# prime ideal factorization is unique

###### Theorem.

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

 $I=\mathfrak{p}_{1}\mathfrak{p}_{2}\cdots\mathfrak{p}_{m}=\mathfrak{q}_{1}% \mathfrak{q}_{2}\cdots\mathfrak{q}_{n}$

are two factorizations of $I$ into a product  of prime ideals   . Then $m=n$ and, up to reordering of the factors, $\mathfrak{p}_{k}=\mathfrak{q}_{k}$ ($k=1,2,\ldots,n$).

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 $\mathfrak{p}_{k}$, $\mathfrak{q}_{k}$ is partially ordered by inclusion, there must be a minimal element. After reordering, without loss of generality we may suppose that it is $\mathfrak{p}_{1}$. Then

 $\mathfrak{q}_{1}\mathfrak{q}_{2}\cdots\mathfrak{q}_{n}\subseteq\mathfrak{p}_{1},$

so $n\geq 1$. Furthermore, as $\mathfrak{p}_{1}$ is prime, this implies that $\mathfrak{q}_{k}\subseteq\mathfrak{p}_{1}$ for some $k$. After reordering the factors, we can take $k=1$, so that $\mathfrak{q}_{1}\subseteq\mathfrak{p}_{1}$.

As $\mathfrak{p}_{1}$ is minimal  among the prime factors, we have $\mathfrak{q}_{1}=\mathfrak{p}_{1}$. Also, $\mathfrak{p}_{1}$ is a factor of the invertible ideal $I$ and so is itself invertible. Therefore, it can be cancelled from the products,

 $\mathfrak{p}_{2}\cdots\mathfrak{p}_{m}=\mathfrak{q}_{2}\cdots\mathfrak{q}_{n}.$

The induction hypothesis gives $m=n$ and, after reordering, $\mathfrak{p}_{k}=\mathfrak{q}_{k}$ for $k=2,\ldots,n$. ∎

Title prime ideal factorization is unique PrimeIdealFactorizationIsUnique 2013-03-22 18:34:24 2013-03-22 18:34:24 gel (22282) gel (22282) 9 gel (22282) Theorem msc 13A15 msc 13F05 DedekindDomain FractionalIdeal PrimeIdeal FundamentalTheoremOfIdealTheory