The Galois Correspondence

A field extension$K/F$ records that $K$ is a larger field containing $F$. Galois theory studies its symmetries: the ways $K$ can be rearranged that leave $F$ untouched. These symmetries form a group, and for a well-behaved class of extensions the internal structure of that group reproduces, exactly and in reverse, the lattice of fields lying between $F$ and $K$. A question about fields (which subfields exist, which are themselves symmetric over $F$) becomes a question about a finite group (which subgroups exist, which are normal). The insolvability of the quintic and the three classical construction problems are settled by reading group structure off the extension.

Automorphisms and the group they form

Composition of automorphisms is again an automorphism, and the inverse of an isomorphism is an isomorphism, so $\mathrm{Aut}(K)$ is a group under composition. The identity map $1$ is always present. Every automorphism sends $1\mapsto 1$ and $0\mapsto 0$, hence fixes the entire prime subfield generated by $1$. So $\mathbb{Q}$ and ${\mathbb{F}}_p$ admit only the identity: $\mathrm{Aut}(\mathbb{Q})=\{1\}$ and $\mathrm{Aut}({\mathbb{F}}_p)=\{1\}$.

The single fact that makes these groups computable is that an automorphism cannot move a root of a polynomial anywhere except to another root of the same polynomial.

The proof is a one-line consequence of $\sigma$ being a ring homomorphism that fixes $F$. If ${\sum }_i{a}_i{\alpha }^i=0$ with $a_i\in F$, then applying $\sigma$ and using $\sigma (a_i)=a_i$ gives ${\sum }_i{a}_i(\sigma \alpha {)}^i=0$, so $\sigma \alpha$ satisfies the same equation.¹ Two corollaries follow immediately:

• Automorphisms are determined by their action on generators. If $K=F({\alpha }_1,\dots ,{\alpha }_k)$, then $\sigma \in \mathrm{Aut}(K/F)$ is determined by the values $\sigma {\alpha }_1,\dots ,\sigma {\alpha }_k$, each of which is a root of the corresponding minimal polynomial.
• $\mathrm{Aut}(K/F)$ is finite for finite extensions. There are finitely many generators, each with finitely many conjugate roots, so only finitely many candidate maps exist.

Not every assignment of generators to conjugate roots yields an automorphism — the generators may satisfy algebraic relations that any automorphism must respect. This is the recurring subtlety in computing a Galois group by hand.

First examples

Fixed fields and the Galois connection

The construction runs both ways. From a subfield we produced a group of automorphisms; from a group of automorphisms we recover a subfield.

That $K^H$ is closed under the field operations follows because each $h$ is a homomorphism: if $h$ fixes $\alpha$ and $\beta$ then it fixes $\alpha \pm \beta$, $\alpha \beta$, and ${\alpha }^{-1}$. The two constructions — subfield $⇝$ fixing group and subgroup $⇝$ fixed field — are inclusion-reversing.

A larger field imposes more conditions to fix, so fewer automorphisms qualify; a larger group of automorphisms fixes fewer elements. This pair of arrows between the poset of subfields and the poset of subgroups is a Galois connection. The content of the theory is that, restricted to the right extensions, the connection is a bijection.

Bounding the automorphism count

The example $\mathbb{Q}(\sqrt[3]{2})$ showed the automorphism group can be smaller than the degree. It is never larger. The bound comes from counting extensions of an isomorphism to a splitting field.

The proof is an induction on $[E:F]$ using the extension theorem for isomorphisms of base fields: an isomorphism $\phi :F\to F^{\prime }$ extends to at most $[E:F]$ isomorphisms of the splitting fields, and to exactly $[E:F]$ when the irreducible factors have distinct roots.² Taking $F=F^{\prime }$ and $\phi =1$ specializes the count to automorphisms. The same argument, refined in the next section, gives $|\mathrm{Aut}(K/F)|\le [K:F]$ for every finite extension.

Artin's theorem

The counting bound goes one direction. The reverse — that a finite group of automorphisms is the full automorphism group of its fixed field, and the degree equals the group order — is a theorem of Artin, and it rests on a lemma about the independence of characters.

An embedding $\sigma :K\hookrightarrow L$ restricts to a character $K^{\times }\to L^{\times }$, and this character carries all the information about $\sigma$ (only $0$ is omitted, and $\sigma (0)=0$). So the following theorem applies directly to field automorphisms.

The proof takes a minimal-length dependence relation ${\sum }_{i=1}^m{a}_i{\chi }_i=0$, picks $g_0$ with ${\chi }_1(g_0)\ne {\chi }_m(g_0)$, evaluates the relation at $g_{0}g$, and subtracts ${\chi }_m(g_0)$ times the original. The ${\chi }_m$ term cancels while the ${\chi }_1$ term remains with nonzero coefficient, producing a shorter relation, contradicting minimality.⁴ The application to fields is Artin's theorem.

Both inequalities are proved by manufacturing a linear dependence among the ${\sigma }_i$ where none can exist. If $n>[K:F]$, a homogeneous system with more unknowns than equations has a nonzero solution, which assembles into a dependence of the ${\sigma }_i$ over $K$ — impossible by independence of characters. If $n<[K:F]$, one takes $n+1$ elements independent over $F$ and, using that $G$ is closed under composition, again extracts a shorter-than-minimal dependence.⁵ Three corollaries follow.

• Sharp bound. $|\mathrm{Aut}(K/F)|\le [K:F]$ for any finite extension, with equality if and only if $F$ is the fixed field of $\mathrm{Aut}(K/F)$. So $K/F$ is Galois exactly when $F=K^{\mathrm{Aut}(K/F)}$.
• No hidden automorphisms. If $G$ is a finite group of automorphisms with fixed field $F$, then $\mathrm{Aut}(K/F)=G$ exactly; $K/F$ is Galois with Galois group $G$.
• Distinct groups, distinct fixed fields. Different finite subgroups of $\mathrm{Aut}(K)$ have different fixed fields.

The last corollary is the injectivity that the Fundamental Theorem upgrades to a bijection: taking fixed fields is a one-to-one map from subgroups to intermediate fields.

Characterizations of a Galois extension

Artin's theorem also delivers the converse of the counting bound and, with it, a structural characterization.

The forward direction shows that for any $\alpha \in K$, the distinct images $\{\sigma \alpha :\sigma \in G\}$ are the roots of a polynomial $f(x)={\prod }_i(x-{\alpha }_i)$ whose coefficients, being symmetric under $G$, lie in the fixed field $F$. Since the minimal polynomial of $\alpha$ divides $f$ and $f$ divides it, they are equal — so the minimal polynomial is separable and splits in $K$.⁶ This yields four interchangeable descriptions.

A normal extension is one that, containing a single root of an irreducible polynomial, contains all of them; a separable extension has minimal polynomials with distinct roots. An extension fails to be Galois the moment one irreducible polynomial has a root in $K$ but not all of them; $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ fails on this very point.

The Fundamental Theorem

Let $K/F$ be Galois with group $G=\mathrm{Gal}(K/F)$.

The bijection is the two Artin corollaries: distinct subgroups have distinct fixed fields (injective), and every intermediate field $E$ is the fixed field of $\mathrm{Aut}(K/E)$ because $K$ is the splitting field of the same separable polynomial viewed over $E$ (surjective). The degree statements are Artin applied to $K/E$ and the tower law.

The normality clause has an operational reading: conjugating $E$ by $\sigma \in G$ yields $\sigma (E)=K^{\sigma H{\sigma }^{-1}}$, so $E$ is stable under all of $G$ exactly when $\sigma H{\sigma }^{-1}=H$ for all $\sigma$, i.e. $H⊴G$. When $E/F$ is not Galois, the elements of $G$ still act, but they carry $E$ to conjugate fields isomorphic to it rather than back to itself.

A worked correspondence: $\mathbb{Q}(\sqrt{2},\sqrt{3})$

The field $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the splitting field of $(x^2-2)(x^2-3)$, hence Galois of degree $4$ over $\mathbb{Q}$. An automorphism sends $\sqrt{2}\mapsto \pm \sqrt{2}$ and $\sqrt{3}\mapsto \pm \sqrt{3}$ independently, giving four maps, all of which are automorphisms. Writing $\sigma$ for the sign flip on $\sqrt{2}$ and $\tau$ for the sign flip on $\sqrt{3}$,

the Klein four-group. Every subgroup of an abelian group is normal, so every intermediate field is Galois over $\mathbb{Q}$. The three subgroups of order $2$ correspond to the three quadratic subfields.

The composite $\sigma \tau$ fixes $\sqrt{6}=\sqrt{2}\sqrt{3}$ because it flips both factors. Displaying the two lattices side by side, inverted so that inclusions match, shows the bijection concretely.

A non-abelian correspondence: the splitting field of $x^3-2$

The splitting field of $x^3-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{2},\rho )$, where $\rho =e^{2\pi i/3}$ is a primitive cube root of unity. It has degree $6$, and the Galois group permutes the three roots $\sqrt[3]{2},\rho \sqrt[3]{2},{\rho }^2\sqrt[3]{2}$ faithfully, so $\mathrm{Gal}(\mathbb{Q}(\sqrt[3]{2},\rho )/\mathbb{Q})\cong S_3$. Generators are $\sigma$ (a $3$-cycle of the roots, sending $\sqrt[3]{2}\mapsto \rho \sqrt[3]{2}$ and fixing $\rho$) and $\tau$ (complex conjugation, fixing $\sqrt[3]{2}$ and sending $\rho \mapsto {\rho }^2$), with ${\sigma }^3={\tau }^2=1$ and $\tau \sigma ={\sigma }^2\tau$.

The subgroups of $S_3$ dictate the intermediate fields. Complex conjugation $\tau =({\alpha }_2{\alpha }_3)$ fixes the real root ${\alpha }_1=\sqrt[3]{2}$, and the $3$-cycle $\sigma =({\alpha }_1{\alpha }_2{\alpha }_3)$ generates the unique normal subgroup $⟨\sigma ⟩=A_3$. The full correspondence has six matched pairs.

The three normal subgroups $1,A_3,S_3$ match the three fields Galois over $\mathbb{Q}$: $K$ itself, the quadratic $\mathbb{Q}(\rho )=\mathbb{Q}(\sqrt{-3})$ with $\mathrm{Gal}(\mathbb{Q}(\rho )/\mathbb{Q})\cong S_3/A_3\cong \mathbb{Z}/2$, and the base $\mathbb{Q}$. The three conjugate order-$2$ subgroups match the three cubic fields $\mathbb{Q}(\sqrt[3]{2}),\mathbb{Q}(\rho \sqrt[3]{2}),\mathbb{Q}({\rho }^2\sqrt[3]{2})$ — isomorphic but distinct, permuted by $G$ as it conjugates the subgroups. None is Galois over $\mathbb{Q}$, since each holds one root of $x^3-2$ and misses the other two.