a polynomial of degree n over a field has at most n roots
Lemma (cf. factor theorem).
Let R be a commutative ring with identity and let p(x)∈R[x] be a polynomial
with coefficients in R. The element a∈R is a root of p(x) if and only if (x-a) divides p(x).
Proof.
Theorem.
Let F be a field and let p(x) be a non-zero polynomial in F[x] of degree n≥0. Then p(x) has at most n roots in F (counted with multiplicity).
Proof.
We proceed by induction. 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 a∈F. Then, by the lemma above, there exist a polynomial q(x) such that:
p(x)=(x-a)⋅q(x). |
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 b≠a 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 algebra states that if F is algebraically closed
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 |
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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 |