Ring is Noetherian if it admits a faithful finitely generated module with ACC on submodules generated by ideals

Solution 1:

Edit (12/31): This result turns out to have a bit of history. It was published in 1973, though there were other formulations around prior to that [1]. One finds that this result has been reproduced in Matsumura's textbook, Commutative Ring Theory [2]. The method of proof is essentially the same as what I had answered prior to looking it up, with the first step proceeding in basically the same way. The second step is done a little differently, so I provide both in this answer. Since the approaches are similar in content, I made the answer community wiki.

First step: Let $IM$ be maximal such that $M/IM$ is not Noetherian as an $A$-module. Let $I_0=\operatorname{Ann}(M/IM)$. Then, $I_0M=IM$ and $M/I_0M$ is a faithful $A/I_0$-module. Thus, replacing $M$ with $M/I_0M$ and $A$ with $A/I_0$, we may assume $M/JM$ is Noetherian for all nonzero ideals $J$ of $A$, but $M$ is not Noetherian. We will arrive at a contradiction by showing that this $A$ is Noetherian (and thus this $M$ is Noetherian).

Second step: (The approach in the references) By Zorn's Lemma, we get a maximal submodule $N\subset M$ such that $M/N$ is a faithful $A$-module. We may assume $M/N$ is not Noetherian. So, replacing $M$ with $M/N$, we have

(a) $M$ is not Noetherian

(b) If $I$ is a nonzero ideal of $A$, then $M/IM$ is Noetherian

Let $N$ be any submodule of $M$. We will prove $N$ is finitely generated. If $N=0$, then that is the case. Suppose $N\ne 0$. By statement (c), there exists $0\ne a\in A$ such that $aM\subset N$. Because $N/aM$ is a submodule of $M/aM$, which is Noetherian, $N/aM$ is finitely generated. Furthermore, if $x_1,\dots,x_n$ generate $M$, then $ax_1,\dots,ax_n$ generate $aM$. So, $aM$ is finitely generated. Therefore, $N$ is finitely generated. It follows that $M$ is Noetherian, violating statement (a). This is the desired contradiction.

(A different approach) Any ascending chain of submodules $M_i$ of $M$ such that $M_i\neq M_{i+1}$ for all $i$ has the property that $M/M_i$ is faithful for all $i$ (otherwise, $M_i\supset IM$ eventually and we know $M/IM$ is Noetherian, contradiction). Thus, if $N$ is a maximal submodule of $M$ such that $M/N$ is a faithful $A$-module, then $M/N$ is Noetherian. Hence, $A$ is Noetherian. This is a contradiction.