Primitive Element Theorem

lean4-prooffield-theoryvisualization
K=F(θ) for some θK(K/F finite separable)K = F(\theta) \text{ for some } \theta \in K \quad (K/F \text{ finite separable})

Statement

Let K/FK/F be a finite separable field extension. Then there exists a primitive element θK\theta \in K such that K=F(θ)K = F(\theta).

Equivalently, KK is a simple extension: there is a single algebraic element whose adjunction to FF recovers all of KK.

In Mathlib: Field.exists_primitive_element states this for [FiniteDimensional F K] and [Algebra.IsSeparable F K].

Visualization

Example: Q(2,3)\mathbb{Q}(\sqrt{2}, \sqrt{3}) over Q\mathbb{Q}.

The extension has degree 4. We claim θ=2+3\theta = \sqrt{2} + \sqrt{3} is a primitive element.

Step 1 — compute powers of θ\theta:

nnθn\theta^n
0011
112+3\sqrt{2} + \sqrt{3}
225+265 + 2\sqrt{6}
33112+9311\sqrt{2} + 9\sqrt{3}

Step 2 — the minimal polynomial of θ=2+3\theta = \sqrt{2} + \sqrt{3} over Q\mathbb{Q}:

(θ25)2=46=24    θ410θ2+25=24    θ410θ2+1=0(\theta^2 - 5)^2 = 4 \cdot 6 = 24 \implies \theta^4 - 10\theta^2 + 25 = 24 \implies \theta^4 - 10\theta^2 + 1 = 0

So minQ(θ)=x410x2+1\min_\mathbb{Q}(\theta) = x^4 - 10x^2 + 1, which is irreducible over Q\mathbb{Q} (it has four distinct real roots ±2±3\pm\sqrt{2} \pm \sqrt{3}, and factors over no quadratic subfield).

Step 3 — since [Q(θ):Q]=4=[Q(2,3):Q][\mathbb{Q}(\theta):\mathbb{Q}] = 4 = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}], we have Q(2,3)=Q(2+3)\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3}).

Key roots of x410x2+1x^4 - 10x^2 + 1:

2+3,23,2+3,23\sqrt{2} + \sqrt{3},\quad \sqrt{2} - \sqrt{3},\quad -\sqrt{2} + \sqrt{3},\quad -\sqrt{2} - \sqrt{3}

Proof Sketch

  1. Infinite base field case. If FF is infinite, choose cFc \in F such that θ=α+cβ\theta = \alpha + c\beta is a primitive element for K=F(α,β)K = F(\alpha, \beta) (two-element case). The set of "bad" values of cc (where F(α+cβ)KF(\alpha + c\beta) \subsetneq K) is finite — it equals the set of ratios (σ(α)α)/(βσ(β))(\sigma(\alpha) - \alpha)/(\beta - \sigma(\beta)) for automorphisms σ\sigma fixing α+cβ\alpha + c\beta but not β\beta. Since FF is infinite, a good cc exists.
  2. Finite base field case. The multiplicative group K×K^\times of a finite field is cyclic; any generator is a primitive element.
  3. Induction. Apply the two-generator case repeatedly to handle K=F(α1,,αn)K = F(\alpha_1, \ldots, \alpha_n).

Connections

The Primitive Element Theorem is used in the proof of the Fundamental Theorem of Galois TheoryFundamental Theorem of Galois Theory{intermediate fields}{subgroups of Gal(K/F)}\{\text{intermediate fields}\} \longleftrightarrow \{\text{subgroups of } \text{Gal}(K/F)\}Bijection between intermediate fields and subgroups of the Galois groupRead more → to ensure Galois extensions are simple. It also appears in the study of Constructible NumbersConstructible NumbersRegular n-gon constructible    n=2ap1p2pk\text{Regular } n\text{-gon constructible} \iff n = 2^a \cdot p_1 \cdot p_2 \cdots p_kWhich regular n-gons can be constructed with compass and straightedge?Read more →: the tower of quadratic extensions used to characterize constructibility consists of successive simple extensions.

Lean4 Proof

/-- Primitive Element Theorem: a finite separable extension has a single generator. -/
theorem primitive_element (F K : Type*) [Field F] [Field K] [Algebra F K]
    [FiniteDimensional F K] [Algebra.IsSeparable F K] :
     θ : K, F⟮θ⟯ = ⊤ :=
  Field.exists_primitive_element F K