Unique factorization and Noetherian implies every prime ideal is generated by finitely many prime elements

Statement

Suppose $R$ is a unique factorization domain that is also a Noetherian ring; in particular, $R$ is a Noetherian unique factorization domain. Then, every prime ideal in $R$ is generated by finitely many prime elements, in other words, it can be expressed as a finitely generated ideal where all the elements of the generating set are prime elements.

Related facts

Related facts about unique factorization domains

• Unique factorization implies every nonzero prime ideal contains a prime element
• Unique factorization and one-dimensional iff principal ideal

Related facts for irreducible elements instead of prime elements

• Integral domain satisfying ACCP implies every nonzero prime ideal contains an irreducible element
• Noetherian domain implies every nonzero prime ideal is generated by finitely many irreducible elements

Proof

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

To prove: $P=(p_{1},p_{2},\dots ,p_{n})$ for primes $p_{i}$ 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 fact (1)). We show that if $(p_{1},p_{2},\dots ,p_{i})\neq P$, there exists a prime $p_{i+1}\in P\setminus \{p_{1},p_{2},\dots ,p_{i}\}$:
   1. For this, pick $a\in P\setminus (p_{1},p_{2},\dots ,p_{i})$. Clearly, $a$ has a prime factorization in $R$ (use fact (2) to argue that the irreducible factors are indeed prime).
   2. Since $P$ is prime, at least one of the prime factors of $a$ is in $P$. Call this prime 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 $p_{i}$ prime, 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 $n$.