# The annihilator induces a module [duplicate]

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.

## marked as duplicate by rschwieb ring-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 26 '15 at 0:03

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\$.