【例題1】の解法手順（RT2&RT4）
EX1-RT2-1\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*}
\[ Remainder \ Theorem \] \[ \qquad (1) \quad \alpha^3+3\alpha+1=0 \\ \qquad (2) \quad \beta^2+\alpha\beta+\alpha^2+3=0\\ \qquad (3) \quad \alpha+\beta +\gamma=0\]
\[ \quad divide \ V(x) \ by \ (1),(2),(3) \] \[ \qquad \qquad \Downarrow \] \[g_{0}(x)=x^6+18x^4+81x^2+135 \]
\[ \qquad g_0(x):minimal \ polynomial \ of \ v \ on \ F_0=Q(\omega) \]
\begin{align*} \begin{pmatrix} 1\\v\\v^2\\v^3\\v^4\\v^5 \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & 0 & 0 & 0\\ 0 & -1 & -2 & 0 & 0 & 0\\ -3 & 0 & 0 & 3 & 3 & 0\\ 3 & 3 & 24 & 0 & 0 & -6\\ 9 & -9 & 0 & -45 & -45 & 0\\ -72 & -9 & -288 & 9 & -9 & 90\end{pmatrix} \begin{pmatrix} 1\\ \beta \\ \alpha \\ \alpha\beta \\ \alpha^2 \\ \alpha^2\beta \end{pmatrix} \end{align*}
\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*}
（覚書1：RT2-4の計算の仕組み）
（覚書2：Lagrange補間式を使って根を求める計算法）
\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？）RT2-1 問題の設定
【例題1】RT2の計算の主眼は、対称式を剰余の定理を使って一気に基礎体の多項式に変換する方式です。
この計算方式の提唱者は以下の方です。
「退職後は素人数学者」氏 「可解な代数方程式のガロア理論に基づいた解法」
https://ikumi.que.jp/blog/wp-content/uploads/2018/09/galois-solution.pdf
この章は同氏の方式そのものを使わせていただきました。
【例題1】は次の３次方程式 \(f(x)\)の根を求める事です。
RT1-1で様々な数学用語を説明をしたので、RT2ではごく簡単に説明するだけにします。
\begin{equation} \setCounter{0} f(x)=x^3+3x+1=0 \quad \in F_0[x]\\ \{ \ \alpha,\beta,\gamma \ \} :roots \ of \ F(x) \end{equation}
\begin{align} v=1\cdot\alpha+2\cdot\beta+3\cdot\gamma \end{align}
RT2-2の計算
対称群 \(S_3=\{ \ \sigma_1,\sigma_2,...,\sigma_6 \ \}\) の元を以下の様に定義します。
\begin{align} \begin{split} \sigma_{1}=\begin{pmatrix} 1&2&3 \\ 1&2&3 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \alpha&\beta&\gamma \end{pmatrix} \quad \sigma_{2}=\begin{pmatrix} 1&2&3 \\ 1&3&2 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \alpha&\gamma&\beta \end{pmatrix} \\ \sigma_{3}=\begin{pmatrix} 1&2&3 \\ 2&1&3 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \beta&\alpha&\gamma \end{pmatrix} \quad \sigma_{4}=\begin{pmatrix} 1&2&3 \\ 2&3&1 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \beta&\gamma&\alpha \end{pmatrix} \\ \sigma_{5}=\begin{pmatrix} 1&2&3 \\ 3&1&2 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \gamma&\alpha&\beta \end{pmatrix} \quad \sigma_{6}=\begin{pmatrix} 1&2&3 \\ 3&2&1 \end{pmatrix}=\begin{pmatrix} \alpha&\beta&\gamma \\ \gamma&\beta&\alpha \end{pmatrix} \end{split} \end{align}
上記置換操作による \(v\) に対する計算例 \(\sigma_4(v)\) を示しておきます。
