The Fundamental Theorem of Galois Theory

⚡ Key Takeaways

The Fundamental Theorem of Galois Theory (FTGT) establishes an order-reversing bijection between the intermediate fields of a finite Galois extension $L/K$ and the subgroups of its Galois group $\mathrm{Gal}(L/K)$ . In words: the lattice of fields between $K$ and $L$ is the mirror image of the lattice of subgroups of the symmetry group of the extension. Where it applies: finite (and with care, infinite) Galois extensions of arbitrary fields; central to solvability of polynomials, algebraic number theory, and arithmetic geometry. Who/when: envisioned by Évariste Galois (1830–1832), made rigorous by Dedekind, Weber, Emil Artin (1940s), and Krull (infinite case, 1928).

💡 The Core Insight

Symmetry organizes structure. The automorphisms of $L$ fixing $K$ form a group whose subgroup lattice is a faithful, inverted photograph of the field lattice. Bigger groups fix smaller fields; bigger fields are fixed by smaller groups. The dimension counts and the symmetry counts coincide exactly ( $|\mathrm{Gal}| = [L:K]$ ) only in the Galois case — that exact equality is what makes the correspondence a bijection rather than a mere injection.

DefinitionGalois Extension and Galois Group

Let $L/K$ be a field extension.

$\mathrm{Aut}(L/K)$ is the group of field automorphisms $\sigma: L \to L$ with $\sigma|_K = \mathrm{id}_K$ , under composition.
For a subset $S \subseteq \mathrm{Aut}(L/K)$ , the fixed field is

1

\quad L^{S} = \{\, x \in L : \sigma(x) = x \ \text{for all } \sigma \in S \,\}.

Derivation hint: $L^S$ is closed under $+,\cdot,(-)^{-1}$ because each $\sigma$ is a field homomorphism, so the fixed set inherits all field operations.

$L/K$ is finite Galois iff it is finite, normal (every irreducible $f \in K[x]$ with a root in $L$ splits in $L$ ), and separable (the minimal polynomial of every $\alpha \in L$ has no repeated roots). In that case we write $G = \mathrm{Gal}(L/K) = \mathrm{Aut}(L/K)$ .

The defining numerical criterion (Artin) is:

2

\quad L/K \text{ Galois} \iff |\mathrm{Aut}(L/K)| = [L:K] \iff L^{\mathrm{Aut}(L/K)} = K.

Derivation hint: in general $|\mathrm{Aut}(L/K)| \le [L:K]$ with equality forcing $K$ to be the full fixed field; Artin's lemma gives the reverse implication.

TheoremFundamental Theorem of Galois Theory

Let $L/K$ be a finite Galois extension with group $G = \mathrm{Gal}(L/K)$ , $|G| = [L:K] = n$ . Define

3

\quad \Phi: \{\,M : K \subseteq M \subseteq L\,\} \longrightarrow \{\,H : H \le G\,\}, \qquad \Phi(M) = \mathrm{Gal}(L/M),

Derivation hint: $L/M$ is automatically Galois for any intermediate $M$ since normality and separability descend to the top of the tower.

4

\quad \Psi: \{\,H : H \le G\,\} \longrightarrow \{\,M : K \subseteq M \subseteq L\,\}, \qquad \Psi(H) = L^{H}.

Derivation hint: $\Psi$ is well-defined because $L^H$ is a field containing $K = L^G \subseteq L^H$ . Then:

(1) Bijection. $\Phi$ and $\Psi$ are mutually inverse:

5

\quad L^{\mathrm{Gal}(L/M)} = M \quad\text{and}\quad \mathrm{Gal}(L/L^{H}) = H.

Derivation hint: $L/M$ Galois gives the first identity from criterion [2]; Artin's lemma applied to $H$ acting on $L$ gives the second.

(2) Order-reversal. $M_1 \subseteq M_2 \iff \Phi(M_2) \subseteq \Phi(M_1)$ .

(3) Degree–index match.

6

\quad [L:M] = |\mathrm{Gal}(L/M)|, \qquad [M:K] = [G : \mathrm{Gal}(L/M)].

Derivation hint: the first is criterion [2] for $L/M$ ; the second is the tower law $[L:K]=[L:M][M:K]$ combined with Lagrange's index formula.

(4) Normality correspondence. $M/K$ is normal (equivalently Galois) $\iff H = \mathrm{Gal}(L/M)$ is normal in $G$ . In that case restriction induces an isomorphism

7

\quad \mathrm{Gal}(M/K) \;\cong\; G\big/\mathrm{Gal}(L/M).

