Is This Field Extension Galois?

[thomtyrrell]

The claim is:

Let K be a subfield of the complex numbers, and let X be a finite set of complex numbers. If the elements of X are permuted by any automorphism of the complex numbers that fixes K, then the field obtained by adjoining the elements of X to K is a finite, Galois extension of K.

The definitions of Galois extension that I have learned do not seem to yield an easy proof. Any ideas?

 

[DeadWolfe]

What definition are you using? Show that this extension is normal is really not so bad. (hint: start with any polynomial which has a root, and find an automorphism taking it to other roots...)

 

[gel]

First, you need to show that every element of X is algebraic over K.
if x was not algebraic over K, then there would be an automorphism taking it to x+a (any a in K), so X would have to contain {x+a:a in K}, which is infinite.

 

[thomtyrrell]

Ah, I believe it makes sense now...

1. The extension is finite for otherwise there would be an infinite number of automorphisms of K(X) that fix K (Correct me if I'm wrong, but one way to see this would be to pick a basis for K(X) over K, and then determine an automorphism by specifying a permutation of the basis and extending that to K(X) by "linearity"). Extending these to $$\mathbb{C}$$, we would have an infinite number of automorphisms of $$\mathbb{C}$$ that fix K and all act differently on K(X), which is a contradiction.
2. The extension is separable as K is perfect (characteristic 0)
3. (Normality) If f is an irreducible polynomial with coefficients in K and a root in K(X), we can construct an automorphism of the complex numbers that fixes K and maps that root to any other root of f. Then, by assumption, it follows that all the roots of f lie in K(X).

Let $$\alpha$$ be the root of f in K(X) and let $$\beta$$ be another root of f (not necessarily in K(X)). To construct such an automorphism, we could, for instance, start by extending the identity on K to an isomorphism of $$K(\alpha)$$ and $$K(\beta)$$ which maps $$\alpha$$ to $$\beta$$, extend that iso. to an automorphism of the splitting field of f, and then extend that to an automorphism of $$\mathbb{C}$$

I was unaware that we could extend any automorphism of a subfield of $$\mathbb{C}$$ to an automorphism of $$\mathbb{C}$$; that seemed to be what was holding me back. Thanks for the suggestions.

 

[gel]

Yes - except for this bit

(Correct me if I'm wrong, but one way to see this would be to pick a basis for K(X) over K, and then determine an automorphism by specifying a permutation of the basis and extending that to K(X) by "linearity").

Permutations of a basis set do not extend to automorphisms in many (most) cases.
If K->K(X) was infinite, then K->K(x) would be infinite for some x in X. Then see my previous post to contruct an infinite number of automorphisms of K(x) fixing K.

 

[thomtyrrell]

Indeed. Thanks again for your help.