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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?

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  • $\begingroup$ Just nitpicking, but there is no such thing as the $7$-th primitive root of unity, there are many. $\endgroup$ – Servaes Jun 1 '14 at 19:01
  • $\begingroup$ You're right ;) $\endgroup$ – user72870 Jun 1 '14 at 19:02
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    $\begingroup$ And the Galois theory stuff is relatively simple in this case; what does $\operatorname{Gal}(\Bbb{Q}(\zeta_7)/\Bbb{Q})$ look like? $\endgroup$ – Servaes Jun 1 '14 at 19:02
  • $\begingroup$ I know that it is simple using Galois theory. My teacher asked me the question at the exam, but we didn't study Galois theory in that course, so I'd like to know if there is an other way to solve the problem. Now I know Galois but it's too late :) $\endgroup$ – user72870 Jun 1 '14 at 19:10
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    $\begingroup$ Look at the equation $x^6+x^5+x^4+ x^3+x^2+x+1=0$. The polynomial is palindromic. Rewrite the equation as $x^3+x^2+x+1+x^{-1}+x^{-2}+x^{-3}=0$, and let $u=x+x^{-1}$. $\endgroup$ – André Nicolas Jun 1 '14 at 19:18
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Note that if $\omega = \zeta_{7} + \zeta_{7}^{-1}$, then $$ \omega^{3} + \omega^{2} - 2 \omega - 1 = 0.\tag{three} $$


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).

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    $\begingroup$ The question is how did you determine this? Did you use Galois theory to find the element and the corresponding minimal polynomial, or did you use another method. $\endgroup$ – Marc Jun 1 '14 at 19:10
  • $\begingroup$ @JyrkiLahtonen, "+1" to you, please make it into an answer. $\endgroup$ – Andreas Caranti Jun 1 '14 at 19:18
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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 $$ [K:\Bbb{Q}]=\frac{[\Bbb{Q}(\zeta_7):\Bbb{Q}]}{[K(\zeta_7):K]}=\frac62=3. $$

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    $\begingroup$ Of course Galois theory tells us immediately that the intersection $\Bbb{Q}(\zeta_7)\cap\Bbb{R}$ is a cubic extension of the rationals. In that sense this is `cheating'. But to an extent I feel that this does answer OP's question. After all, those cosines may have been looked at the same time as the roots of unity. Or they will be when I get a chance to lecture Galois theory in two years time :-) $\endgroup$ – Jyrki Lahtonen Jun 1 '14 at 19:33
  • $\begingroup$ By the way, this is the starting point for Gauss' construction of the 17-gon with ruler and compass. He didn't know Galois theory, of course. $\endgroup$ – egreg Jun 1 '14 at 19:38
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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.

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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.

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