# 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 $$n\in \mathbb{N}\}.$$

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.

• Dear Luc Guyot, thank you for your answer. Probably, you use statment 2 or 3 of [1, Corollary 3.9] to represented $I$ as an irredundant intersection of positive powers of maximal ideals of $R$. But I cannot undrestant how other condition of statment 2 or 3 of [1, Corollary 3.9] are satisfied? Please, if it is possible, explain it more. – user140640 May 16 at 5:35
• @user140640 I do not address the case of an arbitrary almost Dedekind ring $R$. I consider only those $R$ such that $R/I$ has at least one finitely generated maximal ideal for every non-zero proper ideal of $R$. I added a line to make this restriction clear. – Luc Guyot May 16 at 6:52