including when either group is trivial.

Step 1.1 is precisely injectivity of $G\mapsto K^G$; its contrapositive says that distinct finite automorphism groups have distinct fixed fields.

Artin's theorem gives $\mathrm{\mid }\mathrm{Aut}(K\mathrm{/}F)\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }=[K:E]$.

The tower formula gives $[K:F]=[K:E][E:F]=\mathrm{\mid }\mathrm{Aut}(K\mathrm{/}F)\mathrm{\mid }[E:F]$, proving the divisibility. Degree one and a trivial automorphism group both give divisor $1$.

For the implication from condition 1 to condition 2, choose a finite generating family for $K\mathrm{/}F$. By [L2], normality makes $K$ the splitting field of the product of their distinct minimal polynomials; separability makes each factor separable, so their distinct product is separable. If $K=F$, the empty product $1$ has splitting field $F$.

For the implication from condition 2 to condition 3, let $K$ be the splitting field of the separable $f\in F[x]$. Then $K$ is generated over $F$ by the roots of $f$, and the minimal polynomial of each root divides $f$ and therefore has no repeated root, so every generator is separable over $F$; by [L3], $K\mathrm{/}F$ is separable and the full-degree criterion in the Given gives $[K:F{]}_s=[K:F]$. Every $F$-embedding of $K$ into an algebraic closure permutes the roots of $f$ and hence maps $K$ onto itself, so the $[K:F{]}_s$ embeddings are precisely the elements of $G$ and $\mathrm{\mid }G\mathrm{\mid }=[K:F]$.

For the implication from condition 3 to condition 4, [L1] gives $[K:K^G]=\mathrm{\mid }G\mathrm{\mid }=[K:F]$. Since $F\subseteq K^G\subseteq K$, the tower law forces $[K^G:F]=1$, hence $K^G=F$.

For the implication from condition 4 to condition 1, [L4] makes $G$ finite, so [L1] applies to $H=G$ and gives $[K:K^G]=\mathrm{\mid }G\mathrm{\mid }$; with $K^G=F$ this reads $\mathrm{\mid }G\mathrm{\mid }=[K:F]$. The bound in [L4] then gives $[K:F]=\mathrm{\mid }G\mathrm{\mid }\le [K:F{]}_s\le [K:F]$, so $[K:F{]}_s=[K:F]$ and the full-degree criterion in the Given makes $K\mathrm{/}F$ separable. For $\alpha \in K$, the orbit polynomial $q_{\alpha }(x)={\prod }_{\beta \in G\alpha }(x-\beta )$ has distinct roots in $K$ and coefficients fixed by $G$, hence in $K^G=F$. The minimal polynomial of $\alpha$ divides $q_{\alpha }$, while every orbit element is one of its roots; separability makes $q_{\alpha }$ divide that minimal polynomial. They are therefore equal, so every minimal polynomial over $F$ splits in $K$ and $K\mathrm{/}F$ is normal. This also covers $\alpha =0$ and the degree-one extension.

Normality of $K\mathrm{/}F$ descends through the intermediate field, so $K\mathrm{/}E$ is normal.

For $\alpha \in K$, its minimal polynomial over $E$ divides its separable minimal polynomial over $F$, since the latter lies in $E[x]$ and vanishes at $\alpha$. A divisor of a separable polynomial is separable, so $K\mathrm{/}E$ is separable. This includes $\alpha =0$.

The extension $K\mathrm{/}E$ is finite because a finite $F$-basis spans it over $E$; together with steps 1.1 and 1.2 this makes $K\mathrm{/}E$ finite Galois. Both endpoints are included: $E=F$ recovers the hypothesis and $E=K$ gives the degree-one extension.

Choose a finite generating family $K=F({\alpha }_1,\dots ,{\alpha }_r)$. By [L1], the normal closure $L$ is finite over $F$ and is the splitting field of the product of the minimal polynomials of the generators. If $r=0$, then $K=F$ and $L=F$. Repeated minimal polynomials may be removed from the product.

Each generator is separable over $F$, so every root of its minimal polynomial is separable over $F$. The field $L$ is generated by those roots, hence is separable; it is normal by construction, so it is finite Galois.

If $M$ is a subfield of $\mathrm{\Omega }$ containing $K$ and Galois over $F$, then $M\mathrm{/}F$ is normal. The normal closure is the intersection of all normal subextensions containing $K$, so $L\subseteq M$. Thus $L$ is the required minimal Galois overfield.

For the subgroup-to-field-to-subgroup direction, Artin applied to $H$ gives $\mathrm{Gal}(K\mathrm{/}{K}^H)=\mathrm{Aut}(K\mathrm{/}{K}^H)=H$. This includes $H=\{1\}$, whose fixed field is $K$, and $H=G$.

For the field-to-subgroup-to-field direction, put $H=\mathrm{Gal}(K\mathrm{/}E)$. Since $K\mathrm{/}E$ is finite Galois, [L2] gives $\mathrm{\mid }H\mathrm{\mid }=[K:E]$. Artin gives $[K:K^H]=\mathrm{\mid }H\mathrm{\mid }=[K:E]$, while $E\subseteq K^H$; the tower law forces $E=K^H$. This includes $E=K$ and $E=F$.

If $H_1\subseteq H_2$, then every element fixed by $H_2$ is fixed by $H_1$, so $K^{H_2}\subseteq K^{H_1}$; the reverse map is likewise inclusion-reversing. Artin gives $[K:K^H]=\mathrm{\mid }H\mathrm{\mid }$, while [L2] applied to $K\mathrm{/}F$ gives $[K:F]=\mathrm{\mid }G\mathrm{\mid }$, so the tower and Lagrange formulas give $[K^H:F]=\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }H\mathrm{\mid }=[G:H]$. Together with steps 1.1 and 1.2 these statements prove the claimed bijections and degree formulas.

