Let $R$ be a ring, $A$ a left $R$-module and $a ∈ A.$ Denote $O_a = \{r ∈ R : ra = 0\}.$ Show that $O_a$ is an ideal of $R.$
We have to check three conditions:
$1. \forall x,y \in O_a,\ x-y \in O_a\\ (x-y)a=xa-ya=0+0=0$
$2.\forall x \in O_a,\ \forall p \in R,\ px \in O_a\\ (px)a=p(xa)=p\cdot 0=0$
How to check remaining condition?
$\forall x \in O_a,\ \forall p \in R,\ xp\in O_a\\$
For non commutative $R$ I think (but I'm not 100% sure) that this is not true, for example take
$R=M_2(\mathbb{R})$ and $a=\begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix}$ then $\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}\in Ann(a)$ since $$\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}\begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix}=\begin{pmatrix} 0 & 0\\ 0 & 0\end{pmatrix}$$
but $\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix}\not\in Ann(a)$
this means that $Ann(a)$ is a left-ideal in $M_2(\mathbb{R})$