2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user140640 May 16 at 5:35
  • $\begingroup$ @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. $\endgroup$ – Luc Guyot May 16 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.