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

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!

• Related. A point being that you might have avoided the mistakes pointed out by KCd by writing down the generators in simpler forms. The quadratic extensions are $\Bbb{Q}(i)$, $\Bbb{Q}(\sqrt2)$ and $\Bbb{Q}(\sqrt{-2})$. All subfields of $\Bbb{Q}(\zeta_8)$ actually. There is the real subfield $\Bbb{Q}(\cos(\pi/8))=\Bbb{Q}(\sqrt{2+\sqrt2})$. Well done identifying the last field as $\Bbb{Q}(\zeta-\zeta^{-1})$. Commented Aug 14, 2023 at 5:56
• Anyway, four out of the six intermediate fields are subfields of $\Bbb{Q}(\zeta_8)$, so first working out the eighth cyclotomic field seems to be a logical starting point. Commented Aug 14, 2023 at 5:58
• Above $\zeta-\zeta^{-1}=i\sqrt{2-\sqrt2}=2i\sin(\pi/8)$. Why would that field contain $\zeta^4=i$? Commented Aug 14, 2023 at 6:03
• I can't shake the feeling that you had seen the lattice of subgroups of the elementary 2-abelian group of order 8, and wanted to match that willy-nilly :-) Commented Aug 14, 2023 at 6:20
• @JyrkiLahtonen I drew two extra vertical lines in the lattice. One is on the left and the other is on the right. Would everything else be correct if those are removed? Commented Aug 14, 2023 at 15:26

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 $$2$$.
• The main point was the extraneous inclusions, an obvious +1. More discussion about the choice of $5$ vs. $3$ as a generators. May be you can add your reasons to that question as well? That would be welcome! Commented Aug 14, 2023 at 6:16
• @JyrkiLahtonen the reasons you gave in that other post are what I would have said (mainly the 2nd and 3rd points you make), so I don't have anything new to add there. I've never heard anyone ask before about using $3$ instead of $5$ to describe generating sets of units modulo powers of $2$, so the premise that using $3$ looks more "logical" than using $5$ is completely unpersuasive.
• Yes, there are two extra vertical lines in the lattices that shouldn't be there. Would everything else be correct if those are removed? Also, why are $\zeta\mapsto\zeta^{-1}$ and $\zeta\mapsto\zeta^5$ much nicer generators of $(\mathbb Z/16)^\times)$? They all look similar to me. Commented Aug 14, 2023 at 15:25