# Intermediate Fields of $\mathbb{Q}(\zeta_{16})$ and Corresponding Splitting Polynomials

I'm trying to find all intermediate fields $$\mathbb{Q} \subset E \subset \mathbb{Q}(\zeta_{16})$$ and in particular, the polynomial that each of these fields split. I know that $$\text{Gal}(\mathbb{Q}(\zeta_{16})/\mathbb{Q}) \simeq (\mathbb{Z}/16\mathbb{Z})^*$$ where $$r \in (\mathbb{Z}/16\mathbb{Z})^*$$ corresponds to the automorphism $$\zeta_{16} \mapsto \zeta_{16}^r$$. I've listed all the distinct, proper, non-trivial subgroups:

• $$\langle 7 \rangle = \{1,7\}$$
• $$\langle 9 \rangle = \{1,9\}$$
• $$\langle 15 \rangle = \{1,15\}$$
• $$\langle 3 \rangle = \{1,3,9,11\}$$
• $$\langle 5 \rangle = \langle 13 \rangle = \{1,5,9,13\}$$
• $$\langle 11 \rangle = \{1,11,15,5\}$$
• $$\{1,7,9,15\} \simeq K_4$$

The corresponding fixed fields is just a linear combination of powers of $$\zeta_{16}$$ given by elements of the subgroup. For example, the fixed field of $$\langle 3 \rangle$$ is $$\mathbb{Q}(\zeta_{16} + \zeta_{16}^3 + \zeta_{16}^9 + \zeta_{16}^{11})$$.

My main question is how to find the minimal polynomial that each of the intermediate fields split. I know the direct way is to compute powers and find a way to get them to add to 0 using the relation that $$1 + \zeta_{16} + \zeta_{16}^2 +\cdots \zeta_{16}^{15} = 0$$. This method is fine for smaller roots of unity like $$\zeta_7$$ and $$\zeta_8$$ as I've learned. However, an element like $$\zeta_{16} + \zeta_{16}^3 + \zeta_{16}^9 + \zeta_{16}^{11}$$ is cumbersome to work with, and I haven't been successful in finding some combination of powers that ultimately cancel.

• It may be simpler to think of this field as a tower of square root extensions. You do know that $\Bbb{Q}(\zeta_8)=\Bbb{Q}(i,\sqrt2)$? By the half-angle formulas $$2\cos(\pi/8)=\sqrt{2+2\cos(\pi/4)}=\sqrt{2+\sqrt2}.$$ Commented Mar 21, 2023 at 5:29
• Commented Mar 21, 2023 at 5:36
• @JyrkiLahtonen Later in the problem, I'm expected to find $2\cos(\pi/8) = \zeta_{16} + \zeta_{16}^{-1}$ using the polynomial I find. Is there another approach I can take? Commented Mar 21, 2023 at 5:45
• That's not very difficult to find, considering that$$\zeta_{16}^{\pm 1}=\cos(\pi/8)\pm i\sin(\pi/8)$$by elementary unit circle considerations. Add them together, and you get what you have there. Commented Mar 21, 2023 at 7:21
• Another approach might be to observe that as $\zeta:=\zeta_{16}$ is a zero of $x^8+1$, we also have $$\zeta^4+\zeta^{-4}=0.\qquad(*)$$ To find the minimal polynomial of $\zeta+\zeta^{-1}$ you should write the left hand side of $(*)$ as a quartic polynomial $P(x)$ evaluated at $x=\zeta+\zeta^{-1}$. If you get stuck, take a look at this old answer of mine. Anyway, you should get a quartic that has only even degree terms. You can find the roots of such polynomials with the quadratic formula. Commented Mar 21, 2023 at 8:56