Techniques for Solving Solvable Equations via Galois Theory

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[7-6] Cyclic extension \(F_4/F_3\): computation of the minimal polynomial \(g_4(x)\)

Following the calculation procedure described in earlier chapters, we can compute \(g_4(x)\) straightforwardly.

[step1] LRT (Lagrange Resolvent transformation)

\begin{align} &\left\{ \begin{array}{l} h_0=(x-v_1) \\ h_1=(x-v_{16}) \\ \end{array} \right. \\ \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 (6.1) and carrying out the computation, we obtain the following. From here on, all computations are done \( (\mathrm{mod}\,g_3(v)) \ \rightarrow \ (\mathrm{mod}\,B_3(a_3)) \ \rightarrow \ (\mathrm{mod}\,B_2(a_2)) \ \rightarrow \ (\mathrm{mod}\,B_1(a_1)) \).

\begin{align} h_0&=(x-v_1)=x-v \\ h_1&=(x-v_{16})=x+v+{a_3}+\tfrac{{a_2}}{2}+\tfrac{{a_1}}{4}+\tfrac{1}{8} \\ \notag \\ \end{align}

Substituting (6.3)(6.4) into (6.2) yields \(\{ t_0,t_1\}\).

\begin{align} t_0&=x+\tfrac{8a_3+4a_2+2a_1+1}{16} \quad \in \ F_3[x]\\ t_1&=-v-\tfrac{8a_3+4a_2+2a_1+1}{16} \quad \in \ F_3(v) \\ \end{align}

From this we see that \(\bigl[\ t_0 \in F_3[x], \ t_1 \in F_3(v) \ \bigr]\). Using \(t_1\), we now introduce a new adjoint element \(a_4\).

[step2] The binomial equation \(B_4(x)=0\) and construction of a new adjoint element \(a_4\)
\begin{align} t_1 \ \in F_3(v) \quad \Rightarrow \quad \left\{ \begin{array}{l} \tilde{t_1}=a_4\ \in F_4 \\ B_4(x)=x^2-A_4=0 \\ a_4=\sqrt {A_4} \ \in F_3(a_4) \equiv F_4\\ \end{array} \right. \notag \\ \end{align}

Since the cyclic extension has degree 2, let us compute \(t_1^2\). Then, as in (6.8), \(t_1^2\) becomes a value in \(F_3\); denote this by \(A_4\). Using \(A_4\), we define the binomial equation \(B_4(x)=0\) that the newly introduced element \(a_4\) must satisfy. We define \(a_4\) as a radical of this binomial equation. By adjoining \(a_4\) to \(F_3\), we obtain the extension field \(F_4\).

\begin{align} t_1&= -v-\tfrac{8a_3+4a_2+2a_1+1}{16} \quad \in F_3(v)\\ t_1^2&=\tfrac{{a_1} \left( 4 {a_3}-2\right) +\left( 8 {a_2}+2\right) {a_3}+4 {a_2}-17}{32} \equiv A_4 \quad \in \ F_3 \\ &\qquad \Downarrow \notag \\ B_4(x)&\equiv x^2-A_4=0 \quad \rightarrow \quad a_4=\sqrt{A_4} \ \in \ F_4 \equiv F_3(a_4)\\ &\qquad \Downarrow \notag \\ \tilde{t_1} &\equiv a_4 \quad \in F_4 \\ \end{align}

With this new adjoint element \(a_4\), as in (6.10), \(t_1\) is expressed as a number in the extension field \(F_4\). Since \(t_1\) is now a number in \(F_4\), 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_3(x)=\tilde{h_0} \cdot \tilde{h_1} \\ g_4(x) \equiv \tilde{h_0} \ \in \ F_4[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 \(g_4(x)\) over the extension field \(F_3\). Furthermore, since \(g_4(x)\) is now a linear polynomial in \(v\), the equation \([ \ g_4(x)=0 \ ]\) finally gives the value of \(v\).
Substituting this value of \(v\) into (1.4) yields all roots of \(\Phi_{17}=0\).

\begin{align} &\tilde{h_0}=t_0+\tilde{t_1} \equiv g_4(x) \\ &g_4(x)=x+{a_4}+\tfrac{{a_3}}{2}+\tfrac{{a_2}}{4}+\tfrac{{a_1}}{8}+\tfrac{1}{16} \\ \notag \\ &\therefore \ v=-\biggl[{a_4}+\tfrac{{a_3}}{2}+\tfrac{{a_2}}{4}+\tfrac{{a_1}}{8}+\tfrac{1}{16}\biggr] \quad \in F_4 \\ \end{align}

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