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Given a maximal ideal $M$ in a non-Noetherian commutative ring $R$, I'm trying to determine whether or not there can exist infinite strictly ascending chains of ideals of $R$ contained in $M$. I know that since $R$ is not Noetherian:

  • There exists an infinite strictly ascending chain of ideals $I_1\subset I_2\subset \dotsb$ contained in $R$, and
  • There exists an ideal $I$ of $R$ that is not finitely generated.

First I wanted to try supposing that there exists an infinite strictly ascending chain of ideals of $R$ contained in $M$ and derive a contradiction. Somehow I should use the facts that $M$ is maximal and that $R$ is not Noetherian. The maximality of $M$ tells me that there are no proper ideals of $R$ properly containing $M$, but I don't see how it can tell me anything about ideals of $R$ contained inside $M$. Are there any suggestions to see how these facts fit together?

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Let $R$ have maximal ideal $M$. As modules, $R_R$ is Noetherian iff $(R/M)_R$ and $M_R$ are Noetherian. $(R/M)_R$ is a simple module, so it is always Noetherian.

If $R_R$ is to be non-Noetherian, then $M_R$ has to be non-Noetherian. An infinite strictly ascending chain of submodules contained in $M$ is an infinite strictly ascending chain of ideals of $R$.

Generalizing this, you should be able to conclude that for any ideal $I$ such that $(R/I)_R$ is a Noetherian module, $I_R$ is not Noetherian.

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  • $\begingroup$ Could it possibly be this easy? I think the question sounded tricky when I read it, but when I thought about it this way I couldn't see what, if anything, is wrong with it. $\endgroup$ – rschwieb Mar 7 '16 at 16:34

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