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 |
|---|---|
| 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 |