Unity of Ordered Integral Domain is Strictly Positive
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Theorem
Let $\struct {D, +, \times \le}$ be an ordered integral domain whose unity is $1_D$.
Then:
- $\map P {1_D}$
where $P$ is the (strict) positivity property.
Proof
We have by definition of the unity that:
- $\forall a \in D: 1_D \times a = a = a \times 1_D$
This particularly applies to $1_D$ itself:
- $1_D = 1_D \times 1_D$
But then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:
- $\map P {1_D \times 1_D} \implies \map P {1_D}$
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $7 \ \text{(ii)}$