7

Is there a classification of the algebraically closed fields that have maximal proper subfields ?

And if an algebraically closed field has a maximal proper subfield, is that subfield unique ?

Summarizing the answers, an algebraically closed field has a maximal subfield if and only if its characteristic is zero and such a maximal subfield is never unique.

tomasz
  • 557
  • 4
    Can you tell us if your questions are based on any specific examples? Concerning the second question, a proper subfield with finite codimension must have codimension 2 (like R inside of C). Conjugating an automorphism of order 2 will usually give you a different automorphism of order 2, corresponding to a different subfield of codimension 2. – KConrad Jan 22 '12 at 19:14

2 Answers2

15

If $F$ is a maximal proper subfield of a field $K$, then $K=F(x)$ for any $x\in K\setminus F$. Next, $x$ must be algebraic over $F$ (otherwise $F\subsetneq F(x^2)\subsetneq F(x)\subset K$). So $K$ is finite over $F$, and if $K$ is algebraically closed it is well known (cf. KConrad's comment) that $F$ is a real closed field and $K=F(\sqrt{-1})$.

Laurent Moret-Bailly
  • 19,274
8

This occurs iff the field has characteristic 0. By KConrad's comment, being characteristic 0 is certainly a necessary condition. Conversely, given an algebraically closed field K of characteristic 0, we can use Zorn's Lemma to find a maximal ordered subfield F. Since K is algebraically closed, F must be real closed. But also K must be algebraic over F or else we could pick a transcendental element and adjoin it to F (make it infinitely larger than all elements of F, i.e., use lexicographic ordering). Hence K must be of degree 2 over F and thus F is a maximal proper subfield.

Also by KConrad's comment, this is never unique; just apply an automorphism of K that takes, e.g., $\sqrt[3]{2}$ to $\omega \sqrt[3]{2}$, where $\omega$ is a primitive cube root of unity.

Harry Altman
  • 2,575