Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Proposition in Galois Theory

  1. Aug 9, 2008 #1
    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?
     
  2. jcsd
  3. Aug 9, 2008 #2
    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...)
     
  4. Aug 9, 2008 #3

    gel

    User Avatar

    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.
     
  5. Aug 9, 2008 #4
    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 [tex]\mathbb{C}[/tex], we would have an infinite number of automorphisms of [tex]\mathbb{C}[/tex] 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 [tex]\alpha[/tex] be the root of f in K(X) and let [tex]\beta[/tex] 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 [tex]K(\alpha)[/tex] and [tex]K(\beta)[/tex] which maps [tex]\alpha[/tex] to [tex]\beta[/tex], extend that iso. to an automorphism of the splitting field of f, and then extend that to an automorphism of [tex]\mathbb{C}[/tex]

    I was unaware that we could extend any automorphism of a subfield of [tex] \mathbb{C} [/tex] to an automorphism of [tex] \mathbb{C} [/tex]; that seemed to be what was holding me back. Thanks for the suggestions.
     
  6. Aug 9, 2008 #5

    gel

    User Avatar

    Yes - except for this bit
    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.
     
  7. Aug 10, 2008 #6
    Indeed. Thanks again for your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A Proposition in Galois Theory
  1. Galois Theory (Replies: 1)

  2. Galois Theory (Replies: 2)

Loading...