Techniques for Solving Solvable Equations via Galois Theory

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

We now come to the final stage of the computation.

[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_0(x)=\tilde{h_0} \cdot \tilde{h_1} \\ g_1(x) \equiv \tilde{h_0} \ \in \ F_1[x] \end{array} \right. \\ \end{align}

Substituting \(t_0\) from (7.6) and \(\tilde{t_1}\) from (7.13) into the Inverse Lagrange resolvent transformation (8.1), we obtain \(\{\tilde{h_0},\tilde{h_1}\}\). We attach tildes to \(\{\tilde{h_0},\tilde{h_1}\}\) to distinguish them from \(\{h_0,h_1\}\).
Moreover, \(\tilde{h_0}\) contains the factor \((x-v)\). Hence it is natural to regard \(\tilde{h_0}\) as the minimal polynomial \(g_2(x)\) of \(v\) over \(F_2\). Summarizing the results, we have:

\begin{align} &B_2(x)= \ x^2-A_2=0 \qquad A_2=\frac{135 {a_1}-11700}{2} \ \in \ F_1 \notag \\ &a_2=\sqrt{A_2} \quad \Rightarrow \quad F_2 \equiv F_1(a_2) \notag \\ \notag \\ &\qquad \Downarrow \notag \\ \notag \\ &minimal \ polynomial: \ \tilde{h_0}=t_0+\tilde{t_1} \equiv g_2(x) \qquad deg(g_2(x))=5 \notag \\ \end{align}

\begin{align} \notag \\ g_2(x)&=x^5+15x^4+110x^3+\biggl(\frac{a_1}{2}+450 \biggr)x^2+\biggl(3a_1+1009 \biggr)x+\frac{29a_1}{2}-195\notag \\ &+ a_2 \biggl( {{x}^{2}}+\frac{{a_1} x}{20}+11 x+\frac{43 {a_1}}{200}+\frac{251}{10} \biggr) \quad \in F_2[x]\\ \end{align}

[Supplement 2] Constituents of the Frobenius Group

The Frobenius group is formed from a cyclic group \(\varLambda\) of order \(5\) together with five cyclic groups \(\{T_1,T_2,T_3,T_4,T_5\}\) of order \(4\). The numbers below mean, for example, that \([43]\equiv \rho_{43}\); they are automorphisms in the Galois group—keep this in mind.

\begin{align} &\varLambda=[\bbox[#FFAAAA]{43,117,90,52,1}]=[\rho_{43},\rho_{117},\rho_{90},\rho_{51},\rho_{1}] \notag \\ &T_1=[\bbox[#FFE680]{18,8,23},1]=[\rho_{18},\rho_{8},\rho_{23},\rho_{1}] \notag \\ &T_2=[\bbox[#A3FFA3]{70,30,110},1] \quad T_3=[\bbox[#A3FFFF]{33,61,73},1] \quad T_4=\bbox[#A3A3FF]{40,95,99},1] \quad T_5=[\bbox[#FFA3FF]{59,108,80},1] \notag \\ \end{align}

In particular, if we denote the elements \(\{\rho_{43},\rho_{18}\}\) of the cyclic groups \(\{\varLambda,T_1\}\) by the symbols \(\{\lambda,\tau\}\), respectively, then it is clear that \(\{\varLambda,T_1\}\) have the form of cyclic groups:

\begin{align} \rho_{43}&\equiv \lambda & &\Rightarrow & \varLambda&=[43,117,90,52,1]=[\lambda,\lambda^2,\lambda^3,\lambda^4,\lambda^5] \notag \\ \rho_{18} &\equiv \tau & &\Rightarrow & T_1&=[18,8,23,1]=[\tau,\tau^2,\tau^3,\tau^4] \notag \\ \end{align}

First, construct the product table \(\{\varLambda \times T_1\}\) in [Table8-3].
In the first row of [Table8-3] we see the sequence \([1,18,8,23]\). That sequence is placed in the central square of [Fig8-a]. Furthermore, in the first column of the product table there is the sequence \([1,43,117,90,51]\), which is the cyclic group \(\varLambda\) of order \(5\). That sequence lies on the pentagon above the entry \([1]\) of the central square. In the same way, the three sequences of five elements in columns 2 through 4 are placed on the remaining three pentagons. We thus obtain the following table and figure.

