The Fundamental Theorem of Galois Theory

Why can the roots of $x^5 - x - 1$ not be expressed by radicals? Why can the roots of $x^5 - 1$ ? The answer hides inside a dictionary. This dictionary translates questions about fields into questions about groups. There, finiteness and combinatorics make hard problems tractable. This dictionary is the Fundamental Theorem of Galois Theory. It is arguably the single most influential correspondence in algebra. The idea is that a finite Galois extension $K/F$ carries a rich internal structure. Fields sit between $F$ and $K$ . Naively one might expect these intermediate fields to be hard to classify. There could be infinitely many, with no way to organize them. The theorem shows the opposite. They correspond exactly, bijectively, and order-reversingly. This correspondence is to the subgroups of a single finite group. That group is the Galois group $\mathrm{Gal}(K/F)$ . The big picture is that field theory questions become group theory questions. Does this equation have a solution by radicals? Is this field embeddable in that one? What are all the fields between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_n)$ ? These become: is this group solvable? Does it have a subgroup of this index? What is the subgroup lattice of $(\mathbb{Z}/n\mathbb{Z})^\times$ ? Group theory is combinatorial and computable in ways that field theory alone is not. So the correspondence is a genuine engine for solving classical problems. This article states the correspondence with full precision. It proves it from the defining properties of Galois extensions. The essential inputs are the Primitive Element Theorem and Artin's theorem on fixed fields. It works two asymmetric examples in full detail. It shows exactly which hypothesis is load-bearing — finiteness, normality, separability — by breaking each one.

Key takeaways

For a finite Galois extension $K/F$ , intermediate fields $F \subseteq E \subseteq K$ correspond bijectively to subgroups $H \le \mathrm{Gal}(K/F)$ .
The correspondence is inclusion-reversing: bigger fields go with smaller groups, and conversely.
Degrees translate directly into group indices: $[K:E] = |H|$ and $[E:F] = [\mathrm{Gal}(K/F):H]$ .
An intermediate field $E$ is Galois over $F$ exactly when its corresponding subgroup $H$ is normal in $\mathrm{Gal}(K/F)$ , and then $\mathrm{Gal}(E/F) \cong \mathrm{Gal}(K/F)/H$ .
Every step of the proof rests on two pillars: the Primitive Element Theorem (finite separable extensions are simple) and Artin's Lemma on fixed fields of finite automorphism groups.
Key idea: a finite Galois extension packages all of its intermediate field structure into the combinatorics of one finite group.

Where this lives today

The correspondence is the computational backbone of the theory of solvability by radicals: the Galois group of $x^5-x-1$ over $\mathbb{Q}$ is the full symmetric group $S_5$ , which is not solvable because $A_5$ is simple, and the correspondence translates "no radical formula" directly into "no composition series of the subgroup lattice with abelian factors." In number theory, the correspondence applied to cyclotomic extensions $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ (with Galois group $(\mathbb{Z}/n\mathbb{Z})^\times$ ) organizes all subfields of $\mathbb{Q}(\zeta_n)$ — including quadratic subfields governing quadratic reciprocity — as fixed fields of subgroups of a completely explicit abelian group. In algebraic geometry and topology, the same correspondence appears as the classification of covering spaces of a topological space by subgroups of its fundamental group, an exact formal analogue proved by an essentially identical fixed-point argument (deck transformations replace field automorphisms). In cryptography, understanding which finite fields $\mathbb{F}_{p^n}$ sit inside which others, and the Frobenius automorphism structure governing them, is Galois theory over finite fields applied directly to constructing and analyzing elliptic-curve and pairing-based cryptosystems. In computer algebra, algorithms for computing Galois groups of polynomials over $\mathbb{Q}$ rely on the correspondence to certify which subfields of a splitting field exist and to bound possible group structures via resolvent polynomials.

:::quiz[If $[K:F] = 24$ and $K/F$ is Galois, and $H \le \mathrm{Gal}(K/F)$ has order 6, what is $[K^H : F]$ ?]

6
4 ✓
24
18

:::quiz[In the $D_4$ example ( $K$ = splitting field of $x^4-2$ ), why is $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}$ not Galois?]

