Exercise 8.2.1

Proof.

(a)

Let $f=x^3+x^2-2x-1\in \mathbb{Q}[x]$.

The polynomial ${\Phi }_7=\frac{x^7-1}{x-1}=x^6+x^5+x^4+x^3+x^2+x+1$ has the roots $e^{\frac{2\mathit{ik\pi }}{7}},1\le k\le 6$.

As ${\Phi }_7(0)\ne 0$, for all $z\in ℂ$,

$${\Phi }_7(z)=0⟺\left(z^3+\frac{1}{z^3}\right)+\left(z^2+\frac{1}{z^2}\right)+\left(z+\frac{1}{z}\right)+1=0.$$

Writing $u=z+\frac{1}{z}$, we obtain $u^2=z^2+\frac{1}{z^2}+2$, that is $z^2+\frac{1}{z^2}=u^2-2$.

$u^3=z^3+\frac{1}{z^3}+3(z+\frac{1}{z})$, so $z^3+\frac{1}{z^3}=u^3-3u$.

Therefore

$$\begin{array}{llll}\hfill {\Phi }_7(z)=0& ⟺\exists <!--\; FUNCTION\; APPLICATION\; -->u\in ℂ,u=z+\frac{1}{z}\mathrm{and}(u^3-3u)+(u^2-2)+u+1=0& \hfill & \\ \hfill & ⟺\exists <!--\; FUNCTION\; APPLICATION\; -->u\in ℂ,u=z+\frac{1}{z}\mathrm{and}f(u)=u^3+u^2-2u-1=0& \hfill & \\ \hfill & & \hfill \end{array}$$

Applying this equivalence to $z=e^{\frac{2\mathit{ik\pi }}{7}},1\le k\le 3$, we obtain

$$f(2\cos <!--\; FUNCTION\; APPLICATION\; -->(2\mathit{k\pi }∕7))=f({\zeta }_7^k+{\zeta }_7^{-k})=0,k=1,2,3.$$

These 3 roots of $f$ are distinct, since the function $\cos <!--\; FUNCTION\; APPLICATION\; -->$ is strictly decreasing on $[0,\pi ]$.

Therefore

$$\begin{array}{llll}\hfill f=x^3+x^2-2x-1& =(x-2\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕7))(x-2\cos <!--\; FUNCTION\; APPLICATION\; -->(4\pi ∕7))(x-2\cos <!--\; FUNCTION\; APPLICATION\; -->(6\pi ∕7))& \hfill & \\ \hfill & =(x-{\zeta }_7-{\zeta }_7^{-1})(x-{\zeta }_7^2-{\zeta }_7^{-2})(x-{\zeta }_7^3-{\zeta }_7^{-3})& \hfill & \end{array}$$

(b)

$L$ is the splitting field of $f$ over $\mathbb{Q}$, so by definition $$L=\mathbb{Q}({\zeta }_7+{\zeta }_7^{-1},{\zeta }_7^2+{\zeta }_7^{-2},{\zeta }_7^3+{\zeta }_7^{-3}).$$

Therefore $L\supset \mathbb{Q}$, and as the three roots of $f$ lie in $\mathbb{Q}[{\zeta }_7]$,

$$\mathbb{Q}\subset L\subset \mathbb{Q}({\zeta }_7).$$

As ${\zeta }_7^7=1\in \mathbb{Q},\mathbb{Q}({\zeta }_7)$ is by defintion a radical extension of $\mathbb{Q}$, so $\mathbb{Q}\subset L$ is a solvable extension.