Annihilator of Ring Always Contains Zero

Theorem

Let $\struct {R, +, \times}$ be a ring.

Let $\map {\mathrm {Ann} } R$ be the annihilator of $R$.

Then $0 \in \map {\mathrm {Ann} } R$.

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.

$\blacksquare$