prime ideal factorization is unique
The following theorem shows that the decomposition of an (integral) invertible ideal into its prime factors
is unique, if it exists. This applies to the ring of integers
in a number field or, more generally, to any Dedekind domain
, in which every nonzero ideal is invertible.
Theorem.
Let I be an invertible ideal in an integral domain R, and that
I=𝔭1𝔭2⋯𝔭m=𝔮1𝔮2⋯𝔮n |
are two factorizations of I into a product of prime ideals
. 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.
Proof.
We use induction 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
𝔮1𝔮2⋯𝔮n⊆𝔭1, |
so n≥1. Furthermore, as 𝔭1 is prime, this implies that 𝔮k⊆𝔭1 for some k. After reordering the factors, we can take k=1, so that 𝔮1⊆𝔭1.
As 𝔭1 is minimal 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,
𝔭2⋯𝔭m=𝔮2⋯𝔮n. |
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 |