\begin{align*} &f(x)=3x^3-3x+1 \quad \{\alpha,\beta,\gamma\}: \ roots \ of \ f(x)\\ &Primitive \ element \quad v=1\cdot\alpha+2\cdot \beta+3\cdot\gamma \end{align*}
\[\qquad The \ system \ of \ equations\]
\[ \qquad \left\{ \begin{array}{l} \alpha^3-3\alpha+1=0\\ \beta^2+\alpha\beta+\alpha^2-3=0\\ \alpha+\beta+\gamma=0\\ v-(\alpha+2\beta+3\gamma)=0 \end{array} \right.\\ \qquad \qquad \qquad \Downarrow \]
\[\qquad Elimination \ Theory\]
\[ \qquad V(v)= v^6-18v^4+81v^2-81 \\ \qquad \qquad =\left( {{v}^{3}}-9 v-9\right) \, \left( {{v}^{3}}-9 v+9\right) \]
\[g_{0}(x)=x^3-9x-9 \]
\[ g_0(x):minimal \ polynomial \ of \ v \ on \ F_0=Q(\omega) \]
\[Factorization \ of \ f(x) \ on \ F_0(v)\] \[\quad "maxima's \ function \ "\] \[\qquad factor(f(x),g_0(v))\]
\begin{align*} \alpha&=-\frac{{{v}^{2}}}{3}+2 & \beta&=\frac{2 {{v}^{2}}}{3}-v-4\\ \gamma&=-\frac{{{v}^{2}}}{3}+v+2 & & \end{align*}
\begin{align*} v_{1}&=v & v_{2}&={{v}^{2}}-v-6\\ v_{3}&=-{{v}^{2}}+2 v+6 & v_{4}&=-{{v}^{2}}+v+6\\ v_{5}&={{v}^{2}}-2 v-6 & v_{6}&=-v \end{align*}
\begin{align*} &g_0(v_i)=0 \quad for \ (i=1,4,5) \\ &\qquad \qquad \Downarrow\ \\ &A_3: Galois \ group \ of \ F(x) \\ &\qquad composition \ series \quad A_3 \rhd \{e\} \end{align*}
\begin{align*} &g_1(x) : \ minimal \ polynomial \ of \ v\\ &g_{1}(x)=x-\frac{{{a}_{1}^{2}} \omega }{3}-\frac{2 {{a}_{1}^{2}}}{3}+{a_1} \in F_{1}[x] \\ \end{align*}
\begin{align*} \\ F_1=F_0(a_1) \quad Here \ B_1&={{a}_{1}^{3}}-3\omega +3 =0, \\ \Omega&=\omega^2+\omega+1=0 \end{align*}
\begin{align*} v=&\frac{{{a}_{1}^{2}} \omega }{3}+\frac{2 {{a}_{1}^{2}}}{3}-{a_1} \\ \\ \alpha=&\frac{{a_1} \omega }{3}-\frac{{{a}_{1}^{2}}}{3}+\frac{2 {a_1}}{3}\\ \beta=&-\frac{{{a}_{1}^{2}} \omega }{3}-\frac{2 {a_1} \omega }{3}-\frac{{a_1}}{3}\\ \gamma=&\frac{{{a}_{1}^{2}} \omega }{3}+\frac{{a_1} \omega }{3}+\frac{{{a}_{1}^{2}}}{3}-\frac{{a_1}}{3}\\ \\ Here &\quad B_1={{a}_{1}^{3}}-3\omega +3 =0,\\ &\quad \Omega=\omega^2+\omega+1=0 \end{align*}
\begin{align} \setCounter{56} t_1^3&=\frac{1}{27}\left(( 2v^2-3 v-12) \omega +v^2-3v-6 \right)^3 =3\omega -3 \quad \in F_0 \\ t_2^3&=\frac{1}{27}\left( ( 2v^2-3v-12) \omega + v^2-6 \right)^3 =-3\omega -6 \quad \in F_0 \\ \end{align}
