Annihilator of Ring Always Contains Zero

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Theorem

Let $\left({R, +, \times}\right)$ be a ring.

Let $\operatorname{Ann} \left({R}\right)$ be the annihilator of $R$.


Then $0 \in \operatorname{Ann} \left({R}\right)$.


Proof

We have by definition of integral multiple that:

$\forall r \in R: 0 \cdot r = 0_R$

where $0_R$ is the zero of $R$.

Hence the result by definition of annihilator.

$\blacksquare$