[Table8-3] Product table \(\varLambda \times T_1\)
\(\) \(T_1\) \(1\) \(18\) \(8\) \(23\)

\(\varLambda\) \(\) \(\tau^4\) \(\tau\) \(\tau^2\) \(\tau^3\)

\(1\) \(\lambda^5\) \(1\) \(18\) \(8\) \(23\)

\(43\) \(\lambda\) \(43\) \(33\) \(30\) \(40\)

\(117\) \(\lambda^2\) \(117\) \(99\) \(108\) \(110\)

\(90\) \(\lambda^3\) \(90\) \(80\) \(95\) \(73\)

\(52\) \(\lambda^4\) \(52\) \(70\) \(61\) \(59\)

[Table8-3] Product table \(\varLambda \times T_1\)
\(\)	\(T_1\)	\(1\)	\(18\)	\(8\)	\(23\)
\(\varLambda\)	\(\)	\(\tau^4\)	\(\tau\)	\(\tau^2\)	\(\tau^3\)
\(1\)	\(\lambda^5\)	\(1\)	\(18\)	\(8\)	\(23\)
\(43\)	\(\lambda\)	\(43\)	\(33\)	\(30\)	\(40\)
\(117\)	\(\lambda^2\)	\(117\)	\(99\)	\(108\)	\(110\)
\(90\)	\(\lambda^3\)	\(90\)	\(80\)	\(95\)	\(73\)
\(52\)	\(\lambda^4\)	\(52\)	\(70\)	\(61\)	\(59\)

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Next, construct the product table \(\{T_1 \times \varLambda\}\) in [Table8-4].
The first row of the product table contains the sequence \([1,43,117,90,52]\). That sequence is arranged on the central circular band of the diagram of \(F_{20}\) in [Fig8-b]. The first column contains the sequence \([1,18,8,23]\); attached to the entry \([1]\) on the central circle is a small circle carrying four sequences. Likewise, the small circles corresponding to the four sequences in columns 2 through 5 are attached in order to the central circle. In this way the upper-left figure shows that every element of the Frobenius group \(F_{20}\) appearing in [Table8-4] is placed in the diagram. The other three figures in [Fig8-b] visualize how the number of group elements decreases along the composition series of \(F_{20}\).

[Table8-4] Product table \(T_1 \times \varLambda\)
\(\) \(\varLambda\) \(1\) \(43\) \(117\) \(90\) \(52\)

\(T_1\) \(\) \(\lambda^5\) \(\lambda\) \(\lambda^2\) \(\lambda^3\) \(\lambda^4\)

\(1\) \(\tau^4\) \(1\) \(43\) \(117\) \(90\) \(52\)

\(18\) \(\tau\) \(18\) \(80\) \(33\) \(70\) \(99\)

\(8\) \(\tau^2\) \(8\) \(61\) \(95\) \(108\) \(30\)

\(23\) \(\tau^3\) \(23\) \(110\) \(59\) \(40\) \(73\)

[Table8-4] Product table \(T_1 \times \varLambda\)
\(\)	\(\varLambda\)	\(1\)	\(43\)	\(117\)	\(90\)	\(52\)
\(T_1\)	\(\)	\(\lambda^5\)	\(\lambda\)	\(\lambda^2\)	\(\lambda^3\)	\(\lambda^4\)
\(1\)	\(\tau^4\)	\(1\)	\(43\)	\(117\)	\(90\)	\(52\)
\(18\)	\(\tau\)	\(18\)	\(80\)	\(33\)	\(70\)	\(99\)
\(8\)	\(\tau^2\)	\(8\)	\(61\)	\(95\)	\(108\)	\(30\)
\(23\)	\(\tau^3\)	\(23\)	\(110\)	\(59\)	\(40\)	\(73\)

Although drawing [Fig8-a] and [Fig8-b] does not in itself achieve anything remarkable, I hope they help in understanding the Frobenius group \(F_{20}\).

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