$\mathbb{Q}(\zeta_7)$ subextension of degree $3$

Question

Let $\zeta_7$ be a $7$-th primitive root of unity. Is there a way to determine a subextension of $\mathbb{Q}(\zeta_7)$ that has degree $3$, without making use of Galois theory stuff?

Answer 1

Note that if $\omega = \zeta_{7} + \zeta_{7}^{-1}$, then $$ \omega^{3} + \omega^{2} - 2 \omega - 1 = 0.\tag{three} $$

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Addendum

To address the question in the comment, set $z = \zeta_{7}$. You have $$ z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = 0. $$ You want some combination of powers of $z$ to satisfy a polynomial of degree $3$. So rewrite this as $$ z^3 + z^2 + z + 1 + z^{-1} + z^{-2} + z^{-3} = 0.\tag{egal} $$ It is tempting to take $\omega = z + z^{-1}$, so that $$ \omega^{3} = z^{3} + 3 z + 3 z^{-1} + z^{-3}. $$ Now there is only one choice of the coefficients of $\omega^{2}, \omega, 1$ to get (egal) from this, and this leads to (three).

Answer 2

A complementary point of view to the answer by Andreas Caranti is to observe that the number $\omega=\zeta_7+\zeta_7^{-1}\in\mathbb{Q}(\zeta_7)$ is real (actually $\omega=2\cos(2\pi/7)$). Thus $K=\Bbb{Q}(\omega)$ is a proper subfield of $\Bbb{Q}(\zeta_7)$. But also the polynomial $$ p(x)=(x-\zeta_7)(x-\zeta_7^{-1})=x^2-\omega x+1 $$ has its coefficients in the subfield $K$. Thus it has to be the minimal polynomial of $\zeta_7$, and $[K(\zeta_7):K]=2$. Thus the tower of extensions $\Bbb{Q}\subset K\subset \Bbb{Q}(\zeta_7)=K(\zeta_7)$ implies that $$ [K:\Bbb{Q}]=\frac{[\Bbb{Q}(\zeta_7):\Bbb{Q}]}{[K(\zeta_7):K]}=\frac62=3. $$

Answer 3

We give trigonometric identities a workout.

Let $\theta=\frac{2\pi}{7}$. Note that $\cos(3\theta)=\cos(2\pi-3\theta)=\cos(4\theta)$.

Now use the fact that $\cos(3\theta)=4\cos^3\theta-3\cos\theta$ and $\cos(4\theta)=2(2\cos^2\theta-1)^2-1$.

Setting $t=\cos\theta$ we get $8t^4-4t^3-8t^2 +3t+1=0$. This has the uninteresting root $t=1$. Divide. We get $8t^3+4t^2-4t-1=0$. The polynomial is an irreducible cubic.

Remark: The construction of the regular heptagon, like the construction of the regular nonagon (trisecting the $60^\circ$ angle) requires the construction of a root of an irreducible cubic. Of course it cannot be done by straightedge and compass, but for example similar verging constructions work.

Answer 4

This answer is incompletely justified and theorised, and may depend on Galois Theory stuff. Note that $7-1=6$ is divisible by $3$, so an extension of degree $3$ is possible.

The non-zero cubes modulo $7$ are $1, 8\equiv 1, 27\equiv -1, 64 \equiv 1, 125 \equiv -1, 216 \equiv -1$ - since they are $+1, -1$ we take $\zeta + \zeta ^{-1}$.

If we wanted an extension of degree $2$ we'd use the squares and get $\zeta+\zeta^2+\zeta^4$

When it comes to the construction of the $17$-gon this provides a useful mnemonic.