$v$-ideal (or divisorial ideal) which is not invertible

Question

An ideal $I$ of an integral domain $D$ is said to be invertible if there exists a fractional ideal $J$ such that $IJ=D$. In this case, $J$ is unique, and can be denoted by $I^{-1}$. It can also be proved that $I^{-1}=\{x\in K\ |\ xI\subseteq D\}$ where $K$ is the field of fraction of $D$. Denote $I_v=(I^{-1})^{-1}$. An ideal $I$ is called a $v$-ideal or a divisorial ideal if $I_v=I$. Of course, an invertible ideal is a $v$-ideal, so a $v$-ideal can be seen as a generalization of an invertible ideal.

A well-known example of a non-invertible ideal is $\langle x,2\rangle$ as an ideal of the integral domain $\mathbb Z[x]$. It can also be proved that it is not a $v$-ideal either.

My question is: is there any basic example of a $v$-ideal which is not invertible? Thanks in advance!

Answer 1

Let $D$ be a domain with fraction field $K$. Recall that an element $k \in K$ is said to be almost integral if there exists an ideal $I$ of $D$ such that $k I \subseteq I$, equivalently if there exists $a \in D$ such that $ak^n \in D$ for all $n \in \mathbb{N}$. A domain is said to be completely integrally closed when every almost integral element of $K$ is already in $D$. Recall also that a fractional ideal $I$ is said to be $v$-invertible when $(II^{-1})_v = D$.

It is a very good exercise to show that a domain is completely integrally closed iff every (divisorial) ideal is $v$-invertible. Perhaps you could try to solve this before reading on.

This fact (and its proof) tip us off to the following construction:

Let $D$ be a domain with fraction field $K$ and $k \in K \setminus D$ almost integral. Let $I = \sum_{n=0}^\infty k^nD$, which is a fractional ideal. Then $I_v$ is divisorial but not $v$-invertible, and a fortiori not invertible.

Indeed, a fractional ideal $I$ is $v$-invertible iff $(I_v : I_v) = D$, but by construction of our $I$ we have $kI \subseteq I$, therefore $k \in (I_v : I_v)$ but $k \notin D$.

Basic examples of domains that are not completely integrally closed come from valuation domains of Krull dimension greater than $1$, or non-normal Noetherian domains.

Some remarks on rings that do have every divisorial ideal invertible, and why you might have had trouble coming up with examples:

In $\mathbb{Z}[x]$ it actually is the case that every divisorial ideal is invertible. For a direct study of the property in question, you might be interested in the paper On Generalized Dedekind Domains of M. Zafrullah from 1986, where such domains were called $G$-Dedekind.

A couple classes of rings having divisorial ideals invertible are (1) Dedekind rings (rings in which every ideal is invertible) (2) GCD domains in which every (possibly infinite) set of elements has a GCD. In this case the $v$-closure of an ideal is its GCD, so divisorial ideals are even principal. Theorem 1.9 of the cited paper also shows that the property of divisorial ideals being invertible ascends from $D$ to $D[x]$.

Answer 2

Here is a simple concrete example of an divisorial ideal which is not invertible. For $D$, let $K$ be a field and $D=k[T^2,T^3]$ be the subring of $k[T]$ generated by $T^2$ and $T^3$. (In other words $D$ is the ring of polynompal $P(T)$ satisfying $P'(T)=0$, and $\text{spec} D$ is the usual cusp curve $y^2=x^3$.) The fraction ring of the domain $D$ is $k(T)$.

Now the maximal ideal $M=(T^2,T^3)$ of $D$ is divisorial, but not invertible.

To see that $M$ is divisorial, recall that a well-known characterization of divisorial is "intersection of principal fractional ideals" (cf. Bourbaki, Commutative Algebra, beginning of chapter VII.) Now observe that $M$ is the intersection of two principal fractional ideals of $D$, namely $D$ and $T^{-1}D$.

To see that $M$ is not invertible, just assume it is and write $I$ its inverse, so $IM=D$. Since $1$ in $D$, $I$ must contain $T^{-k}$ for some $k \geq 2$, and since $I$ is a $D$-ideal, it must contain either $T^{-3}$ or $T^{-2}$. But in both case, $IM$ contains $T$, a contradiction since $T \not \in D$.