a polynomial of degree n over a field has at most n roots

Lemma (cf. factor theorem).

Let R be a commutative ring with identityPlanetmathPlanetmathPlanetmath and let p(x)R[x] be a polynomialMathworldPlanetmathPlanetmathPlanetmath with coefficients in R. The element aR is a root of p(x) if and only if (x-a) divides p(x).


Let F be a field and let p(x) be a non-zero polynomial in F[x] of degree n0. Then p(x) has at most n roots in F (counted with multiplicityMathworldPlanetmath).


We proceed by inductionMathworldPlanetmath. The case n=0 is trivial since p(x) is a non-zero constant, thus p(x) cannot have any roots.

Suppose that any polynomial in F[x] of degree n has at most n roots and let p(x)F[x] be a polynomial of degree n+1. If p(x) has no roots then the result is trivial, so let us assume that p(x) has at least one root aF. Then, by the lemma above, there exist a polynomial q(x) such that:


Hence, q(x)F[x] is a polynomial of degree n. By the induction hypothesis, the polynomial q(x) has at most n roots. It is clear that any root of q(x) is a root of p(x) and if ba is a root of p(x) then b is also a root of q(x). Thus, p(x) has at most n+1 roots, which concludes the proof of the theorem. ∎

Note: The fundamental theorem of algebraMathworldPlanetmath states that if F is algebraically closedMathworldPlanetmath then any polynomial of degree n has exactly n roots (counted with multiplicity).

Title a polynomial of degree n over a field has at most n roots
Canonical name APolynomialOfDegreeNOverAFieldHasAtMostNRoots
Date of creation 2013-03-22 15:09:01
Last modified on 2013-03-22 15:09:01
Owner alozano (2414)
Last modified by alozano (2414)
Numerical id 5
Author alozano (2414)
Entry type Theorem
Classification msc 13P05
Classification msc 11C08
Classification msc 12E05
Related topic Root
Related topic FactorTheorem
Related topic PolynomialCongruence
Related topic EveryPrimeHasAPrimitiveRoot
Related topic CongruenceOfArbitraryDegree