Techniques of Solving Equations à la Galois
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[8-5] Cyclic Extension \(F_1/F_0\): Computing the Minimal Polynomial \(g_1(x)\)
The computations in this section correspond to the green block on the second tier of Fig.8-2.
The Galois group we use here is the quotient \(F_{20}/D_5\), which is the cyclic group of order \(2\), \(C_2\).
Writing \(C_2\cong \{\kappa_1,\kappa_2\}\), the two elements \(\{\kappa_1,\kappa_2\}\) are cosets, each consisting of ten elements of \(F_{20}\) as in (5.2). Accordingly, \(\{h_0,h_1\}\) are given by (5.3).
Step1 LRT (Lagrange resolvent transformation)
\begin{align} Gal(F_1/F_0)&=F_{20}/D_5 \cong C_2=\{\kappa_1,\kappa_2\} \\ \end{align}
\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 \\ \notag \\ &\left\{ \begin{array}{l} \kappa_1=\{ \ \rho_{1},\rho_{8},\rho_{30},\rho_{43},\rho_{52},\rho_{61},\rho_{90},\rho_{95},\rho_{108},\rho_{117} \ \}=D_5 \\ \kappa_2=\{ \ \rho_{18},\rho_{23},\rho_{33},\rho_{40},\rho_{59},\rho_{70},\rho_{73},\rho_{80},\rho_{99},\rho_{110} \ \} \\ \end{array} \right. \\ \end{align}
\begin{align} \notag \\ &\left\{ \begin{array}{l} h_0&=\displaystyle \prod_{\rho_i \ \in \ \kappa_1 }(x-\rho_i(v)) \\ &=(x-v_{1})(x-v_{8})(x-v_{30})(x-v_{43})(x-v_{52}) \\ &\times (x-v_{61})(x-v_{90})(x-v_{95})(x-v_{108})(x-v_{117}) \\ \\ h_1&=\displaystyle \prod_{\rho_i \ \in \ \kappa_2}(x-\rho_i(v)) \\ &=(x-v_{18})(x-v_{23})(x-v_{33})(x-v_{40})(x-v_{59}) \\ &\times (x-v_{70})(x-v_{73})(x-v_{80})(x-v_{99})(x-v_{110}) \\ \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 (5.3) the polynomial expressions for \(v\) computed in (4.7) of the previous section. Be sure to reduce \( (\mathrm{mod}\,g_0(v)) \) at every step.
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^{10}+30x^{9}+455x^{8}+ca_7x^{7}+ca_6x^{6}+...+ca_{1}x+ca_{0} \\ h_1=x^{10}+30x^{9}+455x^{8}+cb_7x^{7}+cb_6x^{6}+...+cb_{1}x+cb_{0} \\ \end{array} \right. \\ \end{align}
\begin{align} t_0&=x^{10}+30x^9+445x^8+4200x^7+27618x^6+128880x^5+425780x^4 \notag \\ &\qquad +974400x^3+1505321x^2+1402050x+2434705 \quad \in \ F_0[x]\\ \notag \\ t_1&=cd_7x^7+cd_6x^6+....+cd_2x^2+cd_1x+cd_0 \quad \in \ F_0(v)[x] \\ &\quad \bigl[ \ cd_i=\frac{1}{2}(ca_i-cb_i) \quad i=(7,6,..,1,0) \ \bigr] \notag \\ \end{align}
[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_0(v)[x]\\
\\
cd_m \ \in F_0(v) \\
q_1(x) \ \in F_0[x]\\
\end{array}
\right.
\quad \Rightarrow \quad
\left\{
\begin{array}{l}
\tilde{t_1}=a_1 \cdot q_1(x) \ \in F_1[x] \\
\\
B_1(x)=x^2-A_1=0 \\
a_1=\sqrt {A_1} \ \in F_0(a_1) \equiv F_1\\
\end{array}
\right. \notag \\
\end{align}
In (5.7) the leading coefficient of \(t_1\) at \(x^7\) is \(cd_7\). To match our earlier notation, we rename it \(cd_m\) (“m” for the maximum degree). Compute the inverse \(cd_m^{-1}\) and multiply it into \(t_1\).
The result is a monic polynomial \(q_1(x)\) which lies in \(F_0[x]\).
\begin{align} \notag \\ cd_m&=cd_7=\biggl(-\frac{2994097902733737901556757860993463670104796000}{45625916224520146215289291201786941800571678106841887201} {{v}^{19}}+......\biggr. \notag \\ &\qquad \qquad +\biggl. \frac{3351265726041833366041493442624195077062227076230024974550}{45625916224520146215289291201786941800571678106841887201}\biggr) \ \in \ F_0(v) \\ \notag \\ cd_m^{-1}&= \biggl( -\frac{7485244756834344753891894652483659175261990 }{593136910918761900798760785623230243407431815388944533613}{{v}^{19}} ......\biggr. \notag \\ &\qquad \qquad +\frac{5155793424679743640063836065575684733941887809584653807}{365007329796161169722314329614295534404573424854735097608}\biggl. \biggr) \notag \\ \notag \\ \end{align}
\begin{align} q_1&=cd_m^{-1} \cdot t_1 \notag \\ &=x^7+21x^6+229x^5+\frac{2955}{2}x^4+5999x^3+15426x^2+25625x-\frac{6333}{5} \ \in \ F_0[x] \\ \end{align}
Looking at the two ends of (5.10), we can also say that \(cd_m\) is a root of the binomial equation \(B_1(x)=0\) in (5.11).
Thus we re-define \(cd_m\) as the radical of \(B_1(x)=0\) and denote it by the new symbol \(a_1\). By adjoining \(a_1\) to the base field \(F_0\), we generate the new extension \(F_1\).
\begin{align} &cd_m^2=5200 \equiv A_1 \ \in \ F_0\\ \notag \\ &\left\{ \begin{array}{l} B_1(x)=x^2-A_1=0 \\ a_1 \equiv \sqrt{A_1} =\sqrt{5200} \ \in \ \bbox[#FFFF00]{ F_1 \equiv F_0(a_1) }\\ \end{array} \right. \\ \end{align}
\begin{align} &cd_m \ \in \ F_0(v) & &\Rightarrow & &a_1 \ \in \ F_1 \\ &t_1=cd_m \cdot q_1(x) \ \in F_0(v)[x] & &\Rightarrow & &\tilde{t_1}=a_1 \cdot q_1(x) \ \in F_1[x] \\ \end{align}
\begin{align} \notag \\ \tilde{t_1}=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}