Its degree is not a power of 2
$x^4-2$ is separable, contradicting normality
Its corresponding subgroup $\langle\tau\rangle$ is not normal in $D_4$ ✓
$\mathbb{Q}(\sqrt[4]2)$ is not a subfield of $K$

:::faq[Is the correspondence still a bijection if we only assume $K/F$ is normal but allow $[K:F]$ to be infinite?] Not without modification. For infinite (algebraic) Galois extensions, $\mathrm{Gal}(K/F)$ is given the Krull topology, a profinite topology in which basic open subgroups are stabilizers of finite subextensions. The correspondence then holds between intermediate fields and closed subgroups of $\mathrm{Gal}(K/F)$ , not all abstract subgroups; e.g. for $\mathrm{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p) \cong \hat{\mathbb{Z}}$ , the abstract subgroup $\mathbb{Z} \subset \hat{\mathbb{Z}}$ is dense but not closed and does not correspond to any intermediate field. The finite case in this article is the setting where every subgroup is automatically closed (finite groups carry only the discrete topology), so no such subtlety arises.

Ask the tutor: Can you walk through why the compositum and intersection formulas in part 5 require separately checking two containments each, rather than following immediately from the bijection alone?

K/F

Intuition

Think of $\mathrm{Gal}(K/F)$ as a group of symmetries of $K$ that "see" $F$ as the featureless background — every element of $F$ is invisible to every symmetry, fixed in place. An intermediate field $E$ is a partial refinement of that background: more structure than $F$ is now visible and must be preserved, so fewer symmetries can preserve it. The bigger $E$ is, the more rigid the constraint, and the smaller the group of symmetries respecting it. This is exactly why the correspondence reverses inclusions: enlarging the field shrinks the stabilizing group, just as fixing more points of a geometric figure shrinks its symmetry group. The reverse direction says the same thing from the group side. A subgroup $H \le G$ picks out exactly those elements of $K$ that every symmetry in $H$ leaves alone — the fixed field $K^H$ . A bigger subgroup imposes more constraints (more symmetries that all must fix a point), so fewer points survive, and $K^H$ shrinks. The theorem's content is that this back-and-forth loses no information at all: starting from $E$ , fixing it, and then taking the fixed field of the resulting group returns exactly $E$ ; and dually for subgroups. The normal subgroup clause has its own geometric meaning. A subgroup $H$ is normal in $G$ exactly when conjugation by any $\sigma \in G$ sends $H$ to itself, which happens exactly when $\sigma$ permutes the fixed field $K^H$ among itself — i.e. $\sigma$ restricted to $E = K^H$ is again an automorphism of $E$ fixing $F$ . When this holds for every $\sigma \in G$ , the field $E$ is itself Galois over $F$ , and the "leftover" symmetries of $E$ over $F$ are captured exactly by the quotient $G/H$ .

Figure

The subgroup lattice of

\mathrm{Gal}(K/\mathbb{Q})

for

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

mirrored against the field lattice, inclusion-reversed

K/F

H

Worked examples

Example 1: $K = \mathbb{Q}(\sqrt2, \sqrt3)$ over $F = \mathbb{Q}$ . This is the splitting field of $(x^2-2)(x^2-3)$ over $\mathbb{Q}$ , hence normal; it is separable since $\mathrm{char}\,\mathbb{Q} = 0$ . So $K/\mathbb{Q}$ is Galois. A short computation shows $[K:\mathbb{Q}] = 4$ : $[\mathbb{Q}(\sqrt2):\mathbb{Q}] = 2$ , and $\sqrt3 \notin \mathbb{Q}(\sqrt2)$ (else $\sqrt6 = \sqrt2\sqrt3 \in \mathbb{Q}(\sqrt2)$ would force $\sqrt3 = a + b\sqrt2$ for rationals $a,b$ , giving $3 = a^2+2b^2 + 2ab\sqrt2$ , forcing $ab=0$ ; if $b=0$ , $3=a^2$ impossible in $\mathbb{Q}$ ; if $a=0$ , $3 = 2b^2$ impossible in $\mathbb{Q}$ ), so $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)] = 2$ . $G = \mathrm{Gal}(K/\mathbb{Q})$ has order 4, generated by

