Suppose that $A$ is a ring, $M$ is a Noetherian $A$-module, and $\operatorname{Ann}M=0$. I must prove that $A$ is Noetherian.
I tried to prove it by contradiction, assuming that $A$ isn't Noetherian. Then there is a chain $$I_1\subsetneq I_2\subsetneq\dots\subsetneq I_i \subsetneq\dots $$ of ideals in $A$ that doesn't stabilize. If I take the following chain of submodules of $M$ $$I_1M\subseteq I_2M\subseteq\dots\subseteq I_iM \subseteq\dots $$ there is an $r$ such that $I_{r+n}M=I_rM$ for any natural number $n$. But from this equality I don't know how to recover an element $a$ such that $aM=0$. I thought to involve the ideal $\bigcup_iI_i$, but I still don't see a link to the annihilator. Can you give me only a hint?