Exercise 12.1.5

Proof.

(a)

If $y$ is some root of the Ferrari resolvent, then $x_i,x_j$ are the roots of $$x^2-\frac{{\sigma }_1}{2}x+\frac{y}{2}=+\sqrt{y+\frac{{\sigma }_1^2}{4}-{\sigma }_2}\left(x+\frac{\frac{-{\sigma }_1}{2}y+{\sigma }_3}{2(y+\frac{{\sigma }_1^2}{4}-{\sigma }_2)}\right).$$

The product $x_i{x}_j$ is given by

$$x_i{x}_j=\frac{y}{2}-\sqrt{y+\frac{{\sigma }_1^2}{4}-{\sigma }_2}\left(\frac{\frac{-{\sigma }_1}{2}y+{\sigma }_3}{2(y+\frac{{\sigma }_1^2}{4}-{\sigma }_2)}\right).$$

Similarly $x_k,x_l$ are the roots of

$$x^2-\frac{{\sigma }_1}{2}x+\frac{y}{2}=-\sqrt{y+\frac{{\sigma }_1^2}{4}-{\sigma }_2}\left(x+\frac{\frac{-{\sigma }_1}{2}y+{\sigma }_3}{2(y+\frac{{\sigma }_1^2}{4}-{\sigma }_2)}\right).$$

and the product $x_k{x}_l$ is given by

$$x_k{x}_l=\frac{y}{2}+\sqrt{y+\frac{{\sigma }_1^2}{4}-{\sigma }_2}\left(\frac{\frac{-{\sigma }_1}{2}y+{\sigma }_3}{2(y+\frac{{\sigma }_1^2}{4}-{\sigma }_2)}\right).$$

Adding these two formulas, we obtain

$$x_i{x}_j+x_k{x}_l=y.$$

(b)

Using $y_1=x_1{x}_2+x_3{x}_4$, and setting $$t_1=x_1+x_2-x_3-x_4,$$

then

$$\begin{array}{llll}\hfill & y_1+\frac{{\sigma }_1^2}{4}-{\sigma }_2& \hfill & \\ \hfill & =x_1{x}_2+x_3{x}_4+\frac{1}{4}{(x_1+x_2+x_3+x_4)}^2-(x_1{x}_2+x_1{x}_3+x_1{x}_4+x_2{x}_3+x_2{x}_4+x_3{x}_4)& \hfill & \\ \hfill & =\frac{1}{4}\left[x_1^2+x_2^2+x_3^2+x_4^2-2(x_1{x}_2+x_1{x}_3+x_1{x}_4+x_2{x}_3+x_2{x}_4+x_3{x}_4)+4(x_1{x}_2+x_3{x}_4)\right]& \hfill & \\ \hfill & =\frac{1}{4}\left[x_1^2+x_2^2+x_3^2+x_4^2+2x_1{x}_2+2x_3{x}_4-2(x_1{x}_3+x_1{x}_4+x_2{x}_3+x_2{x}_4)\right]& \hfill & \\ \hfill & =\frac{1}{4}\left[{(x_1+x_2)}^2+{(x_3+x_4)}^2-2(x_1+x_2)(x_3+x_4)\right]& \hfill & \\ \hfill & =\frac{1}{4}{(x_1+x_2-x_3-x_4)}^2& \hfill & \\ \hfill & =\frac{t_1^2}{4}& \hfill & \end{array}$$

We choose the square root such that

$$\sqrt{y_1+\frac{{\sigma }_1^2}{4}-{\sigma }_2}=\frac{t_1}{2}.$$

Then the quadratic equation with $y=y_1$ and the plus sign is

$$x^2-\frac{{\sigma }_1}{2}x+\frac{y_1}{2}=+\sqrt{y_1+\frac{{\sigma }_1^2}{4}-{\sigma }_2}\left(x+\frac{\frac{-{\sigma }_1}{2}{y}_1+{\sigma }_3}{2(y_1+\frac{{\sigma }_1^2}{4}-{\sigma }_2)}\right),$$

which gives

$$x^2-\left(\frac{{\sigma }_1}{2}+\frac{t_1}{2}\right)x+\frac{y_1}{2}+\frac{1}{2t_1}({\sigma }_1{y}_1-2{\sigma }_3).$$

Let $u,v$ be the roots of this equation, and $S=u+v,P=\mathit{uv}$ be the sum and product of these roots. Then

$$\begin{array}{llll}\hfill S& =\frac{{\sigma }_1}{2}+\frac{t_1}{2}& \hfill & \\ \hfill & =\frac{1}{2}(x_1+x_2+x_3+x_4+x_1+x_2-x_3-x_4)& \hfill & \\ \hfill & =x_1+x_2& \hfill & \end{array}$$

$$\begin{array}{llll}\hfill P& =\frac{y_1}{2}+\frac{1}{2t_1}({\sigma }_1{y}_1-2{\sigma }_3)& \hfill & \\ \hfill & =\frac{y_1}{2}+\frac{1}{2t_1}\left[(x_1+x_2+x_3+x_4)(x_1{x}_2+x_3{x}_4)-2(x_1{x}_2{x}_3+x_1{x}_2{x}_4+x_1{x}_3{x}_4+x_2{x}_3{x}_4)\right]& \hfill & \\ \hfill & =\frac{y_1}{2}+\frac{1}{2t_1}\left[x_1^2{x}_2+x_1{x}_2^2+x_3^2{x}_4+x_3{x}_4^2-x_1{x}_3{x}_4-x_2{x}_3{x}_4-x_1{x}_2{x}_3-x_1{x}_2{x}_4\right]& \hfill & \\ \hfill & =\frac{y_1}{2}+\frac{1}{2t_1}(x_1+x_2-x_3-x_4)(x_1{x}_2-x_3{x}_4)& \hfill & \\ \hfill & =\frac{1}{2}(x_1{x}_2+x_3{x}_4+x_1{x}_2-x_3{x}_4)& \hfill & \\ \hfill & =x_1{x}_2& \hfill & \end{array}$$ Thus $u,v$ are the roots of $x^2-\mathit{Sx}+P=(x-x_1)(x-x_2)$, so $\{u,v\}=\{x_1,x_2\}$.

$x_1,x_2$ are the roots of (12.11) with the plus sign, so $x_3,x_4$ are the roots of (12.11) with the minus sign.