PID implies one-dimensional

This article gives the statement and possibly, proof, of an implication relation between two integral domain properties. That is, it states that every integral domain satisfying the first integral domain property (i.e., principal ideal domain) must also satisfy the second integral domain property (i.e., one-dimensional domain)
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Statement

A principal ideal domain is a one-dimensional domain. In other words, in a principal ideal domain, every nonzero prime ideal is a maximal ideal.

Related facts

• Principal ideal ring implies one-dimensional
• Unique factorization and one-dimensional iff principal ideal
• Unique factorization implies every nonzero prime ideal contains a prime element
• Unique factorization and finite-dimensional implies every prime ideal is generated by a set of primes of size at most the dimension

Facts used

1. every proper ideal is contained in a maximal ideal

Proof

Given: A principal ideal domain $R$, a nonzero prime ideal $P$ of $R$.

To prove: $P$ is a maximal ideal of $R$.

Proof: Suppose $P$ is not maximal in $R$. Then, by fact (1), there exists a maximal ideal $Q$ of $R$ properly containing $P$. Since $R$ is a principal ideal domain, there exist $q\in Q,p\in P$ such that $(q)=Q,(p)=P$. In particular, we can write $p=qr$ for some $r\in R$.

Note that since $P$ is prime, and $q\notin P$ because $P$ is properly contained in $Q$, we have $r\in P$. Thus, $r=px$ for some $x\in R$. This gives $p=p(qx)$, giving $p(1-qx)=0$. Since $R$ is an integral domain, this forces either $p=0$ (a contradiction to $P$ being nonzero) or $qx=1$ (a contradiction to $Q$ being a proper ideal). Thus, we have the required contradiction.