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.

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