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For any field $K$ of characteristic zero, its prime subfield $Q(K)$ has a natural ordering inherited from $\mathbb{Q}$. I am interested in finding out to what extent (if at all) this natural ordering can be extended to larger subfields of $K$, in a natural way. (Of course, many constraints on $K$ may be required for this to be natural).

My motivating example is $\mathbb{C}$, and I want to generalise the role that $\mathbb{R}$ plays as the 'natural' choice of ordered subfield of $\mathbb{C}$. One can check that $\mathbb{R}$ has a unique structure as an ordered field and that $\mathbb{C}$ cannot be realised as an ordered field. Since there are no fields strictly between $\mathbb{R}$ and $\mathbb{C}$, we have that $\mathbb{R}$ is maximal as an ordered subfield of $\mathbb{C}$. Note that the ordering of proper subfields of $\mathbb{R}$ is not necessarily unique.

There are certainly maximal ordered subfields for any $K$ (Zorn's Lemma), but no guarantee of uniqueness. Any ideas?

user829347
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    Are you sure you can apply Zorn so easily in the last paragraph? Why should the union of a chain of orderable fields be orderable? (I do believe the result is right, but I think it takes more argument.) – Noah Schweber May 30 '21 at 21:10
  • I used Zorn's Lemma to prove the existence of maximal ordered subfields i.e. with a specified ordering on each. I'm not sure if it works for orderable subfields yet – user829347 May 30 '21 at 21:30
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    Zorn gives a field $A\subseteq K$ and an ordering $\trianglelefteq$ on $A$ such that there is no proper field extension $A\subsetneq A'\subseteq K$ and ordering $\trianglelefteq'$ on $A'$ such that $\trianglelefteq'\upharpoonright A=\trianglelefteq$. But there might be a different ordering on $A$ which doesn't have this maximality property. So "maximal ordered subfield" is a bit of a slippery notion. – Noah Schweber May 30 '21 at 21:33
  • @Noah I agree that it is a bit of a slippery notion and I guess the main thing I am trying to determine is whether any kind of 'natural' construction of this sort can be salvaged so I don't have to simply have to 'choose' an ordered subfield in my work. I'm beginning to doubt that this will be possible. – user829347 May 30 '21 at 21:40
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    It may help if you say a bit about the nature of your work. – Noah Schweber May 30 '21 at 22:43
  • I'm trying to generalise the notion of an 'absolute value' on a field $K$, which is real valued, to one that takes values in some ordered subfield of $K$. Right now I am just simply choosing a specific ordered subfield $F(K)$ and defining the norm as taking values in $F(K)$. – user829347 May 30 '21 at 23:17
  • So ideally I'd like to have some more meaningful/natural choice of $F(K)$ to work with to improve on the work I've already done. Ultimately I'm trying to find a minimal set of conditions (both topological and algebraic) to ensure that such a 'normed field' $K$ is isomorphic to either $\mathbb{C}$ or $\mathbb{C}_p$, the complex $p$-adic numbers. – user829347 May 30 '21 at 23:21

1 Answers1

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There are automorphisms of $\mathbb{C}$ which do not send reals to reals; consequently, there are subfields of $\mathbb{C}$ which are isomorphic to $\mathbb{R}$ (hence are maximal orderable subfields of $\mathbb{C}$) but are not equal to $\mathbb{R}$.

In fact, I believe that there are in fact maximal uniquely orderable subfields of $\mathbb{C}$ which are non-Archimedean according to their unique order, hence a fortiori not isomorphic to $\mathbb{R}$ (let alone equal to $\mathbb{R}$); however, I don't have a complete argument for this at the moment. Below is my incomplete argument - the gap is at the end:

First, we isolate a strengthening of unique orderability: say that a field $F$ is nice iff $F$ has characteristic $0$ and for each nonzero $x\in F$ exactly one of $x$ and $-x$ has a square root in $F$. Every nice field is uniquely orderable, with unique ordering given by $$a\le b\leftrightarrow \exists x(x^2+a=b).$$

Now it's not hard to show that there is a countable nice field which is non-Archimedean according to its unique ordering. For example, we can prove this via compactness + downward Lowenheim-Skolem after observing that niceness is given by a conjunction of first-order sentences. Fix such a field $F$.

The algebraic closure $\overline{F}$ of $F$ is again countable. This means that $\overline{F}$, and hence $F$ itself, embeds into $\mathbb{C}$: for algebraically closed fields $A,B$ of the same characteristic, there is an embedding of $A$ into $B$ iff the transcendence dimension of $A$ is $\le$ that of $B$. (To prove this, just pick transcendence bases of $A$ and $B$.) So WLOG we can assume that $F$ is literally a subfield of $\mathbb{C}$.

A natural idea at this point is to apply Zorn's lemma to the poset of uniquely orderable subfields of $\mathbb{C}$ which contain $F$. However, it's not obvious to me that this poset satisfies the hypothesis of Zorn's lemma: why should the union of a chain of uniquely orderable fields be uniquely orderable?

Instead, we turn to niceness again. Let $\mathbb{P}$ be the poset of nice subfields of $\mathbb{C}$ which contain $F$ as a subfield. By Zorn, $\mathbb{P}$ has a maximal element $G$. However, we're not quite done: while $G$ is a maximal nice subfield of $\mathbb{C}$, it's not immediately obvious that $G$ is a maximal uniquely orderable subfield of $\mathbb{C}$. Keep in mind that there are uniquely orderable subfields of $\mathbb{C}$ which are not nice (e.g. $\mathbb{Q}$)!

