Techniques of Solving Equations à la Galois
Profile
Name: scruta \(\quad\)
Daily life: mowing
Revision history
1st upload: 2023/06/17
revision2 : 2023/07/27
revision3 : 2024/12/22
revision4 : 2025/09/14
Contact
Copyright © 2023 scruta
[8-6] Cyclic Extension \(F_1/F_0\): Computing the Minimal Polynomial \(g_1(x)\) (continued)
We now arrive at 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 (5.5) and \(\tilde{t_1}\) from (5.13) into the Inverse Lagrange resolvent transformation (6.1), we obtain \(\{\tilde{h_0},\tilde{h_1}\}\). We add 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 \(g_1(x)\) of \(v\) over \(F_1\). Summarizing the computation, we obtain the following.
\begin{align} &B_1(x)= \ x^2-A_1=0 \qquad A_1=5200 \ \in \ F_0 \notag \\ &a_1=\sqrt{A_1} \quad \Rightarrow \quad F_1 \equiv F_0(a_1) \notag \\ \notag \\ &\qquad \Downarrow \notag \\ \notag \\ &minimal \ polynomial: \ \tilde{h_0}=t_0+\tilde{t_1} \equiv g_1(x) \qquad deg(g_1(x))=10 \notag \\ \end{align}
\begin{align} \notag \\ g_1(x)&=x^{10}+30x^9+445x^8++4200x^7+27618x^6+128880x^5 \notag \\ & +425780x^4+974400x^3+1505321x^2+1402050x+2434705 \notag \\ &+a_1 \biggl( x^7+21x^6+229x^5+\frac{2955}{2}x^4+5999x^3+15426x^2+25625x-\frac{6333}{5} \biggr) \in \ F_1[x]\\ \end{align}
[Supplement 1] How to Find a Composition Series of the Frobenius Group
There are two ways to compute a composition series.(M1) Compute the commutator subgroup series of \(F_{20}\). This is the simplest, but note that doing so yields \([\ F_{20} \ \triangleright \ C_{5} \ \triangleright \ e \ ]\) and misses the dihedral group \(D_5\).
(M2) Compute the conjugacy classes of \(F_{20}\). This method is common in representation theory. It is a bit roundabout, nevertheless it does not miss \(D_5\) as in (M1), and it reveals how all constituents of the composition series are formed.
Here we use method (M2) based on conjugacy classes to find a composition series.
First we recall the Frobenius group of order \(20\) from (3.14). For simplicity we write the automorphisms \(\rho_i\) as plain numbers \([i]\).
\begin{align} F_{20}&=\{ \ \rho_{1},\rho_{8},\rho_{18},\rho_{23},\rho_{30},\rho_{33},\rho_{40},\rho_{43},\rho_{52},\rho_{59}, \notag \\ &\qquad \rho_{61},\rho_{70},\rho_{73},\rho_{80},\rho_{90},\rho_{95},\rho_{99},\rho_{108},\rho_{110},\rho_{117} \ \} \notag \\ \Downarrow \notag \\ F_{20}&=[1,8,18,23,30,33,40,43,52,59,61,70,73,80,90,95,99,108,110,117] \notag \\ \end{align}
| \(class\) | \(\sharp \ \varLambda_i \) | \(elements \ \rho_j\) |
|---|---|---|
| \({\varLambda_1}\) | \(1\) | \([1]\) |
| \({\varLambda_2}\) | \(4\) | \([43,52,90,117]\) |
| \({\varLambda_3}\) | \(5\) | \([8,30,61,95,108]\) |
| \({\varLambda_4}\) | \(5\) | \([18,33,70,80,99]\) |
| \({\varLambda_5}\) | \(5\) | \([23,40,59,73,110]\) |
| \({i \backslash j}\) | \(\varLambda_1\) | \({\varLambda_2}\) | \({\varLambda_3}\) | \({\varLambda_4}\) | \({\varLambda_5}\) |
|---|---|---|---|---|---|
| \({\varLambda_1}\) | \(\varLambda_1\) | \({\varLambda_2}\) | \({\varLambda_3}\) | \({\varLambda_4}\) | \({\varLambda_5}\) |
| \({\varLambda_2}\) | \(\varLambda_2\) | \({4\varLambda_1+3\varLambda_2}\) | \({4\varLambda_3}\) | \({4\varLambda_4}\) | \({4\varLambda_5}\) |
| \({\varLambda_3}\) | \(\varLambda_3\) | \({4\varLambda_3}\) | \({5\varLambda_1+5\varLambda_2}\) | \({5\varLambda_5}\) | \({5\varLambda_4}\) |
| \({\varLambda_4}\) | \(\varLambda_4\) | \({4\varLambda_4}\) | \({5\varLambda_5}\) | \({5\varLambda_3}\) | \({5\varLambda_1+5\varLambda_2}\) |
| \({\varLambda_5}\) | \(\varLambda_5\) | \({4\varLambda_5}\) | \({5\varLambda_4}\) | \({5\varLambda_1+5\varLambda_2}\) | \({5\varLambda_3}\) |
From the table of products \(\varLambda_i \times \varLambda_j\) in [Table8-2], we find three things:
(1) The single entry in the upper-left \(1\times1\) block contains only \(\{\varLambda_1\}\) (trivial).
(2) The four entries in the upper-left \(2\times2\) block contain only \(\{\varLambda_1,\varLambda_2\}\) (ignore coefficients such as 3, 4).
(3) The nine entries in the upper-left \(3\times3\) block contain only \(\{\varLambda_1,\varLambda_2,\varLambda_3\}\) (again ignore coefficients such as 3, 4, 5).
This means that the normal subgroups \(\{e,C_5,D_5\}\) of the Frobenius group \(F_{20}\) are formed from the following combinations of conjugacy classes.
\begin{align} e=\varLambda_1 \qquad C_5=\varLambda_1+\varLambda_2 \qquad D_5=\varLambda_1+\varLambda_2+\varLambda_3 \notag \\ \end{align}
\begin{align} F_{20}&=\varLambda_1+\varLambda_2+\varLambda_3+\varLambda_4+\varLambda_5 \notag \\ &=[1,8,18,23,30,33,40,43,52,59,61,70,73,80,90,95,99,108,110,117] \notag \\ \notag \\ D_5&=\varLambda_1+\varLambda_2+\varLambda_3=[1,8,30,43,52,61,90,95,108,117] \notag \\ \notag \\ C_5&=\varLambda_1+\varLambda_2=[1,43,52,90,117] \notag \\ \notag \\ e&=\varLambda_1=[1] \notag \\ \end{align}