Techniques for Solving Solvable Equations via Galois Theory

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[7-5] Cyclic extension \(F_3/F_2\): computation of the minimal polynomial \(g_3(x)\)

Following the calculation procedure described in earlier chapters, we can compute \(g_3(x)\) straightforwardly.

[step1] LRT (Lagrange Resolvent transformation)

\begin{align} h_0&=\prod_{\rho_i \in \ C_2}\rho_i(x-v)=(x-v_1)(x-v_{16}) \\ h_1&=\prod_{\rho_i \in \ (C_{4}-C_2)}\rho_i(x-v)=(x-v_4)(x-v_{12}) \\ \end{align} \begin{align} \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} \\ \end{align}

Substituting the polynomial expressions in \(v\) from (1.4) into the \(\{h_0,h_1\}\) of (5.1)(5.2) and carrying out the computation, we obtain the following. From here on, all computations are done \( (\mathrm{mod}\,g_2(v)) \ \rightarrow \ (\mathrm{mod}\,B_2(a_2)) \ \rightarrow \ (\mathrm{mod}\,B_1(a_1))\).

\begin{align} h_0&=(x-v_1)(x-v_{16}) \notag \\ &=x^2+\frac{1}{4}\biggl( 4v^3+(4a_2+2a_1+1)v^2+((2a_1-1)a_2 \notag \\ &\qquad \qquad +2a_1+3)v+4a_2+2a_1+1 \biggr)x+1 \\ h_1&=(x-v_4)(x-v_{12}) \notag \\ &=x^2-\frac{x}{4}\biggl(4v^3+(4a_2+2a_1+1)v^2+((2a_1-1)a_2+2a_1+3)v\biggr)+1 \\ \notag \\ \end{align}

Substituting (5.4)(5.5) into (5.3) yields \(\{ t_0,t_1\}\).

\begin{align} t_0&=x^2+\frac{(4a_2+2a_1+1)x}{8}+1 \quad \in \ F_2[x] \\ t_1&=\frac{x}{8}\biggl( 8v^3+(8a_2+4a_1+2)v^2+((4a_1-2)a_2+4a_1+6)v \notag \\ &\qquad \qquad +4a_2+2a_1+1 \biggr) \quad \in \ F_2(v)[x] \\ \end{align}

From this we see that \(\bigl[\ t_0 \in F_2[x], \ t_1 \in F_2(v)[x] \ \bigr]\).
Let \(cd_m\) denote the leading coefficient of \(t_1\). Using \(cd_m\), we introduce a new adjoint element \(a_3\).

[step2] The binomial equation \(B_3(x)=0\) and construction of a new adjoint element \(a_3\)
\begin{align} & \left\{ \begin{array}{l} t_1=cd_m \cdot q_3(x) \ \in F_2(v)[x] \\ cd_m \ \in F_2(v) \\ q_3(x)=x \ \in F_2[x] \\ \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} \tilde{t_1}=a_3 \cdot q_3(x)=a_3x \ \in F_3[x] \\ B_3(x)=x^2-A_3=0 \\ a_3=\sqrt {A_3} \ \in F_2(a_3) \equiv F_3\\ \end{array} \right. \notag \\ \end{align}

Since the cyclic extension has degree 2, let us compute \(cd_m^2\). Then, as in (5.9), \(cd_m^2\) becomes a value in \(F_2\); denote this by \(A_3\). Using \(A_3\), we define the binomial equation \(B_3(x)=0\) that the newly introduced element \(a_3\) must satisfy. We define \(a_3\) as a radical of this binomial equation. By adjoining \(a_3\) to \(F_2\), we obtain the extension field \(F_3\).

\begin{align} cd_m&=\frac{\biggl( 8v^3+(8a_2+4a_1+2)v^2+...+4a_2+2a_1+1 \biggr) }{8} \ \in \ F_2(v)\\ cd_m^2&=-\frac{{a_1} \left( 4 {a_2}+6\right) -6 {a_2}-17}{16} \equiv A_3 \quad \in \ F_2 \\ &\qquad \Downarrow \notag \\ B_3(x)&\equiv x^2-A_3=0 \quad \rightarrow \quad a_3=\sqrt{A_3} \ \in \ F_3 \equiv F_2(a_3)\\ &\qquad \Downarrow \notag \\ \tilde{t_1} &\equiv a_3 \cdot q_3(x)= a_3x \quad \in F_3[x] \\ \end{align}

With this new adjoint element \(a_3\), as in (5.11), \(t_1\) is expressed as a polynomial over the extension field \(F_3\). Since \(t_1\) is now a polynomial over the extension field \(F_3\), we denote it by \(\tilde{t_1}\) rather than \(t_1\).

[step3] 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_2(x)=\tilde{h_0} \cdot \tilde{h_1} \\ g_3(x) \equiv \tilde{h_0} \ \in \ F_3[x] \end{array} \right. \notag \\ \end{align}

Since \(h_0\) originally contains the factor \((x-v)\), it follows that \(\tilde{h_0}\) is the new minimal polynomial \(g_3(x)\) over the extension field \(F_3\).

\begin{align} \tilde{h_0}&=t_0+\tilde{t_1} \equiv g_3(x) \\ \notag \\ g_3(x)&=x^2+\biggl(a_3+\frac{a_2}{2}+\frac{a_1}{4}+\frac{1}{8}\biggr)x+1 \\ \end{align}

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