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I wanted to solve the following problem:

Let $R$ be a commutative ring with $1$ and $M$ a Noetherian $R$-module. Let $I=\operatorname{Ann}(M)$. Show that $R/I$ is a Noetherian ring.

Now to show that an increasing chain of ideals in $R/I$ is stationary, it is enough to show that an increasing chain of ideals in $R$ containing $I$ is stationary. So let $I \subseteq I_1 \subseteq I_2 \subseteq \cdots $ be an increasing chain of ideals in $R$. Then $0=IM\subseteq I_1M\subseteq I_2M\subseteq \cdots$ is an increasing chain of submodules in $M$.

Now, let $I_mM=I_{m+1}M$ for some $m \in \Bbb N$ and let $a\in I_{m+1}$. Define a map $\lambda_a:M\to M$ by $\lambda_a(m)=am \space \space \forall m\in M$. Then $\lambda_a(M) \subseteq I_{m+1}M=I_{m}M$. And as $M$ is Noetherian, it is finitely generated and thus, $\exists a_1,a_2,\ldots,a_n\in I_{m}$ such that $$\lambda_a^{n}+a_1\lambda_a^{n-1}+\cdots+ a_{n-1}\lambda_a+a_n=0.$$

Thus $a^n+a_1a^{n-1}+\cdots+a_n \in \operatorname{Ann}(M)=I\subseteq I_m$. Now as the $a_i$'s are in $I_m$, this shows $a^n\in I_m$ and hence $a\in \sqrt{I_m}$.

As $a\in I_{m+1}$ was arbitrary, we have $I_{m+1}\subseteq \sqrt{I_m} \implies \sqrt{I_{m+1}}\subseteq \sqrt{\sqrt{I_m}}=\sqrt{I_m}.$ Further, $I_m\subseteq I_{m+1}\implies \sqrt {I_{m}}\subseteq\sqrt{I_{m+1}}$ and thus $\sqrt{I_{m}}= \sqrt{I_{m+1}}$.

Now as $M$ is Noetherian, the chain of submodules $0=IM\subseteq I_1M\subseteq I_2M\subseteq \cdots$ becomes stationary and thus, using what we proved just now, for the chain of ideals $I \subseteq I_1 \subseteq I_2 \subseteq \cdots $ in $R$, the chain of radicals $\sqrt I \subseteq \sqrt{I_1} \subseteq \sqrt{I_2} \subseteq \cdots $ becomes stationary.

This is where I am stuck. I can't show from here that the chain of ideals $I \subseteq I_1 \subseteq I_2 \subseteq\cdots $ must be stationary. Any hints would be welcome.

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The $R$-module $M$ is finitely generated, say by $x_1, \dots, x_n$. The mapping $a \mapsto (ax_1, \dots, ax_n)$ from $R$ to $M^n$ induces an embedding of $R/I$ into $M^n$, hence $R/I$ is a Noetherian $R$-module.

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  • $\begingroup$ @AlexWertheim Yes, every submodule of a Noetherian module is Noetherian. $\endgroup$ – user404961 Feb 5 '17 at 7:15
  • $\begingroup$ Yes, you're correct; my mistake. I was confusing this with the fact that a subring of a Noetherian ring need not be Noetherian, but this isn't at all what you're claiming. Sorry about that! +1 :) $\endgroup$ – Alex Wertheim Feb 5 '17 at 7:17
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Does radical ideals $\sqrt{I_1}\subseteq\sqrt{I_2}\cdots$ becoming stationary imply ideals $I_1\subseteq I_2\cdots$ becoming stationary?

No. It is easy to find examples of non-Noetherian local rings with nil radicals . In such rings, the radical of every proper ideal is the maximal ideal, so this is a rather extreme case of your desired outcome failing, since the chain of radicals is constant for any given increasing chain of ideals in $R$.

One of the examples in the link above is $F[x^{1/2},x^{1/4},x^{1/8},\ldots]/(x)$ for a field $F$ (in case the link above changes for some reason.)

Your original already has an answer on the site. One strategy is to note that $M$ is a faithful Noetherian $R/ann(M)$ module, and that when a ring has a faithful Noetherian module, it is also Noetherian.

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