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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?

Robert Shore
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Display Name
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  • 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!) – Noah Schweber Apr 14 '22 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 $\Bbb{R}(x_1,\ldots,x_n)$. – reuns Apr 14 '22 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.) – Erik D Apr 15 '22 at 00:40
  • @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. – Noah Schweber Apr 15 '22 at 04:15
  • @NoahSchweber: I see, yes of course it does. That's funny! – Erik D Apr 15 '22 at 05:37
  • Hi, Display Name. I noticed that you haven't accepted answers to any of your questions on this site. When you receive an answer which resolves your question, you should "accept" it by clicking on the green check mark next to the answer. This marks the question as resolved and also rewards the answerer with reputation. If you receive answer which don't resolve your question, you can follow up with comments or by editing your question. I'm not making this request because I want the reputation - I have enough. I just wanted to let you know that it's the polite thing to do. Thanks! – Alex Kruckman May 15 '22 at 20:46

1 Answers1

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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 $a<b$ 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 $R_L\not\cong R_{L'}$.
Alex Kruckman
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