# Automorphisms of the hyperreals over the rationals and nontrivial automorphism groups

A classic result says the automorphism group of $\mathbb{R}$ (over $\mathbb{Q}$) is trivial. The proof is simple: every automorphism preserves squares, and hence fixes the positive reals, so it must be order preserving. Since it must fix $\mathbb{Q}$, and $\mathbb{Q}$ is dense in $\mathbb{R}$, if any real number were not fixed, this would yield a contradiction.

In larger real-closed fields where $\mathbb{Q}$ is not dense, automorphisms are still order preserving, but the argument that they are trivial does not work. I haven't found any examples of a non-trivial automorphism of a real-closed field, but I also can't prove they don't exist. So, the first question I'd like to ask is whether all automorphism groups of real-closed fields are trivial.

If they aren't all trivial, then I want to know what, if anything, we can say. To ask a less vague question, can we classify the automorphisms of the hyperreals? How many are there, and what is their groups structure like?

• The field of real Puiseux series is real-closed, and has an obvious automorphism $\sum a_rt^r\mapsto \sum a_r t^{2r}$. – YCor Dec 31 '16 at 7:45
• There also exist real-closed fields with no nontrivial field automorphism and that are not subfields of the real field (S. Shelah, Models with second order properties IV. A general method and eliminating diamonds, 1983 sciencedirect.com/science/article/pii/0168007283900131) – YCor Dec 31 '16 at 7:47
• The field $K$ of algebraic real numbers is a real closure of $\mathbf{Q}$, so by the uniqueness of real closures up to isomorphism it admits an automorphism extending one of any subfield. For example, any non-trivial finite Galois extension of number fields $F'/F$ with $F'$ having a real embedding (e.g., $F=\mathbf{Q}$ and $F'$ real quadratic field or $\mathbf{Q}(\zeta + 1/\zeta)$ for a primitive $n$th root of unity $\zeta$ with $n\ge 7$) admits $[F':F]$ automorphisms over $F$, all extending to $K$ upon choosing an embedding of $F'$ into $K$. Hence, ${\rm{Aut}}(K)$ is huge (even uncountable). – nfdc23 Dec 31 '16 at 13:32
• @nfdc23 No, the field of real algebraic numbers has no nontrivial automorphism, by the same argument as for R. The uniqueness of real closures only gives that K has an automorphism extending an order-preserving one of any subfield, but those are all trivial. Only nonarchimedean real-closed fields may admit nontrivial automorphisms. – Emil Jeřábek Dec 31 '16 at 14:50
• @nfdc23 I don't think your argument works. The point is that a real closure of an ordered field has a unique ordering whose positive elements are precisely the sums of squares, and since an automorphism will then preserve the ordering, its restriction to the set of roots $r_1 < r_2 < \ldots < r_n$ of an $f$ must take $r_i$ to $r_i$. See Lang's Algebra, chapter XI section 2. – Todd Trimble Dec 31 '16 at 14:50

This note complement Joel's, who pointed out that every first order theory with an infinite model has a model with many automorphisms (this was first proved by Ehrenfeucht and Mostowski in their famous 1956 paper on indiscernibles). Joel also outlined why saturated models and countable computably saturated ones have nontrivial automorphisms.

It is well-known that if $\cal{M}$ is computably saturated and countable, then as soon as we know that some element $m$ of $\cal{M}$ is not first order definable in $\cal{M}$, then we can build an automorphism of $\cal{M}$ that moves $m$. Indeed, we can build continuum-many such automorphisms. This makes for a fun exercise.

Furthermore, it was shown by F. Körner (1998) that if $\cal{M}$ is a countable structure in a countable signature, then there is an elementary extension of $\cal{M}$ that carries an automorphism that is "maximal", i.e., it moves every nonalgebraic element (she showed that $\cal{M}$ can be chosen as any arithmetically saturated elementary extension of M).

In the above, an element $m$ of $\cal{M}$ is said to be algebraic if there is a formula $\phi$ with one free variable such that $\phi(m)$ holds in $\cal{M}$ and $\phi$ has only finitely many solutions in $\cal{M}$.

Körner's theorem was generalized to all structures (whose universe and signature can have any cardinality) by G. Duby in 2003, with a different, more powerful proof technique.

Here are the References:

Friederike Körner, Automorphisms moving all non-Algebraic points and an application to NF, The Journal of Symbolic Logic,Vol. 63, No. 3 (Sep., 1998), pp. 815-830.

Grégory Duby, Automorphisms with only infinite orbits on non-algebraic elements, Archive for Mathematical Logic, July 2003, Volume 42, Issue 5, pp 435–447.

• Hi Ali, if one takes the simple-minded hyperreal field $\mathbb R^{\mathbb N}/\mathcal F$ what can one say about its order-preserving automorphisms? How does it depend on the ultrafilter? – Mikhail Katz Feb 13 '18 at 14:59

As a general principle, every first-order theory with infinite models, such as the theory of real-closed fields, will have models with rich automorphism groups.

The general reason is that one can take highly saturated models and build the automorphisms with the back-and-forth method.

I claim that every real-closed field has an elementary extension to a real-closed field with non-trivial automorphisms.

This is easy to see if the GCH happens to hold, since in this case we have an abundance of saturated models. The original field extends elementarily to a saturated and hence homogeneous model, which has plenty of automorphisms.

A similar argument works directly with countable models, without any need for GCH, since we can use computable saturation in place of full saturation. Every countable model extends elementarily to a computably saturated model, which has many automorphism by the back-and-forth construction.

But in fact, we don't need any GCH assumption for the general case. Take any real-closed field $F$ and write down the theory of what it would be like to have an elementary extension of $F$ to a real-closed field $F'$ with a nontrivial automorphism $\pi:F'\to F'$. This is a theory in the language with $\pi$ and with constants for every element of $F$. In the forcing extension in which the GCH holds, the theory has a model, since in that forcing extension we have saturated models. So the theory must be consistent in our original universe, and therefore we have the desired model there as desired.

The same argument shows that every infinite model of any first-order theory has elementary extensions to models with automorphisms.

• Is it possible to rewrite your penultimate paragraph so as to make no reference to forcing extensions, but instead appeal to pure model theory? – Todd Trimble Dec 31 '16 at 18:49
• One could argue like this: that theory (the elementary diagram of $F$ plus the assertion that $\pi$ is a nontrivial automorphism) is consistent, because every finite part of the theory is consistent, since any finite subtheory makes reference to only finitely many particular elements of $F$, which generate a finite subfield, and then reduce to the computable saturation argument in the countable case. So, no forcing. I think almost any model theory book will prove the existence of sufficiently homogeneous models, and that is what I'm doing. – Joel David Hamkins Dec 31 '16 at 19:12
• Oh sure; thanks. I rather prefer that formulation. – Todd Trimble Dec 31 '16 at 19:14
• Correction: I should have said finitely-generated subfield; of course, it is countably infinite. – Joel David Hamkins Dec 31 '16 at 21:03