Field Extensions — Mathematics Knowledge System

Overview

A field extension L/K (also written K ⊆ L) occurs when a field L contains a smaller field K, making L a vector space over K; the dimension of this vector space is called the degree [L:K]. Extensions built by adjoining a root of an irreducible polynomial, K(α), are the basic building blocks, and the tower law ([L:K] = [L:M][M:K] for K ⊆ M ⊆ L) lets degrees be computed by breaking an extension into stages. This machinery, developed by Galois and later formalized by Dedekind and Artin, is the language in which classical questions like 'can an angle be trisected with straightedge and compass' get precise answers.

Intuition

Adjoining a root α of an irreducible polynomial to K is like fixing the one thing K was missing (a solution to that polynomial equation) and taking the smallest field that contains both K and this new element — every element of K(α) then becomes a K-linear combination of powers of α, up to the degree of α's minimal polynomial. The Tower Law says degrees multiply across stages exactly like dimensions multiply for iterated vector spaces, which is the key counting tool behind classical impossibility results: constructible numbers must live in an extension of ℚ of degree a power of 2, so trisecting a generic angle (which needs a degree-3 extension) is provably impossible with straightedge and compass.

Formal Definition

Definition

Let K ⊆ L be fields. L is a vector space over K, and:

[L:K] = \dim_K L \quad \text{(the degree of the extension)}

Degree of L over K

\alpha \text{ algebraic over } K \iff \exists\, 0 \neq p(x) \in K[x] \text{ with } p(\alpha)=0

Algebraic element

K(\alpha) \cong K[x]/(m_\alpha(x)), \qquad [K(\alpha):K] = \deg(m_\alpha)

Simple extension via the minimal polynomial m_α

[L:K] = [L:M]\cdot[M:K] \quad \text{for } K \subseteq M \subseteq L

Tower Law

Worked Examples

The minimal polynomial of √2 over ℚ is x²−2, which is irreducible over ℚ (by the rational root theorem, no rational number squares to 2).
$m_{\sqrt{2}}(x) = x^2 - 2$
The degree of the extension equals the degree of the minimal polynomial.
$[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$
A basis is {1, √2}: every element of ℚ(√2) is uniquely a+b√2 for a,b ∈ ℚ.
$\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2} : a,b \in \mathbb{Q}\}$

Answer: [ℚ(√2):ℚ] = 2, with basis {1, √2}.

Practice Problems

Difficulty 6/10

Find [ℚ(∛2):ℚ] and give a ℚ-basis.

Difficulty 7/10

Explain, using the Tower Law, why an angle constructible by straightedge and compass must lie in a field extension of ℚ of degree a power of 2, and why trisecting 60° is impossible.

Difficulty 6/10

Is π algebraic or transcendental over ℚ, and what does this mean for [ℚ(π):ℚ]?

Quiz

Summary

A field extension L/K makes L a K-vector space of dimension [L:K]; K(α) has degree equal to the degree of α's minimal polynomial.
The Tower Law [L:K]=[L:M][M:K] lets extension degrees be computed stage by stage, e.g. [ℚ(√2,√3):ℚ]=4.
Straightedge-and-compass constructions only reach extensions of degree a power of 2 — the reason angle trisection and cube duplication are provably impossible in general.

References

WebsiteWikipedia — Field extension