# Show that $ann(M)$ is an ideal in $R$. [duplicate]

If $$_{R}M$$ is a left $$R$$-module of a ring $$R$$, show that $$ann(M)=\left \{ r\in R:rM=0 \right \}$$ is an ideal in $$R$$.

I have used the fact that the left annihilator is a left ideal in $$R$$ to prove that $$ann(M)$$ is a subring of $$R$$ and also a left ideal in $$R$$. But how do I show $$ann(M)$$ is a right ideal in $$R$$? Do I use the right annihilator for this? For example, if $$r\in R$$ and $$b\in ann(M)$$, then $$bM=0$$. This implies $$rbM=0$$ or $$rb\in ann(M)$$. But how do I show that $$brM=0$$?

## marked as duplicate by rschwieb abstract-algebra 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 4 at 13:26

• As Melody shows below: it comes down to $rM\subseteq M$, so $b(rM)=0$. – Dave Apr 4 at 2:29
Let $$m\in M$$, $$r\in R,$$ and $$b\in\text{ann}(M).$$ Then $$rm\in M,$$ so $$b(rm)=0.$$ Since $$m\in M$$ was arbitrary we have $$brM=0.$$
• Excuse me for my beginner knowledge, but may I ask why $rm\in M$? – numericalorange Apr 4 at 2:33
• That's from the definition of left $R$ module. Since $M$ is a left module there is a map $\cdot:R\times M\to M.$ This map is denoted typically by $r\cdot m$ or $rm$ for all $r\in R$ and $m\in M.$ – Melody Apr 4 at 2:46