Absolute Value on Ordered Integral Domain is Strictly Positive except when Zero

Theorem
Let $\left({D, +, \times}\right)$ be an ordered integral domain.

For all $a \in D$, let $\left \vert{a}\right \vert$ denote the absolute value of $a$.

Then $\left \vert{a}\right \vert$ is always positive except when $a = 0$.

Proof
Let $P$ be the positivity property on $D$, and let $<$ be the ordering induced by it.

From the trichotomy law, exactly one of three possibilities holds for any $ a \in D$:

$(1): \quad P \left({a}\right)$:

In this case $0 < a$ and so $\left \vert{a}\right \vert = a$.

So:
 * $P \left({a}\right) \implies P \left({\left \vert{a}\right \vert}\right)$

$(2): \quad P \left({-a}\right)$:

In this case $a < 0$ and so $\left \vert{a}\right \vert = -a$.

So:
 * $P \left({-a}\right) \implies P \left({\left \vert{a}\right \vert}\right)$

$(3): \quad a = 0$:

In this case $\left \vert{a}\right \vert = a$.

So:
 * $a = 0 \implies \left \vert{a}\right \vert = 0$.

Hence the result.

Also see

 * Absolute Value is Bounded Below by Zero