TITLE: Example of a domain where all irreducibles are primes and that is not a GCD domain
QUESTION [9 upvotes]: 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 $\land$ $P(R))$ ?

REPLY [2 votes]: 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 $%
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$ be the ring of rational numbers and let $X,Y$ be two indeterminates over $%
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.$ Construct the two dimensional regular local ring $R=%
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\lbrack \lbrack X,Y]]$ and for $p$ a prime element set $D=%
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_{(p)}+(X,Y)%
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\lbrack \lbrack X,Y]].$ 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:
https://lohar.com/researchpdf/star%20potent%20domains%20and%20star%20homog%20ideals.pdf
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 condutor 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]=[Commun. 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 $%
a,b\in D\backslash \{0\}$, $((aD\cap bD)(aD+bD))_{v}=abD.$