a polynomial of degree over a field has at most roots
Lemma (cf. factor theorem).
Let be a commutative ring with identity and let be a polynomial with coefficients in . The element is a root of if and only if divides .
Proof.
Theorem.
Let be a field and let be a non-zero polynomial in of degree . Then has at most roots in (counted with multiplicity).
Proof.
We proceed by induction. The case is trivial since is a non-zero constant, thus cannot have any roots.
Suppose that any polynomial in of degree has at most roots and let be a polynomial of degree . If has no roots then the result is trivial, so let us assume that has at least one root . Then, by the lemma above, there exist a polynomial such that:
Hence, is a polynomial of degree . By the induction hypothesis, the polynomial has at most roots. It is clear that any root of is a root of and if is a root of then is also a root of . Thus, has at most roots, which concludes the proof of the theorem. ∎
Note: The fundamental theorem of algebra states that if is algebraically closed then any polynomial of degree has exactly roots (counted with multiplicity).
Title | a polynomial of degree over a field has at most 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 |