Annihilator of Ring Always Contains Zero
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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.
$\blacksquare$