\sigma: \sqrt2 \mapsto -\sqrt2,\ \sqrt3 \mapsto \sqrt3, \qquad \tau: \sqrt2\mapsto\sqrt2,\ \sqrt3\mapsto-\sqrt3,

which commute and each have order 2, so $G = \{1,\sigma,\tau,\sigma\tau\} \cong \mathbb{Z}/2 \times \mathbb{Z}/2$ (the Klein four-group). Since $G$ is abelian, every subgroup is normal, so by part 4, every intermediate field is Galois over $\mathbb{Q}$ . The proper nontrivial subgroups are $\langle\sigma\rangle$ , $\langle\tau\rangle$ , $\langle\sigma\tau\rangle$ , each of order 2, hence index 2. Their fixed fields, each of degree 2 over $\mathbb{Q}$ by part 3, are found by inspection: $\sigma\tau$ sends $\sqrt2 \mapsto -\sqrt2, \sqrt3\mapsto-\sqrt3$ , so it fixes $\sqrt6 = \sqrt2\sqrt3$ . Thus

K^{\langle\sigma\rangle} = \mathbb{Q}(\sqrt3), \qquad K^{\langle\tau\rangle} = \mathbb{Q}(\sqrt2), \qquad K^{\langle\sigma\tau\rangle} = \mathbb{Q}(\sqrt6).

Together with $K^{\{1\}} = K$ and $K^G = \mathbb{Q}$ , this exhausts all five subgroups of $G$ and all five intermediate fields, matching the lattice in the figure above — note the correspondence really does biject the five subgroups with the five fields, with no extra intermediate fields hiding anywhere, exactly as the theorem guarantees.

Example 2: $K =$ splitting field of $x^4 - 2$ over $F = \mathbb{Q}$ , an asymmetric non-abelian case. Let $\alpha = \sqrt[4]{2}$ (real fourth root) and $i = \sqrt{-1}$ . The roots of $x^4-2$ are $\alpha, i\alpha, -\alpha, -i\alpha$ , so $K = \mathbb{Q}(\alpha, i)$ , the splitting field, hence Galois over $\mathbb{Q}$ (normal by construction, separable since $\mathrm{char} = 0$ ). Degree count: $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$ since $x^4-2$ is irreducible over $\mathbb{Q}$ by Eisenstein at $p=2$ . Also $i \notin \mathbb{Q}(\alpha)$ since $\mathbb{Q}(\alpha) \subset \mathbb{R}$ , so $[\mathbb{Q}(\alpha,i):\mathbb{Q}(\alpha)] = 2$ ( $i$ has minimal polynomial $x^2+1$ over $\mathbb{Q}(\alpha)$ , irreducible since $\mathbb{Q}(\alpha)$ is real). Hence $[K:\mathbb{Q}] = 8$ . $G = \mathrm{Gal}(K/\mathbb{Q})$ has order 8. Define $\sigma, \tau \in G$ by

\sigma(\alpha) = i\alpha,\ \sigma(i) = i; \qquad \tau(\alpha) = \alpha,\ \tau(i) = -i.

One checks $\sigma^4 = 1$ , $\tau^2 = 1$ , and $\tau\sigma\tau^{-1} = \sigma^{-1}$ (compute $\tau\sigma\tau(\alpha) = \tau\sigma(\alpha) = \tau(i\alpha) = -i\alpha = \sigma^{-1}(\alpha) = \sigma^3(\alpha)$ , and similarly on $i$ ), so $G \cong D_4$ , the dihedral group of order 8 (symmetries of a square), with presentation $\langle \sigma,\tau \mid \sigma^4=\tau^2=1,\ \tau\sigma\tau=\sigma^{-1}\rangle$ . $D_4$ has ten subgroups: $\{1\}$ , the center $\langle\sigma^2\rangle$ (order 2), four other order-2 subgroups $\langle\tau\rangle, \langle\sigma^2\tau\rangle, \langle\sigma\tau\rangle,\langle\sigma^3\tau\rangle$ , three subgroups of order 4 ( $\langle\sigma\rangle$ , $\langle\sigma^2,\tau\rangle$ , $\langle\sigma^2,\sigma\tau\rangle$ ), and $G$ itself. By part 3, these correspond to intermediate fields of degree $8,4,4,4,4,4,2,2,2,1$ over $\mathbb{Q}$ respectively (degree $= [G:H]$ ). Some computed fixed fields, using the definitions of $\sigma,\tau$ :

