Techniques of Solving Equations à la Galois
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[8-9] Cyclic Extension \(F_3/F_2\): Computing the Minimal Polynomial \(g_3(x)\)
The computations in this section correspond to the green block on the second tier of Fig.8-4.
The Galois group used here is the quotient \(C_5/e\), namely the cyclic group \(C_5\) of order \(5\).
Since the two–digit subscripts \(\{\rho_i\}\) of the elements of \(C_5\) are cumbersome, we set \(\rho \equiv \sigma_{43}\). One can then verify—though the calculation is a bit involved—that \(\rho\) is a generator of the cyclic group \(C_5\), as in (9.2). Consequently, \(\{h_0,h_1,h_2,h_3,h_4\}\) are given by (9.3).
\begin{align} &Gal(F_3/F_2)=C_5/E \cong C_5=\{ \rho_{1},\rho_{43},\rho_{52},\rho_{90},\rho_{117} \} \\ \notag \\ &\rho \equiv \rho_{43}, \quad \rho^2=\rho_{117}, \quad \rho^3=\rho_{90}, \quad \rho^4=\rho_{52}, \quad \rho^5=e=\rho_1 \\ \end{align}
\begin{align} &\qquad \qquad \Downarrow \notag \\ &\left\{ \begin{array}{l} h_0 \equiv (x-v)=(x-v_1) \\ h_1=\rho(h_0)=(x-v_{43}) \\ h_2=\rho(h_1)=\rho^2(h_0)=(x-v_{117}) \\ h_3=\rho(h_2)=\rho^3(h_0)=(x-v_{90}) \\ h_4=\rho(h_3)=\rho^4(h_0)=(x-v_{52}) \end{array} \right. \\ \notag \\ \end{align}
\begin{align} &\begin{bmatrix} t_0 \\ t_1 \\ t_2 \\ t_3 \\ t_4 \end{bmatrix} =\frac{1}{5} \begin{bmatrix} 1&1&1&1&1 \\ 1&\zeta&\zeta^2&\zeta^3&\zeta^4\\ 1&(\zeta^2)&(\zeta^2)^2&(\zeta^2)^3&(\zeta^2)^4\\ 1&(\zeta^3)&(\zeta^3)^2&(\zeta^3)^3&(\zeta^3)^4\\ 1&(\zeta^4)&(\zeta^4)^2&(\zeta^4)^3&(\zeta^4)^4 \end{bmatrix} \cdot \begin{bmatrix} h_0 \\ h_1 \\ h_2 \\ h_3 \\ h_4 \end{bmatrix} \\ \notag \\ &\qquad \qquad Z=\zeta^4+\zeta^3+\zeta^2+\zeta+1=0 \notag \\ \end{align}
Substituting into the \(v_i\) in (9.3) the polynomial expression of \(v\) defined in (4.7) yields (9.5). Each \(h_i\) is linear in \(x\); however, since the constants other than that of \(h_0\) are complicated, we denote them by \(\{cb_0,cc_0,cd_0,ce_0\}\). We abbreviate only \(cb_0\) explicitly.
\begin{align} h_0&=x-v, \quad h_1=x+cb_0, \quad h_2=x+cc_0, \quad h_3=x+cd_0, \quad h_4=x+ce_0\\ \end{align}
\begin{align} \notag \\ cb_0&=\frac{1842652131618713141165321 {a_1} {a_2} {{v}^{4}}}{274783653925128983257243171440}+.....-\frac{20972115705844723956430642351}{2289863782709408193810359762} \notag \\ \end{align}
\begin{align} t_0&=x+3 \\ \notag \\ t_1&=-\frac{347828893813523464908459 {a_1} {a_2} {{v}^{4}} {{\zeta }^{3}}}{228986378270940819381035976200}-\frac{745259948412736253397035 {a_2} {{v}^{4}} {{\zeta }^{3}}}{6869591348128224581431079286}+... \notag \\ &\qquad \qquad ............ \notag \\ &+\frac{2942247769237028894958880059 {a_1}}{228986378270940819381035976200}-\frac{6676879355962104443052886159}{5724659456773520484525899405} \\ \notag \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} \rho(t_0)=t_0=x+3 \ \in \ F_0[x] \\ \rho(t_1)=\zeta^{-1}t_1 \\ \rho(t_2)=\zeta^{-2}t_2 \\ \rho(t_3)=\zeta^{-3}t_3 \\ \rho(t_4)=\zeta^{-4}t_4 \\ \end{array} \right. \qquad \Rightarrow \quad \left\{ \begin{array}{l} \rho(t_1^4 \cdot t_1)=t_1^4 \cdot t_1 \\ \rho(t_1^3 \cdot t_2)=t_1^3 \cdot t_2 \\ \rho(t_1^2 \cdot t_3)=t_1^2 \cdot t_3 \\ \rho(t_1^1 \cdot t_4)=t_1 \cdot t_4 \\ \end{array} \right. \\ \notag \\ &\quad \therefore \ t_1^{5-i} \cdot t_i \ \in \ F_0[x] \ (=F_0) \quad (i=1,2,3,4) \quad (\ \because \ \zeta^{-5}=1 \ ) \end{align}
We indicate the reason why these formulas hold. As an example, we compute \(\rho(t_2)\) on the left-hand side of (9.8). Using (9.3) we can transform as follows.
\begin{align} e.g. \quad \rho(t_2)&=\frac{1}{5}\rho \left (h_0+\zeta^{2\cdot 1}h_1+\zeta^{2\cdot 2}h_2+\zeta^{2\cdot 3}h_3+\zeta^{2\cdot 4}h_4 \right) \notag \\ &=\frac{1}{5} \left (\rho(h_0)+\zeta^{2\cdot 1}\rho(h_1)+\zeta^{2\cdot 2}\rho(h_2) +\zeta^{2\cdot 3}\rho(h_3)+\zeta^{2\cdot 4}\rho(h_4) \right) \notag \\ &=\frac{1}{5} \left (h_1+\zeta^{2\cdot 1}h_2+\zeta^{2\cdot 2}h_3 +\zeta^{2\cdot 3}h_4+\\zeta^{2\cdot 4}h_0 \right) \notag \\ &=\frac{\zeta^{-2}}{5} \left (\zeta^{2\cdot 1}h_1+\zeta^{2\cdot 2}h_2 +\zeta^{2\cdot 3}h_3+\\zeta^{2\cdot 4}h_4+ \zeta^{2\cdot 5}h_0 \right) \notag \\ &=\zeta^{-2}t_2 \notag \\ \notag \\ & \therefore \ \rho(t_2)=\zeta^{-2}t_2 \\ \notag \\ e.g. \quad \rho(t_1^3 \cdot t_2)&=\rho(t_1^3) \cdot \rho(t_2)=\rho(t_1)^3 \cdot \rho(t_2) \notag \\ &=\bigl(\zeta^{-1}t_1 \bigr)^3 \cdot \bigl(\zeta^{-2}t_2 \bigr)= \zeta^{-3} \cdot \zeta^{-2} \cdot t_1^3 \cdot t_2 =t_1^3 \cdot t_2 \notag \\ \notag \\ & \therefore \ \rho(t_1^3 \cdot t_2)=t_1^3 \cdot t_2 \ \in F_2[x](=F_2) \\ \end{align}