Order Embedding between Quotient Fields is Unique

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

Let $K, L$ be totally ordered fields of quotients of $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ respectively.

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

Then there exists exactly one order embedding $\psi: K \to L$ extending $\phi$.

Also:


 * $\forall x \in R, y \in R_{\ne 0}: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$

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

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


 * $\forall x_1, x_2 \in R, y_1, y_2 \in R_{> 0}: \dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff \dfrac {\map \phi {x_1}} {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} }$

Let $x_1 / y_1 \le x_2 / y_2$, where $y_1, y_2 \in R_{> 0}$.

As $y_1, y_2 \in R_{> 0}$, it follows that $0 < y_1 \circ_1 y_2$ and $0 < 1 / \paren {y_1 \circ_1 y_2}$.

We also have:
 * $0 < \map \phi {y_1} \circ_2 \map \phi {y_2} = \map \phi {y_1 \circ_1 y_2}$

Therefore:


 * $x_1 \circ_1 y_2 = \frac {x_1} {y_1} \circ_1 \paren {y_1 \circ y_2} \le \dfrac {x_2} {y_2} \circ_1 \paren {y_1 \circ_1 y_2} = x_2 \circ_1 y_1$

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

Then:


 * $\dfrac {x_1} {y_1} = x_1 \circ_1 y_2 \circ_1 \paren {\dfrac 1 {y_1 \circ_1 y_2} } \le x_2 \circ_1 y_1 \circ_1 \paren {\dfrac 1 {y_1 \circ_1 y_2} } = \dfrac {x_2} {y_2}$

That is, we have:


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

Similarly:


 * $\dfrac {\map \phi {x_1} } {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} } \iff \map \phi {x_1} \circ_2 \map \phi {y_2} \le \map \phi {x_2} \circ_2 \map \phi {y_1}$

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

Therefore:
 * $x_1 \circ_1 y_2 \le x_2 \circ_1 y_1 \iff \map \phi {x_1 \circ_1 y_2} \le \map \phi {x_2 \circ_1 y_1}$

The result follows.