Techniques of Solving Equations à la Galois
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1st upload: 2023/06/17
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[8-10] Cyclic Extension \(F_3/F_2\): Computing the Minimal Polynomial \(g_3(x)\) (continued)
We now compute the actual values appearing on the right-hand side of (9.8). We begin by finding a new adjunction and the associated binomial equation.
[step2] Binomial equation \(B_3(x)=0\) and generation of a new adjunction \(a_3\)
\begin{align}
t_1=cd_m \ \in F_2(v)
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
\tilde{t_1}=a_3\ \in F_3[x] \\
\\
B_3(x)=x^5-A_3=0 \\
a_3=\sqrt[5]{A_3} \ \in F_2(a_3) \equiv F_3\\
\end{array}
\right. \notag \\
\end{align}
First compute \(t_1^5\). As is clear from (9.7), \(t_1\) is a constant in \(F_2(v)\). Thus the highest degree of \(t_1\) is \(0\), and following our previous notation we write \(t_1=cd_m\) in this section as well. The result is (10.1), showing that \(cd_m^5\) lies in \(F_2\).
Observing the two sides of (10.1), we see that \(cd_m\) is a root of the binomial equation \(B_3(x)=0\) in (10.2). We therefore re-define \(cd_m\) as the radical of \(B_3(x)=0\) by introducing the new symbol \(a_3\); adjoining \(a_3\) to the base field \(F_2\) produces the new extension \(F_3\).
We also compute the inverse \(A_3^{-1}\), which will be needed later. Intermediate steps are omitted due to their complexity, but it is important to reduce modulo \(g_2(v)\) and \(Z\).
\begin{align} cd_m^5&=t_1^5=-\frac{7 {a_1} {a_2} {{\zeta }^{3}}}{2500}+\frac{77 {a_2} {{\zeta }^{3}}}{750}+ ..... +\frac{61 {a_1}}{100}-\frac{41356}{625} \equiv A_3 \ \in \ F_2 \\ \notag \\ &\left\{ \begin{array}{l} B_3(x) \equiv x^5-A_3=0 \\ a_3 \equiv \sqrt[5]{A_3} \ \in F_3 \equiv F_2(a_3) \\ \end{array} \right. \\ \notag \\ A_3^{-1}&=\frac{37633875 {a_1} {a_2} {{\zeta }^{3}}}{2638659584}+\frac{2274606875 {a_2} {{\zeta }^{3}}}{1978994688}+ ..... +\frac{56094429125}{329832448} \\ \end{align}
\begin{align} t_2&=t_1^5 \cdot t_2 \cdot t_1^{-5}=t_1^2 \cdot (t_1^3 \cdot t_2)\cdot A_3^{-1}=a_3^2 \cdot (t_1^3 \cdot t_2) \cdot A_3^{-1} \notag \\ t_3&=a_3^3 \cdot (t_1^2 \cdot t_3)\cdot A_3^{-1} \\ t_4&=a_3^4 \cdot (t_1 \cdot t_4) \cdot A_3^{-1} \notag \\ \notag \\ &\qquad \qquad \Downarrow \notag \\ \notag \\ t_1^3 \cdot t_2&=\frac{41 {a_1} {a_2} {{\zeta }^{3}}}{25000}+\frac{791 {a_2} {{\zeta }^{3}}}{3750}+.....-\frac{253 {a_2}}{1875}-\frac{14 {a_1}}{125}-\frac{224}{125} \notag \\ t_1^2 \cdot t_3&=-\frac{3 {a_1} {a_2} {{\zeta }^{3}}}{10000}-\frac{103 {a_2} {{\zeta }^{3}}}{1500}+.....+\frac{4 {a_2}}{375}+\frac{11 {a_1}}{200}-\frac{259}{250} \\ t_1 \cdot t_4&=\frac{36 {{\zeta }^{3}}}{25}+\frac{36 {{\zeta }^{2}}}{25}-\frac{32}{25} \notag \\ \notag \\ &\qquad \qquad \Downarrow \notag \\ \notag \\ \end{align}
\begin{align} t_0&=x+3 \notag \\ \tilde{t_1}&=a_3 \in \ F_3=F_2(a_3) \\ \notag \\ \tilde{t_2}&=\frac{623 {a_1} {a_2} {{a}_{3}^{2}} {{\zeta }^{3}}}{154880}+\frac{5873 {a_2} {{a}_{3}^{2}} {{\zeta }^{3}}}{23232}+....+\frac{1655 {a_1} {{a}_{3}^{2}}}{15488}+\frac{28231 {{a}_{3}^{2}}}{3872} \notag \\ \tilde{t_3}&=\frac{17 {a_1} {a_2} {{a}_{3}^{3}} {{\zeta }^{3}}}{61952}+\frac{1985 {a_2} {{a}_{3}^{3}} {{\zeta }^{3}}}{511104}+....-\frac{27875 {a_1} {{a}_{3}^{3}}}{85184}+\frac{6145 {{a}_{3}^{3}}}{1331} \notag \\ \tilde{t_4}&=-\frac{343575 {a_1} {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{59969536}+...-\frac{18398125 {a_1} {{a}_{3}^{4}}}{29984768}-\frac{506204825 {{a}_{3}^{4}}}{7496192} \notag \\ \end{align}
Since all terms are now expressed as elements of \(F_3[x]\), we can apply the ILRT (Inverse Lagrange resolvent transformation) in (10.7) to obtain the minimal polynomial \(g_3(x)\).
