prime ideal factorization is unique

The following theoremMathworldPlanetmath shows that the decomposition of an (integral) invertible ideal into its prime factorsMathworldPlanetmathPlanetmath is unique, if it exists. This applies to the ring of integersMathworldPlanetmath in a number field or, more generally, to any Dedekind domainMathworldPlanetmath, in which every nonzero ideal is invertible.


Let I be an invertible ideal in an integral domain R, and that


are two factorizations of I into a productPlanetmathPlanetmath of prime idealsPlanetmathPlanetmathPlanetmath. Then m=n and, up to reordering of the factors, pk=qk (k=1,2,,n).

Here we allow the case where m or n is zero, in which case such an empty product is taken to be the full ring R.


We use inductionMathworldPlanetmath on m+n. First, the case with m+n=0 is trivial, so suppose that m+n>0. As the set of prime ideals 𝔭k, 𝔮k is partially ordered by inclusion, there must be a minimal element. After reordering, without loss of generality we may suppose that it is 𝔭1. Then


so n1. Furthermore, as 𝔭1 is prime, this implies that 𝔮k𝔭1 for some k. After reordering the factors, we can take k=1, so that 𝔮1𝔭1.

As 𝔭1 is minimalPlanetmathPlanetmath among the prime factors, we have 𝔮1=𝔭1. Also, 𝔭1 is a factor of the invertible ideal I and so is itself invertible. Therefore, it can be cancelled from the products,


The induction hypothesis gives m=n and, after reordering, 𝔭k=𝔮k for k=2,,n. ∎

Title prime ideal factorization is unique
Canonical name PrimeIdealFactorizationIsUnique
Date of creation 2013-03-22 18:34:24
Last modified on 2013-03-22 18:34:24
Owner gel (22282)
Last modified by gel (22282)
Numerical id 9
Author gel (22282)
Entry type Theorem
Classification msc 13A15
Classification msc 13F05
Related topic DedekindDomain
Related topic FractionalIdeal
Related topic PrimeIdeal
Related topic FundamentalTheoremOfIdealTheory