Order Embedding between Quotient Fields is Unique

Theorem
Let $\left({R_1, +_1, \circ_1, \le_1}\right)$ and $\left({S, +_2, \circ_2, \le_2}\right)$ be totally ordered integral domains.

Let $K, L$ be totally ordered quotient fields of $\left({R_1, +_1, \circ_1, \le_1}\right)$ and $\left({S, +_2, \circ_2, \le_2}\right)$ respectively.

Let $\phi: R \to S$ be a order monomorphism.

Then there is one and only one order monomorphism $\psi: K \to L$ extending $\phi$.

Also:


 * $\displaystyle \forall x \in R, y \in R^*: \psi \left({\frac x y}\right) = \frac {\phi \left({x}\right)} {\phi \left({y}\right)}$

If $\phi: R \to S$ is an order isomorphism, then so is $\psi$.

Proof
By Quotient Field is Unique, all we need to show is:


 * $\displaystyle \forall x_1, x_2 \in R, y_1, y_2 \in R_+^*: \frac {x_1} {y_1} \le \frac {x_2} {y_2} \iff \frac {\phi \left({x_1}\right)} {\phi \left({y_1}\right)} \le \frac {\phi \left({x_2}\right)} {\phi \left({y_2}\right)}$


 * Let $x_1 / y_1 \le x_2 / y_2$, where $y_1, y_2 \in R_+^*$.

As $y_1, y_2 \in R_+^*$, it follows that $0 < y_1 \circ_1 y_2$ and $0 < 1 / \left({y_1 \circ_1 y_2}\right)$.

We also have:
 * $0 < \phi \left({y_1}\right) \circ_2 \phi \left({y_2}\right) = \phi \left({y_1 \circ_1 y_2}\right)$

Therefore:


 * $\displaystyle x_1 \circ_1 y_2 = \frac {x_1} {y_1} \circ_1 \left({y_1 \circ y_2}\right) \le \frac {x_2} {y_2} \circ_1 \left({y_1 \circ_1 y_2}\right) = x_2 \circ_1 y_1$

Conversely, let $x_1 \circ_1 y_2 \le x_2 \circ_1 y_1$. Then:


 * $\displaystyle \frac {x_1} {y_1} = x_1 \circ_1 y_2 \circ_1 \left({\frac 1 {y_1 \circ_1 y_2}}\right) \le x_2 \circ_1 y_1 \circ_1 \left({\frac 1 {y_1 \circ_1 y_2}}\right) = \frac {x_2} {y_2}$

That is, we have:


 * $\displaystyle \frac {x_1} {y_1} \le \frac {x_2} {y_2} \iff x_1 \circ_1 y_2 \le x_2 \circ_1 y_1$

Similarly:


 * $\displaystyle \frac {\phi \left({x_1}\right)} {\phi \left({y_1}\right)} \le \frac {\phi \left({x_2}\right)} {\phi \left({y_2}\right)} \iff \phi \left({x_1}\right) \circ_2 \phi \left({y_2}\right) \le \phi \left({x_2}\right) \circ_2 \phi \left({y_1}\right)$

Now $\phi: R \to S$ is an order monomorphism.

Therefore:
 * $x_1 \circ_1 y_2 \le x_2 \circ_1 y_1 \iff \phi \left({x_1 \circ_1 y_2}\right) \le \phi \left({x_2 \circ_1 y_1}\right)$

The result follows.