Show that $ann(M)$ is an ideal in $R$.

Question

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 $brM=0$?

Answer 1

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 $brM=0.$