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There are counterexamples to order isomorphisms of ordered fields being field isomorphisms, see Is the multiplicative structure of a totally ordered field unique?. However, Wikipedia suggests that for real closed fields order isomorphism implies isomorphism: "If the continuum hypothesis holds, all real closed fields with cardinality the continuum and having the $η_1$ property are order isomorphic. This unique field $F$ can be defined by means of an ultrapower... This is the most commonly used hyperreal number field in non-standard analysis, and its uniqueness is equivalent to the continuum hypothesis."

How exactly does one recover operations from order if $F$ is real closed?

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    $\begingroup$ Your title doesn't capture wiki's statement accurately. Anyway, in the list of the references given on the wiki page, there is a paper with a proof: math.hmc.edu/~henriksen/publications/… $\endgroup$
    – hot_queen
    Aug 7, 2014 at 20:15

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What you quoted from Wikipedia is correct, but your more general statement, which you claim Wikipedia "suggests", is not. Two real-closed fields can be order-isomorphic without being isomorphic as fields. For example, let one of the two fields be the field of real algebraic numbers. Then choose a transcendental number, say $e$, and let the second field consist of all the real numbers that are algebraic over $\mathbb Q(e)$. These two fields are isomorphic as ordered sets, because all countable dense linear orders are isomorphic (a theorem of Cantor), but they are not isomorphic as fields because they have different transcendence degrees.

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  • $\begingroup$ I guess two real closed subfields of R can never be isomorphic unless they are identical in which case the isomorphism is identity. So we can even have the transcendence degrees same without having an isomorphism. Although, I am not sure about the example (smallest real closed fields containing e and $\pi$, respectively) given in the Erdos et al paper listed on wiki. $\endgroup$
    – hot_queen
    Aug 7, 2014 at 21:49
  • $\begingroup$ @hot_queen Your guess is right. The point is that a field-isomorphism between real-closed fields automatically preserves the order too, because the positive elements are just the nonzero squares. Since an isomorphism has to fix all the rational numbers, the two isomorphic fields must fill exactly the same cuts in the rational number line. So, if they're subfields of the reals, they're identical. $\endgroup$
    – Andreas Blass
    Aug 7, 2014 at 21:55
  • $\begingroup$ Thanks for the explanation. Do you know if it is known that e and $\pi$ are relatively transcendental (which seems to be the "obvious and well known") according to page 546 here: math.hmc.edu/~henriksen/publications/… $\endgroup$
    – hot_queen
    Aug 7, 2014 at 21:58
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    $\begingroup$ @Conifold Under CH, all $\aleph_1$-saturated real-closed fields of the cardinality of the continuum are isomorphic as ordered fields. I vaguely recall having seen somewhere that, for real-closed fields, the $\eta_1$-property (which amounts to $\aleph_1$-saturation as an order) implies $\aleph_1$-saturation as an ordered field, but I'm not at all sure about that. (I think "hyperreal" is usually used for broader classes of fields, which may vary from one author to another.) $\endgroup$
    – Andreas Blass
    Aug 7, 2014 at 22:32
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    $\begingroup$ @Conifold This is a very late comment, but the fact that $\kappa$-saturation of the underlying order implies $\kappa$-saturation of the real closed field follows from o-minimality: to check saturation, it suffices to check that all $1$-types over small sets of parameters are realized, and by o-minimality, every formula with parameters and $1$ free variable is equivalent to one in the order language, so realizing general $1$-types reduces to realizing $1$-types in the order language. $\endgroup$
    – Alex Kruckman
    Jul 26 at 3:16

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