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I will write down the definitions first and then what I have done.

Order isomorphic:

$A$ and $B$ are ordered integral domains. They are order isomorphic if $\exists$ a bijection $f: A \to B$ such that $$f(x+y) = f(x) + f(y)$$ $$f(xy) = f(x)f(y)$$ $$x < y \Rightarrow f(x) < f(y)$$

Let $A$ and $B$ be our ordered fields with least upper bound property.

Since they are fields, they are also integral domains.

Since they are fields, each one has an additive identity and a multiplicative identity. Let these four elements be $0_A, 1_A \in A$ and $0_B, 1_B \in B$

Any field that contains the integers contains the rationals as a subfield (I couldn't prove this). Also, I am assuming that this fields contain the integers, so they contain the rationals as well.

Can I set the following function? $f: A \to B$ such that $$f(0_A) \to 0_B; f(1_A) \to 1_B; f(q1_A) = q1_B : q \in \mathbb{Q}$$

These function is linear, but I don't see where can I use the least upper bound properties of these two sets.

I don't know how to continue. Any help is appreciated! :)

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  • $\begingroup$ If $f:\mathbb Z\to Z $is a ring-isomorphism, where $Z$ is a subring of a field, extend the domain of $f$ to $\mathbb Q$ by putting $f(a/b)=f(a)/f(b) $ for $a,b\in \mathbb Z$ with $b\ne 0$. $\endgroup$ – DanielWainfleet Apr 17 '17 at 4:20
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There is another name for an ordered field that satisfies the least upper bound property -- the system of real numbers. So we are trying to construct an order isomorphism between any two systems of real numbers, say $\mathbb{R}_A$ and $\mathbb{R}_B$, in other words, the system of real numbers is unique.

To extend the map $f(q1_A)=q1_B$, $q\in\mathbb{Q}$, for $r_A\in\mathbb{R}_A$, let $$S=\{q\in\mathbb{Q}|q1_A<r_A\}.$$ Then define $$f(r_A):=\sup\{q1_B|q\in S\}$$ where the supreme is taken in the system $\mathbb{R}_B$.

Then you can try to check that this map $f$ indeed defines an order isomorphism between $\mathbb{R}_A$ and $\mathbb{R}_B$.

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