$\langle \sigma \rangle$ (order 4, index 2, and normal since it has index 2): fixes $i$ (as $\sigma(i)=i$ ) but moves $\alpha$ , so $K^{\langle\sigma\rangle} = \mathbb{Q}(i)$ , degree 2. Since $\langle\sigma\rangle \trianglelefteq D_4$ , part 4 predicts $\mathbb{Q}(i)/\mathbb{Q}$ is Galois — true, as it is the splitting field of $x^2+1$ — with $\mathrm{Gal}(\mathbb{Q}(i)/\mathbb{Q}) \cong D_4/\langle\sigma\rangle \cong \mathbb{Z}/2$ , matching the obvious fact that $\mathrm{Gal}(\mathbb{Q}(i)/\mathbb{Q})$ has order 2.
$\langle \tau \rangle$ (order 2, non-normal: check $\sigma\tau\sigma^{-1} = \sigma^2\tau \ne \tau$ ): fixes $\alpha$ , so $K^{\langle\tau\rangle} = \mathbb{Q}(\alpha)$ , degree 4. Since $\langle\tau\rangle$ is not normal, part 4 predicts $\mathbb{Q}(\alpha)/\mathbb{Q}$ is not Galois — correct, since $x^4-2$ has roots $\pm\alpha,\pm i\alpha$ and only $\pm\alpha$ lie in $\mathbb{Q}(\alpha) \subset \mathbb{R}$ , so $x^4-2$ does not split in $\mathbb{Q}(\alpha)$ , i.e. $\mathbb{Q}(\alpha)/\mathbb{Q}$ fails normality exactly as predicted.
$\langle \sigma^2,\tau\rangle$ (order 4, index 2, hence normal): $\sigma^2(\alpha) = -\alpha$ , $\tau(\alpha)=\alpha$ , so this subgroup fixes $\alpha^2 = \sqrt2$ but not $i$ or $\alpha$ ; $K^{\langle\sigma^2,\tau\rangle} = \mathbb{Q}(\sqrt2)$ , degree 2, and (index 2, hence normal) is Galois over $\mathbb{Q}$ with quotient $\mathbb{Z}/2$ , matching that $\mathbb{Q}(\sqrt2)/\mathbb{Q}$ is visibly the splitting field of $x^2-2$ .

This example shows the correspondence at work in a genuinely non-abelian setting, with normal and non-normal subgroups both present and both correctly predicting Galois versus non-Galois intermediate fields.

F = \mathbb{Q}

Scope, edge cases & common traps

Situation	What happens
$K/F$ finite Galois	Full correspondence holds exactly as stated.
$K/F$ separable but not normal	$\mathrm{Aut}(K/F)$ can be far too small (even trivial); no correspondence.
$K/F$ normal but not separable (positive characteristic)	Fixed field of $\mathrm{Aut}(K/F)$ need not equal $F$ ; correspondence fails (inseparable "extra" subfields are invisible to automorphisms).
$K/F$ infinite Galois (e.g. $\overline{\mathbb{Q}}/\mathbb{Q}$ )	A correspondence still holds, but only with closed subgroups in the Krull topology on $\mathrm{Gal}(K/F)$ ; arbitrary subgroups can fail to be closed and then fail to correspond to any intermediate field.
Composite / intersection of intermediate fields	Correctly computed via intersection / subgroup-generation of the corresponding subgroups (part 5), not via naive subgroup intersection/union confusion.

The mathematician behind it

Origins · 1 minÉvariste Galois and how this was discovered

ÉG

Évariste Galois

1811–1832 · Restoration-era France, self-taught in the margins of the École Normale

Galois developed the group-theoretic viewpoint on polynomial solvability as a teenager, submitted his ideas to the French Academy multiple times, and was rejected or ignored by Cauchy, Fourier, and Poisson. He died at twenty from wounds sustained in a duel, having spent the night before frantically writing down his mathematical ideas, including the sketch of what would become the correspondence bearing his name.

History

- 1829–1831: Galois submits memoirs on the solvability of equations by radicals to the French Academy of Sciences; they are lost or rejected.

- 1832: Galois dies after a duel; his papers, including the notion of a normal subgroup and the criterion for solvability by radicals, are left with his friend Auguste Chevalier.

