Exercise 9.2.10

Proof. Let $\zeta ={\zeta }_{11}=e^{2\mathit{i\pi }∕11}$, and $\eta =(2,1)=\zeta +{\zeta }^{-1}=2\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕11)$. The powers of 2 modulo 11 are $1,2,2^2=4,2^3=8,2^4=5,2^5=-1$, so the order of $[2]$ in ${(\mathbb{Z}∕11\mathbb{Z})}^{\text{∗}}$ is 10, so 2 is a primitive root modulo 11.

As ${\Phi }_{11}(\zeta )=1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4+{\zeta }^5+{\zeta }^6+{\zeta }^7+{\zeta }^8+{\zeta }^9+{\zeta }^{10}=0$, we obtain by multiplication by ${\zeta }^{-5}$ :

Write $u_n={\zeta }^n+{\zeta }^{-n}$. As

we obtain for all $n\in \mathbb{N}$

Therefore

The equality (1) gives

So $\eta$ is a root of $f=x^5+x^4-4x^3-3x^2+3x+1\in \mathbb{Q}[x]$.

By Proposition 9.2.6 (b), the fixed field $L_2$ of ${\tilde{H}}_2$ corresponding to $H_2=\{-1,1\}$ is $L_2=\mathbb{Q}(\eta )$, and $[L_2:\mathbb{Q}]=5$ by Proposition 9.2.1. (as $\mathbb{Q}\subset \mathbb{Q}(\zeta )$ is a Galois extension, $[\mathbb{Q}(\eta ):\mathbb{Q}]=|\mathrm{Gal}(\mathbb{Q}(\eta )∕\mathbb{Q})|=(G:{\tilde{H}}_2)=({(\mathbb{Z}∕11\mathbb{Z})}^{\text{∗}}:\{-1,1\})=5$).

The minimal polynomial $g$ of $\eta$ over $\mathbb{Q}$ divides $f$, and has degree 5, so $g=f$.

Using the other form of the minimal polynomial given in Proposition 9.2.6(a), we obtain that

is the minimal polynomial of $\eta ={\zeta }_{11}+{\zeta }_{11}^{-1}$ over $\mathbb{Q}$. □