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 be an invertible ideal in an integral domain , and that
are two factorizations of into a product of prime ideals. Then and, up to reordering of the factors, ().
Here we allow the case where or is zero, in which case such an empty product is taken to be the full ring .
Proof.
We use induction on . First, the case with is trivial, so suppose that . As the set of prime ideals , is partially ordered by inclusion, there must be a minimal element. After reordering, without loss of generality we may suppose that it is . Then
so . Furthermore, as is prime, this implies that for some . After reordering the factors, we can take , so that .
As is minimal among the prime factors, we have . Also, is a factor of the invertible ideal and so is itself invertible. Therefore, it can be cancelled from the products,
The induction hypothesis gives and, after reordering, for . ∎
Title | prime ideal factorization is unique |
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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 |