# Thread: Ruler and compass construction 7-gon

1. ## 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

2. Originally Posted by Niall101
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 $\displaystyle \phi(7) = 6$ is not a power of two.

3. Ok thanks a million for the reply. I have the rest of my notes here so I will try and make sense of that.

4. 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 )

5. Originally Posted by Niall101
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 $\displaystyle 7$-gon is constructible then the central angles are constructible. Thus, $\displaystyle 2\pi/7$ is a constructible angle. But this means $\displaystyle \cos (2\pi/7)$ is a constructible number. Note that $\displaystyle \tfrac{1}{2}(w+w^{-1}) = \cos (2\pi /7)$. Thus, $\displaystyle \cos (2\pi /7) \in \mathbb{Q}(w)$ and $\displaystyle \mathbb{Q}(\cos(2\pi/7)) \not = \mathbb{Q}(w)$. Since $\displaystyle w$ is root of $\displaystyle x^2 - 2\cos(2\pi/n)x+1 \in \mathbb{Q}(\cos(2\pi/7))$ it means $\displaystyle [\mathbb{Q}(w):\mathbb{Q}(\cos(2\pi/7))]=2$. But since $\displaystyle \cos (2\pi/7)$ is assumed constructible it means $\displaystyle [\mathbb{Q}(\cos(2\pi/7)):\mathbb{Q}] = 2^n$ i.e. power of two. Therefore, $\displaystyle [\mathbb{Q}(w):\mathbb{Q}] = [\mathbb{Q}(w):\mathbb{Q}(\cos(2\pi/7))][\mathbb{Q}(\cos(2\pi /7)):\mathbb{Q}] = 2\cdot 2^n = 2^{n+1}$ i.e. a power of two. But it should be a known fact that $\displaystyle [\mathbb{Q}(w):\mathbb{Q}] = \phi (7) = 6\not = 2^m$. We have a contradiction.

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