If $M$ is an $R$-module and $S\subset M$, then does $\text{Ann}(S) \subset \text{Ann}(rS)$ for all nonzero $r\in R$. Let $M$ be an $R$-module, $S\subset M$ any subset, and $r\in R$ nonzero. I'm trying to show that $\text{Ann}(S) \subset\text{Ann}(rS)$. Suppose that $s\in\text{Ann}(S)$ and $m\in S$. Then, if I can show that $sm = 0 = srm$, I'm done.
If $R$ is commutative, then $srm = rsm = r0 = 0$, and so $s\in \text{Ann}(rS)$, but I don't see why this is true if $R$ is not commutative.
 A: The answer is no, $\text{Ann}(S)\not\subseteq\text{Ann}(rS)$ in general. Another way to phrase the question is: $\text{Ann}(S)$ a right ideal of $R$? Indeed, for the equivalence, it is clear that it is a subring, and $ar\in \text{Ann}(S)$ for all $a\in \text{Ann}(S)$ and $r\in R$ is equivalent to $a\in \text{Ann}(rS)$ for all $a$ and $r$, i.e. our desired result.
We know it is a left ideal, but it is not a right ideal in general. Indeed, there is a general result that any maximal left ideal is the annihilator of some element of a simple module, so the existence of maximal left ideals that are not right ideals finishes the proof.
So you won't have to search for the result, let me give an example of such a left ideal. Consider $R=M_2(\mathbb{R})$, and
$$S=\lbrace \begin{pmatrix} 0 & 0 \\
1 & 1 
\end{pmatrix}\rbrace.$$
Then $\text{Ann}(S)$ is exactly the matrices with right column the zero vector, but though this is a maximal left ideal, it is not a right ideal. Indeed, we see that
$$\begin{pmatrix} a & 0 \\
b & 0
\end{pmatrix}\begin{pmatrix} 0 & 1 \\
1 & 0 
\end{pmatrix} = \begin{pmatrix} 0 & a \\
0 & b 
\end{pmatrix},$$
and so if $$r=\begin{pmatrix} 0 & 1 \\
1 & 0 
\end{pmatrix},$$
$\text{Ann}(S)\not\subseteq\text{Ann}(rS)$.
