Non-Trivial Annihilator Contains Positive Integer

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.

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.

$\blacksquare$