Unique factorization and one-dimensional iff principal ideal

This article gives a proof/explanation of the equivalence of multiple definitions for the term principal ideal domain

View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for an integral domain:

• It is a unique factorization domain as well as a one-dimensional domain: every nonzero prime ideal in it is maximal.
• It is a principal ideal domain.

Facts used

1. PID implies UFD
2. PID implies one-dimensional
3. Principal ideal ring iff every prime ideal is principal
4. Unique factorization implies every irreducible is prime

Proof

Principal ideal domain implies unique factorization domain and one-dimensional

This follows from facts (1) and (2).

Unique factorization and one-dimensional implies principal ideal domain

By fact (3), it suffices to show that every prime ideal is principal. In fact, it suffices to show that every nonzero prime ideal is principal, because the zero ideal is principal.

Given: A unique factorization domain $R$ where every nonzero prime ideal is maximal.

To prove: Every nonzero prime ideal in $R$ is principal.

Proof: Suppose $P$ is a nonzero prime ideal in $R$. Since $P$ is nonzero, there exists $0\neq a\in P$. Since $R$ is a unique factorization domain, we can factor $a$ as a product of irreducibles, say:

$a=up_{1}p_{2}\dots p_{r}$

with $u$ a unit and $p_{i}$ prime.

Note that the factorization is nontrivial since $a$ is not a unit (if $a$ were a unit, $P=R$ would not be prime). Thus, $p_{1}p_{2}\dots p_{r}\in P$, and since $P$ is prime, there exists some $p_{i}\in P$. But then, since $R$ is a unique factorization domain, fact (4) tells us that $p_{i}$ is prime. Thus, the ideal $(p_{i})$ is a nonzero prime ideal of $R$ contained in $P$. Thus, by assumption, $(p_{i})$ is maximal, so $(p_{i})=P$, so $P$ is principal.