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$\DeclareMathOperator\hal{hal}$A field isomorphism $\phi:F\rightarrow G$ is a bijection such that (i) $\phi(x+y)=\phi(x)+\phi(y)$ and (ii) $\phi(xy)=\phi(x)\phi(y)$, where $F$ and $G$ are ordered fields. If $F=G$, then $\phi$ is called an automorphism. For $\mathbb{R}$, the only automorphism is the identity map. But for extensions of $\mathbb{R}$, like the hyperreals $^*\mathbb{R}$, there are many non-trivial automorphisms, if one assumes the Continuum Hypothesis.

(I understand that there are many hyperreal fields, as there are many ultrafilters to choose from. But for the purposes of this question, any arbitrary hyperreal field derived via the ultrapower construction with $\mathbb{R}$ embedded will do.)

Let $\hal(r)$ be the halo of $r$, where $r\in{^*\mathbb{R}}$ and is a real number. That is, $\hal(r)$ is the set of all $x\in{^*\mathbb{R}}$ such that $|x-r|<\frac{1}{n}$ for all $n\in\mathbb{N}$. Fix a $y\in \hal(r)$ such that $y>r$. For each $z$ such that $z\in \hal(r)$ and $z>y$, is there an automorphism $\phi$ such that the reals are fixed (i.e. for all real $q\in{^*\mathbb{R}}$, $\phi(q)=q$) and $\phi(y)=z$? Assume the Continuum Hypothesis.

Posted here but got no answer: https://math.stackexchange.com/questions/4937027/automorphism-on-the-hyperreals

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  • $\begingroup$ You pick $z$ relative to $r$, but then later say "for all real $r$", and so your quantifiers are a little confusing. Could you make it clear? You want $r<y<z$ to be infinitesimally close for a given real $r$, and then you want to move $y$ to $z$ and fix just $r$, or fix all reals? $\endgroup$ Commented Jun 24 at 13:08
  • $\begingroup$ Ah my bad. I should have used a different letter. I want the latter, that is, to fix all reals, including r. $\endgroup$
    – phst
    Commented Jun 24 at 13:18
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    $\begingroup$ Under CH, the hyperreals are saturated, and because this is a real-closed field, the types are particularly simple, and so any two infinitesimals are automorphic. The question is whether this is still true while fixing all reals. $\endgroup$ Commented Jun 24 at 13:21
  • $\begingroup$ I think that's good enough for me. For the purposes of my research, as long as there is an automorphism that fixes at least all the rationals and moves $y$ to $z$ for each $z$, it'll be fine. $\endgroup$
    – phst
    Commented Jun 24 at 13:33
  • $\begingroup$ Every field automorphism fixes the rationals. $\endgroup$ Commented Jun 24 at 14:12

1 Answer 1

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Yes.

Lemma 1 Let $K$ be a real-closed field. Let $F(x)$ be a parameter-free 1-variable 1st order sentence of field theory. Suppose that in $K$, $F$ defines a singleton, i.e. $\{x\in K:F(x)\}$ is a singleton $\{u\}$. Then $u$ is algebraic.

Proof. Let $K_0$ be the subfield of algebraic numbers. Since $K_0$ and $K$ are elementarily equivalent, $\{x\in K_0:F(x)\}$ is also a singleton, hence necessarily $\{u\}$. $\Box$

Lemma 2 Let $K$ be a real-closed field. Let $F(x)$ be a parameter-free 1-variable sentence of field theory. Then $\{x\in K:F(x)\}$ is a Boolean combination of intervals with algebraic bounds.

Proof. By quantifier elimination (of real-closed fields in the language of ordered fields), this is equivalent to a Boolean combination of solutions of finite systems of polynomial (in)equalities with integral coefficients. $\Box$.

Lemma 3 For $x\in\mathbf{R}$ and $r,r'>0$ with $r,r'<1/n$ for all $n$, the pairs $(K,x+r)$ and $(K,x+r')$ are elementary equivalent.

Proof. The formulas of Lemma 2 have the same evaluation at $x+r$ and $x+r'$.

This is already sufficient, under CH, to prove that there is an automorphism sending $x+r$ to $x+r'$ (when $K$ is $\omega_1$-saturated). If $x$ is algebraic, we are done, since it is fixed by all automorphisms.

Lemma 3 was an appetizer. What we need is thus:

Lemma 4. The triples $(K,x,x+r)$ and $(K,x,x+r')$ are elementary equivalent, i.e., the pairs $(x,x+r)$ and $(x,x+r')$ satisfy the same order first-order parameter-free formulas.

Again, such a formula is equivalent to a Boolean combination of polynomial inequalities ($P(u,v)>0$, $<0$, $=0$, $\ge 0$, $\le 0$, $\neq 0$) with integral coefficients in two variables. Such an inequality holds for $(x,x+r)$ iff it holds for $(x,x+r')$. Whence the result. $\Box$

Hence, under CH and when $K$ is $\omega_1$-saturated, there is an automorphism mapping $(x,x+r)$ to $(x,x+r')$, i.e., fixing $x$ and mapping $x+r$ to $x+r'$.

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  • $\begingroup$ This may be a silly question, as this area isn't my expertise. But what happens if $K$ is $\kappa$-saturated, where $\kappa>\omega_1$? Will all the reals $x$ still be fixed and we can send $x+r$ to $x+r^{'}$ for any positive $r^{'}$? $\endgroup$
    – phst
    Commented Jun 24 at 13:58
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    $\begingroup$ I'm not so familiar with these maths; my understanding is that two EE $\omega_1$-saturated structures of cardinal $\aleph_1$ are isomorphic. My argument, if I'm correct, also shows under CH, $\mathrm{Aut}(^*\mathbf{R})$ is transitive on the halo of $x$ for every non-algebraic real $x$ (in particular, it fixes $x$ iff $x$ is algebraic). I think it also shows that it is transitive on the set of positive infinite numbers ($\bigcap_{n>0}[n,+\infty[$) $\endgroup$
    – YCor
    Commented Jun 24 at 14:07

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