Characteristic of Ordered Integral Domain is Zero

Theorem
Let $$\left({D, +, \circ}\right)$$ be a totally 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 as defined in Power of an Element.

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

Proof

 * By Properties of an Ordered Ring $$(5)$$:
 * $$\forall n \in \Z^*_+: n \cdot 1_D > 0$$

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 $$D$$.