# The annihilator induces a module [duplicate]

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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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## 1 Answer

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\$.