\begin{align} \setCounter{59} &t_1 \cdot t_2=\frac{1}{9}\left( ( 2v^2-3 v-12) \omega +v^2-3v-6 \right) \left( ( 2v^2-3v-12) \omega + v^2-6 \right) \notag \\ \notag \\ &= -\frac{4 {{v}^{4}} {{\omega }^{2}}}{9}+\frac{4 {{v}^{3}} {{\omega }^{2}}}{3}+\frac{13 {{v}^{2}} {{\omega }^{2}}}{3} -8 v {{\omega }^{2}}-16 {{\omega }^{2}}-\frac{4 {{v}^{4}} \omega }{9} \notag \\ &+\frac{4 {{v}^{3}} \omega }{3}+\frac{13 {{v}^{2}} \omega }{3} -8 v \omega -16 \omega -\frac{{{v}^{4}}}{9}+\frac{{{v}^{3}}}{3}+\frac{4 {{v}^{2}}}{3}-2 v-4 = \ 3 \ \in F_0 \\ \end{align}
\begin{align} &t_1^3=3\omega -3 \equiv A_1 \qquad t_2^3=-3\omega -6 \qquad t_1 \cdot t_2=3 \\ \notag \\ &\therefore \quad \{ \ t_1^3, \ t_2^3, \ t_1t_2 \ \} \quad \in \ F_0 \end{align}
\begin{align} &B_1=t_1^3-A_1=a_1^3-A_1 \qquad \therefore t_1=\sqrt[3]{A_1}\equiv a_1\\ \notag \\ & \tilde{t_1}=a_1 \quad \in F_0(a_1)=F_1\\ \end{align}
\begin{align} t_2=\frac{t_1 \cdot t_2}{t_1}=\frac{t_1^2\cdot (t_1t_2)}{t_1^2\cdot t_1} =\frac{t_1^2\cdot (t_1t_2)}{t_1^3}=\frac{a_1^2 \cdot (3)}{A_1} =\frac{3a_1^2}{A_1} \end{align}
\begin{align} A_1^{-1} = & d_0+d_1\omega \\ \notag \\ A_1 \cdot A_1^{-1} = & (3\omega -3 ) \cdot (d_0+d_1\omega)=3d_1\omega^2-3(d_0+d_1)\omega-3d_0 \notag \\ =&(3d_0-6d_1)\omega-3d_1-3d_0=1 \quad (mod \ \varOmega) \\ \notag \\ \therefore \ [ \ d_0 =-\frac{2}{9} &\ , \ d_1=-\frac{1}{9} \ ] \quad \Rightarrow \quad A_1^{-1}=-\frac{2}{9}-\frac{\omega}{9} \\ \end{align}
\begin{align} t_0=x \qquad t_1=a_1 \qquad t_2=-\frac{a_1^2(\omega+2)}{3} \\ \end{align}
\begin{align} \begin{bmatrix} \tilde{h_0} \\ \tilde{h_1} \\ \tilde{h_2} \end{bmatrix} &= \begin{bmatrix} 1&1&1 \\ 1&\omega^2&\omega\\ 1&\omega&\omega^2 \end{bmatrix} \cdot \begin{bmatrix} t_0 \\ \tilde{t_1} \\ \tilde{t_2} \end{bmatrix} = \begin{bmatrix} t_0+\tilde{t_1}+\tilde{t_2}\\ t_0+\omega^2 \tilde{t_1}+\omega \tilde{t_2}\\ t_0+\omega \tilde{t_1}+\omega^2 \tilde{t_2} \end{bmatrix} \notag \\ \notag \\ &= \begin{bmatrix} x+a_1-\frac{a_1^2(\omega+2)}{3} \\ x+a_1\omega^2-\frac{a_1^2\omega(\omega+2)}{3} \\ x+a_1\omega-\frac{a_1^2\omega^2(\omega+2)}{3} \end{bmatrix}\\ \end{align}
\begin{align} &g_1(x)=x+a_1-\frac{a_1^2(\omega+2)}{3} \quad \in F_1[x]=F_0(a_1)[x]\\ \notag \\ &B_1=a_1^3-A_1=0 \qquad A_1= 3\omega-3 \qquad \therefore \ a_1=\sqrt[3]{A_1}\\ \end{align}
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1st upload: 2023/06/17
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