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Let $M$ be a Noetherian $A$-module where $A$ is a commutative ring with identity. Let $I_1\leq I_2\leq\cdots$ be an ascending chain of ideals of $A$. Then $I_1M\leq I_2M\leq\cdots$ is an ascending chain of submodules of $M$. Since $M$ Noetherian, $I_1M\leq I_2M\leq\cdots$ terminates. Does this imply $I_1\leq I_2\leq\cdots$ terminates hence $A$ is Noetherian? Why am I wrong?

If we let $\text{Ann}{\ M}=\{a\in A\mid am=0 \text{ for any }m\in M\}\leq I_1\leq I_2\leq\cdots$ be an ascending chain of ideals of $A$, then why in this case, $I_1M\leq I_2M\leq\cdots$ terminates can imply $I_1\leq I_2\leq\cdots$ terminates hence $A/\text{Ann}(M)$ Noetherian?

I am confused. Thank you a lot!

Shiquan
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    Take $0$ module. – user52045 Nov 19 '13 at 02:52
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    To have this implication you need $I\subsetneq J$ iff $MI\subsetneq MJ$. This is only the case if $Ann(M)=O$. – user52045 Nov 19 '13 at 03:16
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    @Ren You can't prove in this way that $A/Ann(M)$ is noetherian. Anyway, it's easy to give examples of noetherian modules over a non-noetherian ring. –  Nov 19 '13 at 08:05
  • Dear @user : I think you can show that $A/Ann(M)$ is Noetherian for a nonzero Noetherian $A$ module $M$. I would appreciate feedback if you think I've made a mistake in my solution. Thanks. – rschwieb Nov 19 '13 at 14:55

1 Answers1

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A counterexample

Let $R$ be your favorite commutative ring which isn't Noetherian, and let $I$ be any maximal ideal of it. Then $M=R/I$ is a simple (and so certainly Noetherian) module for this not-Noetherian ring.

Why am I wrong?

The wrong link is the idea that the chain of $I_iM$ stabilizing should imply that the chain of $I_i$ should stabilize. It just isn't so. In the example above, there are two cases that could occur. If $\cup I_i\subseteq I$, then $I_iM=\{0\}$ for all $i$, and this chain has certainly stabilized. On the other hand, if $\cup I_i\nsubseteq I$ then at some index $j$, $I_iM=M$ for all $i\geq j$, and the chain has again stabilized. But $I_i$ could be any nonstablizing chain in $R$ that you like.

The second part

As was mentioned in the comments, it's true that if $R$ has a faithful Noetherian module $M$, ($Ann(M)=\{0\}$) then $R$ is Noetherian. (The counterexample I gave is not a faithful $R$ module if $I\neq 0$.)

The idea of the proof is that since $M$ is finitely generated, you can inject $R$ into $M^n$ for some positive integer $n$. This makes $R$ a submodule of the Noetherian module $M^n$, hence $R$ is also Noetherian.

This particular fact has appeared in answers before. See especially these two: https://math.stackexchange.com/a/258141/29335 https://math.stackexchange.com/a/110219/29335

It can be checked that the $R/Ann(M)$ submodules of $M$ coincide with the $R$ submodules of $M$, so $M$ is a Noetherian $R/Ann(M)$ module, and moreover it is a faithful $R/Ann(M)$ module. By the above this means that $R/Ann(M)$ is in fact a Noetherian ring.

So that's why in the second part a chain based at $Ann(M)$ will in fact stabilize.

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