\begin{align} \sigma_4(v)&=\sigma_4(\alpha+2\beta+3\gamma) =\sigma_4(\alpha)+2\sigma_4(\beta)+3\sigma_4(\gamma) =\beta+2\gamma+3\alpha \notag \\ \end{align}
\begin{align} &\left\{ \begin{array}{l} \sigma_{1}(v)&=v_{1}=\alpha+2\beta+3\gamma \qquad \sigma_{2}(v)=v_{2}=\alpha+2\gamma+3\beta \\ \sigma_{3}(v)&=v_{3}=\beta+2\alpha+3\gamma \qquad \sigma_{4}(v)=v_{4}=\beta+2\gamma+3\alpha \\ \sigma_{5}(v)&=v_{5}=\gamma+2\alpha+3\beta \qquad \sigma_{6}(v)=v_{6}=\gamma+2\beta+3\alpha \end{array} \right. \\ \end{align}
\begin{align} \sigma_3(v_4)&=\sigma_3(\beta+2\gamma+3\alpha ) =\sigma_3(\beta)+2\sigma_3(\gamma)+3\sigma_3(\alpha)\\ &=\alpha+2\gamma+3\beta=v_2 \end{align}
| \( i \backslash j \) | \(\sigma_i(v_1)\) | \(\sigma_i(v_2)\) | \(\sigma_i(v_3)\) | \(\sigma_i(v_4)\) | \(\sigma_i(v_5)\) | \(\sigma_i(v_6)\) |
|---|---|---|---|---|---|---|
| \(\sigma_1\) | \(v_1\) | \(v_2\) | \(v_3\) | \(v_4\) | \(v_5\) | \(v_6\) |
| \(\sigma_2\) | \(v_2\) | \(v_1\) | \(v_5\) | \(v_6\) | \(v_3\) | \(v_4\) |
| \(\sigma_3\) | \(v_3\) | \(v_4\) | \(v_1\) | \(v_2\) | \(v_6\) | \(v_5\) |
| \(\sigma_4\) | \(v_4\) | \(v_3\) | \(v_6\) | \(v_5\) | \(v_1\) | \(v_2\) |
| \(\sigma_5\) | \(v_5\) | \(v_6\) | \(v_2\) | \(v_1\) | \(v_4\) | \(v_3\) |
| \(\sigma_6\) | \(v_6\) | \(v_5\) | \(v_4\) | \(v_3\) | \(v_2\) | \(v_1\) |
次に、この \([ \ v_1,v_2,...,v_6 \ ]\) を使って、次の多項式 \(V(x)\) を考えます。
\begin{align} V(x)=&(x-v_{1})(x-v_{2})(x-v_{3})(x-v_{4})(x-v_{5})(x-v_{6})\\ \end{align}
\begin{align} V(x)=&\sigma_1(V(x)) = \sigma_2(V(x)) = \sigma_3(V(x)) \notag \\ =& \sigma_4(V(x)) = \sigma_5(V(x)) = \sigma_6(V(x)) \\ \notag \\ &\therefore\quad V(x) \in F_0[x] \end{align}
\begin{align} V(x)=&36 {{\gamma }^{6}}+288 \beta {{\gamma }^{5}}+288 \alpha {{\gamma }^{5}}-132 x {{\gamma }^{5}}+863 {{\beta }^{2}} {{\gamma }^{4}}+1802 \alpha \beta {{\gamma }^{4}} \notag \\ &-828 x \beta {{\gamma }^{4}}+863 {{\alpha }^{2}} {{\gamma }^{4}}-828 x \alpha {{\gamma }^{4}}+193 {{x}^{2}} {{\gamma }^{4}}+1226 {{\beta }^{3}}{{\gamma }^{3}} \notag \\ &+4030 \alpha {{\beta }^{2}} {{\gamma }^{3}}-1860 x {{\beta }^{2}} {{\gamma }^{3}}+4030 {{\alpha }^{2}} \beta {{\gamma }^{3}}-3888 x \alpha \beta {{\gamma }^{3}} \notag \\ &+910 {{x}^{2}} \beta {{\gamma }^{3}}+1226 {{\alpha }^{3}} {{\gamma }^{3}}-1860 x {{\alpha }^{2}} {{\gamma }^{3}}+910 {{x}^{2}} \alpha {{\gamma }^{3}}-144 {{x}^{3}} {{\gamma }^{3}} \notag \\ &+863 {{\beta }^{4}} {{\gamma }^{2}}+4030 \alpha {{\beta }^{3}} {{\gamma }^{2}}-1860 x {{\beta }^{3}} {{\gamma }^{2}}+6378 {{\alpha }^{2}} {{\beta }^{2}} {{\gamma }^{2}} \notag \\ &-6156 x \alpha {{\beta }^{2}} {{\gamma }^{2}}+1443 {{x}^{2}} {{\beta }^{2}} {{\gamma }^{2}}+4030 {{\alpha }^{3}} \beta {{\gamma }^{2}}-6156 x {{\alpha }^{2}} \beta {{\gamma }^{2}} \notag \\ &+3024 {{x}^{2}} \alpha \beta {{\gamma }^{2}}-480 {{x}^{3}} \beta {{\gamma }^{2}}+863 {{\alpha }^{4}} {{\gamma }^{2}}-1860 x {{\alpha }^{3}} {{\gamma }^{2}} \notag \\ &+1443 {{x}^{2}} {{\alpha }^{2}} {{\gamma }^{2}}-480 {{x}^{3}} \alpha {{\gamma }^{2}}+58 {{x}^{4}} {{\gamma }^{2}} +288 {{\beta }^{5}} \gamma +1802 \alpha {{\beta }^{4}} \gamma \notag \\ &-828 x {{\beta }^{4}} \gamma +4030 {{\alpha }^{2}}{{\beta }^{3}} \gamma -3888 x \alpha {{\beta }^{3}} \gamma +910 {{x}^{2}} {{\beta }^{3}} \gamma +4030 {{\alpha }^{3}} {{\beta }^{2}} \gamma \notag \\ &-6156 x {{\alpha }^{2}} {{\beta }^{2}} \gamma +3024 {{x}^{2}} \alpha {{\beta }^{2}} \gamma -480 {{x}^{3}} {{\beta }^{2}} \gamma +1802 {{\alpha }^{4}} \beta \gamma \notag \\ &-3888 x {{\alpha }^{3}} \beta \gamma +3024{{x}^{2}} {{\alpha }^{2}} \beta \gamma -1008 {{x}^{3}} \alpha \beta \gamma +122 {{x}^{4}} \beta \gamma +288 {{\alpha }^{5}} \gamma \notag \\ &-828 x {{\alpha }^{4}} \gamma +910 {{x}^{2}} {{\alpha }^{3}} \gamma -480 {{x}^{3}} {{\alpha }^{2}} \gamma +122 {{x}^{4}} \alpha \gamma -12 {{x}^{5}} \gamma \notag \\ &+36 {{\beta }^{6}}+288 \alpha {{\beta }^{5}} -132 x {{\beta }^{5}}+863 {{\alpha }^{2}} {{\beta }^{4}}-828 x \alpha {{\beta }^{4}}+193 {{x}^{2}} {{\beta }^{4}} \notag \\ &+1226 {{\alpha }^{3}} {{\beta }^{3}}-1860 x {{\alpha }^{2}} {{\beta }^{3}} +910 {{x}^{2}} \alpha {{\beta }^{3}}-144 {{x}^{3}} {{\beta }^{3}}+863 {{\alpha }^{4}} {{\beta }^{2}} \notag \\ &-1860 x {{\alpha }^{3}} {{\beta }^{2}}+1443 {{x}^{2}} {{\alpha }^{2}} {{\beta }^{2}}-480 {{x}^{3}} \alpha {{\beta }^{2}}+58 {{x}^{4}} {{\beta }^{2}}+288 {{\alpha }^{5}} \beta \notag \\ &-828 x {{\alpha }^{4}} \beta +910 {{x}^{2}} {{\alpha }^{3}} \beta -480 {{x}^{3}} {{\alpha }^{2}} \beta +122 {{x}^{4}} \alpha \beta -12 {{x}^{5}} \beta \notag \\ &+36 {{\alpha }^{6}}-132 x {{\alpha }^{5}}+193 {{x}^{2}} {{\alpha }^{4}}-144 {{x}^{3}} {{\alpha }^{3}} +58 {{x}^{4}} {{\alpha }^{2}}-12 {{x}^{5}} \alpha +{{x}^{6}} \end{align}
TR1の計算方式は、古典的方法で、\(\{\alpha,\beta,\gamma\}\) の対称式を基本対称式 \(\{e_1,e_2,e_3\}\) に変換して、 \(F_0\) 上の式 \(V(x)\) を求めましたが、次節では全く違った方式で計算してみます。
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