Cyclotomic fields: Determining the fixed field

Question

Let $K=\mathbb Q$ and $L=\mathbb Q(\zeta_8)$, where $\zeta_8$ is a primitve $8th$ root of unity.

I have to determine the Galois group of this extension, the subgroups of it and the associated intermediate fields.

It was not hard to determine the Galois group. We have $Gal(L/K)\cong\mathbb Z_8^{*}\cong\mathbb Z_2\times \mathbb Z_2$

So the Galois group has exactly $4$ subgroups, namely the generated groups of the following automorphisms:

$\sigma_1(\zeta)=\zeta$

$\sigma_2(\zeta)=\zeta^3$

$\sigma_3(\zeta)=\zeta^5$

$\sigma_4(\zeta)=\zeta^7$

So the non trivial subgroups are $<\sigma_2>,<\sigma_3>\text{ and }<\sigma_4>$.

Now its not hard to see that $\mathbb Q(\zeta^1+\zeta^3)\subseteq L^{<\sigma_2>}$

But if we want to show equality, we have to show that $[\mathbb Q(\zeta^1+\zeta^3):\mathbb Q]=2$

Does the following argument work?: $\zeta^1+\zeta^3$ is real if and only if $\overline{\zeta^1}=\zeta^3$, but $\overline{\zeta^1}=\zeta^7$, so $\zeta^1+\zeta^3\in \mathbb C/\mathbb R$, hence the degree over $\mathbb Q$ must be greater than $1$, so it must be $2$.

Is this okay?

However, what we should do with the fixed field of $\sigma_4$?. We know that $\mathbb Q(\zeta^1+\zeta^7)\subseteq L^{<\sigma_4>}$ But the generator is real. How we can show equality?

Answer 1

You have made a good start, so I give some extended hints. Ask for more if needed.

For handling $\zeta^1+\zeta^7$ it may be best to use the explicit formula (and Moivre) $$\zeta=e^{2\pi i/8}=\cos\frac\pi4+i\sin\frac\pi4=\frac{1+i}{\sqrt2}.$$

Actually using Moivre will give you another useful viewpoint to the fixed field of $\langle\sigma_2\rangle$ as well :-)

With $\zeta^5$ (or $\langle \sigma_3\rangle$) the sum $\zeta^1+\zeta^5$ won't help you because it vanishes. What about the product $\zeta^5\zeta^1$?