Associated primes and Noetherian condition

Question

Let $A$ be a commutative ring, $M$ an $A$-module, and $N\subset M$ a submodule. Consider the following two sets: $$\Omega:=\{\mathfrak{p}\in\operatorname{Spec}A \ | \ \mathfrak{p}=(N:m) \ \mbox{for some} \ m\in M\}$$ $$\sqrt{\Omega}:=\{\mathfrak{p}\in\operatorname{Spec}A \ | \ \mathfrak{p}=\sqrt{(N:m)} \ \mbox{for some} \ m\in M\}$$ $(N:m)$ means $\{a\in A \ | \ am\in N\}$. Clearly $\Omega\subset\sqrt{\Omega}$. I know from my knowledge of primary decomposition that the two sets are equal if $M$ is Noetherian. Many things in the theory of primary decomposition work if either $A$ or $M$ is Noetherian, so I have a suspicion that the two sets may be equal if $A$ is Noetherian instead of $M$, but I can't seem to prove it. Is it true at all? I can show that if $A$ is Noetherian, then $\bigcup\Omega=\bigcup\sqrt{\Omega}$.

Answer 1

If $\mathfrak{p}=\sqrt{(N:m)}$, then $\mathfrak p$ is minimal over $(N:m)$, that is, $\mathfrak p$ is weakly associated. But for noetherian rings weakly associated primes are associated. (More details here.)