Exercise 12.1.18

Proof.

(a)

The Ferrari resolvent $𝜃(y)$ is given by Exercise 4: $$𝜃(y)=y^3-{\sigma }_2{y}^2+\left({\sigma }_1{\sigma }_3-4{\sigma }_4\right)y-{\sigma }_1^2{\sigma }_4-{\sigma }_3^2+4{\sigma }_2{\sigma }_4.$$

As $f=x^4+2x^2-4x+2\in \mathbb{Q}[x]$, ${\sigma }_1=0,{\sigma }_2=2,{\sigma }_3=4,{\sigma }_4=2$, so

$$𝜃(y)=y^3-2y^2-8y.$$

(b)

We use the root $y_1=0$ of the Ferrari resolvent in (12.11) $$x^2-\frac{{\sigma }_1}{2}x+\frac{y_1}{2}=\pm \sqrt{y_1+\frac{{\sigma }_1^2}{4}-{\sigma }_2}\left(x+\frac{\frac{-{\sigma }_1}{2}{y}_1+{\sigma }_3}{2(y_1+\frac{{\sigma }_1^2}{4}-{\sigma }_2)}\right),$$

Here ${\sigma }_1=0,{\sigma }_2=2,{\sigma }_3=4,{\sigma }_4=2$, therefore $y_1+\frac{{\sigma }_1^2}{4}-{\sigma }_2=-2$, so the roots of $f$ are the solutions of

$$x^2=\pm \sqrt{-2}(x-1),$$

(More directly, the equation is

$$x^4=-2x^2+4x-2=-2(x^2-2x+1)=-2{(x-1)}^2={[\sqrt{-2}(x-1)]}^2,$$

so

$$x^2=\pm \sqrt{-2}(x-1).)$$

The roots of $f$ are the roots of

$$x^2-i\sqrt{2}x+i\sqrt{2}\text{or}{x}^2+i\sqrt{2}x-i\sqrt{2}.$$

$$\begin{array}{llll}\hfill x^2-i\sqrt{2}x+i\sqrt{2}& ={\left(x-i\frac{\sqrt{2}}{2}\right)}^2+\frac{1}{2}+i\sqrt{2}& \hfill & \\ \hfill & ={\left(x-i\frac{\sqrt{2}}{2}\right)}^2-\frac{1}{4}\left(-2-4i\sqrt{2}\right)& \hfill & \\ \hfill & ={\left(x-i\frac{\sqrt{2}}{2}\right)}^2-{\left(\frac{1}{2}\sqrt{-2-4i\sqrt{2}}\right)}^2& \hfill & \\ \hfill & =\left(x-i\frac{\sqrt{2}}{2}-\frac{1}{2}\sqrt{-2-4i\sqrt{2}}\right)\left(x-i\frac{\sqrt{2}}{2}+\frac{1}{2}\sqrt{-2-4i\sqrt{2}}\right),& \hfill & \end{array}$$

and similarly

$$\begin{array}{llll}\hfill x^2+i\sqrt{2}x-i\sqrt{2}& =\left(x+i\frac{\sqrt{2}}{2}-\frac{1}{2}\sqrt{-2+4i\sqrt{2}}\right)\left(x+i\frac{\sqrt{2}}{2}+\frac{1}{2}\sqrt{-2+4i\sqrt{2}}\right).& \hfill & \end{array}$$

so the roots of $f$ are

$$i\frac{\sqrt{2}}{2}+\frac{1}{2}\sqrt{-2-4i\sqrt{2}},i\frac{\sqrt{2}}{2}-\frac{1}{2}\sqrt{-2-4i\sqrt{2}},-i\frac{\sqrt{2}}{2}+\frac{1}{2}\sqrt{-2+4i\sqrt{2}},-i\frac{\sqrt{2}}{2}-\frac{1}{2}\sqrt{-2+4i\sqrt{2}}$$

