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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.

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  • $\begingroup$ 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 $$2\cos(\pi/8)=\sqrt{2+2\cos(\pi/4)}=\sqrt{2+\sqrt2}.$$ $\endgroup$
    – Jyrki Lahtonen
    Commented Mar 21, 2023 at 5:29
  • $\begingroup$ Related. $\endgroup$
    – Jyrki Lahtonen
    Commented Mar 21, 2023 at 5:36
  • $\begingroup$ @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? $\endgroup$
    – Dalop
    Commented Mar 21, 2023 at 5:45
  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Commented Mar 21, 2023 at 7:21
  • $\begingroup$ 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. $\endgroup$
    – Jyrki Lahtonen
    Commented Mar 21, 2023 at 8:56

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