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

Statement
Suppose $$R$$ is a fact about::unique factorization domain that is also a fact about::Noetherian ring; in particular, $$R$$ is a fact about::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 fact about::finitely generated ideal where all the elements of the generating set are fact about::prime elements.

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 $$(p_1, p_2, \dots, p_i) \ne P$$, there exists a prime $$p_{i+1} \in P \setminus \{ p_1, p_2, \dots, p_i \}$$:
 * 2) 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).
 * 3) Since $$P$$ is prime, at least one of the prime factors of $$a$$ is in $$P$$. Call this prime factor $$p_{i + 1}$$.
 * 4) 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.
 * 5) 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 $$n$$.