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Update. The question has been recently answered in the positive by David Marker and Charles Steinhorn (as in indicated in Marker's answer). Note that Remark 3 below is now expanded by reference to a paper of Mekler and Shelah.

Question. Is there a countable rigid non-Archimedean real closed field?

Background:

  1. As usual, a structure is said to be rigid if the only automorphism of the structure is the identity map.

  2. It is well-known that any Archimedean ordered field is rigid; the argument is the same as that used by Darboux in 1880 to prove that the field of real numbers is rigid. Note that the Archimedean ordered fields are precisely those ordered fields that are isomorphic a subfield of the field of real numbers.

  3. In 1983 Shelah showed that its consistent with the axioms of set theory (ZFC) that there is an uncountable rigid non-Archimedean closed field. Here is the reference: *Shelah, Saharon, Models with second order properties. IV. A general method and eliminating diamonds. Ann. Pure Appl. Logic 25 (1983), no. 2, 183–212. A decade later Mekler & Shelah in this paper showed in ZFC the compactness of the extension of first order logic by a quantifier over isomorphisms between (definable) ordered fields; as explained on page 2 of the article, this result can be used to show the existence, within ZFC, of arbitrarily large rigid RCFs (thanks to Biran Falk-Dotan for the information about the Mekler-Shelah paper).

  4. Using machinery from the metamathematics of arithmetic, one can build (many) countable rigid non-Archimedean ordered fields. More specifically: it is well-known that countable nonstandard rigid models of PA exist (this is independently due to Gaifman and Ehrenfeucht). Let $\cal{M}$ be such a model, and let $\mathbb{Q}^{\cal{M}}$ be the field of rationals as computed in $\cal{M}$. By a classical theorem of Julia Robinson, $\cal{M}$ and $\mathbb{Q}^{\cal{M}}$ are bi-interpretable, therefore $\mathbb{Q}^{\cal{M}}$ is also rigid. One can use the same type of argument to show that there are (many) rigid non-Archimedean ordered fields in every infinite cardinality.

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    $\begingroup$ @JoelDavidHamkins I thought about that strategy but real closed fields tend to be very poor in their power to interpret due to lack of coding machinery (and very well-behaved definable sets), for example even though the standard model of arithmetic can interpret the field of algebraic numbers (which forms a real closed field), the latter cannot interpret the former, since the theory of algebraic numbers is the same as the theory of real closed fields, which is a decidable theory, but of the course the Turing degree of theory of the standard model of arithmetic is the $\omega$-th jump of $0$. $\endgroup$
    – Ali Enayat
    Commented Feb 21, 2018 at 1:56
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    $\begingroup$ The real closure of a countable model of PA is recursively saturated, thus homogeneous. It will have lots of automorphisms. $\endgroup$
    – Emil Jeřábek
    Commented Feb 21, 2018 at 9:27
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    $\begingroup$ Given a countable non-Archimedean real-closed field $F$, with an infinitesimal $t$, can we prove that some automorphism of $F$ extends the ring automorphism $\mathbb{Q}[t]\to\mathbb{Q}[2t]$? That would answer the posted question in the negative. $\endgroup$
    – user44143
    Commented Oct 5, 2022 at 20:37
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    $\begingroup$ @MattF. I don't know the answer to your question. Why not pose it as a new question ?(with a pointer this question). $\endgroup$
    – Ali Enayat
    Commented Oct 6, 2022 at 21:24
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    $\begingroup$ @MattF. I very much doubt this is true in general (though the following is not a finished proof). Let $K$ be the completion of the real closure of $\mathbb Q(t)$ with $t>0$ infinitesimal; $K$ can be described as the field of Levi-Civita series $\sum_{q\in\mathbb Q}a_qt^q$ where $a_q$ are real algebraic, and $\{q:a_q\ne0\}$ has no accumulation point other than $+\infty$. Let $u=\sum_qa_qt^q$ be such a series, and $F$ be the real closure of $\{t,u\}$ in $K$. Then if $F$ has an automorphism mapping $t$ to $2t$, it must map $u$ to $u'=\sum_qa_q2^qt^q$. But if $u$ is sufficiently complicated, ... $\endgroup$
    – Emil Jeřábek
    Commented Oct 7, 2022 at 15:01

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Charles Steinhorn and I have answered this question positively by constructing a rigid non-archimedean real closed field of transcendence degree 2. Our preprint is now posted on arxiv. https://arxiv.org/abs/2407.00542v2

Here is the basic idea. Suppose $a_1,\dots,a_n$ is a transcendecne base for $K$. A necessary and sufficient condition for the rigidity of $K$ is that $a_1,\dots,a_n$ is the unique realization of the type tp$(a_1,\dots,a_n)$--a non-trivial automorphism would move the transcence base to another realization of the type and, on the other hand, if there were another realization of the type using some basic o-minimality and the uniqueness of real closures there would be an automorphism taking one realization to another.

If $n=1$ this is impossible as there is a trancendental infinite $a$ and $a$ and $a+1$ realize the same type so there is an automorphism sending $a$ to $a+1$.

For larger $n$ we need to construct $(a,b)$ such that if $F:K^2\rightarrow K^2$ is any function definable over the real algebraic number $F(a,b)$ does not realize the type of $a,b$. We show this is possble for $n=2$ though we expect this is true for all finite $n\ge 2$.

To the best of our knowledge this is the first construction of a rigid non-archimedean real closed field that does not use extra set theoretic assumptions.

ADDED COMMENT: Our knowledge was incomplete. We thank Biran Falk Dotan for pointing us to the 1993 paper Some Compact Logics--results in ZFC, where Mekler and Shelah show in ZFC--that there are arbitrarily large rigid real closed fields. See https://arxiv.org/abs/math/9301204

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    $\begingroup$ Very cool, I skimmed through this earlier today. Not being much of an algebraist, I had to go and check that finite transcendence degree over a countable field implied countable. Nowhere in the paper do you explicitly say your field is countable, which is a very minor point, but from a reader-ergonomics point of view, it might be nice to word your theorem in a way that explicitly uses the wording of the original question (eg "K has tr.deg. 2 and hence is countable"). Please forgive the noob comment! $\endgroup$
    – David Roberts ♦
    Commented Jul 2 at 12:21

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