most recent 30 from http://mathoverflow.net 2013-05-23T09:19:27Z http://mathoverflow.net/feeds/question/16512 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16512/extensions-of-fields-with-lots-of-symmetry Mariano Suárez-Alvarez 2010-02-26T13:48:47Z 2010-02-26T15:07:27Z

<p>[Cute question heard elsewhere]</p> <p>Is there a nice characterization of extensions of fields $K/k$ such that whenever $E/k$ and $E'/k$ are subextensions and $\sigma:E\to E'$ is an isomorphism over $k$, there is a $\tilde\sigma\in\mathrm{Aut}(K/k)$ such that $\tilde\sigma|_E=\sigma$?</p> <p>Normal extensions and those without proper subextensions have that property. On the other hand, $\mathbb Q(\sqrt[4]{2})/\mathbb Q$ doesn't.</p>

http://mathoverflow.net/questions/16512/extensions-of-fields-with-lots-of-symmetry/16516#16516 Georges Elencwajg 2010-02-26T15:07:27Z 2010-02-26T15:07:27Z

<p>Dear Mariano, I can't characterize the extensions in your interesting question, but here is a class of examples.</p> <p>Take for $k$ an algebraically closed field and for $K$ any algebraically closed extension of transcendence degree one.Then if you give yourself a $k$- morphism $\sigma :E \to E'$ it extends to an endomorphism $\tilde \sigma : K \to K$, since $K$ is algebraic over $E$ and algebraically closed. This extended morphism is surjective, hence an automorphism, because $\tilde \sigma (K) \subset K$ is algebraic and $ \tilde \sigma (K)$ is algebraically closed. Since $k$ is fixed under $\tilde \sigma$ you get your result.</p> <p>I suppose this construction might be generalized somewhat.</p>