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The isomorphism between two complete ordered fields is unique.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!


My attempt:

Let $\mathfrak{R}=\langle \Bbb R,<,+,\cdot,0,1 \rangle,\mathfrak{A}=\langle A,\prec,\oplus,\odot,0',1' \rangle$ be complete ordered fields where $\Bbb R$ is the set of real numbers. Let $\mathfrak{B}=\langle B,\prec,\oplus,\odot,0',1' \rangle$ be the smallest subfield of $\mathfrak{A}$ and $\mathfrak{Q}=\langle \Bbb Q,<,+,\cdot,0,1 \rangle$.

Lemma 1: $\mathfrak{R}$ is isomorphic to $\mathfrak{A}$.

Lemma 2: $\mathfrak{Q}$ is uniquely isomorphic to $\mathfrak{B}$.

By Lemma 1, let $\Phi:\Bbb R \to A,\Psi:\Bbb R \to A$ be isomorphisms between $\mathfrak{R}$ and $\mathfrak{A}$. By Lemma 2, let $f:\Bbb Q \to B$ be the unique isomorphism between $\mathfrak{Q}$ and $\mathfrak{B}$.

Let $X \subseteq \Bbb Q$ be bounded from above and $\sup,\sup'$ supremums w.r.t $<,\prec$ respectively. We next prove that $\Phi(\sup X) = \sup' f[X]$.

$\forall x\in X: x \le \sup X \implies \forall x\in X: \Phi(x) \preccurlyeq \Phi(\sup X) \implies \forall x\in X: f(x) \preccurlyeq \Phi(\sup X) \implies \sup' f[X] \preccurlyeq \Phi(\sup X).$

  • Assume the contrary that $\sup' f[X] \prec \Phi(\sup X)$. Since $B$ is dense in $A$, there exists $b\in B$ such that $\sup' f[X] \prec b \prec \Phi(\sup X)$. Then there exists $p\in \Bbb Q$ such that $f(p)=b$. Thus $\sup' f[X] \prec f(p)=\Phi(p) \prec \Phi(\sup X).$

  • We have $\Phi(p) \prec \Phi(\sup X) \implies p<\sup X \implies p<p'$ for some $p'\in X \implies$ $f(p) \prec f(p')$ for some $p'\in X$ $\implies f(p) \prec \sup' f[X]$. This is a contradiction.

Hence $\Phi(\sup X)=\sup' f[X]$. Similarly, $\Psi(\sup X)=\sup' f[X]$.

Let $X_x=\{p\in\Bbb Q \mid p<x\} \subseteq \Bbb Q$. Since $\Bbb Q$ is dense in $\Bbb R$, $x=\sup X_x$ for all $x\in\Bbb R$. Then $\Phi(x)=\Phi(\sup X_x)=\sup' f[X_x]=\Psi(\sup X_x)=\Psi(x)$ for all $x\in\Bbb R$. It follows that $\Phi=\Psi$.

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  • $\begingroup$ You should use that a complete ordered field has a unique ordering, because an element is nonnegative if and only if it has a square root. $\endgroup$ – egreg Jan 17 '19 at 0:05
  • $\begingroup$ Hi @egreg, DanielWainfleet has utilized your idea and posted it as an answer below. I am reading his answer. Have you seen any error in my proof? $\endgroup$ – Abstract Analysis Jan 17 '19 at 0:10
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This is an extended comment on the uniqueness of the isomorphism.

(1a). Let $F, G$ be sub-fields of $\Bbb R$ such that $\psi:F\to G$ is a field-isomorphism. ($\psi$ is not assumed to preserve order.) If $\forall x\in F\,(0\le x\implies \sqrt x\in F) $ then $\psi=id_F$ and $G=F.$

Proof: $\Bbb Q\subset F$ and $\psi|_{\Bbb Q}=id_{\Bbb Q}$ so for any $x\in F$ and any $q\in \Bbb Q$ we have $$x\ge q\iff \psi(x)-q=\psi(x)-\psi(q)=(\psi(\sqrt {x-q}))^2\ge 0\iff$$ $$\iff \psi(x)\ge q$$ so $\Bbb Q \cap (-\infty,x]=\Bbb Q\cap (-\infty,\psi(x)],$ so $x=\psi(x).$

(1b).In particular, letting $F=\Bbb R$ in (1a), the only sub-field of $\Bbb R$ that is field-isomorphic to $\Bbb R$ is $\Bbb R$ itself, and the only field-isomorphism of $\Bbb R$ to $\Bbb R$ is $id_{\Bbb R}.$

(2). If $ B $ is a field and $\psi_1, \psi_2$ are field-isomorphisms from $\Bbb R$ to $B$ then by (1b), $\psi_2^{-1}\psi_1=id_{\Bbb R},$ so $\psi_1=\psi_2.$

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  • $\begingroup$ A proper sub-field of $\Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $\Bbb R$. For example $\{a+b\sqrt 2\,:a,b\in \Bbb Q\}.$ For $a,b\in \Bbb Q$ let $a+b\sqrt 2\,>^*0\iff a-b\sqrt 2<0.$ $\endgroup$ – DanielWainfleet Jan 16 '19 at 23:06
  • $\begingroup$ I think you meant $\psi$ rather than $f$. Please check if my reasoning is correct: For all $x\in F$ and $q\in\Bbb Q$: $\psi(x)-q=\psi(x)-\psi(q)$ [since $\psi|_{\Bbb Q}=\text{id}_{\Bbb Q}$] $=\psi(x-q)$. Then $x \ge q \iff x-q \ge 0 \iff x-q$ $=\sqrt {x-q} \cdot \sqrt {x-q} \iff \psi(x-q)=\psi(\sqrt {x-q} \cdot \sqrt {x-q})=\psi(\sqrt {x-q})\cdot \psi(\sqrt {x-q})=$ $(\psi(\sqrt {x-q}))^2 \ge 0 \iff \psi(x)-q=\psi(x-q) \ge 0$. To sum up, $x\ge q \iff \psi(x)-q \ge 0$ $\iff \psi(x)\ge q$.[...] $\endgroup$ – Abstract Analysis Jan 17 '19 at 1:54
  • $\begingroup$ [...] It follows that $\{q\in\Bbb Q \mid q\le x\}=\{q\in\Bbb Q \mid q\le \psi(x)\}$. Hence $\sup \{q\in\Bbb Q \mid q\le x\}=$ $\sup \{q\in\Bbb Q \mid q\le \psi(x)\}$ and thus $x=\psi(x)$. As a result, the isomorphism between two fields is unique. $\endgroup$ – Abstract Analysis Jan 17 '19 at 1:54
  • $\begingroup$ Perfectly correct. And $f$ was a typo for $\psi$. $\endgroup$ – DanielWainfleet Jan 17 '19 at 13:58
  • $\begingroup$ Thank you so much for your verification! Have you seen any error in my proof? $\endgroup$ – Abstract Analysis Jan 17 '19 at 14:04

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