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! :)

  • $\begingroup$ 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

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 $\mathbb{R}_B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.