Unity of Ordered Integral Domain is Strictly Positive

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 $7 \ \text{(ii)}$