Unique factorization implies every nonzero prime ideal contains a prime element

Statement

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

Related facts

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. 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 $P\neq 0$, there exists a nonzero element $a\in P$. Factorizing $a$ in $R$ yields $a=up_{1}p_{2}\dots p_{r}$ 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, $up_{1}p_{2}\dots p_{r}\in P\implies p_{i}\in P$ for at least one $i$. Fact (1) now tells us that $p_{i}$ is prime, so we have a prime element in $P$.