Moreover

$$\begin{array}{llll}\hfill {(a+\mathit{ib})}^2=-2-4i\sqrt{2}& ⟺a^2+b^2=|-2-4i\sqrt{2}|=6,a^2-b^2=-2,\mathit{ab}<0& \hfill & \\ \hfill & ⟺a+\mathit{ib}=\pm (\sqrt{2}-2i)& \hfill & \end{array}$$

so

$$\sqrt{-2-4i\sqrt{2}}=\pm (\sqrt{2}-2i),\sqrt{-2+4i\sqrt{2}}=\pm (\sqrt{2}+2i).$$

The roots of $f$ are $x_1,x_2,x_3=\overline{x_1},x_4=\overline{x_2}$, where

$$\begin{array}{llll}\hfill x_1& =\frac{\sqrt{2}}{2}+i\left(\frac{\sqrt{2}}{2}-1\right),& \hfill & \\ \hfill x_2& =-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}-1\right).& \hfill & \end{array}$$

Note: $x_1,x_2,x_3,x_4\in \mathbb{Q}(i,\sqrt{2})$, so $\mathbb{Q}(x_1,x_2,x_3,x_4)\subset \mathbb{Q}(i,\sqrt{2})$.

$\sqrt{2}=x_1+\overline{x_1}=x_1+x_3\in \mathbb{Q}(x_1,x_2,x_3,x_4)$ and $i=-\frac{1}{2}(x_1+x_2)\in \mathbb{Q}(x_1,x_2,x_3,x_4)$. Therefore the splitting field of $f$ over $\mathbb{Q}$ is $L=\mathbb{Q}(i,\sqrt{2})$.

The Galois group is $\mathrm{Gal}(L∕\mathbb{Q})=⟨\sigma ,\tau ⟩$, where $\sigma (\sqrt{2})=-\sqrt{2}),\sigma (i)=i$, and $\tau$ is the complex conjugation. As permutation group, ${\mathrm{Gal}}_{\mathbb{Q}}(f)=⟨(12)(34),(13)(24)⟩\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ has order 4.

(c)

The Euler’s solution gives the roots $$\alpha =\frac{1}{4}\left({\sigma }_1+{𝜀}_1\sqrt{4y_1+{\sigma }_1^2-4{\sigma }_2}+{𝜀}_2\sqrt{4y_2+{\sigma }_1^2-4{\sigma }_2}+{𝜀}_3\sqrt{4y_3+{\sigma }_1^2-4{\sigma }_2}\right),$$

where ${\sigma }_1=0,{\sigma }_2=2$ and $y_1=0,y_2,y_3$ are the roots of

$$y^3-2y^2-8y=y(y^2-2y-8)=y(y-4)(y+2),$$

so $y_1=0,y_2=4,y_3=-2$.

Therefore

$$\begin{array}{llll}\hfill \alpha & =\frac{1}{4}({𝜀}_1\sqrt{-8}+{𝜀}_2\sqrt{8}+{𝜀}_3\sqrt{-16})& \hfill & \\ \hfill & ={𝜀}_{1}i\frac{\sqrt{2}}{2}+{𝜀}_2\frac{\sqrt{2}}{2}+{𝜀}_{3}i& \hfill & \end{array}$$

Moreover ${𝜀}_i=\pm 1$ satisfy

$$t_1{t}_2{t}_3={𝜀}_1{𝜀}_2{𝜀}_3(i\sqrt{8})(\sqrt{8})4i={\sigma }_1^3-4{\sigma }_1{\sigma }_2+8{\sigma }_3=8{\sigma }_3=32,$$

so ${𝜀}_3=-{𝜀}_1{𝜀}_2$. We obtain the four roots

$$\begin{array}{cc}{x}_1=\frac{\sqrt{2}}{2}+i\left(\frac{\sqrt{2}}{2}-1\right),\hfill & x_3=\overline{x_1}=\frac{\sqrt{2}}{2}-i\left(\frac{\sqrt{2}}{2}-1\right),\hfill \\ x_2=-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}-1\right),\hfill & x_4=\overline{x_2}=-\frac{\sqrt{2}}{2}-i\left(-\frac{\sqrt{2}}{2}-1\right)\hfill \\ \hfill \end{array}$$

The formulas are NOT surprisingly different.