Can a polynomial of degree n have more than n roots?

Let’s start with some definitions.

Let R be a ring and f∈R [X] a polynomial of degree n≥0, then f=a0+a1X+…+anXⁿ where a0,a1,…,an are elements of R. An element α∈R is called a root of f if f(α)=0, that is, if a0+a1α+…+anαⁿ=0.

For example, the polynomial X²-2 ∈ ℝ[X] has as its only roots √2 and -√2. Or X-5 ∈ ℝ[X] has as its only root 5.

Try to take any polynomial f with coefficients in ℝ and you will see that the number of real roots is always ≤ the degree of f. In general, this is true for the following more general result:

THEOREM:

Let R be a domain and f∈R[X]. Then the number of roots in R of f is ≤ the degree of f.

Proof:

Let us do induction on the degree n of f. If n=0 it is trivially true since f has no roots in R. If n=1 then f=aX+b with a,b in R. If then α and β are zeros of f we will have aα+b=aβ+b from which a(α-β)=0 and since a≠0 we have α=β. Let us now assume that n-1 is true. Then if α is a root of f in R we will have by Ruffini’s theorem that f=(x-α)g with g∈R[X] of degree n-1. It follows that the zeros of f are the zeros of g that are ≤ n-1 plus α, that is, in total ≤ n. □

The crucial point of the proof is the case n=1. It is in fact in that passage that we use that R is a domain. If in fact R is not a domain things are not as nice as they may seem. For example, consider the polynomial 3X in ℤ₆. The zeros of this polynomial in Z₆ are 0 and 2 so it is greater than one (and in fact Z₆ is not a domain). So yes, if the question is “Can a polynomial of degree n have more than n roots?” then the answer is yes as seen in the example above. At this point the following question arises spontaneously: Is there a bound for the number of zeros of a polynomial with coefficients in a generic commutative ring?