Divisorial ideal ($v$-ideal) of $A=\mathbb{Z}+X\mathbb{Q}[[X]]$.

Question

I know that the domain $R$ is Mori if and only if for every nonzero ideal $I$ of $R$, $I_v=(a_1,\dots,a_n)_v$ for some $a_1,\dots,a_n\in I$.

My goal is to find an example satisfying :

For any $v$-ideal (= divisorial ideal) $I$ of a domain $R$, $I_v=(a_1,\dots,a_n)_v$ for some $a_1,\dots,a_n\in I$, but $R$ is not a Mori.

Let $A=\mathbb{Z}+X\mathbb{Q}[[X]]$. How to check the hypothesis? I know $A$ is not a Mori, but it's difficult to find the $v$-ideals of $A$.

Answer 1

In this post I will

(1) Briefly explore the condition on a domain that every ideal is $v$-finite

(2) Show why $\mathbb{Z} + x\mathbb{Q}[[x]]$ fails this condition and

(3) Give a classic counterexample to distinguish between rings in which ideals are $t$-finite (Mori) and rings in which they are just $v$-finite. In fact, we will see that even having $v$-principal ideals is not even to imply the Mori property.

For convenience (I don't think this property is named in the literature), let's say

A domain $D$ has property (&) if every ideal of $D$ is $v$-finite.

Property (&) has a strong interaction with other important and well-studied properties. In an old note, for example (sorry I do not have reference on hand), M. Zafrullah observes the following:

Let $D$ be a domain with property (&). Then $D$ is completely integrally closed iff it is a PVMD, i.e. if every finitely generated ideal is $t$-invertible.

Proof:

Recall that being completely integrally closed (CIC) is equivalent to every ideal being $v$-invertible. If $D$ is CIC, then fix a finitely generated ideal $I$. We have that $(II^{-1})_v = D$ by CIC assumption. Since $I^{-1}$ is a fractional divisorial ideal, we can write $I^{-1} = J_v$ for some finitely generated fractional ideal $J$ by (&) assumption. Then $(II^{-1})_t = (IJ_t)_t = (IJ)_t = (IJ)_v = D$, where we have used the fact that $(IJ)_t = (IJ)_v$ because $IJ$ is finitely generated. Conversely, if $D$ is a PVMD, then fix an ideal $I$. We have that $I_v = J_v$ for some finitely generated ideal $J$ by (&) assumption. Then $(JJ^{-1})_t = D$ by PVMD assumption. Of course $(IJ^{-1})_v = (JJ^{-1})_v = D$, so $I$ is indeed $v$-invertible. $\square$

In practice this can be helpful in showing that a domain must have non-$v$-finite ideals without having to pursue explicit constructions.

Next I'd like to record an observation about the relationship between the $v$-operation and greatest common denominators (GCDs).

Lemma: Let $D$ be a domain and $I$ an ideal of $D$. If $I_v = aD$ then $a$ is a GCD of $I$. If $D$ is a GCD domain, then the converse holds too.

Proof:

Let $a_\alpha$ be a set of generators of $I$. Since $a_iD \subseteq I \subseteq I_v = aD$, it's clear that $a$ divides $a_\alpha$. Additionally, if $b \mid a_\alpha$ for all $\alpha$, then $1/b \in I^{-1} = 1/a$, hence $b$ divides $a$, and indeed $a$ is a GCD of the set of $a_\alpha$.

Conversely, suppose that $a$ is a GCD of the $a_\alpha$. If $b \in I_v$, then $bI^{-1} \subseteq D$, and in particular, since $1/a \in I^{-1}$, it's clear that $a$ divides $b$. What remains to be shown is that $a \in I_v$. Suppose that $\frac{c}{d}D$ is a principal fractional ideal containing $I$. If we take $D$ to be a GCD domain, then we can assume that $c,d$ are relatively prime, so that in fact $I \subseteq cD$. Then since $a$ is a GCD of $I$, $a$ is in $c$. Hence $a$ is in every principal fractional ideal containing $I$, $a \in I_v$.
$\square$

In a GCD domain, the $v$-closure of any finitely generated ideal is principally generated by the GCD of the ideal's generators, so the above observation immediately leads to

Corollary: A GCD-domain has property (&) iff every set of elements has a GCD.

