\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*}
\begin{align*} v_{1}=\alpha+2\beta+3\gamma \qquad v_{2}=\alpha+2\gamma+3\beta \\ v_{3}=\beta+2\alpha+3\gamma \qquad v_{4}=\beta+2\gamma+3\alpha\\ v_{5}=\gamma+2\alpha+3\beta \qquad v_{6}=\gamma+2\beta+3\alpha \end{align*} \begin{align*} V(x)=&(x-v_{1})(x-v_{2})(x-v_{3})\\ \times&(x-v_{4})(x-v_{5})(x-v_{6}) \end{align*}
\[V(x, \ \alpha,\beta,\gamma): symmetric \ function \ in \{\alpha,\beta,\gamma\} \] \[\qquad \qquad \Downarrow\] \[V(x, \ e_{1},e_{2},e_{3})\] \[\{e_1,e_2,e_3\}: elementary \ symmetric \ functions\]
\[g_{0}(x)=x^6+18x^4+81x^2+135 \]
\[ \qquad g_0(x):minimal \ polynomial \ of \ v \ on \ F_0=Q(\omega) \]
\begin{align*} P_{\alpha}(x)=V(x)&\cdot \big( \frac{\gamma }{x-{v_6}}+\frac{\gamma }{x-{v_5}}+\frac{\beta }{x-{v_4}}\\ &+\frac{\beta }{x-{v_3}}+\frac{\alpha }{x-{v_2}}+\frac{\alpha }{x-{v_1}}\big)\\ \end{align*} \[\alpha=\left.\frac{P_\alpha(x)}{V'(x)}\right|_{x=v} \quad The \ same \ holds \ for \ \beta\ and \ \gamma \]
\begin{eqnarray*} \left\{ \begin{array}{l} \alpha&=\frac{{{v}^{4}}+15 {{v}^{2}}-9 v+36}{18}\\ \beta&=-\frac{{{v}^{4}}+15 {{v}^{2}}+36}{9}\\ \gamma&=\frac{{{v}^{4}}+15 {{v}^{2}}+9 v+36}{18}\\ \end{array} \right. \end{eqnarray*}
\begin{align*} v_{1}&=v & v_{2}&=\frac{-v^4}{6}-\frac{5v^2}{2}+\frac{v}{2}-6\\ v_{3}&=\frac{v^4}{6}+\frac{5v^2}{2}+\frac{v}{2}+6 & v_{4}&=\frac{v^4}{6}+\frac{5v^2}{2}-\frac{v}{2}+6\\ v_{5}&=-\frac{v^4}{6}-\frac{5v^2}{2}-\frac{v}{2}-6 & v_{6}&=-v \end{align*}
\begin{align*} &g_0(v_i)=0 \quad for \ (i=1,2,..,6) \\ &\qquad \qquad \Downarrow\ \\ &S_3: Galois \ group \ of \ f(x) \\ &\qquad composition \ series \ S_3 \rhd A_3 \rhd \{e\} \end{align*}
\[ g_{1}(x)=x^3+9x+a_{1} \in F_{1}[x] \]
\[\quad g_1(x):minimal \ polynomial \ of \ v \ on \ F_1=F_0(a_1)\\ \quad Here \ \ B_1=a_{1}^2 +135=0 \]
\[g_{2}(x)=x+{{a}_{2}^{2}}\, \left( -\frac{\omega }{3}+\frac{{a_1}}{18}-\frac{1}{6}\right) +{a_2} \in F_{2}[x]\]
\[ \quad g_2(x):minimal \ polynomial \ of \ v \ on \ F_2=F_0(a_1,a_2)\\ \quad Here \quad B_2=a_2^3-\frac{6 \omega +{a_1}+3}{2}=0, \ \Omega=\omega^2+\omega+1=0 \]
\begin{align*} v=&\frac{{{a}_{2}^{2}} \omega }{3}-\frac{{a_1} {{a}_{2}^{2}}}{18}+\frac{{{a}_{2}^{2}}}{6}-{a_2} \\ \\ \alpha=&-\frac{{a_1} \left( {{a}_{2}^{2}} \omega -{{a}_{2}^{2}}\right) +\left( 9 {{a}_{2}^{2}}-18 {a_2}\right) \omega +9 {{a}_{2}^{2}}-36 {a_2}}{54}\\ \beta=&\frac{{a_1} \left( 2 {{a}_{2}^{2}} \omega +{{a}_{2}^{2}}\right) -36 {a_2} \omega +9 {{a}_{2}^{2}}-18 {a_2}}{54}\\ \gamma=&-\frac{{a_1} \left( {{a}_{2}^{2}} \omega +2 {{a}_{2}^{2}}\right) +\left( -9 {{a}_{2}^{2}}-18 {a_2}\right) \omega +18 {a_2}}{54}\\ \\ Here &\quad B_1=a_{1}^2 +135=0,\\ &B_2=a_2^3-\frac{6 \omega +{a_1}+3}{2}=0, \\ &\Omega=\omega^2+\omega+1=0 \end{align*}
(覚書:冪根の実態は?)
