Total Ordering on Field of Quotients is Unique

Theorem
Let $$\left({K, +, \circ}\right)$$ be a quotient field of a totally ordered integral domain $$\left({D, +, \circ; \le}\right)$$.

Then there is one and only one total ordering $$\le'$$ on $$K$$ which is compatible with its ring structure and induces on $$D$$ its given total ordering $$\le$$.

That ordering is the one defined by:

$$P = \left\{{\frac x y \in K: x \in D_+, y \in D_+^*}\right\}$$

Proof
First, note that from Divided By a Positive in Quotient Field:

$$\forall z \in K: \exists x, y \in R: z = \frac x y, y \in R_+^*$$

Now we show that $$P$$ satistfies conditions (P1) to (P4) of Positive Elements of Ordered Ring.


 * From Addition of Division Products and Product of Division Products, it is clear that $$P$$ satisfies (P1) and (P3).


 * Now we need to establish (P2).

Let $$z \in P \cap \left({-P}\right)$$.

Then $$z \in P$$ and $$z \in \left({-P}\right)$$. Thus:

$$\exists x_1, x_2 \in D_+, y_1, y_2 \in D_+^*: z = x_1 / y_1, -z = x_2 / y_2$$

So $$x_1 / y_1 = - \left({x_2 / y_2}\right) \Longrightarrow x_1 \circ y_2 = - \left({x_2 \circ y_1}\right)$$.

But $$0 \le x_1 \circ y_2$$ and $$- \left({x_2 \circ y_1}\right) \le 0$$.

So $$x_1 \circ y_2 = 0$$. As $$0 < y_2$$, this means $$x = 0$$ and therefore $$z = 0$$.

Thus (P2) has been established.


 * Now to show that (P4) holds.

Let $$z = x / y$$ where $$x \in D, y \in D_+$$.

Suppose $$0 \le x$$. Then $$z \in P$$.

However, suppose $$x < 0$$. Then $$0 < \left({-x}\right)$$ so $$-z = \left({-x}\right) / y \in P$$.

Thus $$z = - \left({-z}\right) \in -P$$.

So $$P \cup \left({-P}\right) = K$$.

So by Positive Elements of Ordered Ring, the relation $$\le'$$ on $$K$$ defined by $$P$$ is a total ordering on $$K$$ compatible with its ring structure.


 * Now we need to show that the ordering induced on $$D$$ by $$\le'$$ is indeed $$\le$$.

Let $$z \in D_+$$. Then $$z = z / 1_D \in P$$, as $$1_D \in D_+^*$$.

Thus $$D_+ \subseteq P$$ and $$D_+ \subseteq D$$ so $$D_+ \subseteq D \cap P$$ from Intersection Largest.

Conversely, let $$z \in D \cap P$$. Then

$$\exists x \in D_+, y \in D_+^*: z = x / y$$

If $$x = 0$$ then $$z = 0$$, and if $$0 < x$$ then as $$z \circ y = x$$ and $$0 < y$$, it follows that $$0 < z$$ by item 1 of Properties of a Totally Ordered Ring.

So $$\forall z \in D: 0 \le z \iff z \in P$$.

Thus it follows that $$z \in D \cap P \Longrightarrow z \in D_+$$, i.e. $$D \cap P \subseteq D_+$$.

Thus $$D_+ = D \cap P$$.

By item 2 of Properties of an Ordered Ring, we have:

$$x \le y \iff 0 \le y + \left({-x}\right)$$

and thus it follows that the ordering induced on $$D$$ by $$\le'$$ is $$\le$$.


 * Now we need to show uniqueness.

Let $$\preceq$$ be a total ordering on $$K$$ which is compatible with its ring structure and induces on $$D$$ the ordering $$\le$$.

Let $$Q = \left\{{z \in K: 0 \preceq z}\right\}$$.

We now show that $$Q = P$$.

Let $$x \in D_+, y \in D_+^*$$.

Then $$0 \preceq x$$ and $$0 \prec 1 / y$$ from item 4 of Properties of a Totally Ordered Ring.

Thus by compatibility with ring structure, $$0 \preceq x / y$$.

Hence $$P \subseteq Q$$.

Conversely, let $$z \in Q$$.

Let $$z = x / y$$ where $$x \in D, y \in D_+^*$$.

Then $$x = z \circ y$$ and by compatibility with ring structure, $$0 \preceq x$$.

Thus $$0 \le x$$ and hence $$z \in P$$, and so $$Q \subseteq P$$.

So $$Q = P$$.

Therefore, by item 2 of Properties of an Ordered Ring, it follows that $$\preceq$$ is the same as $$\le$$.