It is known that every PID is a UFD.
Is it true that every element of a commutative principal ideal ring (PIR) or rng that is not zero and not a unit is a product of a finite number of irreducible elements (not necessarily unique)?
I am trying to reproduce the proof for the Theorem 3.2.
from D.D. Anderson and Silvia Valdes-Leon,
Factorization in Commutative Rings with Zero Divisors
https://projecteuclid.org/euclid.rmjm/1181072068,
using the following definitions:
A non-unit $a$ of a ring or an rng $R$ is irreducible if $a = b \cdot c \implies a \sim b \lor a \sim c$;
Two elements $a$ and $b$ of $R$ are assotiates ($a \sim b$) if their principal ideals are equal:
$R \cdot a + \mathbb Za = R \cdot b + \mathbb Zb$.
My attempt to prove the main statement:
Let $S$ be the set of all elements of a PIR(rng) that are not irreducible and not products of irreducibles;
Assuming $S$ is not empty, $s_1$ is an element in $S$, $(s_1)$ is the principal ideal of $s_1$;
Creating an ascending chain of principal ideals starting from $(s_1)$ in the following way:
For an element $s_i$ searching for an element $s_{i+1}$ in $S$ such that $(s_i) \subsetneq (s_{i+1})$;
Any strictly ascending chain of ideals in a PIR(rng) is finite,
therefore there is a maximal ideal $(s_n)$ in the chain $(s_1) \subsetneq (s_2) \subsetneq ... \subsetneq (s_n)$;
$s_n$ is in $S$, therefore $s_n = a \cdot b$, where $s_n \nsim a$ and $s_n \nsim b$;
$s_n \in (a)$, and $s_n \nsim a \implies (s_n) \subsetneq (a)$;
$s_n \in (b)$, and $s_n \nsim b \implies (s_n) \subsetneq (b)$;
$a$ and $b$ are not in $S$, otherwise $(s_n)$ is not maximal in the chain;
Therefore, $a$ and $b$ are irreducible or products of irreducibles;
But then $s_n = a \cdot b$ is also a product of irreducibles.
Contradiction.
Therefore, $S$ is empty.
However, the proof in the mentioned article states that it uses the fact that an associate of an irreducible element is irreducible.
This fact is not used in my version of the proof.
Is there a mistake in my logic?
