Subfield Lattice for $\mathbb Q(\zeta_{16})$

Question

Let $\zeta_{16}=e^{2\pi i/16}$, then we know that its minimal polynomial is $\Phi_{16}=x^8+1$. Furthermore, $$ \operatorname{Gal}(\mathbb Q(\zeta_{16})/\mathbb Q)=(\mathbb Z/16)^\times=C_2\times C_4. $$ We can choose the generators of $C_2\times C_4$: take $(1,0)$ to be $\zeta\mapsto\zeta^7$ and $(0,1)$ to be $\zeta\mapsto\zeta^3$. Here, any automorphism of $\mathbb Q(\zeta_{16})$ is determined by the image of $\zeta$.

Now I want to draw the lattice of subfields of this extension, which corresponds to the subgroup lattice of $C_2\times C_4$. Here is my result.

I have two questions:

1. Are these lattices correct?

2. I basically used brute force to search for elements that are fixed under the various subgroups and calculate their minimal polynomials, and then use the degrees of extensions to deduce that they're the correct subfield. But this method is laborious. Is there an easier algorithm?

Any help would be appreciated. Thank you in advance!

Answer 1

Why did you choose the generators of $(\mathbf Z/16\mathbf Z)^\times$, regarded as $C_2 \times C_4$, so that $(1,0)$ is $\zeta \mapsto \zeta^7$ and $(0,1)$ is $\zeta \mapsto \zeta^3$?

The nicest cyclic decomposition of $(\mathbf Z/16\mathbf Z)^\times$ is $\langle -1 \bmod 16\rangle \times \langle 5 \bmod 16 \rangle$, with the first subgroup being cyclic of order $2$ and the second subgroup being cyclic of order $4$. Thus it seems most natural to let $(1,0)$ correspond to $\zeta \mapsto \zeta^{-1}$ and $(0,1)$ correspond to $\zeta \mapsto \zeta^5$.

Your upside-down subgroup diagram for $C_2 \times C_4$ has mistakes. As an example, the cyclic subgroup $\langle (0,1)\rangle$ with order $4$ has just one subgroup of order $2$, but you wrote two subgroups of $\langle(0,1)\rangle$ with order $2$. As another example, $\langle (1,1)\rangle$ is cyclic of order $4$, so it too has just one subgroup of order $2$, but you gave $\langle (1,1)\rangle$ two subgroups of order $2$.