Derivation hint: restriction $\sigma \mapsto \sigma|_M$ is well-defined precisely when $\sigma(M)=M$ for all $\sigma$ , which is conjugation-invariance of $H$ ; the kernel is $\mathrm{Gal}(L/M)$ .

(5) Conjugation. For any $\sigma \in G$ , $\sigma(M)$ corresponds to the conjugate subgroup $\sigma \mathrm{Gal}(L/M)\sigma^{-1}$ .

ProofProof of the Fundamental Theorem

The engine is Artin's Lemma: if $H$ is a finite group of automorphisms of a field $L$ , then $[L : L^{H}] = |H|$ .

Step 1 — Artin's Lemma (linear independence of characters). Distinct automorphisms $\sigma_1,\dots,\sigma_m$ are linearly independent over $L$ as functions $L^\times \to L^\times$ (Dedekind). For $H = \{\sigma_1,\dots,\sigma_m\}$ put $F = L^{H}$ . One shows $[L:F] \le m$ : any $m+1$ elements of $L$ are $F$ -linearly dependent, by solving a homogeneous system whose coefficient matrix has rows $(\sigma_i(x_j))$ and exploiting a minimal nontrivial dependence to land in $F$ . Conversely $[L:F] \ge |H|$ since the $m$ distinct embeddings of $L$ over $F$ are linearly independent. Hence

8

\quad [L : L^{H}] = |H|.

Derivation hint: equality is squeezed from two inequalities — the upper from a counting/linear-algebra argument, the lower from independence of characters forcing at least $|H|$ embeddings.

Step 2 — $\Psi$ then $\Phi$ recovers $H$ . Let $F=L^H$ . Then $H \subseteq \mathrm{Gal}(L/F)$ trivially. By Artin, $[L:F]=|H|$ , while $|\mathrm{Gal}(L/F)| \le [L:F]$ . So $|\mathrm{Gal}(L/F)| \le |H|$ , forcing $\mathrm{Gal}(L/L^H)=H$ . This proves the second identity in [5].

Step 3 — $\Phi$ then $\Psi$ recovers $M$ . For intermediate $M$ , $L/M$ is normal and separable (subextension at the top of a normal separable tower), hence Galois, so by criterion [2], $L^{\mathrm{Gal}(L/M)} = M$ . This is the first identity in [5]. Steps 2–3 establish the bijection (1).

Step 4 — Order reversal and degrees. If $M_1 \subseteq M_2$ , any automorphism fixing $M_2$ fixes $M_1$ , so $\mathrm{Gal}(L/M_2) \subseteq \mathrm{Gal}(L/M_1)$ . Equation [6] follows from [8] with $H=\mathrm{Gal}(L/M)$ and the tower law.

Step 5 — Normal subgroups. Fix $\sigma \in G$ and intermediate $M$ with $H=\mathrm{Gal}(L/M)$ . Then for $\tau \in G$ , $\tau$ fixes $\sigma(M)$ pointwise iff $\sigma^{-1}\tau\sigma$ fixes $M$ , i.e.

9

\quad \mathrm{Gal}\big(L/\sigma(M)\big) = \sigma\, \mathrm{Gal}(L/M)\, \sigma^{-1}.

Derivation hint: $\tau(\sigma m)=\sigma m \iff (\sigma^{-1}\tau\sigma)(m)=m$ ; translate the pointwise-fixing condition through $\sigma$ . Thus $\sigma(M)=M$ for all $\sigma$ (i.e. $M/K$ normal) iff $H$ is invariant under all conjugations, i.e. $H \trianglelefteq G$ .

Step 6 — The quotient. When $H \trianglelefteq G$ , every $\sigma \in G$ satisfies $\sigma(M)=M$ , so restriction $\rho:\sigma \mapsto \sigma|_M$ maps $G \to \mathrm{Gal}(M/K)$ . Its kernel is exactly $\{\sigma : \sigma|_M=\mathrm{id}\} = \mathrm{Gal}(L/M)=H$ . Surjectivity follows because every $K$ -automorphism of $M$ extends to $L$ (normality of $L/K$ ). The first isomorphism theorem yields [7]. $\blacksquare$

∎

mathematician · Évariste Galois | 1811–1832

Galois, dead in a duel at twenty, never wrote "the Fundamental Theorem" — he wrote of the group of permutations of the roots that leaves all rational relations invariant. His 1831 memoir, rejected and nearly lost, contained the decisive idea: solvability by radicals corresponds to a chain of subgroups with abelian quotients (a solvable group). The modern lattice-theoretic statement, the fixed-field formalism, and the clean bijection are due to later hands — Dedekind introduced field automorphisms abstractly, Heinrich Weber systematized "Galois groups," and Emil Artin reorganized the entire theory around linear algebra and Artin's lemma, dethroning the primitive-element argument and making the proof characteristic-free and conceptually transparent.

