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This question already has an answer here:

Let $R$ be a ring, and $M$ an $R$-leftmodule. Let $\operatorname{Ann}_R(M)$ be the annihilator of M, meaning that $r m = 0 \space\space\space\space \forall r \in \operatorname{Ann}_R(M), m \in M$.

Let $I \subseteq \operatorname{Ann}_R(M)$ be a two-sided ideal. Show that M is naturally an $R/I$-module.

Thanks in advance! I'm not that used to annihilators, so any help would be appreciated.

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marked as duplicate by rschwieb ring-theory Apr 26 '15 at 0:03

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You have to prove that if $r\equiv s \mod I$, then $rm=sm$ for any $m\in $M$.

That is equivalent to $(r-s)m=0$, which is by definition since $r\equiv s\mod I\iff r-s\in I\subseteq\operatorname{Ann}_AM$.

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