Number Theory: An Introduction to Mathematics

3 Polynomials 99

We now consider the connection between polynomials in the sense of algebra
(polynomial forms) and polynomials in the sense of analysis (polynomial functions).
LetKbeafieldandf∈K[t]:

f=a 0 +a 1 t+···+antn.

If we replace ‘t’byc∈Kwe obtain an element ofK, which we denote byf(c):

f(c)=a 0 +a 1 c+···+ancn.

A rapid procedure (‘Horner’s rule’) for calculatingf(c)is to use the recurrence rela-
tions

f 0 =an, fj=fj− 1 c+an−j (j= 1 ,...,n).

It is readily shown by induction that

fj=ancj+an− 1 cj−^1 +···+an−j,

and hencef(c)=fnis obtained with justnmultiplications andnadditions.
It is easily seen thatf=g+himpliesf(c)=g(c)+h(c),andf=ghimplies
f(c)=g(c)h(c). Thus the mappingf→f(c)is a ‘homomorphism’ ofK[t]intoK.
A simple consequence is the so-calledremainder theorem:

Proposition 14Let K be a field and c∈K. If f∈K[t],then

f=(t−c)g+f(c),

for some g∈K[t].
In particular, f is divisible by t−c if and only if f(c)= 0.

Proof We already know that there existq,r∈K[t] such that

f=(t−c)q+r, |r|≤ 1.

Thusr∈Kand the homomorphism properties imply thatf(c)=r.

We s a y t h a tc∈Kis arootof the polynomialf∈K[t]iff(c)=0.

Proposition 15Let K be a field. If f∈K[t]is a polynomial of degree n≥ 0 ,then f
has at most n distinct roots in K.

Proof Iffis of degree 0, thenf=cis a nonzero element ofKandfhas no roots.
Suppose now thatn≥1 and the result holds for polynomials of degree less thann.If
cis a root offthen, by Proposition 14,f=(t−c)gfor someg∈K[t]. Sinceghas
degreen−1, it has at mostn−1 roots. But every root offdistinct fromcis a root
ofg. Hencefhas at mostnroots.

We consider next properties of the integral domainR[t], whenRis an integral
domain rather than a field (e.g.,R=Z). The famous Pythagorean proof that

√

2is
irrational is considerably generalized by the following result: