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Consider the field $\mathbb{Q}[\sqrt{5}]$. This can be made into an ordered field in two different ways:

  1. Via the usual order $<$ inherited from $\mathbb{R}$
  2. Or via the "alternative" order $\prec$ defined by $$r\prec s \iff \overline{r} < \overline{s} $$ where $\overline{r}$ denotes the conjugate, i.e. $\overline{a+b\sqrt{5}}=a-b\sqrt{5}$.

More broadly, if $F$ is any field, $\sigma$ an automorphism of $F$, and $<$ an order on $F$, we can define another order $\prec$ by $a\prec b \iff \sigma(a) < \sigma(b)$. If $(F,<)$ satisfies the axioms for an ordered field, than so does $(F,\prec)$.

My question:

Is there an example of a field $F$ that can be made into an ordered field in two different ways, where one of the orders is not induced from the other one by an automorphism as described above?

In other words, I am looking for a field that can be made into an ordered field in two different ways that are "really different", i.e. not equivalent up to an automorphism of the underlying field $F$.

Eric Wofsey
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mweiss
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4 Answers4

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Consider $F=\mathbb{Q}(x)$. You can order this field in uncountably many different ways. For instance, for every transcendental $\alpha\in\mathbb{R}$, the isomorphism $F\to\mathbb{Q}(\alpha)\subset\mathbb{R}$ sending $x$ to $\alpha$ induces an ordering of $F$, and this ordering is different for each $\alpha$. Since $F$ has only countably many automorphisms (an automorphism is determined by where it sends $x$), these give uncountably many orderings that are not related by automorphisms.

Eric Wofsey
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  • Very nice, thank you! – mweiss Dec 23 '16 at 04:59
  • Is the reason for requiring $\alpha$ to be transcendental that otherwise the isomorphism would result in a division by zero? – kasperd Dec 24 '16 at 11:22
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    If $\alpha$ is not transcendental then there's some polynomial of $\alpha$ which is zero. That polynomial of $x$ would then map to $0$, so the map would not be injective. – Ben Millwood Dec 24 '16 at 11:47
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Here's a smaller example: let $f$ be an irreducible quartic polynomial with four real roots. If $\alpha$ is any root of $f$, then $\mathbb{Q}(\alpha)$ can be ordered in four different ways, corresponding to the four real embeddings of $\alpha$

However, most such polynomials $f$ will have Galois group $A_4$, and $\mathbb{Q}(\alpha)$ will have no automorphisms at all.

  • It would be nice to have an example here. Maybe $f(x) = x^4 - 2$ and $\alpha = \sqrt[4]{2}$. – Caleb Stanford Dec 24 '16 at 15:25
  • @6005 (three years late, oh well) $x^4-2$ has only two real roots, and they are actually related by an automorphism, which can be seen for instance from the Galois extension $\mathbb Q(\sqrt[4]{2})/\mathbb Q(\sqrt{2})$ – Wojowu Sep 11 '19 at 10:12
  • @Wojowu Ah, right, so in this example, even though there are four possible embeddings of $\mathbb{Q}(\alpha)$ in $\mathbb{C}$, there are only two possible embeddings in $\mathbb{R}$. So we would need a different example. – Caleb Stanford Sep 11 '19 at 16:54
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After reading the excellent answers by Eric Worfsey and Hurkyl, I realized that I already knew another answer to my own question -- somehow it had escaped me when I posted the question, but I think it is worth posting the answer here.

Let $F=\mathbb{Q}(x)$. Any element of $F$ can be written in the form $$f=\frac{a_n x^n + \cdots +a_0}{b_m x^m + \cdots +b_0}$$ with $a_n, b_m \ne 0$ and where all coefficients are integers. Let $P \subset F$ be defined by the condition $f\in P \iff \frac{a_n}{b_m}> 0$, and define $f \prec g \iff g-f \in P$. Then $(F, \prec)$ is an ordered field.

Now contrast this order with the order on $F$ proposed by Eric Worfsey: in that order, one simply chooses a transcendental real number $\alpha$, maps $x \mapsto \alpha$, and uses the order $<$ inherited from the reals.

These two orders are "really different". The order $<$ obtained by mapping $x \mapsto \alpha$ embeds $\mathbb{Q}(x)$ as an ordered subfield of $\mathbb{R}$, but with respect to the order $\prec$ the field $\mathbb{Q}(x)$ is non-archimedean. Specifically, for all $q \in \mathbb{Q}^+$, we have $q \prec x$ and $0 \prec \frac{1}{x} \prec q$; in other words, $x$ is "infinite" and $\frac{1}{x}$ is "infinitesimal" with respect to the rationals in this order.

Community
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mweiss
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edit: This answer is incorrect, but I'm leaving it in place temporarily in case it can be fixed.

While the other answers do provide orderings that cannot be transformed into each other by field automorphisms, one might also ask whether the orderings could be transformed into each other by any order isomorphism, field-preserving or not.

Now, any ordered field is a dense linear order without endpoints. A countable dense linear order without endpoints is isomorphic to $\mathbb Q$. So (at the time of writing) all the existing answers, being countable, are order-isomorphic. (Note in particular that the Archimedean property depends on the field structure as well as the order structure, so need not be preserved by general order isomorphisms.)

That's enough to satisfy the original question, but can we have two orders on the same field with different order types?

We can answer this question with a variation of the $\mathbb Q(\alpha)$ example. We can choose transcendental numbers $\alpha_i$ such that $\mathbb R = \mathbb Q(\alpha_i : i \in I)$, where $|I|$ is the transcendence degree of $\mathbb R$ over $\mathbb Q$.

We can then order $\mathbb Q(x_i : i \in I)$, polynomials over $\mathbb Q$ in $|I|$-many variables, by mapping $x_i$ to $\alpha_i$ for each $i$. The result will be order-isomorphic to $\mathbb R$, of course.

But we can also map $x_i$ to $\alpha_{f(i)}$, where $f : I \to I$ is some injective, non-surjective function. Then the result will be order-isomorphic to $\mathbb R$ with some "holes", with some transcendental numbers missing. Importantly, and unlike the countable case, the ordering can "detect" this removal, because $\mathbb R$-with-holes is not complete: there are sets bounded above but with no supremum. (In the countable case there were so many holes already that a couple of extra ones wouldn't make any difference.)

So, polynomials over $\mathbb Q$ with sufficiently many unknowns can actually be ordered in ways not only such that field automorphisms can't transform one ordering into the other, but the orderings actually have different order types altogether.

Ben Millwood
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  • Your example doesn't work because $\mathbb{R}$ is not purely transcendental over $\mathbb{Q}$. In fact, $\mathbb{R}$ has only one ordering compatible with the field structure (since every positive element is a square). – Eric Wofsey Dec 24 '16 at 19:26
  • Hm, you're right :/ I'll leave the answer up for now, but if neither of us can think of an easy fix, I'll delete it tomorrow (and maybe ask it as a new question instead) – Ben Millwood Dec 24 '16 at 19:30
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    You can get an example that works by picking a set $S$ of uncountably many algebraically independent elements of $\mathbb{R}$, and ordering $\mathbb{Q}(S)$ either with the usual ordering, or with a well-ordering of $S$ (extending the order to $\mathbb{Q}(S)$ by saying that if $s,t\in S$ and $s<t$, then $t$ is "infinitely larger" than $s$). These won't be order-isomorphic because there is no uncountable well-ordered subset of $\mathbb{R}$. – Eric Wofsey Dec 24 '16 at 19:32