unique factorization and ideals in ring of integers

Theorem.  Let O be the maximal orderMathworldPlanetmathPlanetmath, i.e. the ring of integersMathworldPlanetmath of an algebraic number field.  Then O is a unique factorization domainMathworldPlanetmath if and only if O is a principal ideal domainMathworldPlanetmath.

Proof.1o¯. Suppose that O is a PID.

We first state, that any prime numberMathworldPlanetmath π of O generates a prime idealMathworldPlanetmathPlanetmathPlanetmath (π) of O.  For if  (π)=𝔞𝔟,  then we have the principal idealsMathworldPlanetmathPlanetmathPlanetmath𝔞=(α)  and 𝔟=(β).  It follows that  (π)=(αβ), i.e.  π=λαβ  with some λO,  and since π is prime, one of α and β must be a unit of O.  Thus one of 𝔞 and 𝔟 is the unit ideal O, and accordingly (π) is a maximal idealMathworldPlanetmath of O, so also a prime ideal.

Let a non-zero element γ of O be split to prime number factors πi, ϱj in two ways:  γ=π1πr=ϱ1ϱs.  Then also the principal ideal (γ) splits to principal prime ideals in two ways:  (γ)=(π1)(πr)=(ϱ1)(ϱs).  Since the prime factorizationMathworldPlanetmath of ideals is unique, the   (π1),,(πr)  must be, up to the , identical with  (ϱ1),,(ϱs)  (and  r=s).  Let  (π1)=(ϱj1).  Then π1 and ϱj1 are associatesMathworldPlanetmath of each other; the same may be said of all pairs  (πi,ϱji).  So we have seen that the factorization in O is unique.

2o¯. Suppose then that O is a UFD.

Consider any prime ideal 𝔭 of O.  Let α be a non-zero element of 𝔭 and let α have the prime factorization π1πn.  Because 𝔭 is a prime ideal and divides the ideal product (π1)(πn), 𝔭 must divide one principal ideal  (πi)=(π).  This means that  π𝔭.  We write  (π)=𝔭𝔞, whence  π𝔭  and  π𝔞.  Since O is a Dedekind domainMathworldPlanetmath, every its ideal can be generated by two elements, one of which may be chosen freely (see the two-generator property).  Therefore we can write


We multiply these, getting  𝔭𝔞=(π2,πγ,πδ,γδ),  and so  γδ𝔭𝔞=(π).  Thus  γδ=λπ  with some λO.  According to the unique factorization, we have  π|γ  or  π|δ.

The latter alternative means that  δ=δ1π (with  δ1O),  whence  𝔞=(π,δ1π)=(π)(1,δ1)=(π)(1)=(π);  thus we had  𝔭𝔞=(π)=𝔭(π)  which would imply the absurdity  𝔭=(1).  But the former alternative means that  γ=γ1π (with  γ1O),  which shows that


In other words, an arbitrary prime ideal 𝔭 of O is principal.  It follows that all ideals of O are principal. Q.E.D.

Title unique factorization and ideals in ring of integers
Canonical name UniqueFactorizationAndIdealsInRingOfIntegers
Date of creation 2015-05-06 15:32:53
Last modified on 2015-05-06 15:32:53
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 17
Author pahio (2872)
Entry type Theorem
Classification msc 13B22
Classification msc 11R27
Synonym equivalence of UFD and PID
Related topic ProductOfFinitelyGeneratedIdeals
Related topic PIDsAreUFDs
Related topic NumberFieldThatIsNotNormEuclidean
Related topic DivisorTheory
Related topic FundamentalTheoremOfIdealTheory
Related topic EquivalentDefinitionsForUFD