Algebraically closed fields with proper maximal subfields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:44:53Z http://mathoverflow.net/feeds/question/86395 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86395/algebraically-closed-fields-with-proper-maximal-subfields Algebraically closed fields with proper maximal subfields tomasz 2012-01-22T19:11:27Z 2012-01-22T22:55:43Z <p>Is there a classification of the algebraically closed fields that have maximal proper subfields ? </p> <p>And if an algebraically closed field has a maximal proper subfield, is that subfield unique ? </p> <p>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. </p> http://mathoverflow.net/questions/86395/algebraically-closed-fields-with-proper-maximal-subfields/86399#86399 Answer by Laurent Moret-Bailly for Algebraically closed fields with proper maximal subfields Laurent Moret-Bailly 2012-01-22T20:09:12Z 2012-01-22T20:09:12Z <p>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 $K=F(\sqrt{-1})$.</p> http://mathoverflow.net/questions/86395/algebraically-closed-fields-with-proper-maximal-subfields/86400#86400 Answer by Harry Altman for Algebraically closed fields with proper maximal subfields Harry Altman 2012-01-22T20:41:20Z 2012-01-22T20:41:20Z <p>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.</p> <p>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.</p>