Non-Trivial Annihilator Contains Positive Integer
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Theorem
Let $\left({R, +, \times}\right)$ be a ring with unity.
Let $A = \operatorname{Ann} \left({R}\right)$ be the annihilator of $R$.
Let $a \in A$ such that $a \ne 0$.
Then $A$ contains at least one strictly positive integer.
Proof
Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$.
First we note that:
\(\ds a\) | \(\in\) | \(\ds \operatorname{Ann} \left({R}\right)\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \cdot 1_R\) | \(=\) | \(\ds 0_R\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \left({-a}\right) \cdot 1_R\) | \(=\) | \(\ds 0_R\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\left({a \cdot 1_R}\right)\) | \(=\) | \(\ds 0_R\) |
So:
- $a \in A \implies -a \in A$
But if $a \ne 0$ then either $a > 0$ or $-a > 0$.
That is, either $a$ or $-a$ is positive.
So, if $\operatorname{Ann} \left({F}\right)$ contains at least one non-zero element, it contains at least one strictly positive integer.
$\blacksquare$