Exercise 9.2.3

Proof.

(a)

Let $\zeta$ be any 7th primitive root of unity (i.e. $\zeta ={\zeta }_7^i,i=1,\cdots ,6$).

Then $1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4+{\zeta }^5+{\zeta }^6=0$, and division by ${\zeta }^3$ gives

$$\begin{array}{lll}\hfill {\zeta }^{-3}+{\zeta }^3+{\zeta }^{-2}+{\zeta }^2+\zeta +{\zeta }^{-1}+1=0.& & \hfill \text{(1)}\end{array}$$

Write $\eta =\zeta +{\zeta }^{-1}$. Then

$$\begin{array}{llll}\hfill {\eta }^2& ={\zeta }^2+{\zeta }^{-2}+2,& \hfill & \\ \hfill {\eta }^3& ={\zeta }^3+{\zeta }^{-3}+3(\zeta +{\zeta }^{-1}).& \hfill & \end{array}$$

Therefore

$$\begin{array}{llll}\hfill {\zeta }^2+{\zeta }^{-2}& ={\eta }^2-2,& \hfill & \\ \hfill {\zeta }^3+{\zeta }^{-3}& ={\eta }^3-3\eta .& \hfill & \end{array}$$

By (1), $({\eta }^3-3\eta )+({\eta }^2-2)+\eta +1=0$, so

$$\begin{array}{lll}\hfill {\eta }^3+{\eta }^2-2\eta -1=0.& & \hfill \text{(2)}\end{array}$$

Applying the equality (2) to ${\zeta }_7,{\zeta }_7^2,{\zeta }_7^3$, we obtain that ${\eta }_1={\zeta }_7+{\zeta }_7^{-1},{\eta }_2={\zeta }_7^2+{\zeta }_7^{-2},{\eta }_3={\zeta }_7^3+{\zeta }_7^{-3}$ are roots of

$$f=x^3+x^2-2x-1.$$

As the minimal polynomial of ${\zeta }_7$ over $\mathbb{Q}$ is ${\Phi }_7$ of degree 6, the list $(1,{\zeta }_7,{\zeta }_7^2,{\zeta }_7^3,{\zeta }_7^4,{\zeta }_7^5)$ is linearly independent over $\mathbb{Q}$, thus also the list obtained by multiplication by ${\zeta }_7$, so $({\zeta }_7,{\zeta }_7^2,{\zeta }_7^3,{\zeta }_7^4,{\zeta }_7^5,{\zeta }_7^6)$ is a linearly independent list, therefore ${\eta }_1={\zeta }_7+{\zeta }_7^6,{\eta }_2={\zeta }_7^2+{\zeta }_7^5,{\eta }_3={\zeta }_7^3+{\zeta }_7^4$ are linearly independent, so are a fortiori distinct. Therefore

$$f=x^3+x^2-2x-1=(x-{\eta }_1)(x-{\eta }_2)(x-{\eta }_3).$$

${\eta }_1,{\eta }_2,{\eta }_3$ are the three distinct roots of $f$.

(b)

$f$ has no root in $\mathbb{Q}$. Indeed, if $\alpha =p∕q,p\wedge q=1$ was such a root, we would have the equality $$p^3+p^{2}q-2p{q}^2-q^3=0,$$

which implies, since $p\wedge q=1$, that $p\mid 1,q\mid 1$, so $\alpha =\pm 1$, but neither $1$, nor $-1$ is a root of $f$.

Since $f$ has no root in $\mathbb{Q}$ and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=3$, $f$ is irreducible over $\mathbb{Q}$. So $f$ is the minimal polynomial of ${\eta }_1$ over $\mathbb{Q}$, and also of ${\eta }_2,{\eta }_3$, which are so conjugate of ${\eta }_1$ over $\mathbb{Q}$. Moreover

$$[\mathbb{Q}({\eta }_1):\mathbb{Q}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=3.$$

Let $\tau$ be the complex conjugation restricted to $\mathbb{Q}({\zeta }_7)$. As $\tau ({\zeta }_7)={\bar{\zeta }}_7={\zeta }_7^{-1}\in \mathbb{Q}({\zeta }_7)$, $\tau$ is an automorphism of $\mathbb{Q}({\zeta }_7)$ which fixes the elements of $\mathbb{Q}$, so $\tau \in \mathrm{Gal}(\mathbb{Q}({\zeta }_7)∕\mathbb{Q})$, and ${\tau }^2=e$, therefore $\{e,\tau \}={\tilde{H}}_2$ is the unique subgroup of $G=\mathrm{Gal}(\mathbb{Q}({\zeta }_7)∕\mathbb{Q})$ of order 2.

