Question

4) Suppose that u e R. Then u is said to be constructible if there exists a sequence F0, FI, . . . , Fk of subfields of R so that F。= Q, u e Fe and [F, :F,-1] = 2 for i = 1, k. (a) Show that V2+V1+ v5 is constructible. b) Show that cos(/9) is not constructible.

Answer 1

a) $x=\sqrt{2+\sqrt{1+\sqrt{5}}}\Rightarrow x^2=2+\sqrt{1+\sqrt{5}}$

So $(x^2-2)=\sqrt{1+\sqrt{5}}\Rightarrow (x^2-2)^2=1+\sqrt{5}$

So $((x^2-2)^2-1)^2=5$ which is a polynomial of degree $2\times 2\times 2=8$

b) If $\cos(\pi/9)$ is constructible then so is

sin (π/9)

$z=\cos\left(\frac{\pi}{9} \right )+i\sin\left(\frac{\pi}{9} \right )$

Then $z^9=\cos\left(\frac{\pi}{9}\times 9 \right )+i\sin\left(\frac{\pi}{9}\times 9 \right )$

$z^9=\cos(\pi)+i\sin(\pi )=-1$

Meaning $z^9+1=0$ so $(z^3+1)(z^6-z^3+1)=0$

The minimal polynomial is $z^6-z^3+1=0$

As the minimal polynomial of $z=\cos\left(\frac{\pi}{9} \right )+i\sin\left(\frac{\pi}{9} \right )$ is of order 6, the minimal polynomial of $\cos(\pi/9)$ cannot be a power of 2. So this means $\cos(\pi/9)$ is not constructible