I read that the set of rational functions with rational coefficients forms an ordered field, yet it is non-archimedean. I tried googling this, but I don't think I understood the solution.

  1. How does one define an order on rational functions of the form $\mathbb{Q}(x)=p(x)/q(x)$?
  2. How do you show that $\mathbb{Q}(x)$ is non-archimedean? That there is no natural number $n$, such that $n>\mathbb{Q}(x)$? Do I substitute numerical values of $x$ and show that $\mathbb{Q}(x)$ is unbounded or something?
  • $\begingroup$ Aside: "most" orderings on $\mathbb{Q}(x)$ are archimedean. For example, the map sending $x \to \pi$ is an isomorphism $\mathbb{Q}(x) \to \mathbb{Q}(\pi)$. Therefore, we can define $f < g$ iff $f(\pi) < g(\pi)$ to get an archimedean ordering. Any real transcendental can take the place of $\pi$. $\endgroup$
    – user14972
    Sep 9, 2016 at 13:56
  • $\begingroup$ @Hurkyl, I haven't taken a rigorous course in abstract algebra before, so isomorphism is sort of new to me. I will try to understand your comment. :) $\endgroup$
    – Quasar
    Sep 11, 2016 at 3:29

5 Answers 5


Consider the following order: let $f(x)=p(x)/q(x)$ be a rational function with

\begin{aligned}p(x)=a\cdot x^n &+ \text{ terms of degree less than $n$}\\ q(x)=b\cdot x^m &+ \text{ terms of degree less than $m$}\end{aligned}

where of course $a,b \in \mathbb{Q}$. Our order says that $f > 0$ if and only if $\frac{a}{b} >0$. Notice this defines the order throughout the field; if one wishes to determine whether $f_1 > f_2$, write the difference $f_1-f_2$ as a single rational function and determine whether it is $>0$, $=0$ or $<0$.

Now, this totally ordered field is not Archimedean. Indeed, consider the rational functions $f(x)=x$ and $g(x) = 1$. No matter how large you choose $n \in \mathbb{N}$, $f(x)>n\cdot g(x)$, because $\big(f-n\cdot g\big)(x)=x-n$ and the leading coefficient of $\big(f-n\cdot g\big)$ is $1$, which is positive.

  • $\begingroup$ Does the above non-Archimedean ordered field satisfy nested interval property? $\endgroup$
    – MAS
    Sep 25, 2018 at 2:38

The order is "eventual domination": $f(x)\geq g(x)$ iff for all sufficiently large $q\in\mathbb{Q}$, $f(q)\geq g(q)$. It takes a bit of work to show that this really is a total order on rational functions. For a more explicit version of this definition, if $f(x)=\frac{ax^n+\dots}{bx^m+\dots}$ (where the omitted terms have lower degree), then $f(x)\geq 0$ iff $\frac{a}{b}\geq 0$. To determine whether $f(x)\geq g(x)$, you then just write $h(x)=f(x)-g(x)$ in the form $\frac{ax^n+\dots}{bx^m+\dots}$ to determine whether $h(x)\geq 0$.

To see that this order is non-archimedean, just observe that $x>n$ for all $n\in\mathbb{Z}$. Indeed, taking $f(x)=x$ and $g(x)=n$, then $f(q)\geq g(q)$ for all $q\geq n$. In fact, it can be shown that the eventual domination order is the unique ordering on $\mathbb{Q}(x)$ compatible with the field structure for which $x>n$ for all $n\in\mathbb{Z}$. That is, if you declare that $x>n$ for all $n\in\mathbb{Z}$, then the entire rest of the ordering can be deduced from the ordered field axioms. The intuition is that if $x$ is infinitely large, then $f(x)$ behaves like $f(q)$ for very large $q$, so you can determine whether $f(x)\geq g(x)$ by comparing $f(q)$ and $g(q)$ for large $q$.


For polynomials $f(x),g(x)\in\mathbb{Q}[x]$ define $f<g$ if and only if $f\ne g$ and the leading coefficient of $g-f$ is positive.

It's easy to see that this defines a (strict) order relation on $\mathbb{Q}[x]$ compatible with the operations, in the sense that

  1. for every $f\in\mathbb{Q}[x]$ it holds exactly one among $f>0$, $f=0$, or $0>f$;

  2. if $f,g,h\in\mathbb{Q}[x]$ and $f<g$, then $f+h<g+h$

  3. if $f,g,h\in\mathbb{Q}[x]$, $f<g$ and $0<h$, then $fh<gh$.

