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Let $\mathfrak{M}$ be a model of set theory, and consider HOD (the hereditarily ordinal definable elements) of $\mathfrak{M}$. Let $K$ be any algebraically closed field in HOD of zero characteristic and any cardinality.

Is there always a real closed subfield $R$ of $K$ of index 2 such that $R \in $ HOD?

Under what condition $R$ is archimedean? Or rather, can we always find an archimedean real closed subfield $R$ of $K$ such that $R \in$ HOD?

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  • $\begingroup$ Sure. HOD is a transitive model of ZFC, ZFC proves that every algebraically closed field of characteristic 0 has an index 2 real-closed subfield, and being algebraically closed or real-closed is absolute. $\endgroup$ Dec 11 '13 at 13:00
  • $\begingroup$ Anticipating your line of thought, note that even in the full Solovay model, where every set of reals in HOD has the Baire property, it doesn't mean that HOD itself knows that. That is, HOD is still a model of choice and there are sets that HOD thinks are not Baire measurable. $\endgroup$
    – Asaf Karagila
    Dec 11 '13 at 14:08
  • $\begingroup$ Yes, I am aware of that. Thanks anyway. $\endgroup$
    – user38200
    Dec 11 '13 at 15:22
  • $\begingroup$ The edited question follows the same principle. ZFC proves that $R$ can be taken archimedean if and only if $|K|\le2^\omega$, so the answer is positive iff HOD thinks that the cardinality of $K$ is $\le2^\omega$. $\endgroup$ Dec 11 '13 at 20:44
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The new question added a few minutes ago can be answered by the same idea as in Emil's comment. The following paragraph is provable in ZFC and therefore true in HOD:

For any cardinal $\kappa\leq\mathfrak c$ (the cardinal of the continuum), there is a real-closed subfield of $\mathbb R$ with transcendence degree $\kappa$, and by adjoining $i$ to this field we get an algebraically closed field of transcendence degree $\kappa$. Since all algebraically closed fields of characteristic 0 and transcendence degree $\kappa$ are isomorphic, they all have real-closed Archimedean subfields of index 2. On the other hand, an algebraically closed field of transcendence degree $\kappa>\mathfrak c$ cannot have such a subfield, because the subfield would have cardinality $>\mathfrak c$ and would therefore admit no embedding in the reals, so it can't be archimedean.

Using this information in HOD and the fact that archimedeanness is absolute between HOD and the universe (essentially because the natural numbers are absolute), we get that an HOD algebraically closed field of characteristic 0 has an archimedean real-closed HOD subfield of index 2 if and only if its transcendence degree in HOD (or equivalently its cardinality in HOD) is at most the cardinal that HOD considers to be the cardinality of the continuum.

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  • $\begingroup$ And what is the cardinality of continuum (in $\mathfrak{M}$) as percieved by HOD? $\endgroup$
    – user38200
    Dec 11 '13 at 21:00
  • $\begingroup$ @user38200 I'm not aware of any simpler description of that cardinal $\mathfrak c^{(HOD)}$. In particular, it could be larger than $\mathfrak c^{(\mathfrak M)}$ or smaller or equal to it. $\endgroup$ Dec 11 '13 at 23:00
  • $\begingroup$ To avoid confusion, it can only be larger than $\mathfrak c^{(\mathfrak M)}$ on the grounds that it fails to be a cardinal in $\mathfrak M$. One always has that the cardinality of $\mathfrak c^{(HOD)}$ in $\mathfrak M$ is at most $\mathfrak c^{(\mathfrak M)}$ (as every real in HOD is a real in $\mathfrak M$). This is consistent with the observation that if a field has a degree-2 archimedean real-closed subfield in HOD, it also has one (namely, the same one) in $\mathfrak M$. $\endgroup$ Dec 12 '13 at 13:12
  • $\begingroup$ @EmilJeřábek Thanks for adding that clarification; I guess I should have included it in my comment. $\endgroup$ Dec 12 '13 at 17:38

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