Techniques for Solving Solvable Equations via Galois Theory

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[8-7] Cyclic Extension \(F_2/F_1\): Computing the Minimal Polynomial \(g_2(x)\)

The computations in this section correspond to the green block on the second tier of Fig.8-3.

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Composition series of the Frobenius group

The Galois group used here is the quotient \(D_5/C_5\), which is the cyclic group \(C_2\) of order \(2\).
Writing \(C_2\cong \{\kappa_1,\kappa_2\}\), the two elements \(\{\kappa_1,\kappa_2\}\) are cosets, each consisting of five elements of \(D_5\) as in (7.2). Accordingly, \(\{h_0,h_1\}\) are given by (7.3).

Step1 LRT (Lagrange resolvent transformation)

\begin{align} Gal(F_2/F_1)&=D_5/C_5 \cong C_2=\{\kappa_1,\kappa_2\} \\ \notag \\ &\left\{ \begin{array}{l} \kappa_1=\{\rho_{1},\rho_{43},\rho_{52},\rho_{90},\rho_{117}\} \cong C_5 \\ \kappa_2=\{\rho_{8},\rho_{30},\rho_{61},\rho_{95},\rho_{108}\} \\ \end{array} \right. \\ \notag \\ \end{align}

\begin{align} &\left\{ \begin{array}{l} h_0&=\displaystyle \prod_{\rho_i \ \in \ \kappa_1 }(x-\rho_i(v)) =(x-v_{1})(x-v_{43})(x-v_{52})(x-v_{90})(x-v_{117}) \\ h_1&=\displaystyle \prod_{\rho_i \ \in \ \kappa_2}(x-\rho_i(v)) =(x-v_{8})(x-v_{30})(x-v_{61})(x-v_{95})(x-v_{108}) \\ \end{array} \right. \\ \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}

Substitute into the \(v_i\) in (7.3) the polynomial expression of \(v\) defined in (4.7), and carry out the computations, always reducing \( (\mathrm{mod}\,g_1(v)) \).
Here the coefficients \(\{ca_i,cb_i\}\) of \(x^i\) in \(\{h_0,h_1\}\) become very complicated polynomials in \(v\), so we omit them.

\begin{align} &\left\{ \begin{array}{l} h_0=x^{5}+15^{4}+110x^{3}+ca_2x^{2}+ca_{1}x+ca_{0} \\ h_1=x^{5}+15^{4}+110x^{3}+cb_2x^{2}+cb_{1}x+cb_{0} \\ \end{array} \right. \\ \end{align}

Substitute \(\{h_0,h_1\}\) into the LRT (7.4) to obtain \(\{t_0,t_1\}\).

\begin{align} t_0&=x^5+15x^4+110x^3+\bigl(\frac{a_1}{2}+450 \bigr)x^2+\bigl(3a_1+1009 \bigr)x+\frac{29a_1}{2}-195 \\ \notag \\ t_1&=cd_2x^2+cd_1x+cd_0 \quad \in \ F_0(v)[x] \qquad \bigl[ \ cd_i=\frac{1}{2}(ca_i-cb_i) \quad i=(2,1,0) \ \bigr] \\ \end{align}

We now compute a new adjunction.

[Step2] The binomial equation \(B_1(x)=0\) and creation of a new adjunction \(a_1\)
\begin{align} &\left\{ \begin{array}{l} t_1=cd_m \cdot q_1(x) \ \in F_1(v)[x]\\ \\ cd_m \ \in F_1(v) \\ q_1(x) \ \in F_1[x]\\ \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} \tilde{t_1}=a_1 \cdot q_1(x) \ \in F_2[x] \\ \\ B_1(x)=x^2-A_1=0 \\ a_1=\sqrt {A_1} \ \in F_0(a_1) \equiv F_2\\ \end{array} \right. \notag \\ \end{align}

In (7.7) the leading coefficient of \(t_1\) at \(x^2\) is \(cd_2\). To match our earlier notation, we rename it \(cd_m\) (“m” for the maximum degree). Compute its inverse \(cd_m^{-1}\) and multiply it into \(t_1\).
The result is a monic polynomial \(q_2(x)\) lying in \(F_1[x]\).

\begin{align} dc_m&=\biggl[-\frac{5856606375 }{3587124922224482}a_1v^9-\frac{301046879875 {{v}^{9}}}{1793562461112241}+...-\frac{1122878979268800}{9390379377551} \biggr] \notag \\ \notag \\ dc_m^{-1}&=\biggl[ \frac{1479555083}{746121983822692256}a_1v^9+\frac{19069802845 {{v}^{9}}}{129136497200081352}+....+\frac{11758409218607}{112684552530612}\biggr] \notag \\ \notag \\ q_2&=dc_m^{-1} \cdot t_1 ={{x}^{2}}+\biggl(\frac{a_1}{20}+11 \biggr) x+\frac{43 {a_1}}{200}+\frac{251}{10} \quad \in F_1[x] \\ \notag \\ \end{align}

Next compute \(cd_m^2\). As shown in (7.9), it lies in \(F_0\). We denote this value by \(A_2\).
Looking at the two ends of (7.9), we can also say that \(cd_m\) is a root of the binomial equation \(B_2(x)=0\) in (7.10).
Thus we re-define \(cd_m\) as the radical of \(B_2(x)=0\) and denote it by the new symbol \(a_2\). By adjoining \(a_2\) to the base field \(F_1\), we generate the new extension \(F_2\).

\begin{align} cd_m^2&=\frac{135 {a_1}-11700}{2} \equiv A_2 \ \in \ F_1 \\ \notag \\ & \left\{ \begin{array}{l} B_2 \equiv a_2^2-\frac{135 {a_1}-11700}{2}=0 \\ a_2 \equiv \sqrt{A_1} \quad \in F_2 \equiv F_1(a_2) \\ \end{array} \right. \\ \end{align}

Using the newly introduced number \(a_2\), we can now define \( a_2 \cdot q_2 \) as a polynomial over \(F_2\), as shown in (7.13). To distinguish it from the original \(t_1\), we deliberately write \(\tilde{t_1}\) with a tilde.

\begin{align} &cd_m \ \in \ F_1(v) & &\Rightarrow & &a_1 \ \in \ F_2 \\ &t_1=cd_m \cdot q_1(x) \ \in F_1(v)[x] & &\Rightarrow & &\tilde{t_1}=a_1 \cdot q_1(x) \ \in F_2[x] \\ \end{align}

\begin{align} \notag \\ \tilde{t_1}& \equiv a_2 \biggl( {{x}^{2}}+\frac{{a_1} x}{20}+11 x+\frac{43 {a_1}}{200}+\frac{251}{10} \biggr) \ \in F_2[x] \\ \notag \\ \end{align}

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