What is the Galois group of x^4+1 over Q when e^ipi/4 is adjoined?

[Daveyboy]

This thing splits if we adjoin e^ipi/4.
Let \zeta=e^ipi/4 =\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}
so x⁴+1=

(x-\zeta)(x-\zeta²)(x-\zeta³)(x-\zeta⁴).

Then I want to permute these roots so the Galois group is just S₄.

But, Q(\zeta)=Q(i,\sqrt{2}) and [Q(i,\sqrt{2}):Q]=4 (degree)

I have the theorem that Galois group \leq degree of splitting field over base field.

Since |S₄|=24 something is wrong, but what I can not find what is wrong with the logic.

 

[Dick]

This isn't really in my field of expertise, but zeta, zeta^2, zeta^3 and zeta^4 aren't algebraically independent over the rationals, now are they? If you choose for example to map zeta->zeta^2, you really don't have any choices about how to map the other roots. The symmetry group of the automorphisms is not S4. I think that's the answer. Do you agree?

 

[VKint]

The Galois group is actually \mathbf{V}, the Klein four-group. You know that the Galois group has to have order 4, since the extension \mathbb{Q}(\zeta) is Galois over \mathbb{Q}. There are only two isomorphism types for groups of order four, i.e., the Klein four-group and the cyclic group of order four. You can distinguish between them by the fact that \mathbf{V} has two elements of order 2, while C_4 only has one. Thus, all you need to do is construct two elements of the Galois group having order 2. In any extension involving complex numbers, you know that complex conjugation is an automorphism of order two. To get another one, invoke the theorem that says that the Galois group acts transitively on the roots of any irreducible polynomial.

Edit: \mathbf{V} actually has three elements of order 2. Constructing two of them still suffices, however, since C_4 only has one.

 

[Daveyboy]

Nice, conjugate, that makes sense. I ruled out Klein group to quickly!