# Real closed fields in HOD

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?

• 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. Commented Dec 11, 2013 at 13:00
• 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. Commented Dec 11, 2013 at 14:08
• Yes, I am aware of that. Thanks anyway. Commented Dec 11, 2013 at 15:22
• 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 $\le2^\omega$. Commented Dec 11, 2013 at 20:44

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.
• And what is the cardinality of continuum (in $\mathfrak{M}$) as percieved by HOD? Commented Dec 11, 2013 at 21:00
• @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. Commented Dec 11, 2013 at 23:00
• 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 $\mathfrak M$. Commented Dec 12, 2013 at 13:12