Let $\mathbb Q(\alpha)/\mathbb Q$ be an algebraic exntension with $\alpha=\sqrt{2+\sqrt{2}}$.
1) Show that the extension is a galois extension (normal and separable)
2) Show that $Gal(\mathbb Q(\alpha)/\mathbb Q)\cong \mathbb Z_4$
3) Find all fields between $\mathbb Q$ and $\mathbb Q(\alpha)$
My attempt:
1) One can see that $f(x)=(x^2-2)^2-2=X^4-4x^2+2$ is a polynomial with $f(\alpha)=0$. By Eisenstein's theorem with p=2 it is irreducible over $\mathbb Q$, hence its the minimal polynomial of $\alpha$.
The roots of $f$ are $z_{1,2}=\pm\sqrt{2+\sqrt{2}}$ and $z_{3,4}=\pm\sqrt{2-\sqrt{2}}$. So the splitting field of $f$ is $\mathbb Q(z_1,z_2,z_3,z_4)$.But $z_2,z_3$ and $z_4$ can be written in terms of $z_1$:
$z_2=0-z_1$, $z_3=\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}$, $z_4=-z_3$, hence we have actually $\mathbb Q(z_1,z_2,z_3,z_4)=\mathbb Q(z_1)=\mathbb Q(\alpha)$.
We have shown that $\mathbb Q(\alpha)$ is normal and separabel.
2) We have $[\mathbb Q(\alpha):\mathbb Q]=deg(f)=4$. So its sufficient to find 4 automorphisms with: 1 automorphism has order 2, 2 have order 4 and 1 has order 1.
$id: z_i \mapsto z_i$
$\tau_1: z_1 \mapsto z_3$
with $\tau_1(\sqrt{2})=\tau_1(\sqrt{2})\tau_1(\sqrt{2})-2$ and $\sqrt{2\pm\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2\mp\sqrt{2}}}$ all the other images are determined by the image of $z_1$, we get $z_2\mapsto z_4$, $z_3\mapsto z_2$, $z_4\mapsto z_1$
Ans similarly:
$\tau_2:z_1 \mapsto z_4$, $\tau_2:z_2 \mapsto z_3$, $\tau_2:z_3 \mapsto z_1$, $\tau_2:z_4 \mapsto z_2$
$\tau_3:z_1 \mapsto z_2$, $\tau_3:z_2 \mapsto z_1$, $\tau_3:z_3 \mapsto z_4$, $\tau_3:z_4 \mapsto z_3$
Because of $ord(id)=1, ord(\tau_{1,2})=4$ and $ord(\tau_3)=2$, we have $Gal(\mathbb Q(\alpha)/\mathbb Q)\cong \mathbb Z_2$
3) Only the generated subgroup of $\overline{2}\in \mathbb Z_2$ is not trivial. Its corresponding automorphism is $\tau_3$ which fixes $2+\sqrt{2}$ and $2-\sqrt{2}$, hence a non-trivial field is $\mathbb Q(\sqrt{2})$.
I hope that someone can go trough it and look for mistakes or tips. I am open for any other idea :)
Especially in 3) I am a bit unsure whether my solution is correct or not.
You need more detail in (1) about how you write $z_3$ in terms of $z_1$. The most important point is that $\sqrt{2} \in \mathbb{Q}(z_1)$.
In (2), you just need to show that there is an automorphism $\tau$ with $\tau^2 \ne \operatorname{id}$. So you just need to check that $\tau_1(z_3) \ne z_1$. I don't understand your calculation there. To begin with, $\tau_1(\sqrt{2}) \ne \tau_1(\sqrt{2})\tau_1(\sqrt{2})-2$.
However, the relations $\sqrt{2} = z_1^2 - 2$ and $-\sqrt{2} = z_3^2 - 2$ show that $\tau_1(\sqrt{2}) = -\sqrt{2}$. Now the relation $z_1 z_3 = \sqrt{2}$ should allow you to conclude that $\tau_1(z_3) \ne z_1$.
For (3), your argument isn't wrong, but since there is exactly one intermediate subgroup in $\mathbb{Z}_4$, there is exactly one intermediate field. All you need to do is identify it. Since $\mathbb{Q}(\sqrt{2})$ is a subfield of $\mathbb{Q}(\alpha)$ of degree $2$ over $\mathbb{Q}$, it has to be the intermediate field you're looking for.