Now it's easy to see that $0<f$ if and only if $-f<0$, so any element of $\mathbb{Q}(x)$ can be written as a quotient $f(x)/g(x)$ where $0<g$.

Define, for $f_1(x)/g_1(x),f_2(x)/g_2(x)\in\mathbb{Q}(x)$ with $0<g_1$ and $0<g_2$, $$ \frac{f_1(x)}{g_1(x)}<\frac{f_2(x)}{g_2(x)} \quad\text{if and only if}\quad f_1(x)g_2(x)<f_2(x)g_1(x) $$ and prove that this defines a (strict) order relation with the same properties above, so $\mathbb{Q}(x)$ becomes an ordered field.

Now, of $q\in\mathbb{Q}$, it's obvious that $q<x$ and therefore the order on $\mathbb{Q}(x)$ is not Archimedean. Note also that the order induced on $\mathbb{Q}$ is the usual one.


The ordering that I've seen is "$f(x) > 0$ iff $\exists X\in \Bbb R$ s.t. $x>X$ implies $f(x)>0$", in other words, if $f(x) = \frac{p(x)}{q(x)}$, with $p(x) = a_nx^n + \cdots + a_1x + a_0$ and $q(x) = b_mx^m + \cdots + b_1x + b_0$, then $f(x) > 0$ iff $\frac{a_n}{b_m} > 0$.

As usual, this means that $f>g$ iff $f-g > 0$.

It's not archimedean because if $f(x) = x$ and $g(x) = x^2$, then no matter how many times I add together copies of $f$, we will still have that $g$ is greater.

  • 1
    $\begingroup$ That should be $f>0 \iff (f\ne 0 \land \lim_{x\to \infty}\inf_{y>x}f(y)\geq 0)$. For example if $p=id_R$ and $q=1$ and $f=p/q$ then $f>0$ although $\lim_{x\to \infty}f(x)=0.$ $\endgroup$ Sep 9, 2016 at 9:26
  • 1
    $\begingroup$ That previous comment of mine has $p$ and $q$ in the wrong order. It should say $q=id_R$ and $p=1$. $\endgroup$ Sep 9, 2016 at 9:38
  • $\begingroup$ @user254665 You're right, I didn't think that one through enough. $\endgroup$
    – Arthur
    Sep 9, 2016 at 10:04

A comment on rigor.

For polynomials $p,q$ where $q(x)$ is not identically $0,$ the domain of $p/q$ must exclude the finite set of $x$ for which $q(x)=0.$ If these are the only real numbers excluded from dom $(p/q)$ then we must say that $p_1/q_1\ne p_2/q_2$ when $p_1=q_1=id_R$ and $p_2=q_2=1$ because dom$(p_1/q_1)\ne$ dom $(p_2/q_2)$. Which is not what we want to say.

For polynomials $p_1, q_1,p_2,q_2$ with $q_1\ne 0\ne q_2,$ and any finite $S_1,S_2$ such that $q_1^{-1}\{0\}\subset S_1$ and $q_2^{-1}\{0\}\subset S_2,$ let $$(p_1,q_1,S_2)\sim (p_2,q_2,S_2)\iff \forall x\in (R \backslash (S_1\cup S_2)\; (p_1(x)q_2(x)=p_2(x)q_1(x)).$$ Then for polynomials $p,q$ with $q\ne 0$ we write $p/q$ for the set of all $(p_1,q_1,S_1)$ such that $$(p,q,R \backslash q^{-1}\{0\})\sim (p_1,q_1, S_1).$$

  • 2
    $\begingroup$ In abstract algebra, one usually defines polynomials and rational functions in a formal way. In fact, one pointedly avoids a definition in terms of functions and partial functions, since there aren't enough of them. e.g. there are only 9 partial functions from $\mathbf{F}_2$ to itself, but $\mathbf{F}_2[x]$ has countably infinitely many elements. $\endgroup$
    – user14972
    Sep 9, 2016 at 14:10
  • $\begingroup$ @Hurkyl. Yes of course. But for polynomials on an infinite field of characteristic 0, like Q or R, we can safely ignore this distinction. $\endgroup$ Sep 9, 2016 at 16:23

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