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.

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