8
$\begingroup$

We know that $\mathbb{R}$ and $\mathbb{Q}$ have a unique structure as ordered fields with the usual order, and that $\mathbb{C}$ cannot be realised as an ordered field. Various non-trivial subfields of $\mathbb{R}$, such as $\mathbb{Q}(\sqrt 2)$, have multiple orderings, but I can't yet completely rule out the possibility of there being other subfields of $\mathbb{C}$ with unique order structure.

Clearly any such subfield cannot contain $bi$ for any $b\in\mathbb{R}^\times$. How can I determine whether there are any other possibilities than $\mathbb{R}$ and $\mathbb{Q}$?

$\endgroup$
3
  • 2
    $\begingroup$ I think you want to say "naturally" ordered instead of uniquely order, simply because there's not just one complete order. $\endgroup$ Commented May 30, 2021 at 19:05
  • 5
    $\begingroup$ @DavideTrono That's not right - $\mathbb{R}$ and $\mathbb{Q}$ are uniquely orderable fields - remember that the ordering must be compatible with the field structure. (Incidentally, while the unique orderability of $\mathbb{R}$ is easy to show, for $\mathbb{Q}$ it takes some thought.) $\endgroup$ Commented May 30, 2021 at 20:50
  • 2
    $\begingroup$ In $\Bbb Q$, you can define a number as positive if and only if it's a non-zero number that's a sum of four squares, which makes it a first-order property in the language of fields of characteristic zero. $\endgroup$ Commented Jun 5, 2021 at 18:02

2 Answers 2

9
$\begingroup$

What about the field $\mathbb{Q} \left ( \sqrt{2}, \sqrt{\sqrt{2}}, \sqrt{\sqrt{\sqrt{2}}}, \ldots \right )$?

Now every element we've added is a square of something (it's the square of the next number we add), so each element we add must be positive. This eliminates the choice we have in ordering, and so the ordering should be unique.

This is an adaptation of the usual proof that $\mathbb{R}$ has a unique ordering: There we use the fact that every positive number has a square root, while here we just enforce this property for a generating set.

As an aside, we need to do something aggressive like this (adding infinitely many elements), since any $\mathbb{Q}(\alpha)$ admits an ordering for each real root of the minimal polynomial of $\alpha$.

Edit:

I thought harder about my aside: That means $\mathbb{Q} \left ( \sqrt[3]{2} \right )$ will work too, since there's only one way to embed it into $\mathbb{R}$.


I hope this helps ^_^

$\endgroup$
5
  • $\begingroup$ What do you mean by unique ordering? By axiom of choice every set $X$ has an ordering such that $X$ is well ordered, while $\mathbb{R}$ is far from being well ordered with the usual order. $\endgroup$ Commented May 30, 2021 at 19:09
  • 7
    $\begingroup$ @DavideTrono a unique ordering as an ordered field. $\endgroup$ Commented May 30, 2021 at 19:10
  • 5
    $\begingroup$ @DavideTrono -- we're discussing orderings compatible with the field structure. See here for a definition, and here for the "usual proof" I'm referring to. $\endgroup$ Commented May 30, 2021 at 19:12
  • 8
    $\begingroup$ As $\Bbb{Q}(\root3\of2)$ works so do its conjugate fields! They are isomorphic! In the same vein (assuming the axiom of choice), $\Bbb{C}$ has infinitely many subfields isomorphic to $\Bbb{R}$. As an example of a different kind I think that the field $K$ of (compass&ruler) constructible numbers also works. Simply because if $\alpha$ is a positive constructible number, its square root is constructible. Hence $\alpha$ is forced to be positive in any ordering of $K$. $\endgroup$ Commented May 30, 2021 at 19:21
  • $\begingroup$ @JyrkiLahtonen -- Fantastic addition! Iirc you answered another question about ordered rings the other day, right? Nice to see you again ^_^ $\endgroup$ Commented May 30, 2021 at 19:26
9
$\begingroup$

An ordered field satisfying "For every $x$, either $x$ or $-x$ has a square root" is uniquely orderable and this ordering is definable from the algebraic structure alone (set $a\le b$ iff $b-a$ has a square root). Since this property is expressible by a first-order sentence, we have a model-theoretic way to generate uniquely orderable subfields of $\mathbb{R}$, namely elementary extensions: if $A\preccurlyeq\mathbb{R}$ (indeed if $A$ is a $\forall\exists$-elementary substructure of $\mathbb{R}$) then $A$ is uniquely orderable. Combining this observation with the downward Lowenheim-Skolem theorem, we get results like the following:

  • There is a countable uniquely orderable subfield of $\mathbb{R}$ containing $\pi$.

  • Suppose the continuum hypothesis fails. Then there is a uniquely orderable subfield $A$ of $\mathbb{R}$ of cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$.

Of course results like this can be gotten without model theory, but it's neat to observe that model theory provides a simple machine for generating them.


Perhaps a bit more interestingly, there are non-Archimedean uniquely orderable fields of every infinite cardinality (we can construct one by hand, or if we're feeling lazy we can use compactness). Since any field of characteristic $0$ and cardinality $\le 2^{\aleph_0}$ can be embedded into $\mathbb{C}$ (note that the algebraic closure will again have cardinality $\le 2^{\aleph_0}$, and the isomorphism type of an algebraically closed field is determined by its characteristic and transcendence dimension), this means that there are uniquely orderable non-Archimedean subfields of $\mathbb{C}$.

$\endgroup$
1
  • $\begingroup$ Thank you very much for this. I'm a little bit lost with the model theory references as I have no familiarity with this topic whatsoever. $\endgroup$
    – user829347
    Commented May 30, 2021 at 21:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .