# Finding a Noetherian faithful module

Let $$M$$ be a faithful $$A$$-module with the property that the modules of the form $$IM$$ for some ideal $$I\subseteq A$$ satisfy the ascending chain condition; also, suppose that $$\{m_1,\dots,m_r\}\subseteq M$$ is a set of generators. I have to find a Noetherian faithful $$A$$-module.

I tried two ways: first considering a $$JM$$ maximal among the submodules of the form $$IM$$ for which $$M/IM$$ is not Noetherian; then considering a $$JM$$ maximal among the submodules of the form $$IM$$ for which $$M/IM$$ is faithful. (In both cases $$(0)\subset A$$ satisfies the condition, so there is also a maximal submodule among those satisfying it). However I don't see how too continue, admitting that one of those path is the correct one. I think that an answer could be the one to this question (Ring is Noetherian if it admits a faithful finitely generated module with ACC on submodules generated by ideals), putting together first step and the second approach below. However I don't understand why we can make the reduction in the first step, nor the argument used in the second approach. Could you give me a clarification? Thanks

Perhaps I should spell out the argument given in the linked answer, and you can let me know which step(s) you query.

(1) Let $$M$$ be as given in your question. If $$M$$ is Noetherian, then as $$M$$ is also faithful, it will give you an answer to your question. Suppose that $$M$$ is not Noetherian. We will deduce a contradiction.

(2) By the ACC condition given, we may pick a maximal submodule of the form $$IM\subseteq M$$ where $$I\subset A$$ an ideal, and $$M/IM$$ not Noetherian.

(3) Let $$I_0={\rm Ann}(M/IM)$$. Then $$I\subseteq I_0$$ and $$I_0M\subseteq IM$$ by definition of annihilator. Thus $$IM=I_0M$$. Let $$M'=M/I_0M$$. Then $$M'$$ is faithful as a module over $$A'=A/I_0$$.

(4) $$M'$$ is not Noetherian over $$A$$ by construction. It has the same submodules when regarded as a module over $$A'$$, so it is not Noetherian over $$A'$$ either.

(5) However for any ideal $$0\neq J\subset A'$$, we have $$M'/JM'=M/KM$$ for some ideal of $$A$$: $$K\supsetneq I_0\supset I$$. Here $$K$$ is the preimage of $$J$$ under the quotient map $$A\to A/I_0$$. By the maximality condition on $$I$$, we have that $$M/KM$$ is Noetherian over $$A$$, so $$M'/JM'$$ is Noetherian over $$A'$$.

(6) Suppose that $$N\subseteq M'$$ is a submodule (over $$A'$$). Either $$M'/N$$ is faithful, or it is not, in which case $$N\supset iM'$$ for some $$0\neq i\in A'$$, so $$M'/N$$ is a quotient of $$M'/iM'$$, hence Noetherian by (5).

(7) Given a nested chain of submodules $$\{N_i\subseteq M'|i\in \mathcal{I}\}$$, with $$M'/N_i$$ faithful over $$A'$$, let $$N$$ be the union of the $$N_i$$. If $$aM'\subseteq N$$, for some $$0\neq a\in A'$$, then for $$s=1,2,\cdots,r$$, we have $$am_s\in N$$, so we have $$am_s\in N_{i_s}$$ for some $$i_s\in \mathcal{I}$$. Let $$i$$ denote the supremum of $$i_1,i_2,\cdots,i_r$$. Then we have $$aM\subseteq N_i$$, contradicting the assumption that $$M/N_i$$ is faithful for all $$i\in \mathcal{I}$$. Thus $$M'/N$$ is faithful.

(8) By (3) we know that $$M'/\{0\}$$ is faithful. By (7) we know that modules $$N$$ with $$M/N$$ faithful are closed under nested unions. Thus by Zorn's lemma, we may choose $$N$$ maximal such that $$M'/N$$ is faithful.

(9) Let $$M''=M'/N$$. By the maximality of $$N$$, we know that any proper quotient of $$M''$$ is not faithful, hence by (6) such a quotient must be Noetherian.

(10) Let $$\{0\}\subsetneq N_1\subsetneq N_2\subsetneq N_3\subsetneq\cdots$$ be a strictly increasing infinite chain of submodules in $$M''$$. Then $$N_2/N_1\subsetneq N_3/N_1\subsetneq N_4/N_1\subsetneq\cdots$$ is a strictly increasing infinite chain of submodules of $$M''/N_1$$, contradicting (9).

(11) Now we know that $$M''$$ is both faithful and Noetherian (over $$A'$$), we will deduce that $$A'$$ is Noetherian: Suppose we have an infinite strictly increasing chain of ideals $$I_i$$ in $$A'$$. For fixed $$s=1,2,3,\cdots,r$$, we know that the submodules $$I_im_s\subseteq M''$$ will eventually be constant. Thus we have an infinite strictly ascending chain of ideals $$I_i\cap {\rm Ann} (m_s)\subseteq {\rm Ann} (m_s)$$ (Important: Here we regard $$m_s\in M''$$). Repeating this step for $$s=1,2,3,\cdots,r$$, we obtain an infinite strictly ascending chain of ideals inside $${\rm Ann} (m_1)\cap {\rm Ann} (m_2)\cap {\rm Ann} (m_3)\cap\cdots\cap {\rm Ann} (m_r).$$ However this intersection is $$\{0\}$$ as $$M''$$ is faithful over $$A'$$. This gives us a contradiction to the existence of the infinite strictly increasing chain of ideals in $$A'$$. We conclude $$A'$$ is Noetherian.

(12) As $$A'$$ is Noetherian, and $$M'$$ is finitely generated over $$A'$$, we have that $$M'$$ is Noetherian over $$A'$$, contradicting (4).

(13) This is the contradiction we were seeking in (1). Thus we know that $$M$$ is both faithful and Noetherian over $$A$$, answering your question.