Techniques for Solving Solvable Equations via Galois Theory

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[4-6] Computing \(F_1/F_0\): finding the minimal polynomial \(g_1(x)\)

In this section we compute the green part in (Fig4-1).
The flow of the computation is shown in the box below.

As confirmed in the previous section, the Galois group of the Galois extension \(F_1/F_0\) is the cyclic group of order 3, \(A_3/e=A_3 \cong C_3\).
Therefore, the binomial equation for the cyclic extension is cubic.

The computation splits into three stages; we begin with Stage 1. Substitute the \(\{h_0,h_1,h_2\}\) obtained in (6.2) into the LRT (Lagrange Resolvent transformation) in (6.3) to convert them to \(\{t_0,t_1,t_2\}\).

Step 1: LRT (Lagrange Resolvent transformation)
\begin{align} &Gal(F_1/F_0)=A_3/e \equiv C_3=\{\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+v+6) \\ h_2=\rho_5(x-v)=(x-v_5)=x-(v^2-2v-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} =\frac{1}{3} \begin{bmatrix} 3x \\ ( 2v^2-3 v-12) \omega +v^2-3v-6 \\ -( 2v^2-3v-12) \omega - v^2+6 \end{bmatrix} \\ \end{align}

As in [Chapter 2] for the cubic cyclic extension, we compute \(\{\ t_1^3,\ t_2^3,\ t_1\cdot t_2\ \}\).
Keep in mind that we reduce \(( \mod \ g_0(v))\) and \(( \mod \ \varOmega)\).

\begin{align} t_1^3&=\frac{1}{27}\left(( 2v^2-3 v-12) \omega +v^2-3v-6 \right)^3 =3\omega -3 \quad \in F_0 \\ t_2^3&=\frac{1}{27}\left( ( 2v^2-3v-12) \omega + v^2-6 \right)^3 =-3\omega -6 \quad \in F_0 \\ \end{align}

\begin{align} &t_1 \cdot t_2= -\frac{4 {{v}^{4}} {{\omega }^{2}}}{9}+\frac{4 {{v}^{3}} {{\omega }^{2}}}{3}+\frac{13 {{v}^{2}} {{\omega }^{2}}}{3} -8 v {{\omega }^{2}}-16 {{\omega }^{2}}-\frac{4 {{v}^{4}} \omega }{9} \notag \\ &+\frac{4 {{v}^{3}} \omega }{3}+\frac{13 {{v}^{2}} \omega }{3} -8 v \omega -16 \omega -\frac{{{v}^{4}}}{9}+\frac{{{v}^{3}}}{3}+\frac{4 {{v}^{2}}}{3}-2 v-4 = \ 3 \ \in F_0 \\ \notag \\ \end{align}

\begin{align} &\therefore \{ t_1^3=3\omega -3, \quad t_2^3=-3\omega-6, \quad t_1 \cdot t_2=3 \ \} \ \in F_0\\ \end{align}

From this, introducing \(A_1\) as in (6.8), we can regard \(t_1\) as a power root of the cubic binomial equation \(B_1(x)=0\) in (6.9). Taking this root to be \(a_1\) and adjoining it to the base field \(F_0\) generates the extension \(F_0(a_1)=F_1\). Since \(a_1\) lies in \(F_1\), we deliberately write \(\tilde{t_1}\) instead of \(t_1\) in (6.9).

Step 2: the binomial equation \(B_1(x)\) and the new adjoined element \(a_1\)
\begin{align} t_1^3&=3\omega -3 \equiv A_1 \ \in \ F_0 \\ \notag \\ \therefore \ B_1(x)&=x^3-A_1=0 \qquad \tilde{t_1}=a_1 \ \in F_0(a_1)=F_1 \\ \end{align}

Now that \(t_1\) lies in \(F_1\), we also need to express \(t_2\) as an element of \(F_1\). For this we use (6.10).

\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_1} =\frac{3a_1^2}{A_1} \end{align}

Computing \(A_1^{-1}\) gives (6.11), so all of \(\{t_0,\tilde{t_1},\tilde{t_2}\}\) can be written inside the extension field \(F_1\).

\begin{align} &A_1^{-1}=-\frac{2}{9}-\frac{\omega}{9} \\ \notag \\ &t_0=x \qquad \tilde{t_1} =a_1 \qquad \tilde{t_2} =-\frac{a_1^2(\omega+2)}{3} \\ \end{align}

Finally, Stage 3. Using \(\{ \ t_0, \tilde{t_1},\tilde{t_2}\}\), we compute the inverse transformation below, the ILRT (Inverse Lagrange Resolvent transformation). We denote the results with tildes as \(\{ \ \tilde{h_0}, \tilde{h_1},\tilde{h_2} \ \}\).

Step 3: ILRT (Inverse Lagrange Resolvent transformation) \begin{align} &\begin{bmatrix} \tilde{h_0} \\ \tilde{h_1} \\ \tilde{h_2} \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 1&\omega^2&(\omega^2)^2\\ 1&\omega&(\omega^2)\\ \end{bmatrix} \cdot \begin{bmatrix} t_0 \\ \tilde{t_1} \\ \tilde{t_2} \end{bmatrix} = \begin{bmatrix} x+a_1-\frac{a_1^2(\omega+2)}{3} \\ x+a_1\omega^2-\frac{a_1^2\omega(\omega+2)}{3} \\ x+a_1\omega-\frac{a_1^2\omega^2(\omega+2)}{3} \end{bmatrix}\\ \end{align}

The polynomials \(\{ \ h_0,h_1,h_2 \ \}\) live in the simple extension \(F(v)\), as indicated in (6.4). Once we pass to the extension \(F_1=F_0(a_1)\), the \(\{ \ \tilde{h_0},\tilde{h_1},\tilde{h_2} \ \}\) become polynomials over \(F_1\).
Moreover, since \(\tilde{h_0}\) originally had \((x-v)\) as a factor, it becomes the minimal polynomial of \(v\) over the extension \(F_1\). The conclusion of this section is as follows.

\begin{align} &g_1(x)=x+a_1-\frac{a_1^2(\omega+2)}{3} \quad \in F_1[x]=F_0(a_1)[x]\\ \end{align}

[4-7] Solving for the roots \(\{\alpha,\beta,\gamma\}\) of \(f(x)\)

Up to the previous section we have lowered the degree of the minimal polynomial \(g_1(x)\) of \(v\) all the way to a linear polynomial. Therefore, any root of \(g_1(x)=0\) is a value of \(v\).

\begin{align} g_1(x)&=0 \quad \Rightarrow \quad \therefore \ v=-a_1+\frac{a_1^2(\omega+2)}{3} \\ \end{align}

Substituting this value of \(v\) into (3.8) finally produces the roots of \(f(x)\).
Note that in the calculation we reduce \( \mod ( \varOmega ) \ \rightarrow \mod ( B_1)\).

\begin{align} \alpha=&\frac{{a_1} \omega -{{a}_{1}^{2}}+2 {a_1}}{3} \\ \beta=&-\frac{\left( {{a}_{1}^{2}}+2 {a_1}\right) \omega +{a_1}}{3} \\ \gamma=&\frac{\left( {{a}_{1}^{2}}+{a_1}\right) \omega +{{a}_{1}^{2}}-{a_1}}{3}\\ \end{align}

This completes the computational procedure for solving the cubic equation \(f(x)=x^3-3x+1=0\) using Galois theory.

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