⚗️ Galois Theory Laboratory

An Interactive Journey Through Fields, Groups, and Impossibilities

While we can't square the circle or solve the general quintic, the theory that explains why is absolutely beautiful.

💡 Inspired by Alon Amit's wonderful Quora post on real-world applications of Galois theory.

Level 1: Geometric Impossibilities & Symmetries

⭕ The Impossible Construction: Squaring the Circle

Can you construct a square with the same area as a given circle using only compass and straightedge? Try adjusting both the circle radius and the square side to match their areas. You'll find it's impossible because it would require constructing $\sqrt{\pi}$, but $\pi$ is transcendental!

Circle Radius: 1.00 Square Side Length: 1.77

Why Squaring the Circle is Impossible

A circle with radius $r$ has area $A = \pi r^2$. A square with the same area has side length $s = \sqrt{\pi r^2} = r\sqrt{\pi}$.

Constructible numbers are those obtainable from rationals using $+, -, \times, \div, \sqrt{\phantom{x}}$. These form a field with degree $2^k$ over $\mathbb{Q}$.

In 1882, Ferdinand von Lindemann proved that $\pi$ is transcendental—it's not the root of any polynomial with rational coefficients. Therefore, $\sqrt{\pi}$ cannot be constructed, making it impossible to square the circle!

This was the final nail in the coffin for one of antiquity's greatest mathematical challenges.

🔺 The Impossible Construction: Angle Trisection

Ancient Greeks could bisect any angle with compass and straightedge. But trisecting (dividing into three equal parts) an arbitrary angle? Try it yourself! You'll find success at special angles like 90° and 180°, but most angles are impossible because the algebraic degree is 3, not a power of 2.

Choose Angle (degrees): 60°

Constructible Numbers

A number is constructible if it can be obtained from rationals using $+, -, \times, \div, \sqrt{\phantom{x}}$. This means the degree over $\mathbb{Q}$ must be $2^k$.

The Trisection Problem: To trisect angle $\theta$, we need $\cos(\theta/3)$. Using the triple-angle formula: $$\cos(3\alpha) = 4\cos^3(\alpha) - 3\cos(\alpha)$$

Setting $x = \cos(\theta/3)$ and $c = \cos(\theta)$, we get: $$4x^3 - 3x - c = 0$$

For $\theta = 60°$, we get $8x^3 - 6x - 1 = 0$, which is irreducible over $\mathbb{Q}$ with degree 3. Since $3 \neq 2^k$ for any integer $k$, this proves trisection is impossible for a 60° angle using only compass and straightedge!

🎲 Symmetry Groups: The Tetrahedron

Galois theory connects field automorphisms to permutation groups. Explore this by rotating a tetrahedron! The rotation group is $A_4$ (alternating group), with 12 rotational symmetries total: 1 identity + 8 vertex-face rotations + 3 edge-edge rotations.

Select Rotation Axis Type:

Vertex-to-Face Rotation: Axis passes through a vertex and the center of the opposite face. Rotations of 120° and 240° are symmetries. There are 4 such axes (one per vertex), giving 8 non-identity rotations.

Rotation Angle: 0°

Select Which Axis:

Current Permutation:

Identity: (1)(2)(3)(4)

The Symmetry Group A₄ of a Tetrahedron

The rotation group of a tetrahedron is isomorphic to $A_4$, the alternating group on 4 elements. It has order 12 and consists of:

1 identity element (do nothing)
8 rotations about vertex-to-face axes: For each of 4 vertices, rotate 120° or 240° about the axis through that vertex and the center of the opposite face. These generate 3-cycles like (1 2 3).
3 rotations by 180° about edge-to-edge axes: Each axis passes through the midpoints of two opposite edges. These generate products of two 2-cycles like (1 2)(3 4).

Why A₄ and not S₄? Rotations preserve orientation (they're "even" permutations), so we get the alternating group. If we included reflections (mirror symmetries), we'd get the full symmetry group, which is $S_4$ with 24 elements.

Connection to Galois Theory: The Galois group of certain quartic polynomials is exactly $A_4$! For instance, $x^4 + 8x + 12$ over $\mathbb{Q}$ has Galois group $A_4$. This shows how geometric symmetries directly correspond to algebraic symmetries of polynomial roots.

