Noetherian domain implies every prime ideal is generated by finitely many irreducible elements

Statement

In a Noetherian domain (i.e., a Noetherian ring that is also an integral domain) every prime ideal is generated by finitely many irreducible elements.

Related facts

• Unique factorization and Noetherian implies every prime ideal is generated by finitely many prime elements
• Unique factorization implies every nonzero prime ideal contains a prime element
• ACCP implies every nonzero prime ideal contains an irreducible element

Facts used

1. Noetherian implies ACCP
2. ACCP implies every nonzero prime ideal contains an irreducible element
3. ACCP implies every element has a factorization into irreducibles

Proof

Given: A unique factorization domain $R$. A prime ideal $P$ of $R$.

To prove: $P=(p_{1},p_{2},\dots ,p_{n})$ for irreducibles $p_{i}$ and some nonnegative integer $n$.

Proof:

1. We do the construction inductively. Suppose we have a collection of pairwise distinct primes $p_{1},p_{2},\dots ,p_{i}$ in $P$ ($i$ could be zero, which is covered in the general case, but can also be proved separately as shown in facts (1), (2)). We show that if $(p_{1},p_{2},\dots ,p_{i})\neq P$, there exists an irreducible $p_{i+1}\in P\setminus \{p_{1},p_{2},\dots ,p_{i}\}$:
   1. For this, pick $a\in P\setminus (p_{1},p_{2},\dots ,p_{i})$. $a$ has an irreducible factorization in $R$ (this follows from facts (1), (3)).
   2. Since $P$ is prime, at least one of the irreducible factors of $a$ is in $P$. Call this irreducible factor $p_{i+1}$.
   3. If $p_{i+1}\in (p_{1},p_{2},\dots ,p_{i})$, then $a\in (p_{1},p_{2},\dots ,p_{i})$ because $a$ is a multiple of $p_{i+1}$. This contradicts the way we picked $a$. Thus, $p_{i+1}\in P\setminus (p_{1},p_{2},\dots ,p_{i})$, and we have the required induction step.
2. We thus have, for the prime ideal $P$, a strictly increasing chain of ideals $(p_{1})\subset (p_{1},p_{2})\subset \dots$ with $p_{i}$ irreducible, that terminates at $p_{n}$ if and only if $(p_{1},p_{2},\dots ,p_{n})=P$. Since $R$ is Noetherian, the sequence must terminate at some finite stage, forcing $P=(p_{1},p_{2},\dots ,p_{n})$ for some nonnegative integer $n$.