Proving the Theorem

History, other proofs and scope

Two centuries of proofs

Date	Author	Contribution
$1608$	Roth, Girard	conjectural statement: $n$ roots for degree $n$
$1746$	d'Alembert	first attempt, with an analytic gap
$1799$	Gauss	doctoral thesis, topological argument
$1806$	Argand	the proof via the minimum of the modulus
$1816$	Gauss	second and third proofs
$1849$	Gauss	fourth proof, case of complex coefficients

Gauss would give four proofs during his lifetime, the last one fifty years after the first. The French name of the theorem preserves both contributions: d'Alembert for the idea, Gauss for the rigor.

Other routes

Through complex analysis. This is the shortest proof. If $P$ never vanished, the function $\dfrac{1}{P}$ would be holomorphic on the whole of $\mathbb{C}$ and bounded, since $\lvert P \rvert \to +\infty$ . Liouville's theorem would make it constant, and hence $P$ too — which contradicts $\deg P \geq 1$ .

Through topology. We look at the image under $P$ of the circle of radius $r$ . For large $r$ , this curve winds $n$ times around the origin; for $r = 0$ , it is reduced to a point. A root must appear when the winding number changes.

Through algebra and analysis. Laplace's proof, recast by Artin using Galois theory, uses from analysis only the intermediate value theorem, through the fact that every real polynomial of odd degree has a real root. Everything else is field theory.

What the theorem makes possible

Field	Consequence
Reduction of endomorphisms	every complex endomorphism is triangularizable
Integration	partial fraction decomposition of rational functions
Linear differential equations	roots of the characteristic polynomial, hence solutions
Signal processing	factorization of filters into poles and zeros
Arithmetic	irreducibles in $\mathbb{R}[X]$ : degrees $1$ and $2$ only

The first row has the weightiest consequences: it is because the characteristic polynomial always has a root in $\mathbb{C}$ that every endomorphism of a complex space has an eigenvalue, and hence that the Jordan normal form exists.

What the theorem does not give

It provides no explicit root. Abel and Ruffini showed that no general formula in radicals exists beyond degree $4$ , and Galois theory explains why: the permutation group of the roots becomes non-solvable.

In practice, then, roots are computed numerically — Newton's method, the Durand–Kerner algorithm, eigenvalues of the companion matrix — relying on the theorem to know that there really is something to find.

Classic pitfall