# All intermediate sub extensions of $\mathbb{Q} \subseteq K \subseteq \mathbb{Q}(\zeta_8)$.

I know there is a similar question posted on Stack Exchange, however it deals with periods, and I do not understand the solutions provided.

I know that the Galois Group of the field extension $\mathbb{Q}(\zeta_8)/\mathbb{Q}$ is $\mathbb{Z}_2 \times \mathbb{Z}_2$. To find the intermediate field extensions, we look at the subgroups of $\mathbb{Z}_2 \times \mathbb{Z}_2$. These are: $\{ (0,0) \}, \{(0,0), (0,1) \}, \{(0,0), (1,0) \}, \{(0,0), (1,1) \}$ and $\{(0,0), (0,1), (1,0), (1,1) \}$. Furthermore, I'm aware that $\zeta_8$ has minimal polynomial $x^4+1$, which has roots $\pm \zeta_4$ and $\pm \zeta_8$. I need help determining the explicit correspondence between the subgroups and sub extensions.

• I think it might help to identify explicitly what your field is. Since one primitive eighth root of $1$ is $(1+i)/\sqrt2$, you see that its square root is $i$ and the sum of it and its complex conjugate is $\sqrt2$. So you can just work with the field $\Bbb Q(i,\sqrt2\,)$, very easy. – Lubin May 15 '16 at 1:43

You can do this ad hoc. Note we know all the automorphisms already

$$\begin{cases} \zeta_8\mapsto\zeta_8 \\ \zeta_8\mapsto \zeta_8^{-1} \\ \zeta_8\mapsto \zeta_8^{3} \\ \zeta_8\mapsto\zeta_8^5\end{cases}.$$

The first one is the identity which fixes the whole field, so that's not our issue and the whole group clearly fixes the base field so that's not an interesting case either. So let us consider the three quadratic subfields.

We see $\zeta_8+\zeta_8^{-1}$ is fixed by complex conjugation, $\zeta_8\mapsto\zeta_8^{-1}$, but not $\zeta_8\mapsto \zeta_8^3$ or $\zeta_8\mapsto\zeta_8^5$ so it is the fixed field of that automorphism. Similarly $\zeta_8+\zeta_8^3$ is fixed not by complex conjugation, but is fixed by $\zeta_8+\zeta_8^3$. Finally $\zeta_8^2=\zeta_4$ is fixed by $\zeta_8\mapsto\zeta_8^5$ but conjugation makes this $\zeta_8^{6}$ and the second automorphism makes it $\zeta_8^{6}$ as well, neither of which fix it, so this generates the third fixed field. But we can do better.

• For the first one $(\zeta_8+\zeta_8^{-1})^2=\zeta_4+\zeta_4^{-1}+2=2$ so the fixed field of conjugation is just $\Bbb Q(\sqrt 2)$, which is what you expect.
• The second one is $\zeta_8(1+i)=i\sqrt 2$ by factoring out a $\zeta_8$ because squaring this gives $i\cdot 2i=-2$ and so this fixed field is $\Bbb Q(i\sqrt 2).$
• Similarly $\zeta_8^2=\zeta_4$ so the fixed field of the third automorphism is just $\Bbb Q(i)$.
• Forgive my ignorance, but how did you determine that $\zeta_8 \mapsto \zeta_8^{-1}$, $\zeta_8 \mapsto \zeta_8^3$ and $\zeta_8 \mapsto \zeta_8^5$ were the automorphisms and why $\zeta_8 \mapsto \zeta_8^2$ was not an automorphism? – user319128 May 12 '16 at 2:02
• @Elliot those are the other primitive $8^{th}$ roots of unity. Since $\Bbb Q(\zeta_8)\cong \Bbb Q[x]/(x^4+1)$ you know that any automorphism permutes the roots of this polynomial, and those maps do exactly that. Since we know there are $4$ of them, and we have demonstrated exactly $4$ those are all. And clearly $\zeta_4=\zeta_8^2$ is not a root of that polynomial: $\zeta_4^4+1=2\ne 0$. – Adam Hughes May 12 '16 at 2:16
• Thanks, can you be more explicit with your explanation of dot point 2, how you get $\mathbb{Q}(i\sqrt{2})$? – user319128 May 12 '16 at 2:30
• @Elliot I'm not sure which part you mean, I take an element, and square it to get $-2$, that means that thing is $i\sqrt 2$ up to a sign (numbers only have two square roots), and clearly the sign doesn't matter in generating the field. – Adam Hughes May 12 '16 at 2:31
• We know that $\zeta_8 + \zeta_8^3$ is fixed by the automorphism $\zeta_8 \mapsto \zeta_8^3$. Therefore, consider $(\zeta_8 + \zeta_8^3)^2 = \zeta_8^2 + \zeta_8^6 + 2\zeta_8^4 = 2\zeta_4+\zeta_2$. How do we get $i\sqrt{2}$ out of this? – user319128 May 12 '16 at 2:33