### Zeros of a Polynomial

The following demonstrates a couple of simple, yet extremely important properties of polynomials.

In the section on ring homomorphisms we saw the following example:

Let R be any commutative ring, X indeterminate over R, a R. If f = a0 + a1X + ... + anXn R[X], let us define the value of f at a, by

f(a) = a0 + a1a + ... + anan.

Let a:R[X]R be the mapping defined by

a(f ) = f(a)   (f R[X]).

Then a is a homomorphism. Indeed, if

g = b0 + b1X + ... + bnXn + ... R[X]

then

a(f + g) = a((a0+b0)+(a1+b1)X+...+(an+bn)Xn+...)
= (a0+b0)+(a1+b1)a+...+(an+bn)an+...
= (a0 + a1a + ... + anan + ...)
+(b0 + b1a + ... + bnan + ...)
= a(f) + a(g).

Similarly, since

f · g = c0 + c1X + ... + cnXn + ...,

where

ci = a0bi + a1bi-1 + ... +aib0,

we see that

a(f · g) = c0 + c1a + c2a2 + ... + cnan + ...

= (a0 + a1a + ...)·(b0 + b1a + ...) (R is commutative)

=a(f ) · a(g).

Therefore, a is a ring homomorphism. We usually refer to a as "evaluation at a."

In this demonstration, let F be a field, X an indeterminate over F, f F[X]. A zero of f is an element x F such that x(f) = 0. By the definition of a field the ring of F[X] is commutative and is an integral domain.

Clearly (X-x) F[X] Then by Proposition 6 in the section on polynomials, there exist unique polynomials q and r such that

f = (X - x)q + r    such that deg(r) < deg(q)
since x is a zero of f 0 = (X-x)q + r o this implies that either (X - x)q = -r or that r = 0 since deg(r) < deg(q) thus (X-x)q cannot equal -r so r = 0 which means f = (X - x)q.