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.