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.

  • $\begingroup$ A minor simplification for 2): since there are only 2 groups of order 4, it suffices to check there is a single element of order 4. $\endgroup$ – Kimball Jan 23 '15 at 16:35
  • $\begingroup$ exactly, thanks :) did you find anything else? is 3) correct? $\endgroup$ – Marc Jan 23 '15 at 16:59

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.

  • $\begingroup$ Thanks for the answer. For 1): I did not make it clear in my solution above, but the equation $z_3=\sqrt{2+\sqrt{2}}=\sqrt{2}\cdot (\sqrt{2-\sqrt{2}})^{-1}$ already yields, that $\sqrt{2}\in \mathbb Q(z_1,z_2,z_3,z_4)$. For 2) thats a good idea, because there are just 2 possibilities for the galois group if the order is 4 and in $\mathbb Z_2 \times \mathbb Z_2$ every element has order 2 or 1. But sometimes we have to determine all the elements of the galois group, so I want to know if my solution for 2) is correct, also because I want to learn about this topic more (I just started with it) $\endgroup$ – Marc Jan 23 '15 at 23:02
  • $\begingroup$ $(2)$ contains the error I mentioned in my answer. Once you know the action of $\tau_1$ on the roots, you can easily find the action of the other automorphisms, which are $\operatorname{id}$, $\tau_1^2$ and $\tau_1^3$. It is just a matter of calculating powers of a four-cycle. $\endgroup$ – user208259 Jan 23 '15 at 23:56

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