Principal ideal ring implies one-dimensional

Verbal statement
Any principal ideal ring is a one-dimensional ring: its fact about::Krull dimension is at most one. In other words, it cannot have an ascending chain of fact about::prime ideals of length more than one.

Statement with symbols
Suppose $$R$$ is a principal ideal ring. Then, we cannot have a strictly ascending chain of prime ideals in $$R$$ of the form:

$$P_0 \subset P_1 \subset P_2$$

Related facts

 * PID 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

Proof
Given: A principal ideal ring $$R$$.

To prove: There cannot be three prime ideals $$P_0 \subset P_1 \subset P_2$$ with the containment strict.

Proof: Suppose we have such prime ideals. Since $$R$$ is a principal ideal ring, we can choose generators $$p_1, p_2$$ of $$P_1, P_2$$. We then have $$p_1 = q_{12}p_2$$ for some $$q_{12} \in R$$. Since $$P_1$$ is prime and $$p_2 \notin P_1$$, we have $$q_{12} \in P_1$$. Thus, $$q_{12} = p_1x$$ for some $$x \in R$$. This yields:

$$p_1 = p_1xp_2$$

which simplifies to:

$$p_1(1 - xp_2) = 0$$.

Since $$0 \in P_0$$ and $$P_0$$ is prime, we have that either $$p_1 \in P_0$$ (not possible since the containment is strict) or $$xp_2 = 1$$ (not possible since $$P_2$$ is a proper ideal). Thus, we have the required contradiction.