divisorial ideals Let $I$ be an ideal of a domain. Then is there an ideal $J$ properly located between $I$ and $I^{\nu}$? Here $I^{\nu}$ is divisor of $I$.
 A: The question is not quite clear, but in order to avoid its indefinite reappearence here some sort fo answer (info I gave in comments). [I hope noone 'objects' to me answering after voting to close, but the vote expired a long time ago, so.]
Let me first repeat something on the definition of divisorial ideals as it seems not universal known.
For a usual (ring) ideal $I$ of $R$ one defines $I^v = (I^{-1})^{-1}$ where $I^{-1} = (R:I)$. 
Then $I^v$ is again a ring ideal and $(I^v)^v = I^v$. 
Put differently some ring ideals $J$ have the property $J^v= J$, and such a ring ideal is called a divisorial ideal. (The definition can be/is usually extended to fractional ideals.) 
Indeed, for certain domains every ring ideal is a divisorial ideal, Dedekind domains are an example. 
Such domains of course need to be excluded from considerations as in the question. 
Also, $I$ in the question needs to be non-diviorial
Now, even if one excludes this it still can happen that there is no ideal between $I$ and $I^v$. 
For example, for the (maximal) ideal $I=(X,Y)$ in the polynomial ring $K[X,Y]$, $K$ a field, one has 
$I^v$ is the full ring, so $I$ is not disorial, but there can be no ideal between $I$ and $I^v$ since $I$ is maximal.
However, for $J=(X,Y^2)$ one would still have that $J^v$ is the full ring, and now there is some ideal in between namely the $I$ above. 
A: The question is: Let $I$ be an ideal of a domain. Then is
there an ideal $J$ properly located between $I$ and $I^{v }$? Here $I^{
v }$ is divisor of $I$" .
First let me "nitpick" on the question.
Any ideal of a domain $R$ containing an ideal $I$ is a divisor of $I.$ The
ideal $I^{v }$ is a special kind of divisor of $I$ called the
divisorial envelope of the ideal $I.$ Let $K$ be the quotient field of $R.$
Then $I^{-1}=\{x\in K|$ $xI\subseteq R\}=R:I$ and $I^{v
}=(I^{-1})^{-1}.$ Now why is $I^{v }$ a divisorial envelope? Because it can be shown that $I^{v }=\cap _{I\subseteq xR\subseteq K}xR.$ Next
an ideal $J$ is called divisorial if $J=J^{v }.$ Obviously there is no
ideal between $I$ and $I^{v }$ if $I$ happens to be divisorial because 
$(I^{v })^{v}=I^{v}$ for any nonzero fractional ideal $I$ of $R.$
(Here an $R$-submodule $F$ of $K$ is a fractional ideal if $dF\subseteq R$
for some nonzero $d\in R.)$ The function $I\mapsto I^{v}$ on the set $F(R)$
of all nonzero fractional ideals of $R$ is an example of a so-called star
operation known as the $v$-operation. (From the envelope definition it
follows that for every star operation $\ast $ and for every ideal $A\in F(R)$
we have $A^{\ast }\subseteq A^{v}$  For star operations check out sections
32 and 34 of [Gilmer, Multiplicative Ideal Theory, Marcel Dekker, Inc., New
York,1972] or my survey article
[https://link.springer.com/chapter/10.1007/978-1-4757-3180-4_20], or both.
Now there are domains in which every nonzero ideal is divisorial. The
Dedekind domains have been mentioned Now read [W. Heinzer, Integral domains
in which each non-zero ideal is divisorial, Mathematika 15 (1990),
164--170.] for more examples. There are examples of non-divisorial ideals $I$
such that there is no ideal between $I$ and $I^{v}.$ For example ideals such
as maximal ideals $(p,X)$ in $%
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\lbrack X]$ where $p$ is a prime in the the ring of integers $%
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,$ for $(p,X)^{v}=%
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\lbrack X].$ You can also take the maximal ideal $M$ of a non-discrete rank
one valuation domain $(V,M),$ Here $M^{-1}=V$ and so $M^{v}=V$ and so any
ideal $I$ between $M$ and $M^{v}$ is either $M$ or $M^{v}.$ 
Now let's come down to examples of ideals $I$ such that there is an ideal $J$
properly between $I$ and $I^{v}.$ For this we need to bring in another set
of definitions. Define for each nonzero fractional ideal $I$ the ideal $%
I^{t}=\cup \{F_{v}\}$ where $F$ ranges over nonzero finitely generated
sub-ideals of $I.$ It turns out that $I\mapsto I^{t}$ is another example of
a star operation (i.e. for all $A,B$ in $F(R)$ and $c\in K\backslash \{0\},$
(1) $(cA)^{t}=cA^{t},$ $R^{t}=R,$ (2) $A\subseteq A^{t},$ $A\subseteq B$
implies $A^{t}\subseteq B^{t}$ and (3) $(A^{t})^{t}=A^{t}.)$ One distinctive
property of the $t$-operation is that
for each integral $t$-ideal $I$ ($%
I=I^{t})$ there is a (at least one) proper $t$-ideal $M$ maximal among the
integral $t$-ideals containing $I.$ This $M$ is called a maximal $t$-ideal
and is necessarily a prime ideal. If $Max_{t}(R)$ is the set of maximal $t$-ideals of $R$, it can be shown that $R=\cap R_{M}$ where $M$ ranges over
all the maximal $t$-ideals of $R.$
If you have consulted the cited sources, you would know that given a star
operation $\ast $ and $A,B$ in $F(R)$ we have $(AB)^{\ast }=(A^{\ast
}B)^{\ast }=(A^{\ast }B^{\ast })^{\ast }.$ Now let $\ast $ be a star
operation. Call a fractional ideal $A\in F(R)$ $\ast $-invertible if $(AB)^{\ast }=R$ for some $B$ in $F(R).$ It turns out that if $(AB)^{\ast }=R$
then $B^{\ast }=A^{-1}$ and $A^{\ast }=A^{v}.$ It can be shown that if $A$ is $t$-invertible, then there is a finitely generated ideal $B\subseteq A$
such that $A^{t}=B^{t}=B^{v}=A^{v}.$ ( I say that a $t$-invertible
fractional ideal $A$ is strictly $v$-finite.) 
