Number Theory: An Introduction to Mathematics

4 Euclidean Domains 105

Suppose now that (i) and (ii) both hold. Thenδ( 1 )≤δ(a)for anya=0, since
a= 1 a.Furthermore,δ(a)=δ(ae)for any unite,since

δ(a)≤δ(ae)≤δ(aee−^1 )=δ(a).

On the other hand,δ(a)=δ(ae)for somea=0 implies thateis a unit. For from

a=qae+r,δ(r)<δ(ae),

we obtainr=( 1 −qe)a,δ(r)<δ(a), and hence 1−qe=0. In particular,δ(e)=δ( 1 )
if and only ifeis a unit.
The setKof alla∈Rsuch thatδ(a)≤δ( 1 )thus consists of 0 and all units ofR.
Sincea,b∈Kimpliesa−b∈K, it follows thatKis a field. We assume thatK=R,
since otherwise we have the first alternative of the proposition.
Choosex∈R\Kso that

δ(x)= min

a∈R\K

δ(a).

For anya∈R\K,thereexistq 0 ,r 0 ∈Rsuch that

a=q 0 x+r 0 ,δ(r 0 )<δ(x),

i.e.r 0 ∈ K.Thenδ(q 0 )<δ(q 0 x)=δ(a−r 0 )≤δ(a).Ifδ(q 0 )≥δ(x),i.e.if
q 0 ∈R\K,theninthesamewaythereexistq 1 ,r 1 ∈Rsuch that

q 0 =q 1 x+r 1 ,r 1 ∈K,δ(q 1 )<δ(q 0 ).

After finitely many repetitions of this process we must arrive at someqn− 1 ∈ K.
Puttingrn=qn− 1 , we obtain

a=rnxn+rn− 1 xn−^1 +···+r 0 ,

wherer 0 ,...,rn∈ Kandrn=0. Sinceδ(rjxj)=δ(xj)ifrj =0andδ(xj)<
δ(xj+^1 )for everyj, it follows thatδ(a)=δ(xn). Since the representationa=qxn+r
withδ(r)<δ(xn)is unique, it follows thatr 0 ,...,rnare uniquely determined bya.
Defineamapψ:R→K[t]by

ψ(rnxn+rn− 1 xn−^1 +···+r 0 )=rntn+rn− 1 tn−^1 +···+r 0.

Thenψis a bijection and actually an isomorphism, since it preserves sums and prod-
ucts. Furthermoreδ(a)>,=,or<δ(b)according as|ψ(a)|>,=,or<|ψ(b)|.

Some significant examples of principal ideal domains are provided by quadratic
fields, which will be studied in Chapter III. Any quadratic number field has the form
Q(

√

d),whered∈Zis square-free andd=1. The setOdof all algebraic integers in
Q(

√

d)is an integral domain. In the equivalent language of binary quadratic forms, it
was known to Gauss thatOdis a principal ideal domain for nine negative values ofd,
namely

d=− 1 ,− 2 ,− 3 ,− 7 ,− 11 ,− 19 ,− 43 ,− 67 ,− 163.