Principal ideal ring implies one-dimensional

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

Verbal statement

Any principal ideal ring is a one-dimensional ring: its Krull dimension is at most one. In other words, it cannot have an ascending chain of 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, $q_{12}=p_{1}x$ for some $x\in R$. This yields:

$p_{1}=p_{1}xp_{2}$

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.