Techniques for Solving Solvable Equations via Galois Theory

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[2-12] Computing \(F_2/F_1\): finding the minimal polynomial \(g_2(x)\) (1)

In this section we compute the green portion of (Fig. 2-3).
The procedure is exactly the same as in Section [2-9].
The flow of the computation is indicated in the boxed display below.

However, for the present Galois extension \(F_2/F_1\) the Galois group is the cyclic group of order three, \(A_3/e=A_3 \cong C_3\).
Consequently, the binomial equation in this section will be cubic.

As before, the calculation is split into three stages. We begin with Stage 1, following the boxed steps below.

Step 1 LRT (Lagrange Resolvent Transformation)
\begin{align} &Gal(F_2/F_1)=A_3/e \equiv C_2=\{\rho_1,\rho_4,\rho_5\} \\ \notag \\ &\left\{ \begin{array}{l} h_0=\rho_1(x-v)=(x-v_1)=x-v \\ h_1=\rho_4(x-v)=(x-v_4)=x-{{v}^{2}}+\frac{{a_1} v}{6}+\frac{v}{2}-6 \\ h_2=\rho_5(x-v)=(x-v_5)=x+{{v}^{2}}-\frac{{a_1} v}{6}+\frac{v}{2}+6 \\ \end{array} \right. \\ \end{align}

\begin{align} \notag \\ &\begin{bmatrix} t_0 \\ t_1 \\ t_2 \end{bmatrix} =\frac{1}{3} \begin{bmatrix} 1&1&1 \\ 1&\omega&\omega^2\\ 1&(\omega^2)&(\omega^2)^2\\ \end{bmatrix} \cdot \begin{bmatrix} h_0 \\ h_1 \\ h_2 \end{bmatrix} = \begin{bmatrix} x \\ -\omega\!\left(\frac{2 {{v}^{2}}}{3}-\frac{{a_1} v}{9}+4\right)-\frac{{{v}^{2}}}{3}+\frac{{a_1} v}{18}-\frac{v}{2}-2\\ \ \omega\!\left(\frac{2 {{v}^{2}}}{3}-\frac{{a_1} v }{9}+4 \right) +\frac{{{v}^{2}}}{3}-\frac{{a_1} v}{18}-\frac{v}{2}+2 \end{bmatrix}\\ \notag \\ \varOmega&=\omega^2+\omega+1=0 \notag \\ \end{align}

The key point is that all computations are carried out over \(F_1\), so they must be performed \((mod \ g_1(v))\).
With \(\{t_0,t_1,t_2\}\) now obtained in (11.3), we next compute the quantities \(\{t_1^3,t_2^3,t_1 \cdot t_2\}\).

\begin{align} \bbox[#FFFF00]{ t_1^3}&=\left(-\omega\!\left(\frac{2 {{v}^{2}}}{3}-\frac{{a_1} v}{9}+4\right) -\frac{{{v}^{2}}}{3}+\frac{{a_1} v}{18}-\frac{v}{2}-2 \right)^3 \notag \\ & \bbox[#FFFF00]{=3\omega +\frac{{a_1+3}}{2} } \quad \in \ F_1 \qquad (mod \ g_1(v)),(mod \ B_1), (mod \ \varOmega) \\ \notag \\ t_2^3&=\left( \omega\!\left(\frac{2 {{v}^{2}}}{3}-\frac{{a_1} v }{9}+4 \right) +\frac{{{v}^{2}}}{3}-\frac{{a_1} v}{18}-\frac{v}{2}+2 \right)^3 \notag \\ &=-3\omega +\frac{{a_1-3}}{2} \quad \in \ F_1 \qquad (mod \ g_1(v)),(mod \ B_1) ,(mod \ \varOmega) \\ \notag \\ t_1 \cdot t_2&=\left(-\omega\!\left(\frac{2 {{v}^{2}}}{3}-\frac{{a_1} v}{9}+4\right) -\frac{{{v}^{2}}}{3}+\frac{{a_1} v}{18}-\frac{v}{2}-2 \right) \notag \\ &\qquad \times \left( \omega\!\left(\frac{2 {{v}^{2}}}{3}-\frac{{a_1} v }{9}+4 \right) +\frac{{{v}^{2}}}{3}-\frac{{a_1} v}{18}-\frac{v}{2}+2 \right) \notag \\ &=\frac{\left( {{v}^{2}}+12\right)(\omega^2+ \omega) +{{v}^{2}}+3}{3} \qquad (mod \ g_1(v)), (mod \ B_1) \notag \\ &=-3 \ \in F_0 \qquad \qquad (mod \ \varOmega) \\ \end{align}

Although the intermediate expressions look complicated, the steps are very simple: substitute the values of \(\{t_1,t_2\}\) from (11.3) into \(\{t_1^3, t_2^3,t_1 \cdot t_2 \}\), expand, and then reduce in order by \([\ g_1(v) \rightarrow B_1 \rightarrow \varOmega \ ]\).

