Let $R$ be an integral domain (commutative) and let $Q$ be its field of fractions. The definition of a fractional ideal of $R$ is an $R$-submodule $I$ of $Q$ such that there exists an $x \in Q^\times$ such that $xI \subseteq R$.
In order to build more intuition on why the "$R$-submodule of $Q$" condition is part of the definition, I wanted to find examples of an additive subgroup of $Q$ that is not an $R$-submodule of $Q$, i.e. not closed under $R$-multiplication.
If we take $R = \mathbb{Q}$ then $Q \cong \mathbb{Q}$, and any subring of $\mathbb{Q}$ that is not an ideal, e.g. $2\mathbb{Z}$, will do as an example (because it will not be a $\mathbb{Q}$-submodule of $\mathbb{Q}$).
Now, are there any examples (of additive subrings of $Q$ that are not $R$-modules) when the inclusion $R \hookrightarrow Q $ is strict (i.e. $R$ and $Q$ are not the same)?
As suggested by @user26857, an example would be $R = \mathbb{Z}[i]$, $Q = \mathbb{Q}(i)$ and take the subring $I := \mathbb{Z} \subseteq \mathbb{Q}(i)$. This is not an $R$-module because it is not closed under multiplication by $i \in R$. Note that this is a "minimal example" in the sense that it does satisfy the other condition, namely we have $1I \subseteq R$.
This is a special case of the class of examples $R = S[x]$ taking any subring $T \subseteq S \subsetneq Q$ or even $\frac{1}{x}T$ where $x \in S^*$.