Exercise 13.2.17

Proof.

(a)

We eliminate $x$ between the two polynomials $$\begin{array}{llll}\hfill f& =x^3+3x+1,& \hfill & \\ \hfill g& =x^2+\mathit{bx}+(a-y),& \hfill & \end{array}$$

with the resultant ${\mathrm{Res}}_x(f,g)=\det <!--\; FUNCTION\; APPLICATION\; -->(S)$, where

$$S=\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill b\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill a-y\hfill & \hfill b\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 0\hfill & \hfill a-y\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill a-y\hfill \end{array}\right)$$

We can obtain this determinant with Sage:

     R.<a,b,x,y> = QQ[]
     f = x^3 + 3*x + 1
     g = x^2 + b*x + (a-y)
     S = matrix(R,[[1, 0, 1,   0,   0  ],
                   [0, 1, b,   1,   0  ],
                   [3, 0, a-y, b,   1  ],
                   [1, 3, 0,   a-y, b  ],
                   [0, 1, 0,   0,   a-y]])
     R = S.det(); R

$$a^3+3a{b}^2-b^3-3a^{2}y-3b^{2}y+3a{y}^2-y^3-6a^2+3\mathit{ab}+12\mathit{ay}-3\mathit{by}-6y^2+9a-3b-9y+1$$

But it is more easy to call the method "resultant" to obtain the same result:

     res = f.resultant(g,x);res

The list of coefficients of $-{\mathrm{Res}}_x(f,g)$ is given by

     l =[-res.subs(y=0)] +  [-res.coefficient(y^k) for k in range(1,4)]
     l

$$\left[-a^3-3a{b}^2+b^3+6a^2-3\mathit{ab}-9a+3b-1,3a^2+3b^2-12a+3b+9,-3a+6,1\right]$$

We find the equation(13.26):

$$y^3+(6-3a)y^2+(9+3b+3b^2-12a+3a^2)y+P(a,b)=0,$$

where $P(a,b)=-a^3-3a{b}^2+b^3+6a^2-3\mathit{ab}-9a+3b-1$.

(b)

The coefficient of $y^2$ vanishes if $a=2$, and then the coefficient of $y$ vanishes if $0=9+3b+3b^2-12a+3a^2=3b+3b^2-3$: $$b^2+b-1=0$$

so $b=\frac{\sqrt{5}-1}{2}$ is a solution.

If we pick $b=(\sqrt{5}-1)∕2$, then the above cubic equation becomes

$$y^3+\frac{5}{2}\sqrt{5}-\frac{25}{2}=0,$$

so

$$\begin{array}{llll}\hfill y^3& =\frac{25-5\sqrt{5}}{2}& \hfill & \\ \hfill & ={\sqrt{5}}^3\frac{\sqrt{5}-1}{2}.& \hfill & \end{array}$$

By the property of the resultant, if $y$ is evaluated to $y_0={\omega }^k\sqrt{5}\sqrt[3]{\frac{\sqrt{5}-1}{2}},k=0,1,2,\omega =e^{2\mathit{i\pi }∕3}$, then there exists a common root of $f$ and $g$ in $ℂ$, where

$$\begin{array}{llll}\hfill f& =x^3+3x+1& \hfill & \\ \hfill g& =x^2+\frac{\sqrt{5}-1}{2}x+2-y_0.& \hfill & \end{array}$$

(c)

The Euclidean division of $f$ by $g$ gives $$x^3+3x+1=(x^2+\mathit{bx}+a-y)(x-b)+(y+b^2+3-a)x+1+\mathit{ab}-\mathit{by}.$$

If $x_0$ is a common root of $f,g$, then the remainder is 0, so

$$x_0=\frac{b{y}_0-\mathit{ab}-1}{y_0+b^2+3-a},$$

and this gives a formula for the roots $x_0$ of $f$ in terms of the roots $y_0$ of the resultant.

Since $y_0$ is an algebraic number of degree 3 over $\mathbb{Q}$, and $x_0\in \mathbb{Q}(\sqrt{5},y_0)$, there exists some polynomial $p$ of degree 2 such that $x_0=p(y_0)$.

To find this more simple formula for $x_0$, we search the gcd of $f,g$ in the field $\mathbb{Q}\left(\sqrt{5},\sqrt[3]{\frac{\sqrt{5}-1}{2}}\right)$ by the extended Euclid’s algorithm.

This is obtained with the following Sage instructions:

     K.<u>= QQ[sqrt(5)]
     R.<z> = K[]
     res = z^3 - (u-1)/2
     L.<w> = K.extension(res)
     A.<x> = L[]
     f = x^3 + 3*x + 1
     g = x^2 + (u - 1)/2 * x + (2 - u*w)
     gcd(f,g)

$$x+\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)w^2-w$$

We have obtained that

$$\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,g)=x+\frac{\sqrt{5}+1}{2}{w}^2-w,\text{ where }{w}^3=\frac{\sqrt{5}-1}{2}.$$

Since $w^3=\frac{\sqrt{5}-1}{2}$, so $w^2=\frac{\sqrt{5}-1}{2}{w}^{-1}$, $\frac{\sqrt{5}+1}{2}{w}^2=w^{-1}$.

Therefore the roots of $f$ are

$$\begin{array}{llll}\hfill x_0& =w-\frac{\sqrt{5}+1}{2}{w}^2& \hfill & \\ \hfill & =w-w^{-1}& \hfill & \end{array}$$

We can write $w={\omega }^k\sqrt[3]{\frac{\sqrt{5}-1}{2}},k=0,1,2$, then $w^{-1}={\omega }^{2k}\sqrt[3]{\frac{\sqrt{5}+1}{2}}$, where the cubic roots are chosen so that their product is real, equal to 1. Then

$${\omega }^k\sqrt[3]{\frac{\sqrt{5}-1}{2}}-{\omega }^{2k}\sqrt[3]{\frac{\sqrt{5}+1}{2}},k=0,1,2$$

are the roots of $f$. This is identical to the formulas obtained with Cardano’s formulas, with more sweat.