PID implies one-dimensional

Statement
A principal ideal domain is a one-dimensional domain. In other words, in a principal ideal domain, every nonzero fact about::prime ideal is a fact about::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) uses::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.