Let $L_2=L_{⟨\tau ⟩}$ be the fixed field of ${\tilde{H}}_2$. By the Galois Correspondence (see Proposition 9.2.1 and Exercise 2),

$$[L_2:\mathbb{Q}]=(G:H_2)=3.$$

As ${\eta }_1\in ℝ,\tau ({\eta }_1)={\eta }_1$, hence ${\eta }_1\in L_2$, and so $\mathbb{Q}({\eta }_1)\subset L_2$.

Since $[L_2:\mathbb{Q}]=[\mathbb{Q}({\eta }_1):\mathbb{Q}]=3$, then $[L_2:\mathbb{Q}({\eta }_1)]=1$, hence $L_2=\mathbb{Q}({\eta }_1)$.

The fixed field $L_2$ of ${\tilde{H}}_2=\{e,\tau \}$ is $\mathbb{Q}({\eta }_1)$.

(c)

${\eta }_1=2\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕7),{\eta }_2=2\cos <!--\; FUNCTION\; APPLICATION\; -->(4\pi ∕7),{\eta }_3=4\cos <!--\; FUNCTION\; APPLICATION\; -->(6\pi ∕7)$ are the roots of $f=x^3+x^2-2x-1$. We compute these roots with the Cardan’s Formula.

The substitution $x=y-1∕3$ in $f$ gives

$$\begin{array}{llll}\hfill g(y)& =f\left(y-\frac{1}{3}\right)& \hfill & \\ \hfill & ={\left(y-\frac{1}{3}\right)}^3+{\left(y-\frac{1}{3}\right)}^2-2\left(y-\frac{1}{3}\right)-1& \hfill & \\ \hfill & =y^3-y^2+\frac{1}{3}y-\frac{1}{27}+y^2-\frac{2}{3}y+\frac{1}{9}-2y+\frac{2}{3}-1& \hfill & \\ \hfill & =y^3-\frac{7}{3}y-\frac{7}{27}& \hfill & \end{array}$$

(Note: if $\mathrm{\Delta }$ is the discriminant of $f$ or $g$, then $\mathrm{\Delta }=-4p^3-27q^2=-4{\left(-\frac{7}{3}\right)}^3-27{\left(\frac{7}{27}\right)}^2=\frac{1372}{27}-\frac{49}{27}=\frac{1323}{27}=49=7^2$ is the square of an element of $\mathbb{Q}$, hence the Galois group of $f$ is $A_3\simeq \mathbb{Z}∕3\mathbb{Z}$. This shows again that

$$|\mathrm{Gal}(\mathbb{Q}({\eta }_1)∕\mathbb{Q})|=[L_2:\mathbb{Q}]=3.)$$

Let $\alpha$ a root of $g$ (that is to say $\alpha -1∕3$ is a root of $f$). There exist two complex numbers $u,v$ such that $\alpha =u+v,\mathit{uv}=7∕9$. Then

$$\begin{array}{llll}\hfill 0& ={(u+v)}^3-\frac{7}{3}(u+v)-\frac{7}{27}& \hfill & \\ \hfill & =u^3+v^3+\left(3\mathit{uv}-\frac{7}{3}\right)(u+v)-\frac{7}{27}& \hfill & \\ \hfill & =u^3+v^3-\frac{7}{27}& \hfill & \\ \hfill & & \hfill \end{array}$$

So $(u,v)$, which satisfies the condition $\mathit{uv}=7∕9$, is a solution of the system

$$\begin{array}{llll}\hfill u^3+v^3& =\frac{7}{3^3}& \hfill & \\ \hfill u^3{v}^3& =\frac{7^3}{3^6}& \hfill & \end{array}$$

$u^3,v^3$ are so the roots of the equation $x^2-\frac{7}{3^3}x+\frac{7^3}{3^6}$, of discriminant

$$\delta =\frac{7^2}{3^6}-4\frac{7^3}{3^6}=\frac{7^2(-27)}{3^6}=-\frac{7^2}{3^3}=-\frac{49}{27}.$$

$$\begin{array}{llll}\hfill u^3& =\frac{1}{2}\left(\frac{7}{27}+i\sqrt{\frac{49}{27}}\right)=\frac{1}{27}\times \frac{7}{2}\left(1+3i\sqrt{3}\right)& \hfill & \\ \hfill v^3& =\frac{1}{2}\left(\frac{7}{27}-i\sqrt{\frac{49}{27}}\right)=\frac{1}{27}\times \frac{7}{2}\left(1-3i\sqrt{3}\right)& \hfill & \\ \hfill & & \hfill \end{array}$$

As $u^3={\bar{v}}^3$, and $\mathit{uv}=7∕9\in ℝ$, then $v={\omega }^k\bar{u},k=0,1,2$, and so $\mathit{uv}=u\bar{u}{\omega }^k\in ℝ$, therefore ${\omega }^k\in ℝ$ , so $k=0$, which gives $v=\bar{u}$. The set $\{{\eta }_1,{\eta }_2,{\eta }_3\}$ of the three roots of $f$ is so the set $\{-1∕3+u+\bar{u},-1∕3+\mathit{\omega u}+{\omega }^2\bar{u},-1∕3+{\omega }^{2}u+\omega \bar{u}\}$.

