# 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 $p(x)\in R[x]$ be a polynomial with coefficients in $R$. The element $a\in R$ is a root of $p(x)$ if and only if $(x-a)$ divides $p(x)$.

###### Theorem.

Let $F$ be a field and let $p(x)$ be a non-zero polynomial in $F[x]$ of degree $n\geq 0$. 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 $p(x)\in F[x]$ 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 $a\in F$. Then, by the lemma above, there exist a polynomial $q(x)$ such that:

 $p(x)=(x-a)\cdot q(x).$

Hence, $q(x)\in F[x]$ 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 $b\neq a$ 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 APolynomialOfDegreeNOverAFieldHasAtMostNRoots 2013-03-22 15:09:01 2013-03-22 15:09:01 alozano (2414) alozano (2414) 5 alozano (2414) Theorem msc 13P05 msc 11C08 msc 12E05 Root FactorTheorem PolynomialCongruence EveryPrimeHasAPrimitiveRoot CongruenceOfArbitraryDegree