Formanek's proof of Eakin-Nagata theorem I'm looking at the proof of Formanek of  Eakin-Nagata theorem, that is also the proof given by wikipedia. The theorem says that if $A$ is a ring and $M$ a faithful finite $A$-module, such that ($\cdot$) the ascending chain condition holds on the submodules of the form $IM$ (for any ideal $I\subseteq A$), then $A$ is Noetherian. To prove this result, I know that it suffices to show that $M$ is Noetherian.
The first argument is that we can assume that $M$ is not Noetherian and ($*$) for any non-zero submodule $IM$, with $I$ ideal as above, $M/IM$ is Noetherian. This I'm not sure to understand: I think that the idea  is to suppose that $M$ is not Noetherian and look for a contradiction. So we choose an ideal $I_0\subseteq A$ maximal among the ideals $J\subseteq A$ for which $M/JM$ is not Noetherian, and get the  $A$-module $N:=M/I_0M$, that clearly is not Noetherian and satisfies ($*$); also, $N$ is faithful as $A/\operatorname{Ann}N$-module. So if exists a (finite faithful non-Noetherian) module $M$ over $A$ satisfying ($\cdot$), there is a (finite faithful non-Noetherian) module $N$ over a quotient $B$ of $A$ satisfying ($*$). Thus if we get a contradiction with $N$ and $B$, automatically the initial $M$ and $A$ cannot exist (i.e. $M$ must be Noetherian).
Therefore we assume that $M$ is a faithful  finite $A$-module and that satisfies ($*$). Consider the set $S$ of submodules $N\subseteq M$ such that $M/N$ is faithful; if $\{x_{1},\dots ,x_{n}\}$ is a set of generators of $M$, I understand that $M/N$ is faithful if and only if the only $a\in A$ for which $ax_1,\dots,ax_n\in N$ is $0$. So I'll try to justify why we can apply Zorn's lemma. If $N_1\subset\cdots \subset N_i\subset\cdots$ is a chain of submodules in $S$, the module $\bigcup_i N_i$ is again in $S$: in fact if there was $a\in A$ such that $ax_1,\dots,ax_n\in \bigcup_i N_i$, since the generators are finitely many, there would be $j$ for which $ax_1,\dots,ax_n\in N_j$ (meaning $a=0$).
After Zorn's lemma, we are guaranteed that there is a maximal $N_0\in S$. If $M/N_0$ is Noetherian, it is a Noetherian faithful $A$-module, implying $A$ Noetherian and $M$ Noetherian (a contradiction). Hence we can assume that $M/N_0$ is not Noetherian and, similarly to what we have done in the first paragraph, we look for a contradiction with the $A$-module $M/N_0$.
So this $M/N_0$ we found is surely finite, faithful and non-Noetherian. The fact that $M/N_1$ has also the ($*$) property should follow  simply because, if $I\subseteq A$ is an ideal such that $IM\supseteq N_1$, then $(M/N_1)/(IM/N_1)\cong M/IM$ that is Noetherian. Thus we replace $M$ with $M/N_1$, similarly to what we've done in the first paragraph, and assume that $M$ has also the property that, for every submodule $N$, hte module $M/N$ is not faithful; finally we can prove that this $M$ is Noetherian, and get a contradiction (the last argument is already clear so I won't write it).
Is this  proof correct? I know that other questions, related to this proof of Eakin-Nagata theorem have already been made, but every solution that I see is just a few lines long, so I really couldn't understand any of them.
 A: Your argument is clear until the last paragraph of the proof which looks jumbled.  I think what you are trying to say is the following:

So this $M/N_0$ we found is surely finite, faithful and non-Noetherian. Hence we have a chain of submodules in $M$: $$N_0\subsetneq N_1 \subsetneq N_2\subsetneq N_3\subsetneq\cdots\qquad\qquad(1)$$
We know $M/N_1$ is not faithful (by maximality of $N_0$) so there exists an ideal $I\subset A$, with $IM\subseteq N_1$.  Thus: $$M/N_1\cong (M/IM)/(N_1/IM),$$
so $M/N_1$ is a quotient of $M/IM$ which is Noetherian by the (*) property of $M$. Hence $M/N_1$ is Noetherian.
However $(1)$ implies that we have a chain in $M/N_1$:
$$0\subsetneq N_2/N_1\subsetneq N_3/N_1\cdots\qquad\qquad$$
Thus $M/N_1$ is not Noetherian - a contradiction.

Minor comments:

*

*It is clearer if you do not rename $B$ and $N$ as $A$ and $M$ respectively.  You are already using $A$ and $M$ and it gets quite confusing which you mean.  I think this renaming is the source of any doubts you have.


*When justifying Zorn, I would add a line to say that each $ax_r\in N_i$ for some $i$, and as there are finitely many of them we can take $j$ to be the maximum of these $i$.  Also mention that $0\in S$ so $S\neq\emptyset$.


*To deduce $N$ Noetherian from $B$ Noetherian just say "as $N$ is finite".
