Let $R$ be a ring with $1$ and let $M$ be a left $R$-module.
Let $S$ be a subset of $M$. The annihilator of $S$ in $R$ is the subset of the ring $R$ defined to be
\[\Ann_R(S)=\{ r\in R\mid rx=0 \text{ for all } x\in S\}.\]
(If $rx=0, r\in R, x\in S$, then we say $r$ annihilates $x$.)

Suppose that $N$ is a submodule of $M$. Then prove that the annihilator
\[\Ann_R(N)=\{ r\in R\mid rn=0 \text{ for all } n\in N\}\]
of $M$ in $R$ is a $2$-sided ideal of $R$.

To prove $\Ann_R(N)$ is a $2$-sided ideal of $R$, it suffices to prove the following conditions:

For any $r, s \in \Ann_R(N)$, we have $r+s\in \Ann_R(N)$.

For any $r\in R$ and $s\in \Ann_R(N)$, we have $rs\in\Ann_R(N)$.

For any $r\in R$ and $s\in \Ann_R(N)$, we have $sr\in \Ann_R(N)$.

Let $r, s \in \Ann_R(N)$. Then for any $n\in N$, we have
\[rn=0 \text{ and } sn=0.\]
It follows from these identities that
\begin{align*}
(r+s)n=rn+sn=0+0=0
\end{align*}
for any $n\in N$.
Hence $r+s\in \Ann_R(N)$, and condition 1 is met.

To prove condition 2, let $r\in R$ and $s\in \Ann_R(N)$.
For any $n\in N$, we have
\begin{align*}
&(rs)n=r(sn)\\
&=r(0) && (\text{since } s \in \Ann_R(N))\\
&=0.
\end{align*}
Thus, the element $rs$ annihilates all elements $n$ in $N$.
So $rs\in \Ann_R(N)$ and condition 2 is satisfied.
(At this point, we have proved that $\Ann_R(N)$ is a left ideal of $R$.)

To check condition $3$, let $r\in R$ and $s\in \Ann_R(N)$.
We need to prove that $(sr)n=0$ for any $n\in N$.
Since $N$ is a submodule of $M$, the element $rn$ is in $N$.
Since $s\in \Ann_R(N)$, we have $s(rn)=0$.
It follows from the associativity that
\begin{align*}
(sr)n=s(rn)=0
\end{align*}
for any $n\in N$.
Hence $sr\in \Ann_R(N)$ and condition 3 is proved.

Therefore, $\Ann_R(N)$ is a $2$-sided ideal of $R$.
This completes the proof.

Remark.

Note that the proof of condition 1 and 2 shows that $\Ann_R(S)$ is a left ideal of $R$ for any subset $S$ of $M$.
We needed the assumption that $N$ is a submodule of $M$ when we proved condition $3$.

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