Proof of Eakin-Nagata theorem, Matsumura $thm2.6$ Eakin-Nagata theorem,in Matsumura $thm3.6$,I have question.
Let $A$ be a commutative ring and $M$ is a faithful finitely generated module over it.If the ascending condition holds on the submodules of the form $IM$ for ideals $I⊂A$, the module $I/IM$ is noetherian ring.

Proof
It is enough to show that $M$ is a Noetherian module since, in
general, strong texta ring admitting a faithful Noetherian module over
it is a Noetherian ring. Suppose otherwise. By assumption, the set of
all $IM$, where $I$ is an ideal of $A$ such that $M/I$ is not
Noetherian has a maximal element, $J/M$  .
Replacing $M$ and $A$
by$M/JM$ and $A/Ann（M/JM）$, we can assume, for each nonzero submodule
$I⊂A$,the module $M/IM$ is noetherian.

Why can we replace $M$ and $A$ to $M/JM$ and $A/Ann（M/JM）$?
I think the latter two satisfies the hypothesis of the theorem,
but I believe we cannot say every $M$ and $A$ can be denoted like
$M/JM$ and $A/Ann（M/JM）$.
Thank you for your kind help.
 A: The argument is a proof by contradiction. Here is a rough outline of the strategy: Suppose there is a counterexample $M$ over a ring $A$ (meaning that $M$ satisfies the hypotheses of the theorem but is non-Noetherian). Then we can construct from this one counterexample $M$ another counterexample $M'$ over another ring $A'$ that has an additional property (that $M'/IM'$ is Noetherian for any non-zero ideal $I$ of $A'$). The remainder of the proof then derives a contradiction using this additional property.
So, what this shows is that there's no counterexample that has this additional property. But the proof starts by showing: if there is any counterexample whatsoever to the theorem, then there is a counterexample with that additional property. So the theorem can't have any counterexample, which completes the proof by contradiction.
Of course, if this were a direct proof, then the strategy wouldn't make sense, because you'd only be proving something about modules of that particular form. But since it's a proof by contradiction, we just need to start by assuming there's a counterexample and derive a contradiction from this, and it's perfectly fine if one of the intermediate steps is "show that the existence of a counterexample implies the existence of another counterexample, and consider this other counterexample instead".
