# Are there as many real-closed fields of a given cardinality as I think there are?

Let $\kappa$ be an infinite cardinal. Then there exists at least one real-closed field of cardinality $\kappa$ (e.g. Lowenheim-Skolem; or, start with a function field over $\mathbb{Q}$ in $\kappa$ indeterminates, choose an ordering and a real-closure).

But I think there are many more, namely $2^{\kappa}$ pairwise nonisomorphic real-closed fields of cardinality $\kappa$. This is equal to the number of binary operations on a set of infinite cardinality $\kappa$, so is the largest conceivable number.

As for motivation -- what can I tell you, mathematical curiosity is a powerful thing. One application of this which I find interesting is that there would then be $2^{2^{\aleph_0}}$ conjugacy classes of order $2$ subgroups of the automorphism group of the field $\mathbb{C}$.

Addendum: Bonus points (so to speak) if you can give a general model-theoretic criterion for a theory to have the largest possible number of models which yields this result as a special case.

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Hi Pete!

There's been a lot of study of this and similar problems. I believe that Shelah's theorem, from his 1971 paper "The number of non-isomorphic models of an unstable first-order theory" (Israel J. of Math) answers your question about real closed fields in the positive.

The best big result on such questions that I know of is in the 2000 Annals paper "The uncountable spectra of countable theories." by Hart, Hrushovski, Laskowski.

To answer the question on real closed fields specifically (and somewhat cautiously since I'm not a model-theorist):

The theory of real closed fields is a complete first order theory, with countable language. It is an unstable (an easy fact, I think, and explained better on wikipedia than I could explain) theory as well. Hence Shelah's result applies, and the bound $2^\kappa$ is realized as you surmised.

Bonus points should go to Shelah (and perhaps also to Hart, Hrushovski, Laskowski, whose paper mentions the result of Shelah and proves other things) for proving that this bound is realized (for uncountable cardinals), except for theories $T$ which have all of the following properties:

1. $T$ has infinite models.
2. $T$ is superstable.
3. $T$ has prime models over pairs.
4. $T$ does not have the dimensional order property.

I have no clue what the fourth property means. But there are plenty of non-superstable theories to which Shelah's theorem applies, and hence which realize your bound (for uncountable cardinals).

For countable cardinality, I think there are still some open problems about how many non-isomorphic models there can be of a given theory, with cardinality $\aleph_0$.

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Hi, Marty. Funny that the guy who introduced me to model theory in the first place answers my question. I think I must have been dimly aware of some result like this, or I wouldn't have asked. Thanks! – Pete L. Clark Jan 26 '10 at 0:47
Hi Pete. Thought that your question was a good way to start my Math Overflow contributions. (Your question about transcendental Galois theory is still on my mind too) – Marty Jan 27 '10 at 1:12
+1: As a model theorist who is familiar with Shelah's work in this area, I can assure you that this is correct: Shelah's result gives you that this theory has the maximal number of models in any uncountable cardinality. This is simply because the theory is unstable, since any model is infinite and has a definable strict linear ordering on the universe (unstability is actually a somewhat more general condition than this). Shelah's theorem cannot apply to the countable models -- consider the theory of the linear ordering on the rational numbers, which is aleph-0-categorical, yet unstable. – John Goodrick Jan 27 '10 at 2:59
... also, Shelah's theorem is only about theories in a countable language (which is no problem for this example, of course). Theories in uncountable languages are still mysterious to us. – John Goodrick Jan 27 '10 at 3:00
As for the number of countable models, there is really only one open question left (at least for countable theories), which is Vaught's Conjecture: if there are uncountably many countable models, then there are 2^(aleph_0) countable models (i.e. the maximum number). This is still wide open. – John Goodrick Jan 27 '10 at 3:04

For real closed fields this is fairly easy.

First show that for any infinite cardinal k there are 2^k nonisomorphic linear orders of cardinality k

For example if X is a subset of k let A_x be Q+2+Q if x is in k and Q+3+Q if x is not in X. Let L_X be the sum of the A_x for x in k. It is easy to see that L_X is isomorphic to L_Y if and only if X=Y.

If F is a real closed and x and y are infinite element of R we say that x and y are comparable if and only if there are natural numbers m and n such that x is less than y^m and y is less than x^n. The ordering of R induces a linear order L_R of the comparability classes, which we call the ladder of R.

Suppose L is a linear order. Let F be the real algebraic numbers. Let R_L be the real closure of the transcendental extension of the real algebraic numbers F(x_l:l\in L) ordered such that if i is less than j then x_i^n is less than x_j for all n. It's not hard to show that the ladder of R_L is isomorphic to L.

Thus if we start with nonisomorphic orders A and B then the fields R_A and R_B will be nonisomorphic.

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Dave, welcome to MO! You can use LaTeX as you normally would. Be careful with '>' and '<', which can get confused with html tags. – François G. Dorais May 5 '10 at 2:08
Welcome to MO, Dave! – Joel David Hamkins May 5 '10 at 2:16
Bienvenue, Professor Marker. I learned model theory (such as I know it) from your text. – Pete L. Clark May 5 '10 at 2:27

In the countable case, the bound of 2ω is realized, since any countable real-closed field will contain the rational numbers and fill at most countably many cuts in the rationals with LUBs. But we can arrange that any given cut is filled by a real closed subfield of R containing that real. So there must be 2ω many non-isomorphic countaable real closed fields.

In the general case, because the models have an order, you can easily make this order have different cofinalities, by building elementary chains of different lengths. That is, just use your Lowenheim Skolem construction to add another point on top of the previous model, and continue for δ steps. This will produce an elementary extension of size κ whose order has cofinality δ, for any regular δ up to κ. So this gives many more models, but doesn't quite answer your 2κ question. I'm inclined to agree with you and expect that it must be the maximal number for all κ on general grounds.

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Although Marty has answered the question in full, I thought I'd remark that Joel's argument generalizes easily to give 2^k for any k <= 2^{aleph_0}. Namely, choose a transcendence basis B for R over Q, and for each subset S of B with |S|=k, let F_S be the set of real numbers that are algebraic over Q(S). This gives you 2^k real closed fields, and they are all different. – Bjorn Poonen Jan 26 '10 at 3:00
Thanks, Bjorn, that's helpful -- in particular, that's enough to derive the statement about Aut(C) in my question. – Pete L. Clark Jan 26 '10 at 3:14
Thanks, Bjorn. I wonder if this kind of argument can work at higher k? Or do we need to use Shelah's (un)stability theory... – Joel David Hamkins Jan 26 '10 at 12:58
+1: Together with Marty's response, this answers the original question in full (it really is necessary to consider the case of countable models separately). – John Goodrick Jan 27 '10 at 3:08
@Joel: I'm not sure how Bjorn's argument could work for big k, since if k is bigger than 2^{aleph_0} there are no longer enough types to distinguish all 2^k models. The proof of Shelah's theorem that I've read involves building a big class of Ehrenfeucht-Mosowski models over carefully chosen indiscernible sequences, and then using and extremely clever and indirect argument to show that some subcollection of 2^k of these models must be pairwise nonisomorphic. – John Goodrick Jan 27 '10 at 3:17