# A property of associated prime ideals

Let $A$ be a noetherian ring and $M$ a finitely generated module over $A$.

Consider the set of ideals: $S=\{I\mid I=(0:m),\ m\in M,\ m\neq 0\}$.

Since $A$ is a noetherian ring there exists a maximal element of $S$, say, $J=(0:u)$.

Assume $J$ is the only maximal element of $S$.

Is it true that $M=Au$ ?

Edit:

Actually, this question emerges from my own approach to the well-known theorem which states that there exists:

$0\subseteq M_1\subseteq M_2...\subseteq M_r=M$ and prime ideals $P_1,...P_r$ such that: $M_2/M_1\cong A/P_2$ and so on,...

I know $M_1$ have to be $Au_1$ where $(0:u_1)$ is the maximal element of $S$.

If $(0:u_2)$ is another maximal element then I can choose $M_2$ equals $Au_1+Au_2$, continue on this fashion I can get all $M_i$.

The sum $Au_1+Au_2$ is direct sum provided $(0:u_1)$ is not only maximal element.

If $(0:u_1)$ is only maximal element then $M_2$ have to be $M_1$. If I assume the theorem is true so I may say $M$=$M_1$=$Au_1$.

Please give my argument some light.

This is not true at least for trivial reasons: take for example the module $(A/\mathfrak{m})^{\oplus n}$, where $\mathfrak{m} \subseteq A$ is a maximal ideal, and $n$ is any integer bigger than $2$.
More generally, if you consider $(A, \mathfrak{m})$ to be local Noetherian ring, then for any $A$-module $M$ with $\mathfrak{m} \in \mathrm{Ass} M$ is $\mathfrak{m}$ clearly the maximal element of $S$, yet there are many such modules $M$ that are not cyclic (any module having $A/\mathfrak{m}$ as a submodule, basically).
I am not sure if I understand what you are trying to say, but even if $(0:u_1)$ is the only maximal element of $S$, that does not mean that $M=M_1$: Consider for example $A=\mathbb{Z}$ and $M=\mathbb{Z}/p^2\mathbb{Z}$. Then the only filtration that makes sense is $$0 \subseteq p\mathbb{Z}/p^2\mathbb{Z} \subseteq \mathbb{Z}/p^2\mathbb{Z},$$ the set $S$ equals $S=\{ (p^2), (p)\}$, so the only maximal element is $(p)$, the $u_1$ that realizes it is one of the generators of the subgroup of order $p$, and still $M_1=p\mathbb{Z}/p^2\mathbb{Z} \neq M$.
Note also that the equality $M_1=M_2$, as you claim, does not need to hold: just consider $M=\mathbb{Z}/(p)\oplus \mathbb{Z}/(p).$ Then the situation is similar to the one above, except now one can take $u_1=(1, 0), u_2=(0, 1)$ and then we have $$0 \subseteq M_1 \subsetneq M_2=M,$$