# Homework Help: Galois Extension with intermediate field that is not Galois

1. Nov 22, 2016

### PsychonautQQ

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. Nov 22, 2016

### 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?

3. Nov 23, 2016

### PsychonautQQ

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?

4. Nov 23, 2016

### 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.

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