# The order ideal of a finitely generated module over a Dedekind domain

Let $R$ be a Dedekind domain and $M$ be a nonzero finitely generated torsion $R$-module. In Curtis and Reiner's Methods of Representation Theory it states that $M$ has a composition series and if its factors are given by $R/P_i$ with $P_i$ maximal $R$-ideals, then the ideal $\mathrm{ord}(M)=\prod P_i$ is well defined.

I don't know why a composition series for $M$ exists as I'm not sure if $M$ is an Artinian module, and also it's not clear why $\mathrm{ord}(M)$ is well defined.

I'd be very grateful for some help.

A finitely generated torsion module $M$ over a Dedekind domain $R$ is artinian because it is a finitely generated $(R/\operatorname{Ann}M)$-module, and $\operatorname{Ann}M\ne0$. This shows that $M$ is an $R$-module of finite length.
Moreover, by Jordan-Holder theorem one knows that every two composition series are equivalent. But the quotient modules of a composition series are of the form $R/P$, with $P$ a maximal ideal, and then these maximal ideals are uniquely determined by $M$.

• I've been trying to understand why $M$ is artinian if it is finitely generated over $R/\operatorname{Ann} M$ if $\operatorname{Ann}M$ is nonzero. Also I am not sure why the maximal ideals $P$ are uniquely determined. Is it possible that $R/P_1\cong R/P_2$ for $P_1\neq P_2$? regards – eddie Apr 11 '16 at 13:07
• @eddie This holds since the quotient ring $R/\operatorname{Ann}M$ is zero-dimensional and noetherian, that is, artinian. For the next question: no, it's not. – user26857 Apr 11 '16 at 13:07
• For more details see here. – user26857 Apr 11 '16 at 13:15
• Last question: yes, that's what I've meant. First question: you misread my comment; I've talked about the quotient ring $R/I$ where $I=\operatorname{Ann} M$. Such a quotient ring is zero-dimensional whenever $I\ne 0$, and this happens for $R$ is one-dimensional integral domain, so all its non-zero prime ideals are maximal. – user26857 Apr 11 '16 at 13:28
• @eddie 1. Modules which are artinian and noetherian are of finite lenght, so they have a composition series. 2. $\operatorname{Ann}M \neq 0$ because $M$ is finitely generated and torsion. (This has nothing to do with the dimension of the ring.) 3. Why need a reference for showing that $\dim R/I=0$ whenever $\dim R=1$ and $I\ne 0$? The prime ideals of $R/I$ are of the form $P/I$ with $P$ prime in $R$ containing $I$. Since $I\ne 0$ then $P$ is maximal and you are done. – user26857 Apr 11 '16 at 13:47