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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?

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  • $\begingroup$ 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!) $\endgroup$ Apr 14 at 20:43
  • $\begingroup$ For each $n$, let $K_n$ be a maximal real closed subfield of $\overline{\Bbb{C}(x_1,\ldots,x_n)}$ containing $\Bbb{R}(x_1,\ldots,x_n)$. $\endgroup$
    – reuns
    Apr 14 at 20:51
  • $\begingroup$ 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.) $\endgroup$
    – Erik D
    Apr 15 at 0:40
  • $\begingroup$ @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. $\endgroup$ Apr 15 at 4:15
  • $\begingroup$ @NoahSchweber: I see, yes of course it does. That's funny! $\endgroup$
    – Erik D
    Apr 15 at 5:37

1 Answer 1

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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 $R_L\not\cong R_{L'}$.
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