Given a commutative ring $R$ with unity, we define for $I,J\subseteq R$ ideals
- $I\ \vert\ J\iff I\supseteq J$
- $IJ=\{\sum_i a_ib_i:a_i\in I, b_i\in J\}$
For every commutative unitary ring $R$ it follows that $$\exists K \subseteq R\text{ ideal}:J=IK\Rightarrow I\ \vert\ J$$ And, for Dedekind domains, the converse also holds. So my question is
When is the converse not true in the class of commutative unitary rings? For instance, is there any integral domain such that the converse does not hold?
Here's a concrete counterexample: let $R = \mathbb{C}[x, y]$, $I = (x, y)$, and $J = (x)$. It is clear that $I \subseteq J$. However, there does not exist an ideal $K$ such that $IK = J$. Indeed, if we assume the contrary, then $(y)K \subseteq J$, meaning that for all $p \in K$, we have $x \mid py$. We know that $R$ is a UFD and $x$ is an irreducible not dividing $y$, so it follows that $x \mid p$. Since $p$ was chosen arbitrarily in $K$, we have $K \subseteq IJ$, meaning $J = IK \subseteq J$. However, $x \in J$ and $x \not\in IJ$ since all non-zero polynomials in $IJ$ have degree at least $2$.
More generally, if $R$ is any Noetherian integral domain of dimension greater than $1$, then there exist non-zero prime ideals $\frak{p} \subsetneq \frak{q}$. We claim that there exists no ideal $\frak{a}$ such that $\frak{p} = \frak{q}\frak{a}$. Assuming the contrary, since $\frak{q} \not\subseteq \frak{p}$, we have $\frak{a} \subseteq \frak{p}$. It follows that $\frak{p} = \frak{q}\frak{a} \subseteq \frak{q}\frak{p} \subseteq \frak{p}$, so $\frak{p} = \frak{qp}$. This must also hold true in the local ring $R_{\frak{q}}$. By Nakayama's lemma, we yield $\mathfrak{p} = 0$, which is a contradiction.