Example of FD not satisfying ACCP

Question

Let $A$ be an integral domain. We know that if $A$ satisfies ACC on principal ideals then it is a Factorisation Domain. I want to know about the converse, i.e.,

Does there exist a FD not satisfying ACC on principal ideals?

Edit. A Factorisation Domain is an integral domain whose every non-zero non-unit element is a finite product of irreducible elements.

Thank You.

Answer 1

While Gram's original paper is somewhat hard to find, Zaks' paper including examples is available through ResearchGate.

There is at least a description of Grams' example available in this article although I haven't found my copy of Grams' article to compare it to, yet. This paper is new to me, but it looks like it might even contain a couple other constructions of atomic domains that aren't ACCP in its own right.

The justifications of these typically overrun the sensible length of a post, and since at least some are freely available I don't really feel the need to repost the entire detail.

But here is the construction of Grams' original example, at least:

Enumerate the primes in $\mathbb N$ as $\{p_i\}_{i=0}^\infty $ and generate a submonoid $M$ of positive rationals under addition using the elements $\frac{1}{2^ip_i}$ where $i\geq 0$. Then use an indeterminate $X$ and generate an $F$-algebra $A$ using $\{X^m\mid m\in M\}$ with a field $F$. Finally, take the multiplicative set of elements of $A$ with nonzero constant term and localize $A$ at this set. This localization is atomic but fails the ACCP.

Answer 2

Here is another construction from Roitman's paper "Polynomial Extensions of Atomic Domains", which has some additional interesting properties.

For a domain $D$ and $S \subseteq D\setminus(0)$, define $L(D;S) := D[\{X_s,sX_s^{-1}\}_{s \in S}]$, where $\{X_s\}_{s \in S}$ is a family of algebraically independent indeterminates. (Note the similarity here to a Laurent polynomial ring.) Now, starting from any domain $A_0$ that is not a weak GCD domain, recursively define $A_n := L(A_{n-1}; S_{n-1})$ for $n \ge 1$, where $S_{n-1}$ is the set of reducible elements of $A_{n-1}$. Define $A_\infty:= \bigcup_{n=0}^\infty A_n$.

Roitman shows that $A_\infty$ is atomic but $A_\infty[X]$ is not. Since it is well-known that the ACCP property is preserved by polynomial extensions, it follows that $A_\infty$ is atomic but fails the ACCP. Another interesting property of $A_\infty$ related to the "elasticity" of factorizations: every reducible element has arbitrarily long atomic factorizations, but also has an atomic factorization of length 2. Variants of Roitman's construction have since been used to provide a lot of other interesting factorization examples.

Edit: I added in a part in bold that I originally left out. A domain is a weak GCD domain if every pair of nonzero elements $a,b$ has a maximal common divisor c, meaning that $(c)$ is a minimal element of the set of principal ideals containing $(a,b)$. For example, the domain $\mathbb Q[\{Z,X/Z^n,Y/Z^n\}_{n=0}^\infty]$ is not a weak GCD domain, since $X$ and $Y$ have no maximal common divisor. The domain $A_\infty$ is always atomic regardless of which domain you start with; the other properties stated require some sort of assumption about $A_0$.