Stabilising chain of submodules Take $S$ to be a commutative ring with unit, and $N \subset M$ to be $S$-modules. I am interested in the necessary and sufficient condition for it to be the case that for every collection of submodules $\{M_i\}_{i=1,\dots,n}$, such that $M_i \cap N = 0$, they form a stabilising chain,
$$M_1 \subset M_2 \subset \cdots \subset M_{n}.$$
I believe a necessary and sufficient condition is that every $M_i$ such that $M_i\cap N=0$ is finitely generated. I know that if every submodule was finitely generated, then $M$ would be Noetherian, and thus I am guaranteed this stabilising chain.
However, if I am only requiring that $M_i$'s such that $M_i \cap N = 0$ be finitely generated, this does not exclude the possibility of submodules which are not finitely generated to exist in $M$ so $M$ is not being required to be Noetherian, and so I am not guaranteed this stabilising chain. How can I approach this?
EDIT: This answer is completely wrong, so don't bother reading it. I'll leave it up for the record; details are in the comments, but I amateurishly confused the set $M \setminus N \cup \{ 0 \}$ with the submodule that it generates.
Let $M_N$ be the smallest submodule of $M$ containing $\left( M \setminus N \right) \cup \{ 0 \}$. The submodules of $M_N$ are exactly the submodules $M_i \leq M$ s.t. $M_i \cap N = \{ 0 \}$. The property in question is thus equivalent to whether $M_N$ is Noetherian or not. There are two cases.

Trivially, $M_N$ could simply be $M$ itself, i.e. $M$ has a generating set in $M \setminus N$.
$M_N$ could be a proper submodule of $M$. Since $M \setminus M_N = N \setminus \{ 0 \}$, this works out to say that at least one non-identity of element $N$ cannot be written as a linear combination of elements of $M_N$ over $S$.
EDIT: This answer is completely wrong, so don't bother reading it. I'll leave it up for the record; details are in the comments, but I amateurishly confused the set $M \setminus N \cup \{ 0 \}$ with the submodule that it generates.
Let $M_N$ be the smallest submodule of $M$ containing $\left( M \setminus N \right) \cup \{ 0 \}$. The submodules of $M_N$ are exactly the submodules $M_i \leq M$ s.t. $M_i \cap N = \{ 0 \}$. The property in question is thus equivalent to whether $M_N$ is Noetherian or not. There are two cases.

Trivially, $M_N$ could simply be $M$ itself, i.e. $M$ has a generating set in $M \setminus N$.
$M_N$ could be a proper submodule of $M$. Since $M \setminus M_N = N \setminus \{ 0 \}$, this works out to say that at least one non-identity of element $N$ cannot be written as a linear combination of elements of $M_N$ over $S$.