2 Quadratic Fields 145a,b∈N. Consequently there is a least unitε 0 >1. Then, for any unitε>1, there is
a positive integernsuch thatεn 0 ≤ε<εn 0 +^1 .Sinceεε− 0 nis a unit and 1≤εε− 0 n<ε 0 ,
we must actually haveε=εn 0. (The groupUis the direct product of the cyclic group
of order 2 generated by−1 and the infinite cyclic group generated byε 0 .)
As an example, taked=2. Thenε 0 = 1 +√
2 is a unit. Sinceε 0 >1 and all units
greater than 1 have the forma+b
√
2 witha,b∈N, it follows that all units are given
by±ε 0 n(n∈Z).
Having determined the units, we now consider more generally the theory of divis-
ibility in the integral domainJ.Ifα,β∈Jandβis a proper divisor ofα,thenN(β)
is a proper divisor inZofN(α)and hence|N(β)|<|N(α)|. Consequently the chain
condition (#) of Chapter II is satisfied. It follows that any element ofJwhich is nei-
ther zero nor a unit is a product of finitely many irreducibles. Thus it only remains to
determine the irreducibles. However, this is not such a simple matter, as the following
examples indicate.
The ringG of Gaussian integers is a Euclidean domain. However, an ordinary
primepmay or may not be irreducible inG. For example, 2=( 1 +i)( 1 −i)and
neither factor is one of the units± 1 ,±i. On the other hand, 3 has no proper divisor
α=a+biwhich is not a unit, sinceN( 3 )=9andN(α)=a^2 +b^2 =±3hasno
solutions in integersa,b.
Again, consider the ringO− 5 of integers of the fieldQ(
√
− 5 ).Anelementα=
a+b
√
−5ofO− 5 cannot have normN(α)=a^2 + 5 b^2 equal to±2or±3, since the
square of any ordinary integer is congruent to 0,1 or 4 mod 5. It follows that, in the
factorizations
6 = 2 · 3 =( 1 +√
− 5 )( 1 −
√
− 5 ),
all four factors are irreducible and the factorizations are essentially distinct, since
N( 2 )=4,N( 3 )=9andN( 1 ±
√
− 5 )=6. Thus 2 is not a ‘prime’ inO− 5 and
the ‘fundamental theorem of arithmetic’ does not hold.
It was shown by Kummer and Dedekind in the 19th century that uniqueness of
factorization could be restored by considering ideals instead of elements. Any nonzero
proper ideal ofOdcan be represented as a product of finitely many prime ideals and
the representation is unique except for the order of the factors. This result will now be
established.
A nonempty subsetAof a commutative ringRis anidealifa,b∈Aandx,y∈R
implyax+by∈A. For example,Rand{ 0 }are ideals. Ifa 1 ,...,am∈R, then the
set(a 1 ,...,am)of all elementsa 1 x 1 +···+amxmwithxj∈R( 1 ≤j≤m)is an
ideal, the idealgeneratedbya 1 ,...,am. An ideal generated by a single element is a
principal ideal.
IfAandBare ideals inR, then the setABof all finite sumsa 1 b 1 +···+anbn
withaj∈Aandbj∈B(1≤j≤n;n∈N)is also an ideal, theproductofAandB.
For any idealsA,B,Cwe have
AB=BA,(AB)C=A(BC),since multiplication inRis commutative and associative.