- 1846: Joseph Liouville edits and publishes Galois's papers in the Journal de Mathématiques Pures et Appliquées, bringing them to wider attention.

- 1870: Camille Jordan's Traité des substitutions et des équations algébriques systematizes Galois's ideas into recognizably modern group-theoretic language.

- Early 1900s: Richard Dedekind and Emil Artin, building on Steinitz's field theory, give the correspondence its modern field-theoretic formulation via fixed fields, the form presented here.

Does the Fundamental Theorem require

K/F

to be a splitting field of a single polynomial?

Not as an extra hypothesis — it's automatic. A finite Galois extension $K/F$ is by definition normal and separable, and a finite normal separable extension is always the splitting field of some polynomial over $F$ : take the product of minimal polynomials of a finite generating set (which exists since $[K:F]<\infty$ ), removing repeats. Conversely, a splitting field of a separable polynomial over $F$ is automatically Galois. So "finite Galois" and "finite splitting field of a separable polynomial" name the same class of extensions, and the theorem applies to either description without modification.

What exactly breaks if

K/F

is normal but the characteristic is positive and separability fails?

Take $F = \mathbb{F}_p(t)$ and $K = \mathbb{F}_p(t^{1/p})$ , so $K/F$ has degree $p$ but is purely inseparable: the only root of $x^p - t$ is $t^{1/p}$ itself (in characteristic $p$ , $x^p - t = (x-t^{1/p})^p$ ). Every $F$ -automorphism of $K$ must fix this unique root, forcing $\mathrm{Aut}(K/F) = \{1\}$ , so $K^{\mathrm{Aut}(K/F)} = K \ne F$ . Pillar B fails, and with it the whole correspondence: there is exactly one nontrivial intermediate-field question here ("is there a field strictly between $F$ and $K$ ?", answer no, since degree $p$ is prime) but the trivial group offers zero information, illustrating why separability is indispensable, not just normality.

How does the theorem explain why some quintics have no solution by radicals?

A polynomial's roots are expressible by radicals exactly when its splitting field sits inside a tower of extensions each obtained by adjoining a root of some $x^n - a$ , and this happens exactly when the Galois group of the splitting field is a solvable group — one with a subnormal series with abelian factors. The Fundamental Theorem is what makes this translation rigorous: it shows subfields of the splitting field correspond exactly to subgroups of the Galois group, so a radical tower of fields corresponds to a subnormal chain of subgroups with abelian (in fact cyclic) quotients. For $x^5-x-1$ , the Galois group is $S_5$ , which is not solvable because its only proper normal subgroup $A_5$ is simple and nonabelian, so no such chain exists, and by the correspondence no radical tower can produce the splitting field. See Solvability by Radicals for the full argument.

Does the theorem tell us how to actually compute

\mathrm{Gal}(K/F)

for a given polynomial?

Not directly — it tells you what to do with the group once you know it, not how to find it. Determining $\mathrm{Gal}(K/F)$ for an explicit polynomial typically uses auxiliary tools: resolvent polynomials, discriminants (to detect whether the group lies in the alternating group), reduction mod primes (Dedekind's theorem, relating factorization patterns mod $p$ to cycle types in the Galois group acting on roots), and degree/prime constraints from the tower law. The Fundamental Theorem's role begins once the group is known or partially known: it converts every group-theoretic fact you can extract (subgroup lattice, normal subgroups, solvability) into a field-theoretic fact about the specific extension at hand.

Why must

\Phi

and

\Psi

both be checked, rather than just showing one map is a bijection?

Because a priori $\Phi$ could be injective without being surjective, or vice versa, and separately checking $\Psi \circ \Phi = \mathrm{id}$ and $\Phi \circ \Psi = \mathrm{id}$ is exactly what rules both failure modes out simultaneously; this is the standard criterion for two maps between sets to be mutually inverse bijections. In the proof, $\Psi\circ\Phi = \mathrm{id}_\mathcal{I}$ used Pillar B (applied to $K/E$ ), while $\Phi\circ\Psi = \mathrm{id}_\mathcal{S}$ used Pillar A (applied to $H$ acting on $K$ ) — these are logically independent facts about fixed fields, requiring separate arguments, and neither alone would establish that every subgroup and every intermediate field is accounted for.