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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?

Dr. Scotti
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  • $\text{Ann } M = 0$ means that for each $0 \neq a \in A$ there must be some $m \in M$ so that $am \neq 0$ right? – Henno Brandsma Nov 19 '21 at 09:23
  • Yes in fact I tried to prove that if $A$ is not noetherian there's an $a$ such that $am=0$ for every $m$. I tried also not to reason by absurd, but I couldn't conclude anything anyway – Dr. Scotti Nov 19 '21 at 10:01
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    Hint: what if there was an element $m \in M$ such that $a \longmapsto am$ was injective? – Aphelli Nov 19 '21 at 10:11
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    Before proving that there is such an $m $ I try to answer your question. If there is $m\in M$ such that $a\mapsto am$ is injective, then $A$ (as an $A$-module) is isomorphic to a submodule of $M$ (I mean $Am$), and it must be noetherian. – Dr. Scotti Nov 19 '21 at 10:26

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