Unique factorization implies every nonzero prime ideal contains a prime element

Statement
Suppose $$R$$ is a fact about::unique factorization domain and $$P$$ is a nonzero fact about::prime ideal in $$R$$. Then, there exists a fact about::prime element $$p \in P$$.

Breakdown for more general integral domains

 * Prime ideal need not contain a prime element

Applications

 * Unique factorization and one-dimensional iff principal ideal
 * Unique factorization and Noetherian implies every nonzero prime ideal is generated by finitely many prime elements

Related facts replacing prime element by irreducible element

 * 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

Facts used

 * 1) uses::Unique factorization implies every irreducible element is prime

Proof
Given: A unique factorization domain $$R$$. A nonzero prime ideal $$P$$ in $$R$$.

To prove: There exists a prime element $$p \in P$$.

Proof: Since $$P \ne 0$$, there exists a nonzero element $$a \in P$$. Factorizing $$a$$ in $$R$$ yields $$a = up_1p_2 \dots p_r$$ with $$u$$ a unit and $$p_i$$ irreducible. Note that the factorization has at least one prime, otherwise $$(a) = R$$ and $$P = R$$, a contradiction to $$P$$ being prime.

Since $$P$$ is prime, $$up_1p_2 \dots p_r \in P \implies p_i \in P$$ for at least one $$i$$. Fact (1) now tells us that $$p_i$$ is prime, so we have a prime element in $$P$$.