# Isomorphism Classes of Real Closed Subfields of $\Bbb C$

The real line $$\Bbb R$$ is a maximal real closed subfield of the complex plane $$\Bbb C$$. How many such maximal real closed subfields exist(up to isomorphism)? Is there a way to see that there must be at least infinitely many such maximal real closed subfields?

• Note to readers that this is related but not a duplicate, since this question is counting subfields up to isomorphism. (I tripped over this briefly!) Apr 14 at 20:43
• For each $n$, let $K_n$ be a maximal real closed subfield of $\overline{\Bbb{C}(x_1,\ldots,x_n)}$ containing $\Bbb{R}(x_1,\ldots,x_n)$. Apr 14 at 20:51
• Wouldn't any maximal real closed subfield of $\mathbb{C}$ have index two in $\mathbb{C}$? If so, it would be isomorphic to $\mathbb{C}$ by the Artin-Schreier theorem. (I'm not sure about this, just asking.) Apr 15 at 0:40
• @ErikD There are lots of non-Archimedean maximal real closed subfields of $\mathbb{C}$, which are of course not isomorphic to $\mathbb{R}$. See the link in my previous comment. Apr 15 at 4:15
• @NoahSchweber: I see, yes of course it does. That's funny! Apr 15 at 5:37

There are $$2^{2^{\aleph_0}}$$-many maximal real-closed subfields of $$\mathbb{C}$$ up to isomorphism.

Let me first appeal to a "big theorem": Since the theory $$\mathrm{RCF}$$ of real-closed fields is unstable, it has the maximal number ($$2^\kappa$$) of models of cardinality $$\kappa$$ up to isomorphism for every uncountable cardinal $$\kappa$$.

Taking $$\kappa = 2^{\aleph_0}$$, there are $$2^{2^{\aleph_0}}$$-many real closed fields up to isomorphism of cardinality $$2^{\aleph_0}$$. It remains to show that each such field embeds as a maximal real-closed subfield of $$\mathbb{C}$$.

Let $$R$$ be a real-closed field of cardinality $$2^{\aleph_0}$$, and let $$C = R[i]$$ be its algebraic closure. Then $$|C| = 2^{\aleph_0}$$, so there is an isomorphism $$\sigma\colon C\cong \mathbb{C}$$ (since any two uncountable algebraically closed fields of the same characteristic and cardinality are isomorphic). Since $$R$$ is a maximal real-closed subfield of $$C$$, $$\sigma(R)$$ is a maximal real-closed subfield of $$\mathbb{C}$$ which is isomorphic to $$R$$.

Above I appealed to Shelah's "Many Models Theorem" for unstable theories. The proof of this theorem is rather technical, but it's quite a bit easier in the special case of $$\mathrm{RCF}$$, as explained in this MathOverflow post by Dave Marker. Briefly:

1. Show that there are $$2^\kappa$$-many linear orders of cardinality $$\kappa$$ up to isomorphism.
2. For every linear order $$L$$, order the field $$F _L = \mathbb{Q}(\{x_a\mid a\in L\})$$ in such a way that each $$x_a$$ is greater than every rational number, and $$x_a^n < x_b$$ for all $$n\in \mathbb{N}$$ whenever $$a in $$L$$.
3. Let $$R_L$$ be the real closure of $$F_L$$. Show that $$L$$ can be recovered from $$R_L$$ up to isomorphism, so $$L\not\cong L'$$ implies $$R_L\not\cong R_{L'}$$.