Galois Theory — Mathematics Knowledge System

Overview

Galois theory studies field extensions by associating to each extension a group of symmetries, the Galois group, that permutes the roots of a polynomial while fixing the base field. The celebrated Fundamental Theorem of Galois Theory translates questions about intermediate fields into questions about subgroups, and this translation is what finally proved that no general formula using radicals can solve quintic (degree 5) or higher polynomial equations.

Intuition

A polynomial's roots often look indistinguishable from each other algebraically — for x² - 2 = 0, the roots √2 and -√2 satisfy exactly the same polynomial relations over ℚ, so swapping them is a 'symmetry' of the extension ℚ(√2)/ℚ. The Galois group collects all such root-swapping symmetries. Galois's key insight was that a polynomial is solvable by radicals (nested root expressions like the quadratic formula) exactly when its Galois group has a correspondingly simple ('solvable') structure — turning an analytic question about formulas into a purely algebraic question about groups.

Formal Definition

Definition

Let K/F be a field extension. An automorphism of K fixing F pointwise is an F-automorphism of K. The set of all such automorphisms forms the Galois group. When K/F is a finite, normal, and separable ('Galois') extension:

\operatorname{Gal}(K/F) = \{ \sigma : K \to K \mid \sigma \text{ is a field automorphism, } \sigma(a) = a \ \forall a \in F \}

Galois group of K over F

|\operatorname{Gal}(K/F)| = [K:F] \quad \text{when } K/F \text{ is Galois}

Order equals the extension degree

F \subseteq E \subseteq K \ \longleftrightarrow \ \{e\} \leq \operatorname{Gal}(K/E) \leq \operatorname{Gal}(K/F)

Galois correspondence between intermediate fields and subgroups

Notation

Notation	Meaning
$\operatorname{Gal}(K/F)$	The Galois group of the extension K over F
$[K:F]$	Degree of the field extension — dimension of K as an F-vector space
$K^H$	The fixed field of a subgroup H ≤ Gal(K/F): elements of K fixed by every σ ∈ H
$\operatorname{Fix}(\sigma)$	The fixed field of a single automorphism σ

Properties

Fundamental Theorem of Galois Theory

\text{For } K/F \text{ Galois, the maps } E \mapsto \operatorname{Gal}(K/E) \text{ and } H \mapsto K^H \text{ are mutually inverse, inclusion-reversing bijections between intermediate fields and subgroups of } \operatorname{Gal}(K/F).

Normal subgroups correspond to normal extensions

E/F \text{ is a normal (Galois) extension} \iff \operatorname{Gal}(K/E) \trianglelefteq \operatorname{Gal}(K/F), \text{ and then } \operatorname{Gal}(E/F) \cong \operatorname{Gal}(K/F)/\operatorname{Gal}(K/E).

Solvability by radicals

\text{A polynomial is solvable by radicals} \iff \text{its Galois group is a solvable group.}

Insolvability of the quintic

\text{The Galois group of a general degree-5 polynomial is } S_5, \text{ which is not solvable (since } A_5 \text{ is simple and nonabelian), so no radical formula exists for degree } \geq 5.

Applications

Finite field theory built on Galois-theoretic ideas underlies error-correcting codes and cryptographic constructions over 𝔽_{p^n}.

Worked Examples

K/F has degree 2 since √2 satisfies the minimal polynomial x² - 2 over ℚ.
$[K:F] = 2$
Any F-automorphism σ must send √2 to another root of x² - 2 in K, i.e. σ(√2) = ±√2.
$\sigma(\sqrt2) \in \{\sqrt2, -\sqrt2\}$
This gives exactly two automorphisms: the identity, and the map sending √2 ↦ -√2. Both fix ℚ pointwise.
$\operatorname{Gal}(K/F) = \{ \text{id}, \sigma \}$

Answer: Gal(ℚ(√2)/ℚ) ≅ ℤ₂, the cyclic group of order 2, matching [K:F] = 2.

Practice Problems

Difficulty 8/10

Why does the insolvability of A₅ (the alternating group on 5 letters) imply that the general quintic has no formula using radicals?

Difficulty 7/10

Galois theory settles the ancient problem of 'doubling the cube' (constructing ∛2 with compass and straightedge). Why is it impossible?

Common Mistakes

Common Mistake

Assuming every field extension has a well-defined 'Galois group' in the strong correspondence sense.

The Fundamental Theorem requires K/F to be a Galois extension (normal AND separable); for extensions that aren't normal or aren't separable, the neat bijection between subgroups and intermediate fields can fail.

Common Mistake

Thinking 'unsolvable by radicals' means the equation has no solutions.

Every quintic still has 5 roots (in ℂ, by the Fundamental Theorem of Algebra) — insolvability by radicals only means those roots can't be written using a finite nested-root formula in the coefficients.

Quiz

Galois theory's central idea is to associate to each polynomial a ____ and study the polynomial through it.

matrixgroup (its Galois group of symmetries of the roots)graphprobability

The famous consequence of Galois theory for the general quintic (degree-5) polynomial is that:

it has no rootsthere is no general solution by radicals (no quintic formula)it always factorsit has exactly five rational roots

Summary

The Galois group Gal(K/F) consists of field automorphisms of K that fix F pointwise, capturing the symmetries among roots.
The Fundamental Theorem of Galois Theory gives an inclusion-reversing bijection between intermediate fields of K/F and subgroups of Gal(K/F).
Normal subgroups of the Galois group correspond exactly to normal (Galois) intermediate extensions.
A polynomial is solvable by radicals if and only if its Galois group is a solvable group.
Since S₅ is not solvable (via the simple nonabelian group A₅), the general quintic has no radical solution formula — resolving a 350-year-old question.

References

WebsiteWikipedia — Galois theory
BookDummit, D. & Foote, R. Abstract Algebra, 3rd ed., Ch. 14.