Extensions of fields with lots of symmetry

[Cute question heard elsewhere]

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 $\tilde\sigma|_E=\sigma$?

Normal extensions and those without proper subextensions have that property. On the other hand, $\mathbb Q(\sqrt[4]{2})/\mathbb Q$ doesn't.

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I always thought that the only subextension of Q(2^{1/4})/Q was Q(2^{1/2}). Am I wrong? –  Kevin Buzzard Feb 26 '10 at 14:33
$E$ can coincide with $E'$. –  Mariano Suárez-Alvarez Feb 26 '10 at 14:39
Touch\'e ;-) –  Kevin Buzzard Feb 26 '10 at 14:46

1 Answer

Dear Mariano, I can't characterize the extensions in your interesting question, but here is a class of examples.

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.

I suppose this construction might be generalized somewhat.

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Why is $K$ algebraic over $E$? –  Mariano Suárez-Alvarez Feb 26 '10 at 15:13
Take any x in E but not in k. Then x is transcendental over k, because k is algebraically closed. Since K has transcendence degree 1 over k, the singleton {x} is a transcendence basis of K over k and so K is algebraic over k(x), and a fortiori K is algebraic over E. –  Georges Elencwajg Feb 26 '10 at 15:25
Ah. I'd missed the fact that $k$ is algebraically closed. Thanks! –  Mariano Suárez-Alvarez Feb 26 '10 at 16:52