\begin{align} \setCounter{17} P_{\alpha}(x) &\equiv \ V(x)\cdot \big( \frac{\alpha }{x-{v_1}}+\frac{\alpha }{x-{v_2}} \notag \\ &+\frac{\beta }{x-{v_3}}+\frac{\beta }{x-{v_4}} +\frac{\gamma }{x-{v_5}}+\frac{\gamma }{x-{v_6}} \big)\\ \end{align}
\begin{align} P_{\alpha}(x)=&\alpha(x-v_2)(x-v_3)(x-v_4)(x-v_5)(x-v_6) \notag \\ +&\alpha(x-v_1)(x-v_3)(x-v_4)(x-v_5)(x-v_6) \notag \\ +&\beta(x-v_1)(x-v_2)(x-v_4)(x-v_5)(x-v_6) \notag \\ +&\beta(x-v_1)(x-v_2)(x-v_3)(x-v_5)(x-v_6) \notag \\ +&\gamma(x-v_1)(x-v_2)(x-v_3)(x-v_4)(x-v_6) \notag \\ +&\gamma(x-v_1)(x-v_2)(x-v_3)(x-v_4)(x-v_5) \end{align}
\begin{align} \frac{\alpha }{x-{v_1}}&=\sigma_1(\frac{\alpha }{x-v}) & \frac{\alpha }{x-{v_2}}&=\sigma_2(\frac{\alpha }{x-v}) \notag \\ \frac{\beta }{x-{v_3}}&=\sigma_3(\frac{\alpha }{x-v}) & \frac{\beta }{x-{v_4}}&=\sigma_4(\frac{\alpha }{x-v}) \\ \frac{\gamma }{x-{v_5}}&=\sigma_5(\frac{\alpha }{x-v}) & \frac{\gamma }{x-{v_6}}&=\sigma_6(\frac{\alpha }{x-v}) \notag \\ \end{align}
\begin{align} \therefore \quad P_\alpha(x)=V(x)\cdot \Bigl[ \ \sum_{i=1}^6 \sigma_i(\frac{\alpha }{x-v}) \ \Bigr] \end{align}
\begin{align} P_\alpha(x)=18x^4-9x^3+162x^2-81x+405 \qquad \in F_0[x] \end{align}
\begin{align} &P_{\alpha}(v_1)=\alpha(v_1-v_2)(v_1-v_3)(v_1-v_4)(v_1-v_5)(v_1-v_6)\\ \notag \\ &\alpha=\frac{P_{\alpha}(v_1)}{(v_1-v_2)(v_1-v_3)(v_1-v_4)(v_1-v_5)(v_1-v_6)}\\ \end{align}
\begin{align} \alpha&=\left.\frac{P_\alpha(x)}{V'(x)}\right|_{x=v} \\ \notag \\ V'(x)&=6x^5+72x^3+162x\\ \notag \\ \therefore \ \alpha&=\frac{18v^4-9v^3+162v^2-81v+405}{6v^5+72v^3+162v} \\ \end{align}
\begin{align} &P_\beta(x)=V(x)\cdot \left( \sum_{i=1}^6 \sigma_i(\frac{\beta }{x-v})\right) \\ &P_\gamma(x)=V(x)\cdot \left( \sum_{i=1}^6 \sigma_i(\frac{\gamma }{x-v})\right)\\ \notag \\ &P_\beta(x)=18x^3+162x \\ &P_\gamma(x)= -18x^4-9x^3-162x^2-81x-405\\ \notag \\ &\qquad \Downarrow \notag \\ \notag \\ &\beta=\left.\frac{P_\beta(x)}{V'(x)}\right|_{x=v}=\frac{18v^3+162v}{6v^5+72v^3+162v} \\ \notag \\ &\gamma=\left.\frac{P_\gamma(x)}{V'(x)}\right|_{x=v}=\frac{-18v^4-9v^3-162v^2-81v-405}{6v^5+72v^3+162v} \\ \end{align}
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