Let $~z=e^{\frac{2\pi i}{7}}~$ and $~\theta =z+z^2+z^4~$. Then...

Question

Let $~z=e^{\frac{2\pi i}{7}}~$ and $~\theta =z+z^2+z^4~$. Then which of following are correct?

$1. ~~~ \theta \in \mathbb{Q}$

$2. ~~~\theta \in \mathbb{Q}(D)$ for some $D>0$

$3. ~~~\theta \in \mathbb{Q}(D)$ for some $D<0$

$4. ~~~\theta \in i~\mathbb{R}$

I don't know how to start, I didn't got any progress with moreras theorem. I think required some extension field result

Answer 1

$$\theta^2=(z+z^2+z^4)^2 = z^2+z^4+z^8+2z^3+2z^5+2z^6 = z+z^2+2z^3+z^4+2z^5+2z^6$$ $$ \theta^2+\theta = 2(z+z^2+z^3+z^4+z^5+z^6) = 2\left[\frac{z^7-1}{z-1}-1\right] = -2 \tag{A}$$ where $(A)$ implies $\theta=\frac{-1\pm\sqrt{-7}}{2}$. It is not difficult to prove that $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ is positive, hence the imaginary part of $\theta$ is positive and $\theta=\frac{-1+i\sqrt{7}}{2}$.

Answer 2

Hints:

1. Define $\digamma(z)=z^2$, let $\Phi_7$ be the 7th cyclotomic polynomial, and for any endofunction $f$ write $f^{\circ n}$ for $n$-fold composition. Then

$$\begin{align}\digamma^{\circ 3}(z)&=z + z(z-1)\Phi_7(z)\\ \theta(\digamma(z))&=\theta(z)+z(z-1)\Phi_7(z)\text{.} \end{align}$$

2. You can look up the Galois group, Hasse diagram of intermediate fields, and other number-theoretic facts about $\mathbb{Q}(\zeta_7)$ at the LMFDB.

To solve this problem you need to know:

1. what algebraic field extension $\mathrm{e}^{2\pi \mathrm{i}/N}$ generates over $\mathbb{Q}$,
2. the Galois group of the field $\mathbb{Q}(\zeta_N)$ over $\mathbb{Q}$ as well as its action on the generator $\zeta_N$,
3. the Fundamental Theorem of Galois Theory, and
4. the classification of the subgroups of cyclic groups.

You should find that $\theta\in\mathbb{Q}(\sqrt{-7})$.

One can also show directly that $\theta(z)$ satisfies the quadratic equation $\theta(z)^2+\theta(z)+2=(z^2-z+2)\Phi_7(z)$. However, to know in advance that such a relation exists, you really need to use Galois theory.