Techniques for Solving Solvable Equations via Galois Theory

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[3-14] Computing \(g_4(x)\) for \(F_4/F_3\)

We now compute \(g_4(x)\). The green region in Fig. 3-5 indicates this step of the solution procedure.

\(\qquad\qquad\)

The relevant Galois group is \(N/e=N\cong C_2\), a cyclic group of order 2. In this section the elements of \(C_2\) are exactly the elements of \(N\).

\begin{align} &Gal(F_4/F_3)=N/e \cong C_2 \qquad C_2 =\{\rho_{1},\rho_{8}\} \notag \\ \end{align}

The computation of the minimal polynomial \(g_4(x)\) of \(v\) has three steps. We start with Step 1.

[Step 1] LRT (Lagrange Resolvent Transformation)
\begin{align} & h_0=(x-v_1)\\ & h_1=(x-v_8) \notag \\ \notag \\ &\begin{bmatrix} t_0 \\ t_1 \end{bmatrix} =\frac{1}{2} \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} \cdot \begin{bmatrix} h_0 \\ h_1 \end{bmatrix} \qquad (\ t_1:\ \text{Lagrange resolvent}\ ) \end{align}

We examine \(\{h_0,h_1\}\) from the viewpoint of automorphisms. [Table3-18] shows the action of \(\{\rho_1,\rho_8\}\) on \(v_i\), and [Table3-19] the induced action on \(\{h_0,h_1\}\).

\begin{align} &\left\{ \begin{array}{l} h_0=x-v_1\\ h_1=x-v_8 \end{array} \right. \notag \\ \end{align}

\(\quad \Rightarrow \quad\)

[Table3-18] \(\rho_j(v_i)\)
\(\ \)	\(\rho_i(v_1)\)	\(\rho_i(v_8)\)
\(\rho_1\)	\(v_1\)	\(v_8\)
\(\rho_8\)	\(v_8\)	\(v_1\)

\(\quad \Rightarrow \quad\)

[Table3-19] \(\rho_j(h_i)\)
\(\ \)	\(\rho_j(h_0)\)	\(\rho_j(h_1)\)
\(\rho_1\)	\(h_0\)	\(h_1\)
\(\rho_8\)	\(h_1\)	\(h_0\)

By (14.2) we obtain \(\{t_0,t_1\}\) as

\begin{align} t_0=\tfrac{1}{2}(h_0+h_1)=x-(v_1+v_8),\qquad t_1=\tfrac{1}{2}(h_0-h_1)=-(v_1-v_8). \\ \end{align}

The induced action on their constants \((v_1\pm v_8)\) is listed in [Table3-20], and hence the action on \(\{t_0,t_1\}\) is [Table3-21] (note we keep the table here so “fourth column” below is unambiguous).

[Table3-20] \(\rho_j(v_1 \pm v_8)\)
\(\ \)	\(\rho_j(v_1+v_8)\)	\(\rho_j(v_1-v_8)\)
\(\rho_1\)	\(v_1+v_8\)	\(v_1-v_8\)
\(\rho_8\)	\(v_8+v_1\)	\(-(v_1-v_8)\)

\(\quad \Rightarrow \quad\)

[Table3-21] \(\rho_j(t_i)\)
\(\ \)	\(\rho_j(t_0)\)	\(\rho_j(t_1)\)	\(\rho_j(t_1^2)\)
\(\rho_1\)	\(t_0\)	\(t_1\)	\(t_1^2\)
\(\rho_8\)	\(t_0\)	\(-t_1\)	\(t_1^2\)

Substituting the polynomial expressions of \(v\) into (14.3), and taking remainders in the order \([\,\bmod\ g_3(v),\ \bmod\ B_3,\ \bmod\ B_2,\ \bmod\ B_1,\ \bmod\ \Omega\,]\), we obtain

\begin{align} t_0&= x+\frac{a_3}{2} \quad \in F_3[x]\\ t_1&=-v-\frac{a_3}{2} \quad \in F_3(v) \end{align}

As indicated in the fourth column of [Table3-21], \(t_1^2\) is invariant under \(C_2\), so \(t_1^2\in F_2\). A direct computation gives

\begin{align} & \bbox[#FFFF00]{t_1^2}=-\frac{11 {a_1} {{a}_{2}^{2}} \omega }{2595840}+\frac{9 {{a}_{2}^{2}} \omega }{676} +\frac{2 {a_2} \omega }{13}-\frac{23 {a_1} {{a}_{2}^{2}}}{5191680}-\frac{63 {{a}_{2}^{2}}}{5408} +\frac{3 {a_2}}{26} \equiv \bbox[#FFFF00]{A_4} \ \in \ F_2\\ \notag \\ &B_4(x) \equiv x^2-A_4=0,\qquad a_4 \equiv \sqrt{A_4} \quad \Rightarrow \quad \bbox[#FFFF00]{ F_4 \equiv F_3(a_4) } \\ &t_1 \ \Rightarrow\ \tilde{t_1}=a_4 \ \in \ F_4 \end{align}

It turns out that \(t_1\) is a root of the binomial equation \(B_4(x)=0\). We therefore define \(a_4=\sqrt{A_4}\) and adjoin it to obtain the Galois extension \(F_4=F_3(a_4)\). Using this newly introduced \(a_4\), \(t_1\) can be represented as an element over the enlarged field \(F_4\) as in (14.8). To avoid confusion, we deliberately write \(\tilde{t_1}\) instead of \(t_1\) from this point on.

[Step 2] Binomial \(B_4(x)=0\) and the new adjunction \(a_4\)
\begin{align} t_1=-v-\frac{a_3}{2}\ \in F_3(v) \quad \Rightarrow \quad \left\{ \begin{array}{l} B_4(x)=x^2-A_4=0,\\ a_4=\sqrt {A_4} \in F_3(a_4)\equiv F_4,\\ \tilde{t_1}=a_4 \in F_4 \end{array} \right. \notag \\ \end{align}

Finally, substitute \(t_0\) from (14.4) and \(\tilde t_1\) from above into the inverse LRT to compute \(\tilde h_0\).

[Step 3] ILRT (Inverse Lagrange Resolvent Transformation)
\begin{align} \begin{bmatrix}\tilde{h_0} \\ \tilde{h_1}\end{bmatrix} = \begin{bmatrix}1&1\\ 1&-1\end{bmatrix} \cdot \begin{bmatrix}t_0\\ \tilde{t_1}\end{bmatrix} \quad \Rightarrow \quad \left\{ \begin{array}{l} g_3(x)=\tilde{h_0}\cdot \tilde{h_1},\\ g_4(x)\equiv \tilde{h_0} \in F_4[x] \end{array} \right. \end{align}

Since \(h_0=(x-v_1)\), the polynomial \(\tilde h_0\) is the desired minimal polynomial \(g_4(x)\) of \(v\) over \(F_4\):

\begin{align} &\tilde{h_0}=x+\frac{a_3}{2}+a_4 \equiv g_4(x) \\ \notag \\ &B_4(x)=x^2-A_4=0,\quad a_4=\sqrt{A_4}\ \Rightarrow\ F_4 \equiv F_3(a_4)\\ \notag \\ &A_4=-\frac{11 {a_1} {{a}_{2}^{2}} \omega }{2595840}+\frac{9 {{a}_{2}^{2}} \omega }{676} +\frac{2 {a_2} \omega }{13}-\frac{23 {a_1} {{a}_{2}^{2}}}{5191680}-\frac{63 {{a}_{2}^{2}}}{5408} +\frac{3 {a_2}}{26} \end{align}

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