## The annihilator of a submodule is a two-sided ideal

Let $R$ be a ring with 1 and let $M$ be a left $R$-module. Let $N \subseteq M$ be a submodule. We define the annihilator of $N$ in $R$ to be $\mathsf{Ann}_R(N) = \{r \in R \ |\ r \cdot N = 0 \}$. Prove that $\mathsf{Ann}_R(N)$ is a two-sided ideal in $R$.

We need to show that $\mathsf{Ann}_R(N)$ is nonempty, closed under subtraction, and absorbs $R$ from both sides.

Let $n \in N$; certainly $0 \cdot n = 0$. Thus $0 \in \mathsf{Ann}_R(N)$.

Let $n \in N$ and $a,b \in \mathsf{Ann}_R(N)$. Then $(a-b) \cdot n = (a \cdot n) - (b \cdot n) = 0-0 = 0$, so that $a-b \in \mathsf{Ann}_R(N)$.

Let $n \in N$, let $a \in \mathsf{Ann}_R(N)$ and let $r \in R$. Note that $ra \cdot n = r \cdot (a \cdot n)$ $= r \cdot 0 = 0$ and $ar \cdot n = a \cdot (r \cdot n) = 0$, so that $ra,ar \in \mathsf{Ann}_R(N)$.

Thus $\mathsf{Ann}_R(M)$ is a two-sided ideal in $R$.