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Math Help - Ruler and compass construction 7-gon

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    Ruler and compass construction 7-gon

    I'm studying for an exam in rings and fields and I took out this page of my notes to study it the other day but I must have left in in the library. grr I can seam to find the proof of this anywhere online of construct it myself. If anyone could help It would be greatly appreciated. I am almost sure this will be on the exam.

    Thanks
    Niall
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    Quote Originally Posted by Niall101 View Post
    I'm studying for an exam in rings and fields and I took out this page of my notes to study it the other day but I must have left in in the library. grr I can seam to find the proof of this anywhere online of construct it myself. If anyone could help It would be greatly appreciated. I am almost sure this will be on the exam.
    7 gon is not constructible because it \phi(7) = 6 is not a power of two.
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    Ok thanks a million for the reply. I have the rest of my notes here so I will try and make sense of that.
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    Hey thanks I just saw that I have to use the result of

    w=exp(2*Pi*i/7) and A=w + w^6

    Finding the minimal polynomial or w over Q.

    I know this is the polynomial Q[x] of least degree which A is a root.
    Im cant really get any further than that.

    (pity cos the proof using Fermat primes is so much nicer )
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    Quote Originally Posted by Niall101 View Post
    Hey thanks I just saw that I have to use the result of

    w=exp(2*Pi*i/7) and A=w + w^6
    Solution 1: If 7-gon is constructible then the central angles are constructible. Thus, 2\pi/7 is a constructible angle. But this means \cos (2\pi/7) is a constructible number. Note that \tfrac{1}{2}(w+w^{-1}) = \cos (2\pi /7). Thus, \cos (2\pi /7) \in \mathbb{Q}(w) and \mathbb{Q}(\cos(2\pi/7)) \not = \mathbb{Q}(w). Since w is root of x^2 - 2\cos(2\pi/n)x+1 \in \mathbb{Q}(\cos(2\pi/7)) it means [\mathbb{Q}(w):\mathbb{Q}(\cos(2\pi/7))]=2. But since \cos (2\pi/7) is assumed constructible it means [\mathbb{Q}(\cos(2\pi/7)):\mathbb{Q}] = 2^n i.e. power of two. Therefore, [\mathbb{Q}(w):\mathbb{Q}] = [\mathbb{Q}(w):\mathbb{Q}(\cos(2\pi/7))][\mathbb{Q}(\cos(2\pi /7)):\mathbb{Q}] = 2\cdot 2^n = 2^{n+1} i.e. a power of two. But it should be a known fact that [\mathbb{Q}(w):\mathbb{Q}] = \phi (7) = 6\not = 2^m. We have a contradiction.

    Solution 2: Here we are assuming knowledge of Galois theory. Let K = \mathbb{Q}(w). Since \text{Gal}(K/\mathbb{Q}) \simeq \mathbb{Z}_7^{\times} it means the Galois groups is cyclic. Let \{ \theta_1,...,\theta_6 \} by the Galois group so that \theta_k (w) = w^k for 1\leq k\leq 6. It is simple to check that \theta_3 generates the Galois group. Consider the polynomial f(x) = (x - w - w^6)(x - w^3 - w^4)(x - w^2 - w^5). Note that \theta_3 leaves the polynomial fixed, therefore \left< \theta_3 \right> fixes f(x) and therefore the coefficients of f(x) lie in K^{\left< \theta_3 \right>} = K^{\text{Gal}(K/\mathbb{Q})} = \mathbb{Q}. This means f(x) \in \mathbb{Q}[x]. Since degree of w+w^6 cannot be 1 over \mathbb{Q} and it solves f(x) it means the degree of w+w^6 is a divisor of three i.e. it must be three. Thus, the minimal polynomial is f(x). Note that this not a power of two therefore the number is not constructible.
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