Exercise 8.3.8

Proof.

(a)

We know that ${\alpha }_1=x_1+{\omega }^2{x}_2+\omega x_3$, so $$\begin{array}{llll}\hfill {\alpha }_1^3=& {(x_1+{\omega }^2{x}_2+\omega x_3)}^3& \hfill & \\ \hfill =& x_1^3+x_2^3+x_3^3& \hfill & \\ \hfill & +3\omega (x_1^2{x}_3+x_2^2{x}_1+x_3^2{x}_2)& \hfill & \\ \hfill & +3{\omega }^2(x_1^2{x}_2+x_2^2{x}_3+x_3^2{x}_1)& \hfill & \\ \hfill & +6x_1{x}_2{x}_3.& \hfill & \end{array}$$

Thus ${\alpha }_1^3$ is of the form

$${\alpha }_1^3=p+3\mathit{\omega r}+3{\omega }^{2}s+6q,$$

where

$$\begin{array}{llll}\hfill p& =x_1^3+x_2^3+x_3^3={\sigma }_1^3-3{\sigma }_1{\sigma }_2+3{\sigma }_3,& \hfill & \\ \hfill q& =x_1{x}_2{x}_3={\sigma }_3,& \hfill & \\ \hfill r& =x_1^2{x}_3+x_2^2{x}_1+x_3^2{x}_2,& \hfill & \\ \hfill s& =x_1^2{x}_2+x_2^2{x}_3+x_3^2{x}_1.& \hfill & \end{array}$$

Since $r,s$ are fixed by the permutations of $A_3$, they must be of the form $A+B\sqrt{\mathrm{\Delta }},A,B\in K=\mathbb{Q}({\sigma }_1,{\sigma }_2,{\sigma }_3)$, where we choose

$$\begin{array}{llll}\hfill \sqrt{\mathrm{\Delta }}& =(x_2-x_1)(x_3-x_2)(x_3-x_1)& \hfill & \\ \hfill & =(x_1^2{x}_3+x_2^2{x}_1+x_3^2{x}_2)-(x_1^2{x}_3+x_2^2{x}_1+x_3^2{x}_2)& \hfill & \\ \hfill & =r-s.& \hfill & \end{array}$$

The pair $(r,s)$ is thus the solution of the system

$$\begin{array}{llll}\hfill r-s& =\sqrt{\mathrm{\Delta }},& \hfill & \\ \hfill r+s& ={\sigma }_1{\sigma }_2-3{\sigma }_3.& \hfill & \end{array}$$

Therefore

$$\begin{array}{llll}\hfill r& =\frac{1}{2}({\sigma }_1{\sigma }_2-3{\sigma }_3+\sqrt{\mathrm{\Delta }}),& \hfill & \\ \hfill s& =\frac{1}{2}({\sigma }_1{\sigma }_2-3{\sigma }_3-\sqrt{\mathrm{\Delta }}).& \hfill & \end{array}$$

Then

$$\begin{array}{llll}\hfill {\alpha }_1^3=& {(x_1+{\omega }^2{x}_2+\omega x_3)}^3& \hfill & \\ \hfill =& +{\sigma }_1^3-3{\sigma }_1{\sigma }_2+3{\sigma }_3& \hfill & \\ \hfill & +\frac{3}{2}\omega ({\sigma }_1{\sigma }_2-3{\sigma }_3+\sqrt{\mathrm{\Delta }})& \hfill & \\ \hfill & +\frac{3}{2}{\omega }^2({\sigma }_1{\sigma }_2-3{\sigma }_3-\sqrt{\mathrm{\Delta }})& \hfill & \\ \hfill & +6{\sigma }_3& \hfill & \\ \hfill =& {\sigma }_1^3-3{\sigma }_1{\sigma }_2+9{\sigma }_3-\frac{3}{2}({\sigma }_1{\sigma }_2-3{\sigma }_3)+\frac{3}{2}(\omega -{\omega }^2)\sqrt{\mathrm{\Delta }}& \hfill & \\ \hfill =& {\sigma }_1^3-\frac{9}{2}{\sigma }_1{\sigma }_2+\frac{27}{2}{\sigma }_3+\frac{3\sqrt{3}}{2}i\sqrt{\mathrm{\Delta }}& \hfill & \end{array}$$

Therefore

$$\begin{array}{llll}\hfill {\alpha }_1^3& =-\frac{27}{2}q+\frac{3\sqrt{3}}{2}i\sqrt{\mathrm{\Delta }}& \hfill & \\ \hfill & =\frac{27}{2}\left(-q+\sqrt{\frac{-\mathrm{\Delta }}{27}}\right)& \hfill & \end{array}$$

where

$$q=-\frac{2}{27}{\sigma }_1^3+\frac{1}{3}{\sigma }_1{\sigma }_2-{\sigma }_3.$$

So

$$\begin{array}{llll}\hfill {\alpha }_1& =x_1+{\omega }^2{x}_2+\omega x_3& \hfill & \\ \hfill & =3\sqrt[3]{\frac{1}{2}\left(-q+\sqrt{\frac{-\mathrm{\Delta }}{27}}\right)},& \hfill & \end{array}$$

and

$$\begin{array}{llll}\hfill {\beta }_1& =(23)\cdot {\alpha }_1& \hfill & \\ \hfill & =x_1+{\omega }^2{x}_3+\omega x_2& \hfill & \\ \hfill & =3\sqrt[3]{\frac{1}{2}\left(-q-\sqrt{\frac{-\mathrm{\Delta }}{27}}\right)}.& \hfill & \end{array}$$

Indeed the same calculation gives ${\beta }_1$, by the exchange of $x_2$ with $x_3$, which sends $\sqrt{\mathrm{\Delta }}$ on $-\sqrt{\mathrm{\Delta }}$.

(b)

The system of equations $$\begin{array}{llll}\hfill {\sigma }_1& =x_1+x_2+x_3& \hfill & \\ \hfill {\alpha }_1& =x_1+{\omega }^2{x}_2+\omega x_1& \hfill & \\ \hfill {\beta }_1& =x_1+\omega x_2+{\omega }^2{x}_3,& \hfill & \end{array}$$

has for solution

$$\begin{array}{llll}\hfill x_1& =\frac{1}{3}({\sigma }_1+{\alpha }_1+{\beta }_1)& \hfill & \\ \hfill x_2& =\frac{1}{3}({\sigma }_1+\omega {\alpha }_1+{\omega }^2{\beta }_1)& \hfill & \\ \hfill x_3& =\frac{1}{3}({\sigma }_1+{\omega }^2{\alpha }_1+\omega {\beta }_1)& \hfill & \\ \hfill & & \hfill \end{array}$$

And these are the Cardan’s formula for the roots of $\tilde{f}=x^3-{\sigma }_1{x}^2+{\sigma }_{2}x-{\sigma }_3$.