🧠 Intuition

Think of $L$ as a crystal and $G$ as its symmetry group. Each intermediate field $M$ is a substructure you can carve out by demanding invariance under part of the symmetry. Demand invariance under the whole group $G$ : you get the smallest possible field, $K$ (the most symmetric part). Demand invariance under only the trivial subgroup $\{1\}$ : nothing is forced, you get all of $L$ . The order reversal is the natural "more constraints $\Rightarrow$ smaller object" logic: a larger subgroup imposes more symmetry constraints, so its fixed field is smaller. The degree of $L$ over $M$ literally counts how many symmetries you have left after pinning $M$ down — exactly $|\mathrm{Gal}(L/M)|$ . Normality is the subtle part. A subfield $M$ is "stable" — globally fixed as a set, not pointwise — exactly when the symmetries fixing it form a normal subgroup. Normal subgroups are the ones you can quotient by to get a new group; correspondingly normal subfields are the ones that carry their own self-contained Galois theory, with group the quotient.

Figure

The Galois correspondence for

\mathbb{Q}(\sqrt 2,\sqrt 3)/\mathbb{Q}

: fields (left) mirror subgroups (right)

ExampleThe biquadratic field

\mathbb{Q}(\sqrt2,\sqrt3)

Given. $L=\mathbb{Q}(\sqrt2,\sqrt3)$ , $K=\mathbb{Q}$ . The minimal polynomials $x^2-2$ and $x^2-3$ split in $L$ and are separable (char 0), so $L/\mathbb{Q}$ is Galois.

Step 1 — Degree. $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$ and $\sqrt3 \notin \mathbb{Q}(\sqrt2)$ , so $[L:\mathbb{Q}] = 2\cdot 2 = 4$ .

Step 2 — Group. Each automorphism sends $\sqrt2 \mapsto \pm\sqrt2$ , $\sqrt3 \mapsto \pm\sqrt3$ independently, giving $4$ automorphisms. Thus $G \cong \mathbb{Z}/2 \times \mathbb{Z}/2$ , the Klein four-group $V$ . Write $\tau_2:\sqrt2\mapsto-\sqrt2$ (fixes $\sqrt3$ ), $\tau_3:\sqrt3\mapsto-\sqrt3$ (fixes $\sqrt2$ ).

Step 3 — Subgroups → fixed fields.

$\langle\tau_3\rangle$ fixes $\sqrt2$ ⟹ $L^{\langle\tau_3\rangle}=\mathbb{Q}(\sqrt2)$ .
$\langle\tau_2\rangle$ fixes $\sqrt3$ ⟹ $\mathbb{Q}(\sqrt3)$ .
$\langle\tau_2\tau_3\rangle$ fixes $\sqrt2\sqrt3=\sqrt6$ ⟹ $\mathbb{Q}(\sqrt6)$ .

Step 4 — Normality. $V$ is abelian ⟹ every subgroup is normal ⟹ every intermediate field is Galois over $\mathbb{Q}$ (each is a degree-2, hence normal, extension).

\boxed{\;\text{3 nontrivial subgroups} \leftrightarrow \mathbb{Q}(\sqrt2),\ \mathbb{Q}(\sqrt3),\ \mathbb{Q}(\sqrt6);\quad G\cong V_4\;}

⚠️ Galois is not automatic

The bijection requires $L/K$ to be both normal and separable. Either failure breaks the theorem:

Inseparable failure: $K=\mathbb{F}_p(t)$ , $L=K(t^{1/p})$ . Then $x^p-t=(x-t^{1/p})^p$ has a single repeated root, $\mathrm{Aut}(L/K)=\{1\}$ , yet $[L:K]=p$ . The fixed field of the trivial group is all of $L$ , not $K$ , so [5] fails. The map $\Psi$ is not surjective onto $\{K \subseteq M \subseteq L\}$ .
Non-normal failure: $L=\mathbb{Q}(\sqrt[3]2)/\mathbb{Q}$ has $\mathrm{Aut}=\{1\}$ but degree 3; $x^3-2$ does not split.

