# Algebraically closed fields with proper maximal subfields

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.

• 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

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 $K=F(\sqrt{-1})$.
Also by KConrad's comment, this is never unique; just apply an automorphism of K that takes, e.g., $\sqrt{2}$ to $\omega \sqrt{2}$, where $\omega$ is a primitive cube root of unity.