For $x\in K$, the element $x$ is fixed by $\sigma H{\sigma }^{-1}$ exactly when ${\sigma }^{-1}(x)$ is fixed by $H$, exactly when ${\sigma }^{-1}(x)\in E$, and exactly when $x\in \sigma (E)$. Thus $K^{\sigma H{\sigma }^{-1}}=\sigma (E)$, and [L1] gives $\mathrm{Gal}(K\mathrm{/}\sigma (E))=\sigma H{\sigma }^{-1}$.

For the forward direction, if $H⊴G$, then step 1.1 gives $\sigma (E)=E$ for every $\sigma \in G$; every $F$-conjugate of an element of $E$ is obtained by extending its embedding to $K$ and hence lies in $E$, so $E\mathrm{/}F$ is normal, and it is separable as a subextension of the separable extension $K\mathrm{/}F$, hence Galois. For the reverse direction, if $E\mathrm{/}F$ is Galois, normality gives $\sigma (E)=E$ for every $\sigma \in G$, so step 1.1 and [L1] give $\sigma H{\sigma }^{-1}=H$ and [F1] gives $H⊴G$.

In the normal case, restriction $\rho :G\to \mathrm{Gal}(E\mathrm{/}F)$ is defined by step 2.1. Every $F$-automorphism of $E$ extends to an embedding of $K$ in an algebraic closure; normality of $K\mathrm{/}F$ makes the extension an element of $G$, so $\rho$ is surjective. Its kernel consists exactly of the automorphisms fixing $E$, namely $H$ by [L1]. The first isomorphism theorem therefore gives $G\mathrm{/}H\cong \mathrm{Gal}(E\mathrm{/}F)$; for $H=\{1\}$ and $H=G$ this yields the two endpoint quotients.

A finite group has a finite power set, and its subgroups form a subcollection of that power set; hence $G$ has finitely many subgroups.

By [L1], the intermediate fields are in bijection with those subgroups, so there are finitely many. When $K=F$, both collections have one member, and the base and top endpoints coincide.

An automorphism of $K$ fixes $E_1{E}_2$ exactly when it fixes every element of both $E_1$ and $E_2$, exactly when it belongs to both $H_1$ and $H_2$. Therefore $\mathrm{Gal}(K\mathrm{/}{E}_1{E}_2)=H_1\cap H_2$.

An element of $K$ is fixed by $⟨H_1,H_2⟩$ exactly when it is fixed by every element of both generating subgroups, so $K^{⟨H_1,H_2⟩}=K^{H_1}\cap K^{H_2}=E_1\cap E_2$. Applying [L1] gives the second formula. These membership equivalences also cover equal fields, the base and top fields, and trivial or full subgroups.

By [L1], $E$ is the splitting field over $F$ of a separable polynomial $f$. The same polynomial over $L$ remains separable and has splitting field $EL$, so [L1] makes $EL\mathrm{/}L$ finite Galois.

An element of $\mathrm{Gal}(EL\mathrm{/}L)$ permutes the roots of $f$, hence preserves $E$, and its restriction to $E$ fixes $D=E\cap L$. Restriction therefore defines a homomorphism into $\mathrm{Gal}(E\mathrm{/}D)$; it is injective because $EL$ is generated by $E$ and $L$.

Let $H$ be the restriction image. By [L1], an element of $E$ is fixed by $H$ exactly when it is fixed by every automorphism of $EL\mathrm{/}L$, exactly when it lies in $L$; thus $E^H=E\cap L=D$. By [L2], $H=\mathrm{Aut}(E\mathrm{/}{E}^H)=\mathrm{Gal}(E\mathrm{/}D)$, so restriction is surjective and hence an isomorphism. If $E\subseteq L$, both groups are trivial; if $L=F$ or $D=F$, the displayed formula specializes directly.

Restriction sends $\mathrm{Gal}(E_1{E}_2\mathrm{/}F)$ injectively into the product of the two relative Galois groups, since $E_1{E}_2$ is generated by $E_1$ and $E_2$. Both restrictions agree on $D$, so the image lies in the displayed fibre product.

Conversely, let $({\sigma }_1,{\sigma }_2)$ have equal restrictions to $D$. Extend ${\sigma }_2$ to some $\tau \in \mathrm{Gal}(E_1{E}_2\mathrm{/}F)$ using surjectivity of restriction. Then $\rho :={\sigma }_1(\tau {\mathrm{\mid }}_{E_1}{)}^{-1}$ fixes $D$, and [L1] supplies $h\in \mathrm{Gal}(E_1{E}_2\mathrm{/}{E}_2)$ with $h{\mathrm{\mid }}_{E_1}=\rho$. The automorphism $h\tau$ restricts to ${\sigma }_1$ and ${\sigma }_2$, proving that every compatible pair is in the image.

For the forward implication of the last assertion, if $D=F$ then compatibility is automatic and step 2.1 gives the full product. For the reverse implication, if the image is the full product, every pair $(1,{\sigma }_2)$ is compatible, so every ${\sigma }_2\in \mathrm{Gal}(E_2\mathrm{/}F)$ fixes $D$; its fixed field is $F$, hence $D=F$. If $E_1=E_2$, the fibre product is instead the diagonal subgroup, as the formula requires.