Next let $f(R)=\{A\in F(R)|$ $A$ is finitely generated$\}$ and call a domain 
$R$ a Prufer $v$-multiplication domain (PVMD) if every $A\in f(R)$ is $t$-invertible. It was shown by Griffin, [M. Griffin, Some
results on $v$-multiplication rings, Canad. J.
Math.19(1967) 710-722], who introduced the study of these domains, that $R$
is a PVMD if and only if $R_{M}$ is a valuation domain for each maximal $t$ -ideal $M$ of $R.$ So a PVMD is integrally closed, because an intersection
of integrally closed domains is integrally closed. Now it so happens that
most of the well known integrally closed domains are PVMDs (Prufer domains,
Krull domains, GCD domains, integrally closed Noetherian domains etc.).
Another characterizing property of PVMD's is that $R$ is a PVMD if and only
if every two generated nonzero fractional ideal $(a,b)$ of $R$ is $t$-invertible. Griffin (cited above) also called $R$ essential if for some set 
$\mathcal{F}$ of prime ideals $P$ with the property that $R_{P}$ is a
valuation domain we have $R=\cap _{P\in \mathcal{F}}R_{P}$ and asked if an
essential domain is indeed a PVMD. Heinzer and Ohm in [An essential ring
that is not a $v$-multiplication ring, Canad J. Math. 25(1973), 856-851]
showed, giving an example, that there is indeed an essential domain $R$ that
is not a PVMD. Building on this, and using the $D+XD_{S}[X]$ construction I
gave a number of examples of locally GCD domains ($R$ such that $R_{M}$ is a
GCD domain for each maximal ideal $M)$ that were not PVMDs, in [J. Pure
Appl. Algebra 50(1988) 93-107]. (Locally GCD domains are obviously
essential, as a GCD domain, being a PVMD, is essential.)
Reason why I have brought in essential domains is partly because I wanted to
mention some papers of mine and mainly because every finitely generated ideal in
an essential domain is $v$-invertible Proposition 2.1 of [On $v$-domains a
survey, https://link.springer.com/chapter/10.1007/978-1-4419-6990-3_6]
Now comes the crunch. An essential domain that is not a PVMD must have a two
generated nonzero ideal $(a,b)$ such that $((a,b)(a,b)^{-1})^{t}=((a,b)\frac{%
(a)\cap (b)}{ab})^{t}\neq R$, yet $((a,b)\frac{(a)\cap (b)}{ab})^{v}=R.$
Also as $t$ is a star operation we have $((a,b)\frac{(a)\cap (b)}{ab})^{t}\subseteq ((a,b)\frac{(a)\cap (b)}{ab})^{v}=R.$ Since $((a,b)\frac{%
(a)\cap (b)}{ab})^{t}\neq R$, we have $((a,b)\frac{(a)\cap (b)}{ab}%
)^{t}\subsetneq ((a,b)\frac{(a)\cap (b)}{ab})^{v}=R.$ This leads to $((a,b)%
\frac{(a)\cap (b)}{ab})\subseteq ((a,b)\frac{(a)\cap (b)}{ab})^{t}\subsetneq
((a,b)\frac{(a)\cap (b)}{ab})^{v}.$ So if $R$ is an essential domain there
is always an ideal $I$ ($=(((a,b)\frac{(a)\cap (b)}{ab})_{t}$) such that $%
I\subseteq I^{t}\subsetneq I^{v}.$ Now if you want an ideal $I$ such that $%
I\subsetneq I^{t}\subsetneq I^{v},$ you can pick a non-PVMD essential
domains with more than one maximal $t$-ideals, $M,N,$ with say $M$
non-maximal. Letting $x\in N\backslash M$ we have $(x,M)^{t}=R.$ By the
definition of the $t$-operation you have $J\subseteq (x,M)$ such that $%
J^{t}=(x,M)^{t}=R.$ (Again by the definition of the $t$-operation $%
J^{t}=J^{v})$. Now take a non-PVMD essential domain $R$ with $a,b$ such that 
$((a,b)\frac{(a)\cap (b)}{ab})_{t}\neq R$ and with $J\neq R$ such that $%
J_{t}=R$ and set $I=J((a,b)\frac{(a)\cap (b)}{ab}).$ Then you have $%
I\subsetneq I_{t}=(J_{t}((a,b)\frac{(a)\cap (b)}{ab})_{t})_{t}=((a,b)\frac{%
(a)\cap (b)}{ab})_{t}$ $\subsetneq I_{v}=(J((a,b)\frac{(a)\cap (b)}{ab}%
))_{v}=(J_{v}((a,b)\frac{(a)\cap (b)}{ab})_{v})_{v}=R.$ 
Note : Things could have gone slightly easier if I had noted as I did on
page 107 of [ J. Pure Appl. Algebra 50(1988) 93-107] that if $R$ is a
non-PVMD essential domain of a certain kind then so is $R[X].$
A: Take $I=(4,X)$ in $\mathbb{Z}[X]$. Then $I^v=\mathbb{Z}[X]$ and $I \subset (2,X) \subset I^v$
