Techniques of Solving Equations à la Galois
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[7-4] Cyclic extension \(F_2/F_1\): computation of the minimal polynomial \(g_2(x)\)
Following the calculation procedure described in earlier chapters, we can compute \(g_2(x)\) straightforwardly.\begin{align} h_0&=\prod_{\rho_i \in \ C_4}\rho_i(x-v)=(x-v_1)(x-v_4)(x-v_{13})(x-v_{16}) \\ h_1&=\prod_{\rho_i \in \ (C_{8}-C_4)}\rho_i(x-v)=(x-v_2)(x-v_8)(x-v_{9})(x-v_{15}) \\ \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 (4.1)(4.2) and carrying out the computation, we obtain the following. From here on, all computations are done \( (\mathrm{mod}\,g_1(v)) \ \rightarrow \ (\mathrm{mod}\,B_1(a_1))\).
\begin{align} h_0&=(x-v_1)(x-v_4)(x-v_{13})(x-v_{16}) \notag \\ &=x^4+\frac{\left( 4 {{v}^{7}}+\left( 4 {a_1}+6\right) {{v}^{6}}+...+\left( 4 {a_1}+10\right) v+4 {a_1}+4\right) }{2}x^3+...\notag \\ &+\frac{\left( 4 {{v}^{7}}+\left( 4 {a_1}+6\right) {{v}^{6}}+...+\left( 4 {a_1}+10\right) v+4 {a_1}+4\right) }{2}x+1\\ h_1&=(x-v_2)(x-v_8)(x-v_9)(x-v_{15}) \notag \\ &=x^4+\frac{\bigl(-4v^7+(-4a_1-6)v^6+...+(-4a_1-10)v-2a_1-3\bigr)}{2}x^3+... \notag \\ &+\frac{\bigl( -4v^7+(-4a_1-6)v^6+...+(-4a_1-10)v-2a_1-3\bigr)}{2}x+1 \\ \notag \\ \end{align}
\begin{align} t_0&=x^4+\frac{(2a_1+1)}{4}x^3+\frac{(2a_1+7)}{4}x^2+\frac{(2a_1+1)}{4}+1 \quad \in \ F_1[x]\\ t_1&=\frac{\bigl( 8v^7+(8a_1+12)v^6+...+(8a_1+20)v+6a_1+7 \bigr) }{4}x^3 \notag \\ & +\frac{\bigl( (4a_1-2)v^7+(4a_1+14)v^6+...+(8a_1+12)v+2a_1+11 \bigr) }{4}x^2 \notag \\ & +\frac{\bigl( 8v^7+(8a_1+12)v^6+...+(8a_1+20)v+6a_1+7 \bigr) }{4}x \quad \in \ F_1(v)[x] \\ \end{align}
From this we see that \(\bigl[\ t_0 \in F_1[x], \ t_1 \in F_1(v)[x] \ \bigr]\).
Let \(cd_m\) denote the leading coefficient of \(t_1\). Using \(cd_m\), we introduce a new adjoint element \(a_2\).
[step2] The binomial equation \(B_2(x)=0\) and construction of a new adjoint element \(a_2\)
\begin{align}
& \left\{
\begin{array}{l}
t_1=cd_m \cdot q_2(x) \ \in F_1(v)[x] \\
cd_m \ \in F_1(v) \\
q_2(x) \ \in F_1[x] \\
\end{array}
\right.
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
\tilde{t_1}=a_2 \cdot q_2(x) \ \in F_2[x] \\
B_2(x)=x^2-A_2=0 \\
a_2=\sqrt {A_2} \ \in F_1(a_2) \equiv F_2\\
\end{array}
\right. \notag \\
\end{align}
Since the cyclic extension has degree 2, let us compute \(cd_m^2\). Then, as in (4.9), \(cd_m^2\) becomes a value in \(F_1\); denote this by \(A_2\). Using \(A_2\), we define the binomial equation \(B_2(x)=0\) that the newly introduced element \(a_2\) must satisfy. We define \(a_2\) as a radical of this binomial equation. By adjoining \(a_2\) to \(F_1\), we obtain the extension field \(F_2\).
\begin{align} cd_m&=\frac{\bigl( 8v^7+(8a_1+12)v^6+...+(8a_1+20)v+6a_1+7 \bigr) }{4} \ \in \ F_1(v)\\ cd_m^2&=\frac{2 {a_1}+17}{8} \equiv A_2 \quad \in \ F_1 \\ &\qquad \Downarrow \notag \\ B_2(x)&\equiv x^2-A_2=0 \quad \rightarrow \quad a_2=\sqrt{A_2} \ \in \ F_2 \equiv F_1(a_2)\\ \end{align}
We further transform \(t_1\). For this we compute the reciprocal of \(cd_m\). (Methods for computing reciprocals are explained in Chapter 2.) Multiplying \(t_1\) by this \(cd_m^{-1}\) yields \(q_2(x) \in F_1[x]\).
\begin{align} cd_m^{-1}&=-\frac{\bigl( (4a_1-34)v^7+(-28a_1-34)v^6+...+(-24a_1-68)v-22a_1-17)}{34} \\ q_2(x)&=cd_m^{-1} \cdot t_1=x^3+\frac{(2a_1-1)}{4}x^2+x \quad \in \ F_1[x]\\ &\qquad \Downarrow \notag \\ \tilde{t_1} &\equiv a_2 \cdot q_2(x)= a_2 \cdot \biggl(x^3+\frac{(2a_1-1)}{4}x^2+x\biggr) \quad \in F_2[x] \\ \end{align}
Since \(h_0\) originally contains the factor \((x-v)\), it follows that \(\tilde{h_0}\) is the new minimal polynomial \(g_2(x)\) over the extension field \(F_2\).
\begin{align} \tilde{h_0}&=t_0+\tilde{t_1} \equiv g_2(x) \\ \notag \\ g_2(x)&=x^4+\frac{(4a_2+2a_1+1)}{4}x^3+\frac{ \bigl( (2a_1-1)a_2+2a_1+7 \bigr) }{4}x^2 \notag \\ &+\frac{(4a_2+2a_1+1)}{4}x+1 \\ \end{align}