Special submodules over almost Dedekind domains An integral domain $R$ is an almost Dedekind 
domain if for each maximal ideal $m$ of $R$, the ring $R_m$ is a Dedekind 
domain, where $R_m$  is the localization of $R$ at $m$. 

Question: Let $M$ be an $R$-module, where $R$ is an almost Dedekind domain and let $m$ be a maximal ideal of $R$ and there exists $x\in M$ such that $m$ is a minimal prime ideal over $Ann_R(x)$, where $Ann_R(x):=\{r\in R\mid rx=0_M\}$. How can we construct a submodule $N$ of $M$ such that $\sqrt{Ann_R(N)}=m$?

Note: $\sqrt{I}:=\{r\in R\mid r^n\in I$ for some $n\in \mathbb{N}\}.$
 A: Without loss of generality, we can assume that $M = R/I$, $x = 1 +  I$ where $I$ is a proper non-zero ideal of $R$ and $\mathfrak{m}$ is any maximal ideal of $R$ containing $I$ (recall that $R$ is one-dimensional).
We will also assume that 

(Condition IIPM) $I$ can be represented as an irredundant intersection $\bigcap_{e \in E} \mathfrak{m}_e^{n_e}$ of powers of maximal ideals of $R$, with $E$ possibly infinite.

Conditions under which (IIPM) holds, with unique decomposition and for every proper ideal of an almost Dedekind ring, have been investigated in [1, see, e.g Corollary 3.9] and subsequent papers.
One necessary and sufficient condition is that $R/A$ has at least one finitely generated maximal ideal for every proper ideal $A$ of $R$.
We suppose moreover that $\mathfrak{m} = \mathfrak{m}_f$ for some $f \in E$. So, we will only address the question under the additional assumption that $\mathfrak{m}$ appears in a decomposition of $I$ as an irredundant intersection of powers of maximal ideals of $R$.
Set now $J \Doteq \bigcap_{e \in E \setminus \{f \}}\mathfrak{m}_e^{n_e}$ and $K \Doteq (I: J) = \{ r \in R \,\vert \, rJ \subseteq I \}$. Then $ R \supsetneq K \supseteq \mathfrak{m}_f^{n_f}$ so that $\sqrt{K} = \mathfrak{m}_f$ and $\text{Ann}_R(N) = K$ where $N$ is the image of $J$ in $R/I$.

[1] W Heinzer and B. Olberding, "Unique irredundant intersections of completely
irreducible ideals", 2005.