It's at this point that I get stuck: basically, I need to be able to add square roots to an orderable subfield of $\mathbb{C}$ in a way which results in a still-orderable subfield of $\mathbb{C}$, and I don't immediately see how to do this. However, I also don't see an obstacle to this, so I tentatively suspect that this gap can be closed.

Noah Schweber
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  • A sub field of $mathbb C$ isomorphic to $mathbb R$! Is it possible to have an example? – Lawrence Mano May 30 '21 at 21:38
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    @LawrenceMano ... $\mathbb{R}$ itself? – Noah Schweber May 30 '21 at 22:08
  • Forgot to add 'not equal to $mathbb R$' as stated by you. – Lawrence Mano May 30 '21 at 22:28
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    @LawrenceMano Fix a transcendence basis $T$ for $\mathbb{C}$ with $\pi\in T$, take some non-real element $\alpha$ of $T$, and consider the image of $\mathbb{R}$ under the automorphism of $\mathbb{C}$ generated by swapping $\alpha$ and $\pi$ and leaving all other elements of $T$ fixed. (Of course given the unavoidable complexity of transcendence bases this is a bit of an odd notion of "having an example" - offhand I'm not sure whether $\mathbb{C}$ has any "nice" subfields isomorphic to $\mathbb{R}$ other than $\mathbb{R}$ itself.) – Noah Schweber May 30 '21 at 22:30
  • To get an example, take a non-archimedean real-closed field of cardinality $2^{\aleph_0}$. Then its algebraic closure is isomorphic to $\mathbb{C}$. – Eric Wofsey Jun 05 '21 at 18:06
  • @EricWofsey Can the gap at the end of my answer be fixed? (I suspect it can and I'm missing something very simple.) – Noah Schweber Jun 15 '21 at 18:23
  • Yes, if $K$ is an ordered field and $a\in K$ is a positive element then $K[\sqrt{a}]$ has a canonical ordering given by saying $b\sqrt{a}\geq c$ (for positive $b,c\in K$) iff $b^2a\geq c^2$. (This is enough to order all of $K[\sqrt{a}]$ because every element has the form $b\sqrt{a}+c$ for $b,c\in K$.) – Eric Wofsey Jun 15 '21 at 18:54
  • Or more simply, the real closure of $K$ is an ordered extension that has all square roots of positive elements. (But the previous comment is a lemma along the way to proving the existence/basic properties of real closures, so this isn't really a different argument.) – Eric Wofsey Jun 15 '21 at 18:58
  • Also, it is true that the union of a chain of uniquely orderable fields is uniquely orderable. This is because a field $K$ is orderable iff $-1$ is not a sum of squares, and then for any $a\in K$ there is an ordering where $a>0$ iff $-a$ is not a sum of squares. So, $K$ is uniquely orderable iff for each nonzero $a\in K$, exactly one of $a$ and $-a$ is a sum of squares. – Eric Wofsey Jun 15 '21 at 19:01
  • Switching to formally real fields (instead of ordered ones because then there is no need to specify ordering) we can find a maximal formally real field $F$ with $\mathbb {Q} \subset F\subset \mathbb {C} $ and $F(i) =\mathbb{C} $ (proved here ). If such an $F$ is not unique then we have multiple proper subfields $F$ with $F(i) =\mathbb {C} $. This implies existence of distinct automorphisms of $\mathbb {C} $ which send $i$ to $-i$. Is this really possible? – Paramanand Singh Sep 14 '21 at 17:41
  • @ParamanandSingh Yes it is. In fact there are $2^{2^{\aleph_0}}$-many (or $\beth_2$ more snappily) distinct automorphisms of $\mathbb{C}$ swapping $i$ and $-i$ (think about picking a transcendence basis). – Noah Schweber Sep 14 '21 at 17:44
  • @NoahSchweber: Well I can see the argument in your answer using transcendence basis and understand it at some level. But i find the conclusions really surprising and hard to believe. Just one final question : If we add the archimedean property as well, does that lead to a unique $F$ namely $\mathbb {R} $? – Paramanand Singh Sep 14 '21 at 17:47
  • @ParamanandSingh Up to isomorphism, sure, but that's not surprising since all Archimedean ordered fields embed uniquely into $\mathbb{R}$ in the first place. And if we don't allow an isomorphism, then the answer is negative: consider the subfield $f[\mathbb{R}]$ for some $f\in Aut(\mathbb{C})$ satisfying (for example) $f(\pi)\not\in\mathbb{R}$. – Noah Schweber Sep 14 '21 at 17:48
  • Thanks a lot for your responses. Although I still find these automorphisms psychologically disturbing. Maybe it will take time for me to get used to it. And +1 – Paramanand Singh Sep 14 '21 at 17:53
  • Our standard intuition for $\mathbb{C}$ is both algebraic and geometric. Now in the case of $\mathbb{R}$ that's fine, since the geometry can be reconstructed from the algebra to a large extent (specifically, the topology is definable from the ordering, and the ordering is definable from the algebra via $x\le y\iff\exists z(x+z^2=y)$). However, despite being a bigger structure than $\mathbb{R}$, the field of complex numbers is actually significantly less structurally complicated in many ways (model theory comes in handy here). So we may need to decouple algebraic and geometric intuitions. – Noah Schweber Sep 14 '21 at 17:57
  • That said, it's worth noting that $(i)$ we need the axiom of choice to prove that $\mathbb{C}$ has automorphisms (besides the identity and conjugation) and $(ii)$ even assuming choice we can prove that there are no "geometrically nice" (e.g. Borel) interesting automorphisms. So in some sense the algebraic structure of $\mathbb{C}$ almost lets us figure out some geometry ... but not quite, and certainly not in anything like a useful way. – Noah Schweber Sep 14 '21 at 17:58