Extend order on $\mathbb{Q}$ to $\mathbb Q(X)$

Question

I struggle to show that there exists a unique order on $\mathbb Q(X)$ that extends the order on $\mathbb Q$ such that $\forall q \in \mathbb Q, ~X>q$.

Answer 1

One such order is given by the following:

Given a rational function $f$, we have $f>0$ if there is some rational number $x_0$ such that $f(x)>0$ (as rational numbers) for any $x>x_0$.

(And in general, $f> g$ iff $f-g>0$.) Show that this is indeed an order (using your favorite definition), that it extends the standard order on $\Bbb Q$, and that $X>q$ for any rational constant $q$. That gives you existence.

As for uniqueness, consider the case where there are two distinct such orders $<$ and $\prec$. Since they are distinct, there must be somewhere where they disagree. Specifically, there must be two rational functions $f\neq g$ such that $f<g$ and $g\prec f$. Now use the properties that these two orders have in common (they extend the standard order on $\Bbb Q$, and $X$ is larger than any constant) to reach a contradiction.

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Proof of uniqueness:

Take any ordering $>$ fulfilling our properties. For any rational number $q$ we have $X>q$, giving $1>\frac qX$. Now given a monic polynomial $$ X^n+a_{n-1}X^{n-1}+\cdots+a_0 $$ we have \begin{align*} X & >-a_{n-1}+1+1+\cdots+1\\ & >-a_{n-1}+\frac{-a_{n-2}}{X}+\cdots+\frac{-a_0}{X^{n-1}} \end{align*} Multiplying this by the positive $X^{n-1}$ and rearranging makes our monic polynomial necessarily positive in this ordering.

Since all orderings fulfilling our criteria must agree on the fact that monic polynomials are positive, they must agree on the sign of any non-zero polynomial.

Given two non-zero polynomials $\varphi, \psi$, since all orderings agree on their sign, they must also agree on the sign of $\frac\varphi\psi$, which is to say they must agree on the sign of any rational function.

Finally, for any two distinct rational functions $f,g$, since all orderings must agree on the sign of $f-g$, they must agree on which of $f$ and $g$ is larger. Thus all orderings with these properties must agree on every possible comparison and they must therefore be the same ordering.