Why principal ideal ring has Krull dimension at most 1? We know PID which is not has a Krull dimension at most one.
But can we say the same thing with PIR（which is not necessarily integral domain）？
We cannot say in PIR that every nonzero prime ideal is maximal, so I wonder we can say the statement.
 A: A PIR is a Noetherian ring, so Krull's height theorem holds.
Let $M$ be a maximal ideal in the ring: since it is principal, Krull's theorem says it is of height at most $1$. Therefore the whole ring can only be Krull dimension at most $1$.
A: We can reduce to the case of a PID by modding out as follows. Let $R$ be a PIR and suppose $\newcommand{\q}{\mathfrak{q}} \q_0 \subsetneq \q_1 \subseteq \q_2$ is a chain of prime ideals. Then $R/\q_0$ is a PID, and we have the induced chain of primes $0 \subsetneq \q_1/\q_0 \subseteq \q_2/\q_0$. As you pointed out, nonzero primes are maximal in a PID, so $\q_1/\q_0 = \q_2/\q_0$.
Since $R$ is a PIR we have $\q_1 = (\pi_1)$ and $\q_2 = (\pi_2)$ for some $\pi_1, \pi_2 \in R$. By the above we have $(\overline{\pi_1}) = (\overline{\pi_2})$, where the bar denotes the image in $R/\q_0$. Then $\overline{\pi_2} = \overline{r \pi_1}$ for some $r$ in $R$, so $\pi_2 - r \pi_1 \in \q_0$. Thus $\pi_2 \in (\pi_1) + \q_0 \subseteq (\pi_1)$, and hence $\q_1 = (\pi_1) = (\pi_2) = \q_2$.
