Fractional ideals: Is $D$ included in $I'I$?

Question

Let $D$ be a domain and let $F$ be its field of fractions. If $I$ is fractional, we define $I'$ to be the set of elements $c \in F$ such that $cI$ is included in $D$.

It is clear that if $I$ is fractional then $I'I$ is included in $D$. Does the converse holds? So, can we say that $D$ is included in $I'I$?

Would you help me, please? Thank you in advance.

Answer 1

A fractional ideal $I$ is said to be invertible if $I'I=D$.

A trivial counterexample is $I=\{0\}$, where $I'=F$ and $I'I=\{0\}$.

You can also note that if $I'I=D$, then $I$ is finitely generated as $D$-module. Indeed, there exist $x_1,x_2,\dots,x_n\in I$ and $y_1,\dots,y_n\in I'$ such that $$ \sum_{i=1}^n y_ix_i=1 $$ If $x\in I$, then $$ x=\sum_{i=1}^nxy_ix_i $$ and, by assumption, $d_i=xy_i\in D$, so $x=\sum_{i=1}^n d_ix_i$.

So you just need to find a non finitely generated fractional ideal and you're done. For instance, with $D=\mathbb{Z}$, the fractional ideal $I$ generated by $\{1/2^n:n>0\}$ is not finitely generated. What's $I'$ in this case?

Answer 2

How about this example: $D=\Bbb Q[x,y]$ and $I=\langle x,y\rangle$. I reckon $I'=D$.