MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given two real closed fields $R_1$ and $R_2$ such that both have cardinality continuum, archimedean, but not necessarily complete. Assume further that they are back and forth equivalent (in the language of rings).

Under which conditions can we get that they are isomorphic? Isomorphic to $\mathbb{R}$?

share|cite|improve this question
up vote 9 down vote accepted

Any two back-and-forth equivalent archimedean ordered fields (or domains) are isomorphic. (This refers to the language of ordered rings, but in your case, this makes no difference, as the order is definable in the ring structure for real-closed fields).

Archimedean fields are canonically isomorphic to subfields of $\mathbb R$. Back-and-forth equivalence is the same as equivalence in the $L_{\infty,\omega}$ logic, and $F\subseteq\mathbb R$ is uniquely determined by its $L_{\omega_1,\omega}$ theory, because a real $r$ is in $F$ if and only if $F$ realizes the countable type $$p_r(x)=\{q<x<q':q,q'\in\mathbb Q,q<r<q'\}$$ (note that rational constants are definable in the language of rings). This also tells you that an archimedean $F$ is isomorphic to $\mathbb R$ if and only if it realizes all the types above.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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