# Characteristic of Ordered Integral Domain is Zero

## Theorem

Let $\left({D, +, \circ}\right)$ be an ordered integral domain whose zero is $0_D$ and whose unity is $1_D$.

Let $\operatorname{Char} \left({D}\right)$ be the characteristic of $D$.

Then $\operatorname{Char} \left({D}\right) = 0$.

Let $g: \Z \to D$ be the mapping defined as:

- $\forall n \in \Z: g \left({n}\right) = n \cdot 1_D$

where $n \cdot 1_D$ is defined as the $n$th power of $1_D$.

Then $g$ is the only monomorphism from the ordered ring $\Z$ onto the ordered ring $D$.

## Proof

- By Properties of Ordered Ring $(5)$:

- $\forall n \in \Z_{>0}: n \cdot 1_D > 0$

Thus $\operatorname{Char} \left({D}\right) \ne p$ for any $p > 0$.

Hence $\operatorname{Char} \left({D}\right) = 0$ and so $g$ is a monomorphism from $\Z$ into $D$.

Also, if $m < p$, then $p - m \in \Z_+$, so $p \cdot 1_D - m \cdot 1_D > 0_D$.

Hence $g \left({m}\right) < g \left({p}\right)$ and thus by Monomorphism from Total Ordering, $g$ is a monomorphism from $\Z$ into $D$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 24$: Theorem $24.11$