Example of a domain where all irreducibles are primes and that is not a GCD domain

Question

One has the following relations for a domain $R$:

$R$ GCD domain $\Rightarrow$ All irreducible elements are prime

$R$ PID $\Rightarrow$ $(R$ GCD domain $\land$ $R$ statisfies ACCP$)$

$R$ UFD $\Leftrightarrow$ $(R$ GCD domain $\land$ $R$ satisfies ACCP$)$

$R$ satisfies ACCP $\Rightarrow$ $(R$ UFD $\Leftrightarrow$ All irreducible elements are prime$)$

$R$ GCD domain $\Rightarrow$ $(R$ UFD $\Leftrightarrow$ $R$ atomic domain$)$

So what is the difference between a GCD domain and a domain where all irreducible elements are prime? What is the "weakest" predicate $P$ such that

$R$ GCD domain $\Leftrightarrow$ $($All irreducible elements are prime $\land$ $P(R))$ ?

Answer 1

I'm very far from being an expert on these things, but I suspect that the answer might be the trivial (and disappointing?) one: the property which distinguishes GCD domains is ... the existence of GCDs.

Compare the chain of (strict) class inclusions

$\qquad$ UFDs $\subset$ bounded factorization domains $\subset$ ACCP domains $\subset$ atomic domains

to the chain

$\qquad$ UFDs $\subset$ GCD domains $\subset$ Schreier domains $\subset$ domains where irreducibles are prime.

I don't think there is any "interaction" (beyond the inclusion) within either chain. The relations that you quote have to do with interactions between the two chains, and basically they boil down to the fact that the only thing they have in common is "UFDs", i.e.,

$\qquad$ any class from the first chain $\cap$ any class from the second chain = UFDs,

plus the fact that PIDs $\subset$ UFDs.

So if you're looking for some class of domains such that

$\qquad$ domains where irreducibles are prime $\cap$ (??? domains) = GCD domains,

then at least it can't be one of those in the first chain. Perhaps there is some completely different class of domains that will do the trick, but I haven't come across any.

(Noetherian/Bézout/Dedekind/Prüfer/Krull/normal domains won't work either, but it's more complicated to describe how they fit into the above picture.)

Answer 2

A GCD domain must be a Schreier domain: It must be integrally closed, and every nonzero element must be primal (which is stronger than the condition that irreducibles are prime).

On the other hand, there are Schreier domains that are not GCD domains. Following Example 2.10 from The Schreier Property and Gauss’ Lemma, let $S$ be the integral closure of $\mathbb{C}[X]$ in $\overline{\mathbb{C}(X)}$, let $M$ be a maximal ideal of $S$, and let $R=\overline{\mathbb{Q}}+MS_M$. Then $R$ is Schreier but not GCD.

There are several other examples in the literature, including a fairly simple rank 2 monoid domain over a field.

There are (somewhat complicated) characterizations in the literature of GCD domains, the most common of which appears to be a condition for a PVMD (Prüfer v-multiplication domain) to be a GCD domain.

Answer 3

There are several examples of integral domains in which every atom is a prime but which are not GCD.

We can, roughly, put them in the following classes.

(1) Antimatter domains. These are domains in which there is no atom and hence the statement "every atom is a prime" holds vacuously. If you search the net, you will find several references on antimatter domains. One reference is: "Monoid domain construction of antimatter domains", Comm. Algebra 35(10) (2007). Antimatter domains can be GCD domains, such as valuation domains with a non-principal maximal ideal, they can be pre-Schreier and they can be non-pre-Schreier (see e.g. Example 2.11 in "The Schreier property and Gauss Lemma" [Bollettino U. M. I.(8) 10-B (2007), 43-62.]

Another one is example 4.5 of my paper on pre-Schreier domains: [Comm. Algebra 15 (1987) 1895--1920]. This is a one-dimensional quasi local ring that is pre-Schreier, a concept that has often been discussed at this forum. But quickly an element $x$ is primal in a domain $R$ if for $y,z\in R$, $x|yz $ implies that $x=rs$ where $r|y$ and $s|z.$ And $R$ is pre-Schreier if every element of $R$ is primal. Now an irreducible element that is also primal, is a prime and if a one dimensional quasi local domain contains a prime, this prime will be the generator of the maximal ideal.

(2) Non-GCD rings that contain primes: Here's an example that may not be too hard to understand:

Let $\mathbb{Z}$ denote the ring of integers, let $\mathbb{Q}$ be the ring of rational numbers and let $X,Y$ be two indeterminates over $\mathbb{Q}.$ Construct the two dimensional regular local ring $R=\mathbb{Q} \lbrack \lbrack X,Y]]$ and for $p$ a prime element set $D=\mathbb{Z}_{(p)}+(X,Y)\mathbb{Q} \lbrack \lbrack X,Y]].$ This ring $D$ is a quasi local ring with only one atom that is $p$ that divides every nonzero non unit of $D,$ so there is no atom besides $p.$ That $D$ is not a GCD domain follows from the fact that in $D$ every power of $p$ divides both $X$ and $Y,$ but $X$ and $Y$ have no GCD. I mentioned that antimatter domains, i.e., ones that do not have any atoms are automatically domains in which every atom is a prime. Then I mentioned that there are antimatter domains that may not be pre-Schreier hence not GCD and mentioned an example. Now let me mention that there are tons of examples of non-pre-Schreier antimatter domains and they can be found here.

Now these are the babies no predicate will make them into a GCD domain. That means that there is no hope from this angle. On the other hand, let us call a domain $D$ a $v$-finite conductor domain if for each pair of nonzero elements $a,b$ of $D$ we have a finitely generated ideal $B$ such that $aD\cap bD=B_{v}.$ (Here $B_{v}=D:D:B)$ and I showed long ago in Theorem 3.6 of [Z]=[Comm. Algebra 15 (1987) 1895--1920] that a $v$-finite conductor domain that is pre-Schreier must be a GCD domain. So the closest answer the the other part of the question seems to be: A pre-Schreier domain that is also a $v$-finite conductor domain is a GCD domain. That a Schreier PVMD is GCD follows from this result.

Now the proof that a pre-Schreier $v$-finite conductor domain is GCD is as follows: The Schreier property is characterized by: for every pair of nonzero elements $a,b$ in $D$ and for every finite set $x_{1},x_{2},...,x_{n}\subseteq aD\cap bD$ there is $d\in aD\cap bD$ such that $x_{1},x_{2},...,x_{n}\subseteq (d)$ (see part (3) of Theorem 1.1 of [Z]). Now if it so happens that for each pair $a,b\in D\backslash \{0\},$ $% aD\cap bD=(x_{1},x_{2},...,x_{n})_{v}$ for some $x_{1},x_{2},...,x_{n},$ then there is a $d\in aD\cap bD$ such that $x_{1},x_{2},...,x_{n}\subseteq (d).$ But then $x_{1},x_{2},...,x_{n}\subseteq (d)$ implies that $(x_{1},x_{2},...,x_{n})_{v}\subseteq (d)$ and this gives $aD\cap bD=(x_{1},x_{2},...,x_{n})_{v}\subseteq (d)\subseteq aD\cap bD,$ making $aD\cap bD$ principal for each pair $a,b\in D\backslash \{0\}.$

Finally $D$ being $v$-finite conductor is much weaker than $D$ being a PVMD which requires that $D$ is a $v$-finite conductor domain and for each pair $a,b\in D\backslash \{0\}$, $((aD\cap bD)(aD+bD))_{v}=abD.$