## The annihilator of a right ideal is a submodule

Let $R$ be a ring with 1, let $M$ be a left $R$-module, and let $I \subseteq R$ be a right ideal. We define the annihilator of $I$ in $M$ to be $\mathsf{Ann}_M(I) = \{ m \in M \ |\ I \cdot m = 0 \}$. Prove that $\mathsf{Ann}_M(I)$ is a submodule of $M$.

We use the submodule criterion.

Certainly $I \cdot 0 = 0$; then $0 \in \mathsf{Ann}_M(I)$.

Let $x,y \in \mathsf{Ann}_M(I)$, let $r \in R$, and let $a \in I$. Now $ar \in I$ since $I$ is a right ideal. Then $a \cdot (x+r \cdot y) = a \cdot x + ar \cdot y = 0$. Thus $x+ry \in \mathsf{Ann}_M(I)$.

Thus $\mathsf{Ann}_M(I)$ is a submodule of $M$.