Let $K$ be a field, and let $D$ be a subring of $K$ with identity. Let $K^*$ be the multiplicative group of nonzero elements of $K$. The group $U$ of units of $D$ is a subgroup of $K^*$. We take $G=K^*/U$ the factor group with the usual operation.
For $xU,yU \in G$ we say $xU \leq yU$ if and only if $y/x \in D$. We note that $xU \leq yU$ if and only if $Dx$, the $D$-submodule of $K$ generated by $x$, contains $Dy$, the $D$-submodule of $K$ generated by $y$.
The above quote is taken from Gilmer's book, $\textit{Multiplicative Ideal Theory}$ (1972), page 171, in Chapter 3 titled $\underline{Valuation\,\,Theory}$ and section titled $\textit{Groups of Divisibility, Valuations, and Generalizations}$.
I have worked through the first few pages of Hungerford's graduate algebra text about modules to try to understand the above opening paragraph. I am wondering what exactly $Dx$ is.
I know that if we have a ring $R$ and some $R$-module $A$, then the submodule of $A$ generated by some $x \in A$ is the intersection of all submodules of $A$ containing $x$ and further this can be represented by the set $$\{rx+nx \mid r \in R,\, n\in \mathbb{Z}\}.$$ I see "the $D$-submodule of $K$ generated by $x$," and I think of $K$ being the module on $D$ taking just the abelian group $(K,+)$ and then making the module operation as $$D\times K \rightarrow K \text{ by } (d,k)\mapsto dk, \text{ the usual product in $K$}.$$ Then taking the submodule generated by $x$ to be the submodule of $K$, which would just be $$\{ dx+nx \mid d \in D,\, n \in\mathbb{Z}\}.$$ Does this sound about right?
Since the ring $D$ has identity, the $D$-submodule generated by an element $x\in K$ is $Dx=\{dx:d\in D\}$. There is no need to add ``integer multiples" of $x$. The reason Hungerford has that in his general definitions is that he isn't requiring his rings to have identity. You already get integer multiples of $x$ by taking $d=n\cdot 1_D$, where $1_D$ is the multiplicative identity in $D$.