Level 2: Field Extensions & The Fundamental Theorem

🎯 The Unsolvable Quintic (Abel-Ruffini Theorem)

Galois proved that there is no general formula for solving fifth-degree polynomials using radicals. But specific quintics can be solved! Explore different quintics or create your own by adjusting coefficients. Watch for solvability indicators!

Select Preset Quintic:

Abel-Ruffini Theorem & Galois Theory

A polynomial is solvable by radicals if and only if its Galois group is a solvable group. A group is solvable if it has a subnormal series with abelian quotients.

The symmetric group $S_5$ is not solvable because it contains $A_5$ (the alternating group), which is simple and non-abelian. This is why the general quintic cannot be solved by radicals!

Solvable Examples:

$x^5 - 1$: Galois group is cyclic $C_4$ (abelian, hence solvable)
$x^5 + 5x^3 + 5x$: Galois group is the Frobenius group $F_{20}$ (solvable)
$x^5 - 5$: Galois group is $D_5$ (dihedral, solvable)

Unsolvable Examples: Generic quintics have Galois group $S_5$ or $A_5$, which are not solvable. No radical formula exists!

🔗 The Lattice Explorer

The Fundamental Theorem of Galois Theory establishes a correspondence between subfields of a splitting field $L/K$ and subgroups of the Galois group $\text{Gal}(L/K)$. Explore this by examining different polynomials! The lattice diagram shows intermediate fields (going up = larger fields) and corresponding subgroups (going up = smaller subgroups).

Select Polynomial:

Fundamental Theorem of Galois Theory

Let $L/K$ be a Galois extension with Galois group $G = \text{Gal}(L/K)$. Then there is a one-to-one correspondence (bijection) between:

Subgroups $H \subseteq G$
Intermediate fields $K \subseteq E \subseteq L$

Key Properties:

The correspondence reverses inclusion: larger subgroups correspond to smaller fields
$[L:E] = |H|$ and $[E:K] = [G:H]$ (degrees and orders match)
$E/K$ is Galois if and only if $H$ is normal in $G$
When $E/K$ is Galois, $\text{Gal}(E/K) \cong G/H$

Example: For $x^4 - 2$, the splitting field is $\mathbb{Q}(\sqrt[4]{2}, i)$ with Galois group $D_4$ (dihedral group of order 8). The 10 subgroups of $D_4$ correspond exactly to 10 intermediate fields! The lattice diagram visualizes this beautiful correspondence.

Level 3: Finite Fields & Real-World Applications

💿 The Data Correction Chamber

Galois Fields $\text{GF}(2^n)$ power error-correcting codes! Reed-Solomon codes (used in CDs, DVDs, QR codes) work by encoding data as polynomials over finite fields. The minimum distance $d$ between codewords determines correction power: we can correct up to $\lfloor (d-1)/2 \rfloor$ errors. More parity bits = more correction power!

Number of Message Bits: 4 Number of Parity (Redundancy) Bits: 3

Correction Capability:
Can correct up to 1 error(s)
Code rate: 4/7 = 0.57 (efficiency)

Message Value: 1010 Number of Errors to Introduce: 0

Reed-Solomon Codes over Galois Fields

Error-correcting codes work by adding redundancy. A code with $k$ message bits and $r$ parity bits has code rate $R = k/(k+r)$, measuring efficiency.

The minimum Hamming distance $d$ is the smallest number of bit positions in which any two distinct codewords differ. For a systematic linear code with $r$ parity bits checking independent constraints, we typically achieve $d = r + 1$.

Correction Capability: A code with minimum distance $d$ can:

Detect up to $d - 1$ errors
Correct up to $\lfloor (d-1)/2 \rfloor$ errors

Why Galois Fields? Working in $\text{GF}(2^m)$ gives us the algebraic structure needed for efficient encoding and decoding. The automorphisms (Frobenius maps $x \mapsto x^2$) form the Galois group, and this structure enables fast error correction algorithms!

Real Applications: QR codes, CDs/DVDs, satellite communications, and 5G networks all use variants of these codes. They're why you can scan a dirty QR code or play a scratched CD!

Created by Van Warren & Claude (2025)