Can nonstandard fields contain $\mathbb R$ in different ways?

Question

Suppose $e : \mathbb R \to F$ is an elementary embedding in the language of ordered fields. Can there exist an elementary embedding $e' : \mathbb R \to F$ such that $e \not= e'$? Note that it would have to be the case that for some undefinable $r \in \mathbb R$, $e(r) \not= e'(r)$, and $e(r) - e'(r)$ is infinitesimal.

Answer 1

Yes, this is possible. Let $F$ be a nonarchimedean strongly $\omega$-homogeneous real-closed field such that $\mathbb R\subseteq F$ (which exists by model-theoretic general reasons). Fix an infinitesimal element $\epsilon>0$. By quantifier elimination for real-closed fields, $\pi$ and $\pi+\epsilon$ have the same type (the only property of $\pi$ I need here is that it is transcendental), hence by homogeneity, there exists an automorphism $f$ of $F$ such that $f(\pi)=\pi+\epsilon$. Then $f\restriction\mathbb R$ is an (elementary) embedding of $\mathbb R$ in $F$ different from inclusion.

In fact, there is nothing special about $\mathbb R$:

Proposition. For any first-order structure $A$, the following are equivalent:

1. For every structure $B$, there is at most one elementary embedding of $A$ in $B$.

2. All elements of $A$ are definable.

2${}\to{}$1 is obvious. For 1${}\to{}$2, if $a\in A$ is not definable, the type $\{x\ne a\}\cup\{\phi(x):A\models\phi(a)\}$ is consistent, hence there exists an elementary extension $B$ of $A$ and $b\in B$ such that $b\ne a$ and $(B,a)\equiv(B,b)$. By elementarily extending $B$ further if necessary, we may assume $B$ to be strongly $\omega$-homogeneous. Thus, there exists an automorphism $f$ of $B$ such that $f(a)=b$, and then $f\restriction A$ gives a second elementary embedding of $A$ in $B$.

Answer 2

In fact $\mathbb{R}$ is elementarily embedded in several ways in any non-archimedean real-closed field which contains it. The proof is more involved than I thought before writing it, but if you don't know the arguments, I think this will make the situation of your question clearer.

Let $K\supset \mathbb{R}$ be a non-archimedean real-closed field. Let $K^{\prec}$ denote the set of infinitesimal elements in $K$, including zero. Those are elements $\varepsilon \in K$ with $\frac{-1}{n+1}<\varepsilon < \frac{1}{n+1}$ for all $n \in \mathbb{N}$. The set of finite elements in $K$, i.e. elements $a$ with $-n<a<n$ for some $n \in \mathbb{N}$, is denoted $K^{\preceq}$. As a subring of $K$, the set $K^{\preceq}$ is a valuation ring with maximal ideal $K^{\prec}$. I write $k$ for the residue field. It is ordered by setting $a+K^{\prec}<b+K^{\prec}$ iff $a<b$. It is easy to see that $k$ is archimedean. So far this didn't require any assumption besides $K$ being a non-archimedean ordered field.

Any finite element $a \in K$ can be written as $a=r+\varepsilon$ for certain $r \in \mathbb{R}$ and $\varepsilon \in K^{\prec}$. Indeed, the morphism $\mathbb{R} \rightarrow k$ given by $r\mapsto r+K^{\prec}$ must be the identity because $\mathbb{R}$ is the maximal archimedean field.

Now for the proof of my claim. Let $0\neq\varepsilon \in K^{\prec}$ and write $\pi'=\pi+\varepsilon$. The field $\mathbb{Q}(\pi')$ is an archimedean subfield of $K$ containing $\pi'$. Indeed $\pi$ is irrational, so for any polynomial $P \in \mathbb{Q}[X]$, the quantity $P(\pi')=P(\pi)+Q(\varepsilon)$ (where $Q \in \mathbb{Q}[\pi][X]$ is obtained through a Taylor expansion of $P$ at $\pi$) is infinitesimally close to $P(\pi)$ and thus non infinitesimal. This argument generalizes to any archimedean field besides $\mathbb{Q}$ provieded that $\pi$ is replaced by a finite element which is transcendent over it. Thus the set of archimedean subfields of $K$ containing $\pi'$ is non-empety. It is inductive for the inclusion, so by Zorn's lemma, there is a maximal such field, say $R$. This field is real-closed by the initial property of the real closure and the fact that an ordered field is always cofinal in its ordered algebraic extensions. Thus $R$ has no ordered algebraic extension. Asssume for contradiction that there is a bounded set without an upper bound in $R$. The corresponding cut in $R$ is filled in $K$ by an element of the form $\iota+\epsilon$ where $\iota \in \mathbb{R}$ and $\epsilon$ is infinitesimal. Note that $\iota$ lies in the cut as well, so it does not lie in $R$, whence it is transcendant over $R$. The previous argument shows that $R(\iota+\varepsilon)$ is archimedean, contradicting the maximality of $R$.

This proves that $R$ has the LUB property, so it is isomorphic to $\mathbb{R}$. The unique isomorphism $\mathbb{R} \rightarrow R$, seen as an embedding $\mathbb{R} \rightarrow K$, is elementary because the first order theory of $(\mathbb{R},+,\times)$ is model-complete and its models are real-closed fields. By our choice of $\pi'$, it is distinct from the embedding we started with.