Characteristic of Ordered Integral Domain is Zero
Theorem
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 $D$.
Proof
By Properties of Ordered Ring $(5)$:
- $\forall n \in \Z_{>0}: n \cdot 1_D > 0$
Thus:
- $\forall p > 0: \Char D \ne p$
Hence:
- $\Char D = 0$
and so $g$ is a monomorphism from $\Z$ into $D$.
Also, if $m < p$, then $p - m \in \Z_+$, so:
- $p \cdot 1_D - m \cdot 1_D > 0_D$
Hence:
- $\map g m < \map g p$
Thus by Monomorphism from Total Ordering, $g$ is a monomorphism from $\Z$ into $D$.
$\blacksquare$
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We have a new and more axiomatic definition of an ordered integral domain which we may want to bring this into line with. Ultimately it boils down to the fact that an ordering on an integral domain is a lot more rigid than for a ring, such that totality is a consequence of the trichotomy law.
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Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.11$