To identify each root, we must define the determination of $3u=\sqrt[3]{\frac{7}{2}\left(1+3i\sqrt{3}\right)}$. Choose for this cubic root the one which lies in the first quadrant (there exists one and only one such a cubic root since $\mathrm{Arg}(1+3i\sqrt{3})\in [0,\pi ∕2]$), and write $3\bar{u}=\sqrt[3]{\frac{7}{2}\left(1-3i\sqrt{3}\right)}$ its conjugate.

Then

$$\begin{array}{llll}\hfill -\frac{1}{3}+u+\bar{u}& =\frac{1}{3}(-1+3u+3\bar{u})& \hfill & \\ \hfill & =\frac{1}{3}\left(-1+\sqrt[3]{\frac{7}{2}\left(1+3i\sqrt{3}\right)}+\sqrt[3]{\frac{7}{2}\left(1-3i\sqrt{3}\right)}\right)& \hfill & \\ \hfill & & \hfill \end{array}$$

As $\left|\frac{7}{2}\left(1+3i\sqrt{3}\right)\right|=\frac{7}{2}\sqrt{28}={(\sqrt{7})}^3$, then $|3u|=\sqrt{7}$, and $\mathrm{Arg}(3u)\in [0,\pi ∕6]$, therefore $\mathrm{Re}(3u)\ge \sqrt{7}\cos <!--\; FUNCTION\; APPLICATION\; -->(\pi ∕6)=\sqrt{7}\sqrt{3}∕2$, so $2\mathrm{Re}(3u)\ge \sqrt{21}$.

Therefore $\mathrm{Re}(-1+3u+3\bar{u})\ge \sqrt{21}-1>0$.

As ${\eta }_1=2\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕7)$ is the only positive root of $f$,

$${\eta }_1={\zeta }_7+{\zeta }_7^{-1}=2\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕7)=\frac{1}{3}\left(-1+\sqrt[3]{\frac{7}{2}\left(1+3i\sqrt{3}\right)}+\sqrt[3]{\frac{7}{2}\left(1-3i\sqrt{3}\right)}\right)$$

where $\sqrt[3]{\frac{7}{2}\left(1+3i\sqrt{3}\right)}$ is chosen such that

$$\mathrm{Re}\left(\sqrt[3]{\frac{7}{2}\left(1+3i\sqrt{3}\right)}\right)>0,\mathrm{Im}\left(\sqrt[3]{\frac{7}{2}\left(1+3i\sqrt{3}\right)}\right)>0$$

and $\sqrt[3]{\frac{7}{2}\left(1-3i\sqrt{3}\right)}$ is its conjugate.

As ${\zeta }_7$ is a root of $x^2-{\eta }_{1}x+1$, with positive imaginary part, then ${\zeta }_7=\frac{1}{2}\left({\eta }_1+i\sqrt{4-{\eta }_1^2}\right)$, so

$$\begin{array}{llll}\hfill {\zeta }_7& =-\frac{1}{6}+\frac{1}{6}\sqrt[3]{\frac{7}{2}(1+3i\sqrt{3})}+\frac{1}{6}\sqrt[3]{\frac{7}{2}(1-3i\sqrt{3})}& \hfill & \\ \hfill & +\frac{i}{2}\sqrt{4-{\left(\frac{1}{3}-\frac{1}{3}\sqrt[3]{\frac{7}{2}(1+3i\sqrt{3})}-\frac{1}{3}\sqrt[3]{\frac{7}{2}(1-3i\sqrt{3})}\right)}^2}& \hfill & \\ \hfill & =-\frac{1}{6}+\frac{1}{6}\sqrt[3]{\frac{7}{2}(1+3i\sqrt{3})}+\frac{1}{6}\sqrt[3]{\frac{7}{2}(1-3i\sqrt{3})}& \hfill & \\ \hfill & +i\sqrt{1-{\left(\frac{1}{6}-\frac{1}{6}\sqrt[3]{\frac{7}{2}(1+3i\sqrt{3})}-\frac{1}{6}\sqrt[3]{\frac{7}{2}(1-3i\sqrt{3})}\right)}^2}& \hfill & \end{array}$$

with the same cube roots.

(It seems that there is a misprint in (9.11)).