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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 positive powers of maximal ideals of $R$.

Conditions under which (IIPM) holds, with a unique decomposition, 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.

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.

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 positive powers of maximal ideals of $R$.

Conditions under which (IIPM) holds 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 $R$. If this condition is satisfied, the decomposition is moreover unique.

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.

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 positive powers of maximal ideals of $R$.

Conditions under which (IIPM) holds, with a unique decomposition, 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.

Without loss of generality, we can assume that $M = R/I$, $x = 1 + I$ where $I$ is a proper non-zero ideal of $R$.

Let us also assume that

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

We suppose moreover that $\mathfrak{m} = \mathfrak{n}_f$ 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 $R/I$.

Conditions under which (IIPM) holds 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 $R/I$ has at least one finitely generated maximal ideal for every proper ideal $I$ of $R$. If this condition is satisfied, the decomposition is moreover unique.

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

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 positive powers of maximal ideals of $R$.

Conditions under which (IIPM) holds 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 $R$. If this condition is satisfied, the decomposition is moreover unique.

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.