[step3] ILRT (Inverse Lagrange resolvent transformation)
\begin{align}
\begin{bmatrix}
\tilde{h_0}\\
\tilde{h_1} \\
\tilde{h_2} \\
\tilde{h_3} \\
\tilde{h_4}
\end{bmatrix}
&=
\begin{bmatrix}
1&1&1&1&1 \\
1&(\zeta^4)&(\zeta^4)^2&(\zeta^4)^3&(\zeta^4)^4\\
1&(\zeta^3)&(\zeta^3)^2&(\zeta^3)^3&(\zeta^3)^4\\
1&(\zeta^2)&(\zeta^2)^2&(\zeta^2)^3&(\zeta^2)^4\\
1&\zeta&\zeta^2&\zeta^3&\zeta^4
\end{bmatrix}
\cdot
\begin{bmatrix}
t_0\\
\tilde{t_1} \\
\tilde{t_2}\\
\tilde{t_3} \\
\tilde{t_4}
\end{bmatrix} \\
\notag \\
\therefore \ g_3(x) &\equiv \tilde{h_0}=t_0+\tilde{t_1}+\tilde{t_2}+\tilde{t_3}+\tilde{t_4} \\
\notag \\
&=x+3+a_3+\frac{623 {a_1} {a_2} {{a}_{3}^{2}} {{\zeta }^{3}}}{154880}+....-\frac{506204825 {{a}_{3}^{4}}}{7496192} \notag \\
\end{align}
Finally, the value of \(v\) that we sought is a root of the minimal polynomial \(g_3(x)=0\). Substituting the value of \(v\) from (10.9) into (3.9) yields the five roots of \(f(x)\).
\begin{align} g_3(x)=0 \quad \rightarrow \quad \therefore \ v=-3-a_3-\frac{623 {a_1} {a_2} {{a}_{3}^{2}} {{\zeta }^{3}}}{154880}+....+\frac{506204825 {{a}_{3}^{4}}}{7496192} \\ \end{align}
\begin{align} \alpha&=-\frac{126285 {a_1} {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{29984768}-\frac{3832525 {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{11244288}+...-\frac{2405 {{a}_{3}^{2}}}{3872}-\frac{24 {a_3}}{55}-\frac{1}{5} \notag \\ \notag \\ \beta&=-\frac{2916385 {a_1} {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{179908608}-\frac{19619075 {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{14992384}+... +\frac{5217 {{a}_{3}^{2}}}{3872}+\frac{16 {a_3}}{55}-\frac{1}{5} \notag \\ \notag \\ \gamma&=\frac{1224145 {a_1} {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{89954304}+\frac{12345475 {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{11244288}+...-\frac{2257 {{a}_{3}^{2}}}{3872}-\frac{19 {a_3}}{55}-\frac{1}{5} \notag \\ \notag \\ \delta&=-\frac{171735 {a_1} {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{29984768}-\frac{10552175 {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{22488576}+... -\frac{6771 {{a}_{3}^{2}}}{3872}+\frac{26 {a_3}}{55}-\frac{1}{5} \notag \\ \notag \\ \epsilon&=\frac{2256215 {a_1} {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{179908608}+\frac{45909775 {a_2} {{a}_{3}^{4}} {{\zeta }^{3}}}{44977152}+... +\frac{777 {{a}_{3}^{2}}}{484}+\frac{{a_3}}{55}-\frac{1}{5} \notag \\ \end{align}