100 II Divisibility
Proposition 16Let R be a GCD domain and K its field of fractions. Let
f=a 0 +a 1 t+···+antnbe a polynomial of degree n> 0 with coefficients aj∈R( 0 ≤j≤n).Ifc∈Kisa
root of f and c=ab−^1 ,wherea,b∈R and(a,b)= 1 ,thenb|anand a|a 0.
In particular, if f is monic, then c∈R.
Proof We h av e
a 0 bn+a 1 abn−^1 +···+an− 1 an−^1 b+anan= 0.Henceb|anananda|a 0 bn.Since(an,b)=(a,bn)=1, by Proposition 3(v), the result
follows from Proposition 3(ii).
The polynomialt^2 −2 has no integer roots, since 0, 1 ,−1 are not roots and ifc∈Z
andc= 0 , 1 ,−1, thenc^2 ≥4. Consequently, by Proposition 16, the polynomialt^2 − 2
also has no rational roots. It now follows from Proposition 14 thatt^2 −2 is irreducible
inQ[t], since it has no divisors of degree 1.
Proposition 16 was known to Euler (1774) for the caseR=Z. In this case it shows
that to obtain all rational roots of a polynomial with rational coefficients we need test
only a finite number of possibilities, which can be explicitly enumerated. For exam-
ple, ifz∈Z, the cubic polynomialt^3 +zt+1 has no rational roots unlessz=0or
z=−2.
It was shown by Gauss (1801), again for the caseR=Z, that Proposition 16
may itself be considerably generalized. His result may be formulated in the following
way:
Proposition 17Let f,g∈R[t], where R is a GCD domain with field of fractions K.
Then g divides f in R[t]if and only if g divides f in K[t]and the greatest common
divisor of the coefficients of g divides the greatest common divisor of the coefficients
of f.
Proof For any polynomialf∈R[t], letc(f)denote the greatest common divisor of
its coefficients. We say thatfisprimitiveifc(f)=1. We show first that the product
f=ghof two primitive polynomialsg,his again primitive.
Let
g=b 0 +b 1 t+···, h=c 0 +c 1 t+···, f=a 0 +a 1 t+···,and assume on the contrary that the coefficientsaihave a common divisordwhich
is not a unit. Thend does not divide all the coefficientsbj, nor all the coeffi-
cientsck.Letbm,cnbe the first coefficients ofg,hwhich are not divisible byd.
Then
am+n=∑
j+k=m+nbjckandddivides every term on the right, except possiblybmcn. In fact, sinced|am+n,
dmust also dividebmcn. Hence we cannot have both(d,bm)=1and(d,cn)=1.