CounterexampleOrder-reversal must not be dropped, and conjugates need not coincide

In $\mathbb{Q}(\sqrt[3]2,\omega)$ the three subfields $\mathbb{Q}(\omega^k\sqrt[3]2)$ are distinct fields of the same degree 3 over $\mathbb{Q}$ , corresponding to three distinct conjugate order-2 subgroups of $S_3$ . A common error is to assume "same degree ⟹ same field" or "all index-3 subgroups are normal." Both are false here: the lattice genuinely has three separate branches, and none of these subfields is Galois over $\mathbb{Q}$ because the matching subgroups are not normal. Equation [9] is what forces conjugate subgroups to give conjugate (isomorphic but possibly unequal) subfields.

⚠️ Infinite extensions need topology

For an infinite Galois extension (e.g. $\overline{\mathbb{Q}}/\mathbb{Q}$ , or $\overline{\mathbb{F}_p}/\mathbb{F}_p$ ), the naive bijection fails: there are more subgroups than intermediate fields. The fix (Krull) is to topologize $G$ with the Krull (profinite) topology, where a base of open neighborhoods of $1$ is given by $\mathrm{Gal}(L/M)$ for finite $M/K$ . Then the correspondence holds between intermediate fields and closed subgroups, not all subgroups. See Infinite Galois Theory.

connection

The FTGT is a hub linking many fields:

<a class="wiki-link" href="/wiki/math-algebra-solvability-by-radicals">Solvability by Radicals</a>: A polynomial is solvable by radicals iff its Galois group is a solvable group; the chain of subfields obtained by adjoining radicals mirrors a chain of normal subgroups with abelian quotients via [7].
Class Field Theory: describes abelian extensions of number fields — those with abelian Galois group — via the idele class group; the FTGT's normality/quotient structure is the foundation.
Inverse Galois Problem: which finite groups arise as $\mathrm{Gal}(L/\mathbb{Q})$ ? The correspondence makes this a precise question about realizing abstract groups.
Covering Space Theory: the topological analogue — the deck transformation group of a covering corresponds to subgroups of the fundamental group, with normal subgroups giving regular (Galois) covers. This is the prototype of a Galois Category.
Differential Galois Theory: replaces field automorphisms with those commuting with a derivation, yielding algebraic groups; solvability of linear ODEs in closed form mirrors solvability of the differential Galois group.

🔬 Where this appears

⏳ History

- 1770 — Lagrange. Réflexions sur la résolution algébrique des équations: studies how rational functions of roots permute, foreshadowing the "Galois resolvent."

- 1799–1813 — Ruffini. Argues (incompletely) that the general quintic is unsolvable by radicals.

- 1824 — Abel. Gives the first rigorous proof of quintic unsolvability, but without the conceptual group-theoretic framework.

- 1830–1832 — Galois. Introduces the group of an equation and the solvability criterion; memoir submitted to the Académie, rejected/lost. Dies in a duel May 1832 at 20.

- 1846 — Liouville. Publishes Galois's manuscripts in his Journal, finally making the work accessible.

- 1870 — Jordan. Traité des substitutions: first systematic treatment, framing Galois theory through permutation groups.

- 1893–1895 — Weber & Dedekind. Recast the theory abstractly using fields and automorphisms; "Galois group" enters standard language.

- 1928 — Krull. Extends the correspondence to infinite extensions via the profinite topology.

- 1942–1944 — Artin. Reformulates the entire subject around linear independence of characters and fixed fields (Artin's lemma), giving the clean, primitive-element-free proof used in modern graduate courses.

open-problem

Inverse Galois Problem. Is every finite group the Galois group of some extension of $\mathbb{Q}$ ? Known for all solvable groups (Shafarevich) and many simple groups, but open in general. The FTGT defines the question but does not solve it.
Grothendieck's anabelian program. To what extent does the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ , as a profinite group with its action on geometric fundamental groups, determine arithmetic varieties? The Section Conjecture remains open. This generalizes FTGT to a correspondence between covers and group actions (Grothendieck's Galois Theory).
Explicit Langlands reciprocity. The Langlands program seeks to understand all (including non-abelian) Galois representations via automorphic forms — a vast generalization of the abelian case settled by class field theory; large parts are conjectural.
Differential and difference Galois theory extend FTGT to linear ODEs/recurrences (Picard–Vessiot theory), with subtle constraints (algebraic closure of constants) and open inverse problems for differential Galois groups.

The Fundamental Theorem of Galois Theory

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also