98 II Divisibility
then
(q−q 1 )f=r 1 −r, |r 1 −r|<|f|,which is only possible ifq=q 1.
Ideals inK[t] can be defined in the same way as forZand the proof of Lemma 9
remains valid. ThusK[t] is a principal ideal domain and,a fortiori, a GCD domain.
The Euclidean algorithm can also be applied inK[t]inthesamewayasforZand
again, from the sequence of polynomialsf 0 ,f 1 ,...,fNwhich it provides to deter-
mine the greatest common divisorfNoff 0 and f 1 we can obtain polynomialsuk,vk
such that
fk=f 1 uk+f 0 vk ( 0 ≤k≤N).We can actually say more for polynomials than for integers, since if
fk− 1 =qkfk+fk+ 1 , |fk+ 1 |<|fk|,then|fk− 1 |=|qk||fk|and hence, by induction,
|fk− 1 ||uk|=|f 0 |,|fk− 1 ||vk|=|f 1 | ( 1 <k≤N).It may be noted in passing that the Euclidean algorithm can also be applied in the
ringK[t,t−^1 ]ofLaurent polynomials. A Laurent polynomialf=O, with coefficients
from the fieldK,hastheform
f=amtm+am+ 1 tm+^1 +···+antn,wherem,n∈Zwithm≤nandaj∈Kwithaman=0. Thus we can writef=tmf 0 ,
wheref 0 ∈K[t]. Put
|f|= 2 n−m, |O|= 0 ;then the division algorithm for ordinary polynomials implies one for Laurent polyno-
mials: for anyf,g∈K[t,t−^1 ] with f =O,thereexistq,r∈K[t,t−^1 ] such that
g=qf+r,|r|<|f|.
We return now to ordinary polynomials. Thegeneral definition for integral domains
in Section 1 means, in the present case, that a polynomialp∈K[t]isirreducibleif it
has positive degree and if every proper divisor has degree zero.
It follows that any polynomial of degree 1 is irreducible. However, there may exist
also irreducible polynomials of higher degree. For example, we will show shortly that
the polynomialt^2 −2 is irreducible inQ[t]. ForK=C, however, every irreducible
polynomial has degree 1, by the fundamental theorem of algebra (Theorem I.30) and
Proposition 14 below. It follows that, forK =R, every irreducible polynomial has
degree 1 or 2. (For if a real polynomialf(t)has a rootα∈C\R, its conjugateα ̄is
also a root andf(t)has the real irreducible factor(t−α)(t− ̄α).)
It is obvious that the chain condition (#) of Section 1 holds in the integral domain
K[t], since ifgis a proper divisor off,then|g|<|f|. It follows that any polyno-
mial of positive degree can be represented as a product of finitely many irreducible
polynomials and that the representation is essentially unique.