Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

[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.

share|improve this question
    
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
1  
$E$ can coincide with $E'$. –  Mariano Suárez-Alvarez Feb 26 '10 at 14:39
    
Touch\'e ;-) –  Kevin Buzzard Feb 26 '10 at 14:46
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.