I want to find the intermediate extension of $\Bbb Q(\zeta_8)$ over $\Bbb Q$. And for each subfield, I need to find what is the correspondence subgroup of $(\Bbb Z/\Bbb Z_8)^*$ that fixes the field.
The following is what I have tried so far:
- Find the degree of $[\Bbb Q(\zeta_8):\Bbb Q]$.
I calculated the minimial polynomial by the definition of cyclotomic polynomial. It is $x^4+1$.
Knowing this, I find the basis of $\Bbb Q(\zeta_8)/\Bbb Q$ is: $1,\zeta,\zeta^2,\zeta^3$.
$Gal(\Bbb Q(\zeta_8)/\Bbb Q)=\lbrace \sigma_1,\sigma_2, \sigma_3,\sigma_4\rbrace$ where
$\sigma_1:\zeta\mapsto \zeta$
$\sigma_2:\zeta\mapsto \zeta^3$
$\sigma_3:\zeta\mapsto \zeta^5$
$\sigma_4:\zeta\mapsto \zeta^7$
- I want to determine the fixed field via the basis:
$\sigma_3(a+b\zeta+c\zeta^2+d\zeta^3)=a+b\zeta^5+c\zeta^2+d\zeta^7$.
So $\sigma_3$ fixes $\Bbb Q(\zeta^2)=\Bbb Q(i)$.
$\sigma_2(a+b\zeta+c\zeta^2+d\zeta^3)=a+b\zeta^3+c\zeta^6+d\zeta$.
So $\sigma_2$ fixes $\Bbb Q(\alpha)$ where $\alpha$ is of form $b=d,c=0$. Which is $\Bbb Q(i\sqrt{2})$.
Now $\sigma_4(a+b\zeta+c\zeta^2+d\zeta^3)=a+b\zeta^7+c\zeta^6+d\zeta^5$.
I am confused, how could I deal with that?
Thanks for help!
As $\sigma_4^2$ is the identity, any $\alpha+\sigma_4(\alpha)$ lies in the fixed field of $\alpha$. In particular, $\zeta+\sigma_4(\zeta)=\zeta+\zeta^7=\zeta+\zeta^{-1} =e^{\pi i/4}+e^{-\pi i/4}=2\cos(\pi/4)$. What is that?