Let's now examine $\mathbb{Z} + x\mathbb{Q}[[x]]$, and in particular show that it does NOT have property (&).

Let $R = \mathbb{Z} + x\mathbb{Q}[[x]]$. The following hold:

(1) $R$ does not satisfy the ascending chain condition on principal ideals, and a fortiori is not Mori.

(2) $R$ is a GCD-domain

(3) Not every ideal of $R$ is $v$-finite

(4) But every prime ideal of $R$ is $v$-principal.

Proof

(1) Consider the ascending chain of principal ideals $a_nR$ where $a_n = \frac{x}{2^n}$.

(2) I direct you to Theorem 2.11 in this paper of Anderson and Nour el Abidine, The A + XB[X ] and A + XB[[X ]] constructions from GCD-domains. The proof is not hard and consists of using the fact that (in our case) $\mathbb{Q}[[x]]$ and $\mathbb{Z}$ are both GCD-domains to establish putative GCDs in $R$, with three distinct cases emerging depending on whether the elements under consideration have constant terms or not.

(3) I'll first note that we could argue abstractly from some of the lemmas above, e.g. note that $R$ is certainly not completely integrally closed, and since $R$ is a GCD domain and hence a PVMD, one lemma shows it cannot satisfy property (&). But we can also argue directly and produce non $v$-finite ideals quite easily, using our observation about the interaction of property (&) with GCD domains. For example, let $I$ be the ideal generated by the $a_n$ in (1). It obviously doesn't have a GCD, and hence it can't be $v$-finite.

(4) Let $P$ be a prime ideal of $R$. Since $P \cap \mathbb{Z}$ is prime, it's clear that $P$ contains exactly one prime integer $p$ or contains no nonzero element of $\mathbb{Z}$. In the first case, check that $P = pR$ is principal. In the second case, check that $P$ contains $x$. In case $x \in P$, let $a \in P_v$. Then $a \in dD$ for every $d \in R$ such that $P \subseteq dD$. But then then $d$ divides $x$, so that $d$ has a unit for it's lowest coefficient. Hence $a$ must have an element of $\mathbb{Z}$ for it's lowest coefficient, and we conclude $x$ divides $a$. To summarize, every prime ideal $P \subset R$ is either of the form $pR$ where $p \in \mathbb{Z}$, or satisfies $P_v = xR$.

This example is also interesting in that it demonstrates that prime ideals do not witness $v$-finiteness. On the other hand, prime ideals (in particular $t$-prime ideals) DO witness $t$-finiteness. The proof is a good exercise in generalizing classic results — the crux of it is to show that if $*_f$ is any finite character $*$-operation, then (a) the $*_f$-finite ideals form an Oka set, and (b) the set of ideals non-$*_f$-finite $*_f$-ideals is an inductive set under partial order of inclusion which then by (a) has maximal $t$-prime elements.

Although $\mathbb{Z} + x\mathbb{Q}[[x]]$ does not help us to distinguish property (&) from Mori, there is a known class of rings called Pseudo-Dedekind that accomplishes this. Call a domain Pseudo-Dedekind if every divisorial ideal is invertible (recall that invertible ideals are always divisorial). These have also been called Generalized or G-Dedekind in the past, but now that term is usually taken to mean something else. Pseudo-Dedekind domains are equivalently characterized as domains in which divisorial ideals form a group with binary operation the ideal product (in general, we have to set $A *B := (AB)_v$. M. Zafrullah introduced them in this paper called On Generalized Dedekind Domains and the work was expanded upon shortly thereafter by Anderson and Kang. It was Zafrullah who noticed that the classic example of the Ring of Entire Functions is a Pseudo-Dedekind domain that is not Mori.

It's well known that the ring of entire functions is Bezout but not Noetherian. Therefore there are ascending chains of principal ideals that do not stabilize (Recall for Bezout domains, ACCP is equivalent to Noetherian), and Mori fails. Every element of the ring of entire functions factors into a (possibly infinite) product of prime linear factors, so it immediately follows that arbitrary sets have greatest common divisors, which we can produce by taking the product of all of the common prime factors (preserving multiplicity). Since the ring is a GCD domain, property (&) is thus satisfied. Moreover every ideal is $v$-principal, and the ring of entire functions is indeed a Pseudo-Dedekind domain.