Number Theory: An Introduction to Mathematics

102 II Divisibility

whereg=b 0 +b 1 t+···, then by equating constant coefficients we would obtain
a 0 b 0 =1, which is a contradiction. ThusR[t] is not a B ́ezout domain.
As an application of the preceding results we show that ifa 1 ,...,anare distinct
integers, then the polynomial

f=

∏n

j= 1

(t−aj)− 1

is irreducible inQ[t]. Assume, on the contrary, thatf=gh,whereg,h∈Q[t]and
have positive degree. We may suppose without loss of generality thatg∈Z[t]and
that the greatest common divisor of the coefficients ofgis 1. Sincef∈Z[t], it then
follows from Proposition 17 that alsoh∈Z[t]. Thusg(aj)andh(aj)are integers for
everyj.Sinceg(aj)h(aj)=−1, it follows thatg(aj)=−h(aj). Thus the polyno-
mialg+hhas the distinct rootsa 1 ,...,an.Sinceg+hhas degree less thann,thisis
possible only ifg+h=O. Hencef=−g^2. But, since the highest coefficient off
is 1, this is a contradiction.
In general, it is not an easy matter to determine if a polynomial with rational
coefficients is irreducible inQ[t]. However, the followingirreducibility criterion, due
to Eisenstein (1850), is sometimes useful:

Proposition 19If

f(t)=a 0 +a 1 t+···+an− 1 tn−^1 +tn

is a monic polynomial of degree n with integer coefficients such that a 0 ,a 1 ,...,an− 1
are all divisible by some prime p, but a 0 is not divisible by p^2 , then f is irreducible in
Q[t].

Proof Assume on the contrary that f is reducible. Then there exist polynomials
g(t),h(t)of positive degreesl,mwithintegercoefficients such thatf=gh.If

g(t)=b 0 +b 1 t+···+bltl,

h(t)=c 0 +c 1 t+···+cmtm,

thena 0 =b 0 c 0. The hypotheses imply that exactly one ofb 0 ,c 0 is divisible byp. With-
out loss of generality, assume it to beb 0 .Sincepdividesa 1 =b 0 c 1 +b 1 c 0 , it follows
thatp|b 1 .Sincepdividesa 2 =b 0 c 2 +b 1 c 1 +b 2 c 0 , it now follows thatp|b 2. Proceeding
in this way, we see thatpdividesbjfor everyj≤l.But,sinceblcm=1, this yields a
contradiction.

It follows from Proposition 19 that, for any primep,thep-thcyclotomic polyno-
mial

Φp(x)=xp−^1 +xp−^2 +···+ 1

is irreducible inQ[x]. ForΦp(x)=(xp− 1 )/(x− 1 )and, if we putx= 1 +t,the
transformed polynomial