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In terms of the algebraic properties of $\mathbb{C}$, what (if anything) is so special about $\mathbb{R}$ as a subfield?

$\mathbb{R}$ is the subfield of $\mathbb{C}$ consisting of all 'conjugation-invariant' elements, yet complex conjugation is not a natural choice of field automorphism (although it is the unique non-trivial topological field automorphism of $\mathbb{C}$ with the Euclidean topology).

Then there is the order structure: $\mathbb{R}$ is maximal as an orderable/formally real subfield of $\mathbb{C}$. Yet as has been helpfully pointed out in response to my recent question regarding uniquely orderable subfields of $\mathbb{C}$, there are many subfields of $\mathbb{C}$ that are isomorphic to $\mathbb{R}$ that would therefore have an identical order structure.

Is there any intrinsic reason that $\mathbb{R}$ is, algebraically, a 'special' subfield of $\mathbb{C}$ or is it more to do with the historical development of the subject?

Any thoughts on this are most welcome.

user829347
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    There seem to be two distinct questions here: how to define $\mathbb{C}$ without going through $\mathbb{R}$, and how to best characterize $\mathbb{R}$ within $\mathbb{C}$. Which one do you care more about? Re: the first, note that if we just care about the algebraic structure then $\mathbb{C}$ is up to isomorphism the unique algebraically closed field of characteristic $0$ and transcendence dimension $2^{\aleph_0}$, and this makes no reference to metric/topological concepts or to the reals. – Noah Schweber Jun 05 '21 at 17:50
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    If you just know $\mathbb{C}$ as a field (with no additional structure), then you cannot define the subfield $\mathbb{R}$ of it, since $\mathbb{C}$ has field automorphisms that do not map $\mathbb{R}$ to itself. – Eric Wofsey Jun 05 '21 at 17:53
  • I'd say I'm more concerned with the latter out of the two although they seem very closely related to me. For future reference, would it be better to separate such questions out into different posts rather than both in one? – user829347 Jun 05 '21 at 17:53
  • @thewonderfulwizardofoz Yes, it is. (And certainly don't make the question you care less about the title question.) – Noah Schweber Jun 05 '21 at 17:54
  • @Noah Good point. I've changed the title accordingly. – user829347 Jun 05 '21 at 18:01
  • @thewonderfulwizardofoz The first two paragraphs are still rather misdirecting. I would just remove them since they don't seem to relate to what you're really asking, at least as far as I can tell. – Noah Schweber Jun 05 '21 at 18:03
  • @Noah Schweber My main reason for linking the two questions is that if $\mathbb{C}$ can be defined without first defining $\mathbb{R}$ then that would be a good indicator that $\mathbb{R}$ serves no 'special' role as a subfield. If a prior definition of $\mathbb{R}$ is required, then that might provide some insight into why it could be in some way unique as a subfield. – user829347 Jun 05 '21 at 18:13

2 Answers2

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Since you ignore the topological structure on $\mathbb{C}$, you need to realize that - as an abstract field - $\mathbb{C}$ is isomorphic to every other algebraically closed field of characteristic $0$ which has cardinality $c$ (the continuum cardinal). In particular, there is an isomorphism to $\overline{\mathbb{C}(x)}$, but also to $\mathbb{C}_p$ (see here) for example. Inside these abstract fields you do not have any natural copy of $\mathbb{R}$ over which the field has degree $2$.

The answer is therefore "no - when you ignore the topology".

Martin Brandenburg
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    Hmm, it still feels like it might be possible to define $\mathbb R$ up to isomorphism. For example, I would expect that every maximal real closed subfield of $\mathbb C$ is the image of $\mathbb R$ under some isomorphism of $\mathbb C$. – Troposphere Jun 05 '21 at 18:16
  • One thing I was wondering recently is if every maximal formally real/orderable subfield of $\mathbb{C}$ is isomorphic to $\mathbb{R}$, in which case would the isomorphism class of maximal formally real subfields of $\mathbb{C}$ not be a 'natural' choice? – user829347 Jun 05 '21 at 18:21
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    @Troposphere: That's not true, there are non-archimedean maximal real closed subfields of $\mathbb{C}$. (And also archimedean maximal real closed subfields that are not complete.) You could talk about maximal real closed subfields of $\mathbb{C}$ that are complete but then I don't know why you aren't just defining $\mathbb{R}$ separately from $\mathbb{C}$. – Eric Wofsey Jun 05 '21 at 18:22
  • Can you put your general discussion perhaps under the question? It doesn't seem to address my answer. Actually I am about to delete it since I think that the OP wants something else. – Martin Brandenburg Jun 05 '21 at 18:25
  • @EricWofsey: Hmm, right. – Troposphere Jun 05 '21 at 18:25
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Expanding on Martin Brandenburg's answer and Eric Wofsey's commands to same:

Consider $\mathbb C$ as a structure over the signature $(0,1,{+},{\times},R)$ where $R$ is a unary predicate with the intuitive meaning "is a real". Since the positive reals are exactly the squares of nonzero reals, this makes the ordering of reals first-order definable in this structure. Complex conjugation and therefore $z\mapsto |z|$ is also first-order definable.

Using standard tools from model theory (compactness and the Löwenheim-Skolem theorem) we can find an extension of $(\mathbb C,0,1,{+},{\times},\mathbb R)$ which is elementarily equivalent to it (meaning it has the same first-order properties), still has cardinality $2^{\aleph_0}$, but contains a real that is larger than $n$ ($=1+1+\cdots+1$) for every natural number $n$. The "reals" in this structure are still a field, but a non-archimedean one, and therefore manifestly not isomorphic to our usual $\mathbb R$.

However, our extension is isomorphic to $\mathbb C$ as a field -- that is, ignoring the $R$ predicate in the signature -- because it is algebraically closed (a collection of first-order properties that it inherits from $\mathbb C$ due to being an elementary extension), and two algebraically closed fields of the same characteristic and cardinality are necessarily isomorphic.

On the other hand, the non-standard reals of the extended model sit inside their "$\mathbb C$" in the same way as $\mathbb R$ sits inside the actual $\mathbb C$ -- at least as long as "the same way" means according to any first-order description we can possibly write down.

So it's not possible to characterize $\mathbb R$ even up to isomorphism by a first-order statement of how it behaves as a subset of $\mathbb C$.

Can we get around this by using a higher-order description, such as one that quantifies over subfields, symmetries, subgroups of the symmetry group, or generally sets of sets of sets of subsets of the field, etcetera? It seems that no matter what we do, if it is to retain any arguable "algebraic" flavor, it would be something we can express in a first-order form if we first add finite number of new sorts and symbols to our signature to denote subsets, etc. And then the same construction as above will make us a non-archimedean subfield of $\mathbb C$ that the proposed criterion cannot distinguish from $\mathbb R$. (Hmm, this won't work -- there are so many subfields of $\mathbb C$ that if we want to quantify over them, Löwenheim-Skolem will not be able to guarantee that the "numbers" in the extended structure are few enough that they can be isomorphic to our original $\mathbb C$).

You might conclude from this that "$\mathbb R$ is not all that special, once we know $\mathbb C$". However, this feels rather backwards to me.

In most cases where we use $\mathbb C$ in particular (rather than immediately generalize to a wider class of fields), the reason we care about $\mathbb C$ in the first place is that it is the algebraic closure of the $\mathbb R$ we already know and love.

So it's hard to see in which context one would start from $\mathbb C$ and then look for the reals.

Troposphere
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