1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Galois Extension with intermediate field that is not Galois

  1. Nov 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Hey PF, I'm trying to find fields K<M<L such that K:L is Galois but K:M is not.

    2. Relevant equations


    3. The attempt at a solution
    My first idea was let K=Q the field of rational numbers and c be a primitive 6th root of unity, so then Q<Q(c^4)<Q(c). Q:Q(c) Is galois, and I'm hoping that Q<Q(c^4) is not. Then again, would c^4 be a third primitive root of unity? If so then Q<Q(c^4) would be Galois I believe. Right?

    Can anyone help me find something that would work?
     
  2. jcsd
  3. Nov 22, 2016 #2

    fresh_42

    Staff: Mentor

    As long as you deal with powers of two, you will probably find a lot of conjugates. This means that with one root, the other one is found by an automorphism. What if you consider a polynomial with two kinds of roots: a conjugate pair and another one? Which power comes to mind in such a case?
     
  4. Nov 23, 2016 #3
    Powers of 3 come to mind. A splitting field for x^3-1=(x-1)(x^2+x+1). However if c is a primitive third root of unity then Q(c)=Q(c^2) so I do not think that this example will help me find an extension with a non-Galois intermediate field. Perhaps Q(ac) where a is the third real root of some integer n and c is a primitive third root of unity, then Q<Q(a)<Q(ac) where Q:Q(ac) is Galois and Q:(a) is not. the minimal polynomial of ac over Q will be x^3-a, and this polynomial will split in Q(ac).
    But the minimal polynomial of a over Q is also x^3-a and this polynomial will not split in Q(a).

    Are my statements correct?
     
  5. Nov 23, 2016 #4

    fresh_42

    Staff: Mentor

    Would have been easier to read if you just said ##2## instead of ##a## and didn't introduce ##n## because you confused yourself ("a is the third real root of some integer n" and then "x^3-a"), but, yes, you're correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Galois Extension with intermediate field that is not Galois
  1. Galois fields (Replies: 3)

  2. Galois Extensions (Replies: 1)

  3. Galois extensions (Replies: 7)

Loading...