96 II Divisibility
It follows from the remarks at the end of Section 1 that a principal ideal domain
is factorial, i.e. any element which is neither zero nor a unit can be represented as a
product of finitely many irreducibles and the representation is essentially unique.
In the next section we will show that the ringK[t] of all polynomials in one inde-
terminatetwith coefficients from an arbitrary fieldKis a principal ideal domain.
It may be shown that the ring of all algebraic integers is a B ́ezout domain, and
likewise the ring of all functions which are holomorphic in a nonempty connected
open subsetGof the complex planeC. However, neither is a principal ideal domain.
In the former case there are no irreducibles, since any algebraic integerahas the
factorizationa=
√
a·√
a. In the latter casez−ζis an irreducible for anyζ∈G,but
the chain condition is violated. For example, take
an(z)=f(z)/(z−ζ 1 )···(z−ζn),wheref(z)is a non-identically vanishing function which is holomorphic inGand has
infinitely many zerosζ 1 ,ζ 2 ,...inG.
3 Polynomials
In this section we study the most important example of a principal ideal domain other
thanZ, namely the ringK[t] of all polynomials intwith coefficients from an arbitrary
fieldK(e.g.,K=QorC).
The attitude adopted towards polynomials inalgebra is different from that adopted
in analysis. In analysis we regard ‘t’ as a variable which can take different values; in
algebra we regard ‘t’ simply as a symbol, an ‘indeterminate’, on which we can perform
various algebraic operations. Since the concept of function is so pervasive, the alge-
braic approach often seems mysterious at first sight and it seems worthwhile taking the
time to give a precise meaning to an ‘indeterminate’.
Let R be an integral domain(e.g.,R=ZorQ). Apolynomialwith coefficients
fromRis defined to be a sequencef=(a 0 ,a 1 ,a 2 ,...)of elements ofRin which at
most finitely many terms are nonzero. The sum and product of two polynomials
f=(a 0 ,a 1 ,a 2 ,...), g=(b 0 ,b 1 ,b 2 ,...)are defined by
f+g=(a 0 +b 0 ,a 1 +b 1 ,a 2 +b 2 ,...),
fg=(a 0 b 0 ,a 0 b 1 +a 1 b 0 ,a 0 b 2 +a 1 b 1 +a 2 b 0 ,...).It is easily verified that these are again polynomials and that the setR[t] of all polyno-
mials with coefficients fromRis a commutative ring withO=( 0 , 0 , 0 ,...)as zero
element. (By dropping the requirement that at most finitely many terms are nonzero,
we obtain the ringR[[t]] of allformal power serieswith coefficients fromR.)
We define thedegree∂(f)of a polynomialf =(a 0 ,a 1 ,a 2 ,...)=Oto be the
greatest integernfor whichan=0 and we put
|f|= 2 ∂(f), |O|= 0.