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.

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

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

Proof:

We do the construction inductively. Suppose we have a collection of pairwise distinct primes $p_{1},p_{2},\dots ,p_{i}$ $p_{1},p_{2},\dots ,p_{i}$ in $P$ $P$ ( $i$ $i$ 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 $(p_{1},p_{2},\dots ,p_{i})\neq P$ $(p_{1},p_{2},\dots ,p_{i})\neq P$ , there exists a prime $p_{i+1}\in P\setminus \{p_{1},p_{2},\dots ,p_{i}\}$ $p_{i+1}\in P\setminus \{p_{1},p_{2},\dots ,p_{i}\}$ :
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.
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 $n$ .