148 III More on Divisibility
In the terminology of Chapter II,§1, this shows thatany two nonzero ideals inOd
have a greatest common divisor.
In a commutative ringR, an idealA=R,{ 0 }is said to beirreducibleif its only
factors areAandR.Itissaidtobemaximalif the only ideals containingAareAand
R.Itissaidtobeprimeif, wheneverAdivides the product of two ideals, it also divides
at least one of the factors.
By Proposition 17, an ideal inOdis irreducible if and only if it is maximal. As we
saw in§1 of Chapter II, the existence of greatest common divisors implies that an ideal
inOdis irreducible if and only if it is prime. (These equivalences do not hold in all
commutative rings, but they do hold for the ring of all algebraic integers in any given
algebraic number field, and also for the rings associated with algebraic curves.)
Proposition 19A nonzero ideal A inOdhas only finitely many factors.
Proof SinceAA′=lOdfor some positive integerl, any factorBofAis also a factor
oflOdand so containsl. Proposition 14 implies, in particular, thatBis generated
by two elements, sayB =(β 1 ,β 2 ).A fortiori,B =(β 1 ,β 2 ,l)and hence, for any
γ 1 ,γ 2 ∈Od,also
B=(β 1 −lγ 1 ,β 2 −lγ 2 ,l).We can chooseγ 1 ∈Odso that in the representation
β 1 −lγ 1 =a 1 +b 1 ω(a 1 ,b 1 ∈Z)we have 0≤a 1 ,b 1 <l. Similarly we can chooseγ 2 ∈Odso that in the representation
β 2 −lγ 2 =a 2 +b 2 ω(a 2 ,b 2 ∈Z)we have 0≤a 2 ,b 2 <l. It follows that there are at mostl^4 different possibilities for
the idealB.
Corollary 20There exists no infinite sequence{An}of nonzero ideals inOdsuch that,
for every n, An+ 1 divides Anand An+ 1 =An.
In the terminology of Chapter II, this shows thatthe set of all nonzero ideals inOd
satisfies the chain condition(#). Since also the conclusion of Proposition II.1 holds,
the argument in§1 of Chapter II now shows that any nonzero proper ideal inOdis a
product of finitely many prime ideals and the representation is unique apart from the
order of the factors.
It remains to determine the prime ideals. This is accomplished by the following
three propositions.
Proposition 21For each prime ideal P inOdthere is a unique prime number p such
that P divides pOd. Furthermore, for any prime number p there is a prime ideal P in
Odsuch that exactly one of the following alternatives holds: