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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 $\mathbb{R}$?

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1 Answer 1

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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.

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