Fundamental Theorem of Algebra

lean4-proofcomplex-analysisvisualization
degp1,  pC[z]    z0C:p(z0)=0\deg p \geq 1,\; p \in \mathbb{C}[z] \implies \exists z_0 \in \mathbb{C}: p(z_0) = 0

The Fundamental Theorem of Algebra (FTA) makes C\mathbb{C} algebraically closed: every non-constant polynomial with complex coefficients splits completely over C\mathbb{C}. This is the reason that eigenvalue theory, spectral analysis, and partial-fraction decomposition all work without restriction over the complex numbers.

Statement

Let pC[z]p \in \mathbb{C}[z] be a polynomial of degree d1d \geq 1. Then there exists z0Cz_0 \in \mathbb{C} such that p(z0)=0p(z_0) = 0.

By induction, pp factors completely:

p(z)=adk=1d(zzk),z1,,zdC.p(z) = a_d \prod_{k=1}^{d} (z - z_k), \qquad z_1, \ldots, z_d \in \mathbb{C}.

Visualization

Factoring over C\mathbb{C} versus R\mathbb{R} — explicit examples:

p(z)=z41p(z) = z^4 - 1. Over R\mathbb{R}: z41=(z21)(z2+1)=(z1)(z+1)(z2+1)z^4 - 1 = (z^2 - 1)(z^2 + 1) = (z-1)(z+1)(z^2+1), and z2+1z^2 + 1 is irreducible over R\mathbb{R}. Over C\mathbb{C}: z2+1=(zi)(z+i)z^2 + 1 = (z - i)(z + i), giving the complete factorisation

z41=(z1)(z+1)(zi)(z+i).z^4 - 1 = (z-1)(z+1)(z-i)(z+i).

The four roots {1,1,i,i}\{1, -1, i, -i\} lie at the vertices of the unit square in C\mathbb{C}.

p(z)=z2+1p(z) = z^2 + 1. Roots: z=±iz = \pm i. Real axis: no root.

p(z)=z31p(z) = z^3 - 1. Roots: 1,ω,ω21,\, \omega,\, \omega^2 where ω=e2πi/3=12+32i\omega = e^{2\pi i/3} = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i.

  Im
   i ──── × (i, root of z²+1)
   |
───+────────── Re
   |
  -i ──── × (-i, root of z²+1)

The unit circle z=1|z| = 1 contains all roots of unity — the FTA guarantees their existence; complex analysis pinpoints their location.

Proof Sketch

There are several proofs. The slickest uses Liouville's TheoremLiouville's Theoremf:CC entire,  fC    fconstf : \mathbb{C} \to \mathbb{C} \text{ entire},\; |f| \leq C \implies f \equiv \text{const}A bounded entire function on ℂ must be constant — complex differentiability is far more rigid than real differentiabilityRead more →:

Suppose pp has no root in C\mathbb{C}. Then f(z)=1/p(z)f(z) = 1/p(z) is entire. Because p(z)|p(z)| \to \infty as z|z| \to \infty (a degree-dd polynomial grows like adzd|a_d||z|^d), f(z)0f(z) \to 0 at infinity, so ff is bounded. By Liouville, ff is constant, contradicting the assumption that pp is non-constant.

Connections

The FTA is the endpoint of a path starting with completeness: the Cauchy CriterionCauchy Criterionε>0  N:  m,nN    aman<ε\forall\varepsilon>0\;\exists N:\; m,n \geq N \implies |a_m - a_n| < \varepsilonA sequence converges if and only if its terms eventually become arbitrarily close to each other — no candidate limit requiredRead more → ensures C\mathbb{C} is complete; complex differentiability is rigid enough for the LiouvilleLiouville's Theoremf:CC entire,  fC    fconstf : \mathbb{C} \to \mathbb{C} \text{ entire},\; |f| \leq C \implies f \equiv \text{const}A bounded entire function on ℂ must be constant — complex differentiability is far more rigid than real differentiabilityRead more → argument; and together they yield algebraic closure. From the other direction, the FTA implies that every endomorphism of a finite-dimensional complex vector space has an eigenvalue — the starting point for the Jordan normal form. The explicit factorisation of zn1z^n - 1 via roots of unity connects to complex iterationMandelbrot Setzn+1=zn2+cz_{n+1} = z_n^2 + cInteractive visualization of z_{n+1} = z_n^2 + c and its connectednessRead more → (the nn-th roots of unity are fixed points of zznz \mapsto z^n).

Lean4 Proof

import Mathlib.Analysis.Complex.Polynomial.Basic

open Polynomial

/-- Fundamental Theorem of Algebra: every non-constant polynomial over ℂ has a root. -/
theorem fta {p : [X]} (h : 0 < degree p) :  z : , IsRoot p z :=
  Complex.exists_root h