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I was looking for following two examples:

A divisible module $M$ over a commutative domain $R$ such that

  1. $R$ is Noetherian but $M$ is not.
  2. $M$ is Noetherian but $R$ is not.

I observed that if we consider any infinite dimensional vector space $V$ over a field $F$ then it satisfies conditions of example 1.

I am stuck in finding an example satisfying condition 2. Please suggest me how can I construct such module or give example of any such module, if possible.

Dr. Nirbhay Kumar
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1 Answers1

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A trivial example is of course $M=0$ where $R$ can be any non-noetherian domain. But other than that there are not examples for 2:
Let $M$ be a divisible $R$-module and $M\ne0$. Then there is some $m\in M$ with $m\ne0$. By definition for every $r\in R\setminus\{0\}$ there is some $m'$ with $rm'=m$, in particular $rm'\ne0$, so $r\notin\operatorname{Ann}M$. We conclude that $\operatorname{Ann}M=0$, i.e. $M$ is a faithful $R$-module. Now if $M$ is noetherian this will then imply that $R$ is also noetherian because we can embed $R$ (as an $R$-module) in $M^n$ for some $n$, see e.g. this answer.

leoli1
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    Nice: if I ever learned that divisible modules over domains were always faithful, I must have forgotten it. Now that I've read this I think it'll stick, though. – rschwieb Apr 08 '21 at 14:25