# The isomorphism between two complete ordered fields is unique

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 $$\Phi(p) \prec \Phi(\sup X) \implies p<\sup X \implies p for some $$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 $$X_x=\{p\in\Bbb Q \mid p. Since $$\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$$.

• 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. Jan 17 '19 at 0:05
• 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? Jan 17 '19 at 0:10

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 $$\psi_1=\psi_2.$$

• 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 $a+b\sqrt 2\,>^*0\iff a-b\sqrt 2<0.$ Jan 16 '19 at 23:06
• 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, $x\ge q \iff \psi(x)-q \ge 0$ $\iff \psi(x)\ge q$.[...] Jan 17 '19 at 1:54
• [...] 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. Jan 17 '19 at 1:54
• Perfectly correct. And $f$ was a typo for $\psi$. Jan 17 '19 at 13:58
• Thank you so much for your verification! Have you seen any error in my proof? Jan 17 '19 at 14:04