Characteristic of Ordered Integral Domain is Zero

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

Let $\Char D$ be the characteristic of $D$.

Then $\Char D = 0$.

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

$\forall n \in \Z: \map g n = n \cdot 1_D$

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 $D$.


By Properties of Ordered Ring $(5)$:

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


$\forall p > 0: \Char D \ne p$


$\Char D = 0$

and so $g$ is a monomorphism from $\Z$ into $D$.

Also, if $m < p$, then $p - m \in \Z_+$, so:

$p \cdot 1_D - m \cdot 1_D > 0_D$


$\map g m < \map g p$

Thus by Monomorphism from Total Ordering, $g$ is a monomorphism from $\Z$ into $D$.