From (11.4) we learn that \(t_1^3 \in F_1\). Hence, by introducing the number \([ \ A_2=3\omega +\frac{{a_1+3}}{2} \ ]\), we may impose the relation \(t_1^3=A_2\) as in (11.7).
Therefore, we may regard \(t_1\) as a radical (power-root) of the cubic binomial equation \(B_2(x)=0\), as shown in (11.8).
By adjoining this radical as a new number \( a_2 \ (\equiv t_1)\) to the field \(F_1\), we obtain the extension field \(F_2\).

Step 2 Binomial equation \(B_2(x)\) and the new adjoined element \(a_2\)

\begin{align} t_1^3&=3\omega +\frac{1}{2}(a_1+3)=A_2 \ \in \ F_1 \\ \notag \\ \therefore \ B_2(x)&=x^3-A_2=0 \qquad t_1 \equiv a_2=\sqrt[3]{A_2} \ \in \ F_2 \\ \end{align}

Although \(t_1\) can be written directly as \(a_2\) as in (11.8), the question is whether we can express \(t_2\) using \(a_2\) as well.
For this we use the relation (11.6). Please follow the algebra carefully:

\begin{align} t_2=\frac{t_1 \cdot t_2}{t_1}=\frac{t_1^2\cdot (t_1t_2)}{t_1^2\cdot t_1} =\frac{t_1^2\cdot (t_1t_2)}{t_1^3}=\frac{a_1^2 \cdot (-3)}{A_2} =-\frac{3a_2^2}{A_2} \end{align}

The issue in (11.9) is the denominator \(A_2\) in the expression for \(t_2\). Thus, within the extension \(F_2\) we need a polynomial expression for \(A_2^{-1}\). A direct computation yields (11.10) (the concrete steps are given in the supplement).

\begin{align} A_2^{-1}&=\frac{1}{18}+\frac{\omega}{9}-\frac{a_1}{54} \\ \end{align}

Substituting this \(A_2^{-1}\) into (11.9), we can also express \(t_2\) using \(\{\omega,a_1,a_2\}\). Collecting \(\{t_0,t_1,t_2\}\) gives (11.11). Since the previous \(\{t_1,t_2\}\) are now expressed using elements of the extension field \(F_2\), we denote them by \(\{\tilde{t_1},\tilde{t_2}\}\).

\begin{align} t_0=x ,\quad \tilde{t_1}=a_2, \quad \tilde{t_2}=-( \ 3 \ a_2^2 \ )\cdot A_2^{-1}=a_2^2\left(-\frac{\omega }{3}+\frac{{a_1}}{18}-\frac{1}{6}\right) \\ \end{align}

[2-12] Supplement: procedure for computing \(A_2^{-1}\)

Since \(A_2\) is a number in \(F_2=Q(\omega,a_1)\), its inverse \(A_2^{-1}\) should be expressible in the basis \(\{1,\omega,a_1,\omega \cdot a_1\}\).
We therefore assume the form in (12.1). Solving the linear system (12.3) for \(\{d_0,d_1,d_2,d_3\}\) so that \([ \ A_2 \cdot A_2^{-1}=1 \ ]\) yields the desired coefficients.
Remember to reduce at the end modulo \((mod \ g_1(v)), \ (mod \ B_1) ,(mod \ \varOmega)\).

\begin{align} A_2^{-1}&=d_0+d_1\omega+d_2 a_1+d_3 \omega \cdot a_1\\ \notag \\ A_2 \cdot A_2^{-1}&=\left(3\omega +\frac{{a_1+3}}{2} \right) \cdot \left(d_0+d_1\omega+d_2 a_1+d_3 \omega a_1\right) \notag \\ &=D_0+\omega D_1+ a_1 D_2+ \omega a_1 D_3=1 \\ \end{align}

\begin{align} &\left\{ \begin{array}{l} D_0=\left( -\frac{135 {d_2}}{2}-3 {d_1}+\frac{3 {d_0}}{2} \right)=1, \qquad D_1=\left(-\frac{135 {d_3}}{2}-\frac{3 {d_1}}{2}+3 {d_0} \right)=0 \\ D_2=\left(-3 {d_3}+\frac{3 {d_2}}{2}+\frac{{d_0}}{2}\right) =0, \qquad D_3=\left(-\frac{3 {d_3}}{2}+3 {d_2}+\frac{{d_1}}{2} \right)=0 \end{array} \right. \\ \notag \\ &\therefore \quad \left[ \ d_0=\frac{1}{18}, \ d_1=\frac{1}{9}, \ d_2=-\frac{1}{54}, \ d_3=0 \ \right]\\ \notag \\ &\therefore \quad A_2^{-1}=\frac{1}{18}+\frac{\omega}{9}-\frac{a_1}{54} \\ \end{align}

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