Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive

Theorem

Let $\struct {D, +, \times, \le}$ be an ordered integral domain whose zero is $0_D$.

Then:

$\forall x \in D: x \ne 0_D \iff \map P {x \times x}$

where $\map P {x \times x}$ denotes that $x \times x$ has the (strict) positivity property.

That is, the square of any element of an ordered integral domain is (strictly) positive if and only if that element is non-zero.

Proof

Let $x = 0_D$.

Then $x \times x = 0_D \times 0_D = 0_D$ by the properties of the ring zero.

Thus by definition of strict positivity property:

$\neg \map P {0_D \times 0_D}$

$\Box$

Now suppose $x \ne 0_D$.

One of two cases applies:

$\map P x$

$\neg \map P x$

Let $\map P x$.

Then by definition of (strict) positivity:

$\map P {x \times x}$

Now suppose $\neg \map P x$.

Then by the trichotomy law of ordered integral domains:

$\map P {-x}$

Then again by definition:

$\map P {\paren {-x} \times \paren {-x} }$.

But by Product of Ring Negatives:

$\paren {-x} \times \paren {-x} = x \times x$

So again:

$\map P {x \times x}$

Hence the result.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $7 \ \text{(i)}$